TEST BANK FOR An Introduction to Management Science Quantitative Approaches to Decision Making 14e D by StudyGuide

Chapter 1 - Introduction True / False 1. The process of decision making is more limited than that of problem solving. a. True b. False ANSWER: True POINTS: 1 TOPICS: Problem solving and decision making 2. The terms 'stochastic' and 'deterministic' have the same meaning in quantitative analysis. a. True b. False ANSWER: False POINTS: 1 TOPICS: Model development 3. The volume that results in marginal revenue equaling marginal cost is called the break-even point. a. True b. False ANSWER: False POINTS: 1 TOPICS: Problem solving and decision making 4. Problem solving encompasses both the identification of a problem and the action to resolve it. a. True b. False ANSWER: True POINTS: 1 TOPICS: Problem solving and decision making 5. The decision making process includes implementation and evaluation of the decision. a. True b. False ANSWER: False POINTS: 1 TOPICS: Problem solving and decision making 6. The most successful quantitative analysis will separate the analyst from the managerial team until after the problem is fully structured. a. True b. False ANSWER: False POINTS: 1 TOPICS: Quantitative analysis 7. The value of any model is that it enables the user to make inferences about the real situation.

Chapter 1 - Introduction a. True b. False ANSWER: True POINTS: 1 TOPICS: Model development 8. Uncontrollable inputs are the decision variables for a model. a. True b. False ANSWER: False POINTS: 1 TOPICS: Model development 9. The feasible solution is the best solution possible for a mathematical model. a. True b. False ANSWER: False POINTS: 1 TOPICS: Model solution 10. A company seeks to maximize profit subject to limited availability of man-hours. Man-hours is a controllable input. a. True b. False ANSWER: False POINTS: 1 TOPICS: Model development 11. Frederick Taylor is credited with forming the first MS/OR interdisciplinary teams in the 1940's. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 12. To find the choice that provides the highest profit and the fewest employees, apply a single-criterion decision process. a. True b. False ANSWER: False POINTS: 1 TOPICS: Problem solving and decision making 13. The most critical component in determining the success or failure of any quantitative approach to decision making is problem definition. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction ANSWER: True POINTS: 1 TOPICS: Quantitative analysis 14. The first step in the decision making process is to identify the problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 15. All uncontrollable inputs or data must be specified before we can analyze the model and recommend a decision or solution for the problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Quantitative analysis 16. In quantitative analysis, the optimal solution is the mathematically-best solution. a. True b. False ANSWER: True POINTS: 1 TOPICS: Quantitative analysis 17. If you are deciding to buy either machine A, B, or C with the objective of minimizing the sum of labor, material and utility costs, you are dealing with a single-criterion decision. a. True b. False ANSWER: True POINTS: 1 TOPICS: Problem solving and decision making 18. Model development should be left to quantitative analysts; the model user's involvement should begin at the implementation stage. a. True b. False ANSWER: False POINTS: 1 TOPICS: Problem solving and decision making 19. A feasible solution is one that satisfies at least one of the constraints in the problem. a. True b. False ANSWER: False Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction POINTS: 1 TOPICS: Model solution 20. A toy train layout designed to represent an actual railyard is an example of an analog model. a. True b. False ANSWER: False POINTS: 1 TOPICS: Model development Multiple Choice 21. The field of management science a. approaches decision making rationally, with techniques based on the scientific method. b. concentrates on the use of quantitative methods to assist in decision making. c. is another name for decision science and for operations research. d. each of these choices are true. ANSWER: d POINTS: 1 TOPICS: Introduction 22. Identification and definition of a problem a. is the final step of problem solving. b. cannot be done until alternatives are proposed. c. requires consideration of multiple criteria. d. is the first step of decision making. ANSWER: d POINTS: 1 TOPICS: Problem solving and decision making 23. Decision alternatives a. should be identified before decision criteria are established. b. are limited to quantitative solutions c. are evaluated as a part of the problem definition stage. d. are best generated by brain-storming. ANSWER: a POINTS: 1 TOPICS: Problem solving and decision making 24. Decision criteria a. are the ways to evaluate the choices faced by the decision maker. b. are the choices faced by the decision maker. c. must be unique for a problem. d. are the problems faced by the decision maker. ANSWER: a Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction POINTS: 1 TOPICS: Problem solving and decision making 25. In a multicriteria decision problem a. successive decisions must be made over time. b. it is impossible to select a single decision alternative. c. the decision maker must evaluate each alternative with respect to each criterion. d. each of these choices are true. ANSWER: c POINTS: 1 TOPICS: Problem solving and decision making 26. The quantitative analysis approach requires a. mathematical expressions for the relationships. b. the manager's prior experience with a similar problem. c. a relatively uncomplicated problem. ANSWER: a POINTS: 1 TOPICS: Quantitative analysis and decision making 27. A physical model that does not have the same physical appearance as the object being modeled is a. a qualitative model. b. a mathematical model. c. an analog model. d. an iconic model. ANSWER: c POINTS: 1 TOPICS: Model development 28. Inputs to a quantitative model a. must all be deterministic if the problem is to have a solution. b. are uncertain for a stochastic model. c. are a trivial part of the problem solving process. d. are uncontrollable for the decision variables. ANSWER: b POINTS: 1 TOPICS: Model development 29. When the value of the output cannot be determined even if the value of the controllable input is known, the model is a. deterministic. b. analog. c. stochastic. d. digital. ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction TOPICS: Model development 30. The volume that results in total revenue being equal to total cost is the a. profit mix. b. marginal volume. c. marginal cost. d. break-even point. ANSWER: d POINTS: 1 TOPICS: Break-even analysis 31. Management science and operations research both involve a. operational management skills. b. quantitative approaches to decision making. c. scientific research as opposed to applications. d. qualitative managerial skills. ANSWER: b POINTS: 1 TOPICS: Introduction 32. George Dantzig is important in the history of management science because he developed a. the scientific management revolution. b. powerful digital computers. c. World War II operations research teams. d. the simplex method for linear programming. ANSWER: d POINTS: 1 TOPICS: Introduction 33. The first step in problem solving is a. definition of decision variables. b. the identification of a difference between the actual and desired state of affairs. c. determination of the correct analytical solution procedure. d. implementation. ANSWER: b POINTS: 1 TOPICS: Problem solving and decision making 34. Problem definition a. must involve the analyst and the user of the results. b. includes specific objectives and operating constraints. c. must occur prior to the quantitative analysis process. d. each of these choices are true. ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction TOPICS: Quantitative analysis 35. A model that uses a system of symbols to represent a problem is called a. iconic. b. constrained. c. mathematical. d. analog. ANSWER: c POINTS: 1 TOPICS: Model development 36. Which of the following is not one of the commonly used names for the body of knowledge involving quantitative approaches to decision-making? a. efficiency studies b. management science c. business analytics d. operations research ANSWER: a POINTS: 1 TOPICS: Introduction Subjective Short Answer 37. A snack food manufacturer buys corn for tortilla chips from two cooperatives, one in Iowa and one in Illinois. The price per unit of the Iowa corn is $6.00 and the price per unit of the Illinois corn is $5.50. a. Define variables that would tell how many units to purchase from each source. b. Develop an objective function that would minimize the total cost. The manufacturer needs at least 12000 units of corn. The Iowa cooperative can supply up to c. 8000 units, and the Illinois cooperative can supply at least 6000 units. Develop constraints for these conditions. ANSWER: a. Let x1 = the number of units from Iowa Let x2 = the number of units from Illinois b. Min 6x1 + 5.5x2 c. x1 + x 2 ≥ 12000 x1 ≥ 8000 x1 ≥ 6000 POINTS: 1 TOPICS: Model development 38. The relationship d = 5000 − 25p describes what happens to demand (d) as price (p) varies. Here, price can vary between $10 and $50. a. How many units can be sold at the $10 price? How many can be sold at the $50 price? b. Model the expression for total revenue. Consider prices of $20, $30, and $40. Which of these three price alternative will maximize c. total revenue? What are the values for demand and revenue at this price? ANSWER: Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction a. b. c.

For p = 10, d = 4750 For p = 50, d = 3750 TR = p(5000 − 25p) For p = 20, d = 4500, TR = $90,000 For p = 30, d = 4250, TR = $127,500 For p = 40, d = 4000, TR = $160,000 (maximum total revenue)

POINTS: 1 TOPICS: Model development 39. There is a fixed cost of $50,000 to start a production process. Once the process has begun, the variable cost per unit is $25. The revenue per unit is projected to be $45. a. Write an expression for total cost. b. Write an expression for total revenue. c. Write an expression for total profit. d. Find the break-even point. ANSWER: a. C(x) = 50000 + 25x b. R(x) = 45x c. P(x) = 45x − (50000 + 25x) d. x = 2500 POINTS: 1 TOPICS: Break-even analysis 40. An author has received an advance against royalties of $10,000. The royalty rate is $1.00 for every book sold in the United States, and $1.35 for every book sold outside the United States. Define variables for this problem and write an expression that could be used to calculate the number of books to be sold to cover the advance. ANSWER: Let x1 = the number of books sold in the U.S. Let x2 = the number of books sold outside the U.S. 10000 = 1x1 + 1.35x2 POINTS: 1 TOPICS: Break-even analysis 41. A university schedules summer school courses based on anticipated enrollment. The cost for faculty compensation, laboratories, student services, and allocated overhead for a computer class is $8500. If students pay $920 to enroll in the course, how large would enrollment have to be for the university to break even? ANSWER: Enrollment would need to be 10 students. POINTS: 1 TOPICS: Break-even analysis 42. As part of their application for a loan to buy Lakeside Farm, a property they hope to develop as a bed-and-breakfast operation, the prospective owners have projected: Monthly fixed cost (loan payment, taxes, insurance, maintenance) Variable cost per occupied room per night Revenue per occupied room per night a. b. c.

$6000 $ 20 $ 75

Write the expression for total cost per month. Assume 30 days per month. Write the expression for total revenue per month. If there are 12 guest rooms available, can they break even? What percentage of rooms would need to be occupied, on average, to break even?

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Chapter 1 - Introduction ANSWER:

a. b. c.

C(x) = 6000 + 20(30)x (monthly) R(x) = 75(30)x (monthly) Break-even occupancy = 3.64 or 4 occupied rooms per night, so they have enough rooms to break even. This would be a 33% occupancy rate.

POINTS: 1 TOPICS: Break-even analysis 43. Organizers of an Internet training session will charge participants $150 to attend. It costs $3000 to reserve the room, hire the instructor, bring in the equipment, and advertise. Assume it costs $25 per student for the organizers to provide the course materials. a. How many students would have to attend for the company to break even? If the trainers think, realistically, that 20 people will attend, then what price should be b. charged per person for the organization to break even? ANSWER: a. C(x) = 3000 + 25x R(x) = 150x Break-even students = 24 b. Cost = 3000 + 25(20) Revenue = 20p Break-even price = 175 POINTS: 1 TOPICS: Break-even analysis 44. In this portion of an Excel spreadsheet, the user has given values for selling price, the costs, and a sample volume. Give the cell formula for a. cell E12, break-even volume. b. cell E16, total revenue. c. cell E17, total cost. d. cell E19, profit (loss). A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Break-even calculation Selling price per unit

Costs Fix cost Variable cost per unit

8400 4.5

Break-even volume Sample calculation Volume Total revenue Total cost

2000

Profit (loss)

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Chapter 1 - Introduction ANSWER:

a. =E9/(E6-E10) b. =E15*E6 c. =E9+E10*E15 d. =E16-E17 POINTS: 1 TOPICS: Spreadsheets for management science 45. A furniture store has set aside 800 square feet to display its sofas and chairs. Each sofa utilizes 50 sq. ft. and each chair utilizes 30 sq. ft. At least five sofas and at least five chairs are to be displayed. a. Write a mathematical model representing the store's constraints. Suppose the profit on sofas is $200 and on chairs is $100. On a given day, the probability b. that a displayed sofa will be sold is .03 and that a displayed chair will be sold is .05. Mathematically model each of the following objectives: 1. Maximize the total pieces of furniture displayed. 2. Maximize the total expected number of daily sales. 3. Maximize the total expected daily profit. ANSWER: a. 50s + 30c ≤ 800 s≥5 c≥5 b. (1) Max s + c (2) Max .03s + .05c (3) Max 6s + 5c POINTS: 1 TOPICS: Model development 46. A manufacturer makes two products, doors and windows. Each must be processed through two work areas. Work area #1 has 60 hours of available production time per week. Work area #2 has 48 hours of available production time per week. Manufacturing of a door requires 4 hours in work area #1 and 2 hours in work area #2. Manufacturing of a window requires 2 hours in work area #1 and 4 hours in work area #2. Profit is $8 per door and $6 per window. a. Define decision variables that will tell how many units to build (doors and windows) per week. b. Develop an objective function that will maximize total profit per week. c. Develop production constraints for work area #1 and #2. ANSWER: a. Let D = the number of doors to build per week Let N = the number of windows to build per week b. Weekly Profit = 8D + 6W c. 4D + 2W ≤ 60 2D + 4W ≤ 48 POINTS: 1 TOPICS: Model development 47. A small firm builds galvanized swing sets. The investment in plant and equipment is $200,000. The variable cost per swing set is $500. The selling price of the swing set is $1000. How many swing sets would have to be sold for the firm to break even? ANSWER: 400 swing sets POINTS: 1 TOPICS: Break-even analysis Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction 48. A computer rework center has the capacity to rework 300 computers per day. The expected number of computers needing to be reworked per day is 225. The center is paid $26 for each computer reworked. The fixed cost of renting the reworking equipment is $250 per day. Work space rents for $150 per day. The cost of material is $18 per computer and labor costs $3 per computer. What is the break-even number of computers reworked per day? ANSWER: 80 computers POINTS: 1 TOPICS: Break-even analysis 49. To establish a driver education school, organizers must decide how many cars, instructors, and students to have. Costs are estimated as follows. Annual fixed costs to operate the school are $30,000. The annual cost per car is $3000. The annual cost per instructor is $11,000 and one instructor is needed for each car. Tuition for each student is $350. Let x be the number of cars and y be the number of students. a. Write an expression for total cost. b. Write an expression for total revenue. c. Write an expression for total profit. d. The school offers the course eight times each year. Each time the course is offered, there are two sessions. If they decide to operate five cars, and if four students can be assigned to each car, will they break even? ANSWER: a. C(x) = 30000 + 14000x b. R(y) = 350y c. P(x,y) = 350y − (30000 + 14000x) d. Each car/instructor can serve up to (4 students/session)(2 sessions/course)(8 courses/year) = 64 students annually. Five cars can serve 320 students. If the classes are filled, then profit for five cars is 350(320) − (30000 + 14000(5)) = 12000. So, the school can reach the break-even point. POINTS: 1 TOPICS: Break-even analysis 50. Zipco Printing operates a shop that has five printing machines. The machines differ in their capacities to perform various printing operations due to differences in the machines' designs and operator skill levels. At the start of the workday there are five printing jobs to schedule. The manager must decide what the job-machine assignments should be. a. How could a quantitative approach to decision making be used to solve this problem? b. What would be the uncontrollable inputs for which data must be collected? Define the decision variables, objective function, and constraints to appear in the c. mathematical model. d. Is the model deterministic or stochastic? e. Suggest some simplifying assumptions for this problem. ANSWER: A quantitative approach to decision making can provide a systematic way for deciding the a. job-machine pairings so that total job processing time is minimized. How long it takes to process each job on each machine, and any job-machine pairings that b. are unacceptable. Decision variables: one for each job-machine pairing, taking on a value of 1 if the pairing is c. used and 0 otherwise. Objective function: minimize total job processing time. Constraints: each job is assigned to exactly one machine, and each machine be assigned no more than one job. Stochastic: job processing times vary due to varying machine set-up times, variable operator d. performance, and more. e. Assume that processing times are deterministic (known/fixed). POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction TOPICS: Model development 51. Consider a department store that must make weekly shipments of a certain product from two different warehouses to four different stores. a. How could a quantitative approach to decision making be used to solve this problem? b. What would be the uncontrollable inputs for which data must be gathered? What would be the decision variables of the mathematical model? the objective function? the c. constraints? d. Is the model deterministic or stochastic? e. Suggest assumptions that could be made to simplify the model. ANSWER: A quantitative approach to decision making can provide a systematic way to determine a a. minimum shipping cost from the warehouses to the stores. Fixed costs and variable shipping costs; the demand each week at each store; the supplies b. each week at each warehouse. Decision variables--how much to ship from each warehouse to each store; objective c. function--minimize total shipping costs; constraints--meet the demand at the stores without exceeding the supplies at the warehouses. Stochastic--weekly demands fluctuate as do weekly supplies; transportation costs could vary d. depending upon the amount shipped, other goods sent with a shipment, etc. Make the model deterministic by assuming fixed shipping costs per item, demand is constant e. at each store each week, and weekly supplies in the warehouses are constant. POINTS: 1 TOPICS: Model development 52. Three production processes - A, B, and C - have the following cost structure: Process

Fixed Cost per Year

Variable Cost per Unit

A B C

$120,000 90,000 80,000

$3.00 4.00 4.50

a. What is the most economical process for a volume of 8,000 units? How many units per year must be sold with each process to have annual profits of $50,000 if the selling price is $6.95 b. per unit? c. What is the break-even volume for each process? ANSWER: a. C(x) = FC + VC(x) Process A: C(x) = $120,000 + $3.00(8,000) = $144,000 per year Process B: C(x) = $ 90,000 + $4.00(8,000) = $122,000 per year Process C: C(x) = $ 80,000 + $4.50(8,000) = $116,000 per year Process C has the lowest annual cost for a production volume of 8,000 units. b. Q = (profit + FC)/(price - VC) Process A: Q = ($50,000 + $120,000)/($6.95 - $3.00) = 43,038 units Process B: Q = ($50,000 + $ 90,000)/($6.95 - $4.00) = 47,458 units Process C: Q = ($50,000 + $ 80,000)/($6.95 - $4.50) = 53,062 units Process A requires the lowest production volume for an annual profit of $50,000. c. At breakeven, profit (the pretax profits per period) is equal to zero. Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction Q = FC/(price - VC) Process A: Q = $120,000/ ($6.95 - $3.00) = 30,380 units Process B: Q = $ 90,000/ ($6.95 - $4.00) = 30,509 units Process C: Q = $ 80,000/ ($6.95 - $4.50) = 32,654 units Process A has the lowest break-even quantity, while Process B’s is almost as low. POINTS: 1 TOPICS: Break-even analysis 53. Jane Persico, facility engineer at the El Paso plant of Computer Products Corporation (CPC), is studying a process selection decision at the plant. A new printer is to be manufactured and she must decide whether the printer will be autoassembled or manually assembled. The decision is complicated by the fact that annual production volume is expected to increase by almost 50% over three years. Jane has developed these estimates for two alternatives for the printer assembly process:

Annual fixed cost Variable cost per product Estimated annual production (in number of products):

Year 1 Year 2 Year 3

AutoAssembly Process

Manual Assembly Process

$690,000 $29.56

$269,000 $31.69

152,000 190,000 225,000

a. Which production process would be the least-cost alternative in Years 1, 2, and 3? b. How much would the variable cost per unit have to be in Year 2 for the auto-assembly process to justify the additional annual fixed cost for the auto-assembly process over the manual assembly process? ANSWER:

a. C(x) = fixed cost + variable cost(x) Year 1: CA = 690,000 + 29.56(152,000) = $5,183,120 CM = 269,000 + 31.69(152,000) = $5,085,880 (least-cost alternative) Year 2: CA = 690,000 + 29.56(190,000) = $6,306,400 CM = 269,000 + 31.69(190,000) = $6,290,100 (least-cost alternative) Year 3: CA = 690,000 + 29.56(225,000) = $7,341,000 (least-cost alternative) CM = 269,000 + 31.69(225,000) = $7,399,250 b. CA = CM FCA + vA(190,000) = FCM + vM(190,000) 690,000 + v(190,000) = 269,000 + 31.69(190,000) vA = (269,000 + 6,021,100 - 690,000)/190,000 vA = $29.47 (roughly a 0.3% reduction)

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Chapter 1 - Introduction TOPICS: Cost and volume models Essay 54. Should the problem solving process be applied to all problems? ANSWER: Answer not provided. POINTS: 1 TOPICS: Problem solving and decision making 55. Explain the difference between quantitative and qualitative analysis from the manager's point of view. ANSWER: Answer not provided. POINTS: 1 TOPICS: Quantitative analysis and decision making 56. Explain the relationship among model development, model accuracy, and the ability to obtain a solution from a model. ANSWER: Answer not provided. POINTS: 1 TOPICS: Model solution 57. What are three of the management science techniques that practitioners use most frequently? How can the effectiveness of these applications be increased? ANSWER: Answer not provided. POINTS: 1 TOPICS: Methods used most frequently 58. What steps of the problem solving process are involved in decision making? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 59. Give three benefits of model development and an example of each. ANSWER: Answer not provided. POINTS: 1 TOPICS: Model development 60. Explain the relationship between information systems specialists and quantitative analysts in the solution of large mathematical problems. ANSWER: Answer not provided. POINTS: 1 TOPICS: Data preparation 61. Define and contrast the terms feasible solution, infeasible solution and optimal solution. ANSWER: Answer not provided. POINTS: 1 TOPICS: Model solution 62. Define three forms of models and provide an example of each. Cengage Learning Testing, Powered by Cognero

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Chapter 1 - Introduction ANSWER: Answer not provided. POINTS: 1 TOPICS: Model development 63. Explain the difference between controllable and uncontrollable inputs to a mathematical model and provide an example of each. ANSWER: Answer not provided. POINTS: 1 TOPICS: Model development

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Chapter 2 - An Introduction to Linear Programming True / False 1. Increasing the right-hand side of a nonbinding constraint will not cause a change in the optimal solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 2. In a linear programming problem, the objective function and the constraints must be linear functions of the decision variables. a. True b. False ANSWER: True POINTS: 1 TOPICS: Mathematical statement of the RMC Problem 3. In a feasible problem, an equal-to constraint cannot be nonbinding. a. True b. False ANSWER: True POINTS: 1 TOPICS: Graphical solution 4. Only binding constraints form the shape (boundaries) of the feasible region. a. True b. False ANSWER: False POINTS: 1 TOPICS: Graphical solution 5. The constraint 5x1 − 2x2 ≤ 0 passes through the point (20, 50). a. True b. False ANSWER: True POINTS: 1 TOPICS: Graphing lines 6. A redundant constraint is a binding constraint. a. True b. False ANSWER: False POINTS: 1 TOPICS: Slack variables

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Chapter 2 - An Introduction to Linear Programming 7. Because surplus variables represent the amount by which the solution exceeds a minimum target, they are given positive coefficients in the objective function. a. True b. False ANSWER: False POINTS: 1 TOPICS: Slack variables 8. Alternative optimal solutions occur when there is no feasible solution to the problem. a. True b. False ANSWER: False POINTS: 1 TOPICS: Alternative optimal solutions 9. A range of optimality is applicable only if the other coefficient remains at its original value. a. True b. False ANSWER: True POINTS: 1 TOPICS: Simultaneous changes 10. Because the dual price represents the improvement in the value of the optimal solution per unit increase in right-handside, a dual price cannot be negative. a. True b. False ANSWER: False POINTS: 1 TOPICS: Right-hand sides 11. Decision variables limit the degree to which the objective in a linear programming problem is satisfied. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 12. No matter what value it has, each objective function line is parallel to every other objective function line in a problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Graphical solution 13. The point (3, 2) is feasible for the constraint 2x1 + 6x2 ≤ 30. Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming a. True b. False ANSWER: True POINTS: 1 TOPICS: Graphical solution 14. The constraint 2x1 − x2 = 0 passes through the point (200,100). a. True b. False ANSWER: False POINTS: 1 TOPICS: A note on graphing lines 15. The standard form of a linear programming problem will have the same solution as the original problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Surplus variables 16. An optimal solution to a linear programming problem can be found at an extreme point of the feasible region for the problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Extreme points 17. An unbounded feasible region might not result in an unbounded solution for a minimization or maximization problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Special cases: unbounded 18. An infeasible problem is one in which the objective function can be increased to infinity. a. True b. False ANSWER: False POINTS: 1 TOPICS: Special cases: infeasibility 19. A linear programming problem can be both unbounded and infeasible. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming ANSWER: False POINTS: 1 TOPICS: Special cases: infeasibility and unbounded 20. It is possible to have exactly two optimal solutions to a linear programming problem. a. True b. False ANSWER: False POINTS: 1 TOPICS: Special cases: alternative optimal solutions Multiple Choice 21. The maximization or minimization of a quantity is the a. goal of management science. b. decision for decision analysis. c. constraint of operations research. d. objective of linear programming. ANSWER: d POINTS: 1 TOPICS: Introduction 22. Decision variables a. tell how much or how many of something to produce, invest, purchase, hire, etc. b. represent the values of the constraints. c. measure the objective function. d. must exist for each constraint. ANSWER: a POINTS: 1 TOPICS: Objective function 23. Which of the following is a valid objective function for a linear programming problem? a. Max 5xy b. Min 4x + 3y + (2/3)z c. Max 5x2 + 6y2 d. Min (x1 + x2)/x3 ANSWER: b POINTS: 1 TOPICS: Objective function 24. Which of the following statements is NOT true? a. A feasible solution satisfies all constraints. b. An optimal solution satisfies all constraints. c. An infeasible solution violates all constraints. Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming d. A feasible solution point does not have to lie on the boundary of the feasible region. ANSWER: c POINTS: 1 TOPICS: Graphical solution 25. A solution that satisfies all the constraints of a linear programming problem except the nonnegativity constraints is called a. optimal. b. feasible. c. infeasible. d. semi-feasible. ANSWER: c POINTS: 1 TOPICS: Graphical solution 26. Slack a. is the difference between the left and right sides of a constraint. b. is the amount by which the left side of a ≤ constraint is smaller than the right side. c. is the amount by which the left side of a ≥ constraint is larger than the right side. d. exists for each variable in a linear programming problem. ANSWER: b POINTS: 1 TOPICS: Slack variables 27. To find the optimal solution to a linear programming problem using the graphical method a. find the feasible point that is the farthest away from the origin. b. find the feasible point that is at the highest location. c. find the feasible point that is closest to the origin. d. None of the alternatives is correct. ANSWER: d POINTS: 1 TOPICS: Extreme points 28. Which of the following special cases does not require reformulation of the problem in order to obtain a solution? a. alternate optimality b. infeasibility c. unboundedness d. each case requires a reformulation. ANSWER: a POINTS: 1 TOPICS: Special cases 29. The improvement in the value of the objective function per unit increase in a right-hand side is the a. sensitivity value. b. dual price. Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming c. constraint coefficient. d. slack value. ANSWER: b POINTS: 1 TOPICS: Right-hand sides 30. As long as the slope of the objective function stays between the slopes of the binding constraints a. the value of the objective function won't change. b. there will be alternative optimal solutions. c. the values of the dual variables won't change. d. there will be no slack in the solution. ANSWER: c POINTS: 1 TOPICS: Objective function 31. Infeasibility means that the number of solutions to the linear programming models that satisfies all constraints is a. at least 1. b. 0. c. an infinite number. d. at least 2. ANSWER: b POINTS: 1 TOPICS: Alternate optimal solutions 32. A constraint that does not affect the feasible region is a a. non-negativity constraint. b. redundant constraint. c. standard constraint. d. slack constraint. ANSWER: b POINTS: 1 TOPICS: Feasible regions 33. Whenever all the constraints in a linear program are expressed as equalities, the linear program is said to be written in a. standard form. b. bounded form. c. feasible form. d. alternative form. ANSWER: a POINTS: 1 TOPICS: Slack variables 34. All of the following statements about a redundant constraint are correct EXCEPT a. A redundant constraint does not affect the optimal solution. b. A redundant constraint does not affect the feasible region. Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming c. Recognizing a redundant constraint is easy with the graphical solution method. d. At the optimal solution, a redundant constraint will have zero slack. ANSWER: d POINTS: 1 TOPICS: Slack variables 35. All linear programming problems have all of the following properties EXCEPT a. a linear objective function that is to be maximized or minimized. b. a set of linear constraints. c. alternative optimal solutions. d. variables that are all restricted to nonnegative values. ANSWER: c POINTS: 1 TOPICS: Problem formulation 36. If there is a maximum of 4,000 hours of labor available per month and 300 ping-pong balls (x1) or 125 wiffle balls (x2) can be produced per hour of labor, which of the following constraints reflects this situation? a. 300x1 + 125x2 > 4,000 b. 300x1 + 125x2 < 4,000 c. 425(x1 + x2) < 4,000 d. 300x1 + 125x2 = 4,000 ANSWER: b POINTS: 1 37. In what part(s) of a linear programming formulation would the decision variables be stated? a. objective function and the left-hand side of each constraint b. objective function and the right-hand side of each constraint c. the left-hand side of each constraint only d. the objective function only ANSWER: a POINTS: 1 38. The three assumptions necessary for a linear programming model to be appropriate include all of the following except a. proportionality b. additivity c. divisibility d. normality ANSWER: d POINTS: 1 39. A redundant constraint results in a. no change in the optimal solution(s) b. an unbounded solution c. no feasible solution Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming d. alternative optimal solutions ANSWER: a POINTS: 1 40. A variable added to the left-hand side of a less-than-or-equal-to constraint to convert the constraint into an equality is a. a standard variable b. a slack variable c. a surplus variable d. a non-negative variable ANSWER: b POINTS: 1 Subjective Short Answer 41. Solve the following system of simultaneous equations. 6X + 2Y = 50 2X + 4Y = 20 ANSWER: X = 8, Y =1 POINTS: 1 TOPICS: Simultaneous equations 42. Solve the following system of simultaneous equations. 6X + 4Y = 40 2X + 3Y = 20 ANSWER: X = 4, Y = 4 POINTS: 1 TOPICS: Simultaneous equations 43. Consider the following linear programming problem Max s.t.

8X + 7Y 15X + 5Y ≤ 75 10X + 6Y ≤ 60 X+ Y≤8 X, Y ≥ 0

Use a graph to show each constraint and the feasible region. Identify the optimal solution point on your graph. What are the values of X and Y at the b. optimal solution? c. What is the optimal value of the objective function? ANSWER:

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Chapter 2 - An Introduction to Linear Programming

The optimal solution occurs at the intersection of constraints 2 and 3. The point is X = 3, Y = 5. The value of the objective function is 59.

c. POINTS: 1 TOPICS: Graphical solution

44. For the following linear programming problem, determine the optimal solution by the graphical solution method −X + 2Y 6X − 2Y ≤ 3 −2X + 3Y ≤ 6 X+ Y≤3 X, Y ≥ 0 ANSWER: X = 0.6 and Y = 2.4 Max s.t.

POINTS: 1 TOPICS: Graphical solution Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming 45. Use this graph to answer the questions.

Max s.t.

20X + 10Y 12X + 15Y ≤ 180 15X + 10Y ≤ 150 3X − 8Y ≤ 0 X,Y≥0

a. Which area (I, II, III, IV, or V) forms the feasible region? b. Which point (A, B, C, D, or E) is optimal? c. Which constraints are binding? d. Which slack variables are zero? ANSWER: a. Area III is the feasible region b. Point D is optimal c. Constraints 2 and 3 are binding d. S2 and S3 are equal to 0 POINTS: 1 TOPICS: Graphical solution 46. Find the complete optimal solution to this linear programming problem. Min s.t.

5X + 6Y 3X + Y ≥ 15 X + 2Y ≥ 12 3X + 2Y ≥ 24 X,Y≥0 ANSWER:

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Chapter 2 - An Introduction to Linear Programming

The complete optimal solution is POINTS: 1 TOPICS: Graphical solution

X = 6, Y = 3, Z = 48, S1 = 6, S2 = 0, S3 = 0

47. Find the complete optimal solution to this linear programming problem. Max s.t.

5X + 3Y 2X + 3Y ≤ 30 2X + 5Y ≤ 40 6X − 5Y ≤ 0 X,Y≥ 0 ANSWER:

The complete optimal solution is Cengage Learning Testing, Powered by Cognero

X = 15, Y = 0, Z = 75, S1 = 0, S2 = 10, S3 = 90 Page 11

Chapter 2 - An Introduction to Linear Programming POINTS: 1 TOPICS: Graphical solution 48. Find the complete optimal solution to this linear programming problem. Max s.t.

2X + 3Y 4X + 9Y ≤ 72 10X + 11Y ≤ 110 17X + 9Y ≤ 153 X,Y≥0 ANSWER:

The complete optimal solution is POINTS: 1 TOPICS: Graphical solution

X = 4.304, Y = 6.087, Z = 26.87, S1 = 0, S2 = 0, S3 = 25.043

49. Find the complete optimal solution to this linear programming problem. Min s.t.

3X + 3Y 12X + 4Y ≥ 48 10X + 5Y ≥ 50 4X + 8Y ≥ 32 X,Y≥0 ANSWER:

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Chapter 2 - An Introduction to Linear Programming

The complete optimal solution is POINTS: 1 TOPICS: Graphical solution

X = 4, Y = 2, Z = 18, S1 = 8, S2 = 0, S3 = 0

50. For the following linear programming problem, determine the optimal solution by the graphical solution method. Are any of the constraints redundant? If yes, then identify the constraint that is redundant. Max s.t.

X + 2Y X+ Y≤3 X − 2Y ≥ 0 Y≤1 X, Y ≥ 0 ANSWER: X = 2, and Y = 1 Yes, there is a redundant constraint; Y ≤ 1

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Chapter 2 - An Introduction to Linear Programming 51. Maxwell Manufacturing makes two models of felt tip marking pens. Requirements for each lot of pens are given below. Fliptop Model 3 5 5

Plastic Ink Assembly Molding Time

Tiptop Model 4 4 2

Available 36 40 30

The profit for either model is $1000 per lot. a. What is the linear programming model for this problem? b. Find the optimal solution. c. Will there be excess capacity in any resource? ANSWER: a. Let F = the number of lots of Fliptop pens to produce Let T = the number of lots of Tiptop pens to produce Max s.t.

1000F + 1000T 3F + 4T ≤ 36 5F + 4T ≤ 40 5F + 2T ≤ 30 F,T≥0

The complete optimal solution is F = 2, T = 7.5, Z = 9500, S1 = 0, S2 = 0, S3 = 5 There is an excess of 5 units of molding time available.

c. POINTS: 1 TOPICS: Modeling and graphical solution

52. The Sanders Garden Shop mixes two types of grass seed into a blend. Each type of grass has been rated (per pound) according to its shade tolerance, ability to stand up to traffic, and drought resistance, as shown in the table. Type A seed costs $1 and Type B seed costs $2. If the blend needs to score at least 300 points for shade tolerance, 400 points for traffic resistance, and 750 points for drought resistance, how many pounds of each seed should be in the blend? Which targets will be exceeded? How much will the blend cost? Shade Tolerance Traffic Resistance

Type A 1 2

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Type B 1 1 Page 14

Chapter 2 - An Introduction to Linear Programming Drought Resistance 2 5 ANSWER: Let A = the pounds of Type A seed in the blend Let B = the pounds of Type B seed in the blend Min s.t.

1A + 2B 1A + 1B ≥ 300 2A + 1B ≥ 400 2A + 5B ≥ 750 A, B ≥ 0

The optimal solution is at A = 250, B = 50. Constraint 2 has a surplus value of 150. The cost is 350. POINTS: 1 TOPICS: Modeling and graphical solution 53. Muir Manufacturing produces two popular grades of commercial carpeting among its many other products. In the coming production period, Muir needs to decide how many rolls of each grade should be produced in order to maximize profit. Each roll of Grade X carpet uses 50 units of synthetic fiber, requires 25 hours of production time, and needs 20 units of foam backing. Each roll of Grade Y carpet uses 40 units of synthetic fiber, requires 28 hours of production time, and needs 15 units of foam backing. The profit per roll of Grade X carpet is $200 and the profit per roll of Grade Y carpet is $160. In the coming production period, Muir has 3000 units of synthetic fiber available for use. Workers have been scheduled to provide at least 1800 hours of production time (overtime is a possibility). The company has 1500 units of foam backing available for use. Develop and solve a linear programming model for this problem. ANSWER: Let X = the number of rolls of Grade X carpet to make Let Y = the number of rolls of Grade Y carpet to make Max

200X + 160Y

s.t.

50X + 40Y ≤ 3000 25X + 28Y ≥ 1800 20X + 15Y ≤ 1500 X,Y≥0

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Chapter 2 - An Introduction to Linear Programming

The complete optimal solution is X = 30, Y = 37.5, Z = 12000, S1 = 0, S2 = 0, S3 = 337.5 POINTS: 1 TOPICS: Modeling and graphical solution 54. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternate optimal solutions? Explain. Min s.t.

1X + 1Y 5X + 3Y ≤ 30 3X + 4Y ≥ 36 Y≤7 X,Y≥0 ANSWER: The problem is infeasible.

POINTS: 1 TOPICS: Special cases 55. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternate optimal solutions? Explain. Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming Min s.t.

3X + 3Y 1X + 2Y ≤ 16 1X + 1Y ≤ 10 5X + 3Y ≤ 45 X,Y≥0 ANSWER: The problem has alternate optimal solutions.

POINTS: 1 TOPICS: Special cases 56. A businessman is considering opening a small specialized trucking firm. To make the firm profitable, it is estimated that it must have a daily trucking capacity of at least 84,000 cu. ft. Two types of trucks are appropriate for the specialized operation. Their characteristics and costs are summarized in the table below. Note that truck 2 requires 3 drivers for long haul trips. There are 41 potential drivers available and there are facilities for at most 40 trucks. The businessman's objective is to minimize the total cost outlay for trucks. Truck Small Large

Cost $18,000 $45,000

Capacity (Cu. Ft.) 2,400 6,000

Drivers Needed 1 3

Solve the problem graphically and note there are alternate optimal solutions. Which optimal solution: a. uses only one type of truck? b. utilizes the minimum total number of trucks? c. uses the same number of small and large trucks? ANSWER: a. 35 small, 0 large b. 5 small, 12 large c. 10 small, 10 large POINTS: 1 TOPICS: Alternative optimal solutions 57. Consider the following linear program: Max s.t.

60X + 43Y X + 3Y ≥ 9

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Chapter 2 - An Introduction to Linear Programming 6X − 2Y = 12 X + 2Y ≤ 10 X, Y ≥ 0 a. b.

Write the problem in standard form. What is the feasible region for the problem? Show that regardless of the values of the actual objective function coefficients, the optimal c. solution will occur at one of two points. Solve for these points and then determine which one maximizes the current objective function. ANSWER: a. Max 60X + 43Y s.t. X + 3Y − S1 = 9 6X − 2Y = 12 X + 2Y + S3 = 10 X, Y, S1, S3 ≥ 0 b. Line segment of 6X − 2Y = 12 between (22/7,24/7) and (27/10,21/10). c. Extreme points: (22/7,24/7) and (27/10,21/10). First one is optimal, giving Z = 336. POINTS: 1 TOPICS: Standard form and extreme points 58. Solve the following linear program graphically. Max s.t.

5X + 7Y X ≤6 2X + 3Y ≤ 19 X+ Y≤8 X, Y ≥ 0 ANSWER: From the graph below we see that the optimal solution occurs at X = 5, Y = 3, and Z = 46.

POINTS: 1 TOPICS: Graphical solution procedure 59. Given the following linear program: Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming Min s.t.

150X + 210Y 3.8X + 1.2Y ≥ 22.8 Y≥6 Y ≤ 15 45X + 30Y = 630 X, Y ≥ 0

Solve the problem graphically. How many extreme points exist for this problem? ANSWER: Two extreme points exist (Points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760 (Point B).

POINTS: 1 TOPICS: Graphical solution procedure 60. Solve the following linear program by the graphical method. Max s.t.

4X + 5Y X + 3Y ≤ 22 −X + Y ≤ 4 Y≤6 2X − 5Y ≤ 0 X, Y ≥ 0 ANSWER: Two extreme points exist (Points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760 (Point B).

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Chapter 2 - An Introduction to Linear Programming

POINTS: 1 TOPICS: Graphical solution procedure Essay 61. Explain the difference between profit and contribution in an objective function. Why is it important for the decision maker to know which of these the objective function coefficients represent? ANSWER: Answer not provided. POINTS: 1 TOPICS: Objective function 62. Explain how to graph the line x1 − 2x2 ≥ 0. ANSWER: Answer not provided. POINTS: 1 TOPICS: Graphing lines 63. Create a linear programming problem with two decision variables and three constraints that will include both a slack and a surplus variable in standard form. Write your problem in standard form. ANSWER: Answer not provided. POINTS: 1 TOPICS: Standard form 64. Explain what to look for in problems that are infeasible or unbounded. ANSWER: Answer not provided. POINTS: 1 TOPICS: Special cases 65. Use a graph to illustrate why a change in an objective function coefficient does not necessarily lead to a change in the optimal values of the decision variables, but a change in the right-hand sides of a binding constraint does lead to new values. ANSWER: Answer not provided. POINTS: 1 TOPICS: Graphical sensitivity analysis Cengage Learning Testing, Powered by Cognero

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Chapter 2 - An Introduction to Linear Programming 66. Explain the concepts of proportionality, additivity, and divisibility. ANSWER: Answer not provided. POINTS: 1 TOPICS: Notes and comments 67. Explain the steps necessary to put a linear program in standard form. ANSWER: Answer not provided. POINTS: 1 TOPICS: Surplus variables 68. Explain the steps of the graphical solution procedure for a minimization problem. ANSWER: Answer not provided. POINTS: 1 TOPICS: Graphical solution procedure for minimization problems

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution True / False 1. Classical sensitivity analysis provides no information about changes resulting from a change in the coefficient of a variable in a constraint. a. True b. False ANSWER: True POINTS: 1 TOPICS: Changes in constraint coefficients 2. The reduced cost for a positive decision variable is 0. a. True b. False ANSWER: True POINTS: 1 TOPICS: Reduced cost 3. When the right-hand sides of two constraints are each increased by one unit, the objective function value will be adjusted by the sum of the constraints' dual prices. a. True b. False ANSWER: False POINTS: 1 TOPICS: Simultaneous changes 4. If the range of feasibility indicates that the original amount of a resource, which was 20, can increase by 5, then the amount of the resource can increase to 25. a. True b. False ANSWER: True POINTS: 1 TOPICS: Range of feasibility 5. The 100% Rule does not imply that the optimal solution will necessarily change if the percentage exceeds 100%. a. True b. False ANSWER: True POINTS: 1 TOPICS: Simultaneous changes 6. For any constraint, either its slack/surplus value must be zero or its dual price must be zero. a. True b. False ANSWER: True POINTS: 1 TOPICS: Dual price Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 7. A negative dual price indicates that increasing the right-hand side of the associated constraint would be detrimental to the objective. a. True b. False ANSWER: True POINTS: 1 TOPICS: Dual price 8. In order to tell the impact of a change in a constraint coefficient, the change must be made and then the model resolved. a. True b. False ANSWER: True POINTS: 1 TOPICS: Changes in constraint coefficients 9. Decreasing the objective function coefficient of a variable to its lower limit will create a revised problem that is unbounded. a. True b. False ANSWER: False POINTS: 1 TOPICS: Range of optimality 10. The dual price for a percentage constraint provides a direct answer to questions about the effect of increases or decreases in that percentage. a. True b. False ANSWER: False POINTS: 1 TOPICS: Dual price 11. The dual price associated with a constraint is the change in the value of the solution per unit decrease in the right-hand side of the constraint. a. True b. False ANSWER: False POINTS: 1 TOPICS: Interpretation of computer output 12. For a minimization problem, a positive dual price indicates the value of the objective function will increase. a. True b. False ANSWER: False POINTS: 1 TOPICS: Interpretation of computer output--a second example Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 13. There is a dual price for every decision variable in a model. a. True b. False ANSWER: False POINTS: 1 TOPICS: Interpretation of computer output 14. The amount of a sunk cost will vary depending on the values of the decision variables. a. True b. False ANSWER: False POINTS: 1 TOPICS: Cautionary note on the interpretation of dual prices 15. If the optimal value of a decision variable is zero and its reduced cost is zero, this indicates that alternative optimal solutions exist. a. True b. False ANSWER: True POINTS: 1 TOPICS: Interpretation of computer output 16. Any change to the objective function coefficient of a variable that is positive in the optimal solution will change the optimal solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: Range of optimality 17. Relevant costs should be reflected in the objective function, but sunk costs should not. a. True b. False ANSWER: True POINTS: 1 TOPICS: Cautionary note on the interpretation of dual prices 18. If the range of feasibility for b1 is between 16 and 37, then if b1 = 22 the optimal solution will not change from the original optimal solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: Right-hand sides Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 19. The 100 percent rule can be applied to changes in both objective function coefficients and right-hand sides at the same time. a. True b. False ANSWER: False POINTS: 1 TOPICS: Simultaneous changes 20. If the dual price for the right-hand side of a ≤ constraint is zero, there is no upper limit on its range of feasibility. a. True b. False ANSWER: True POINTS: 1 TOPICS: Right-hand sides Multiple Choice 21. To solve a linear programming problem with thousands of variables and constraints a. a personal computer can be used. b. a mainframe computer is required. c. the problem must be partitioned into subparts. d. unique software would need to be developed. ANSWER: a POINTS: 1 TOPICS: Computer solution 22. A negative dual price for a constraint in a minimization problem means a. as the right-hand side increases, the objective function value will increase. b. as the right-hand side decreases, the objective function value will increase. c. as the right-hand side increases, the objective function value will decrease. d. as the right-hand side decreases, the objective function value will decrease. ANSWER: a POINTS: 1 TOPICS: Dual price 23. If a decision variable is not positive in the optimal solution, its reduced cost is a. what its objective function value would need to be before it could become positive. b. the amount its objective function value would need to improve before it could become positive. c. zero. d. its dual price. ANSWER: b POINTS: 1 TOPICS: Reduced cost 24. A constraint with a positive slack value Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution a. will have a positive dual price. b. will have a negative dual price. c. will have a dual price of zero. d. has no restrictions for its dual price. ANSWER: c POINTS: 1 TOPICS: Slack and dual price 25. The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the a. optimal solution. b. dual solution. c. range of optimality. d. range of feasibility. ANSWER: c POINTS: 1 TOPICS: Range of optimality 26. The range of feasibility measures a. the right-hand-side values for which the objective function value will not change. b. the right-hand-side values for which the values of the decision variables will not change. c. the right-hand-side values for which the dual prices will not change. d. each of these choices are true. ANSWER: c POINTS: 1 TOPICS: Range of feasibility 27. The 100% Rule compares a. proposed changes to allowed changes. b. new values to original values. c. objective function changes to right-hand side changes. d. dual prices to reduced costs. ANSWER: a POINTS: 1 TOPICS: Simultaneous changes 28. An objective function reflects the relevant cost of labor hours used in production rather than treating them as a sunk cost. The correct interpretation of the dual price associated with the labor hours constraint is a. the maximum premium (say for overtime) over the normal price that the company would be willing to pay. b. the upper limit on the total hourly wage the company would pay. c. the reduction in hours that could be sustained before the solution would change. d. the number of hours by which the right-hand side can change before there is a change in the solution point. ANSWER: a POINTS: 1 TOPICS: Dual price Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 29. A section of output from The Management Scientist is shown here. Variable 1

Lower Limit 60

Current Value 100

Upper Limit 120

What will happen to the solution if the objective function coefficient for variable 1 decreases by 20? a. Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same. b. The value of the objective function will change, but the values of the decision variables and the dual prices will remain the same. c. The same decision variables will be positive, but their values, the objective function value, and the dual prices will change. d. The problem will need to be resolved to find the new optimal solution and dual price. ANSWER: b POINTS: 1 TOPICS: Range of optimality 30. A section of output from The Management Scientist is shown here. Constraint 2

Lower Limit 240

Current Value 300

Upper Limit 420

What will happen if the right-hand-side for constraint 2 increases by 200? a. Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same. b. The value of the objective function will change, but the values of the decision variables and the dual prices will remain the same. c. The same decision variables will be positive, but their values, the objective function value, and the dual prices will change. d. The problem will need to be resolved to find the new optimal solution and dual price. ANSWER: d POINTS: 1 TOPICS: Range of feasibility 31. The amount the objective function coefficient of a decision variable would have to improve before that variable would have a positive value in the solution is the a. dual price. b. surplus variable. c. reduced cost. d. upper limit. ANSWER: c POINTS: 1 TOPICS: Interpretation of computer output 32. The dual price measures, per unit increase in the right hand side of the constraint, a. the increase in the value of the optimal solution. b. the decrease in the value of the optimal solution. Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution c. the improvement in the value of the optimal solution. d. the change in the value of the optimal solution. ANSWER: d POINTS: 1 TOPICS: Interpretation of computer output 33. Sensitivity analysis information in computer output is based on the assumption of a. no coefficient changes. b. one coefficient changes. c. two coefficients change. d. all coefficients change. ANSWER: b POINTS: 1 TOPICS: Simultaneous changes 34. When the cost of a resource is sunk, then the dual price can be interpreted as the a. minimum amount the firm should be willing to pay for one additional unit of the resource. b. maximum amount the firm should be willing to pay for one additional unit of the resource. c. minimum amount the firm should be willing to pay for multiple additional units of the resource. d. maximum amount the firm should be willing to pay for multiple additional units of the resource. ANSWER: b POINTS: 1 TOPICS: Dual price 35. Which of the following is not a question answered by standard sensitivity analysis information? a. If the right-hand side value of a constraint changes, will the objective function value change? b. Over what range can a constraint's right-hand side value without the constraint's dual price possibly changing? c. By how much will the objective function value change if the right-hand side value of a constraint changes beyond the range of feasibility? d. By how much will the objective function value change if a decision variable's coefficient in the objective function changes within the range of optimality? ANSWER: c POINTS: 1 TOPICS: Interpretation of computer output 36. The cost that varies depending on the values of the decision variables is a a. reduced cost. b. relevant cost. c. sunk cost. d. dual cost. ANSWER: b POINTS: 1 TOPICS: Sunk and relevant costs 37. A cost that is incurred no matter what values the decision variables assume is Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution a. a reduced cost. b. an optimal cost. c. a sunk cost. d. a dual cost. ANSWER: c POINTS: 1 TOPICS: Sunk and relevant costs 38. Sensitivity analysis is often referred to as a. feasibility testing. b. duality analysis. c. alternative analysis. d. postoptimality analysis. ANSWER: d POINTS: 1 TOPICS: Introduction 39. Sensitivity analysis is concerned with how certain changes affect a. the feasible solution. b. the unconstrained solution. c. the optimal solution. d. the degenerative solution. ANSWER: c POINTS: 1 TOPICS: Introduction 40. The dual price for a < constraint a. will always be < 0. b. will always be > 0. c. will be < 0 in a minimization problem and > 0 in a maximization problem. d. will always equal 0. ANSWER: b POINTS: 1 TOPICS: Dual price Subjective Short Answer 41. In a linear programming problem, the binding constraints for the optimal solution are 5X + 3Y ≤ 30 2X + 5Y ≤ 20 a.

Fill in the blanks in the following sentence: As long as the slope of the objective function stays between _______ and _______, the current optimal solution point will remain optimal. Which of these objective functions will lead to the same optimal solution? 1) 2X + 1Y 2) 7X + 8Y 3) 80X + 60Y 4) 25X + 35Y

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution ANSWER:

a. −5/3 and −2/5 b. Objective functions 2), 3), and 4) POINTS: 1 TOPICS: Graphical sensitivity analysis 42. The optimal solution of the linear programming problem is at the intersection of constraints 1 and 2. Max s.t.

2x1 + x2 4x1 + 1x2 ≤ 400 4x1 + 3x2 ≤ 600 1x1 + 2x2 ≤ 300 x1 , x2 ≥ 0

Over what range can the coefficient of x1 vary before the current solution is no longer optimal? b. Over what range can the coefficient of x2 vary before the current solution is no longer optimal? c. Compute the dual prices for the three constraints. ANSWER: a. 1.33 ≤ c1 ≤ 4 b. .5 ≤ c2 ≤ 1.5 c. Dual prices are .25, .25, 0 POINTS: 1 TOPICS: Graphical sensitivity analysis a.

43. The binding constraints for this problem are the first and second. Min

x1 + 2x2

s.t.

x1 + x2 ≥ 300 2x1 + x2 ≥ 400 2x1 + 5x2 ≤ 750 x1 , x2 ≥ 0

Keeping c2 fixed at 2, over what range can c1 vary before there is a change in the optimal solution point? b. Keeping c1 fixed at 1, over what range can c2 vary before there is a change in the optimal solution point? c. If the objective function becomes Min 1.5x1 + 2x2, what will be the optimal values of x1, x2, and the objective function? d. If the objective function becomes Min 7x1 + 6x2, what constraints will be binding? e. Find the dual price for each constraint in the original problem. ANSWER: a. .8 ≤ c1 ≤ 2 b. 1 ≤ c2 ≤ 2.5 c. x1 = 250, x2 = 50, z = 475 d. Constraints 1 and 2 will be binding. a.

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution e. Dual prices are .33, 0, .33 (The first and third values are negative.) POINTS: 1 TOPICS: Graphical sensitivity analysis 44. Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a maximization objective function and all ≤ constraints. Input Section Objective Function Coefficients X Y 4 6 Constraints #1 #2 #3

3 3 1

5 2 1

Avail. 60 48 20

Variables Profit

13.333333 53.333333

4 24

77.333333

Constraint #1 #2 #3

Usage 60 48 17.333333

Slack 1.789E-11 -2.69E-11 2.6666667

Output Section

a. Give the original linear programming problem. b. Give the complete optimal solution. ANSWER: a. Max 4X + 6Y 3X + 5Y ≤ 60 3X + 2Y ≤ 48 1X + 1Y ≤ 20 X,Y≥0 The complete optimal solution is X = 13.333, Y = 4, Z = 73.333, S1 = 0, S2 = 0, S3 = 2.667 s.t.

POINTS: 1 TOPICS: Spreadsheet solution of LPs 45. Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a minimization objective function and all ≥ constraints. Input Section Objective Function Coefficients X 5 Cengage Learning Testing, Powered by Cognero

Y 4 Page 10

Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution Constraints #1 #2 #3

4 2 9

3 5 8

Req'd 60 50 144

Variables Profit

9.6 48

7.2 28.8

76.8

Constraint #1 #2 #3

Usage 60 55.2 144

Slack 1.35E-11 -5.2 -2.62E-11

Output Section

a. Give the original linear programming problem. b. Give the complete optimal solution. ANSWER: a. Min 5X + 4Y 4X + 3Y ≥ 60 2X + 5Y ≥ 50 9X + 8Y ≥ 144 X,Y≥0 The complete optimal solution is X = 9.6, Y = 7.2, Z = 76.8, S1 = 0, S2 = 5.2, S3 = 0 s.t.

POINTS: 1 TOPICS: Spreadsheet solution of LPs 46. Use the spreadsheet and Solver sensitivity report to answer these questions. a. What is the cell formula for B12? b. What is the cell formula for C12? c. What is the cell formula for D12? d. What is the cell formula for B15? e. What is the cell formula for B16? f. What is the cell formula for B17? g. What is the optimal value for x1? h. What is the optimal value for x2? i. Would you pay $.50 each for up to 60 more units of resource 1? Is it possible to figure the new objective function value if the profit on product 1 increases by j. a dollar, or do you have to rerun Solver? A 1 2 3 4 5 6 7

Var. 1 2 3 1

Var. 2 5 1 1

(type) < < >

Avail. 40 30 12

Input Information Constraint 1 Constraint 2 Constraint 3

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 8 Profit 9 10 Output Information 11 Variables 12 Profit 13 14 Resources 15 Constraint 1 16 Constraint 2 17 Constraint 3 18 19

= Total Used

Slack/Surplus

Name Variable 1 Variable 2

Final Reduced Value Cost 8.461538462 0 4.615384615 0

Objective Coefficient 5 4

Final Value

Shadow Price

Constraint Allowable R.H. Side Increase

0.538461538 40

110

1.307692308 30

4.666666667

Sensitivity Report Changing Cells Cell $B$12 $C$12

Allowable Increase 7 8.5

Allowable Decrease 3.4 2.333333333

Constraints Cell $B$15 $B$16 $B$17

Name constraint 1 Used constraint 2 Used constraint 3 Used

13.07692308 0

Allowable Decrease

1.076923077 1E+30

ANSWER:

a. =B8*B11 b. =C8*C11 c. =B12+C12 d. =B4*B11+C4*C11 e. =B5*B11+C5*C11 f. =B6*B11+C6*C11 g. 8.46 h. 4.61 i. yes j. no POINTS: 1 TOPICS: Spreadsheet solution of LPs 47. Use the following Management Scientist output to answer the questions. LINEAR PROGRAMMING PROBLEM MAX 31X1+35X2+32X3 Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution S.T. 1) 3X1+5X2+2X3>90 2) 6X1+7X2+8X3<150 3) 5X1+3X2+3X3<120 OPTIMAL SOLUTION Objective Function Value = 763.333 Variable X1 X2 X3

Value 13.333 10.000 0.000

Reduced Cost 0.000 0.000 10.889

Constraint 1 2 3

Slack/Surplus 0.000 0.000 23.333

Dual Price −0.778 5.556 0.000

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3

Lower Limit 30.000 No Lower Limit No Lower Limit

Current Value 31.000 35.000 32.000

Upper Limit No Upper Limit 36.167 42.889

Current Value 90.000 150.000 120.000

Upper Limit 107.143 163.125 No Upper Limit

RIGHT HAND SIDE RANGES Constraint 1 2 3

Lower Limit 77.647 126.000 96.667

a. Give the solution to the problem. b. Which constraints are binding? c. What would happen if the coefficient of x1 increased by 3? d. What would happen if the right-hand side of constraint 1 increased by 10? ANSWER: a. x1 = 13.33, x2 = 10, x3 = 0, s1 = 0, s2 = 0, s3 = 23.33, z = 763.33 b. Constraints 1 and 2 are binding. c. The value of the objective function would increase by 40. d. The value of the objective function would decrease by 7.78. POINTS: 1 TOPICS: Interpretation of Management Scientist output 48. Use the following Management Scientist output to answer the questions. MIN 4X1+5X2+6X3 S.T. 1) X1+X2+X3<85 2) 3X1+4X2+2X3>280 3) 2X1+4X2+4X3>320 Objective Function Value = 400.000 Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution Variable X1 X2 X3

Value 0.000 80.000 0.000

Reduced Cost 1.500 0.000 1.000

Constraint 1 2 3

Slack/Surplus 5.000 40.000 0.000

Dual Price 0.000 0.000 −1.250

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3

Lower Limit 2.500 0.000 5.000

Current Value 4.000 5.000 6.000

Upper Limit No Upper Limit 6.000 No Upper Limit

Current Value 85.000 280.000 320.000

Upper Limit No Upper Limit 320.000 340.000

RIGHT HAND SIDE RANGES Constraint 1 2 3

Lower Limit 80.000 No Lower Limit 280.000

a. What is the optimal solution, and what is the value of the profit contribution? b. Which constraints are binding? c. What are the dual prices for each resource? Interpret. d. Compute and interpret the ranges of optimality. e. Compute and interpret the ranges of feasibility. ANSWER: a. x1 = 0, x2 = 80, x3 = 0, s1 = 5, s2 = 40, s3 = 0, Z = 400 b. Constraint 3 is binding. c. Dual prices are 0, 0, and −1.25. They measure the improvement in Z per unit increase in each right-hand side. d. 2.5 ≤ c1 < ∞ 0 ≤ c2 ≤ 6 5 ≤ c3 < ∞

As long as the objective function coefficient stays within its range, the current optimal solution point will not change, although Z could. 80 ≤ b1 < ∞ −∞ < b2 ≤ 320 280 ≤ b3 ≤ 340 As long as the right-hand side value stays within its range, the currently binding constraints will remain so, although the values of the decision variables could change. The dual variable values will remain the same.

POINTS: 1 TOPICS: Interpretation of Management Scientist output 49. The following linear programming problem has been solved by The Management Scientist. Use the output to answer the questions. Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution LINEAR PROGRAMMING PROBLEM MAX 25X1+30X2+15X3 S.T. 1) 4X1+5X2+8X3<1200 2) 9X1+15X2+3X3<1500 OPTIMAL SOLUTION Objective Function Value = 4700.000 Variable X1 X2 X3

Value 140.000 0.000 80.000

Reduced Cost 0.000 10.000 0.000

Constraint 1 2

Slack/Surplus 0.000 0.000

Dual Price 1.000 2.333

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3

Lower Limit 19.286 No Lower Limit 8.333

Current Value 25.000 30.000 15.000

Upper Limit 45.000 40.000 50.000

Current Value 1200.000 1500.000

Upper Limit 4000.000 2700.000

RIGHT HAND SIDE RANGES Constraint 1 2 a. b. c. d. e.

Lower Limit 666.667 450.000

Give the complete optimal solution. Which constraints are binding? What is the dual price for the second constraint? What interpretation does this have? Over what range can the objective function coefficient of x2 vary before a new solution point becomes optimal? By how much can the amount of resource 2 decrease before the dual price will change?

What would happen if the first constraint's right-hand side increased by 700 and the second's decreased by 350? ANSWER: a. x1 = 140, x2 = 0, x3 = 80, s1 = 0, s2 = 0, z = 4700 b. Constraints 1 and 2 are binding. Dual price 2 = 2.33. A unit increase in the right-hand side of constraint 2 will increase the c. value of the objective function by 2.33. d. As long as c2 ≤ 40, the solution will be unchanged. e. 1050 The sum of percentage changes is 700/2800 + (−350)/(−1050) < 1 so the solution will not f. change. POINTS: 1 f.

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution TOPICS: Interpretation of Management Scientist output 50. LINDO output is given for the following linear programming problem. MIN 12 X1 + 10 X2 + 9 X3 SUBJECT TO 2) 5 X1 + 8 X2 + 5 X3 >= 60 3) 8 X1 + 10 X2 + 5 X3 >= 80 END LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 80.000000 VARIABLE X1 X2 X3 ROW 2) 3)

VALUE .000000 8.000000 .000000

REDUCED COST 4.000000 .000000 4.000000

SLACK OR SURPLUS 4.000000 .000000

DUAL PRICE .000000 −1.000000

NO. ITERATIONS= 1 RANGES IN WHICH THE BASIS IS UNCHANGED:

VARIABLE X1 X2 X3

ROW 2 3

CURRENT COEFFICIENT 12.000000 10.000000 9.000000 CURRENT RHS 60.000000 80.000000

OBJ. COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE INFINITY 4.000000 5.000000 10.000000 INFINITY 4.000000

RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 4.000000 INFINITY INFINITY 5.000000

a. What is the solution to the problem? b. Which constraints are binding? c. Interpret the reduced cost for x1. d. Interpret the dual price for constraint 2. e. What would happen if the cost of x1 dropped to 10 and the cost of x2 increased to 12? ANSWER: a. x1 = 0, x2 = 8, x3 = 0, s1 = 4, s2 = 0, z = 80 b. Constraint 2 is binding. c. c1 would have to decrease by 4 or more for x1 to become positive. d. Increasing the right-hand side by 1 will cause a negative improvement, or increase, of 1 in Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution this minimization objective function. The sum of the percentage changes is (−2)/(−4) + 2/5 < 1 so the solution would not change.

e. POINTS: 1 TOPICS: Interpretation of LINDO output

51. The LP problem whose output follows determines how many necklaces, bracelets, rings, and earrings a jewelry store should stock. The objective function measures profit; it is assumed that every piece stocked will be sold. Constraint 1 measures display space in units, constraint 2 measures time to set up the display in minutes. Constraints 3 and 4 are marketing restrictions. LINEAR PROGRAMMING PROBLEM MAX 100X1+120X2+150X3+125X4 S.T. 1) X1+2X2+2X3+2X4<108 2) 3X1+5X2+X4<120 3) X1+X3<25 4) X2+X3+X4>50 OPTIMAL SOLUTION Objective Function Value = 7475.000 Variable X1 X2 X3 X4

Value 8.000 0.000 17.000 33.000

Reduced Cost 0.000 5.000 0.000 0.000

Constraint 1 2 3 4

Slack/Surplus 0.000 63.000 0.000 0.000

Dual Price 75.000 0.000 25.000 −25.000

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3 X4

Lower Limit 87.500 No Lower Limit 125.000 120.000

Current Value 100.000 120.000 150.000 125.000

Upper Limit No Upper Limit 125.000 162.500 150.000

Current Value 108.000 120.000 25.000 50.000

Upper Limit 123.750 No Upper Limit 58.000 54.000

RIGHT HAND SIDE RANGES Constraint 1 2 3 4

Lower Limit 100.000 57.000 8.000 41.500

Use the output to answer the questions. a.

How many necklaces should be stocked?

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution b. c. d. e. f. g. h. i. j. k.

Now many bracelets should be stocked? How many rings should be stocked? How many earrings should be stocked? How much space will be left unused? How much time will be used? By how much will the second marketing restriction be exceeded? What is the profit? To what value can the profit on necklaces drop before the solution would change? By how much can the profit on rings increase before the solution would change? By how much can the amount of space decrease before there is a change in the profit? You are offered the chance to obtain more space. The offer is for 15 units and the total price l. is 1500. What should you do? ANSWER: a. 8 b. 0 c. 17 d. 33 e. 0 f. 57 g. 0 h. 7475 i. 87.5 j 12.5 k. 0 l. Say no. Although 15 units can be evaluated, their value (1125) is less than the cost (1500). POINTS: 1 TOPICS: Interpretation of Management Scientist output 52. The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3. Use the output to answer these questions. a. How much of ingredient 1 will be put into the blend? b. How much of ingredient 2 will be put into the blend? c. How much of ingredient 3 will be put into the blend? d. How much resource A is used? e. How much resource B will be left unused? f. What will the profit be? g. What will happen to the solution if the profit from ingredient 2 drops to 4? h. What will happen to the solution if the profit from ingredient 3 increases by 1? i. What will happen to the solution if the amount of resource C increases by 2? What will happen to the solution if the minimum requirement for ingredient 3 increases to j. 15? LINEAR PROGRAMMING PROBLEM MAX 4X1+6X2+7X3 S.T. 1) 3X1+2X2+5X3<120 2) 1X1+3X2+3X3<80 3) 5X1+5X2+8X3<160 4) +1X3>10 Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution OPTIMAL SOLUTION Objective Function Value = 166.000 Variable X1 X2 X3

Value 0.000 16.000 10.000

Reduced Cost 2.000 0.000 0.000

Constraint 1 2 3 4

Slack/Surplus 38.000 2.000 0.000 0.000

Dual Price 0.000 0.000 1.200 −2.600

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3

Lower Limit No Lower Limit 4.375 No Lower Limit

Current Value 4.000 6.000 7.000

Upper Limit 6.000 No Upper Limit 9.600

Current Value 120.000 80.000 160.000 10.000

Upper Limit No Upper Limit No Upper Limit 163.333 20.000

RIGHT HAND SIDE RANGES Constraint 1 2 3 4 ANSWER:

Lower Limit 82.000 78.000 80.000 8.889

a. 0 b. 16 c. 10 d. 44 e. 2 f. 166 g. rerun h. Z = 176 i. Z = 168.4 j. Z = 153 POINTS: 1 TOPICS: Interpretation of Management Scientist output 53. The LP model and LINDO output represent a problem whose solution will tell a specialty retailer how many of four different styles of umbrellas to stock in order to maximize profit. It is assumed that every one stocked will be sold. The variables measure the number of women's, golf, men's, and folding umbrellas, respectively. The constraints measure storage space in units, special display racks, demand, and a marketing restriction, respectively. MAX 4 X1 + 6 X2 + 5 X3 + 3.5 X4 SUBJECT TO 2) 2 X1 + 3 X2 + 3 X3 + X4 <= 120 3) 1.5 X1 + 2 X2 <= 54 4) 2 X2 + X3 + X4 <= 72 Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 5) X2 + X3 >= 12 END OBJECTIVE FUNCTION VALUE 1) 318.00000 VARIABLE X1 X2 X3 X4 ROW 2) 3) 4) 5)

VALUE 12.000000 .000000 12.000000 60.000000

REDUCED COST .000000 .500000 .000000 .000000

SLACK OR SURPLUS .000000 36.000000 .000000 .000000

DUAL PRICE 2.000000 .000000 1.500000 −2.500000

RANGES IN WHICH THE BASIS IS UNCHANGED:

VARIABLE X1 X2 X3 X4

ROW 2 3 4 5

CURRENT COEFFICIENT 4.000000 6.000000 5.000000 3.500000 CURRENT RHS 120.000000 54.000000 72.000000 12.000000

OBJ. COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 1.000000 2.500000 .500000 INFINITY 2.500000 .500000 INFINITY .500000

RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 48.000000 24.000000 INFINITY 36.000000 24.000000 48.000000 12.000000 12.000000

Use the output to answer the questions. a. b. c. d. e. f. g. h.

How many women's umbrellas should be stocked? How many golf umbrellas should be stocked? How many men's umbrellas should be stocked? How many folding umbrellas should be stocked? How much space is left unused? How many racks are used? By how much is the marketing restriction exceeded? What is the total profit? By how much can the profit on women's umbrellas increase before the solution would i. change? j. To what value can the profit on golf umbrellas increase before the solution would change? k. By how much can the amount of space increase before there is a change in the dual price? You are offered an advertisement that should increase the demand constraint from 72 to 86 l. for a total cost of $20. Would you say yes or no? ANSWER: a. 12 Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution b. c. d. e. f. g. h. i. j. k. l.

0 12 60 0 18 0 318 1 6.5 48 Yes. The dual price is 1.5 for 24 additional units. The value of the ad (14)(1.5)=21 exceeds the cost of 20.

POINTS: 1 TOPICS: Interpretation of solution 54. Eight of the entries have been deleted from the LINDO output that follows. Use what you know about linear programming to find values for the blanks. MIN 6 X1 + 7.5 X2 + 10 X3 SUBJECT TO 2) 25 X1 + 35 X2 + 30 X3 >= 2400 3) 2 X1 + 4 X2 + 8 X3 >= 400 END LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 612.50000 VARIABLE X1 X2 X3 ROW 2) 3)

VALUE ________ ________ 27.500000

REDUCED COST 1.312500 ________ ________

SLACK OR SURPLUS ________ ________

DUAL PRICE −.125000 −.781250

NO. ITERATIONS= 2 RANGES IN WHICH THE BASIS IS UNCHANGED:

VARIABLE X1 X2 X3

CURRENT COEFFICIENT 6.000000 7.500000 10.000000

OBJ. COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE _________ _________ 1.500000 2.500000 5.000000 3.571429 RIGHTHAND SIDE RANGES

ROW

CURRENT

ALLOWABLE

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ALLOWABLE Page 21

Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution RHS INCREASE DECREASE 2 2400.000000 1100.000000 900.000000 3 400.000000 240.000000 125.714300 ANSWER: It is easiest to calculate the values in this order. x1 = 0, x2 = 45, reduced cost 2 = 0, reduced cost 3 = 0, row 2 slack = 0, row 3 slack = 0, c1 allowable decrease = 1.3125, allowable increase = infinity POINTS: 1 TOPICS: Interpretation of solution 55. Portions of a Management Scientist output are shown below. Use what you know about the solution of linear programs to fill in the ten blanks. LINEAR PROGRAMMING PROBLEM MAX 12X1+9X2+7X3 S.T. 1) 3X1+5X2+4X3<150 2) 2X1+1X2+1X3<64 3) 1X1+2X2+1X3<80 4) 2X1+4X2+3X3>116 OPTIMAL SOLUTION Objective Function Value = 336.000 Variable X1 X2 X3

Value ______ 24.000 ______

Reduced Cost 0.000 ______ 3.500

Constraint 1 2 3 4

Slack/Surplus 0.000 ______ ______ 0.000

Dual Price 15.000 0.000 0.000 ______

OBJECTIVE COEFFICIENT RANGES Variable X1 X2 X3

Lower Limit 5.400 2.000 No Lower Limit

Current Value 12.000 9.000 7.000

Upper Limit No Upper Limit 20.000 10.500

RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value 1 145.000 150.000 2 ______ ______ 3 ______ ______ 4 110.286 116.000 ANSWER: x3 = 0 because the reduced cost is positive.

Upper Limit 156.667 64.000 80.000 120.000

x1 = 24 after plugging into the objective function Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution The second reduced cost is 0. s2 = 20 and s3 = 22 from plugging into the constraints. The fourth dual price is −16.5 from plugging into the dual objective function, which your students might not understand fully until Chapter 6. The lower limit for constraint 2 is 44 and for constraint 3 is 58, from the amount of slack in each constraint. There are no upper limits for these constraints. POINTS: 1 TOPICS: Interpretation of solution 56. A large sporting goods store is placing an order for bicycles with its supplier. Four models can be ordered: the adult Open Trail, the adult Cityscape, the girl's Sea Sprite, and the boy's Trail Blazer. It is assumed that every bike ordered will be sold, and their profits, respectively, are 30, 25, 22, and 20. The LP model should maximize profit. There are several conditions that the store needs to worry about. One of these is space to hold the inventory. An adult's bike needs two feet, but a child's bike needs only one foot. The store has 500 feet of space. There are 1200 hours of assembly time available. The child's bike need 4 hours of assembly time; the Open Trail needs 5 hours and the Cityscape needs 6 hours. The store would like to place an order for at least 275 bikes. a. Formulate a model for this problem. b. Solve your model with any computer package available to you. c. How many of each kind of bike should be ordered and what will the profit be? d. What would the profit be if the store had 100 more feet of storage space? e. If the profit on the Cityscape increases to $35, will any of the Cityscape bikes be ordered? f. Over what range of assembly hours is the dual price applicable? If we require 5 more bikes in inventory, what will happen to the value of the optimal g. solution? h. Which resource should the company work to increase, inventory space or assembly time? ANSWER: NOTE TO INSTRUCTOR: The problem is suitable for a take-home or lab exam. The student must formulate the model, solve the problem with a computer package, and then interpret the solution to answer the questions. a.

MAX 30 X1 + 25 X2 + 22 X3 + 20 X4 SUBJECT TO 2) 2 X1 + 2 X2 + X3 + X4 <= 500 3) 5 X1 + 6 X2 + 4 X3 + 4 X4 <= 1200 4) X1 + X2 + X3 + X4 >= 275

OBJECTIVE FUNCTION VALUE 1) 6850.0000 VARIABLE X1 X2 X3 X4

VALUE 100.000000 .000000 175.000000 .000000

REDUCED COST .000000 13.000000 .000000 2.000000

ROW 2) 3) 4)

SLACK OR SURPLUS 125.000000 .000000 .000000

DUAL PRICE .000000 8.000000 −10.000000

NO. ITERATIONS= 2 RANGES IN WHICH THE BASIS IS UNCHANGED: Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution

X1 X2 X3 X4

CURRENT COEFFICIENT 30.000000 25.000000 22.000000 20.000000

OBJ. COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE INFINITY 2.500000 13.000000 INFINITY 2.000000 2.000000 2.000000 INFINITY

ROW 2 3 4

CURRENT RHS 500.000000 1200.000000 275.000000

RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE INFINITY 125.000000 125.000000 100.000000 25.000000 35.000000

VARIABLE

Order 100 Open Trails, 0 Cityscapes, 175 Sea Sprites, and 0 Trail Blazers. Profit will be 6850. 6850 No. The $10 increase is below the reduced cost. 1100 to 1325 It will decrease by 50. Assembly time.

d. e. f. g. h. POINTS: 1 TOPICS: Formulation and computer solution

57. A company produces two products made from aluminum and copper. The table below gives the unit requirements, the unit production man-hours required, the unit profit and the availability of the resources (in tons). Product 1 Product 2 Available

Aluminum 1 1 10

Copper 0 1 6

Man-hours 2 3 24

Unit Profit 50 60

The Management Scientist provided the following solution output: Objective Function Value = 540.000 VARIABLE X1 X2 CONSTRAINT 1 2 3

VALUE 6.000 4.000

REDUCED COST 0.000 0.000

SLACK/SURPLUS .000 2.000 0.000

DUAL PRICE 30.000 0.000 10.000

RANGES IN WHICH THE BASIS IS UNCHANGED:

VARIABLE X1 X2

CURRENT COEFFICIENT 50.000 60.000

OBJ. COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 10.000 10.000 15.000 10.000 RIGHTHAND SIDE RANGES

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution CONSTRAINT 1 2 3

CURRENT RHS 10.000 6.000 24.000

ALLOWABLE INCREASE 2.000 INFINITY 2.000

ALLOWABLE DECREASE 1.000 2.000 4.000

What is the optimal production schedule? Within what range for the profit on product 2 will the solution in (a) remain optimal? What is b. the optimal profit when c2 = 70? c. Suppose that simultaneously the unit profits on x1 and x2 changed from 50 to 55 and 60 to 65 respectively. Would the optimal solution change? Explain the meaning of the "DUAL PRICES" column. Given the optimal solution, why d. should the dual price for copper be 0? e. What is the increase in the value of the objective function for an extra unit of aluminum? Man-hours were not figured into the unit profit as it must pay three workers for eight hours f. of work regardless of the number of man-hours used. What is the dual price for man-hours? Interpret. On the other hand, aluminum and copper are resources that are ordered as needed. The unit profit coefficients were determined by: (selling price per unit) - (cost of the resources per g. unit). The 10 units of aluminum cost the company $100. What is the most the company should be willing to pay for extra aluminum? ANSWER: a. 6 product 1, 4 product 2, Profit = $540 b. Between $50 and $75; at $70 the profit is $580 c. No; total % change is 83 1/3% < 100% Dual prices are the shadow prices for the resources; since there was unused copper (because d. S2 = 2), extra copper is worth $0 e. $30 f. $10; this is the amount extra man-hours are worth The shadow price is the "premium" for aluminum -- would be willing to pay up to $10 + $30 g. = $40 for extra aluminum POINTS: 1 TOPICS: Interpretation of solution 58. Given the following linear program: MAX

5x1 + 7x2

s.t.

x1 ≤ 6 2x1 + 3x2 ≤ 19 x1 + x2 ≤ 8 x1 , x2 ≥ 0

The graphical solution to the problem is shown below. From the graph we see that the optimal solution occurs at x1 = 5, x2 = 3, and z = 46.

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution

a. Calculate the range of optimality for each objective function coefficient. b. Calculate the dual price for each resource. ANSWER: a. Ranges of optimality: 14/3 ≤ c1 ≤ 7 and 5 ≤ c2 ≤ 15/2 Summarizing, the dual price for the first resource is 0, for the second resource is 2, and for b. the third is 1 POINTS: 1 TOPICS: Introduction to sensitivity analysis 59. Consider the following linear program: MAX

3x1 + 4x2 ($ Profit)

s.t.

x1 + 3x2 ≤ 12 2x1 + x 2 ≤ 8 x1 ≤ 3 x1 , x2 ≥ 0

The Management Scientist provided the following solution output: OPTIMAL SOLUTION Objective Function Value = 20.000 Variable X1 X2

Value 2.400 3.200

Reduced Cost 0.000 0.000

Constraint

Slack/Surplus

Dual Price

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 1 2 3

0.000 0.000 0.600

1.000 1.000 0.000

OBJECTIVE COEFFICIENT RANGES Variable X1 X2

Lower Limit 1.333 1.500

Current Value 3.000 4.000

Upper Limit 8.000 9.000

Current Value 12.000 8.000 3.000

Upper Limit 24.000 9.000 No Upper Limit

RIGHT HAND SIDE RANGES Constraint 1 2 3

Lower Limit 9.000 4.000 2.400

What is the optimal solution including the optimal value of the objective function? b. Suppose the profit on x1 is increased to $7. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $7? c. If the unit profit on x2 was $10 instead of $4, would the optimal solution change? d. If simultaneously the profit on x1 was raised to $5.5 and the profit on x2 was reduced to $3, would the current solution still remain optimal? ANSWER: a. x1 = 2.4 and x2 = 3.2, and z = $20.00. b. Optimal solution will not change. Optimal profit will equal $29.60. c. Because 10 is outside the range of 1.5 to 9.0, the optimal solution likely would change. Sum of the change percentages is 50% + 40% = 90%. Since this does not exceed 100% the d. optimal solution would not change. POINTS: 1 TOPICS: Interpretation of solution 60. Consider the following linear program: MIN

6x1 + 9x2 ($ cost)

x1 + 2x2 ≤ 8 10x1 + 7.5x2 ≥ 30 x2 ≥ 2 x1, x2 ≥ 0 The Management Scientist provided the following solution output: s.t.

OPTIMAL SOLUTION Objective Function Value = 27.000 Variable X1 X2

Value 1.500 2.000

Reduced Cost 0.000 0.000

Constraint 1

Slack/Surplus 2.500

Dual Price 0.000

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 2 0.000 −0.600 3 0.000 −4.500 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit X1 0.000 X2 4.500 RIGHT HAND SIDE RANGES

Current Value 6.000 9.000

Upper Limit 12.000 No Upper Limit

Constraint 1 2 3

Current Value 8.000 30.000 2.000

Upper Limit No Upper Limit 55.000 4.000

Lower Limit 5.500 15.000 0.000

What is the optimal solution including the optimal value of the objective function? b. Suppose the unit cost of x1 is decreased to $4. Is the above solution still optimal? What is the value of the objective function when this unit cost is decreased to $4? c. How much can the unit cost of x2 be decreased without concern for the optimal solution changing? d. If simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was reduced to $6, would the current solution still remain optimal? If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal e. solution? ANSWER: a. x1 = 1.5 and x2 = 2.0, and the objective function value = 27.00. 4 is within this range of 0 to 12, so the optimal solution will not change. Optimal total cost b. will be $24.00. c. x2 can fall to 4.5 without concern for the optimal solution changing. Sum of the change percentages is 91.7%. This does not exceed 100%, so the optimal solution d. would not change. The right-hand side remains within the range of feasibility, so there is no change in the e. optimal solution. However, the objective function value increases by $4.50. POINTS: 1 TOPICS: Interpretation of solution Essay 61. Describe each of the sections of output that come from The Management Scientist and how you would use each. ANSWER: Answer not provided. POINTS: 1 TOPICS: Interpretation of computer output 62. Explain the connection between reduced costs and the range of optimality, and between dual prices and the range of feasibility. ANSWER: Answer not provided. POINTS: 1 TOPICS: Interpretation of computer output 63. Explain the two interpretations of dual prices based on the accounting assumptions made in calculating the objective function coefficients. Cengage Learning Testing, Powered by Cognero

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Chapter 3 - Linear Programming: Sensitivity Analysis and Interpretation of Solution ANSWER: Answer not provided. POINTS: 1 TOPICS: Dual price 64. How can the interpretation of dual prices help provide an economic justification for new technology? ANSWER: Answer not provided. POINTS: 1 TOPICS: Dual price 65. How is sensitivity analysis used in linear programming? Given an example of what type of questions that can be answered. ANSWER: Answer not provided. POINTS: 1 TOPICS: Sensitivity analysis 66. How would sensitivity analysis of a linear program be undertaken if one wishes to consider simultaneous changes for both the right-hand-side values and objective function. ANSWER: Answer not provided. POINTS: 1 TOPICS: Simultaneous sensitivity analysis

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management True / False 1. Media selection problems can maximize exposure quality and use number of customers reached as a constraint, or maximize the number of customers reached and use exposure quality as a constraint. a. True b. False ANSWER: True POINTS: 1 TOPICS: Media selection 2. Revenue management methodology was originally developed for the banking industry. a. True b. False ANSWER: False POINTS: 1 TOPICS: Revenue management 3. Portfolio selection problems should acknowledge both risk and return. a. True b. False ANSWER: True POINTS: 1 TOPICS: Portfolio selection 4. If an LP problem is not correctly formulated, the computer software will indicate it is infeasible when trying to solve it. a. True b. False ANSWER: False POINTS: 1 TOPICS: Computer solution 5. It is improper to combine manufacturing costs and overtime costs in the same objective function. a. True b. False ANSWER: False POINTS: 1 TOPICS: Make-or-buy decision 6. Production constraints frequently take the form: beginning inventory + sales − production = ending inventory a. True b. False ANSWER: False POINTS: 1 TOPICS: Production scheduling Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management 7. If a real-world problem is correctly formulated, it is not possible to have alternative optimal solutions. a. True b. False ANSWER: False POINTS: 1 TOPICS: Problem formulation 8. To properly interpret dual prices, one must know how costs were allocated in the objective function. a. True b. False ANSWER: True POINTS: 1 TOPICS: Make-or-buy decision 9. A company makes two products from steel; one requires 2 tons of steel and the other requires 3 tons. There are 100 tons of steel available daily. A constraint on daily production could be written as: 2x1 + 3x2 ≤ 100. a. True b. False ANSWER: True POINTS: 1 TOPICS: Production scheduling 10. Double-subscript notation for decision variables should be avoided unless the number of decision variables exceeds nine. a. True b. False ANSWER: False POINTS: 1 TOPICS: Formulation notation 11. Using minutes as the unit of measurement on the left-hand side of a constraint and using hours on the right-hand side is acceptable since both are a measure of time. a. True b. False ANSWER: False POINTS: 1 TOPICS: General 12. Compared to the problems in the textbook, real-world problems generally require more variables and constraints. a. True b. False ANSWER: True POINTS: 1 TOPICS: General Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management 13. For the multiperiod production scheduling problem in the textbook, period n − 1's ending inventory variable was also used as period n's beginning inventory variable. a. True b. False ANSWER: True POINTS: 1 TOPICS: Production scheduling 14. A company makes two products, A and B. A sells for $100 and B sells for $90. The variable production costs are $30 per unit for A and $25 for B. The company's objective could be written as: MAX 190x1 − 55x2. a. True b. False ANSWER: False POINTS: 1 TOPICS: Production scheduling 15. The primary limitation of linear programming's applicability is the requirement that all decision variables be nonnegative. a. True b. False ANSWER: False POINTS: 1 TOPICS: General 16. A decision maker would be wise to not deviate from the optimal solution found by an LP model because it is the best solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: General 17. Data collection for large-scale LP models can be more time-consuming than either the formulation of the model or the development of the computer solution. a. True b. False ANSWER: True POINTS: 1 TOPICS: Production scheduling 18. The media selection model presented in the textbook involves maximizing the number of potential customers reached subject to a minimum total exposure quality rating. a. True b. False ANSWER: False POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management TOPICS: Media selection 19. The marketing research model presented in the textbook involves minimizing total interview cost subject to interview quota guidelines. a. True b. False ANSWER: True POINTS: 1 TOPICS: Marketing research 20. A marketing research firm must determine how many daytime interviews (D) and evening interviews (E) to conduct. At least 40% of the interviews must be in the evening. A correct modeling of this constraint is: -0.4D + 0.6E > 0. a. True b. False ANSWER: True POINTS: 1 TOPICS: Marketing research Multiple Choice 21. Media selection problems usually determine a. how many times to use each media source. b. the coverage provided by each media source. c. the cost of each advertising exposure. d. the relative value of each medium. ANSWER: a POINTS: 1 TOPICS: Media selection 22. To study consumer characteristics, attitudes, and preferences, a company would engage in a. client satisfaction processing. b. marketing research. c. capital budgeting. d. production planning. ANSWER: b POINTS: 1 TOPICS: Marketing research 23. A marketing research application uses the variable HD to represent the number of homeowners interviewed during the day. The objective function minimizes the cost of interviewing this and other categories and there is a constraint that HD ≥ 100. The solution indicates that interviewing another homeowner during the day will increase costs by 10.00. What do you know? a. the objective function coefficient of HD is 10. b. the dual price for the HD constraint is 10. c. the objective function coefficient of HD is −10. d. the dual price for the HD constraint is −10. Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management ANSWER: d POINTS: 1 TOPICS: Marketing research 24. The dual price for a constraint that compares funds used with funds available is .058. This means that a. the cost of additional funds is 5.8%. b. if more funds can be obtained at a rate of 5.5%, some should be. c. no more funds are needed. d. the objective was to minimize. ANSWER: b POINTS: 1 TOPICS: Portfolio selection 25. Let M be the number of units to make and B be the number of units to buy. If it costs $2 to make a unit and $3 to buy a unit and 4000 units are needed, the objective function is a. Max 2M + 3B b. Min 4000 (M + B) c. Max 8000M + 12000B d. Min 2M + 3B ANSWER: d POINTS: 1 TOPICS: Make-or-buy decision 26. If Pij = the production of product i in period j, then to indicate that the limit on production of the company's three products in period 2 is 400, a. P21 + P22 + P23 ≤ 400 b. P12 + P22 + P32 ≤ 400 c. P32 ≤ 400 d. P23 ≤ 400 ANSWER: b POINTS: 1 TOPICS: Production scheduling 27. Let Pij = the production of product i in period j. To specify that production of product 1 in period 3 and in period 4 differs by no more than 100 units, a. P13 − P14 ≤ 100; P14 − P13 ≤ 100 b. P13 − P14 ≤ 100; P13 − P14 ≥ 100 c. P13 − P14 ≤ 100; P14 − P13 ≥ 100 d. P13 − P14 ≥ 100; P14 − P13 ≥ 100 ANSWER: a POINTS: 1 TOPICS: Production scheduling Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management 28. Let A, B, and C be the amounts invested in companies A, B, and C. If no more than 50% of the total investment can be in company B, then a. B ≤ 5 b. A − .5B + C ≤ 0 c. .5A − B − .5C ≤ 0 d. −.5A + .5B − .5C ≤ 0 ANSWER: d POINTS: 1 TOPICS: Portfolio selection 29. Department 3 has 2500 hours. Transfers are allowed to departments 2 and 4, and from departments 1 and 2. If Ai measures the labor hours allocated to department i and Tij the hours transferred from department i to department j, then a. T13 + T23 − T32 − T34 − A3 = 2500 b. T31 + T32 − T23 − T43 + A3 = 2500 c. A3 + T13 + T23 − T32 − T34 = 2500 d. A3 − T13 − T23 + T32 + T34 = 2500 ANSWER: d POINTS: 1 TOPICS: Workforce assignment 30. Modern revenue management systems maximize revenue potential for an organization by helping to manage a. pricing strategies. b. reservation policies. c. short-term supply decisions. d. All of the alternatives are correct. ANSWER: d POINTS: 1 TOPICS: Revenue management 31. Blending problems arise whenever a manager must decide how to a. mix several different asset types in one investment strategy. b. mix two or more resources to produce one or more products. c. combine the results of two or more research studies into one. d. allocate workers with different skill levels to various work shifts. ANSWER: b POINTS: 1 TOPICS: Blending problems 32. In a production scheduling LP, the demand requirement constraint for a time period takes the form a. beginning inventory + production + ending inventory > demand. b. beginning inventory − production + ending inventory = demand. c. beginning inventory + production − ending inventory = demand. d. beginning inventory − production − ending inventory > demand. Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management ANSWER: c POINTS: 1 TOPICS: Production scheduling problems 33. A mutual fund manager must decide how much money to invest in Atlantic Oil (A) and how much to invest in Pacific Oil (P). At least 60% of the money invested in the two oil companies must be in Pacific Oil. A correct modeling of this constraint is a. 0.4A + 0.6P > 0. b. -0.4A + 0.6P > 0. c. 0.6A + 0.4P > 0. d. -0.6A + 0.4P > 0. ANSWER: d POINTS: 1 TOPICS: Portfolio selection 34. The production scheduling problem modeled in the textbook involves capacity constraints on all of the following types of resources except a. material. b. labor. c. machine. d. storage. ANSWER: a POINTS: 1 TOPICS: Production scheduling 35. Which of the operations management applications modeled in the text has an objective function that minimizes the sum of manufacturing costs, purchasing costs, and overtime costs? a. make-or-buy decision b. blending c. production scheduling d. workforce assignment ANSWER: a POINTS: 1 TOPICS: Operations management applications Subjective Short Answer 36. A&C Distributors is a company that represents many outdoor products companies and schedules deliveries to discount stores, garden centers, and hardware stores. Currently, scheduling needs to be done for two lawn sprinklers, the Water Wave and Spring Shower models. Requirements for shipment to a warehouse for a national chain of garden centers are shown below.

March

Shipping Capacity 8000

April

7000

Month

Product Water Wave Spring Shower Water Wave

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Minimum Requirement 3000 1800 4000

Unit Cost to Ship .30 .25 .40

Per Unit Inventory Cost .06 .05 .09 Page 7

Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Spring Shower 4000 .30 .06 Water Wave 5000 .50 .12 Spring Shower 2000 .35 .07 Let Sij be the number of units of sprinkler i shipped in month j, where i = 1 or 2, and j = 1, 2, or 3. Let Wij be the number of sprinklers that are at the warehouse at the end of a month, in excess of the minimum requirement. a Write the portion of the objective function that minimizes shipping costs. An inventory cost is assessed against this ending inventory. Give the portion of the objective b. function that represents inventory cost. There will be three constraints that guarantee, for each month, that the total number of c. sprinklers shipped will not exceed the shipping capacity. Write these three constraints. There are six constraints that work with inventory and the number of units shipped, making d. sure that enough sprinklers are shipped to meet the minimum requirements. Write these six constraints. ANSWER: a. Min .3S11 + .25S21 + .40S12 + .30S22 + .50S13 + .35S 23 b. Min .06W11 + .05W 21 + .09W12 + .06W22 + .12W13 + .07W23 c. S11 + S21 ≤ 8000 S12 + S22 ≤ 7000 S13 + S23 ≤ 6000 d. S11 − W11 = 3000 S21 − W21 = 1800 W11 + S12 − W12 = 4000 W21 + S22 − W 22 = 4000 W12 + S13 − W13 = 5000 W22 + S23 − W23 = 2000 POINTS: 1 TOPICS: Production scheduling May

6000

37. An ad campaign for a new snack chip will be conducted in a limited geographical area and can use TV time, radio time, and newspaper ads. Information about each medium is shown below. Medium TV Radio Newspaper

Cost Per Ad 500 200 400

# Reached 10000 3000 5000

Exposure Quality 30 40 25

If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is $10000, develop the model that will maximize the number reached and achieve an exposure quality if at least 1000. ANSWER: Let T = the number of TV ads Let R = the number of radio ads Let N = the number of newspaper ads Max

10000T + 3000R + 5000N

s.t.

500T + 200R + 400N ≤ 10000 30T + 40R + 25N ≥ 1000 T−R≤4 T, R, N ≥ 0

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management TOPICS: Media selection 38. Information on a prospective investment for Wells Financial Services is given below. 1 3000

2 7000

Period 3 4000

4 Loan Funds Available 5000 Investment Income 110% 112% 113% (% of previous period's investment) Maximum Investment 4500 8000 6000 7500 Payroll Payment 100 120 150 100 In each period, funds available for investment come from two sources: loan funds and income from the previous period's investment. Expenses, or cash outflows, in each period must include repayment of the previous period's loan plus 8.5% interest, and the current payroll payment. In addition, to end the planning horizon, investment income from period 4 (at 110% of the investment) must be sufficient to cover the loan plus interest from period 4. The difference in these two quantities represents net income, and is to be maximized. How much should be borrowed and how much should be invested each period? ANSWER: Let Lt = loan in period t, t = 1,...,4 It = investment in period t, t = 1,...,4 Max

1.1I4 − 1.085L4

s.t.

L1 ≤ 3000 I1 ≤ 4500 L1 − I1 = 100 L2 ≤ 7000 I2 ≤ 8000 L 2 + 1.1I1 − 1.085L1 − I2 = 120 L3 ≤ 4000 I3 ≤ 6000 L 3 + 1.12I2 − 1.085L2 − I3 = 150 L4 ≤ 5000 I4 ≤ 7500 L 4 + 1.13I3 −1.085L3 −I4 = 100 1.10I4 − 1.085L4 ≥ 0 Lt, It ≥ 0

POINTS: 1 TOPICS: Financial planning 39. Tots Toys makes a plastic tricycle that is composed of three major components: a handlebar-front wheel-pedal assembly, a seat and frame unit, and rear wheels. The company has orders for 12,000 of these tricycles. Current schedules yield the following information. Component Front Seat/Frame

Plastic 3 4

Requirements Time 10 6

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Space 2 2

Cost to Manufacture 8 6

Cost to Purchase 12 9 Page 9

Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Rear wheel (each) .5 2 .1 1 3 Available 50000 160000 30000 The company obviously does not have the resources available to manufacture everything needed for the completion of 12000 tricycles so has gathered purchase information for each component. Develop a linear programming model to tell the company how many of each component should be manufactured and how many should be purchased in order to provide 12000 fully completed tricycles at the minimum cost. ANSWER: Let FM = number of fronts made SM = number of seats made WM = number of wheels made FP = number of fronts purchased SP = number of seats purchased WP = number of wheels purchased Min

8FM + 6SM + 1WM + 12FP + 9SP + 3WP

s.t.

3FM + 4SM + .5WM ≤ 50000 10FM + 6SM + 2WM ≤ 160000 2FM + 2SM + .1WM ≤ 30000 FM + FP ≥ 12000 SM + SP ≥ 12000 WM + WP ≥ 24000 FM, SM, WM, FP, SP, WP ≥ 0

POINTS: 1 TOPICS: Make-or-buy decision 40. The Tots Toys Company is trying to schedule production of two very popular toys for the next three months: a rocking horse and a scooter. Information about both toys is given below. Toy Rocking Horse Scooter

Begin. Invty. June 1 25 55

Required Plastic 5 4

Required Time 2 3

Production Cost 12 14

Summer Schedule June July August

Plastic Available 3500 5000 4800

Time Available 2100 3000 2500

Monthly Demand Horse 220 350 600

Production Cost 1 1.2

Monthly Demand Scooter 450 700 520

Develop a model that would tell the company how many of each toy to produce during each month. You are to minimize total cost. Inventory cost will be levied on any items in inventory on June 30, July 31, or August 31 after demand for the month has been satisfied. Your model should make use of the relationship Beginning Inventory + Production − Demand = Ending Inventory for each month. The company wants to end the summer with 150 rocking horses and 60 scooters as beginning inventory for Sept. 1. Don't forget to define your decision variables. ANSWER: Let Pij = number of toy i to produce in month j Sij = surplus (inventory) of toy i at end of month j Min

12P11 + 12P 12 + 12P13 + 14P21 + 14P22 + 14P23 + 1S11 + 1S12 + 1S13 + 1.2S21 +

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management 1.2S22 + 1.2S23 s.t.

P11 − S11 = 195 S11 + P12 − S12 = 350 S12 + P13 − S13 = 600 S13 ≥ 150 P21 − S21 = 395 S21 + P22 − S22 = 700 S22 + P23 − S 23 = 520 S23 ≥ 60 5P11 + 4P21 ≤ 3500 5P12 + 4P22 ≤ 5000 5P13 + 4P23 ≤ 4800 2P11 + 3P21 ≤ 2100 2P12 + 3P22 ≤ 3000 2P13 + 3P23 ≤ 2500 Pij, Sij ≥ 0

POINTS: 1 TOPICS: Production scheduling 41. Larkin Industries manufactures several lines of decorative and functional metal items. The most recent order has been for 1200 door lock units for an apartment complex developer. The sales and production departments must work together to determine delivery schedules. Each lock unit consists of three components: the knob and face plate, the actual lock itself, and a set of two keys. Although the processes used in the manufacture of the three components vary, there are three areas where the production manager is concerned about the availability of resources. These three areas, their usage by the three components, and their availability are detailed in the table. Resource Brass Alloy Machining Finishing

Knob and Plate 12 18 15

Lock 5 20 5

Key (each) 1 10 1

Available 15000 units 36000 minutes 12000 minutes

A quick look at the amounts available confirms that Larkin does not have the resources to fill this contract. A subcontractor, who can make an unlimited number of each of the three components, quotes the prices below. Component Knob and Plate Lock Keys (set of 2)

Subcontractor Cost 10.00 9.00 1.00

Larkin Cost 6.00 4.00 .50

Develop a linear programming model that would tell Larkin how to fill the order for 1200 lock sets at the minimum cost. ANSWER: Let PM = the number of knob and plate units to make PB = the number of knob and plate units to buy LM = the number of lock units to make LB = the number of lock units to buy KM = the number of key sets to make KB = the number of key sets to buy Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Min

6PM + 10PB + 4LM + 9LB + .5KM + 1KB

s.t.

12PM + 5LM + 2KM ≤ 15000 18PM + 20LM + 20KM ≤ 36000 15PM + 5LM + 2KM ≤ 12000 PM + PB ≥ 1200 LM + LB ≥ 1200 KM + KB ≥ 1200 PM, PB, LM, LB, KM, KB ≥ 0

POINTS: 1 TOPICS: Make-or-buy decision 42. G and P Manufacturing would like to minimize the labor cost of producing dishwasher motors for a major appliance manufacturer. Although two models of motors exist, the finished models are indistinguishable from one another; their cost difference is due to a different production sequence. The time in hours required for each model in each production area is tabled here, along with the labor cost. Area A Area B Area C Cost

Model 1 15 4 4 80

Model 2 3 10 8 65

Currently labor assignments provide for 10,000 hours in each of Areas A and B and 18000 hours in Area C. If 2000 hours are available to be transferred from area B to Area A, 3000 hours are available to be transferred from area C to either Areas A or B, develop the linear programming model whose solution would tell G&P how many of each model to produce and how to allocate the workforce. ANSWER: Let P1 = the number of model 1 motors to produce P2 = the number of model 2 motors to produce AA = the number of hours allocated to area A AB = the number of hours allocated to area B AC = the number of hours allocated to area C TBA = the number of hours transferred from B to A TCA = the number of hours transferred from C to A TCB = the number of hours transferred from C to B Min

80P1 + 65P2

s.t.

15P1 + 3P2 − AA ≤ 0 4P1 + 10P2 − AB ≤ 0 4P1 + 8P2 − AC ≤ 0 AA − TBA − TCA = 10000 AB − TCB + TBA = 10000 AC + TCA + TCB = 18000 TBA ≤ 2000 TCA + TCB ≤ 3000 all variables ≥ 0

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management POINTS: 1 TOPICS: Workforce assignment 43. FarmFresh Foods manufactures a snack mix called TrailTime by blending three ingredients: a dried fruit mixture, a nut mixture, and a cereal mixture. Information about the three ingredients (per ounce) is shown below. Ingredient Dried Fruit Nut Mix Cereal Mix

Cost .35 .50 .20

Volume 1/4 cup 3/8 cup 1 cup

Fat Grams 0 10 1

Calories 150 400 50

The company needs to develop a linear programming model whose solution would tell them how many ounces of each mix to put into the TrailTime blend. TrailTime is packaged in boxes that will hold between three and four cups. The blend should contain no more than 1000 calories and no more than 25 grams of fat. Dried fruit must be at least 20% of the volume of the mixture, and nuts must be no more than 15% of the weight of the mixture. Develop a model that meets these restrictions and minimizes the cost of the blend. ANSWER: Let D = the number of ounces of dried fruit mix in the blend N = the number of ounces of nut mix in the blend C = the number of ounces of cereal mix in the blend Min

.35D + .50N + .20C

s.t.

.25D + .375N + C ≥ 3 . 25D + .375N + C ≤ 4 150D + 400N + 50C ≤ 1000 10N + C ≤ 25 .2D − .075N − .2C ≥ 0 −.15D + .85N − .15C ≤ 0 D,N,C≥0

POINTS: 1 TOPICS: Blending 44. The Meredith Ribbon Company produces paper and fabric decorative ribbon which it sells to paper products companies and craft stores. The demand for ribbon is seasonal. Information about projected demand and production for a particular type of ribbon is given. Quarter 1 Quarter 2 Quarter 3 Quarter 4

Demand (yards) 10,000 18,000 16,000 30,000

Production Cost Per Yard .03 .04 .06 .08

Production Capacity (yards) 30,000 20,000 20,000 15,000

An inventory holding cost of $.005 is levied on every yard of ribbon carried over from one quarter to the next. a. Define the decision variables needed to model this problem. The objective is to minimize total cost, the sum of production and inventory holding cost. b. Give the objection function. c. Write the production capacity constraints. Write the constraints that balance inventory, production, and demand for each quarter. d. Assume there is no beginning inventory in quarter 1. To attempt to balance the production and avoid large changes in the workforce, production in e. period 1 must be within 5000 yards of production in period 2. Write this constraint. ANSWER: Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management a. b. c.

Let Pi = the production in yards in quarter i Let Si = the ending surplus (inventory) in quarter i Min .03P1 + .04P2 + .06P3 + .08P4 + .005(S1 + S2 + S3 + S4 ) P1 ≤ 30000 P2 ≤ 20000 P3 ≤ 20000 P4 ≤ 15000 P1 − S1 = 10000 S1 + P2 − S2 = 18000 S2 + P3 − S3 = 16000 S3 + P4 − S4 = 30000 P1 ≤ 30000 P2 ≤ 20000 P3 ≤ 20000 P4 ≤ 15000

POINTS: 1 TOPICS: Production scheduling 45. Island Water Sports is a business that provides rental equipment and instruction for a variety of water sports in a resort town. On one particular morning, a decision must be made of how many Wildlife Raft Trips and how many Group Sailing Lessons should be scheduled. Each Wildlife Raft Trip requires one captain and one crew person, and can accommodate six passengers. The revenue per raft trip is $120. Ten rafts are available, and at least 30 people are on the list for reservations this morning. Each Group Sailing Lesson requires one captain and two crew people for instruction. Two boats are needed for each group. Four students form each group. There are 12 sailboats available, and at least 20 people are on the list for sailing instruction this morning. The revenue per group sailing lesson is $160. The company has 12 captains and 18 crew available this morning. The company would like to maximize the number of customers served while generating at least $1800 in revenue and honoring all reservations. ANSWER: Let R = the number of Wildlife Raft Trips to schedule S = the number of Group Sailing Lessons to schedule Max

6R + 4S

s.t.

R + S ≤ 12 R + 2S ≤ 18 6R ≥ 30 4S ≥ 20 120R + 160S ≥ 1800 R ≤ 10 2S ≤ 12 R,S≥0

POINTS: 1 TOPICS: Scheduling 46. Evans Enterprises has bought a prime parcel of beachfront property and plans to build a luxury hotel. After meeting with the architectural team, the Evans family has drawn up some information to make preliminary plans for construction. Excluding the suites, which are not part of this decision, the hotel will have four kinds of rooms: beachfront non-smoking, beachfront smoking, lagoon view non-smoking, and lagoon view smoking. In order to decide how many of each of the Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management four kinds of rooms to plan for, the Evans family will consider the following information. a.

c. d. e.

After adjusting for expected occupancy, the average nightly revenue for a beachfront nonsmoking room is $175. The average nightly revenue for a lagoon view non-smoking room is $130. Smokers will be charged an extra $15. Construction costs vary. The cost estimate for a lagoon view room is $12,000 and for a beachfront room is $15,000. Air purifying systems and additional smoke detectors and sprinklers ad $3000 to the cost of any smoking room. Evans Enterprises has raised $6.3 million in construction guarantees for this portion of the building. There will be at least 100 but no more than 180 beachfront rooms. Design considerations require that the number of lagoon view rooms be at least 1.5 times the number of beachfront rooms, and no more than 2.5 times that number. Industry trends recommend that the number of smoking rooms be no more than 50% of the number of non-smoking rooms.

Develop the linear programming model to maximize revenue. ANSWER: Let BN = the number of beachfront non-smoking rooms BS = the number of beachfront smoking rooms LN = the number of lagoon view non-smoking rooms LS = the number of lagoon view smoking rooms Max

175BN + 190BS + 130LN + 145LS

s.t.

15000BN + 18000BS + 12000LN + 15000LS ≤ 6,300,000 BN + BS ≥ 100 BN + BS ≤ 180 −1.5BN − 1.5BS + LN + LS ≥ 0 −2.5BN − 2.5BS + LN + LS ≤ 0 − .5BN + BS − .5LN + LS ≤ 0 BN, BS, LN, LS ≥ 0

POINTS: 1 TOPICS: Product mix 47. Super City Discount Department Store is open 24 hours a day. The number of cashiers need in each four hour period of a day is listed below. Period 10 p.m. to 2 a.m. 2 a.m. to 6 a.m. 6 a.m. to 10 a.m. 10 a.m. to 2 p.m. 2 p.m. to 6 p.m. 6 p.m. to 10 p.m.

Cashiers Needed 8 4 7 12 10 15

If cashiers work for eight consecutive hours, how many should be scheduled to begin working in each period in order to minimize the number of cashiers needed? ANSWER: Let TNP = the number of cashiers who begin working at 10 p.m. TWA = the number of cashiers who begin working at 2 a.m. SXA = the number of cashiers who begin working at 6 a.m. TNA = the number of cashiers who begin working at 10 a.m. TWP = the number of cashiers who begin working at 2 p.m. SXP = the number of cashiers who begin working at 6 p.m. Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Min

TNP + TWA + SXA + TNA + TWP + SXP

s.t.

TNP + TWA ≥ 4 TWA + SXA ≥ 7 SXA + TNA ≥ 12 TNA + TWP ≥ 10 TWP + SXP ≥ 15 SXP + TNP ≥ 8 all variables ≥ 0

POINTS: 1 TOPICS: Staff scheduling 48. Winslow Savings has $20 million available for investment. It wishes to invest over the next four months in such a way that it will maximize the total interest earned over the four month period as well as have at least $10 million available at the start of the fifth month for a high rise building venture in which it will be participating. For the time being, Winslow wishes to invest only in 2-month government bonds (earning 2% over the 2-month period) and 3-month construction loans (earning 6% over the 3-month period). Each of these is available each month for investment. Funds not invested in these two investments are liquid and earn 3/4 of 1% per month when invested locally. Formulate and solve a linear program that will help Winslow Savings determine how to invest over the next four months if at no time does it wish to have more than $8 million in either government bonds or construction loans. ANSWER: Define the decision variables gj = amount of new investment in government bonds in month j cj = amount of new investment in construction loans in month j lj = amount invested locally in month j, where j = 1,2,3,4 Define the objective function MAX .02g1 + .02g2 + .02g3 + .02g4 + .06c1 + .06c2 + .06c3 + .06c4 + .0075l1 + .0075l2 + .0075l3 + .0075l4 Define the constraints (1) g1 + c1 + l1 = 20,000,000 (2) g2 + c2 + l2 = 1.0075l1 or g2 + c2 − 1.0075l1 + l2 = 0 (3) g3 + c3 + l3 = 1.02g1 + 1.0075l2 or − 1.02g1 + g3 + c3 − 1.0075l2 + l3 = 0 g4 + c4 + l4 = 1.06c1 + 1.02g2 + 1.0075l3 or − 1.02g2 + g4 − 1.06c1 + c4 − 1.0075l3 + l4 = (4) 0 (5) 1.06c2 + 1.02g3 + 1.0075l4 ≥ 10,000,000 (6) g1 ≤ 8,000,000 (7) g1 + g2 ≤ 8,000,000 (8) g2 + g3 ≤ 8,000,000 (9) g3 + g4 ≤ 8,000,000 (10) c1 ≤ 8,000,000 (11) c1 + c2 ≤ 8,000,000 (12) c1 + c2 + c3 ≤ 8,000,000 (13) c2 + c3 + c4 ≤ 8,000,000 Nonnegativity: gj, cj, lj ≥ 0 for j = 1,2,3,4 Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management The Management Scientist provided the following solution: OBJECTIVE FUNCTION VALUE = 1429213.7987 VARIABLE VALUE G1 8000000.000 G2 0.000 G3 5108613.923 G4 2891386.077 C1 8000000.000 C2 0.000 C3 0.000 C4 8000000.000 I1 4000000.000 I2 4030000.000 I3 7111611.077 I4 4753562.083 POINTS: 1 TOPICS: Portfolio selection

REDUCED COST 0.000 0.000 0.000 0.000 0.000 0.045 0.008 0.000 0.000 0.000 0.000 0.000

49. National Wing Company (NWC) is gearing up for the new B-48 contract. Currently NWC has 100 equally qualified workers. Over the next three months NWC has made the following commitments for wing production: Month May June July

Wing Production 20 24 30

Each worker can either be placed in production or can train new recruits. A new recruit can be trained to be an apprentice in one month. The next month, he, himself, becomes a qualified worker (after two months from the start of training). Each trainer can train two recruits. The production rate and salary per employee is estimated below. Employee Production Trainer Apprentice Recruit

Production Rate (Wings/Month) .6 .3 .4 .05

Salary Per Month $3,000 3,300 2,600 2,200

At the end of July, NWC wishes to have no recruits or apprentices but have at least 140 full-time workers. Formulate and solve a linear program for NWC to accomplish this at minimum total cost. ANSWER: Pi = the number of producers in month i (where i = 1,2,3) Ti = the number of trainers in month i (where i = 1,2) Ai = the number of apprentices in month i (where i = 2,3) Ri = the number of recruits in month i (where i = 1,2) MIN

3000P1 + 3300T1 + 2200R1 + 3000P2 + 3300T2 + 2600A2+2200R 2 + 3000P3 + 2600A3

s.t.

.6P1 + .3T1 +.05R1 ≥ 20 .6P1 + .3T1 + .05R1 + .6P2 + .3T2 + .4A2 + .05R2 ≥ 44 .6P1 + .3T1 + .05R1 + .6P2 + .3T2 + .4A2 + .05R2 + .6P3 + .4A3 ≥ 74 P1 − P2 + T1 − T2 = 0

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management P2 − P3 + T2 + A2 = 0 A2 − R1 = 0 A3 − R2 = 0 2T1 − R1 ≥ 0 2T2 − R2 ≥ 0 P1 + T1 = 100 P3 + A3 ≥ 140 Pj, T j, Aj, Rj ≥ 0 for all j Solution:

P1 = 100, T1 = 0, R1 = 0, P2 = 80, T2 = 20, A2 = 0, R2 = 40, P3 = 100, A3 = 40 Total cost = $1,098,000.

POINTS: 1 TOPICS: Production scheduling 50. The SMM Company, which is manufacturing a new instant salad machine, has $280,000 to spend on advertising. The product is only to be test marketed initially in the Dallas area. The money is to be spent on an advertising blitz during one weekend (Friday, Saturday, and Sunday) in January, and SMM is limited to television advertising. The company has three options available: daytime advertising, evening news advertising and the Super Bowl. Even though the Super Bowl is a national telecast, the Dallas Cowboys will be playing in it, and hence, the viewing audience will be especially large in the Dallas area. A mixture of one-minute TV spots is desired. The table below gives pertinent data:

Daytime Evening News Super Bowl

Cost Per Ad $ 5,000 $ 7,000 $100,000

Estimated New Audience Reached With Each Ad 3,000 4,000 75,000

SMM has decided to take out at least one ad in each option. Further, there are only two Super Bowl ad spots available. There are 10 daytime spots and 6 evening news spots available daily. SMM wants to have at least 5 ads per day, but spend no more than $50,000 on Friday and no more than $75,000 on Saturday. Formulate and solve a linear program to help SMM decide how the company should advertise over the weekend. ANSWER: Define the decision variables x1 = the number of day ads on Friday x2 = the number of day ads on Saturday x3 = the number of day ads on Sunday x4 = the number of evening ads on Friday x5 = the number of evening ads on Saturday x6 = the number of evening ads on Sunday x7 = the number of Super Bowl ads Define the objective function MAX 3000x1 +3000x2 +3000x3 +4000x4 +4000x5 +4000x6 +75000x7 Define the constraints (1) x1 + x2 + x3 ≥ 1 (2) x4 + x5 + x6 ≥ 1 (3) x7 ≥ 1 Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management x1 ≤ 10 x2 ≤ 10 x3 ≤ 10 x4 ≤ 6 x5 ≤ 6 x6 ≤ 6 x7 ≤ 2 x1 + x4 ≥ 5 x 2 + x5 ≥ 5 x3 + x6 + x7 ≥ 5 5000x1 + 7000x4 ≤ 50000 5000x2 + 7000x5 ≤ 75000 5000x1 + 5000x2 + 5000x3 + 7000x4 + 7000x5 + 7000x6 + 100000x7 ≤ 280000 xj ≥ 0 j = 1,...,7 (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)

The Management Scientist provided the following solution: OBJECTIVE FUNCTION VALUE = 197800.000 VARIABLE X1 X2 X3 X4 X5 X6 X7 POINTS: 1 TOPICS: Media selection

VALUE 7.600 5.000 2.000 0.000 0.000 1.000 2.000

REDUCED COST 0.000 0.000 0.000 0.000 0.000 0.000 0.000

51. John Sweeney is an investment advisor who is attempting to construct an "optimal portfolio" for a client who has $400,000 cash to invest. There are ten different investments, falling into four broad categories that John and his client have identified as potential candidates for this portfolio. The following table lists the investments and their important characteristics. Note that Unidyde Equities (stocks) and Unidyde Debt (bonds) are two separate investments, whereas First General REIT is a single investment that is considered both an equities and a real estate investment. Category Equities

Debt

Real Estate Money

Investment Unidyne Corp. Col. Mustard Restaurant First General REIT Metropolitan Electric Unidyne Corp. Lemonville Transit Fairview Apartment Partnership First General REIT T-Bill Account

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Exp. Annual After Tax Return 15.0% 17.0% 17.5% 11.8% 12.2% 12.0% 22.0% (see above) 9.6%

Liquidity Factor 100 100 100 95 92 79 0 (see above) 80

Risk Factor 60 70 75 20 30 22 50 (see above) 0 Page 19

Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Money Market Fund All Saver's Certificate

10.5% 12.6%

100 0

10 0

Formulate and solve a linear program to accomplish John's objective as an investment advisor which is to construct a portfolio that maximizes his client's total expected after-tax return over the next year, subject to a number of constraints placed upon him by the client for the portfolio: 1. Its (weighted) average liquidity factor must be at least 65. 2. The (weighted) average risk factor must be no greater than 55. 3. At most, $60,000 is to be invested in Unidyde stocks or bonds. 4. No more than 40% of the investment can be in any one category except the money category. No more than 20% of the investment can be in any one investment except the money market 5. fund. 6. At least $1,000 must be invested in the money market fund. 7. The maximum investment in All Saver's Certificates is $15,000. 8. The minimum investment desired for debt is $90,000. 9. At least $10,000 must be placed in a T-Bill account. ANSWER: Xj = $ invested in investment j; where j = 1(Uni Eq.), 2(Col. Must.), 3(1st Gen REIT), 4(Met. Elec.), 5(Uni Debt), 6(Lem. Trans.), 7(Fair. Apt.), 8(T-Bill), 9(Money Market), 10(All Saver's) MAX

.15X1 + .17X2 + .175X3 + .118X4 + .122X5 + .12X6 + .22X 7 + .096X8 + .105X9 + .126X10

S.T.

X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 = 400,000 100X1 + 100X2 + 100X3 + 95X4 + 92X5 + 79X6 + 80X8 + 100X9 ≥ 65(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10) 60X1 + 70X2 + 75X3 + 20X4 + 30X5 + 22X6 + 50X7 + 10X9 ≤ 55(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10) X1 + X5 ≤ 60,000 X1 + X2 + X3 ≤ 160,000 X4 + X5 + X6 ≤ 160,000 X3 + X7 ≤ 160,000 X1 ≤ 80,000 X2 ≤ 80,000 X3 ≤ 80,000 X4 ≤ 80,000 X5 ≤ 80,000 X6 ≤ 80,000 X7 ≤ 80,000 X8 ≤ 80,000 X9 ≥ 1,000 X10 ≤ 15,000 X4 + X5 + X6 ≥ 90,000 X8 ≥ 10,000 Xj ≥ 0 j = 1,...,10

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management Solution:

X1 = 0; X2 = 80,000; X3 = 80,000; X4 = 0; X5 = 60,000; X6 = 74,000; X7 = 80,000; X8 = 10,000; X9 = 1,000; X10 = 15,000; Total return = $64,355.

POINTS: 1 TOPICS: Portfolio selection 52. BP Cola must decide how much money to allocate for new soda and traditional soda advertising over the coming year. The advertising budget is $10,000,000. Because BP wants to push its new sodas, at least one-half of the advertising budget is to be devoted to new soda advertising. However, at least $2,000,000 is to be spent on its traditional sodas. BP estimates that each dollar spent on traditional sodas will translate into 100 cans sold, whereas, because of the harder sell needed for new products, each dollar spent on new sodas will translate into 50 cans sold. To attract new customers BP has lowered its profit margin on new sodas to 2 cents per can as compared to 4 cents per can for traditional sodas. How should BP allocate its advertising budget if it wants to maximize its profits while selling at least 750 million cans? ANSWER: X1 = amount invested in new soda advertising X2 = amount invested in traditional soda advertising MAX

X1 + 4X2

S.T.

X1 + X2 ≤ 10,000,000 X1 ≥ 5,000,000 X2 ≥ 2,000,000 50X1 + 100X2 ≥ 750,000,000 X1, X2 ≥ 0

Answer:

spend $5,000,000 on new soda ad, spend $5,000,000 on traditional ad, profit = $25 million

POINTS: 1 TOPICS: Advertising budget allocation 53. Wes Wheeler is the production manager of Wheeler Wheels, Inc. Wes has just received orders for 1,000 standard wheels and 1,250 deluxe wheels next month and for 800 standard and 1,500 deluxe wheels the following month. All orders are to be filled. The cost of producing standard wheels is $10 and deluxe wheels is $16. Overtime rates are 50% higher. There are 1,000 hours of regular time and 500 hours of overtime available each month. It takes .5 hour to make a standard wheel and .6 hour to make a deluxe wheel. The cost of storing one wheel from one month to the next is $2. Wes wants to develop a two-month production schedule for standard and deluxe wheels. Formulate this production planning problem as a linear program. ANSWER: Define the decision variables We want to determine the production levels, xj, as follows: Month 1 Month 2 Reg. Time Overtime Reg. Time Overtime Standard x1 x2 x5 x6 Deluxe x3 x4 x7 x8 Also let: y1 = number of standard wheels stored from month 1 to month 2 Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management y2 = number of deluxe wheels stored from month 1 to month 2 Define the objective function Minimize total production and storage costs: Min (cost per wheel)(number of wheels produced) + 2y1 + 2y2 Min 10x1 + 15x2 + 16x3 + 24x4 + 10x5 + 15x6 + 16x7 + 24x8 + 2y1 + 2y2 Define the constraints Standard Wheel Production Month 1 = (Requirements) + (Amount Stored) (1) x1 + x2 = 1,000 + y1 or x1 + x2 - y1 = 1,000 Deluxe Wheel Production Month 1 = (Requirements) + (Amount Stored) (2) x3 + x4 = 1,250 + y2 or x3 + x4 - y2 = 1,250 Standard Wheel Production Month 2 = (Requirements) - (Amount Stored) (3) x5 + x6 = 800 - y1 or x5 + x6 + y1 = 800 Deluxe Wheel Production Month 2 = (Requirements) - (Amount Stored) (4) x7 + x8 = 1,500 - y2 or x7 + x8 + y2 = 1,500 Regular Hours Used Month 1 < Regular Hours Available Month 1: (5) .5x1 + .6x3 < 1000 Overtime Hours Used Month 1 < Overtime Hours Available Month 1: (6) .5x2 + .6x4 < 500 Regular Hours Used Month 2 < Regular Hours Available Month 2: (7) .5x5 + .6x7 < 1000 Overtime Hours Used Month 2 < Overtime Hours Available Month 2: (8) .5x6 + .6x8 < 500 POINTS: 1 TOPICS: Production scheduling 54. Comfort Plus Inc. (CPI) manufactures a standard dining chair used in restaurants. The demand forecasts for quarter 1 (January-March) and quarter 2 (April-June) are 3700 chairs and 4200 chairs, respectively. CPI has a policy of satisfying all demand in the quarter in which it occurs. The chair contains an upholstered seat that can be produced by CPI or purchased from DAP, a subcontractor. DAP currently charges $12.50 per seat, but has announced a new price of $13.75 effective April 1. CPI can produce the seat at a cost of $10.25. CPI can produce up to 3800 seats per quarter. Seats that are produced or purchased in quarter 1 and used to satisfy demand in quarter 2 cost CPI $1.50 each to hold in inventory, but maximum inventory cannot exceed 300 seats. Formulate this problem as a linear programming problem. ANSWER: x1 = number of seats produced by CPI in quarter 1, x2 = number of seats purchased from DAP in quarter 1, x3 = number of seats carried in inventory from quarters 1 to 2, x4 = number of seats produced by CPI in quarter 2, and x5 = number of seats purchased from DAP in quarter 2. Min 10.25x1 + 12.5x2 + 1.5x3 + 10.25x4 + 13.75x5 (costs) s.t. x1 + x2 - x3 > 3700 (quarter 1 demand) x3 + x4 + x5 > 4200 (quarter 2 demand) Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management x1 < 3800 (CPI's production capacity in quarter 1) x4 < 3800 (CPI’s production capacity in quarter 2) x3 < 300 (inventory capacity) POINTS: 1 TOPICS: Make-or-buy decision 55. Target Shirt Company makes three varieties of shirts: Collegiate, Traditional and European. These shirts are made from different combinations of cotton and polyester. The cost per yard of unblended cotton is $5 and for unblended polyester is $4. Target can receive up to 4,000 yards of raw cotton and 3,000 yards of raw polyester fabric weekly. The table below pertinent data concerning the manufacture of the shirts. Shirt Collegiate

Total Yards 1.00

Traditional

1.20

European

.90

Fabric Requirement At least 50% cotton No more than 20% polyester As much as 80% polyester

Weekly Contracts 500

Weekly Demand 600

Selling Price $14.00

650

850

$15.00

280

675

$18.00

Formulate and solve this blending problem as a linear program. ANSWER: Formulation Decision variables Not only must we decide how many shirts to make and how much fabric to purchase, we also need to decide how much of each fabric is blended into each shirt. Let, sj = the total number of shirt style j produced fi = the number of yards of material i purchased xij = yards of fabric i blended into shirt style j where i = 1 (cotton) or 2 (polyester) and j = 1 (collegiate), 2(traditional), or 3 (European) Objective function Maximize the overall profit. To determine the profit function, subtract the cost of purchasing the fabric from the shirt sales revenue. Thus the objective function is: Max 14s1 + 15s2 + 18s3 - 5f1 − 4f2 Constraints Definition of Total Number of Shirts of Each Style Total Number of each style = (Total Yardage used in making the style) / (Yardage/Shirt) (1) Collegiate: s1 = (x11 + x21) / 1 >>> s1 − x11 − x21 = 0 (2) Traditional: s2 = (x12 + x22) / 1.2 >>> 1.2s2 − x12 − x22 = 0 (3) European: s3 = (x13 + x23) / .9 >>> .95s3 − x13 − x23 = 0 Definition of Total Yardage of Materials Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management (4) Cotton: (5) Polyester:

f1 = x11 + x12 + x13 f2 = x21 + x22 + x23

>>> >>>

f1 − x11 − x12 − x13 = 0 f2 − x21 − x22 − x23 = 0

Weekly Availability of the Resources (6) Cotton: f1 < 4000 (7) Polyester: f2 < 3000 Meet Weekly Contracts (8) Collegiate: s1 > 500 (9) Traditional: s2 > 650 (10) European: s3 > 280 Do Not Exceed Weekly Demand (11) Collegiate: s1 < 600 (12) Traditional: s2 < 850 (13) European: s3 < 675 Fabric Requirements Collegiate At Least 50% Cotton: (Total yds. of Cotton in Colleg. Shirts) > [(.5(1.00) yds./shirt)(number of colleg. shirts)] (14) x11 > .5s1 >>> x11 − .5s1 > 0 Traditional At Most 20% Polyester: (Total yds. Polyester in Tradit. Shirts) < [.2(1.20) yds./shirt)(number of tradit. shirts)] (15) x22 < .24s2 >>> x22 − .24s2 < 0 European At Most 80% Polyester: (Total yds. Polyester in Europ. Shirts) < [.8(.90) yds./shirt)(number of Europ. shirts)] (16) x23 < .72s3 >>> x23 − .72s3 < 0 Nonnegativity of variables sj, fi, xij > 0 for i = 1,2 and j = 1,2,3. Solution Total profit = $23,152.50 Cotton collegiate yds. = 300.0, cotton traditional yds. = 816.0, cotton European yds. = 121.5, polyester collegiate yds. = 300.0, polyester traditional yds. = 204.0, polyester European yds = 486.0, collegiate shirts = 600, traditional shirts = 850, European shirts = 675, cotton yards = 1237.5, polyester yards = 990.0 POINTS: 1 TOPICS: Blending problem Essay 56. Discuss the need for the use of judgment or other subjective methods in mathematical modeling. ANSWER: Answer not provided. POINTS: 1 TOPICS: Media selection 57. What benefits exist in using linear programming for production scheduling problems? ANSWER: Answer not provided. Cengage Learning Testing, Powered by Cognero

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Chapter 4 - Linear Programming Applications in Marketing, Finance, and Operations Management POINTS: 1 TOPICS: Production scheduling 58. Describe some common feature of multiperiod financial planning models. ANSWER: Answer not provided. POINTS: 1 TOPICS: Financial planning 59. Why should decision makers who are primarily concerned with marketing or finance or production know about linear programming? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 60. Discuss several resource allocation problems that can be modeled by varying the workforce assignment model. ANSWER: Answer not provided. POINTS: 1 TOPICS: Workforce assignment model 61. Discuss at least three additional considerations that might be included in media selection LP models more complex than the one covered in the text. ANSWER: Answer not provided. POINTS: 1 TOPICS: Marketing applications: media selection 62. Describe three situations where variations of the workforce assignment model could be applied. ANSWER: Answer not provided. POINTS: 1 TOPICS: Workforce assignment model 63. Interpret the dual price for any two of the following constraints: 1) the available funds in a portfolio selection model, 2) the cash requirement due at the beginning of a year in a financial planning model, and 3) the manufacturing capacity in a make-or-buy decision model. ANSWER: Answer not provided. POINTS: 1 TOPICS: Solution interpretation in LP applications

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Chapter 5 - Advanced Linear Programming Applications True / False 1. Revenue management methodology was originally developed for the banking industry. a. True b. False ANSWER: False POINTS: 1 TOPICS: Revenue management 2. The goal of portfolio models is to create a portfolio that provides the best balance between risk and return. a. True b. False ANSWER: True POINTS: 1 TOPICS: Portfolio models and asset allocation 3. DEA is used to measure the relative efficiency of two (or more) units in different industries, such as a bank and a school. a. True b. False ANSWER: False POINTS: 1 TOPICS: Data envelopment analysis 4. If the inputs of the composite unit in DEA are greater than the inputs for an individual unit, then the composite is more efficient. a. True b. False ANSWER: False POINTS: 1 TOPICS: Data envelopment analysis 5. It is possible for DEA to show all operating units to be relatively inefficient. a. True b. False ANSWER: False POINTS: 1 TOPICS: Data envelopment analysis 6. DEA does not necessarily identify the operating units that are relatively efficient. a. True b. False ANSWER: True POINTS: 1 TOPICS: Data envelopment analysis Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications 7. DEA will show all but one operating unit to be relatively inefficient in the case where an operating unit producing the most of every output and also consumes the least of every input. a. True b. False ANSWER: True POINTS: 1 TOPICS: Data envelopment analysis 8. For a two-person, zero-sum, mixed-strategy game, it is necessary to solve the LP for each player in order to learn both players' optimal strategies. a. True b. False ANSWER: False POINTS: 1 TOPICS: Game theory 9. If a pure strategy solution exists for a two-person, zero-sum game, it is the optimal solution to the game. a. True b. False ANSWER: True POINTS: 1 TOPICS: Game theory 10. In a two-person, zero-sum, pure-strategy game, there is no advantage to either player to switch from its strategy even if one of the players discovers the other player's strategy in advance. a. True b. False ANSWER: True POINTS: 1 TOPICS: Game theory 11. In portfolio models, risk is minimized by diversification. a. True b. False ANSWER: True POINTS: 1 TOPICS: Portfolio models and asset allocation 12. Revenue management methodology can be applied in the case of nonperishable assets. a. True b. False ANSWER: True POINTS: 1 TOPICS: Revenue management 13. In a revenue management model, dual prices tell reservation agents the revenue loss associated with overbooking each Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications ODIF. a. True b. False ANSWER: False POINTS: 1 TOPICS: Revenue management 14. In game theory, the player seeking to maximize the value of the game selects a maximax strategy. a. True b. False ANSWER: False POINTS: 1 TOPICS: Game theory: competing for market share 15. The analysis of a two-person, zero-sum game begins with checking to see whether a pure strategy exists. a. True b. False ANSWER: True POINTS: 1 TOPICS: Game theory 16. Revenue management methodology enables an airline to maximize the number of full-fare seats it sells on each flight. a. True b. False ANSWER: False POINTS: 1 TOPICS: Revenue management Multiple Choice 17. Revenue management methodology was originally developed for a. a cruise line. b. an airline. c. a car rental company. d. a hotel chain. ANSWER: b POINTS: 1 TOPICS: Revenue management 18. The overall goal of portfolio models is to create a portfolio that provides the best balance between a. short-term and long-term investments. b. gains and losses. c. risk and return. d. liquidity and stability. ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications POINTS: 1 TOPICS: Portfolio models and asset allocation 19. The composite unit in DEA a. has as output a weighted average of the outputs of the individual units. b. has as input a weighted average of the inputs of the individual units. c. has outputs greater than or equal to the outputs of any individual unit. d. All of the alternatives are correct. ANSWER: d POINTS: 1 TOPICS: Data envelopment analysis 20. Modern revenue management systems maximize revenue potential for an organization by helping to manage a. pricing strategies. b. reservation policies. c. short-term supply decisions. d. All of the alternatives are correct. ANSWER: d POINTS: 1 TOPICS: Revenue management 21. A DEA linear programming model involving 4 input measures and 3 output measures will have a. 7 constraints. b. 8 constraints. c. 12 constraints. d. 13 constraints. ANSWER: b POINTS: 1 TOPICS: Data envelopment analysis 22. In general, every DEA linear programming model will include a constraint that requires the weights for the operating units to sum to a. 1. b. the number of operating units. c. 100. d. any arbitrary value. ANSWER: a POINTS: 1 TOPICS: Data envelopment analysis 23. For a two-person, zero-sum, mixed-strategy game, it is necessary to solve the LP for only one of the players to learn the optimal strategies for both players. a. 1. b. the number of operating units. c. 100. Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications d. any arbitrary value. ANSWER: a POINTS: 1 TOPICS: Game theory 24. For a two-person, zero-sum, mixed-strategy game, each player selects its strategy according to a. what strategy the other player used last. b. a fixed rotation of strategies. c. a probability distribution. d. the outcome of the previous game. ANSWER: c POINTS: 1 TOPICS: Game theory 25. If it is optimal for both players in a two-person, zero-sum game to select one strategy and stay with that strategy regardless of what the other player does, the game a. has more than one equilibrium point. b. will have alternating winners. c. will have no winner. d. has a pure strategy solution. ANSWER: d POINTS: 1 TOPICS: Game theory 26. A linear programming application used to measure the relative efficiency of operating units with the same goals and objectives is a. game theory. b. asset allocation. c. data envelopment analysis. d. revenue management. ANSWER: c POINTS: 1 TOPICS: Data envelopment analysis 27. In data envelopment analysis, the percentage of an individual operating unit's resources that are available to the composite operating unit is the a. efficiency index. b. saddle point. c. maximin. d. minimax. ANSWER: a POINTS: 1 TOPICS: Data envelopment analysis 28. To develop a portfolio that provides the best return possible with a minimum risk, the linear programming model will Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications have an objective function which a. minimizes the maximum risk. b. minimizes total risk. c. maximizes return and minimizes risk. d. maximizes the minimum return. ANSWER: d POINTS: 1 TOPICS: Portfolio models and asset allocation 29. Revenue management methodology enables an airline to a. maximize the number of full-fare seats it sells on each flight. b. minimize the number of discount-fare seats it sells on each flight. c. increase the average number of passengers per flight. d. decrease the number of overbooked flights. ANSWER: c POINTS: 1 TOPICS: Revenue management Subjective Short Answer 30. The Eastern Washington County School Corporation is interested in comparing educational performance at four elementary schools and has hired you to prepare a DEA model to do so. After detailed conversations with the corporation administrative staff and the building principals, you have isolated the following input and output measurements: Input Measures Average classroom size Percent of students on reduced price lunch Average number of parent volunteer hours per week Data is collected for each school on each measure Archer

Output Measures Percent of children in remedial classes Average first grade score on standard test Average fifth grade score on standard test

School Hayes Ralston

Creekside

Inputs Classroom size 21 28 32 20 % reduced lunch 10 8 25 2 Avg. vol. hours 30 46 15 64 Outputs % remediation 15 12 19 3 1st grade score 83 84 62 85 5th grade score 76 72 53 79 Develop the DEA model that would evaluate the efficiency of Ralston Elementary School. ANSWER: Let A = the weight applied to inputs and outputs for Archer Elementary H = the weight applied to inputs and outputs for Hayes Elementary R = the weight applied to inputs and outputs for Ralston Elementary C = the weight applied to inputs and outputs for Creekside Elementary Min s.t.

E A+H+R+C=1 .15A + .12H + .19R + .03C ≥ .19

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Chapter 5 - Advanced Linear Programming Applications 83A + 84H + 62R + 85C ≥ 62 76A + 72H + 53R + 79C ≥ 53 21A + 28H + 32R + 20C ≤ 32E .10A + .08H + .25R + .02C ≤ .25E 30A + 46H + 15R + 64C ≤ 15 POINTS: 1 TOPICS: Data envelopment analysis 31. Mountainside State Park has four visitor centers. To study the operation of these centers, a DEA model has been developed that compares inputs (size, number of staff, weekly hours of operation) and outputs (% of visitors attending educational program, daily sales in gift shop). The computer solution is shown below. What can you conclude about the efficiency of the North center? LINEAR PROGRAMMING PROBLEM Min s.t.

1E+0wn+0ws+0we+0ww 1) −400E+400wn+1200ws+2400we+1500ww<0 2) −3E+3wn+6ws+10we+7ww<0 3) −56E+56wn+108ws+92we+108ww<0 4) −49E+49wn+83ws+56we+72ww>0 5) −38E+38wn+425ws+1200we+630ww>0 6) +1wn+1ws+1we+1ww=1

OPTIMAL SOLUTION Objective Function Value =

1.000

Variable E wn ws we ww

Reduced Cost 0.000 0.000 0.000 0.000 0.000

Value 1.000 1.000 0.000 0.000 0.000

Constraint Slack/Surplus Dual Price 1 0.000 0.008 2 0.000 0.000 3 0.000 0.001 4 0.000 −0.034 5 0.000 −0.013 6 0.000 −1.000 ANSWER: Because E = 1, the composite visitors' center requires as much input as the North center does. There is no evidence that the North center is inefficient. POINTS: 1 TOPICS: Data envelopment analysis 32. The output shows the solution to a DEA model where facilities in Seaview (S), Farmington (F), Lewiston (L), and San Domingo (D) are compared. The inputs, in order, are number of machines, size of work force, and goodness of location. The outputs, in order, are production, quality rating, and on-time completion percentage. The model examines the efficiency of Lewiston. MIN

0S+0F+0L+0D+1E

S.T.

1) 1S+1F+1L+1D=1

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Chapter 5 - Advanced Linear Programming Applications 2) 5S+22F+36L+15D−36E<0 3) 400S+1500F+3150L+1060D−3150E<0 4) 24S+13F+32L+17D−32E<0 5) 800S+2900F+1860L+1700D+0E>1860 6) 95S+92F+83L+94D+0E>83 7) 83S+85F+90L+91D+0E>90 OPTIMAL SOLUTION Objective Function Value =

0.510

Variable S F L D E

Value 0.000 0.167 0.000 0.833 0.510

Reduced Cost 0.385 0.000 0.490 0.000 0.000

Constraint 1 2 3 4 5 6 7

Slack/Surplus 0.000 2.208 474.479 0.000 40.000 10.667 0.000

Dual Price 1.365 0.000 0.000 0.031 0.000 0.000 −0.021

Is the Lewiston plant efficient? Why or why not? If not, which plants should it emulate in order to improve? b. How much more production does the composite facility provide than the Lewiston site? c. What is the quality rating for the composite facility? ANSWER: No. The composite plant requires less input than Lewiston to obtain the output. Lewiston a. should emulate Farmington and San Domingo. b. 40 c. 83 + 10.667 = 93.667 POINTS: 1 TOPICS: Data envelopment analysis a.

33. Consider a two-person, zero-sum game where the payoffs listed below are the winnings for Player A. Identify the pure strategy solution. What is the value of the game? Player A Strategies a1 a2 a3

b1 5 1 7

Player B Strategies b2 5 6 2

b3 4 2 3

ANSWER: Optimal pure strategies: Player A uses strategy a1; Player B uses strategy b3. Value of game: Gain of 4 for Player A; loss of 4 for Player B. POINTS: 1 TOPICS: Game theory Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications 34. Consider a two-person, zero-sum game where the payoffs listed below are the winnings for Company X. Identify the pure strategy solution. What is the value of the game? Company X Strategies x1 x2 x3

Company Y Strategies y1 y2 y3 3 5 9 8 4 3 7 6 7

ANSWER: Optimal pure strategies: Company X uses strategy x3; Company Y uses strategy y2. Value of game: Gain of 6 for Company X; loss of 6 for Company Y. POINTS: 1 TOPICS: Game theory 35. Consider the following two-person, zero-sum game. Payoffs are the winnings for Company X. Formulate the linear program that determines the optimal mixed strategy for Company X. Company X Strategies x1 x2 x3 ANSWER:

Company Y Strategies y1 y2 y3 4 3 9 2 5 1 6 1 7

Max

GAINA

s.t.

4PA1 + 2PA2 + 6PA3 − GAINA ≥ 0 3PA1 + 5PA2 + 1PA3 − GAINA ≥ 0 9PA1 + 1PA2 + 7PA3 − GAINA ≥ 0 PA1 + PA2 + PA3 = 1 PA1, PA2, PA3, GAINA ≥ 0

(Strategy B1) (Strategy B2) (Strategy B3) (Probabilities must sum to 1) (Non-negativity)

POINTS: 1 TOPICS: Game theory 36. Shown below is the solution to the linear program for finding Player A's optimal mixed strategy in a two-person, zerosum game. OBJECTIVE FUNCTION VALUE =

3.500

VARIABLE PA1 PA2 PA3 GAINA

VALUE 0.050 0.600 0350 3.500

REDUCED COSTS 0.000 0.000 0.000 0.000

CONSTRAINT 1 2 3 4

SLACK/SURPLUS 0.000 0.000 0.000 0.000

DUAL PRICES −0.500 −0.500 0.000 3.500

What is Player A's optimal mixed strategy?

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Chapter 5 - Advanced Linear Programming Applications b. What is Player B's optimal mixed strategy? c. What is Player A's expected gain? d. What is Player B's expected loss? ANSWER: a. Player A's optimal mixed strategy:

Player B's optimal mixed strategy:

Player A's expected gain: 3.500

Use strategy A1 with .05 probability Use strategy A2 with .60 probability Use strategy A3 with .35 probability Use strategy B1 with .50 probability Use strategy B2 with .50 probability Do not use strategy B3

d. Player B's expected loss: 3.500 POINTS: 1 TOPICS: Game theory 37. Portfolio manager Max Gaines needs to develop an investment portfolio for his conservative clients. His task is to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the portfolio provides the best return possible with a minimum risk. Formulate the maximin linear program. Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond ANSWER:

Year 1 22.37 14.88 19.45 13.79 7.29

Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 6.46 18.61 10.52 18.04 5.91 11.33 −2.07 8.05 9.18

Year 4 −3.19 5.25 −1.94 6.85 3.92

Max M −M + 22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB ≥ 0 −M + 26.73IS + 18.61LC + 18.04MC + 11.33SC + 8.05IB ≥ 0 −M + 6.46IS + 10.52LC + 5.91MC − 2.07SC + 9.18IB ≥ 0 −M − 3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB ≥ 0 IS + LC + MC + SC + IB = 1

(Scenario 1) (Scenario 2) (Scenario 3) (Scenario 4) (Total proportion must equal 1)

M, IS, LC, MC, SC, IB ≥ 0

(Non-negativity)

POINTS: 1 TOPICS: Portfolio models and asset allocation 38. Portfolio manager Max Gaines needs to develop an investment portfolio for his clients who are willing to accept a moderate amount of risk. His task is to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the portfolio provides an annual return of no less than 3%. Formulate the appropriate linear program. Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend

Year 1 22.37 14.88 19.45 13.79

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Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 6.46 18.61 10.52 18.04 5.91 11.33 −2.07

Year 4 −3.19 5.25 −1.94 6.85 Page 10

Chapter 5 - Advanced Linear Programming Applications Intermediate Bond 7.29 8.05 9.18 ANSWER: Max 13.093IS + 12.315LC + 10.365MC + 7.475SC + 7.110IB

3.92

22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB ≥ 3 26.73IS + 18.61LC + 18.04MC + 11.33SC + 8.05IB ≥ 3 6.46IS + 10.52LC + 5.91MC − 2.07SC + 9.18IB ≥ 3 −3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB ≥ 3 IS + LC + MC + SC + IB = 1

(Scenario 1) (Scenario 2) (Scenario 3) (Scenario 4) (Total proportion must equal 1)

IS, LC, MC, SC, IB ≥ 0

(Non-negativity)

POINTS: 1 TOPICS: Portfolio models and asset allocation 39. Hervis Car Rental in Austin, TX has 50 high-performance Shelby-H Mustangs in its rental fleet. These cars will be in greater demand than usual during the last weekend in July when the Central Texas Mustang Club holds its annual rally in Austin. At times like this, Hervis uses a revenue management system to determine the optimal number of reservations to have available for the Shelby-H cars. Hervis has agreed to have at least 60% of its Shelby-H Mustangs available for rally attendees at a special rate. Although many of the rally attendees will request a Saturday and Sunday two-day package, some attendees may select a Saturday only or a Sunday only reservation. Customers not attending the rally may also request a Saturday and Sunday two-day package, or make a Saturday only or Sunday only reservation. Thus, six types of reservations are possible. The cost for each type of reservation is shown here.

Rally Regular

Two-Day Package 125 $150

Saturday Only $75 $85

Sunday Only $65 $75

The anticipated demand for each type of reservation is as follows:

Rally Regular

Two-Day Package 20 10

Saturday Only 10 20

Sunday Only 15 25

Hervis Car Rental would like to determine how many Shelby-H Mustangs to make available for each type of reservation in order to maximize total revenue. a. Define the decision variables. b. Formulate a linear programming model for this revenue management application. ANSWER: a. RATW = number of reservations available for rally attendees for two days RGTW = number of reservations available for regular customers for two days RASA = number of reservations available for rally attendees for Saturday only RGSA = number of reservations available for regular customers for Saturday only RASU = number of reservations available for rally attendees for Sunday only RGSU = number of reservations available for regular customers for Sunday only b. Max

125RATW + 155RGTW + 75RASA + 90RGSA + 65RASU + 80RGSU

s.t. RATW + RETW + RASA + RGSA ≤ 50 Cengage Learning Testing, Powered by Cognero

(Shelby-H Saturday Capacity) Page 11

Chapter 5 - Advanced Linear Programming Applications RATW + RETW + RASU + RGSU ≤ 50

RATW ≤ 20 RGTW ≤ 10 RASA ≤ 10 RGSA ≤ 20 RASU ≤ 15 RGSU ≤ 15

(Shelby-H Sunday Capacity) (60% Saturday Capacity for Rally attendees) (60% Sunday Capacity for Rally attendees) (Two-Day Package Rally Demand) (Two-Day Package Regular Demand) (Saturday-Only Rally Demand) (Saturday-Only Regular Demand) (Sunday-Only Rally Demand) (Sunday-Only Regular Demand)

RATW,RGTW,RASA,RGSA,RASU,RGSU ≥ 0

(Non-negativity)

RATW + RASA ≥ 30 RATW + RASU ≥ 30

POINTS: 1 TOPICS: Revenue management 40. LeapFrog Airways provides passenger service for Indianapolis, Baltimore, Memphis, Austin, and Tampa. LeapFrog has two WB828 airplanes, one based in Indianapolis and the other in Baltimore. Each morning the Indianapolis based plane flies to Austin with a stopover in Memphis, and the Baltimore based plane flies to Tampa with a stopover in Memphis. Both planes have a coach section with a 120-seat capacity. LeapFrog uses two fare classes: a discount fare D class and a full fare F class. Leapfrog's products, each referred to as an origin destination itinerary fare (ODIF), are listed below with their fares and forecasted demand. ODIF 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Origin Indianapolis Indianapolis Indianapolis Indianapolis Indianapolis Indianapolis Baltimore Baltimore Baltimore Baltimore Baltimore Baltimore Memphis Memphis Memphis Memphis

Destination Memphis Austin Tampa Memphis Austin Tampa Memphis Austin Tampa Memphis Austin Tampa Austin Tampa Austin Tampa

Fare Class D D D F F F D D D F F F D D F F

ODIF Code IMD IAD ITD IMF IAF ITF BMD BAD BTD BMF BAF BTF MAD MTD MAF MTF

Fare 175 275 285 395 425 475 185 315 290 385 525 490 190 180 310 295

Demand 44 25 40 15 10 8 26 50 42 12 16 9 58 48 14 11

Develop a linear programming model for LeapFrog's problem situation and determine how many seats LeapFrog should allocate to each ODIF. ANSWER: Define the decision variables There are 16 variables, one for each ODIF. IMD = number of seats allocated to Indianapolis-Memphis-Discount class IAD = number of seats allocated to Indianapolis-Austin-Discount class ITD = number of seats allocated to Indianapolis-Tampa-Discount class IMF = number of seats allocated to Indianapolis-Memphis-Full Fare class IAF = number of seats allocated to Indianapolis-Austin-Full Fare class Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications ITF = number of seats allocated to Indianapolis-Tampa-Full Fare class BMD = number of seats allocated to Baltimore-Memphis-Discount class BAD = number of seats allocated to Baltimore-Austin-Discount class BTD = number of seats allocated to Baltimore-Tampa-Discount class BMF = number of seats allocated to Baltimore-Memphis-Full Fare class BAF = number of seats allocated to Baltimore-Austin-Full Fare class BTF = number of seats allocated to Baltimore-Tampa-Full Fare class MAD = number of seats allocated to Memphis-Austin-Discount class MTD = number of seats allocated to Memphis-Tampa-Discount class MAF = number of seats allocated to Memphis-Austin-Full Fare class MTF = number of seats allocated to Memphis-Tampa-Full Fare class Define the objective function Maximize total revenue: Max (fare per seat for each ODIF) x (number of seats allocated to the ODIF). Max 175IMD + 275IAD + 285ITD + 395IMF + 425IAF + 475ITF + 185BMD + 315BAD + 290BTD + 385BMF + 525BAF + 490BTF + 190MAD + 180MTD + 310MAF + 295MTF Define the constraints There are 4 capacity constraints, one for each flight leg: (1) IMD + IAD + ITD + IMF + IAF + ITF  120 (Indianapolis-Memphis leg) (2) BMD + BAD + BTD + BMF + BAF + BTF  120 (Baltimore-Memphis leg) (3) IAD + IAF + BAD + BAF + MAD + MAF  120 (Memphis-Austin leg) (4) ITD + ITF + BTD + BTF + MTD + MTF  120 (Memphis-Tampa leg) There are 16 demand constraints, one for each ODIF: (5) IMD  44 (11) BMD  26 (17) MAD  58 (6) IAD  25 (12) BAD  50 (18) MTD  48 (7) ITD  40 (13) BTD  42 (19) MAF  14 (8) IMF  15 (14) BMF  12 (20) MTF  11 (15) BAF < 16 (9) IAF  10 (10) ITF  8 (16) BTF  9 Nonnegativity constraints: IMD, IAD, ITD, ..., MTF  0 OBJECTIVE FUNCTION VALUE = 94735.000 VARIABLE IMD IAD ITD IMF IAF ITF BMD BAD BTD BMF BAF BTF MAD MTD MAF MTF

VALUE 44.000 3.000 40.000 15.000 10.000 8.000 26.000 50.000 7.000 12.000 16.000 9.000 27.000 45.000 14.000 11.000

CONSTRAINT

SLACK/SURPLUS

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REDUCED COSTS 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 DUAL PRICE Page 13

Chapter 5 - Advanced Linear Programming Applications 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.000 0.000 0.000 0.000 0.000 22.000 0.000 0.000 0.000 0.000 0.000 0.000 35.000 0.000 0.000 0.000 31.000 3.000 0.000 0.000

85.000 110.000 190.000 180.000 90.000 0.000 20.000 310.000 150.000 210.000 75.000 15.000 0.000 275.000 225.000 200.000 0.000 0.000 120.000 115.000

POINTS: 1 Essay 41. List several industries in which revenue management has been applied and list several different types of decisions being made with the aid of the methodology. ANSWER: Answer not provided. POINTS: 1 TOPICS: Revenue management 42. List several types of organizations with multiple operating units where data envelopment analysis might be applied and give examples of possible inputs and outputs for each organization. ANSWER: Answer not provided. POINTS: 1 TOPICS: Data envelopment analysis 43. Describe the steps used to determine when a two-person, zero-sum game has a pure-strategy solution. ANSWER: Answer not provided. POINTS: 1 TOPICS: Game theory 44. Write a summary of the DEA approach and explain how you would interpret the solution. ANSWER: Answer not provided. POINTS: 1 TOPICS: Data envelopment analysis 45. Explain the differences between the LP formulations for a conservative portfolio and moderate-risk portfolio. ANSWER: Answer not provided. POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 5 - Advanced Linear Programming Applications TOPICS: Portfolio models and asset allocation 46. Explain the difference between a pure strategy and a mixed strategy in game theory. ANSWER: Answer not provided. POINTS: 1 TOPICS: Game theory 47. Explain the logic of a DEA model. On what basis can the facility being evaluated be judged as relatively inefficient? ANSWER: Answer not provided. POINTS: 1 TOPICS: Data envelopment analysis

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Chapter 6 - Distribution and Network Models True / False 1. Whenever total supply is less than total demand in a transportation problem, the LP model does not determine how the unsatisfied demand is handled. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 2. Converting a transportation problem LP from cost minimization to profit maximization requires only changing the objective function; the conversion does not affect the constraints. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 3. A transportation problem with 3 sources and 4 destinations will have 7 decision variables. a. True b. False ANSWER: False POINTS: 1 TOPICS: Transportation problem 4. If a transportation problem has four origins and five destinations, the LP formulation of the problem will have nine constraints. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 5. The capacitated transportation problem includes constraints which reflect limited capacity on a route. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 6. When the number of agents exceeds the number of tasks in an assignment problem, one or more dummy tasks must be introduced in the LP formulation or else the LP will not have a feasible solution. a. True b. False ANSWER: False POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models TOPICS: Assignment problem 7. A transshipment constraint must contain a variable for every arc entering or leaving the node. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transshipment problem 8. The shortest-route problem is a special case of the transshipment problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Shortest-route problem 9. Transshipment problem allows shipments both in and out of some nodes while transportation problems do not. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation and transshipment problems 10. A dummy origin in a transportation problem is used when supply exceeds demand. a. True b. False ANSWER: False POINTS: 1 TOPICS: Transportation problem 11. When a route in a transportation problem is unacceptable, the corresponding variable can be removed from the LP formulation. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 12. In the LP formulation of a maximal flow problem, a conservation-of-flow constraint ensures that an arc's flow capacity is not exceeded. a. True b. False ANSWER: False POINTS: 1 TOPICS: Maximal flow problem Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models 13. The maximal flow problem can be formulated as a capacitated transshipment problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Maximal flow problem 14. The direction of flow in the shortest-route problem is always out of the origin node and into the destination node. a. True b. False ANSWER: True POINTS: 1 TOPICS: Shortest-route problem 15. A transshipment problem is a generalization of the transportation problem in which certain nodes are neither supply nodes nor destination nodes. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transshipment problem 16. The assignment problem is a special case of the transportation problem in which all supply and demand values equal one. a. True b. False ANSWER: True POINTS: 1 TOPICS: Assignment problem 17. A transportation problem with 3 sources and 4 destinations will have 7 variables in the objective function. a. True b. False ANSWER: False POINTS: 1 TOPICS: Assignment problem 18. Flow in a transportation network is limited to one direction. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transportation problem 19. In a transportation problem with total supply equal to total demand, if there are four origins and seven destinations, and there is a unique optimal solution, the optimal solution will utilize 11 shipping routes. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models a. True b. False ANSWER: False POINTS: 1 TOPICS: Transportation problem 20. In the general assignment problem, one agent can be assigned to several tasks. a. True b. False ANSWER: True POINTS: 1 TOPICS: Assignment problem 21. In a capacitated transshipment problem, some or all of the transfer points are subject to capacity restrictions. a. True b. False ANSWER: False POINTS: 1 TOPICS: Transshipment problem Multiple Choice 22. The problem which deals with the distribution of goods from several sources to several destinations is the a. maximal flow problem b. transportation problem c. assignment problem d. shortest-route problem ANSWER: b POINTS: 1 TOPICS: Transportation problem 23. The parts of a network that represent the origins are a. the capacities b. the flows c. the nodes d. the arcs ANSWER: c POINTS: 1 TOPICS: Transportation problem 24. The objective of the transportation problem is to a. identify one origin that can satisfy total demand at the destinations and at the same time minimize total shipping cost. b. minimize the number of origins used to satisfy total demand at the destinations. c. minimize the number of shipments necessary to satisfy total demand at the destinations. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models d. minimize the cost of shipping products from several origins to several destinations. ANSWER: d POINTS: 1 TOPICS: Transportation problem 25. The number of units shipped from origin i to destination j is represented by a. xij. b. xji. c. cij. d. cji. ANSWER: a POINTS: 1 TOPICS: Transportation problem 26. Which of the following is not true regarding the linear programming formulation of a transportation problem? a. Costs appear only in the objective function. b. The number of variables is (number of origins) x (number of destinations). c. The number of constraints is (number of origins) x (number of destinations). d. The constraints' left-hand side coefficients are either 0 or 1. ANSWER: c POINTS: 1 TOPICS: Transportation problem 27. The difference between the transportation and assignment problems is that a. total supply must equal total demand in the transportation problem b. the number of origins must equal the number of destinations in the transportation problem c. each supply and demand value is 1 in the assignment problem d. there are many differences between the transportation and assignment problems ANSWER: c POINTS: 1 TOPICS: Assignment problem 28. In the general linear programming model of the assignment problem, a. one agent can do parts of several tasks. b. one task can be done by several agents. c. each agent is assigned to its own best task. d. one agent is assigned to one and only one task. ANSWER: d POINTS: 1 TOPICS: Assignment problem 29. The assignment problem is a special case of the a. transportation problem. b. transshipment problem. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models c. maximal flow problem. d. shortest-route problem. ANSWER: a POINTS: 1 TOPICS: Assignment problem 30. Which of the following is not true regarding an LP model of the assignment problem? a. Costs appear in the objective function only. b. All constraints are of the ≥ form. c. All constraint left-hand side coefficient values are 1. d. All decision variable values are either 0 or 1. ANSWER: b POINTS: 1 TOPICS: Assignment problem 31. The assignment problem constraint x31 + x32 + x33 + x34 ≤ 2 means a. agent 3 can be assigned to 2 tasks. b. agent 2 can be assigned to 3 tasks. c. a mixture of agents 1, 2, 3, and 4 will be assigned to tasks. d. there is no feasible solution. ANSWER: a POINTS: 1 TOPICS: Assignment problem 32. Arcs in a transshipment problem a. must connect every node to a transshipment node. b. represent the cost of shipments. c. indicate the direction of the flow. d. All of the alternatives are correct. ANSWER: c POINTS: 1 TOPICS: Transshipment problem 33. Constraints in a transshipment problem a. correspond to arcs. b. include a variable for every arc. c. require the sum of the shipments out of an origin node to equal supply. d. All of the alternatives are correct. ANSWER: b POINTS: 1 TOPICS: Transshipment problem 34. In a transshipment problem, shipments a. cannot occur between two origin nodes. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models b. cannot occur between an origin node and a destination node. c. cannot occur between a transshipment node and a destination node. d. can occur between any two nodes. ANSWER: d POINTS: 1 TOPICS: Transshipment problem 35. Consider a shortest route problem in which a bank courier must travel between branches and the main operations center. When represented with a network, a. the branches are the arcs and the operations center is the node. b. the branches are the nodes and the operations center is the source. c. the branches and the operations center are all nodes and the streets are the arcs. d. the branches are the network and the operations center is the node. ANSWER: c POINTS: 1 TOPICS: Shortest-route problem 36. The shortest-route problem finds the shortest-route a. from the source to the sink. b. from the source to any other node. c. from any node to any other node. d. from any node to the sink. ANSWER: b POINTS: 1 TOPICS: Shortest-route problem 37. Consider a maximal flow problem in which vehicle traffic entering a city is routed among several routes before eventually leaving the city. When represented with a network, a. the nodes represent stoplights. b. the arcs represent one way streets. c. the nodes represent locations where speed limits change. d. None of the alternatives is correct. ANSWER: b POINTS: 1 TOPICS: Maximal flow problem 38. We assume in the maximal flow problem that a. the flow out of a node is equal to the flow into the node. b. the source and sink nodes are at opposite ends of the network. c. the number of arcs entering a node is equal to the number of arcs exiting the node. d. None of the alternatives is correct. ANSWER: a POINTS: 1 TOPICS: Maximal flow problem Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models 39. If a transportation problem has four origins and five destinations, the LP formulation of the problem will have a. 5 constraints b. 9 constraints c. 18 constraints d. 20 constraints ANSWER: b POINTS: 1 TOPICS: Transportation problem 40. Which of the following is not a characteristic of assignment problems? a. costs appear in the objective function only b. the RHS of all constraints is 1 c. the value of all decision variables is either 0 or 1 d. the signs of constraints are always < ANSWER: d POINTS: 1 TOPICS: Assignment problem 41. The network flows into and out of demand nodes are what makes the production and inventory application modeled in the textbook a a. shortest-route model. b. maximal flow model. c. transportation model d. transshipment model ANSWER: d POINTS: 1 TOPICS: A production and inventory application Subjective Short Answer 42. Write the LP formulation for this transportation problem.

ANSWER: Min

5X1A + 6X1B + 4X2A + 2X2B + 3X3A + 6X3B + 9X 4A + 7X4B

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Chapter 6 - Distribution and Network Models s.t.

X1A + X1B ≤ 100 X2A + X2B ≤ 200 X3A + X3B ≤ 150 X4A + X4B ≤ 50 X1A + X2A + X3A + X4A = 250 X1B + X 2B + X3B + X4B = 250 all Xij ≥ 0

POINTS: 1 TOPICS: Transportation problem 43. Draw the network for this transportation problem. Min

2XAX + 3XAY + 5XAZ+ 9XBX + 12XBY + 10XBZ

XAX + XAY + XAZ ≤ 500 X BX + XBY + XBZ ≤ 400 XAX + XBX = 300 XAY + XBY = 300 XAZ + XBZ = 300 Xij ≥ 0 ANSWER: s.t.

POINTS: 1 TOPICS: Transportation problem 44. Canning Transport is to move goods from three factories to three distribution centers. Information about the move is given below. Give the network model and the linear programming model for this problem. Source A B C

Supply 200 100 150

Destination X Y Z

Demand 50 125 125

Destination Y

Shipping costs are: Source

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Chapter 6 - Distribution and Network Models A B C

3 2 5 9 10 -5 6 4 (Source B cannot ship to destination Z)

ANSWER:

Min

3XAX + 2XAY + 5XAZ + 9XBX + 10XBY + 5XCX + 6XCY + 4XCZ

s.t.

XAX + XAY + XAZ ≤ 200 XBX + XBY ≤ 100 XCX + XCY + XCZ ≤ 150 XDX + XDY + XDZ ≤ 50 XAX + XBX + XCX + XDX = 250 X AY + XBY + XCY + XDY = 125 XAZ + XBZ + XCZ + XDZ = 125 Xij ≥ 0

POINTS: 1 TOPICS: Transportation problem 45. The following table shows the unit shipping cost between cities, the supply at each source city, and the demand at each destination city. The Management Scientist solution is shown. Report the optimal solution. Source St. Louis Evansville Bloomington Demand

Terre Haute 8 5 3 150

Destination Indianapolis Ft. Wayne 6 12 5 10 2 9 60 45

South Bend 9 8 10 45

Supply 100 100 100

TRANSPORTATION PROBLEM ***************************** OBJECTIVE: MINIMIZATION SUMMARY OF ORIGIN SUPPLIES ******************************** ORIGIN SUPPLY -------------------Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models 1 2 3

100 100 100

SUMMARY OF DESTINATION DEMANDS *************************************** DESTINATION DEMAND ------------------------------1 150 2 60 3 45 4 45 SUMMARY OF UNIT COST OR REVENUE DATA ********************************************* FROM TO DESTINATION ORIGIN 1 2 3 4 -------------------------1 8 6 12 9 2 5 5 10 8 3 3 2 9 10 OPTIMAL TRANSPORTATION SCHEDULE **************************************** SHIP FROM TO DESTINATION ORIGIN 1 2 3 4 -------------------------1 0 10 45 45 2 100 0 0 0 3 50 50 0 0 TOTAL TRANSPORTATION COST OR REVENUE IS 1755 ANSWER: Ship 10 from St. Louis to Indianapolis, 45 from St. Louis to Ft. Wayne, 45 from St. Louis to South Bend, 100 from Evansville to Terre Haute, 50 from Bloomington to Terre Haute, and 50 from Bloomington to Indianapolis. The total cost is 1755. POINTS: 1 TOPICS: Transportation problem 46. After some special presentations, the employees of the AV Center have to move projectors back to classrooms. The table below indicates the buildings where the projectors are now (the sources), where they need to go (the destinations), and a measure of the distance between sites. Source Baker Hall Tirey Hall Arena Demand

Business 10 12 15 12

Education 9 11 14 20

Destination Parsons Hall 5 1 7 10

Holmstedt Hall 2 6 6 10

Supply 35 10 20

a. If you were going to write this as a linear programming model, how many decision variables would there be, and how many constraints would there be? The solution to this problem is shown below. Use it to answer the questions b - e. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models TRANSPORTATION PROBLEM ***************************** OPTIMAL TRANSPORTATION SCHEDULE **************************************** FROM TO DESTINATION FROM ORIGIN 1 2 ------------------- ----------1 12 20 2 0 0 3 0 0

3 -----0 10 0

4 -----3 0 7

TOTAL TRANSPORTATION COST OR REVENUE IS 358 NOTE: THE TOTAL SUPPLY EXCEEDS THE TOTAL DEMAND BY 13 ORIGIN ---------3

EXCESS SUPPLY ----------------------13

b. How many projectors are moved from Baker to Business? c. How many projectors are moved from Tirey to Parsons? d. How many projectors are moved from the Arena to Education? e. Which site(s) has (have) projectors left? ANSWER: a. 12 decision variables, 7 constraints b. 12 c. 10 d. 0 e. Arena POINTS: 1 TOPICS: Transportation problem 47. Show both the network and the linear programming formulation for this assignment problem. Task Person 1 2 3

A 9 12 11

B 5 6 6

C 4 3 5

D 2 5 7

ANSWER:

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Chapter 6 - Distribution and Network Models

Let

Xij = 1 if person i is assigned to job j = 0 otherwise

Min

9X1A + 5X1B + 4X1C + 2X1D + 12X2A + 6X2B + 3X2C + 5X2D + 11X3A + 6X3B + 5X3C + 7X3D

s.t.

X1A + X1B + X1C + X1D ≤ 1 X2A + X2B + X2C + X2D ≤ 1 X 3A + X3B + X3C + X3D ≤ 1 X4A + X4B + X4C + X4D ≤ 1 X1A + X2A + X3A + X4A = 1 X1B + X2B + X3B + X4B = 1 X1C + X2C + X3C + X4C = 1 X1D + X2D + X3D + X4D = 1

POINTS: 1 TOPICS: Assignment problem 48. Draw the network for this assignment problem. Min

10x1A + 12x1B + 15x1C + 25x1D + 11x2A + 14x2B + 19x2C + 32x2D + 18x3A + 21x3B + 23x3C + 29x3D + 15x4A + 20x4B + 26x4C + 28x4D

s.t.

x1A + x1B + x1C + x1D = 1 x2A + x2B + x2C + x2D = 1 x3A + x3B + x3C + x3D = 1 x4A + x4B + x4C + x4D = 1 x1A + x2A + x3A + x4A = 1 x1B + x 2B + x3B + x4B = 1 x1C + x 2C + x3C + x4C = 1

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Chapter 6 - Distribution and Network Models x1D + x2D + x3D + x4D = 1 ANSWER:

POINTS: 1 TOPICS: Assignment problem 49. A professor has been contacted by four not-for-profit agencies that are willing to work with student consulting teams. The agencies need help with such things as budgeting, information systems, coordinating volunteers, and forecasting. Although each of the four student teams could work with any of the agencies, the professor feels that there is a difference in the amount of time it would take each group to solve each problem. The professor's estimate of the time, in days, is given in the table below. Use the computer solution to see which team works with which project.

Team Budgeting A 32 B 38 C 41 D 45 ASSIGNMENT PROBLEM ************************ OBJECTIVE: MINIMIZATION

Projects Information Volunteers 35 15 40 18 42 25 45 30

Forecasting 27 35 38 42

SUMMARY OF UNIT COST OR REVENUE DATA ********************************************* TASK AGENT 1 2 3 4 -------------------------1 32 35 15 27 2 38 40 18 35 3 41 42 25 38 4 45 45 30 42 OPTIMAL ASSIGNMENTS COST/REVENUE ************************ *************** ASSIGN AGENT 3 TO TASK 1 41 ASSIGN AGENT 4 TO TASK 2 45 ASSIGN AGENT 2 TO TASK 3 18 ASSIGN AGENT 1 TO TASK 4 27 Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models ----------------------------------------------TOTAL COST/REVENUE 131 ANSWER: Team A works with the forecast, Team B works with volunteers, Team C works with budgeting, and Team D works with information. The total time is 131. POINTS: 1 TOPICS: Assignment problem 50. Write the linear program for this transshipment problem.

ANSWER: Min

3x16 + 2x14 + 3x15 + 5x24 + 6x25 + 2x32 + 8x 34 + 10x35 + 5x46 + 9x47 + 12x56 + 15x57

s.t.

x16 + x14 + x35 ≤ 500 x24 + x25 − x23 ≤ 400 x32 + x34 + x35 ≤ 300 x46 + x47 − (x14 + x24 + x34) = 0 x56 + x57 − (x15 + x25 + x35) = 0 x16 + x46 + x56 = 600 x56 + x57 = 600

POINTS: 1 TOPICS: Transshipment problem 51. Peaches are to be transported from three orchard regions to two canneries. Intermediate stops at a consolidation station are possible. Orchard Riverside Sunny Slope Old Farm

Supply 1200 1500 2000

Station Waterford Northside

Cannery Sanderson Millville

Capacity 2500 3000

Shipment costs are shown in the table below. Where no cost is given, shipments are not possible. Where costs are shown, shipments are possible in either direction. Draw the network model for this problem. R Riverside Sunny Side

SS 1

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W 5 4

S 3

5 Page 15

Chapter 6 - Distribution and Network Models Old Farm Waterford Northside Sanderson Millville ANSWER:

3 2

2 5

4 9 2

POINTS: 1 TOPICS: Transshipment problem 52. RVW (Restored Volkswagens) buys 15 used VW's at each of two car auctions each week held at different locations. It then transports the cars to repair shops it contracts with. When they are restored to RVW's specifications, RVW sells 10 each to three different used car lots. There are various costs associated with the average purchase and transportation prices from each auction to each repair shop. Also there are transportation costs from the repair shops to the used car lots. RVW is concerned with minimizing its total cost given the costs in the table below. a. Given the costs below, draw a network representation for this problem. Repair Shops S1 S2 Auction 1

550

500

Auction 2

600

450

S1 S2

Used Car Lots L1

250

300

500

350

650

450

b. Formulate this problem as a transshipment linear programming model. ANSWER:

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Chapter 6 - Distribution and Network Models

Denote A1 as node 1, A2 as node 2, S1 as node 3, S2 as node 4, L1 as node 5, L2 as node 6, and L3 as node 7 Min

50X13 + 500X14 + 600X23 + 450X24 + 250X35 + 300X36 + 500X37 + 350X45 + 650X46 + 450X47

s.t.

X13 + X14 ≤ 15 X23 + X24 ≤ 15 X13 + X23 − X35 − X36 − X37 = 0 X14 + X24 − X45 − X46 − X47 = 0 X35 + X45 = 10 X36 + X46 = 10 X37 + X4 = 10 Xij ≥ 0 for all i,j

POINTS: 1 TOPICS: Transshipment problem 53. Consider the network below. Formulate the LP for finding the shortest-route path from node 1 to node 7.

ANSWER: Min

10X12 + 12X13 + 4X24 + 8X25 + 7X35 + 9X36 + 4X 42 + 3X45

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Chapter 6 - Distribution and Network Models + 6X47 + 8X52 + 7X53 + 3X54 + 4X57 + 9X63 + 3X67 s.t.

X12 + X13 =1 −X12 + X24 + X25 − X42 − X52 =0 −X13 + X35 + X36 − X53 − X63 =0 −X24 + X42 + X45 + X47 − X54 =0 −X25 − X35 − X45 + X52 + X53 + X57 = 0 −X36 + X63 + X67 =0 X47 + X57 + X67 =1 Xij ≥ 0 for all i,j

POINTS: 1 TOPICS: Shortest-route problem 54. Consider the following shortest-route problem involving six cities with the distances given. Draw the network for this problem and formulate the LP for finding the shortest distance from City 1 to City 6. Path 1 to 2 1 to 3 2 to 4 2 to 5 3 to 4 3 to 5 4 to 6 5 to 6 ANSWER:

Distance 3 2 4 5 3 7 6 2

Min

3X12 + 2X13 + 4X24 + 5X25 + 3X34 + 7X35 + 4X42 + 3X43 + 6X46 + 5X 52 + 7X53 + 2X56

s.t.

X12 + X13 =1 −X12 + X24 + X25 − X42 − X52 = 0 −X13 + X34 + X35 − X 43 − X53 = 0 −X24 − X34 + X42 + X43 + X46 = 0 −X25 − X35 + X52 + X53 + X56 = 0 X46 + X56 =1 Xij ≥ 0 for all i,j

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Chapter 6 - Distribution and Network Models POINTS: 1 TOPICS: Shortest-route problem 55. A beer distributor needs to plan how to make deliveries from its warehouse (Node 1) to a supermarket (Node 7), as shown in the network below. Develop the LP formulation for finding the shortest route from the warehouse to the supermarket.

ANSWER: Min

3X12 + 3X15 + 12X16 + 5X23 + 5X32 + 6X34 + 6X 43 + 4X46 + 5X47 + 8X56 + 4X64 + 8X65 + 3X67

s.t.

X12 + X15 + X16 =1 −X 12 + X23 =0 −X23 + X32 + X34 =0 −X34 + X43 + X46 + X47 − 4X64 =0 −X15 + X56 =0 −X16 − X46 − X56 + X64 + X65 + X67 = 0 X47 + X67 =1 Xij ≥ 0 for all i,j

POINTS: 1 TOPICS: Shortest-route problem 56. Consider the following shortest-route problem involving seven cities. The distances between the cities are given below. Draw the network model for this problem and formulate the LP for finding the shortest route from City 1 to City 7. Path 1 to 2 1 to 3 1 to 4 2 to 3 2 to 5 3 to 4 3 to 5 3 to 6 4 to 6 5 to 7 6 to 7 ANSWER:

Distance 6 10 7 4 5 5 2 4 8 7 5

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Chapter 6 - Distribution and Network Models

Min

6X12 + 10X13 + 7X14 + 4X23 + 5X25 + 4X32 + 5X34 + 2X35 + 4X36 + 5X43 + 8X46 + 5X52 + 2X53 + 7X57 + 4X63 + 8X64 + 5X67

s.t.

X12 + X13 + X14 =1 −X12 + X23 + X25 − X32 − X52 =0 −X13 − X23 + X32 + X34 + X35 + X36 − X43 − X53 − X63 = 0 −X14 − X34 + X43 + X46 − X64 =0 −X25 − X35 + X52 + X53 + X57 =0 −X36 − X46 + X63 + X64 + X67 =0 X57 + X67 =1 Xij ≥ 0 for all i,j

POINTS: 1 TOPICS: Shortest-route problem 57. The network below shows the flows possible between pairs of six locations. Formulate an LP to find the maximal flow possible from Node 1 to Node 6.

ANSWER: Min X61 s.t. X12 + X13 + X15 − X61 =0 X23 + X 24 − X12 − X32 =0 X32 + X34 + X35 − X13 − X23 − X53 = 0 Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models X43 + X46 − X24 − X34 X53 + X56 − X15 − X35 X61 − X36 − X46 − X56 X12 ≤ 18 X23 ≤ 9 X32 ≤ 4 X43 ≤ 10 X53 ≤ 8

=0 =0 =0 X13 ≤ 20 X24 ≤ 15 X34 ≤ 10 X46 ≤ 14 X56 ≤ 10 Xij ≥ 0 for all i,j

X15 ≤ 10 X35 ≤ 8

POINTS: 1 TOPICS: Maximal flow problem 58. A network of railway lines connects the main lines entering and leaving a city. Speed limits, track reconstruction, and train length restrictions lead to the flow diagram below, where the numbers represent how many cars can pass per hour. Formulate an LP to find the maximal flow in cars per hour from Node 1 to Node F.

ANSWER: Min XF1 s.t. X12 + X15 + X16 − XF1 =0 X23 + X24 − X 12 =0 X34 − X23 =0 X48 + X4F − X24 − X34 − X84 = 0 X57 − X15 =0 X67 + X69 − X16 − X76 =0 X76 + X79 − X57 − X67 =0 X84 + X89 + X8F − X48 − X98 = 0 X98 + X9F − X69 − X79 − X89 = 0 XF1 − X4F − X8F − X9F =0 X12 ≤ 500 X15 ≤ 300 Cengage Learning Testing, Powered by Cognero

X16 ≤ 600 Page 21

Chapter 6 - Distribution and Network Models X23 ≤ 300 X34 ≤ 150 X48 ≤ 400 X57 ≤ 400 X67 ≤ 300 X76 ≤ 200 X84 ≤ 200 X98 ≤ 300

X24 ≤ 400 X4F ≤ 600 X69 ≤ 500 X79 ≤ 350 X89 ≤ 300 X9F ≤ 500 Xij ≥ 0 for all i,j

X8F ≤ 450

POINTS: 1 TOPICS: Maximal flow problem 59. Fodak must schedule its production of camera film for the first four months of the year. Film demand (in 1,000s of rolls) in January, February, March and April is expected to be 300, 500, 650 and 400, respectively. Fodak's production capacity is 500 thousand rolls of film per month. The film business is highly competitive, so Fodak cannot afford to lose sales or keep its customers waiting. Meeting month i 's demand with month i +1's production is unacceptable. Film produced in month i can be used to meet demand in month i or can be held in inventory to meet demand in month i +1 or month i +2 (but not later due to the film's limited shelflife). There is no film in inventory at the start of January. The film's production and delivery cost per thousand rolls will be $500 in January and February. This cost will increase to $600 in March and April due to a new labor contract. Any film put in inventory requires additional transport costing $100 per thousand rolls. It costs $50 per thousand rolls to hold film in inventory from one month to the next. a. Modeling this problem as a transshipment problem, draw the network representation. b. Formulate and solve this problem as a linear program. ANSWER:

Define the decision variables: xij = amount of film "moving" between node i and node j Define objective: MIN 600x15 + 500x18 + 600x26 + 500x29 + 700x 37 + 600x310 + 600x411 + 50x59

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Chapter 6 - Distribution and Network Models + 100x510 + 50x610 + 100x611 + 50x711 Define the constraints: Amount (1000's of rolls) of film produced in January: Amount (1000's of rolls) of film produced in February: Amount (1000's of rolls) of film produced in March: Amount (1000's of rolls) of film produced in April: Amount (1000's of rolls) of film in/out of January inventory: Amount (1000's of rolls) of film in/out of February inventory: Amount (1000's of rolls) of film in/out of March inventory: Amount (1000's of rolls) of film satisfying January demand: Amount (1000's of rolls) of film satisfying February demand: Amount (1000's of rolls) of film satisfying March demand: Amount (1000's of rolls) of film satisfying April demand:

x15 + x18 ≤ 500 x26 + x29 ≤ 500 x37 + x310 ≤ 500 x411 ≤ 500 x15 − x59 − x510 = 0 x26 − x610 − x611 = 0 x37 − x711 = 0 x18 = 300 x29 + x59 = 500 x310 + x510 + x610 = 650 x411 + x611 + x711 = 400

Non-negativity of variables: xij ≥ 0, for all i and j. The Management Scientist provided the following solution: Objective Function Value = 1045000.000 VARIABLE X15 X18 X26 X29 X37 X310 X411 X59 X510 X610 X611 X711

VALUE 150.000 300.000 0.000 500.000 0.000 500.000 400.000 0.000 150.000 0.000 0.000 0.000

REDUCED COST 0.000 0.000 100.000 0.000 250.000 0.000 0.000 0.000 0.000 0.000 150.000 0.000

POINTS: 1 TOPICS: Production and inventory application 60. Find the maximal flow from node 1 to node 7 in the following network.

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Chapter 6 - Distribution and Network Models

ANSWER: Decision variables: xij = amount of flow from node i to node j Objective function: Maximize the flow through the network: Max x71 Constraints: Conservation of flow for each node: (1) x12 + x13 + x14 − x 71 = 0 (2) x24 + x25 − x12 − x42 − x52 = 0 (3) x34 + x36 − x13 − x43 = 0 (4) x42 + x 43 + x45 + x46 + x47 − x14 − x24 − x34 − x54 − x64 = 0 (5) x52 + x54 + x57 − x25 − x45 = 0 (6) x64 + x67 − x36 − x46 = 0 (7) x71 − x47 − x57 − x67 = 0 Capacity for each arc: (8) x12 ≤ 4 (14) x36 ≤ 6 (20) (9) x13 ≤ 3 (15) x42 ≤ 3 (21) (10) x14 ≤ 4 (16) x43 ≤ 5 (22) (11) x24 ≤ 2 (17) x45 ≤ 3 (23) (12) x25 ≤ 3 (18) x46 ≤ 1 (24) (13) x34 ≤ 3 (19) x47 ≤ 3 Nonnegativity: All xij ≥ 0

(node 1) (node 2) (node 3) (node 4) (node 5) (node 6) (node 7) x52 ≤ 3 x54 ≤ 4 x57 ≤ 2 x64 ≤ 1 x67 ≤ 5

The LP was solved using The Management Scientist. Two solutions are given below. Objective Function Value = 10.000 Variable x12 x13 x14 x24 x25 x36

Solution 1 3.000 3.000 4.000 1.000 2.000 5.000

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Solution 2 3.000 3.000 4.000 1.000 2.000 4.000 Page 24

Chapter 6 - Distribution and Network Models x43 2.000 x46 0.000 x47 3.000 x57 2.000 x67 5.000 x71 10.000 POINTS: 1 TOPICS: Maximal flow problem

1.000 1.000 3.000 2.000 5.000 10.000

61. A foreman is trying to assign crews to produce the maximum number of parts per hour of a certain product. He has three crews and four possible work centers. The estimated number of parts per hour for each crew at each work center is summarized below. Solve for the optimal assignment of crews to work centers. Work Center WC1 WC2 WC3 WC4 Crew A 15 20 18 30 Crew B 20 22 26 30 Crew C 25 26 27 30 ANSWER: OBJECTIVE FUNCTION VALUE = 82.000 VARIABLE AA1 AA2 AA3 AA4 AB1 AB2 AB3 AB4 AC1 AC2 AC3 AC4 AD1 AD2 AD3 AD4

VALUE 0.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 0.000

REDUCED COST 12.000 0.000 0.000 1.000 2.000 4.000 0.000 0.000 12.000 0.000 0.000 1.000 2.000 4.000 0.000 0.000

CONSTRAINT SLACK/SURPLUS 1 0.000 2 0.000 3 0.000 4 0.000 5 0.000 6 0.000 7 0.000 8 0.000 An optimal solution is: Cengage Learning Testing, Powered by Cognero

DUAL PRICE 18.000 23.000 24.000 −1.000 1.000 2.000 3.000 12.000

Page 25

Chapter 6 - Distribution and Network Models Crew Crew A Crew B Crew C ----------

Work Center WC4 WC3 WC2 WC1 Total Parts Per Hour

Parts/Hour 30 26 26 --82

POINTS: 1 TOPICS: Assignment problem 62. A plant manager for a sporting goods manufacturer is in charge of assigning the manufacture of four new aluminum products to four different departments. Because of varying expertise and workloads, the different departments can produce the new products at various rates. If only one product is to be produced by each department and the daily output rates are given in the table below, which department should manufacture which product to maximize total daily product output? (Note: Department 1 does not have the facilities to produce golf clubs.) Department 1 2 3 4

Baseball Bats 100 100 110 85

Tennis Rackets 60 80 75 50

Golf Clubs X 140 150 100

Racquetball Rackets 80 100 120 75

Formulate this assignment problem as a linear program. ANSWER: xi represent all possible combinations of departments and products For example: x1 = 1 if Department 1 is assigned baseball bats; = 0 otherwise x2 = 1 if Department 1 is assigned tennis rackets; = 0 otherwise x5 = 1 if Department 2 is assigned baseball bats; = 0 otherwise x15 = 1 if Department 4 is assigned golf clubs; = 0 otherwise (Note: x3 is not used because Dept.1/golf clubs is not a feasible assignment) Min Z = 100x1+60x2+80x4+100x5+80x6+140x7+100x8+110x9 +75x10+150x11+120x12+85x13+50x14+100x15+75x16 S.T. x1 + x2 + x4 = 1 x5 + x6 + x7 + x8 = 1 x9 + x10 + x11 + x12 = 1 x13 + x14 + x15 + x16 = 1 x1 + x5 + x9 + x13 = 1 x2 + x6 + x10 + x14 = 1 x7 + x11 + x15 = 1 x4 + x8 + x12 + x16 = 1 POINTS: 1 TOPICS: Assignment problem 63. A clothing distributor has four warehouses which serve four large cities. Each warehouse has a monthly capacity of 5,000 blue jeans. They are considering using a transportation LP approach to match demand and capacity. The following table provides data on their shipping cost, capacity, and demand constraints on a per-month basis. Develop a linear programming model for this problem. Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models Warehouse A B C D City Demand

City E City F City G City H .53 .21 .52 .41 .31 .38 .41 .29 .56 .32 .54 .33 .42 .55 .34 .52 2,000 3,000 3,500 5,500

ANSWER: Xij = each combination of warehouse i and city j Min .53XAE + .21XAF + .52XAG + .41XAH + .31XBE + .38XBF + .41XBG + .29XBH + .56XCE + .32XCF + .54XCG + .33XCH + .42XDE + .55XDF + .34XDG + .52XDH st XAE + XAF + XAG + XAH  5,000 XAE + XBE + XCE + XDE = 2,000 XBE + XBF + XBG + XBH  5,000 XAF + XBF + XCF + XDF = 3,000 XCE + XCF + XCG + XCH  5,000 XAG + XBG + XCG + XDG = 3,500 XDE + XDF + XDG + XDH  5,000 XAH + XBH + XCH + XDH = 5,500 POINTS: 1 TOPICS: Transportation problem 64. A computer manufacturing company wants to develop a monthly plan for shipping finished products from three of its manufacturing facilities to three regional warehouses. It is thinking about using a transportation LP formulation to exactly match capacities and requirements. Data on transportation costs (in dollars per unit), capacities, and requirements are given below. Plant A B C Requirement

Warehouse 1 2 3 2.41 1.63 2.09 3.18 5.62 1.74 4.12 3.16 3.09 8,000 2,000 3,000

Capacities 4,000 6,000 3,000

a. How many variables are involved in the LP formulation? b. How many constraints are there in this problem? c. What is the constraint corresponding to Plant B? d. What is the constraint corresponding to Warehouse 3? ANSWER: The problem formulation is shown below. Use it to answer the questions a - d. Xij = each combination of plant i and warehouse j Min 2.41XA1 + 1.63XA2 + 2.09XA3 + 3.18XB1 + 5.62XB2 + 1.74XB3 + 4.12XC1 + 3.16XC2 + 3.09XC3 st XA1 + XA2 +XA3 = 4,000 (capacities) XA1 + XB1 +XC1 = 8,000 (requirements) XB1 + XB2 + XB3 = 6,000 XA2 + XB2 +XC2 = 2,000 XC1 +XC2 + XC3 = 3,000 XA3 + XB3 +XC3 = 3,000 a. 9 variables b. 6 constraints c. XB1 + XB2 + XB3 = 6,000 d. XA3 + XB3 +XC3 = 3,000 POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 6 - Distribution and Network Models TOPICS: Transportation problem Essay 65. Explain how the general linear programming model of the assignment problem can be modified to handle problems involving a maximization function, unacceptable assignments, and supply not equally demand. ANSWER: Answer not provided. POINTS: 1 TOPICS: Assignment problem 66. Define the variables and constraints necessary in the LP formulation of the transshipment problem. ANSWER: Answer not provided. POINTS: 1 TOPICS: Transshipment problem 67. Explain what adjustments can be made to the transportation linear program when there are unacceptable routes. ANSWER: Answer not provided. POINTS: 1 TOPICS: Transportation problem 68. Is it a coincidence to obtain integer solutions to network problems? Explain. ANSWER: Answer not provided. POINTS: 1 TOPICS: Network problems 69. How is the assignment linear program different from the transportation model? ANSWER: Answer not provided. POINTS: 1 TOPICS: Transportation and assignment problems 70. Define the variables and constraints necessary in the LP formulation of the maximal flow problem. ANSWER: Answer not provided. POINTS: 1 TOPICS: Maximal flow problem 71. How is the shortest-route problem like the transshipment problem? ANSWER: Answer not provided. POINTS: 1 TOPICS: Shortest-route problem

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Chapter 7 - Integer Linear Programming True / False 1. The LP Relaxation contains the objective function and constraints of the IP problem, but drops all integer restrictions. a. True b. False ANSWER: True POINTS: 1 TOPICS: LP relaxation 2. In general, rounding large values of decision variables to the nearest integer value causes fewer problems than rounding small values. a. True b. False ANSWER: True POINTS: 1 TOPICS: LP relaxation 3. The solution to the LP Relaxation of a minimization problem will always be less than or equal to the value of the integer program minimization problem. a. True b. False ANSWER: False POINTS: 1 TOPICS: Graphical solution 4. If the optimal solution to the LP relaxation problem is integer, it is the optimal solution to the integer linear program. a. True b. False ANSWER: True POINTS: 1 TOPICS: LP relaxation 5. Slack and surplus variables are not useful in integer linear programs. a. True b. False ANSWER: False POINTS: 1 TOPICS: Capital budgeting 6. A multiple choice constraint involves selecting k out of n alternatives, where k ≥ 2. a. True b. False ANSWER: False POINTS: 1 TOPICS: Multiple choice constraint Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 7. In a model involving fixed costs, the 0 - 1 variable guarantees that the capacity is not available unless the cost has been incurred. a. True b. False ANSWER: True POINTS: 1 TOPICS: Fixed costs 8. If x1 + x2 ≤ 500y1 and y1 is 0 - 1, then if y1 is 0, x1 and x2 will be 0. a. True b. False ANSWER: True POINTS: 1 TOPICS: Distribution system design 9. The constraint x1 + x2 + x3 + x4 ≤ 2 means that two out of the first four projects must be selected. a. True b. False ANSWER: False POINTS: 1 TOPICS: k out of n alternatives constraint 10. The constraint x1 − x2 = 0 implies that if project 1 is selected, project 2 cannot be. a. True b. False ANSWER: False POINTS: 1 TOPICS: Conditional and corequisite constraints 11. The product design and market share optimization problem presented in the textbook is formulated as a 0-1 integer linear programming model. a. True b. False ANSWER: True POINTS: 1 TOPICS: Product design and market share optimization problem 12. The objective of the product design and market share optimization problem presented in the textbook is to choose the levels of each product attribute that will maximize the number of sampled customers preferring the brand in question. a. True b. False ANSWER: True POINTS: 1 TOPICS: Product design and market share optimization problem Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 13. If a problem has only less-than-or-equal-to constraints with positive coefficients for the variables, rounding down will always provide a feasible integer solution. a. True b. False ANSWER: True POINTS: 1 TOPICS: Rounding to obtain an integer solution 14. Dual prices cannot be used for integer programming sensitivity analysis because they are designed for linear programs. a. True b. False ANSWER: True POINTS: 1 TOPICS: A cautionary note about sensitivity analysis 15. Some linear programming problems have a special structure that guarantees the variables will have integer values. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction to integer linear programming 16. Generally, the optimal solution to an integer linear program is less sensitive to the constraint coefficients than is a linear program. a. True b. False ANSWER: False POINTS: 1 TOPICS: A cautionary note about sensitivity analysis 17. The classic assignment problem can be modeled as a 0-1 integer program. a. True b. False ANSWER: True POINTS: 1 TOPICS: Applications involving 0-1 variables 18. If Project 5 must be completed before Project 6, the constraint would be x5 − x6 ≤ 0. a. True b. False ANSWER: False POINTS: 1 TOPICS: Conditional and corequisite constraints 19. If the LP relaxation of an integer program has a feasible solution, then the integer program has a feasible solution. Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming a. True b. False ANSWER: False POINTS: 1 TOPICS: Rounding to obtain an integer solution 20. Multiple choice constraints involve binary variables. a. True b. False ANSWER: True POINTS: 1 TOPICS: Multiple choice and mutually exclusive constraints 21. Most practical applications of integer linear programming involve only 0 -1 integer variables. a. True b. False ANSWER: True POINTS: 1 TOPICS: 0 - 1 integer variables 22. Integer linear programs are harder to solve than linear programs. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction Multiple Choice 23. Which of the following is the most useful contribution of integer programming? a. finding whole number solutions where fractional solutions would not be appropriate b. using 0-1 variables for modeling flexibility c. increased ease of solution d. provision for solution procedures for transportation and assignment problems ANSWER: b POINTS: 1 TOPICS: Introduction 24. In a model, x1 ≥ 0 and integer, x2 ≥ 0, and x3 = 0, 1. Which solution would not be feasible? a. x1 = 5, x2 = 3, x3 = 0 b. x1 = 4, x2 = .389, x3 = 1 c. x1 = 2, x2 = 3, x3 = .578 d. x1 = 0, x2 = 8, x3 = 0 ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming POINTS: 1 TOPICS: Introduction 25. Rounded solutions to linear programs must be evaluated for a. feasibility and optimality. b. sensitivity and duality. c. relaxation and boundedness. d. each of these choices are true. ANSWER: a POINTS: 1 TOPICS: LP relaxation 26. Rounding the solution of an LP Relaxation to the nearest integer values provides a. a feasible but not necessarily optimal integer solution. b. an integer solution that is optimal. c. an integer solution that might be neither feasible nor optimal. d. an infeasible solution. ANSWER: c POINTS: 1 TOPICS: Graphical solution 27. The solution to the LP Relaxation of a maximization integer linear program provides a. an upper bound for the value of the objective function. b. a lower bound for the value of the objective function. c. an upper bound for the value of the decision variables d. a lower bound for the value of the decision variables ANSWER: a POINTS: 1 TOPICS: Graphical solution 28. The graph of a problem that requires x1 and x2 to be integer has a feasible region a. the same as its LP relaxation. b. of dots. c. of horizontal stripes. d. of vertical stripes. ANSWER: b POINTS: 1 TOPICS: Graphical solution 29. The 0-1 variables in the fixed cost models correspond to a. a process for which a fixed cost occurs. b. the number of products produced. c. the number of units produced. d. the actual value of the fixed cost. Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming ANSWER: a POINTS: 1 TOPICS: Fixed costs 30. Sensitivity analysis for integer linear programming a. can be provided only by computer. b. has precisely the same interpretation as that from linear programming. c. does not have the same interpretation and should be disregarded. d. is most useful for 0 - 1 models. ANSWER: c POINTS: 1 TOPICS: Sensitivity analysis 31. Let x1 and x2 be 0 - 1 variables whose values indicate whether projects 1 and 2 are not done or are done. Which answer below indicates that project 2 can be done only if project 1 is done? a. x1 + x2 = 1 b. x1 + x2 = 2 c. x1 − x2 ≤ 0 d. x1 − x2 ≥ 0 ANSWER: d POINTS: 1 TOPICS: Conditional and corequisite constraints 32. Let x1 , x2 , and x3 be 0 - 1 variables whose values indicate whether the projects are not done (0) or are done (1). Which answer below indicates that at least two of the projects must be done? a. x1 + x2 + x3 ≥ 2 b. x1 + x2 + x3 ≤ 2 c. x1 + x2 + x3 = 2 d. x1 − x2 = 0 ANSWER: a POINTS: 1 TOPICS: k out of n alternatives constraint 33. If the acceptance of project A is conditional on the acceptance of project B, and vice versa, the appropriate constraint to use is a a. multiple-choice constraint. b. k out of n alternatives constraint. c. mutually exclusive constraint. d. corequisite constraint. ANSWER: d POINTS: 1 TOPICS: Modeling flexibility provided by 0-1 integer variables Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 34. In an all-integer linear program, a. all objective function coefficients must be integer. b. all right-hand side values must be integer. c. all variables must be integer. d. all objective function coefficients and right-hand side values must be integer. ANSWER: c POINTS: 1 TOPICS: Types of integer linear programming models 35. To perform sensitivity analysis involving an integer linear program, it is recommended to a. use the dual prices very cautiously. b. make multiple computer runs. c. use the same approach as you would for a linear program. d. use LP relaxation. ANSWER: b POINTS: 1 TOPICS: A cautionary note about sensitivity analysis 36. Modeling a fixed cost problem as an integer linear program requires a. adding the fixed costs to the corresponding variable costs in the objective function. b. using 0-1 variables. c. using multiple-choice constraints. d. using LP relaxation. ANSWER: b POINTS: 1 TOPICS: Applications involving 0-1 variables 37. Most practical applications of integer linear programming involve a. only 0-1 integer variables and not ordinary integer variables. b. mostly ordinary integer variables and a small number of 0-1 integer variables. c. only ordinary integer variables. d. a near equal number of ordinary integer variables and 0-1 integer variables. ANSWER: a POINTS: 1 TOPICS: Applications involving 0-1 variables 38. Assuming W1, W2 and W3 are 0 -1 integer variables, the constraint W1 + W2 + W3 < 1 is often called a a. multiple-choice constraint. b. mutually exclusive constraint. c. k out of n alternatives constraint. d. corequisite constraint. ANSWER: b POINTS: 1 TOPICS: Mutually exclusive constraints Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 39. Which of the following applications modeled in the textbook does not involve only 0 - 1 integer variables? a. supply chain design b. bank location c. capital budgeting d. product design and market share optimization ANSWER: a POINTS: 1 TOPICS: 0 - 1 integer variables Subjective Short Answer 40. Solve the following problem graphically. Max s.t.

5X + 6Y 17X + 8Y ≤ 136 3X + 4Y ≤ 36 X, Y ≥ 0 and integer

Graph the constraints for this problem. Indicate all feasible solutions. Find the optimal solution to the LP Relaxation. Round down to find a feasible integer b. solution. Is this solution optimal? c. Find the optimal solution. ANSWER: a. The feasible region is those integer values in the space labeled feasible region.

Optimal LP relaxed occurs at X = 5.818, Y = 4.636, with Z = 56.909. Rounded down solution occurs at X = 5, Y = 4, Z= 49. Optimal solution is at X = 4, Y = 6, and Z = 56.

c. POINTS: 1 TOPICS: Graphical solution

41. Solve the following problem graphically. Max s.t.

X + 2Y 6X + 8Y ≤ 48

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Chapter 7 - Integer Linear Programming 7X + 5Y ≥ 35 X, Y ≥ 0 Y is integer a.

Graph the constraints for this problem. Indicate all feasible solutions. Find the optimal solution to the LP Relaxation. Round down to find a feasible integer b. solution. Is this solution optimal? c. Find the optimal solution. ANSWER: The feasible region consists of the portions of the horizontal lines that lie within the area a. labeled F. R.

The optimal relaxed solution is at X = 1.538, Y = 4.846 where Z = 11.231. The rounded solution is X = 1.538, Y = 4. The optimal solution is at X = 2.667, Y = 4, Z = 10.667.

c. POINTS: 1 TOPICS: Graphical solution

42. Solve the following problem graphically. Min s.t.

6X + 11Y 9X + 3Y ≥ 27 7X + 6Y ≥ 42 4X + 8Y ≥ 32 X, Y ≥ 0 and integer

Graph the constraints for this problem. Indicate all feasible solutions. Find the optimal solution to the LP Relaxation. Round up to find a feasible integer solution. b. Is this solution optimal? c. Find the optimal solution. ANSWER: a. The feasible region is the set of integer points in the area labeled feasible region.

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Chapter 7 - Integer Linear Programming

The optimal relaxed solution is at X = 4.5, Y = 1.75, and Z = 46.25. The rounded solution is X = 5, Y = 2. The optimal solution is at X = 6, Y = 1, and Z = 47.

c. POINTS: 1 TOPICS: Graphical solution

43. Consider a capital budgeting example with five projects from which to select. Let xi = 1 if project i is selected, 0 if not, for i = 1,...,5. Write the appropriate constraint(s) for each condition. Conditions are independent. a. Choose no fewer than three projects. b. If project 3 is chosen, project 4 must be chosen. c. If project 1 is chosen, project 5 must not be chosen. d. Projects cost 100, 200, 150, 75, and 300 respectively. The budget is 450. e. No more than two of projects 1, 2, and 3 can be chosen. ANSWER: a. x1 + x2 + x3 + x4 + x5 ≥ 3 b. x3 − x4 ≤ 0 c. x1 + x5 ≤ 1 d. 100x1 + 200x2 + 150x3 + 75x4 + 300x5 ≤ 450 e. x1 + x2 + x3 ≤ 2 POINTS: 1 TOPICS: Capital budgeting 44. Grush Consulting has five projects to consider. Each will require time in the next two quarters according to the table below. Project A B C D E

Time in first quarter 5 3 7 2 15

Time in second quarter 8 12 5 3 1

Revenue 12000 10000 15000 5000 20000

Revenue from each project is also shown. Develop a model whose solution would maximize revenue, meet the time Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming budget of 25 in the first quarter and 20 in the second quarter, and not do both projects C and D. ANSWER: Let A = 1 if project A is selected, 0 otherwise; same for B, C, D, and E Max s.t.

12000A + 10000B + 15000C + 5000D + 20000E 5A + 3B + 7C + 2D + 15E ≤ 25 8A + 12B + 5C + 3D + 1E ≤ 20 C+D≤1

POINTS: 1 TOPICS: Capital budgeting 45. The Westfall Company has a contract to produce 10,000 garden hoses for a large discount chain. Westfall has four different machines that can produce this kind of hose. Because these machines are from different manufacturers and use differing technologies, their specifications are not the same. Machine 1 2 3 4

Fixed Cost to Set Up Production Run 750 500 1000 300

Variable Cost Per Hose 1.25 1.50 1.00 2.00

Capacity 6000 7500 4000 5000

a. This problem requires two different kinds of decision variables. Clearly define each kind. b. The company wants to minimize total cost. Give the objective function. c. Give the constraints for the problem. d. Write a constraint to ensure that if machine 4 is used, machine 1 cannot be. ANSWER: a. Let Pi = the number of hoses produced on machine i Ui = 1 if machine i is used, = 0 otherwise b. Min 750U1 + 500U2 + 1000U3 + 300U4 + 1.25P 1 + 1.5P2 + P3 + 2P4 c. P1 ≤ 6000U1 P2 ≤ 7500U2 P3 ≤ 4000U3 P4 ≤ 5000U4 P1 + P2 + P3 + P4 ≥ 10000 c. U1 + U4 ≤ 1 POINTS: 1 TOPICS: Fixed costs 46. Hansen Controls has been awarded a contract for a large number of control panels. To meet this demand, it will use its existing plants in San Diego and Houston, and consider new plants in Tulsa, St. Louis, and Portland. Finished control panels are to be shipped to Seattle, Denver, and Kansas City. Pertinent information is given in the table. Shipping Cost to Destination Sources San Diego Houston Tulsa St. Louis

Construction Cost ------350,000 200,000

Seattle

Denver

Kansas City

Capacity

5 10 9 12

7 8 4 6

8 6 3 2

2,500 2,500 10,000 10,000

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Chapter 7 - Integer Linear Programming Portland

480,000 Demand

4 3,000

10 8,000

11 9,000

10,000

Develop a model whose solution would reveal which plants to build and the optimal shipping schedule. ANSWER: Let Pij = the number of panels shipped from source i to destination j Bi = 1 if plant i is built, = 0 otherwise (i = 3, 4, 5) Min

350000B3 + 200000B4 + 480000B5 + 5P11 + 7P12 + 8P13 + 10P21 + 8P22 + 6P23 + 9P31 + 4P32 + 3P33 + 12P41 + 6P42 + 2P43 + 4P51 + 10P52 + 11P53

s.t.

P11 + P12 + P13 ≤ 2500 P21 + P22 + P23 ≤ 2500 P31 + P32 + P33 ≤ 10000B3 P41 + P42 + P43 ≤ 10000B4 P51 + P52 + P53 ≤ 10000B5 P11 + P21 + P31 + P41 + P51 = 3000 P12 + P22 + P32 + P42 + P52 = 8000 P13 + P23 + P33 + P43 + P53 = 9000

POINTS: 1 TOPICS: Distribution system design 47. Simplon Manufacturing must decide on the processes to use to produce 1650 units. If machine 1 is used, its production will be between 300 and 1500 units. Machine 2 and/or machine 3 can be used only if machine 1's production is at least 1000 units. Machine 4 can be used with no restrictions. Machine 1 2 3 4

Fixed cost 500 800 200 50

Variable cost 2.00 0.50 3.00 5.00

Minimum Production 300 500 100 any

Maximum Production 1500 1200 800 any

(HINT: Use an additional 0 - 1 variable to indicate when machines 2 and 3 can be used.) ANSWER: Let Ui = the number of units made by machine i Si = 1 if machine i is used (requiring a set-up), = 0 otherwise K = 1 if machine 1 produces at least 1000 units, = 0 otherwise Min s.t.

500S1 + 2U1 + 800S2 + 5U2 + 200S3 + 3U3 + 50S4 + 5U4 U1 ≥ 300S1 U1 ≤ 1500S1 U1 ≥ 1000K S2 ≤ K S3 ≤ K U2 ≥ 500S2 U2 ≤ 1200S2

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Chapter 7 - Integer Linear Programming U3 ≥ 100S3 U3 ≤ 800S3 U4 ≤ 1650S4 U1 + U 2 + U3 + U4 = 1650 POINTS: 1 TOPICS: Distribution system design 48. Your express package courier company is drawing up new zones for the location of drop boxes for customers. The city has been divided into the seven zones shown below. You have targeted six possible locations for drop boxes. The list of which drop boxes could be reached easily from each zone is listed below. Zone Can Be Served By Locations: Downtown Financial 1, 2, 5, 6 Downtown Legal 2, 4, 5 Retail South 1, 2, 4, 6 Retail East 3, 4, 5 Manufacturing North 1, 2, 5 Manufacturing East 3, 4 Corporate West 1, 2, 6 Let xi = 1 if drop box location i is used, 0 otherwise. Develop a model to provide the smallest number of locations yet make sure that each zone is covered by at least two boxes. ANSWER: Min Σxi s.t. x1 + x2 + x5 + x6 ≥ 2 x2 + x4 + x5 ≥ 2 x1 + x2 + x4 + x6 ≥ 2 x3 + x4 + x5 ≥ 2 x1 + x 2 + x5 ≥ 2 x3 + x4 ≥ 2 x1 + x2 + x6 ≥ 2 POINTS: 1 TOPICS: Applications of integer linear programming 49. Consider the problem faced by a summer camp recreation director who is trying to choose activities for a rainy day. Information about possible choices is given in the table below. Category Art

Music Sports Computer

Activity 1 - Painting 2 - Drawing 3 - Nature craft 4 - Rhythm band 5 - Relay races 6 - Basketball 7 - Internet 8 - Creative writing 9 - Games

Time (minutes) 30 20 30 20 45 60 45 30 40

Popularity with Campers 4 5 3 5 2 1 1 4 1

Popularity with Counselors 2 2 1 5 1 3 1 3 2

Give a general definition of the variables necessary in this problem so that each activity can

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Chapter 7 - Integer Linear Programming be considered for inclusion in the day's schedule. The popularity ratings are defined so that 1 is the most popular. If the objective is to keep the b. campers happy, what should the objective function be? Write constraints for these restrictions: c. At most one art activity can be done. d. No more than two computer activities can be done. e. If basketball is chosen, then the music must be chosen. f. At least 120 minutes of activities must be selected. g. No more than 165 minutes of activities may be selected. h. To keep the staff happy, the counselor rating should be no higher than 10. ANSWER: a. Let xi = 1 if activity i is chosen, 0 if not, for i = 1, ... , 9 b. Max 4x1 + 5x2 + 3x3 + 5x4 + 2x5 + 1x6 + 1x7 + 4x8 + 1x9 c. x1 + x2 + x3 ≤ 1 d. x7 + x8 + x9 ≤ 2 e. x6 ≤ x4 f. 30x1 + 20x2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≥ 120 g. 30x1 + 20x 2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≤ 165 h. 2x1 + 2x2 + 1x3 + 5x4 + 1x5 + 3x6 + 1x7 + 3x8 + 2x9 ≤ 10 POINTS: 1 TOPICS: Applications of integer programming 50. Tower Engineering Corporation is considering undertaking several proposed projects for the next fiscal year. The projects, the number of engineers and the number of support personnel required for each project, and the expected profits for each project are summarized in the following table: Project Engineers Required Support Personnel Required Profit ($1,000,000s)

1 20 15 1.0

2 55 45 1.8

3 47 50 2.0

4 38 40 1.5

5 90 70 3.6

6 63 70 2.2

Formulate an integer program that maximizes Tower's profit subject to the following management constraints: 1) Use no more than 175 engineers 2) Use no more than 150 support personnel 3) If either project 6 or project 4 is done, both must be done 4) Project 2 can be done only if project 1 is done 5) If project 5 is done, project 3 must not be done and vice versa 6) No more than three projects are to be done. ANSWER: Max P1 + 1.8P2 + 2P3 + 1.5P4 + 3.6P5 + 2.2P6 s.t. 20P1 + 55P2 + 47P 3 + 38P4 + 90P5 + 63P6 ≤ 175 15P1 + 45P2 + 50P3 + 40P4 + 70P5 + 70P6 ≤ 150 P4 − P6 = 0 P1 − P2 ≥ 0 P3 + P5 ≤ 1 P1 + P2 + P3 + P4 + P5 + P6 ≤ 3 Pi = 0 or 1 Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming POINTS: 1 TOPICS: Applications of integer programming 51. Given the following all-integer linear program: Max s.t.

3x1 + 2x2 3x1 + x2 ≤ 9 x1 + 3x2 ≤ 7 −x1 + x2 ≤ 1 x1, x2 ≥ 0 and integer

Solve the problem as a linear program ignoring the integer constraints. Show that the optimal solution to the linear program gives fractional values for both x1 and x2. b. What is the solution obtained by rounding fractions greater than of equal to 1/2 to the next larger number? Show that this solution is not a feasible solution. c. What is the solution obtained by rounding down all fractions? Is it feasible? d. Enumerate all points in the linear programming feasible region in which both x1 and x2 are integers, and show that the feasible solution obtained in (c) is not optimal and that in fact the optimal integer is not obtained by any form of rounding. ANSWER: a. From the graph on the next page, the optimal solution to the linear program is x1 = 2.5, x2 = 1.5, z = 10.5. b. By rounding the optimal solution of x1 = 2.5, x2 = 1.5 to x1 = 3, x2 = 2, this point lies outside the feasible region. c. By rounding the optimal solution down to x1 = 2, x2 = 1, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives z = 8. d. There are eight feasible integer solutions in the linear programming feasible region with z values as follows: a.

1. 2. 3. 4. 5. 6. 7. 8.

x1 0 1 2 3 0 1 2 1

x2 0 0 0 0 1 1 1 2

z 0 3 6 9 2 5 8 7

optimal

part (c) solution

x1 = 3, x2 = 0 is the optimal solution. Rounding the LP solution (x1 = 2.5, x2 = 1.5) would not have been optimal.

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Chapter 7 - Integer Linear Programming

POINTS: 1 TOPICS: Rounding to obtain an integer solution 52. Tom's Tailoring has five idle tailors and four custom garments to make. The estimated time (in hours) it would take each tailor to make each garment is listed below. (An 'X' in the table indicates an unacceptable tailor-garment assignment.) Garment Wedding gown Clown costume Admiral's uniform Bullfighter's outfit

1 19 11 12 X

2 23 14 8 20

Tailor 3 20 X 11 20

4 21 12 X 18

5 18 10 9 21

Formulate and solve an integer program for determining the tailor-garment assignments that minimize the total estimated time spent making the four garments. No tailor is to be assigned more than one garment and each garment is to be worked on by only one tailor. ANSWER: Define the decision variables: xij = 1 if garment i is assigned to tailor j; = 0 otherwise. Number of decision variables = [(number of garments)(number of tailors)] − (number of unacceptable assignments) = [4(5)] − 3 = 17. Define the objective function: Minimize total time spent making garments: MIN 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21 + 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33 + 9x35 + 20x42 + 20x43 + 18x44 + 21x45 Define the constraints: Exactly one tailor per garment: No more than one garment per tailor: 1) x11 + x12 + x13 + x14 + x15 = 1 5) x11 + x21 + x31 ≤ 1 2) x21 + x22 + x 24 + x25 = 1 6) x12 + x22 + x32 + x42 ≤ 1 3) x31 + x32 + x33 + x35 = 1 7) x13 + x33 + x43 ≤ 1 4) x42 + x43 + x44 + x45 = 1 8) x14 + x24 + x44 ≤ 1 9) x15 + x25 + x35 + x45 ≤ 1 Nonnegativity: xij ≥ 0 for i = 1,..,4 and j = 1,..,5 Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming Optimal Solution: Assign wedding gown to tailor 5 Assign clown costume to tailor 1 Assign admiral uniform to tailor 2 Assign bullfighter outfit to tailor 4 Total time spent = 55 hours POINTS: 1 TOPICS: Assignment problem 53. Market Pulse Research has conducted a study for Lucas Furniture on some designs for a new commercial office desk. Three attributes were found to be most influential in determining which desk is most desirable: number of file drawers, the presence or absence of pullout writing boards, and simulated wood or solid color finish. Listed below are the partworths for each level of each attribute provided by a sample of 7 potential Lucas customers. Consumer 1 2 3 4 5 6 7

0 5 18 4 12 19 6 9

File Drawers 1 2 26 20 11 5 16 22 8 4 9 3 15 21 6 3

Pullout Writing Boards Present Absent 18 11 12 16 7 13 18 9 4 14 8 17 13 5

Finish Simul. Wood 17 15 11 22 30 20 16

Solid Color 10 26 19 14 19 11 28

Suppose the overall utility (sum of part-worths) of the current favorite commercial office desk is 50 for each customer. What is the product design that will maximize the share of choices for the seven sample participants? Formulate and solve, using Lindo or Excel, this 0 - 1 integer programming problem. ANSWER: Define the decision variables: There are 7 lij decision variables, one for each level of attribute. lij = 1 if Lucas chooses level i for attribute j; 0 otherwise. There are 7 Yk decision variables, one for each consumer in the sample. Yk = 1 if consumer k chooses the Lucas brand, 0 otherwise. Define the objective function: Maximize the number of consumers preferring the Lucas brand desk. Max Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 Define the constraints: There is one constraint for each consumer in the sample. 5l11 + 26l21 + 20l31 + 18l12 + 11l22 + 17l13 + 10l23 − 50Y1 ≥ 1 18l11 + 11l21 + 5l31 + 12l12 + 16l22 + 15l13 + 26l23 − 50Y2 ≥ 1 4l11 + 16l21 + 22l31 + 7l12 + 13l22 + 11l13 + 19l23 − 50Y3 ≥ 1 12l11 + 8l21 + 4l31 + 18l12 + 9l22 + 22l13 + 14l23 − 50Y4 ≥ 1 19l11 + 9l21 + 3l31 + 4l12 + 14l22 + 30l13 + 19l23 − 50Y5 ≥ 1 6l11 + 15l21 + 21l31 + 8l12 + 17l22 + 20l13 + 11l23 − 50Y6 ≥ 1 9l11 + 6l21 + 3l31 + 13l12 + 5l22 + 16l13 + 28l23 − 50Y7 ≥ 1 There is one constraint for each attribute. l11 + l21 + l31 = 1 l12 + l22 = 1 l13 + l23 = 1 Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming Optimal Solution: Lucas should choose these product features: 1 file drawer (I21 = 1) No pullout writing boards (I22 = 1) Simulated wood finish (I13 = 1) Three sample participants would choose the Lucas design: Participant 1 (Y1 = 1) Participant 5 (Y5 = 1) Participant 6 (Y6 = 1) POINTS: 1 TOPICS: Product design and market share optimization 54. Kloos Industries has projected the availability of capital over each of the next three years to be $850,000, $1,000,000, and $1,200,000, respectively. It is considering four options for the disposition of the capital: (1) Research and development of a promising new product (2) Plant expansion (3) Modernization of its current facilities (4) Investment in a valuable piece of nearby real estate Monies not invested in these projects in a given year will NOT be available for following year's investment in the projects. The expected benefits three years hence from each of the four projects and the yearly capital outlays of the four options are summarized in the table below in $1,000,000's. In addition, Kloos has decided to undertake exactly two of the projects, and if plant expansion is selected, it will also modernize its current facilities. Capital Outlay Options New Product R&D Plant Expansion Modernization Real Estate

Year 1

Year 2

Year 3

.35 .50 .35 .50

.55 .50 .40 0

.75 0 .45 0

Projected Benefits 5.2 3.6 3.2 2.8

Formulate and solve this problem as a binary programming problem. ANSWER: Max 5.2X1 + 3.6X2 + 3.2X3 + 2.8X4 s.t. .35X1 + .50X2 + .35X3 + .50X4 ≤ 0.85 (First Year) .55X1 + .50X2 + .40X3 ≤ 1.00 (Second Year) .75X 1 + .45X3 ≤ 1.20 (Third Year) X1 + X2 + X3 + X4 = 2 X2 + X3 ≥0 Xi = 0 or 1 Solution:

X1 = 1, X2 = 0, X3 = 1, X4 = 0, Total projected benefits = $8.4 million.

POINTS: 1 TOPICS: Capital budgeting Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 55. Given the following all-integer linear program: Max s. t.

15x1 + 2x2 7x1 + x2 < 23 3x1 - x2 < 5 x1, x2 > 0 and integer

a. Solve the problem as an LP, ignoring the integer constraints. b. What solution is obtained by rounding up fractions greater than or equal to 1/2? Is this the optimal integer solution? c. What solution is obtained by rounding down all fractions? Is this the optimal integer solution? Explain. d. Show that the optimal objective function value for the ILP is lower than that for the optimal LP. e. Why is the optimal objective function value for the ILP problem always less than or equal to the corresponding LP's optimal objective function value? When would they be equal? Comment on the MILP's optimal objective function compared to the corresponding LP & ILP. ANSWER: a. x1 = 2.8, x2 = 3.4, Obj. function = 48.8 b. x1 = 3, x2 = 3 ---> infeasible c. x1 = 2, x2 = 3 ---> feasible but not optimal d. Optimal x1 = 2, x2 = 9, Obj. function = 48 < 48.8 e. Additional integer constraints restrict the feasible region further; optimal solution values: ILP < Mixed ILP < LP. POINTS: 1 TOPICS: LP relaxation 56. A business manager for a grain distributor is asked to decide how many containers of each of two grains to purchase to fill its 1,600 pound capacity warehouse. The table below summarizes the container size, availability, and expected profit per container upon distribution. Grain A B

Container Size 500 lbs. 600 lbs.

Containers Available 3 2

Container Profit $1,200 $1,500

a. Formulate as a linear program with the decision variables representing the number of containers purchased of each grain. Solve for the optimal solution. b. What would be the optimal solution if you were not allowed to purchase fractional containers? c. There are three possible results from rounding an LP solution to obtain an integer solution: (1) the rounded optimal LP solution will be the optimal IP solution; (2) the rounded optimal LP solution gives a feasible, but not optimal IP solution; (3) the rounded optimal LP solution is an infeasible IP solution. For this problem (i) round down all fractions; (ii) round up all fractions; (iii) round off (to the nearest integer) all fractions (NOTE: Two of these are equivalent.) Which result above (1, 2, or 3) occurred under each rounding method? ANSWER: a. 4/5 container of Grain A, 2 containers of Grain B, $3960 b. 2 containers of Grain A, 1 container of Grain B, $3900 c. (i) Round down all fractions: For (0,2) the result is (feasible but not optimal) (ii) Round up all fractions: For (1,2) the result is (infeasible) (iii) Round off all fractions: For (1,2) the result is (infeasible) POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming TOPICS: LP relaxation 57. Given the following all-integer linear programming problem: Max 3x1 + 10x2 s. t.

2x1 + x2 < 5 x1 + 6x2 < 9 x1 - x2 > 2 x1, x2 > 0 and integer

a. Solve the problem graphically as a linear program. b. Show that there is only one integer point and it is optimal. c. Suppose the third constraint was changed to x1 - x2 > 2.1. What is the new optimal solution to the LP? ILP? ANSWER: a. x1 = 7/3, x2 = 1/3, obj. function = 31/3 b. x1 = 2, x2 = 0, obj. function = 6 c. L.P. -- optimal solution changes slightly; ILP -- infeasible POINTS: 1 TOPICS: LP relaxation Essay 58. The use of integer variables creates additional restrictions but provides additional flexibility. Explain. ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 59. Why are 0 - 1 variables sometimes called logical variables? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 60. Give a verbal interpretation of each of these constraints in the context of a capital budgeting problem. a. x1 − x2 ≥ 0 b. x1 − x2 = 0 c. x1 + x2 + x3 ≤ 2 ANSWER: Answer not provided. POINTS: 1 TOPICS: Captial budgeting 61. Explain how integer and 0-1 variables can be used in an objective function to minimize the sum of fixed and variable costs for production on two machines. ANSWER: Answer not provided. POINTS: 1 TOPICS: Distribution system design Cengage Learning Testing, Powered by Cognero

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Chapter 7 - Integer Linear Programming 62. Explain how integer and 0-1 variables can be used in a constraint to enable production. ANSWER: Answer not provided. POINTS: 1 TOPICS: Distribution system design 63. List and explain four types of constraints involving 0-1 integer variables only. ANSWER: Answer not provided. POINTS: 1 TOPICS: Modeling flexibility provided by 0-1 integer variables

Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models True / False 1. A nonlinear optimization problem is any optimization problem in which at least one term in the objective function or a constraint is nonlinear. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 2. A function is quadratic if its nonlinear terms have a power of 4. a. True b. False ANSWER: False POINTS: 1 TOPICS: An unconstrained problem 3. Nonlinear programming algorithms are more complex than linear programming algorithms. a. True b. False ANSWER: True POINTS: 1 TOPICS: A constrained problem 4. Many linear programming algorithms such as the simplex method optimize by examining only the extreme points of the feasible region. a. True b. False ANSWER: True POINTS: 1 TOPICS: A constrained problem 5. Nonlinear optimization problems can have only one local optimal solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: Capital budgeting 6. A feasible solution is a global optimum if there are no other feasible solutions with a better objective function value in the immediate neighborhood. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models 7. A feasible solution is a global optimum if there are no other feasible points with a better objective function value in the feasible region. a. True b. False ANSWER: True POINTS: 1 TOPICS: Local and global optima 8. For a typical nonlinear problem, duals price are relatively insensitive to small changes in right-hand side values. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima 9. The interpretation of the dual price for nonlinear models is different than the interpretation of the dual price for linear models. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima 10. In the case of functions with multiple local optima, most nonlinear optimization software methods can get stuck and terminate at a local optimum. a. True b. False ANSWER: True POINTS: 1 TOPICS: Local and global optima 11. For a minimization problem, a point is a global minimum if there are no other feasible points with a smaller objective function value. a. True b. False ANSWER: True POINTS: 1 TOPICS: Local and global optima 12. There are nonlinear applications in which there is a single local optimal solution that is also the global optimal solution. a. True b. False ANSWER: True POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models TOPICS: Local and global optima 13. Functions that are convex have a single local maximum that is also the global maximum. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima 14. The function f (X, Y) = X 2 + Y 2 has a single global minimum and is relatively easy to minimize. a. True b. False ANSWER: True POINTS: 1 TOPICS: Local and global optima 15. The problem of maximizing a concave quadratic function over a linear constraint set is relatively difficult to solve. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima 16. Each point on the efficient frontier is the maximum possible risk, measured by portfolio variance, for the given return. a. True b. False ANSWER: False POINTS: 1 TOPICS: Markowitz portfolio model 17. Any feasible solution to a blending problem with pooled components is feasible to the problem with no pooling. a. True b. False ANSWER: True POINTS: 1 TOPICS: Blending - the pooling problem 18. Any feasible solution to a blending problem without pooled components is feasible to the problem with pooled components. a. True b. False ANSWER: False POINTS: 1 TOPICS: Blending - the pooling problem Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models 19. When components (or ingredients) in a blending problem must be pooled, the number of feasible solutions is reduced. a. True b. False ANSWER: True POINTS: 1 TOPICS: Blending - the pooling problem 20. The value of the coefficient of imitation, q, in the Bass model for forecasting adoption of a new product cannot be negative. a. True b. False ANSWER: False POINTS: 1 TOPICS: Forecasting adoption of a new product 21. The Markowitz mean-variance portfolio model presented in the text is a convex optimization problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Markowitz portfolio model 22. Because most nonlinear optimization codes will terminate with a local optimum, the solution returned by the codes will be the best solution. a. True b. False ANSWER: False POINTS: 1 TOPICS: Local and global optima 23. It is possible for the optimal solution to a nonlinear optimization problem to lie in the interior of the feasible region. a. True b. False ANSWER: True POINTS: 1 TOPICS: Nonlinear optimization models Multiple Choice 24. Which of the following is incorrect? a. A global optimum is a local optimum in a nonlinear optimization problem. b. A local maximum is a global maximum in a concave nonlinear optimization problem. c. A global minimum is a local minimum in a convex nonlinear optimization problem. d. A local optimum is a global optimum in a nonlinear optimization problem. ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models TOPICS: Local and global optima 25. The measure of risk most often associated with the Markowitz portfolio model is the a. portfolio average return. b. portfolio minimum return. c. portfolio variance. d. portfolio standard deviation. ANSWER: c POINTS: 1 TOPICS: Markowitz portfolio model 26. An investor can pick the mean-variance tradeoff that he or she is most comfortable with by looking at a graph of the a. feasible region. b. pooled components. c. rolling horizon. d. efficient frontier. ANSWER: d POINTS: 1 TOPICS: Markowitz portfolio model 27. Which of the following is not a parameter of the Bass model for forecasting adoption of a new product? a. the coefficient of innovation b. the coefficient of interaction c. the coefficient of imitation d. the estimated number of people to eventually adopt the new product ANSWER: b POINTS: 1 TOPICS: Forecasting adoption of a new product 28. When the number of blending components exceeds the number of storage facilities, the number of feasible solutions to the blending problem a. is reduced. b. is increased. c. is unchanged. d. is zero. ANSWER: a POINTS: 1 TOPICS: Blending - the pooling problem 29. In the Bass model for forecasting the adoption of a new product, the objective function a. minimizes the sum of forecast errors. b. minimizes the sum of squared forecast errors. c. maximizes the number of adoptions. d. maximizes the number of adoptions and imitations. ANSWER: b Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models POINTS: 1 TOPICS: Forecasting adoption of a new product 30. Which of the following is not true regarding a concave function? a. It is bowl-shaped down. b. It is relatively easy to maximize. c. It has multiple local maxima. d. It has a single global maximum. ANSWER: c POINTS: 1 TOPICS: Local and global optima 31. A convex function is a. bowl-shaped up. b. bowl-shaped down. c. elliptical in shape. d. sinusoidal in shape. ANSWER: a POINTS: 1 TOPICS: Local and global optima 32. If the coefficient of each squared term in a quadratic function is positive, the function is a. concave. b. convex. c. elliptical. d. sinusoidal. ANSWER: b POINTS: 1 TOPICS: Local and global optima 33. Components that share a storage facility are called a. constrained components. b. indexed components. c. blended components. d. pooled components. ANSWER: d POINTS: 1 TOPICS: Blending - the pooling problem 34. The key idea behind constructing an index fund is to choose a portfolio of securities that a. is a mix of growth-oriented and income-oriented stocks. b. minimizes risk without sacrificing liquidity. c. mimics the performance of a broad market index. d. balances short-term and long-term investments. ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models POINTS: 1 TOPICS: Constructing an index fund 35. Components are referred to as pooled if they a. are shared by two or more customers b. have common ingredients c. share a storage facility d. are interchangeable ANSWER: c POINTS: 1 TOPICS: Blending: the pooling problem Subjective Short Answer 36. Investment manager Max Gaines has several clients who wish to own a mutual fund portfolio that matches, as a whole, the performance of the S&P 500 stock index. His task is to determine what proportion of the portfolio should be invested in each of the five mutual funds listed below so that the portfolio most closely mimics the performance of the S&P 500 index. Formulate the appropriate nonlinear program. Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond S&P 500 Index

Year 1 26.73 18.61 18.04 11.33 8.05 21.00

Annual Returns (Planning Scenarios) Year 2 Year 3 22.37 6.46 14.88 10.52 19.45 15.91 13.79 −2.07 7.29 9.18 19.00 12.00

Year 4 −3.19 5.25 −1.94 6.85 3.92 4.00

ANSWER: IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap blend MC = proportion of portfolio invested in mid-cap blend SC = proportion of portfolio invested in small-cap blend IB = proportion of portfolio invested in intermediate bond R1 = portfolio return for scenario 1 (year 1) R2 = portfolio return for scenario 2 (year 2) R3 = portfolio return for scenario 3 (year 3) R4 = portfolio return for scenario 4 (year 4) Min

(R1 − 21)2 + (R2 − 19)2 + (R3 − 12)2 + (R4 − 4)2 26.73IS + 18.61LC + 18.04MC + 11.33SC + 8.05IB = R1 22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB = R2 6.46IS + 10.52LC + 15.91MC − 2.07SC + 9.18IB = R3 −3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB = R4 IS + LC + MC + SC + IB = 1 IS, LC, MC, SC, IB ≥ 0

POINTS: 1 TOPICS: Constructing an index fund 37. Financial planner Minnie Margin has a substantial number of clients who wish to own a mutual fund portfolio that matches, as a whole, the performance of the Russell 2000 index. Her task is to determine what proportion of the portfolio should be invested in each of the five mutual funds listed below so that the portfolio most closely mimics the performance of the Russell 2000 index. Formulate the appropriate nonlinear program. Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models

Mutual Fund International Stock Large-Cap Value Mid-Cap Value Small-Cap Growth Short-Term Bond Russell 2000 Index

Year 1 22.37 15.48 17.42 23.18 9.26 20.00

Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 4.86 19.64 11.50 20.07 −4.97 12.36 3.25 8.81 6.15 22.00 8.00

Year 4 2.17 −5.25 −1.69 3.81 4.04 −2.00

ANSWER: IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap value MC = proportion of portfolio invested in mid-cap value SC = proportion of portfolio invested in small-cap growth SB = proportion of portfolio invested in short-term bond R1 = portfolio return for scenario 1 (year 1) R2 = portfolio return for scenario 2 (year 2) R3 = portfolio return for scenario 3 (year 3) R4 = portfolio return for scenario 4 (year 4) Min

(R1 − 20)2 + (R2 − 22)2 + (R3 − 8)2 + (R4 + 2)2 22.37IS + 15.48LC + 17.42MC + 23.18SC + 9.26SB = R1 26.73IS + 19.64LC + 20.07MC + 12.36SC + 8.81SB = R2 4.86IS + 11.50LC − 4.97MC + 3.25SC + 6.15SB = R3 2.17IS − 5.25LC − 1.69MC + 3.81SC + 4.04SB = R4 IS + LC + MC + SC + SB = 1 IS, LC, MC, SC, SB ≥ 0

POINTS: 1 TOPICS: Constructing an index fund 38. Investment manager Max Gaines wishes to develop a mutual fund portfolio based on the Markowitz portfolio model. He needs to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the variance of the portfolio is minimized subject to the constraint that the expected return of the portfolio be at least 4%. Formulate the appropriate nonlinear program. Annual Returns (Planning Scenarios) Mutual Fund Year 1 Year 2 Year 3 International Stock 26.73 22.37 6.46 Large-Cap Blend 18.61 14.88 10.52 Mid-Cap Blend 18.04 19.45 15.91 Small-Cap Blend 11.33 13.79 −2.07 Intermediate Bond 8.05 7.29 9.18 ANSWER: IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap blend MC = proportion of portfolio invested in mid-cap blend SC = proportion of portfolio invested in small-cap blend IB = proportion of portfolio invested in intermediate bond R1 = portfolio return for scenario 1 (year 1) R2 = portfolio return for scenario 2 (year 2) R3 = portfolio return for scenario 3 (year 3) R4 = portfolio return for scenario 4 (year 4) Cengage Learning Testing, Powered by Cognero

Year 4 −3.19 5.25 −1.94 6.85 3.92

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Chapter 8 - Nonlinear Optimization Models AR = expected (average) portfolio return Min

.25(R1 − AR)2 + .25(R2 − AR)2 + .25(R3 − AR)2 + .25(R4 − AR)2 26.73IS + 8.61LC + 18.04MC + 11.33SC + 8.05IB = R1 22.37IS + 4.88LC + 19.45MC + 13. 79SC + 7.29IB = R2 6.46IS + 10.52LC + 15.91MC − 2.07SC + 9.18IB = R3 −3.19IS + 5.25LC − 1.69MC + 3.81SC + 4.04IB = R4 IS + LC + MC + SC + IB = 1 .25R1 + .25R2 + .25R3 + .25R4 = AR AR ≥ 4 IS, LC, MC, SC, IB ≥ 0

POINTS: 1 TOPICS: Markowitz portfolio model 39. Financial planner Minnie Margin wishes to develop a mutual fund portfolio based on the Markowitz portfolio model. She needs to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the variance of the portfolio is minimized subject to the constraint that the expected return of the portfolio be at least 5%. Formulate the appropriate nonlinear program. Annual Returns (Planning Scenarios) Mutual Fund Year 1 Year 2 Year 3 International Stock 22.37 26.73 4.86 Large-Cap Value 15.48 19.64 11.50 Mid-Cap Value 17.42 20.07 −4.97 Small-Cap Growth 23.18 12.36 3.25 Short-Term Bond 9.26 8.81 6.15 ANSWER: IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap value MC = proportion of portfolio invested in mid-cap value SC = proportion of portfolio invested in small-cap growth SB = proportion of portfolio invested in short-term bond R1 = portfolio return for scenario 1 (year 1) R2 = portfolio return for scenario 2 (year 2) R3 = portfolio return for scenario 3 (year 3) R4 = portfolio return for scenario 4 (year 4) AR = expected (average) portfolio return Min

Year 4 2.17 −5.25 −1.69 3.81 4.04

.25(R1 − AR)2 + .25(R2 − AR)2 + .25(R3 − AR)2 + .25(R4 − AR)2 22.37IS + 15.48LC + 17.42MC + 23.18SC + 9.26SB = R1 26.73IS + 19.64LC + 20.07MC + 12.36SC + 8.81SB = R2 4.86IS + 11.50LC − 4.97MC + 3.25SC + 6.15SB = R3 2.17IS − 5.25LC − 1.69MC + 3.81SC + 4.04SB = R4 IS + LC + MC + SC + SB = 1 .25R1 + .25R2 + .25R3 + .25R4 = AR AR ≥ 5 IS, LC, MC, SC, SB ≥ 0

POINTS: 1 TOPICS: Markowitz portfolio model 40. Shampooch is a mobile dog grooming service firm that has been quite successful developing a client base in the Dallas Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models area. The firm plans to expand to other cities in Texas during the next few years. Shampooch would like to use its Dallas subscription data shown below to develop a model for forecasting service subscriptions in cities where it might expand. The first step is to estimate values for p (coefficient of innovation) and q (coefficient of imitation). Formulate the appropriate nonlinear program. Month 1 2 3 4 5 6 7 8 9 ANSWER:

Subscribers St 0.42 2.56 3.81 4.47 3.54 2.61 1.90 0.83 0.59 Min

Cum. Subscribers Ct 0.42 2.98 6.79 11.26 14.80 17.41 19.31 20.14 20.73

E12 + E22 + E32 + E42 + E52 + E62 + E72 + E82 + E92 F1 = pm F2 = (p + q(0.42/m)) (m − 0.42) F3 = (p + q(2.98/m)) (m − 2.98) F4 = (p + q(6.79/m)) (m − 6.79) F5 = (p + q(11.26/m)) (m − 11.26) F6 = (p + q(14.80/m)) (m − 14.80) F7 = (p + q(17.41/m)) (m − 17.41) F8 = (p + q(19.31/m)) (m − 19.31) F9 = (p + q(20.14/m)) (m − 20.14) E1 = F1 − 0.42 E2 = F2 − 2.56 E3 = F3 − 3.81 E4 = F4 − 4.47 E5 = F5 − 3.54 E6 = F6 − 2.61 E7 = F7 − 1.90 E8 = F8 − 0.83 E9 = F9 − 0.59

POINTS: 1 TOPICS: Forecasting adoption of a new product 41. Cutting Edge Yard Care is a residential and commercial lawn service company that has been in business in the Atlanta metropolitan area for almost one year. Cutting Edge would like to use its Atlanta service subscription data below to develop a model for forecasting service subscriptions in other metropolitan areas where it might expand. The first step is to estimate values for p (coefficient of innovation) and q (coefficient of imitation). Formulate the appropriate nonlinear program. Month 1

Subscribers St 0.53

Cum. Subscribers Ct 0.53

Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models 2 3 4 5 6 7 8 9 10 ANSWER:

2.04 3.37 4.85 5.14 3.90 2.26 1.43 0.98 0.42 Min

2.57 5.94 10.79 15.93 19.83 22.09 23.52 24.50 24.92

E12 + E22 + E32 + E42 + E52 + E62 + E72 + E82 + E92 + E102 F1 = pm F2 = (p + q(0.53/m)) (m − 0.53) F3 = (p + q(2.57/m)) (m − 2.57) F4 = (p + q(5.94/m)) (m − 5.94) F5 = (p + q(10.79/m)) (m − 10.79) F6 = (p + q(15.93/m)) (m − 15.93) F7 = (p + q(19.83/m)) (m − 19.83) F8 = (p + q(22.09/m)) (m − 22.09) F9 = (p + q(23.52/m)) (m − 23.52) F10 = (p + q(24.50/m)) (m − 24.50) E1 = F1 − 0.53 E2 = F2 − 2.04 E3 = F3 − 3.37 E4 = F4 − 4.85 E5 = F 5 − 5.14 E6 = F6 − 3.90 E7 = F7 − 2.26 E8 = F8 − 1.43 E9 = F9 − 0.98 E10 = F10 − 0.42

POINTS: 1 TOPICS: Forecasting adoption of a new product 42. MegaSports, Inc. produces two high-priced metal baseball bats, the Slugger and the Launcher, that are made from special aluminum and steel alloys. The cost to produce a Slugger bat is $100, and the cost to produce a Launcher bat is $120. We can not assume that MegaSports will sell all the bats it can produce. As the selling price of each bat model -Slugger and Launcher -- increases, the quantity demanded for each model goes down. Assume that the demand, S, for Slugger bats is given by S = 640 − 4PS and the demand, L, for Launcher bats is given by L = 450 − 3PL where PS is the price of a Slugger bat and PL is the price of a Launcher bat. The profit contributions are PS S − 100S for Slugger bats and PL L − 120L for Launcher bats. Develop the total profit contribution function for this problem. ANSWER: Solving S = 640 − 4PS for PS we get: PS = 160 − 1/4S Substituting 160 − 1/4S for PS in PS S − 100S we get: Cengage Learning Testing, Powered by Cognero

60S − 1/4 S 2 Page 11

Chapter 8 - Nonlinear Optimization Models Solving L = 450 − 3PL for PL we get: PL = 150 − 1/3L Substituting 150 − 1/3L for PL in PL L − 120L we get: 30L − 1/3L2 Total Profit Contribution = 60S − 1/4S2 + 30L − 1/3L2 POINTS: 1 TOPICS: An unconstrained production problem 43. Pacific-Gulf Oil Company is faced with the problem of refining three petroleum components into regular and premium gasoline in order to maximize profit. Components 1 and 2 are pooled in a single storage tank and component 3 has its own storage tank. Regular and premium gasolines are made from blending the pooled components and component 3. Prices per gallon for the two products and three components, as well as product specifications, are listed below. Regular gasoline Premium gasoline Component 1 Component 2 Component 3 Product Regular gasoline

Premium gasoline

Price Per Gallon $2.80 3.10 2.40 2.50 2.75 Specifications At most 25% component 1 At least 40% component 2 At most 30% component 3 At least 30% component 1 At most 50% component 2 At least 25% component 3

The maximum number of gallons available for each of the three components is 4000, 8000, and 8000, respectively. Formulate a nonlinear program to determine: 1) what percentages of component 1 and component 2 should be used in the pooled mixture, and 2) how to make regular and premium gasoline by blending the mixture of components 1 and 2 from the pooling tank with component 3. ANSWER: y1 = gallons of component 1 in the pooling tank y2 = gallons of component 2 in the pooling tank xpr = gallons of pooled components 1 and 2 in regular gasoline xpp = gallons of pooled components 1 and 2 in premium gasoline x3r = gallons of component 3 in regular gasoline x3p = gallons of component 2 in premium gasoline Max

2.8(x pr + x3 r) + 3.1(x pp + x3 p) − 2.4y1 − 2.5y2 − 2.75(x3 r + x3 p) y1 + y2 = x pr + x pp (y1 /(y1 + y2)) x pr ≤ .25(x pr + x3 r) (y2 /(y1 + y2)) x pr ≥ .4(x pr + x3 r) x3 r ≤ .3(x pr + x3 r) (y1 /(y1 + y2)) x pp ≥ .3(x pp + x3 p) (y2 /(y1 + y2)) x pp ≤ .5(x pp + x3 p) x3 p ≥ .25(x pp + x3 p) y1 ≤ 4000 y2 ≤ 8000

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Chapter 8 - Nonlinear Optimization Models x3 r + x3 p ≤ 8000 x pr + x3 r ≤ 8000 x pr , x pp , x3 r , x3 p , y1 , y2 ≥ 0 POINTS: 1 TOPICS: Blending - the pooling problem 44. Skooter's Skateboards produces two models of skateboards, the FX and the ZX. Skateboard revenue (in $l,000s) for the firm is nonlinear and is stated as (number of FXs)(5 − 0.2 number of FXs) + (number of ZXs)(7 − 0.3 number of ZXs). Skooter's has 80 labor-hours available per week in its paint shop. Each FX requires 2 labor-hours to paint and each ZX requires 3 labor-hours. Formulate this nonlinear production planning problem to determine how many FX and ZX skateboards should be produced per week at Scooter's. ANSWER: FX = number of FX skateboards to produce per week ZX = number of ZX skateboards to produce per week Max

5FX − 0.2FX 2 + 7ZX − 0.3ZX 2

s.t.

2FX + 3ZX ≤ 80 FX, ZX ≥ 0

POINTS: 1 TOPICS: A constrained production problem 45. Native Customs sells two popular styles of hand-sewn footwear: a sandal and a moccasin. The cost to make a pair of sandals is $18, and the cost to make a pair of moccasins is $24. The demand for these two items is sensitive to the price, and historical data indicate that the monthly demands are given by S = 400 − 10P1 and M = 450 − 15P2 , where S = demand for sandals (in pairs), M = demand for moccasins (in pairs), P1 = price for a pair of sandals, and P2 = price for a pair of moccasins. To remain competitive, Native Customs must limit the price (per pair) to no more than $60 and $75 for its sandals and moccasins, respectively. Formulate this nonlinear programming problem to find the optimal production quantities and prices for sandals and moccasins that maximize total monthly profit. ANSWER: Max 22S − 1 /10 S 2 + 6M − 1/15 M 2 s.t.

S + 10P1 = 400 M − 15P2 = 450 P1 ≤ 60 P2 ≤ 75 S, M, P1, P2 ≥ 0

POINTS: 1 TOPICS: A constrained production problem Essay 46. Explain how the local minimum, local maximum, local optimum, global minimum, global maximum, and global optimum relate to one another in nonlinear optimization problems. ANSWER: Answer not provided. POINTS: 1 TOPICS: Local and global optima Cengage Learning Testing, Powered by Cognero

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Chapter 8 - Nonlinear Optimization Models 47. Explain how the parameters required for the Bass new-product adoption model can be estimated when no historical data are available for the new product. ANSWER: Answer not provided. POINTS: 1 TOPICS: Forecasting adoption of a new product 48. Describe how the Markowitz portfolio model can be modified to account for upper and lower bounds being placed on the amount of an asset type invested in the portfolio. ANSWER: Answer not provided. POINTS: 1 TOPICS: Markowitz portfolio model 49. Provide several examples of both nonlinear objective functions and nonlinear constraints. ANSWER: Answer not provided. POINTS: 1 TOPICS: A constrained problem 50. Discuss the essence of the pooling problem in terms of the circumstances, objective, and constraint ANSWER: Answer not provided. POINTS: 1 TOPICS: Blending - the pooling problem 51. Describe the difference between a convex function and a concave function and give an example of each. ANSWER: Answer not provided. POINTS: 1 TOPICS: Local and global optima

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Chapter 9 - Project Scheduling: PERT/CPM True / False 1. Critical activities are those that can be delayed without delaying the entire project. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 2. PERT and CPM are applicable only when there is no dependence among activities. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 3. A path through a project network must reach every node. a. True b. False ANSWER: False POINTS: 1 TOPICS: Critical path 4. A critical activity can be part of a noncritical path. a. True b. False ANSWER: True POINTS: 1 TOPICS: Critical path 5. When activity times are uncertain, an activity's most likely time is the same as its expected time. a. True b. False ANSWER: False POINTS: 1 TOPICS: Variability in project completion time 6. The earliest finish time for the final activity is the project duration. a. True b. False ANSWER: True POINTS: 1 TOPICS: Critical path 7. The length of time an activity can be delayed without affecting the project completion time is the slack. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM a. True b. False ANSWER: True POINTS: 1 TOPICS: Critical path 8. When activity times are uncertain, total project time is normally distributed with mean equal to the sum of the means of all of the critical activities. a. True b. False ANSWER: True POINTS: 1 TOPICS: Variability in project completion time 9. Crashing refers to an unanticipated delay in a critical path activity that causes the total time to exceed its limit. a. True b. False ANSWER: False POINTS: 1 TOPICS: Crashing activity times 10. Constraints in the LP models for crashing decisions are required to compare the activity's earliest finish time with the earliest finish time of each predecessor. a. True b. False ANSWER: True POINTS: 1 TOPICS: Crashing activity times 11. The project manager should monitor the progress of any activity with a large time variance even if the expected time does not identify the activity as a critical activity. a. True b. False ANSWER: True POINTS: 1 TOPICS: Variability in project completion time 12. The variance in the project completion time is the sum of the variances of all activities in the project. a. True b. False ANSWER: False POINTS: 1 TOPICS: Variability in project completion time 13. The latest finish time for an activity is the largest of the latest start times for all activities that immediately follow the activity. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM a. True b. False ANSWER: False POINTS: 1 TOPICS: Earliest and latest times 14. The earliest start time for an activity is equal to the smallest of the earliest finish times for all its immediate predecessors. a. True b. False ANSWER: False POINTS: 1 TOPICS: Earliest and latest times 15. The linear programming model for crashing presented in the textbook assumes that any portion of the activity crash time can be achieved for a corresponding portion of the activity crashing cost. a. True b. False ANSWER: True POINTS: 1 TOPICS: Linear programming model for crashing 16. All activities on a critical path have zero slack time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Critical path 17. The difference between an activity's earliest finish time and latest finish time equals the difference between its earliest start time and latest start time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Earliest and latest times 18. It is possible to have more than one critical path at a time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Critical path 19. Activities require time to complete while events do not. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM b. False ANSWER: True POINTS: 1 TOPICS: Critical path 20. Precedence relationships among activities is critical in CPM analysis but not in PERT. a. True b. False ANSWER: False POINTS: 1 TOPICS: Critical path 21. The normal distribution tends to be a better approximation of the distribution of total time for shorter projects where the critical path has relatively few activities. a. True b. False ANSWER: False POINTS: 1 TOPICS: Variability in project completion time Multiple Choice 22. PERT and CPM a. are most valuable when a small number of activities must be scheduled. b. have different features and are not applied to the same situation. c. do not require a chronological relationship among activities. d. have been combined to develop a procedure that uses the best of each. ANSWER: d POINTS: 1 TOPICS: Introduction 23. Which is not a significant challenge of project scheduling? a. deadlines exist. b. activities are independent. c. many employees could be required. d. delays are costly. ANSWER: b POINTS: 1 TOPICS: Introduction 24. Arcs in a project network indicate a. completion times. b. precedence relationships. c. activities. d. the critical path. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM ANSWER: b POINTS: 1 TOPICS: Critical path 25. The critical path a. is any path that goes from the starting node to the completion node. b. is a combination of all paths. c. is the shortest path. d. is the longest path. ANSWER: d POINTS: 1 TOPICS: Critical path 26. The earliest start time rule a. compares the starting times of all activities for successors of an activity. b. compares the finish times for all immediate predecessors of an activity. c. determines when the project can begin. d. determines when the project must begin. ANSWER: b POINTS: 1 TOPICS: Critical path 27. Activities following a node a. can begin as soon as any activity preceding the node has been completed. b. have an earliest start time equal to the largest of the earliest finish times for all activities entering the node. c. have a latest start time equal to the largest of the earliest finish times for all activities entering the node. d. None of the alternatives is correct. ANSWER: b POINTS: 1 TOPICS: Critical path 28. Activities G, P, and R are the immediate predecessors for activity W. If the earliest finish times for the three are 12, 15, and 10, then the earliest start time for W a. is 10. b. is 12. c. is 15. d. cannot be determined. ANSWER: c POINTS: 1 TOPICS: Critical path 29. Activities K, M and S immediately follow activity H, and their latest start times are 14, 18, and 11. The latest finish time for activity H a. is 11. b. is 14. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM c. is 18. d. cannot be determined. ANSWER: a POINTS: 1 TOPICS: Critical path 30. When activity times are uncertain, a. assume they are normally distributed. b. calculate the expected time, using (a + 4m + b)/6. c. use the most likely time. d. calculate the expected time, using (a + m + b)/3. ANSWER: b POINTS: 1 TOPICS: Uncertain activity times 31. To determine how to crash activity times a. normal activity costs and costs under maximum crashing must be known. b. shortest times with crashing must be known. c. realize that new paths may become critical. d. All of the alternatives are true. ANSWER: d POINTS: 1 TOPICS: Crashing activity times 32. Slack equals a. LF − EF. b. EF − LF. c. EF − LS. d. LF − ES. ANSWER: b POINTS: 1 TOPICS: Determining the critical path 33. Activities with zero slack a. can be delayed. b. must be completed first. c. lie on a critical path. d. have no predecessors. ANSWER: c POINTS: 1 TOPICS: Determining the critical path 34. In deciding which activities to crash, one must a. crash all critical activities. b. crash largest-duration activities. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM c. crash lowest-cost activities. d. crash activities on the critical path(s) only. ANSWER: d POINTS: 1 TOPICS: Crashing activity times 35. For an activity with more than one immediate predecessor activity, which of the following is used to compute its earliest finish (EF) time? a. the largest EF among the immediate predecessors. b. the average EF among the immediate predecessors. c. the largest LF among the immediate predecessors. d. the difference in EF among the immediate predecessors. ANSWER: a POINTS: 1 TOPICS: Determining the critical path 36. Which of the following is always true about a critical activity? a. LS = EF. b. LF = LS. c. ES = LS. d. EF = ES. ANSWER: c POINTS: 1 TOPICS: Determining the critical path 37. For an activity with more than one immediate successor activity, its latest-finish time is equal to the a. largest latest-finish time among its immediate successors. b. smallest latest-finish time among its immediate successors. c. largest latest-start time among its immediate successors. d. smallest latest-start time among its immediate successors. ANSWER: d POINTS: 1 TOPICS: Determining the critical path 38. Which of the following is a general rule for crashing activities? a. Crash only non-critical activities. b. Crash activities with zero slack. c. Crash activities with the greatest number of predecessors. d. Crash the path with the fewest activities. ANSWER: b POINTS: 1 TOPICS: Crashing activity times 39. To calculate an activity’s latest finish time, you should consider its a. predecessors’ latest finish times Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM b. predecessors’ latest start times c. successors’ earliest start times d. successors’ latest start times ANSWER: d POINTS: 1 TOPICS: Earliest/latest start/finish times 40. A critical activity is a. an activity that consumes no time but shows precedence between events. b. a milestone accomplishment within the project. c. an activity with zero slack. d. the beginning of an event. ANSWER: c POINTS: 1 TOPICS: Critical path 41. The main difference between CPM and PERT is a. the use of different activity time estimates. b. PERT analysis is less expensive to conduct. c. PERT lends itself to computerization while CPM networks must be constructed manually. d. CPM integrates time and cost performance while PERT is based solely on time performance. ANSWER: a POINTS: 1 TOPICS: Uncertain activity times 42. In PERT, the activity duration time is equal to the a. pessimistic time. b. optimistic time. c. most likely time. d. mean duration. ANSWER: d POINTS: 1 TOPICS: Uncertain activity times Subjective Short Answer 43. From this schedule of activities, draw the PERT/CPM network. Activity A B C D E F G

Immediate Predecessor --A B B A C, D E, F

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Chapter 9 - Project Scheduling: PERT/CPM ANSWER:

POINTS: 1 TOPICS: PERT/CPM networks 44. From this PERT/CPM network, determine the list of activities and their predecessors.

ANSWER: Activity A B C D E F G H

Immediate Predecessor ----A A, B C, D D E F, G

POINTS: 1 TOPICS: PERT/CPM networks 45. A cookie recipe gives the following numbered steps. 1. Preheat oven. 2. Grease cookie sheets. 3. Cream shortening and sugar. 4. Add eggs and flavoring. 5. Measure and sift dry ingredients. 6. Add dry ingredients to mixture. 7. Drop by spoonfuls onto sheets and bake for 10 minutes. Although the steps are numbered, they do not always reflect immediate precedence relationships. Develop a table that lists the immediate predecessors for each activity. ANSWER: Immediate Activity Predecessor 1 --2 --Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM 3 4 5 6 7

--3 --4, 5 1, 2, 6

POINTS: 1 TOPICS: Precedence relationships 46. A senior MIS design class project team has developed the following schedule of activities for their project, using their best estimate of completion times. Both written and oral reports are required. Draw the project network. Can they complete the project in the 38 class days remaining until the end of the semester? Activity A. Find client B. Write prospectus C. Obtain approval from client and professor D. Complete programming E. Do industry background research F. Write final paper G. Write oral report ANSWER:

Time 4 2 3 12 10 6 5

Immediate Predecessor --A B C --D, E D, E

and the critical path is A-B-C-D-F POINTS: 1 TOPICS: Critical path 47. A project network is shown below. Use a forward and a backward pass to determine the critical path, and then fill out the table below. Activity times are in weeks.

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Chapter 9 - Project Scheduling: PERT/CPM

Precedence Activity Critical Activity ES LS EF LF Slack Activities Time (weeks) Path? A B C D E F G H I Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in fewer than 25 weeks more or less than 50%? ANSWER: Precedence Activity Critical Activity ES LS EF LF Slack Activities Time (weeks) Path? A -5 0 0 5 5 0 Yes B -4 0 1 4 5 1 C -8 0 0 8 8 0 Yes D A, B 4 5 5 9 9 0 Yes E B, C 12 8 8 20 20 0 Yes F D 10 9 9 19 19 0 Yes G A, F 3 19 19 22 22 0 Yes H E, F 2 20 20 22 22 0 Yes I G, H 5 22 22 27 27 0 Yes The probability is less than 50% because 25 weeks is less than the mean time of 27 weeks. POINTS: 1 TOPICS: Variability in project completion time 48. A project network is shown below. Use a forward and a backward pass to determine the critical path, and then fill out the table below. Activity times are in weeks.

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Chapter 9 - Project Scheduling: PERT/CPM

Precedence Activity Critical Activity ES LS EF LF Slack Activities Time (weeks) Path? A B C D E F G H I Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in more than 28 weeks more or less than 50%? ANSWER: Precedence Activity Critical Activity ES LS EF LF Slack Activities Time (weeks) Path? A -5 0 13 5 18 13 B -4 0 14 4 18 14 C -8 0 0 8 8 0 Yes D A, B 4 5 18 9 22 13 E C 12 8 8 20 20 0 Yes F C 10 8 10 18 20 2 G D, E 3 20 22 23 25 2 H E 2 20 23 22 25 3 I E, F 5 20 20 25 25 0 Yes The probability is less than 50% because 28 weeks is more than the mean time of 25 weeks. POINTS: 1 TOPICS: Variability in project completion time 49. Use the following network of related activities with their duration times (weeks) to complete a row for each activity under the column headings below.

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Chapter 9 - Project Scheduling: PERT/CPM

Activity A B C D E F G H ANSWER:

Immediate Predecessors

Activity A B C D E F G H

Activity Time (weeks)

Precedence Activities --B A, C B B D, E F

Activity Time (weeks) 8 2 5 3 7 5 12 10

ES 0 0 2 8 2 2 11 8

LS 0 1 3 8 4 7 11 13

EF 8 2 7 11 9 8 23 18

Slack

Critical Path?

LF 8 3 8 11 11 13 23 23

Slack 0 1 1 0 2 5 0 5

Critical Path? Yes

Yes

CRITICAL PATH: A-D-G PROJECT COMPLETION TIME = 23 POINTS: 1 TOPICS: Critical path 50. Use the following network of related activities with their duration times (weeks) to complete a row for each activity under the column headings below.

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Chapter 9 - Project Scheduling: PERT/CPM

Activity A B C D E F G H I J ANSWER:

Immediate Predecessors

Activity A B C D E F G H I J

Activity Time (weeks)

Precedence Activities -A A A B B C D E F, G, H

Slack

Activity Time (weeks) 2 4 3 7 6 5 8 5 4 6

ES 0 2 2 2 6 6 5 9 12 14

LS 0 5 3 2 10 9 6 9 16 14

EF 2 6 5 9 12 11 13 14 16 20

LF 2 9 6 9 16 14 14 14 20 20

Critical Path?

Slack 0 3 1 0 4 3 1 0 4 0

Critical Path? Yes

Yes

Yes Yes

CRITICAL PATH: A-D-H-J PROJECT COMPLETION TIME = 20 POINTS: 1 TOPICS: Critical path 51. Given the following network with activities and times estimated in days,

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Chapter 9 - Project Scheduling: PERT/CPM

Activity

Optimistic

A B C D E F G H I J K

2 1 6 5 3 8 4 3 5 12 1

Most Probable 5 3 7 12 4 9 6 6 7 13 3

Pessimistic 6 7 10 14 5 12 8 8 12 14 4

a. What are the critical path activities? b. What is the expected time to complete the project? c. What is the probability the project will take more than 28 days to complete? ANSWER: a. and b. Activity Expected Time Variance A 4.67 0.44 B 3.33 1.00 C 7.33 0.44 D 11.17 2.25 E 4.00 0.11 F 9.33 0.44 G 6.00 0.44 H 5.83 0.69 I 7.50 1.36 J 13.00 0.11 K 2.83 0.25 Activity A B

Precedence Activities -----

Time (days) ES 4.67 0.00 3.33 0.00

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LS 4.67 6.00

EF 4.67 3.33

LF 9.33 9.33

Slack 4.67 6.00

Critical Path?

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Chapter 9 - Project Scheduling: PERT/CPM C D E F G H I J K

--A A A, B C C E, F, G H D, J

7.33 11.17 4.00 9.33 6.00 5.83 7.50 13.00 2.83

0.00 4.67 4.67 4.67 7.33 7.33 14.00 13.17 26.17

0.00 14.67 14.67 9.33 12.67 7.33 18.67 13.17 26.17

7.33 8.67 8.67 14.00 13.33 13.17 21.50 26.17 29.00

7.33 18.67 18.67 18.67 18.67 13.17 26.17 26.17 29.00

0.00 10.33 10.00 4.67 5.33 0.00 4.67 0.00 0.00

Yes

Yes Yes Yes

CRITICAL PATH: C-H-J-K EXPECTED PROJECT COMPLETION TIME = 29 VARIANCE OF PROJECT COMPLETION TIME = 1.5 The probability that the project will take more than 28 days is P(z > (28 − 29) / 1.22) = P(z > −.82) = .7939

c. POINTS: 1 TOPICS: Critical path with uncertain times

52. The critical path for this network is A - E - F and the project completion time is 22 weeks.

Normal Time 12 14 8 5 4 6 10

Activity A B C D E F G

Crash Time 8 10 8 3 3 5 8

Normal Cost 8,000 5,000 10,000 6,000 5,000 9,000 5,000

Crash Cost 12,000 7,500 10,000 8,000 7,000 12,000 8,000

If a deadline of 17 weeks is imposed, give the linear programming model for the crashing decision. ANSWER: Let Ei = the earliest finish time for activity i Let Ci = the amount to crash activity i Min s.t.

1000CA + 625CB + 1000CD + 2000CE + 3000CF + 1500CG EA ≥ 0 + 12 − CA

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Chapter 9 - Project Scheduling: PERT/CPM EB ≥ 0 + 14 − CB EC ≥ 0 + 8 − CC ED ≥ EA + 5 − CD EE ≥ EA + 4 − CE EF ≥ EE + 6 − CF EF ≥ EC + 6 − CF EG ≥ EC + 10 − CG EA ≤ 4 EB ≤ 4 ED ≤ 2 EE ≤ 1 EF ≤ 1 EG ≤ 2 POINTS: 1 TOPICS: Crashing activity times 53. For the project represented below, determine the earliest and latest start and finish times for each activity as well as the expected overall completion time.

Activity Duration ES EF LS LF Slack A 4 B 3 C 4 D 2 E 5 F 2 G 5 H 6 ANSWER: [Activity, ES, EF, LS, LF, Slack] [A, 0, 4, 0, 4, 0]; [B, 4, 7, 4, 7, 0]; [C, 4, 8, 6, 10, 2]; [D, 7, 9, 8, 10, 1]; [E, 7, 12, 7, 12, 0] [F, 9, 11, 10, 12, 1]; [G, 12, 17, 12, 17, 0]; [H, 9, 15, 11, 17, 2] POINTS: 1 TOPICS: Determining the critical path 54. Consider the following PERT/CPM network with estimated times in weeks. The project is scheduled to begin on May 1. Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM

The three-time estimate approach was used to calculate the expected times and the following table gives the variance for each activity: Activity A B C D

Variance 1.1 0.5 1.2 0.8

Activity E F G H

Variance 0.3 0.6 0.6 1.0

a. Give the expected project completion date and the critical path. b. By what date are you 99% sure the project will be completed? ANSWER: a. 16 weeks = Aug. 21 b. About Sept. 22 (20.66 weeks from May 1) POINTS: 1 TOPICS: Variability in project completion time 55. A project consists of five activities. Naturally the paint mixing precedes the painting activities. Also, both ceiling painting and floor sanding must be done prior to floor buffing. Optimistic Time (hr) 3 1 0.5 1 1

Activity Floor sanding Floor buffing Paint mixing Wall painting Ceiling painting

Most Probable Time (hr) 4 2 1 2 5.5

Pessimistic Time (hr) 5 3 1.5 9 7

a. Construct the PERT/CPM network for this problem. b. What is the expected completion time of this project? c. What is the probability that the project can be completed within 9 hr.? ANSWER:

b. c. POINTS: 1

8 hrs. .8264

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Chapter 9 - Project Scheduling: PERT/CPM TOPICS: Variability in project completion time 56. Consider a project that has been modeled as follows: Activity A B C D E F G H I J K L

Immediate Predecessors ----A A A B,C B,C E,F E,F E,F D,H G,J

Duration (hr) 7 10 4 30 7 12 15 11 25 6 21 25

Draw the PERT/CPM network for this project and determine project's expected completion time and its critical path. Can activities E and G be performed simultaneously without delaying the minimum project b. completion time? c. Can one person perform A, G, and I without delaying the project? d. By how much can activities G and L be delayed without delaying the entire project? How much would the project be delayed if activity G was delayed by 7 hours and activity L e. was delayed by 4 hours? Explain. ANSWER: a.

Expected Completion Time = 58; Critical Path = A - D - K b. Yes c. Yes d. Slack on G = 7 hours; Slack on L = 4 hours e. 4 hours; the slack times are not independent as G and L are on the same path. POINTS: 1 TOPICS: Determining the critical path 57. Joseph King has ambitions to be mayor of Williston, North Dakota. Joe has determined the breakdown of the steps to the nomination and has estimated normal and crash costs and times for the campaign as follows (times are in weeks).

Activity Solicit Volunteers

Time 6

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Normal Cost $5,000

Time 4

Crash Cost $ 9,000

Immediate Predecessors --Page 19

Chapter 9 - Project Scheduling: PERT/CPM B. C. D. E. F. G. H.

Initial "Free" Exposure Raise Money Organize Schedule Hire Advertising Firm Arrange TV Interview Advertising Campaign Personal Campaigning

3 9 4 2 3 5 7

$4,000 $4,000 $1,000 $1,500 $4,000 $7,000 $8,000

3 6 2 1 1 4 5

$ 4,000 $10,000 $ 2,000 $ 2,000 $ 8,000 $12,000 $20,000

--A A B B C, E D, F

Joe King is not a wealthy man and would like to organize a 16-week campaign at minimum cost. Write and solve a linear program to accomplish this task. ANSWER: Define Xi = earliest finish time for activity i Yi = the amount of time activity i is crashed MIN

2000YA + 2000YC + 500YD + 500YE + 2000YF + 5000YG + 6000Y H

S.T.

XA ≥ 0 + (6 − YA) XB ≥ 0 + 3 XC ≥ XA + (9 − YC) XD ≥ XA + (4 − YD) XE ≥ XB + (2 − YE) XF ≥ XB + (3 − YF) XG ≤ 16 XG ≥ XC + (5 − YG) XG ≥ XE + (5 − YG) XH ≥ XD + (7 − YH) XH ≥ XF + (7 − YH) XH ≤ 16 YA ≤ 2 YC ≤ 3 YD ≤ 2 YE ≤ 1 YF≤2 YG ≤ 1 YH ≤ 2 Xi, Yj ≥ 0 for all i

Solution:

XA = 4, XB = 6, XC = 11, XD = 9, XE = 11, XF = 9, XG = 16, XH = 16, YA = 2, YC = 2, YD = 0, YE = 0, YF = 0, YG = 0, YH = 0, Total crash cost = $8,000.

POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM TOPICS: Crashing activity times 58. Marcy Fetter, a staff analyst at the Los Angeles plant of Computer Products Corporation, is assigned to the team that is developing the process design for producing an RFID sensor. The corporate planning group in San Jose, California has contacted her and has asked how confident the design group is about completing the project in 60 days. She has developed these estimated time durations in days for the project: Immediate Predecessor Activities -A A A B C,E B B,D B,D F.G.H

Activity A B C D E F G H I J

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

10 6 10 9 5 10 12 18 10 8

12 10 15 9 6 12 14 21 15 10

15 14 20 18 8 13 16 24 20 14

a. Compute the expected time and variance for each activity. b. Determine the critical path and the expected duration of the project. c. What is the probability that the project will take longer than 58 days to complete? d. Which path in the project network offers the greatest risk of overrunning a new deadline of 56 days? ANSWER: a. Activity Optimistic Most Likely Pessimistic Activity Expected Activity Time (to) Time (tm) Time (tp) Variance (Vt) Time (te) A

12.17

.694

10.00

1.778

15.00

2.778

10.50

.694

6.17

.250

11.83

.250

14.00

.444

21.00

1.000

15.00

2.778

10.33

1.000

b. Path A-C-F-J

Length of Path (days) 12.17 + 15 + 11.83 + 10.33

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= 49.33 Page 21

Chapter 9 - Project Scheduling: PERT/CPM A-B-E-F-J A-B-G-J A-B-H-J A-B-I

12.17+ 10 + 6.17 + 11.83 + 10.33 12.17 + 10 + 14 + 10.33 12.17 + 10 + 21 + 10.33 12.17 + 10 + 15

= 50.50 = 46.50 = 53.50 = 37.17

A-D-H-J

12.17 + 10.5 + 21 + 10.33

= 54.00

A-D-I

12.17 + 10.5 + 15

= 37.67

The critical path is A-D-H-J; its duration is 54 days. c. For path A-D-H-J: SVt = .694 + .694 + 1.000 + 1.000 = 3.388 days z = (x – m)/st = (58 – 54)/1.84 = 2.17 P(D > 58) = 1 - .9850 = .0150 or about 1.5% (There is a 98.5% chance of the project being completed within 58 days.) d. The longest paths (the ones offering the greatest risk of exceeding 56 days) are A-B-E-F-J, A-B-H-J, and A-D-H-J. The variances for these paths are: Path

Length of Path (days)

A-C-F-J A-B-E-F-J A-B-G-J A-B-H-J A-B-I A-D-H-J A-D-I

49.33 50.50 46.50 53.50 37.17 54.00 37.67

Variance of Path (days)

.694 + 1.778 + 1.0 + 1.0 = 4.472 .694 + .694 + 1.0 + 1.0 = 3.388

Path A-B-H-J offers a risk roughly as great as path A-D-H-J. Path A-B-H-J: z = (56 – 53.5)/ = 2.5/2.115 = 1.18 P(D > 56) = .1190 or about a 12% chance Path A-D-H-J: z = (56 – 54.0)/ = 2.0/1.841 = 1.09 P(D > 56) = .1380 or about a 14% chance Both path need to be managed closely in order to complete the project in no more than 56 days. POINTS: 1 TOPICS: Variability in project completion time 59. A project has the following activities, precedence relationships, and time estimates in weeks:

Activity

Immediate Predecessor Activities

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

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Chapter 9 - Project Scheduling: PERT/CPM E

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

Activity Expected Time (te)

Activity Variance (Vt)

2.778

0.444

30.83

6.250

24.83

.694

19.50

1.361

20.33

.111

b. Path

Length of Path (weeks)

B-D-G B-E-F A-C

10 + 15 + 20.33 10 + 24.83 + 19.5 20 + 30.83

= 45.33 = 54.33* = 50.83

c. SVt = 2.499 weeks; ; z = 1.06; P(D > 56) = .1446 POINTS: 1 TOPICS: Variability in project completion time 60. Three paths of a PERT network have these mean durations and variances in weeks:

Path

Mean Duration

Variance

2.75

5.50

1.20

Which path offers the greatest risk of overrunning a contract deadline of 48 weeks? ANSWER: Path 1: st = 1.658; z = 1.809; P(D > 48) = 3.5% Path 2: st = 2.345; z = 1.706; P(D > 48) = 4.4% Path 3: st = 1.095; z = 1.826; P(D > 48) = 3.4% Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM Path 2 offers the greatest risk (but only by a small margin) POINTS: 1 TOPICS: Variability in project completion time 61. A project has the following activities, durations, costs, and precedence relationships:

Activity

Present Duration (Weeks)

Accelerated Duration (Weeks)

Immediate Predecessor Activities

Present Cost

Accelerated Cost

$11,000

$15,000

20,000

25,000

9,000

20,000

25,000

30,000

20,000

35,000

20,000

30,000

15,000

25,000

D,F

12,000

18,000

G,H

10,000

15,000

Develop a cost-time trade-off analysis. Detail the steps that you would use to accelerate or crash the project to its minimum duration at the lowest cost. Determine each step's cost and the duration of the project. ANSWER:

Activity

Present Duration (Weeks)

Accelerated Duration (Weeks)

Present Cost

Accelerated Cost

Crashing Cost Per Week

$11,000

$15,000

$4,000

20,000

25,000

2,500

9,000

20,000

2,750

25,000

30,000

2,500

20,000

35,000

3,000

20,000

30,000

5,000

15,000

25,000

5,000

12,000

18,000

6,000

10,000

15,000

5,000

Path

Length of Path (weeks)

A-C-E A-D-H-I B-F-H-I B-G-I

10 + 10 + 15 = 10 + 20 + 9 + 7 = 15 + 17 + 9 + 7 = 15 + 12 + 7 = Activity Crashed

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35 46 48 34

35 46 47 33

35 46 46 32

35 45 45 31

35 44 44 31

35 43 43 31

35 42 42 31

D,F

D,F Page 24

Chapter 9 - Project Scheduling: PERT/CPM Additional Cost For Crashing

2,500

5,000

6,000

7,500

Project is reduced from 48 to 42 weeks at an additional cost of $31,000. POINTS: 1 TOPICS: Crashing activity times 62. National Oil Company (NATOCO) must plan the shutdown of its Houston refinery for routine preventive maintenance. Each hour of downtime is lost production time and is very costly, so NATOCO wants the maintenance project completed in 22 hours. The PERT network below shows the precedence relationships of the activities involved in the project. The table gives the activity times and costs under normal operations and maximum crashing.

NATOCO wants to know the minimum cost of completing the maintenance project within the 22-hour period. Formulate and solve a linear program that will yield this information. Activity A B C D E F G H I J K L M ANSWER:

Normal Time (hr) 2 4 1 4 6 10 8 2 5 12 7 11 4

Normal Cost $2,000 3,000 1,500 5,300 5,400 6,000 4,800 2,800 4,500 6,000 7,000 8,800 1,000

Crash Time (hr) 1.5 3 1 2.5 5 8 5 1 4 6 4 9 1

Crash Cost $3,000 3,500 1,500 8,000 7,000 9,000 9,900 2,900 5,000 9,600 9,700 9,200 7,000

m = (Normal Time) - (Time Under Maximum Crashing) k = ([Cost Under Maximum Crashing] - [Normal Cost])/m Activity

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k Page 25

Chapter 9 - Project Scheduling: PERT/CPM A B C D E F G H I J K L M

0.5 1 0 1.5 1 2 3 1 1 6 3 2 3

2000 500 0 1800 1600 1500 1700 100 500 600 900 200 2000

LP Formulation Let xA= earliest finish time for Activity A, and yA = amount of time Activity A is crashed. Min 2000yA + 500yB + 1800yD + 1600yE + 1500yF + 1700yG + 100yH + 500yI + 600yJ + 900yK + 200yL + 2000yM Subject to: (1) The project must be completed within 22 hours: xK < 22, xL < 22, xM < 22 (2) The amount an activity is crashed cannot exceed its maximum crashing: yA < 0.5 yB < 1 yD < 1.5 yE < 1 yF < 2 yG < 3 yH < 1 yI < 1 yJ < 6 yK < 3 yL < 2 yM < 3 (3) For each activity: (Earliest Finish Time) > (Earliest Finish Time of Preceding Activity) + [(Normal Activity Time) − (Amount of Time the Activity is crashed)] xA > 0 + (2 – yA) or xB > 0 + (4 – yB) or xC > 0 + (1 – 0) or

xA + yA > 2 xB + yB > 4 xC > 1

xD > xA + (4 – yD) or xD + yD – xA > 4 xE > xB + (6 – yE) or xE + yE – xB > 6 xG > xB + (8 – yG) or xG + yG – xB > 8 xH > xB + (2 – yH) or xH + yH – xB > 2 xI > xC + (5 – yI) or xI + yI – xC > 5 xJ > xC + (12 – yJ) or xJ + yJ – xC > 12 Cengage Learning Testing, Powered by Cognero

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Chapter 9 - Project Scheduling: PERT/CPM xF > xD + (10 – yF) or xF > xE + (10 – yF) or xL > xG + (11 – yL) or

xF + yF – xD > 10 xF + yF – xE > 10 xL + yL – xG > 11

xK > xF + (7 – yK) or xK > xI + (7 – yK) or xM > xH + (4 – yM) or xM > xJ + (4 – yM) or

xK + yK – xF > 7 xK + yK – xI > 7 xM + yM – xH > 4 xM + yM – xJ > 4

(4) Non-negativity of the variables: xi > 0 for i = A, B, C, …., M yj > 0 for j = A, B, D, …., M Solution OBJECTIVE FUNCTION VALUE = VARIABLE XA XB XC XD XE XF XG XH XI XJ XK XL XM YA YB YD YE YF YG YH YI YJ YK YL YM

VALUE 5.000 3.000 1.000 9.000 9.000 18.000 11.000 13.000 18.000 13.000 22.000 22.000 17.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 0.000 0.000 3.000 0.000 0.000

4700.000

REDUCED COSTS 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 2000.000 0.000 1800.000 100.000 0.000 1500.000 100.000 500.000 600.000 0.000 0.000 2000.000

Summary: Crash activity B one hour, activity F one hour, and activity K three hours. The preventive maintenance project will be completed in 22 hours at a crashing cost of $4700. POINTS: 1 TOPICS: LP model for crashing Essay Cengage Learning Testing, Powered by Cognero

Page 27

Chapter 9 - Project Scheduling: PERT/CPM 63. Name at least three managerial situations where answers are provided by project management solutions. ANSWER: POINTS: 1 TOPICS: Introduction 64. Explain how and why all predecessor activities must be considered when finding the earliest start time. ANSWER: POINTS: 1 TOPICS: Earliest and latest times 65. Explain how and why all successor activities must be considered when finding the latest finish time. ANSWER: POINTS: 1 TOPICS: Earliest and latest times 66. Once the earliest and latest times are calculated, how is the critical path determined? ANSWER: POINTS: 1 TOPICS: Critical path 67. Why should projects be monitored after the critical path is found? ANSWER: POINTS: 1 TOPICS: Critical path 68. Suppose that, after analyzing a project network, the project manager finds the project duration unacceptable. Discuss the options the manager might have for reducing the duration. ANSWER: POINTS: 1 TOPICS: Critical path

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Chapter 10 - Inventory Models True / False 1. To be considered as inventory, goods must be finished and waiting for delivery. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 2. When demand is independent, it is not related to demand for other components or items produced by the firm. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 3. Constant demand is a key assumption of the EOQ model. a. True b. False ANSWER: True POINTS: 1 TOPICS: EOQ model 4. In the EOQ model, the average inventory per cycle over many cycles is Q/2. a. True b. False ANSWER: True POINTS: 1 TOPICS: EOQ model 5. The single-period inventory model is most applicable to items that are perishable or have seasonal demand. a. True b. False ANSWER: True POINTS: 1 TOPICS: When-to-order decision 6. The time between placing orders is the lead time. a. True b. False ANSWER: False POINTS: 1 TOPICS: When-to-order decision 7. If the optimal production lot size decreases, average inventory increases. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models a. True b. False ANSWER: True POINTS: 1 TOPICS: Economic production lot-size model 8. If an item's per-unit backorder cost is greater than its per-unit holding cost, no intentional shortage should be planned. a. True b. False ANSWER: False POINTS: 1 TOPICS: Inventory model with planned shortages 9. When quantity discounts are available, order an amount from the highest discount category. a. True b. False ANSWER: False POINTS: 1 TOPICS: Quantity discounts for the EOQ model 10. When there is probabilistic demand in a multi-period model, the inventory level will not decrease smoothly and can fall below 0. a. True b. False ANSWER: True POINTS: 1 TOPICS: Order quantity, reorder point model with probabilistic demand 11. In the periodic review model, the order quantity at each review period must be sufficient to cover demand for the review period plus the demand for the following lead time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Periodic review model with probabilistic demand 12. Periodic review systems require smaller safety stock levels than corresponding continuous review systems. a. True b. False ANSWER: False POINTS: 1 TOPICS: Periodic review model with probabilistic demand 13. The cost of overestimating demand is usually harder to determine than the cost of underestimating demand. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models b. False ANSWER: False POINTS: 1 TOPICS: Single-period inventory model with probabilistic demand 14. The terms "inventory on hand" and "inventory position" have the same meaning. a. True b. False ANSWER: False POINTS: 1 TOPICS: The when-to-order decision 15. The EOQ model is insensitive to small variations or errors in the cost estimates. a. True b. False ANSWER: True POINTS: 1 TOPICS: Sensitivity analysis for the EOQ model 16. At the optimal order quantity for the quantity discount model, the sum of the annual holding and ordering costs is minimized. a. True b. False ANSWER: False POINTS: 1 TOPICS: Quantity discounts for the EOQ model 17. As lead time for an item increases, the cycle time increases. a. True b. False ANSWER: False POINTS: 1 TOPICS: The when-to-order decision 18. An assumption in the economic production lot size model is that there is storage capacity to hold the entire production lot. a. True b. False ANSWER: False POINTS: 1 TOPICS: Economic production lot-size model 19. If lead time is longer than the review period, the order quantity at any review point is the amount needed for the inventory on hand plus all outstanding orders to reach the replenishment level. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models ANSWER: True POINTS: 1 TOPICS: Periodic review model with probabilistic demand 20. For the periodic review inventory model presented in the textbook, it is assumed that a special replenishment order will be placed in the event of a stockout between review points. a. True b. False ANSWER: False POINTS: 1 TOPICS: Periodic review model with probabilistic demand Multiple Choice 21. Inventory a. is held against uncertain usage so that a supply of items is available if needed. b. constitutes a small part of the cost of doing business. c. is not something that can be managed effectively. d. All of the alternatives are correct. ANSWER: a POINTS: 1 TOPICS: Introduction 22. Inventory models in which the rate of demand is constant are called a. fixed models. b. deterministic models. c. JIT models. d. requirements models. ANSWER: b POINTS: 1 TOPICS: Introduction 23. The EOQ model a. determines only how frequently to order. b. considers total cost. c. minimizes both ordering and holding costs. d. All of the alternatives are correct. ANSWER: b POINTS: 1 TOPICS: EOQ model 24. Which cost would not be considered part of a holding cost? a. cost of capital b. shipping cost c. insurance cost Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models d. warehouse overhead ANSWER: b POINTS: 1 TOPICS: EOQ model 25. For inventory systems with constant demand and a fixed lead time, a. the reorder point = lead-time demand. b. the reorder point > lead-time demand. c. the reorder point < lead-time demand. d. the reorder point is unrelated to lead-time demand. ANSWER: a POINTS: 1 TOPICS: When-to-order decision 26. Safety stock a. can be determined by the EOQ formula. b. depends on the inventory position. c. depends on the variability of demand during lead time. d. is not needed if Q* is the actual order quantity. ANSWER: c POINTS: 1 TOPICS: EOQ model 27. The economic production lot size model is appropriate when a. demand exceeds the production rate. b. there is a constant supply rate for every period, without pause. c. ordering cost is equivalent to the production setup cost. d. All of the alternatives are correct. ANSWER: c POINTS: 1 TOPICS: Economic production lot-size model 28. The maximum inventory with backorders is a. Q b. Q − S c. S d. (Q − S) / 2 ANSWER: b POINTS: 1 TOPICS: Inventory model with planned shortages 29. Annual purchase cost is included in the total cost in a. the EOQ model. b. the economic production lot size model. c. the quantity discount model. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models d. all inventory models. ANSWER: c POINTS: 1 TOPICS: Quantity discounts for the EOQ model 30. In the single-period inventory model with probabilistic demand, a. surplus items are not allowed to be carried in future inventory. b. co = cu. c. probabilities are used to calculate expected losses. d. All of the alternatives are correct. ANSWER: c POINTS: 1 TOPICS: Single-period inventory model 31. For the inventory model with planned shortages, the optimal order quantity results in a. annual holding cost = annual ordering cost. b. annual holding cost = annual backordering cost. c. annual ordering cost = annual holding cost + annual backordering cost. d. annual ordering cost = annual holding cost − annual backordering cost. ANSWER: c POINTS: 1 TOPICS: Inventory model with planned shortages 32. The definition of service level used in this chapter is a. the percentage of all demand that can be satisfied from inventory. b. the percentage of all order cycles that do not experience a stockout. c. the percentage of demand during the lead-time period that can be satisfied from inventory d. None of the alternatives is correct. ANSWER: b POINTS: 1 TOPICS: Inventory models with probabilistic demand 33. Periodic review inventory systems a. are less subject to stockouts than corresponding continuous review systems. b. require larger safety stock levels than corresponding continuous review systems. c. have constant order quantities. d. make the coordination of orders for multiple products more difficult. ANSWER: b POINTS: 1 TOPICS: Periodic review model with probabilistic demand 34. Inventory position is defined as a. the amount of inventory on hand in excess of expected demand. b. the amount of inventory on hand. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models c. the amount of inventory on hand plus the amount of inventory on order. d. None of the alternatives is correct. ANSWER: c POINTS: 1 TOPICS: The when-to-order decision 35. A firm that is presently using the Economic Order Quantity model and is planning to switch to the Economic Production Lot-Size model can expect a. the Q* to increase b. the maximum inventory level to increase. c. the order cycle to decrease. d. annual holding cost to be less than annual setup cost. ANSWER: a POINTS: 1 TOPICS: Economic production lot-size model 36. Which of the following is not implied when average inventory is Q/2, where Q is the order quantity? a. An entire order quantity arrives at one time. b. The previous order quantity is entirely depleted when the next order arrives. c. An order quantity is depleted at a uniform rate over time. d. Backorders are permitted. ANSWER: d POINTS: 1 TOPICS: EOQ model 37. For the EOQ model, which of the following relationships is incorrect? a. As the order quantity increases, the number of orders placed annually decreases. b. As the order quantity increases, annual holding cost increases. c. As the order quantity increases, annual ordering cost increases. d. As the order quantity increases, average inventory increases. ANSWER: c POINTS: 1 TOPICS: EOQ model 38. The objective of the EOQ with quantity discounts model is to a. determine the minimum order quantity required for the maximum discount. b. balance annual ordering and holding costs. c. minimize annual purchase cost. d. minimize the sum of annual carrying, holding, and purchase costs. ANSWER: d POINTS: 1 TOPICS: Quantity discounts for the EOQ model 39. When the reorder point r exceeds Q*, the difference is a. safety stock b. one or more outstanding orders Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models c. surplus inventory d. backorders ANSWER: b POINTS: 1 TOPICS: Economic order quantity model 40. Inventory position is defined as the amount of inventory on hand plus the amount a. on order b. promised to customers c. on reserve d. to be returned to suppliers ANSWER: a POINTS: 1 TOPICS: The when-to-order decision 41. For the EOQ model, cycle time is the time between a. placing successive orders b. placing and receiving an order c. stocking out and receiving an order d. receiving and storing an order ANSWER: a POINTS: 1 TOPICS: The when-to-order decision Subjective Short Answer 42. Show the total cost expression and calculate the EOQ for an item with holding cost rate 18%, unit cost $8.00, annual demand of 40000, and ordering cost of $48. ANSWER: TC = (1/2)Q(.18)(8) + 40000(48)/Q Q* = 1633 POINTS: 1 TOPICS: EOQ model 43. Demand for a popular athletic shoe is nearly constant at 800 pairs per week for a regional division of a national retailer. The cost per pair is $54. It costs $72 to place an order, and annual holding costs are charged at 22% of the cost per unit. The lead time is two weeks. a. What is the EOQ? b. What is the reorder point? c. What is the cycle time? d. What is the total annual cost? ANSWER: a. Q* = 710.1 b. r = 1600 c. T = 710.1/800 = .8876 weeks or .8876(7) = 6.2 days d. TC = .5(710.1)(.22)(54) + (41600/710.1)(72) = 8435.99 POINTS: 1 TOPICS: EOQ model 44. The Super Discount store (open 24 hours a day, every day) sells 8-packs of paper towels, at the rate of approximately 420 packs per week. Because the towels are so bulky, the annual cost to carry them in inventory is estimated at $.50. The cost to place an order for more is $20 and it takes four days for an order to arrive. a. Find the optimal order quantity. b. What is the reorder point? Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models c. How often should an order be placed? ANSWER: a. Q* = 1322 b. r = (60)(4) = 240 c. T = 365(1322)/21840 = 22 days POINTS: 1 TOPICS: EOQ model 45. An office supply store open 5 days a week must determine the best inventory policy for boxes of copier paper. Weekly demand is nearly constant at 250 boxes and when orders are placed, then entire shipment arrives at once. The cost per box is $22 and the inventory holding cost is 30%. Orders are placed at a cost of $40 each, including preparation time and communication charges, and the lead time is 2 days. a. Find the optimal order quantity. b. What is the reorder point? c. How often should an order be placed? d. What is the cycle time? ANSWER: a. Q* = 397 b. TC = .5(397)(6.6) + (13000/397)(40) = 2620 c. r = 100 d. Cycle = 7.94 days POINTS: 1 TOPICS: EOQ model 46. Zip Games purchases blank DVD disks onto which it copies its software for sale through its mail order operation. A disk costs Zip $.20. Processing an order for more disks cost $15. Zip uses 60000 disks annually, and the company has a 25% cost of capital. a. Find the optimal order quantity. b. How many orders are placed annually? c. How frequently will orders be placed? ANSWER: a. Q* = 6000 b. 10 orders placed annually c. every 36.5 days POINTS: 1 TOPICS: EOQ model 47. Kellam Images prints snack food bags on long rolls of plastic film. The plant operates 250 days a year. The daily production rate is 6000 bags, and the daily demand is 3500 bags. The cost to set up the design for printing is $300. The holding cost is estimated at 2 cents per bag. a. What is the recommended production lot size? b. If there is a five-day lead time to set up the line, what is the recommended reorder point? ANSWER: a. Q* = 250,998 b. 17500 POINTS: 1 TOPICS: Economic production lot-size model 48. Henderson Furniture sells reproductions of 18th century furniture. For a particular table, the assumptions of the inventory model with backorders are valid. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models D = 200 tables per year I = 25% per year C = $800 per table Co = $80 per order Cb = $50 per table per year The store is open 250 days a year. What are the values for order quantity and number of planned backorders that will minimize total cost? b. What is the maximum inventory? c. What is the cycle time? d. What is total cost? ANSWER: a. Q* = 28.28 S* = 28.28(200/250) = 22.62 b. 28 − 23 = 5 tables c. Cycle time = (28/200)250 = 35 days d. TC = 89.29 + 571.43 + 473.32 = 1134.04 (using whole numbers of tables for Q and S) POINTS: 1 TOPICS: Inventory model with planned shortages a.

49. The Tiernan Gallery and Art Museum distributes to its visitors a printed guide to its collections. There are about 18000 visitors per year. Holding costs for the brochures are 20% and it costs $30 to place an order with the printer. The printer has offered the following discount schedule: Category 1 2 3

Order Size 0 - 1499 1500 - 2999 3000 and over

Unit Cost $2.50 $2.20 $1.80

How many brochures should be printed at a time? ANSWER: EOQ3 = 1732.1 (raise to 3000) EOQ2 = 1566.7 (feasible, so no reason to compute EOQ1) TC(1566.7) = 689.35 TC(3000) = 720.00 So the gallery should order 1566.7 or 1567 brochures at a time. POINTS: 1 TOPICS: Quantity discounts for the EOQ model 50. A weekly sports magazine publishes a special edition for the World Series. The sales forecast is for the number of copies to be normally distributed with mean 800,000 copies and standard deviation 60,000 copies. It costs $.35 to print a copy, and the newsstand price is $1.95. Unsold copies will be scrapped. How many copies should be printed? ANSWER: P(demand < Q*) = 1.6/1.95 = .8205 Q* = 800000 + .92(60000) = 855,200 POINTS: 1 TOPICS: Single-period inventory model with probabilistic demand 51. The Fitness Shop is considering ordering a special model exercise machine. Each unit will cost the shop $410 and it will sell for $750. Any units not sold at the regular price will be sold at the year-end model clearance for $340. Assume that demand follows a normal probability distribution with μ = 20 and σ = 6. What is the recommended order quantity? Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models ANSWER: cu = 340, co = 70 P(demand ≤ Q*) = .8293 Q* = 20 + .95(6) = 25.7 so order 26 POINTS: 1 TOPICS: Single-period inventory model with probabilistic demand 52. Daily demand for packages of five videotapes at a warehouse store is found to be normally distributed with mean 50 and standard deviation 5. When the store orders more tapes, the ordering cost is $42 and the orders take 4 days to arrive. Each pack of tapes costs $7.20 and there is a 24% annual holding cost for inventory. Assume the store is open 360 days a year. a. What is the EOQ? If the store wants the probability of stocking out to be no more than 5%, and demand each b. day is independent of the day before, what reorder point should be set? c. How much of your reorder point in part b) is safety stock? ANSWER: a. Q* = 866.025 b. DDLT is N(200,10) r = 200 + 1.645(10) = 216.45 c. 16.45 POINTS: 1 TOPICS: Order quantity, reorder point model with probabilistic demand 53. A gourmet food store uses a one-week periodic review system for its supply of coffee beans. There is a five-day lead time for orders, and the store will allow two stockouts per year. a. What is the probability of a stockout associated with each replenishment decision? What is the replenishment level if demand during the review and lead-time periods is b. normally distributed with mean 120 pounds and standard deviation 8 pounds? c. How many pounds of beans should be ordered if there are 42 pounds of beans on hand? ANSWER: a. 2/52 = .0385 b. M = 120 + 1.77 (8) = 134.16 c. 134 − 40 = 94 POINTS: 1 TOPICS: Periodic review model with probabilistic demand 54. Chez Paul Restaurant orders special Styrofoam "doggy bags" for its customers once a month and lead time is one week. Weekly demand for doggy bags is approximately normally distributed with an average of 120 bags and a standard deviation of 25. Chez Paul wants at most a 3% chance of running out of doggy bags during the replenishment period. If he has 150 bags in stock when he places an order, how many additional bags should he order? What is the safety stock in this case? ANSWER: Order 555 doggy bags. Safety stock is 105 bags. POINTS: 1 TOPICS: Periodic review model with probabilistic demand 55. Every year in early October Steven King buys pumpkins of one size from a farmer in Maine and then hires an artist to carve bewitching faces in them. He then tries to sell them at his produce stand in a public market in Boston. The farmer charges Steven $2.50 per pumpkin and the artist is paid $2.00 per carved pumpkin. Steven sells a carved pumpkin for $8.00. Any pumpkins not sold by 5:00 p.m. on Halloween are donated to Steven's favorite children's hospital. Steven pays the artist $0.75 per pumpkin to rush the pumpkins to the hospital for the youngsters to enjoy. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models Steven estimates the demand for his pumpkins this season to be uniformly distributed within a range of 30 to 70. a. How many pumpkins should Steven have available for sale? Based on your answer to (a), what is the probability that Steven will be short five or more b. pumpkins? ANSWER: a. 46 pumpkins b. .525 POINTS: 1 TOPICS: Single-period inventory model with probabilistic demand 56. Bank Drugs sells Jami Michelle lipstick. The Jami Michelle Company offers a 6% discount on orders of at least 500 tubes, a 10% discount on orders of at least 1,000 tubes, a 12% discount on orders of at least 1,800 tubes and a 15% discount on orders at least 2,500 tubes. Bank sells an average of 40 tubes of Jami Michelle lipstick weekly. The normal price paid by Bank drugs is $1 per tube. If it costs Bank $30 to place an order, and Bank's annual holding cost rate is 27%, determine the optimal order policy for Bank Drugs. ANSWER: Order 1000 tubes at a time POINTS: 1 TOPICS: Quantity discounts for the EOQ model 57. Amazing Bakers sells bread to 40 supermarkets. It costs Amazing $1,250 per day to operate its plant. The profit per loaf of bread sold in the supermarket is $.025. Any unsold bread is returned to the Amazing Thrift Store to be sold at a loss of $.015. If sales follow a normal distribution with μ = 70,000 and σ = 5,000 per day, how many a. loaves should Amazing bake daily? Amazing is considering a different sales plan for which the profit per loaf of bread sold in b. the supermarket is $.03 and the loss per loaf bread returned is $.018. If μ = 60,000 and σ = 4,000 per day, how many loaves should Amazing bake daily? ANSWER: a. 71,600 loaves b. 61,280 loaves POINTS: 1 TOPICS: Single-period inventory model with probabilistic demand 58. A lawn and garden shop that is open for business seven days a week orders bags of grass seed every OTHER Monday. Lead time for seed orders is 5 days. On Monday, at ordering time, a clerk found 112 bags of seed in stock, and so he ordered 198 bags. Daily demand for grass seed is normally distributed with a mean of 15 bags and a standard deviation of four bags. The manager would like to know what the probability is that a grass seed stockout will occur before the next order arrives. ANSWER: z = 1.43, so Pr(stockout) = .0764 POINTS: 1 TOPICS: Periodic review model with probabilistic demand 59. Kelly's Service Station does a large business in tune-ups. Demand has been averaging 210 spark plugs per week. Holding costs are $.01 per plug per week and reorder costs are estimated at $10 per order. Kelly does not want to be out of stock on more than 1% of his orders. There is a one-day delivery time. The standard deviation of demand is five plugs per day. Assume a normal distribution of demand during lead time and a 7-day work Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models week. a. What inventory policy do you suggest for Kelly's station? b. What is the average amount of safety stock for the reorder point in (a)? c. What are the total variable weekly costs including safety stock costs? ANSWER: a. Order 648 every 3.08 weeks when supply reaches 42 b. 12 spark plugs c. $6.60 POINTS: 1 TOPICS: When-to-order decision 60. Non-Slip Tile Company (NST) has been using production runs of 100,000 tiles, 10 times per year to meet the demand of 1,000,000 tiles annually. The set-up cost is $5,000 per run and holding cost is estimated at 10% of the manufacturing cost of $1 per tile. The production capacity of the machine is 500,000 tiles per month. The factory is open 365 days per year. a. What production schedule do you recommend? b. How much is NST losing annually with their present production schedule? c. What is the maximum number of tiles in inventory under the current policy? under the optimal policy? d. What fraction of time is the machine idle (not producing tiles) under the current policy? Under the optimal policy? ANSWER: D = 1,000,000, P = 6,000,000, Ch = .10, Co = 5,000. a. Q = 346,410 D/Q = 2.89 times per year Recommended policy is: produce 346,410 tiles at a time, 2.89 times per year b. Optimal TC = $28,867.51 Current TC = $54,166.67 Difference = $25,299.16 c. Current maximum inventory = 83,333 Optimal maximum inventory = 288,675 d. The machine is producing tiles d/p = 1/6 of the time. The machine is idle 5/6 of the time. Optimal Policy: There are 2.89 cycles per year Each cycle lasts 126.3 days Each run is 346,410 units Each run takes 21.1 days Machine is idle 105.2 days between runs Machine is idle .833 (or 5/6) of time Current Policy: There are 10 cycles per year Each cycle lasts 36.5 days Each run is 100,000 units Each run takes 6.083 days Machine is idle 30.417 days between runs The machine is idle .833 (or 5/6) of time POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models TOPICS: Economic production lot size model 61. Ken Ells, owner and operator of Kennels, Inc. is concerned that the person in charge of ordering dog food is often incurring an order expediting expense because he is waiting too long (letting the inventory level drop too low) before ordering. Past data indicates that demand during lead time (when expediting does not occur) is normally distributed with a mean of 340 pounds and a standard deviation of 45 pounds. Ken wants the probability of running out of dog food to be .03. a. If the current order point is 400 pounds, what is the resulting service level? b. If Ken wants a service level of 97% for dog food, what should be the order point? ANSWER: a. 90.82% b. r = 340 + 1.88(45) = 424.6 or 425 pounds POINTS: 1 TOPICS: The when-to-order decision Essay 62. What are the two most critical behaviors of an inventory system you need to recognize in order to apply the best model? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 63. What compromise must be made in the how-much-to-order decision? ANSWER: Answer not provided. POINTS: 1 TOPICS: How-much-to-order decision 64. Derive and explain the total cost expression for the EOQ model. ANSWER: Answer not provided. POINTS: 1 TOPICS: EOQ model 65. What could be meant by service level? Why is a clear understanding of this term important? ANSWER: Answer not provided. POINTS: 1 TOPICS: When-to-order decision 66. Explain the difference between a periodic and a continuous review system. ANSWER: Answer not provided. POINTS: 1 TOPICS: Periodic review model with probabilistic demand 67. List five assumptions of the EOQ model. ANSWER: Answer not provided. Cengage Learning Testing, Powered by Cognero

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Chapter 10 - Inventory Models POINTS: 1 TOPICS: EOQ model

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Chapter 11 - Waiting Line Models True / False 1. For an M/M/1 queuing system, if the service rate, µ, is doubled, the average wait in the system, W, is cut in half. a. True b. False ANSWER: False POINTS: 1 TOPICS: Operating characteristics 2. A waiting line situation where every customer waits in the same line before being served by the same server is called a single server waiting line. a. True b. False ANSWER: False POINTS: 1 TOPICS: Single-channel waiting line 3. Use of the Poisson probability distribution assumes that arrivals are not random. a. True b. False ANSWER: False POINTS: 1 TOPICS: Distribution of arrivals 4. Queue discipline refers to the assumption that a customer has the patience to remain in a slow moving queue. a. True b. False ANSWER: False POINTS: 1 TOPICS: Queue discipline 5. For all waiting lines, P0 + Pw = 1. a. True b. False ANSWER: False POINTS: 1 TOPICS: Operating characteristics 6. Before waiting lines can be analyzed economically, the arrivals' cost of waiting must be estimated. a. True b. False ANSWER: True POINTS: 1 TOPICS: Economic analysis

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Chapter 11 - Waiting Line Models 7. In a multiple channel system it is more efficient to have a separate waiting line for each channel. a. True b. False ANSWER: False POINTS: 1 TOPICS: Multiple-channel waiting line 8. Little's flow equations indicate that the relationship of L to Lq is the same as that of W to Wq. a. True b. False ANSWER: False POINTS: 1 TOPICS: General relationships for waiting line models 9. If some maximum number of customers is allowed in a queuing system at one time, the system has a finite calling population. a. True b. False ANSWER: False POINTS: 1 TOPICS: Other waiting line models 10. When blocked customers are cleared, an important decision is how many channels to provide. a. True b. False ANSWER: True POINTS: 1 TOPICS: Multiple-channel model with Poisson arrivals, arbitrary service times, and no waiting line 11. If service time follows an exponential probability distribution, approximately 63% of the service times are less than the mean service time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Distribution of service times 12. Queue discipline refers to the manner in which waiting units are arranged for service. a. True b. False ANSWER: True POINTS: 1 TOPICS: Queue discipline 13. Waiting line models describe the transient-period operating characteristics of a waiting line. Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models a. True b. False ANSWER: False POINTS: 1 TOPICS: Steady-state operation 14. For a single-channel waiting line, the utilization factor is the probability that an arriving unit must wait for service. a. True b. False ANSWER: True POINTS: 1 TOPICS: Operating characteristics 15. When a waiting system is in steady-state operation, the number of units in the system is not changing. a. True b. False ANSWER: False POINTS: 1 TOPICS: Steady-state operation 16. Adding more channels always improves the operating characteristics of the waiting line and reduces the waiting cost. a. True b. False ANSWER: True POINTS: 1 TOPICS: Economic analysis of waiting lines 17. In developing the total cost for a waiting line, waiting cost takes into consideration both the time spent waiting in line and the time spent being served. a. True b. False ANSWER: True POINTS: 1 TOPICS: Economic analysis of waiting lines 18. In waiting line systems where the length of the waiting line is limited, the mean number of units entering the system might be less than the arrival rate. a. True b. False ANSWER: True POINTS: 1 TOPICS: Some general relationships for waiting line models 19. A multiple-channel system has more than one waiting line. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models b. False ANSWER: False POINTS: 1 TOPICS: Multiple-channel waiting line models 20. For an M/M/k system, the average number of customers in the system equals the customer arrival rate times the average time a customer spends waiting in the system. a. True b. False ANSWER: True POINTS: 1 TOPICS: Multiple-channel waiting line model operating characteristics 21. For a single-server queuing system, the average number of customers in the waiting line is one less than the average number in the system. a. True b. False ANSWER: False POINTS: 1 TOPICS: Single-channel waiting line model operating characteristics 22. In waiting line applications, the exponential probability distribution indicates that approximately 63 percent of the service times are less than the mean service time. a. True b. False ANSWER: True POINTS: 1 TOPICS: Distribution of service times 23. With no waiting allowed, operating characteristics Lq and Wq are automatically zero regardless of the number of servers. a. True b. False ANSWER: True POINTS: 1 TOPICS: Multiple-channel model with no waiting line 24. Little’s flow equations apply to any waiting line model. a. True b. False ANSWER: True POINTS: 1 TOPICS: Some general realtionships for waiting line models Multiple Choice Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models 25. Decision makers in queuing situations attempt to balance a. operating characteristics against the arrival rate. b. service levels against service cost. c. the number of units in the system against the time in the system. d. the service rate against the arrival rate. ANSWER: b POINTS: 1 TOPICS: Introduction 26. Performance measures dealing with the number of units in line and the time spent waiting are called a. queuing facts. b. performance queues. c. system measures. d. operating characteristics. ANSWER: d POINTS: 1 TOPICS: Introduction 27. If arrivals occur according to the Poisson distribution every 20 minutes, then which is NOT true? a. λ = 20 arrivals per hour b. λ = 3 arrivals per hour c. λ = 1/20 arrivals per minute d. λ = 72 arrivals per day ANSWER: a POINTS: 1 TOPICS: Distribution of arrivals 28. The manner in which units receive their service, such as FCFS, is the a. queue discipline. b. channel. c. steady state. d. operating characteristic. ANSWER: a POINTS: 1 TOPICS: Queue discipline 29. In a waiting line situation, arrivals occur, on average, every 10 minutes, and 10 units can be received every hour. What are λ and μ? a. λ = 10, μ = 10 b. λ = 6, μ = 6 c. λ = 6, μ = 10 d. λ = 10, μ = 6 ANSWER: c POINTS: 1 TOPICS: Structure of a waiting line system Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models 30. Operating characteristics formulas for the single-channel queue do NOT require a. λ ≥ μ. b. Poisson distribution of arrivals. c. an exponential distribution of service times. d. an FCFS queue discipline. ANSWER: a POINTS: 1 TOPICS: Operating characteristics 31. In a multiple channel system a. each server has its own queue. b. each server has the same service rate. c. μ > λ d. All of the alternatives are correct. ANSWER: b POINTS: 1 TOPICS: Multiple channel waiting line 32. Little's flow equations a. require Poisson and exponential assumptions. b. are applicable to any waiting line model. c. require independent calculation of W, L, Wq, and Lq. d. All of the alternatives are correct. ANSWER: b POINTS: 1 TOPICS: General relationships for waiting line models 33. The total cost for a waiting line does NOT specifically depend on a. the cost of waiting. b. the cost of service. c. the number of units in the system. d. the cost of a lost customer. ANSWER: d POINTS: 1 TOPICS: Economic analysis 34. Models with a finite calling population a. have an arrival rate independent of the number of units in the system. b. have a service rate dependent on the number of units in the system. c. use the size of the population as a parameter in the operating characteristics formulas. d. All of the alternatives are correct. ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models TOPICS: Finite calling population 35. Which of the following can NOT be found by the queuing formulas presented in the textbook? a. the probability that no units are in the system. b. the average number of units in the system. c. the maximum time a unit spends in the system. d. the average time a unit spends in the system. ANSWER: c POINTS: 1 TOPICS: Operating characteristics 36. The arrival rate in queuing formulas is expressed as a. the mean time between arrivals. b. the minimum number of arrivals per time period. c. the mean number of arrivals per channel. d. the mean number of arrivals per time period. ANSWER: d POINTS: 1 TOPICS: Distribution of arrivals 37. What queue discipline is assumed by the waiting line models presented in the textbook? a. first-come first-served. b. last-in first-out. c. shortest processing time first. d. No discipline is assumed. ANSWER: a POINTS: 1 TOPICS: Queue discipline 38. For many waiting line situations, the arrivals occur randomly and independently of other arrivals and it has been found that a good description of the arrival pattern is provided by a. a normal probability distribution. b. an exponential probability distribution. c. a uniform probability distribution. d. a Poisson probability distribution. ANSWER: d POINTS: 1 TOPICS: Distribution of arrivals 39. The assumption of exponentially distributed service times indicates that a. 37% of the service times are less than the mean service time. b. 50% of the service times are less than the mean service time. c. 63% of the service times are less than the mean service time. d. service time increase at an exponential rate as the waiting line grows. ANSWER: c Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models POINTS: 1 TOPICS: Distribution of service times 40. Single-booth ticket sales at a theater would be an example of which queuing model? a. single-channel, Poisson service rate distribution, unlimited queue length. b. single-channel, Poisson service rate distribution, limited queue length. c. single-channel, constant service rate distribution, unlimited queue length. d. single-channel, normal service rate distribution, unlimited queue length. ANSWER: a POINTS: 1 TOPICS: Single-channel waiting line 41. The machine repair problem is an application of the M/M/1 model with a. no waiting line. b. arbitrary service times. c. a finite calling population. d. blocked customers cleared. ANSWER: c POINTS: 1 TOPICS: Finite calling populations 42. The equations provided in the textbook for computing operating characteristics apply to a waiting line operating a. at start-up. b. at steady-state. c. at peak-demand times. d. in transition ANSWER: b POINTS: 1 TOPICS: Operating characteristics Subjective Short Answer 43. During summer weekdays, boats arrive at the inlet drawbridge according to the Poisson distribution at a rate of 3 per hour. In a 2-hour period, a. what is the probability that no boats arrive? b. what is the probability that 2 boats arrive? c. what is the probability that 8 boats arrive? ANSWER: a. P(0) = .0025 b. P(2) = .0446 c. P(8) = .1033 POINTS: 1 TOPICS: Distribution of arrivals 44. The time to process a registration at the Sea View Resort follows the exponential distribution and has a mean of 6 minutes. a. What is the probability of a registration time shorter than 3 minutes? Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models b. What is the probability of a registration time shorter than 6 minutes? c. What is the probability of a registration time between 3 and 6 minutes? ANSWER: a. P(time ≤ 3) = 1 − .6065 = .3935 b. P(time ≤ 6) = 1 − .3679 = .6321 c. P(3 ≤ time ≤ 6) = .6321 − .3935 = .2386 POINTS: 1 TOPICS: Distribution of service times 45. The Grand Movie Theater has one box office clerk. On average, each customer that comes to see a movie can be sold its ticket at the rate of 6 per minute. For the theater's normal offerings of older movies, customers arrive at the rate of 3 per minute. Assume arrivals follow the Poisson distribution and service times follow the exponential distribution. a. What is the average number of customers waiting in line? b. What is the average time a customer spends in the waiting line? c. What is the average number of customers in the system? d. What is a customer's average time in the system? e. What is the probability that someone will be buying tickets when an arrival occurs? The Grand has booked the Stars Wars Trilogy and expects more customers. From conversations with other theater owners, it estimates that the arrival rate will increase to 10 per minute. Output is supplied for a two-cashier and a three-cashier system. Number of Channels Arrival Rate Service Rate Probability of No Units in System Average Waiting Time Average Time in System Average Number Waiting Average Number in System Probability of Waiting Probability of 11 in System

2 10 6 .0909 .3788 .5455 3.7879 5.4545 .7576 .0245

3 10 6 .1727 .0375 .2041 .3747 2.0414 .2998 less than .0088

The Grand has space for ten customers to wait indoors to buy tickets. Which system will be better? g. Do you think it is more sensible for them to continue the one-cashier system? ANSWER: a. Lq = .5 b. Wq = .1667 c. L = 1 d. W = .3333 e. Pw = .5 f. The three-cashier system is probably too good and not cost effective. The one cashier system won't work because now the arrival rate is faster than the service g. rate. POINTS: 1 TOPICS: Operating characteristics f.

46. The Arctic Flyers minor league hockey team has one box office clerk. On average, each customer that comes to see a game can be sold a ticket at the rate of 8 per minute. For normal games, customers arrive at the rate of 5 per minute. Assume arrivals follow the Poisson distribution and service times follow the exponential distribution. Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models a. b. c. d. e.

What is the average number of customers waiting in line? What is the average time a customer spends in the waiting line? What is the average number of customers in the system? What is a customer's average time in the system? What is the probability that someone will be buying tickets when an arrival occurs?

The Flyers are playing in the league playoffs and anticipate more fans, estimating that the arrival rate will increase to 12 per minute. Output is supplied for a two-cashier and a three-cashier system. Number of Channels Arrival Rate Service Rate Probability of No Units in System Average Waiting Time Average Time in System Average Number Waiting Average Number in System Probability of Waiting Probability of 7 in System

2 12 8 .1429 .1607 .2857 1.9286 3.4286 .6429 .0381

3 12 8 .2105 .0197 .1447 .2368 1.7368 .2368 .0074

The rink has space for six customers to wait indoors to buy tickets. Which system will be better? g. Do you think it is more sensible for them to continue the one cashier system? ANSWER: a. Lq = 1.04 b. Wq = .2083 c. L = 1.665 d. W = .3333 e. Pw = .625 f. The three-cashier system is probably too good and not cost effective. The one cashier system won't work because now the arrival rate is faster than the service g. rate. POINTS: 1 TOPICS: Operating characteristics f.

47. In a waiting line situation, arrivals occur at a rate of 2 per minute, and the service times average 18 seconds. Assume the Poisson and exponential distributions. a. What is λ? b. What is μ? c. Find probability of no units in the system. d. Find average number of units in the system. e. Find average time in the waiting line. f. Find average time in the system. g. Find probability that there is one person waiting. h. Find probability an arrival will have to wait. ANSWER: a. λ = 3/min. b. μ = 4/min. c. P0 = .25 d. L = 3 e. Wq = .75 min. Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models f. g. h.

W = 1.00 min. P(2) = .140625 Pw = .75

POINTS: 1 TOPICS: Operating characteristics: M/M/1 48. In a waiting line situation, arrivals occur around the clock at a rate of six per day, and the service occurs at one every three hours. Assume the Poisson and exponential distributions. a. What is λ? b. What is μ? c. Find probability of no units in the system. d. Find average number of units in the system. e. Find average time in the waiting line. f. Find average time in the system. g. Find probability that there is one person waiting. h. Find probability an arrival will have to wait. ANSWER: a. λ = 6 b. μ = 8 c. P0 = .25 d. L = 3 e. Wq = .375 f. W = .5 g. P(2) = .1406 h. Pw = .75 POINTS: 1 TOPICS: Operating characteristics: M/M/1 49. The Sea View Resort uses a multiple-channel queue registration system. If the average service time is 8 minutes, there are three registration clerks, and guests arrive at the rate of one every 5 minutes, find a. λ and μ. b. the probability all three clerks are idle. c. the probability a guest will have to wait. d. the average time a customer is in line. e. the average number of customers in line. ANSWER: a. λ = 12 per hour, μ = 7.5 per hour b. P0 = .1872 c. Pw = .2738 d. Wq = .0261 e. Lq = .3129 POINTS: 1 TOPICS: Operating characteristics: M/M/3 50. The post office uses a multiple channel queue, where customers wait in a single line for the first available window. If the average service time is 1 minute and the arrival rate is 7 customers every five minutes, find, when two service windows are open, a. the probability both windows are idle. b. the probability a customer will have to wait. Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models c. the average time a customer is in line. d. the average time a customer is in the post office. ANSWER: a. P0 = .1765 b. Pw = .5765 c. Wq = .1922 (of a five-minute period) d. W = .3922 (of a five-minute period) POINTS: 1 TOPICS: Operating characteristics: M/M/2 51. Two new checkout scanning systems are under consideration by a retail store. Arrivals to the checkout stand follow the Poisson distribution with λ = 2 per minute. The cost for waiting is $18 per hour. The first system has an exponential service rate of 5 per minute and costs $10 per hour to operate. The second system has an exponential service rate of 8 per minute and costs $20 per hour to operate. Which system should be chosen? ANSWER: First system costs .37 per minute, second system costs .43 per minute. Choose first system. POINTS: 1 TOPICS: Economic analysis 52. Circle Electric Supply is considering opening a second service counter to better serve the electrical contractor customers. The arrival rate is 10 per hour. The service rate is 14 per hour. If the cost of waiting is $30 and the cost of each service counter is $22 per hour, then should the second counter be opened? ANSWER: With one window the cost per hour is 97. With two windows the cost per hour is 68.56. Choose two windows. POINTS: 1 TOPICS: Economic analysis 53. For an M/G/1 system with λ = 6 and μ = 9, with σ = .03, find a. the probability the system is idle. b. the average length of the queue. c. the average number in the system. ANSWER: a. P0 = .67 b. Lq = .7153 c. L = 1.3819 POINTS: 1 TOPICS: Operating characteristics: M/G/1 54. For an M/G/1 system with λ = 20 and μ = 35, with σ = .005, find a. the probability the system is idle. b. the average length of the queue. c. the average number in the system. ANSWER: a. P0 = .4286 b. Lq = .3926 c. L = .964 POINTS: 1 TOPICS: Operating characteristics: M/G/1 55. Arrivals at a box office in the hour before the show follow the Poisson distribution with λ = 7 per minute. Service Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models times are constant at 7.5 seconds. Find the average length of the waiting line. ANSWER: Lq = 3.0625 POINTS: 1 TOPICS: Operating characteristics: M/D/1 56. The 8 students in a seminar class must come to the professor's office to turn in a paper and give a 5-minute oral summary. Assume there is a service rate of 10 per hour and adequate time is available for all. The arrival rate for each unit is 5 per hour. What is the probability there is no one in the office or waiting when you come? ANSWER: λ = 5, μ = 10, N = 8 P0 = .0009 POINTS: 1 TOPICS: Finite calling population 57. Andy Archer, Ph.D., is a training consultant for six mid-sized manufacturing firms. On the average, each of his six clients calls him for consulting assistance once every 25 days. Andy typically spends an average of five days at the client's firm during each consultation. Assuming that the time between client calls follows an exponential distribution, determine the following: a. the average number of clients Andy has on backlog b. the average time a client must wait before Andy arrives to it c. the proportion of the time Andy is busy ANSWER: a. Lq = .7094 b. Wq = 4.96 days c. 1 − P0 = .7151 POINTS: 1 TOPICS: Finite calling population 58. The Quick Snap photo machine at the Lemon County bus station takes four snapshots in exactly 75 seconds. Customers arrive at the machine according to a Poisson distribution at the mean rate of 20 per hour. On the basis of this information, determine the following: a. the average number of customers waiting to use the photo machine b. the average time a customer spends in the system c. the probability an arriving customer must wait for service. ANSWER: a. Lq = .15 b. W = .028 hour = 1.7 minutes c. PW = .417 POINTS: 1 TOPICS: Constant service times 59. Quick Clean Rooter cleans out clogged drains. Due to the competitive nature of the drain cleaning business, if a customer calls Quick Clean and finds the line busy, they immediately try another company and Quick Clean loses the business. Quick Clean management estimates that on the average, a customer tries to call Quick Clean every three minutes and the average time to take a service order is 200 seconds. The company wishes to hire enough operators so that at most 4% of its potential customers get the busy signal. a. How many operators should be hired to meet this objective? Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models b. Given your answer to a), what is the probability that all the operators are idle? ANSWER: a. k = 4; PW = .021 b. P0 = .331 POINTS: 1 TOPICS: M/G/k model with blocked customers cleared 60. A company has tool cribs where workmen draw parts. Two men have applied for the position of distributing parts to the workmen. George Fuller is fresh out of trade school and expects a $6 per hour salary. His average service time is 4 minutes. John Cox is a veteran who expects $12 per hour. His average service time is 2 minutes. A workman's time is figured at $10 per hour. Workmen arrive to draw parts at an average rate of 12 per hour. What is the average waiting time a workman would spend in the system under each a. applicant? b. Which applicant should be hired? ANSWER: a. George: W = 20 minutes; John: W = 3 1/3 minutes b. Total Cost: George = $46.00, John = $18.67. Hire John POINTS: 1 TOPICS: Economic analysis of waiting lines 61. The insurance department at Shear's has two agents, each working at a mean speed of 8 customers per hour. Customers arrive at the insurance desk at a mean rate of one every six minutes and form a single queue. Management feels that some customers are going to find the wait at the desk too long and take their business to Word's, Shear's competitor. In order to reduce the time required by an agent to serve a customer Shear's is contemplating installing one of two minicomputer systems: System A which leases for $18 per day and will increase an agent's efficiency by 25%; or, System B which leases for $23 per day and will increase an agent's efficiency by 50%. Agents work 8-hour days. If Shear's estimates its cost of having a customer in the system at $3 per hour, determine if Shear's should install a new minicomputer system, and if so, which one. ANSWER: No system: $49.20; System A: $50.00; System B: $47.50. Install System B POINTS: 1 TOPICS: Economic analysis of waiting lines 62. The postmaster at the Oak Hill Post Office expects the mean arrival rate of people to her customer counter will soon increase by fifty percent due to a large apartment complex being built. Currently, the mean arrival rate is 15 people per hour. The postmaster can serve an average of 25 people per hour. By what percentage must the postmaster's mean service rate increase when the apartment complex is completed in order that the average time spent at the post office remains at its current value? ANSWER: 30 percent POINTS: 1 TOPICS: M/M/1 waiting line model Essay 63. Discuss the importance of the utilization factor in a queuing system and the assumptions made about its value. ANSWER: Answer not provided. POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 11 - Waiting Line Models TOPICS: Operating characteristics 64. How can a system be changed to improve the service rate? ANSWER: Answer not provided. POINTS: 1 TOPICS: Improving the waiting line operation 65. Diagram the servers and arrivals in the single and multiple channel models. Designate the line and the system. ANSWER: Answer not provided. POINTS: 1 TOPICS: Multiple channel waiting line 66. Explain what is meant by the following statement, "operating characteristics are non-optimizing." ANSWER: Answer not provided. POINTS: 1 TOPICS: Operating characteristics 67. Give examples of systems you have seen in which a) blocked arrivals are cleared, and b) there is a finite calling population. ANSWER: Answer not provided. POINTS: 1 TOPICS: Blocked channel; finite population 68. List six steady-state operating characteristics for a single-channel waiting line with Poisson arrivals and exponential service times. ANSWER: Answer not provided. POINTS: 1 TOPICS: Operating characteristics

Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation True / False 1. Simulation is an excellent technique to use when a situation is too complicated to use standard analytical procedures. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 2. Simulation is a trial-and-error approach to problem solving. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 3. The degree of risk is associated with the probability or magnitude of loss. a. True b. False ANSWER: True POINTS: 1 TOPICS: Simulation approach 4. To use Excel to generate a normally distributed random variable, you must know the mean and standard deviation of the distribution and have a random number between 0 and 1. a. True b. False ANSWER: True POINTS: 1 TOPICS: Simulation approach 5. Trials of a simulation show what would happen when values of the probabilistic input change. a. True b. False ANSWER: True POINTS: 1 TOPICS: Simulation approach 6. In a Monte Carlo simulation, each simulation trial is dependent upon the result of a previous trial. a. True b. False ANSWER: False POINTS: 1 TOPICS: Monte Carlo simulation 7. Verification is the process of ensuring that the simulation model provides an accurate representation of the real system. Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation a. True b. False ANSWER: False POINTS: 1 TOPICS: Verification and validation 8. In comparing different policies using simulation, one should use the same set of random numbers whenever possible. a. True b. False ANSWER: True POINTS: 1 TOPICS: Waiting line simulation 9. Validation determines that the computer procedure is operating as it is intended to operate. a. True b. False ANSWER: False POINTS: 1 TOPICS: Verification and validation 10. A discrete-event simulation reviews the status of the system periodically, whether or not an event occurs. a. True b. False ANSWER: True POINTS: 1 TOPICS: Waiting line simulation 11. A static simulation model is used in situations where the state of the system affects how the system changes or evolves over time. a. True b. False ANSWER: False POINTS: 1 TOPICS: Waiting line simulation 12. For any waiting line system, (Average number of units in waiting line) = (Total waiting time) divided by (Total time of simulation). a. True b. False ANSWER: True POINTS: 1 TOPICS: Waiting line simulation 13. The parameters of a simulation model are the controllable inputs. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation b. False ANSWER: False POINTS: 1 TOPICS: Controllable inputs 14. Using simulation to perform risk analysis is like playing out many what-if scenarios by randomly generating values for the probabilistic inputs. a. True b. False ANSWER: True POINTS: 1 TOPICS: Risk analysis 15. Computer-generated random numbers are normally distributed over the interval from 0 to 1. a. True b. False ANSWER: False POINTS: 1 TOPICS: Random numbers and generating probabilistic input values 16. Computer-generated random numbers are not technically random. a. True b. False ANSWER: True POINTS: 1 TOPICS: Random numbers and generating probabilistic input values 17. Simulation is an optimization technique. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 18. Simulation models that must take into account how the system changes or evolves over time are referred to as dynamic simulation models. a. True b. False ANSWER: True POINTS: 1 TOPICS: Waiting line simulation 19. Computer-generated random numbers are normally distributed. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation ANSWER: False POINTS: 1 TOPICS: Random numbers and generating probabilistic input values 20. Each simulation run provides only a sample of how the real system will operate. a. True b. False ANSWER: True POINTS: 1 TOPICS: Advantages and disadvantages of using simulation 21. One disadvantage of simulation is that it is limited in the variety of probability distributions that can be used in modeling a system. a. True b. False ANSWER: False POINTS: 1 TOPICS: Simulation 22. A simulation model provides a convenient experimental laboratory for the real system. a. True b. False ANSWER: True POINTS: 1 TOPICS: Advantages and disadvantages of using simulation 23. Simulation allows the user to specify certain desired results (for example, profit or service level values), and then the necessary model parameters and operating policies are determined. a. True b. False ANSWER: False POINTS: 1 TOPICS: Inventory simulation Multiple Choice 24. A simulation model uses the mathematical expressions and logical relationships of the a. real system. b. computer model. c. performance measures. d. estimated inferences. ANSWER: a POINTS: 1 TOPICS: Introduction 25. Values for the probabilistic inputs to a simulation Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation a. are selected by the decision maker. b. are controlled by the decision maker. c. are randomly generated based on historical information. d. are calculated by fixed mathematical formulas. ANSWER: c POINTS: 1 TOPICS: Introduction 26. A quantity that is difficult to measure with certainty is called a a. risk analysis. b. project determinant. c. probabilistic input. d. profit/loss process. ANSWER: c POINTS: 1 TOPICS: Risk analysis 27. A value for probabilistic input from a discrete probability distribution a. is the value given by the RAND() function. b. is given by matching the probabilistic input with an interval of random numbers. c. is between 0 and 1. d. must be non-negative. ANSWER: b POINTS: 1 TOPICS: Simulation approach 28. The number of units expected to be sold is uniformly distributed between 300 and 500. If r is a random number between 0 and 1, then the proper expression for sales is a. 200(r) b. r + 300 c. 300 + 500(r) d. 300 + r(200) ANSWER: d POINTS: 1 TOPICS: Simulation approach 29. When events occur at discrete points in time a. a simulation clock is required. b. the simulation advances to the next event. c. the model is a discrete-event simulation. d. All of the alternatives are correct. ANSWER: d POINTS: 1 TOPICS: Discrete-event simulation Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation 30. If customer 2 has a service time of 1.6, and if customer 3 has an interarrival time of 1.1 and a service time of 2.3, when will customer 3's service be completed? a. 5.0 b. 3.9 c. 3.4 d. There is not enough information to answer. ANSWER: d POINTS: 1 TOPICS: Waiting line simulation 31. Common features of simulations--generating values from probability distributions, maintaining records, recording data and summarizing results--led to the development of a. Excel and Lotus. b. BASIC, FORTRAN, PASCAL, and C. c. GPSS, SIMSCRIPT, SLAM, and Arena d. LINDO and The Management Scientist ANSWER: c POINTS: 1 TOPICS: Computer implementation 32. In order to verify a simulation model a. compare results from several simulation languages. b. be sure that the procedures for calculations are logically correct. c. confirm that the model accurately represents the real system. d. run the model long enough to overcome initial start-up results. ANSWER: b POINTS: 1 TOPICS: Verification and validation 33. Simulation a. does not guarantee optimality. b. is flexible and does not require the assumptions of theoretical models. c. allows testing of the system without affecting the real system. d. All of the alternatives are correct. ANSWER: d POINTS: 1 TOPICS: Advantages and disadvantages 34. A simulation model used in situations where the state of the system at one point in time does not affect the state of the system at future points in time is called a a. dynamic simulation model. b. static simulation model. c. steady-state simulation model. d. discrete-event simulation model. ANSWER: b Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation POINTS: 1 TOPICS: Static simulation models 35. The process of determining that the computer procedure that performs the simulation calculations is logically correct is called a. implementation. b. validation. c. verification. d. repetition. ANSWER: c POINTS: 1 TOPICS: Other simulation issues 36. Numerical values that appear in the mathematical relationships of a model and are considered known and remain constant over all trials of a simulation are a. parameters. b. probabilistic input. c. controllable input. d. events. ANSWER: a POINTS: 1 TOPICS: Other simulation issues 37. Which of the following statements is INCORRECT regarding the disadvantages of simulation? a. Each simulation run only provides a sample of how the real system will operate. b. The summary of the simulation data only provides estimates about the real system. c. The process of developing a simulation model of a complex system can be time-consuming. d. The larger the number of probabilistic inputs a system has, the less likely a simulation will provide the best approach for studying the system. ANSWER: d POINTS: 1 TOPICS: Advantages and disadvantages of using simulation 38. Which of the following statements is INCORRECT regarding the advantages of simulation? a. Simulation is relatively easy to explain and understand. b. Simulation guarantees an optimal solution. c. Simulation models are flexible. d. A simulation model provides a convenient experimental laboratory for the real system. ANSWER: b POINTS: 1 TOPICS: Advantages and disadvantages of using simulation 39. The word "uniform" in the term "uniform random numbers" means a. all the numbers have the same number of digits. b. if one number is, say, 10 units above the mean, the next number will be 10 units below the mean. Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation c. all the numbers are odd or all are even. d. each number has an equal probability of being drawn. ANSWER: d POINTS: 1 TOPICS: Random numbers and generating probabilistic input values 40. A table of uniformly distributed random numbers should be read a. from left to right. b. from top to bottom. c. diagonally, starting from the top left corner and moving to the bottom right. d. in any consistent sequence. ANSWER: d POINTS: 1 TOPICS: Random numbers and generating probabilistic input values 41. The process of generating probabilistic inputs and computing the value of the output is called a. simulation. b. verification. c. validation. d. implementation. ANSWER: a POINTS: 1 TOPICS: Simulation 42. A graphical tool that helps describe the logic of the simulation model is a a. Gantt chart. b. histogram. c. flowchart. d. stem-and-leaf display. ANSWER: c POINTS: 1 Subjective Short Answer 43. For the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the nearest 10). Units Sold 30 40 50 60 70

Number of Times 8 12 15 10 5

a. Determine the relative frequency for each number of units sold. b. Suppose that the following random numbers were obtained using Excel: .12 .96 .53 .80 .95 .10 .40 .45 .77 .29 Use these random numbers to simulate 10 days of sales. Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation ANSWER: a.

Units Sold 30 40 50 60 70

Relative Frequency .12 .24 .32 .26 .06

Cumulative Frequency .12 .36 .68 .94 1.00

Interval of Random Numbers 0.00 but less than 0.12 0.12 but less than 0.36 0.36 but less than 0.68 0.68 but less than 0.94 0.94 but less than 1.00

b. Random Number Units Sold .12 40 .96 70 .53 50 .80 60 .95 70 .10 30 .40 50 .45 50 .77 60 .29 40 POINTS: 1 TOPICS: Simulation approach: discrete-event 44. The drying rate in an industrial process is dependent on many factors and varies according to the following distribution. Minutes 3 4 5 6 7

Relative Frequency .14 .30 .27 .18 .11

a. Compute the mean drying time. b. Using these random numbers, simulate the drying time for 12 processes. .33 .09 .19 .81 .12 .88 .53 .95 .77 .61 .91 .47 c. What is the average drying time for the 10 processes you simulated? ANSWER: a. 4.82 b. Random Number Drying Time .33 4 .09 3 .19 4 .81 6 .12 3 .88 6 .53 5 .95 7 .77 6 .61 5 .91 7 Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation .47

c. The average is 5.083 POINTS: 1 TOPICS: Simulation approach: discrete-event 45. Greenfields is a mail order seed and plant business. The size of orders is uniformly distributed over the interval from $25 to $80. Use the following random numbers to generate the size of 10 orders. .41 .99 .07 .05 .38 .77 .19 ANSWER: Random Number Order Size 0.41 47.55 0.99 79.45 0.07 28.85 0.05 27.75 0.38 45.90 0.77 67.35 0.19 35.45 0.12 31.60 0.58 56.90 0.60 58.00 POINTS: 1 TOPICS: Simulation approach: uniform distribution

.12

.58

.60

46. The time required to set up lighting for a portrait studio is uniformly distributed between 12 and 20 minutes. Use the following random numbers to generate the setup time for 10 customers. .27 .53 .06 .92 .16 .74 .06 ANSWER: Random Number Time 0.27 14.16 0.53 16.24 0.06 12.48 0.92 19.36 0.16 13.28 0.74 17.92 0.06 12.48 0.29 14.32 0.82 18.56 0.23 13.84 POINTS: 1 TOPICS: Simulation approach: uniform distribution

.29

.82

.23

47. Estimates of the financial information for a new product show the following information: Units Sold 600 800 1000

Probability .35 .45 .20

Fixed cost Variable cost Revenue

$8,000 $6 / unit $22 / unit

Use the random numbers .51, .97, .58, .22, and .16 to simulate five trials. What is the net profit for each trial? Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation ANSWER:

Trial 1 Random Number 0.51 Units sold 800 Revenue 17600 Variable cost 4800 Fixed cost 8000 Net Profit 4800 POINTS: 1 TOPICS: Simulation approach: profit

2 0.97 1000 22000 6000 8000 8000

3 0.58 800 17600 4800 8000 4800

4 0.22 600 13200 3600 8000 1600

5 0.16 600 13200 3600 8000 1600

48. Seventy-five percent of calls arriving at a help line can be handled by the person who answers the phone, but the remaining 25% of them will need to be referred to someone else. Assume that every call requires one minute of attention by the person who answers the phone (either to answer the question or to figure out how the referral should be handled). Calls that are referred need an additional amount of time, as given in the table below. Time Required Probability 3 minutes .25 5 minutes .35 10 minutes .40 Callers are served on a first come, first served basis, and are put on hold until the line is free. Use the random numbers to simulate what happens to 10 callers. (Use the random numbers in order - from left to right, first row first - as you need them.) What percentage of your callers needs to be referred? Of those who had to be referred, what is the average referral time? .82 .39 .85 .32 ANSWER:

.16 .64 Call 1 2 3 4 5 6 7 8 9 10

.79 .90 RN 0.82 0.16 0.79 0.62 0.13 0.04 0.42 0.81 0.85 0.64

.56 .73

.62 .02

Referred? yes no yes no no no no yes yes no

.13 .76

.04 .03

.42 .86

RN 0.39

Time 5

0.56

0.85 0.32

10 5

.81 .67

40% of the callers were referred. The average referral time was 6.25 minutes. POINTS: 1 TOPICS: Simulation approach: two events 49. An airline reservation system first asks customers whether they want to schedule a domestic or an international flight. Sixty-five percent of the reservations are for domestic flights. The time distribution of advance sales is also important, and it is given below. Domestic Flights Make Reservations Rel. Freq. Less than 1 week .25 in advance 1 week to 2 months .35 Cengage Learning Testing, Powered by Cognero

RN Range

International Flights Make Reservations Rel. Freq. Less than 1 week .12 in advance 1 week to 2 months .35

RN Range

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Chapter 12 - Simulation in advance Over 2 months in advance

Flight Type Domestic International

in advance 2 months to 6 months in advance Over 6 months In advance

.40

Rel. Freq. .65 .35

.40 .13

RN Range

Place the appropriate random number ranges in the tables above. Set up and perform a simulation for three customers. Determine whether they want a b. domestic or international flight, and how far in advance the reservation is being made. Use random numbers from this list: .632 .715 .998 .671 .744 .021 ANSWER: Domestic Flights International Flights a. Make Rel. Reservations Freq. Less than 1 week .25 in advance 1 week to 2 months .35 in advance Over 2 months In advance

Make Reservations

Rel. Freq.

RN Range

.00 to < .25

Less than 1 week in advance

.12

.00 to < .12

.25 to < .60

1 week to 2 months in advance

.35

.12 to < .47

.40

.47 to < .87

.13

.87 to < 1.00

.6 to < 1.00

Domestic International

Rel. Freq. .65 .35

.00 to < .65 .65 to < 1.00

Observation 1 2 3

RN .632 .998 .774

Flight Type Domestic International International

Flight Type

.40

RN Range

2 months to 6 months in advance Over 6 months In advance

RN Range

RN Booking .715 Over 2 months .671 2 to 6 months .021 Less than 1 week

POINTS: 1 TOPICS: Simulation approach: two events 50. On a visit to an amusement park you pass someone who has just ridden a roller coaster and asks you for directions to the First Aid Station. Realizing that traffic at the First Aid Station would be something to study with simulation, you gather some information. Two EMTs staff the station, and patients wait and go to the first one available. People coming there can be divided into two groups: those who need something minor (e.g. Tylenol, a band-aid) or those who need more help. Assume those in the first group constitute 25% of the patients and take 5 minutes to have their problem solved. Those in the second group need an uncertain amount of time, as given by a probability distribution. Develop a flowchart for this simulation problem. ANSWER: Let i = patient counter AT(i) = arrival time IAT(i) = interarrival time Start(i) = time patient i begins service ST(i) = service time Finish(i) = time patient leaves Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation CT1 = time that EMT 1 completes current service CT2 = time that EMT 2 completes current service

POINTS: 1 TOPICS: Simulation flowchart 51. Using the spreadsheet below, give the cell address which would have the formula shown. Cell Formula =VLOOKUP(B18,$B$10:$C$12,2) =VLOOKUP(D23,$F$11:$G$14,2) Cengage Learning Testing, Powered by Cognero

Belongs in Cell

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Chapter 12 - Simulation =K19*($I$16-I19) =VLOOKUP(H27,$B$10:$C$12,2) =AVERAGE(L18:L27)

ANSWER: Cell Formula =VLOOKUP(B18,$B$10:$C$12,2) =VLOOKUP(D23,$F$11:$G$14,2) =K19*($I$16-I19) =VLOOKUP(H27,$B$10:$C$12,2) =AVERAGE(L18:L27)

Belongs in Cell C18 E23 L19 I27 L28

POINTS: 1 TOPICS: Simulation with Excel 52. As the owner of a rent-a-car agency you have determined the following statistics: Potential Rental Probability Probability Rentals Daily Duration 0 .10 1 day .50 1 .15 2 day .30 2 .20 3 days .15 3 .30 4 days .05 4 .25 The gross profit is $40 per car per day rented. When there is demand for a car when none is available there is a goodwill loss of $80 and the rental is lost. Each day a car is unused costs you $5 per car. Your firm initially has 4 cars. a. Conduct a 10-day simulation of this business using Row #1 below for demand and Row #2 below for rental length. Row #1: Row #2:

63 59

88 09

55 57

46 87

55 07

69 92

13 29

17 28

36 64

81 36

b. If your firm can obtain another car for $200 for 10 days, should you take the extra car? Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation ANSWER:

a. 10-day profit is $885 b. 10-day profit is $970. Take the extra car. POINTS: 1 TOPICS: Simulation approach: two events 53. Arrivals to a truck repair facility have an interarrival time that is uniformly distributed between 20 and 50 minutes. Service times are normally distributed with mean 30 minutes and standard deviation 10 minutes. Develop a spreadsheet model to simulate the arrival of 100 trucks. Collect information on the time the repair facility is idle and on the average waiting time for trucks. ANSWER: NOTE TO INSTRUCTOR: The problem requires the use of a computer and is suitable for a lab or take-home exam. Representative formulas for cell references are shown in comments.

POINTS: 1 TOPICS: Spreadsheet simulation, uniform and normal distributions 54. Susan Winslow has two alternative routes to travel from her home in Olport to her office in Lewisburg. She can travel on Freeway 5 to Freeway 57 or on Freeway 55 to Freeway 91. The time distributions are as follows: Freeway 5 Relative Time Frequency 5 .30 6 .20 7 .40 8 .10

Freeway 57 Relative Time Frequency 4 .10 5 .20 6 .35 7 .20 8 .15

Cengage Learning Testing, Powered by Cognero

Freeway 55 Relative Time Frequency 6 .20 7 .20 8 .40 9 .20

Freeway 91 Relative Time Frequency 3 .30 4 .35 5 .20 6 .15 Page 15

Chapter 12 - Simulation Do a five-day simulation of each of the two combinations of routes using the random numbers below. Based on this simulation, which routes should Susan take if her objective is to minimize her total travel time? Freeway 5 63 88 55 46 55 Freeway 57 59 09 57 87 07 Freeway 55 71 95 83 44 34 Freeway 91 51 79 09 67 15 ANSWER: Freeways 5-57 have 62 minutes; Freeways 55-91 have 61 minutes; Select 55-91. POINTS: 1 TOPICS: Simulation approach: multiple events 55. Three airlines compete on the route between New York and Los Angeles. Stanton Marketing has performed an analysis of first class business travelers to determine their airline choice. Stanton has modeled this choice as a Markov process and has determined the following transition probabilities. Last Airline A B C

A .50 .30 .10

Next Airline B .30 .45 .35

C .20 .25 .55

a. Show the random number assignments that can be used to simulate the first class business traveler's next airline when her last airline is A, B, and C. b. Assume the traveler used airline C last. Simulate which airline the traveler will be using over her next 25 flights. What percent o her flights are on each of the three airlines? Use the following random numbers, going from left to right, top to bottom. 71 49 75 77 37 ANSWER:

95 88 12 76 46 a.

44 05 59 15 24

A Last A Next B Next C Next

83 56 03 57 85

34 39 29 53 53 B Last

RNs 00 - 49 50 - 79 80 - 99

A Next B Next C Next

C Last RNs 00 - 29 30 - 74 75 - 99

A Next B Next C Next

RNs 00 - 09 10 - 44 45 - 99

20% of flights on Airline A 48% of flights on Airline B 32% of flights on Airline C

POINTS: 1 TOPICS: Dynamic simulation models 56. Attendees at the National Management Science Society (NMSS) Conference register by first standing in line to pay their fees. They then proceed to a designated line based on the first letter of their last name to collect their conference materials. At the conference, it is planned to have three different parallel lines for the collection of materials: one each for people whose last names begin with A-H, I-Q, and R-Z respectively. During each minute of the morning registration period it is anticipated that attendees will arrive to pay their fees according Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation to the following distribution: Number of Arrivals 0 1 2 3

Probability .30 .30 .30 .10

The time to pay one's fees is either one minute or two minutes depending upon whether one uses a check or credit card. The probability of a one-minute time is .60. After paying his fees, an attendee then goes to the correct line for the conference materials. At this year's conference 35% of the attendees have last names beginning with A-H, 36% with last names beginning with I-Q, and 29% with last names beginning with R-Z. The time required to pick up conference materials is fixed at 2 minutes. a. Simulate the waiting line for the first 10 minutes of the morning registration that begins at 8:00 AM. Use the following random numbers to generate: Number of arrivals in any given minute: 71, 95, 83, 44, 34, 49, 88, 56, 05, 39 Registration fee service time: 51, 79, 09, 67, 15, 58, 04, 78, 30, 56 First letter of the last name: 15, 08, 19, 45, 76, 42, 38, 47, 82, 37 b. What is the average size of the waiting line to pay fees (not including the person being served), and the average customer waiting time to pay fees based on this simulation? ANSWER: ARRIVALS FEE PAY CONFERENCE MATERIAL Time RN # Q I Wait Begin RN Time End RN Name Wait Begin End 8:00 71 2 0 1 0 8:00 51 1 8:01 15 A-H 0 8:01 8:03 2 1 8:01 79 2 8:03 08 A-H 0 8:03 8:05 8:01 95 3 1 3 2 8:03 09 1 8:04 19 A-H 1 8:05 8:07 4 3 8:04 67 2 8:06 45 I-Q 0 8:06 8:08 5 5 8:06 15 1 8:07 76 R-Z 0 8:07 8:09 8:02 83 2 3 6 5 8:07 58 1 8:08 42 I-Q 0 8:08 8:10 7 6 8:08 04 1 8:09 38 I-Q 1 8:10 8:12 8:03 44 1 5 8 6 8:09 78 2 8:11 47 I-Q 1 8:12 8:14 8:04 34 1 5 9 7 8:11 30 1 8:12 82 R-Z 0 8:12 8:14 8:05 49 1 5 10 7 8:12 56 1 8:13 37 I-Q 1 8:14 8:16 8:06 88 2 6 11 7 8:13 75 2 8:15 49 I-Q 1 8:16 8:18 12 9 8:15 75 2 8:17 43 I-Q 1 8:18 8:20 8:07 56 1 7 13 10 8:17 05 1 8:18 37 I-Q 2 8:20 8:22 8:08 05 0 7 8:09 39 1 6 14 9 8:18 49 1 8:19 11 A-H 0 8:19 8:21 8:10 6 ----------------------------- Simulation Over ---------------------10 14 51 77 19 b. Avg. Number in Fee Wait Line = 51/10 = 5.1 people Avg. Fee Wait Time = 77/14 = 5.5 minutes POINTS: 1 TOPICS: Simulation approach: multiple events Essay 57. Simulation is to be used to study customer waiting patterns at several branches of an organization. Acknowledging that arrivals and service times follow different distributions over the branches, of what use is the development of a Cengage Learning Testing, Powered by Cognero

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Chapter 12 - Simulation general simulation model? ANSWER: Answer not provided. POINTS: 1 TOPICS: Advantages and disadvantages 58. Why would one want to use a general purpose programming language rather than a spreadsheet to develop a simulation? ANSWER: Answer not provided. POINTS: 1 TOPICS: Computer implementation 59. How are both analysts and managers involved in the validation process? ANSWER: Answer not provided. POINTS: 1 TOPICS: Verification and validation 60. How can historical information be used to create discrete probability distributions? ANSWER: Answer not provided. POINTS: 1 TOPICS: Simulation approach 61. Why is a flowchart useful in simulation? ANSWER: Answer not provided. POINTS: 1 TOPICS: Simulation approach 62. Explain the difference between verification and validation as they relate to a simulation model. ANSWER: Answer not provided. POINTS: 1 TOPICS: Other simulation issues

Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis True / False 1. Sample information with an efficiency rating of 100% is perfect information. a. True b. False ANSWER: True POINTS: 1 TOPICS: Efficiency of sample information 2. States of nature should be defined so that one and only one will actually occur. a. True b. False ANSWER: True POINTS: 1 TOPICS: Structuring the decision process 3. Decision alternatives are structured so that several could occur simultaneously. a. True b. False ANSWER: False POINTS: 1 TOPICS: Structuring the decision problem 4. Square nodes in a decision tree indicate that a decision must be made. a. True b. False ANSWER: True POINTS: 1 TOPICS: Decision trees 5. Circular nodes in a decision tree indicate that it would be incorrect to choose a path from the node. a. True b. False ANSWER: True POINTS: 1 TOPICS: Decision trees 6. Risk analysis helps the decision maker recognize the difference between the expected value of a decision alternative and the payoff that may actually occur. a. True b. False ANSWER: True POINTS: 1 TOPICS: Risk analysis 7. The expected value of an alternative can never be negative. Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis a. True b. False ANSWER: False POINTS: 1 TOPICS: Decision making with probabilities 8. Expected value is the sum of the weighted payoff possibilities at a circular node in a decision tree. a. True b. False ANSWER: True POINTS: 1 TOPICS: Decision making with probabilities 9. EVPI is always greater than or equal to EVSI. a. True b. False ANSWER: True POINTS: 1 TOPICS: Expected value of sample information 10. After all probabilities and payoffs are placed on a decision tree, the decision maker calculates expected values at state of nature nodes and makes selections at decision nodes. a. True b. False ANSWER: True POINTS: 1 TOPICS: Developing a decision strategy 11. A decision strategy is a sequence of decisions and chance outcomes, where the decisions chosen depend on the yet to be determined outcomes of chance events. a. True b. False ANSWER: True POINTS: 1 TOPICS: Decision strategy 12. EVPI equals the expected regret associated with the minimax decision. a. True b. False ANSWER: True POINTS: 1 TOPICS: Minimax regret approach 13. The expected value approach is more appropriate for a one-time decision than a repetitive decision. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis b. False ANSWER: False POINTS: 1 TOPICS: Decision making with probabilities 14. Maximizing the expected payoff and minimizing the expected opportunity loss result in the same recommended decision. a. True b. False ANSWER: True POINTS: 1 TOPICS: Minimax regret approach 15. The expected value of sample information can never be less than the expected value of perfect information. a. True b. False ANSWER: False POINTS: 1 TOPICS: Expected value of sample information 16. The minimum expected opportunity loss provides the best decision, regardless of whether the decision analysis involves minimization or maximization. a. True b. False ANSWER: True POINTS: 1 TOPICS: Opportunity loss 17. The primary value of decision trees is as a useful way of organizing how operations managers think about complex multiphase decisions. a. True b. False ANSWER: True POINTS: 1 TOPICS: Decision trees 18. A high efficiency rating indicates that the sample information is almost as good as perfect information. a. True b. False ANSWER: True POINTS: 1 TOPICS: Efficiency of sample information 19. When the expected value approach is used to select a decision alternative, the payoff that actually occurs will usually have a value different from the expected value. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis b. False ANSWER: True POINTS: 1 TOPICS: Decision making with probabilities 20. The decision alternative with the best expected monetary value will always be the most desirable decision. a. True b. False ANSWER: False POINTS: 1 TOPICS: Meaning of utility 21. When monetary value is not the sole measure of the true worth of the outcome to the decision maker, monetary value should be replaced by utility. a. True b. False ANSWER: True POINTS: 1 TOPICS: Meaning of utility 22. The outcome with the highest payoff will also have the highest utility. a. True b. False ANSWER: True POINTS: 1 TOPICS: Developing utilities for monetary payoffs 23. Expected utility is a particularly useful tool when payoffs stay in a range considered reasonable by the decision maker. a. True b. False ANSWER: False POINTS: 1 TOPICS: Meaning of utility 24. To assign utilities, consider the best and worst payoffs in the entire decision situation. a. True b. False ANSWER: True POINTS: 1 TOPICS: Developing utilities for monetary payoffs 25. A risk avoider will have a concave utility function. a. True b. False ANSWER: True Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis POINTS: 1 TOPICS: Developing utilities for monetary payoffs 26. The expected utility is the utility of the expected monetary value. a. True b. False ANSWER: False POINTS: 1 TOPICS: Expected utility approach 27. The risk premium is never negative for a conservative decision maker. a. True b. False ANSWER: True POINTS: 1 TOPICS: Developing utilities for monetary payoffs 28. The risk neutral decision maker will have the same indications from the expected value and expected utility approaches. a. True b. False ANSWER: True POINTS: 1 TOPICS: Expected monetary value versus expected utility 29. The utility function for a risk avoider typically shows a diminishing marginal return for money. a. True b. False ANSWER: True POINTS: 1 TOPICS: Developing utilities for monetary payoffs 30. The expected monetary value approach and the expected utility approach to decision making usually result in the same decision choice unless extreme payoffs are involved. a. True b. False ANSWER: True POINTS: 1 TOPICS: Utility and decision making 31. A risk neutral decision maker will have a linear utility function. a. True b. False ANSWER: True POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Developing utilities for monetary payoffs 32. Given two decision makers, one risk neutral and the other a risk avoider, the risk avoider will always give a lower utility value for a given outcome. a. True b. False ANSWER: False POINTS: 1 TOPICS: Developing utilities for monetary payoffs 33. When the payoffs become extreme, most decision makers are satisfied with the decision that provides the best expected monetary value. a. True b. False ANSWER: False POINTS: 1 TOPICS: Meaning of utility Multiple Choice 34. The options from which a decision maker chooses a course of action are a. called the decision alternatives. b. under the control of the decision maker. c. not the same as the states of nature. d. All of the alternatives are true. ANSWER: d POINTS: 1 TOPICS: Structuring the decision problem 35. States of nature a. can describe uncontrollable natural events such as floods or freezing temperatures. b. can be selected by the decision maker. c. cannot be enumerated by the decision maker. d. All of the alternatives are true. ANSWER: a POINTS: 1 TOPICS: Structuring the decision problem 36. A payoff a. is always measured in profit. b. is always measured in cost. c. exists for each pair of decision alternative and state of nature. d. exists for each state of nature. ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Payoff tables 37. Making a good decision a. requires probabilities for all states of nature. b. requires a clear understanding of decision alternatives, states of nature, and payoffs. c. implies that a desirable outcome will occur. d. All of the alternatives are true. ANSWER: b POINTS: 1 TOPICS: Decision making without probabilities 38. A decision tree a. presents all decision alternatives first and follows them with all states of nature. b. presents all states of nature first and follows them with all decision alternatives. c. alternates the decision alternatives and states of nature. d. arranges decision alternatives and states of nature in their natural chronological order. ANSWER: d POINTS: 1 TOPICS: Decision trees 39. Which of the methods for decision making best protects the decision maker from undesirable results? a. the optimistic approach b. the conservative approach c. minimum regret d. minimax regret ANSWER: b POINTS: 1 TOPICS: Conservative approach 40. Sensitivity analysis considers a. how sensitive the decision maker is to risk. b. changes in the number of states of nature. c. changes in the values of the payoffs. d. changes in the available alternatives. ANSWER: c POINTS: 1 TOPICS: Sensitivity analysis 41. To find the EVSI, a. use the EVPI to calculate sample information probabilities. b. use indicator probabilities to calculate prior probabilities. c. use prior and sample information probabilities to calculate revised probabilities. d. use sample information to revise the sample information probabilities. ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Expected value of sample information 42. If P(high) = .3, P(low) = .7, P(favorable | high) = .9, and P(unfavorable | low) = .6, then P(favorable) = a. .10 b. .27 c. .30 d. .55 ANSWER: d POINTS: 1 TOPICS: Conditional probability 43. The efficiency of sample information is a. EVSI*(100%) b. EVSI/EVPI*(100%) c. EVwoSI/EVwoPI*(100%) d. EVwSI/EVwoSI*(100%) ANSWER: b POINTS: 1 TOPICS: Efficiency of sample information 44. Decision tree probabilities refer to a. the probability of finding the optimal strategy b. the probability of the decision being made c. the probability of overlooked choices d. the probability of an uncertain event occurring ANSWER: d POINTS: 1 TOPICS: Decision trees 45. For a maximization problem, the conservative approach is often referred to as the a. minimax approach. b. maximin approach. c. maximax approach. d. minimin approach. ANSWER: b POINTS: 1 TOPICS: Decision making without probabilities 46. For a minimization problem, the optimistic approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Decision making without probabilities 47. For a maximization problem, the optimistic approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: c POINTS: 1 TOPICS: Decision making without probabilities 48. For a minimization problem, the conservative approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: a POINTS: 1 TOPICS: Decision making without probabilities 49. In an influence diagram, decision nodes are represented by a. circles or ovals b. squares or rectangles c. diamonds d. triangles ANSWER: b POINTS: 1 TOPICS: Influence diagram 50. Which of the following approaches to decision making requires knowledge of the probabilities of the states of nature? a. minimax regret b. maximin c. expected value d. conservative ANSWER: c POINTS: 1 TOPICS: Decision making with probabilities 51. Decision tree probabilities refer to the probability of a. an uncertain event occurring. b. the decision being made. c. finding an optimal value. d. overlooked choices. ANSWER: a POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Decision making with probabilities 52. Which of the following is not an advantage of using decision tree analysis? a. the ability to see clearly what decisions must be made b. the ability to see clearly in what sequence the decisions must occur c. the ability to see clearly the interdependence of decisions d. the ability to see clearly the future outcome of a decision ANSWER: d POINTS: 1 TOPICS: Decision making with probabilities 53. A decision tree provides a. a heuristic method for analyzing decisions. b. a deterministic approach to decision analysis. c. the absolute value of the decision. d. an objective way of determining the relative value of each decision alternative. ANSWER: d POINTS: 1 TOPICS: Decision trees 54. The approach to determining the optimal decision strategy involves a. a forward (left to right) pass through the decision tree. b. a backward (right to left) pass through the decision tree. c. choosing the outcome of a chance event with the greatest probability. d. choosing the outcome of a chance event with the greatest payoff. ANSWER: b POINTS: 1 TOPICS: Decision strategy 55. The difference between the expected value of an optimal strategy based on sample information and the "best" expected value without any sample information is called the a. information sensitivity. b. expected value of sample information. c. expected value of perfect information. d. efficiency of sample information. ANSWER: b POINTS: 1 TOPICS: Expected value of sample information 56. When consequences are measured on a scale that reflects a decision maker's attitude toward profit, loss, and risk, payoffs are replaced by a. utility values. b. multicriteria measures. c. sample information. d. opportunity loss. Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis ANSWER: a POINTS: 1 TOPICS: Meaning of utility 57. The purchase of insurance and lottery tickets shows that people make decisions based on a. expected value. b. sample information. c. utility. d. maximum likelihood. ANSWER: c POINTS: 1 TOPICS: Meaning of utility 58. The expected utility approach a. does not require probabilities. b. leads to the same decision as the expected value approach. c. is most useful when excessively large or small payoffs are possible. d. requires a decision tree. ANSWER: c POINTS: 1 TOPICS: Expected utility approach 59. Utility reflects the decision maker's attitude toward a. probability and profit. b. profit, loss, and risk. c. risk and regret. d. probability and regret. ANSWER: b POINTS: 1 TOPICS: Meaning of utility 60. Values of utility a. must be between 0 and 1. b. must be between 0 and 10. c. must be nonnegative. d. must increase as the payoff improves. ANSWER: d POINTS: 1 TOPICS: Utility functions 61. If the payoff from outcome A is twice the payoff from outcome B, then the ratio of these utilities will be a. 2 to 1. b. less than 2 to 1. c. more than 2 to 1. d. unknown without further information. Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis ANSWER: d POINTS: 1 TOPICS: Utility functions 62. The probability for which a decision maker cannot choose between a certain amount and a lottery based on that probability is a. the indifference probability. b. the lottery probability. c. the uncertain probability. d. the utility probability. ANSWER: a POINTS: 1 TOPICS: Utility functions 63. A decision maker has chosen .4 as the probability for which he cannot choose between a certain loss of 10,000 and the lottery p(−25000) + (1 − p)(5000). If the utility of −25,000 is 0 and of 5000 is 1, then the utility of −10,000 is a. .5 b. .6 c. .4 d. 4 ANSWER: b POINTS: 1 TOPICS: Utility functions 64. When the decision maker prefers a guaranteed payoff value that is smaller than the expected value of the lottery, the decision maker is a. a risk avoider. b. a risk taker. c. an optimist. d. an optimizer. ANSWER: a POINTS: 1 TOPICS: Risk avoiders versus risk takers 65. A decision maker whose utility function graphs as a straight line is a. conservative. b. a risk taker. c. risk neutral. d. a risk avoider. ANSWER: c POINTS: 1 TOPICS: Risk avoiders versus risk takers 66. When the utility function for a risk-neutral decision maker is graphed (with monetary value on the horizontal axis and utility on the vertical axis), the function appears as Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis a. a convex curve. b. a concave curve. c. an 'S' curve. d. a straight line. ANSWER: d POINTS: 1 TOPICS: Risk avoiders versus risk takers Subjective Short Answer 67. Jim has been employed at Gold Key Realty at a salary of $2,000 per month during the past year. Because Jim is considered to be a top salesman, the manager of Gold Key is offering him one of three salary plans for the next year: (1) a 25% raise to $2,500 per month; (2) a base salary of $1,000 plus $600 per house sold; or, (3) a straight commission of $1,000 per house sold. Over the past year, Jim has sold up to 6 homes in a month. a. Compute the monthly salary payoff table for Jim. b. For this payoff table find Jim's optimal decision using: (1) the conservative approach, (2) minimax regret approach. c. Suppose that during the past year the following is Jim's distribution of home sales. If one assumes that this a typical distribution for Jim's monthly sales, which salary plan should Jim select? Home Sales 0 1 2 3 4 5 6 ANSWER: a.

Number of Months 1 2 1 2 1 3 2 There are three decision alternatives (salary plans) and seven states of nature (the number of houses sold monthly). PAYOFF TABLE Number of Homes Sold 0 1 2 3 4 5 6 Salary Plan I 2500 2500 2500 2500 2500 2500 2500 Salary Plan II 1000 1600 2200 2800 3400 4000 4600 Salary Plan III 0 1000 2000 3000 4000 5000 6000 (1) Conservative Approach (Maximin): Plan I REGRET TABLE 0 1 Salary Plan I 0 0 Salary Plan II 1500 900 Salary Plan III 2500 1500 (2) Minimax Regret Approach: Plan II

Number of Homes Sold 2 3 4 0 500 1500 300 200 600 500 0 0

5 2500 1000 0

6 3500 1400 0

Use the relative frequency method for determining the probabilities. Using the EV approach: EV(Plan I) = 2500, EV(Plan II) = 3050, EV(Plan III) = 3417; Choose Plan III.

POINTS: 1 TOPICS: Decision making with and without probabilities Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis 68. East West Distributing is in the process of trying to determine where they should schedule next year's production of a popular line of kitchen utensils that they distribute. Manufacturers in four different countries have submitted bids to East West. However, a pending trade bill in Congress will greatly affect the cost to East West due to proposed tariffs, favorable trading status, etc. After careful analysis, East West has determined the following cost breakdown for the four manufacturers (in $1,000's) based on whether or not the trade bill passes: Country A Country B Country C Country D

Bill Passes 260 320 240 275

Bill Fails 210 160 240 210

If East West estimates that there is a 40% chance of the bill passing, which country should they choose for manufacturing? b. Over what range of values for the "bill passing" will the solution in part (a) remain optimal? ANSWER: Using EV approach: EV(A) = 230, EV(B) = 224, EV(C) = 240; Choose Country B (lowest a. EV) As long as the probability of the bill passing is less than .455, East West should choose b. Country B. a.

POINTS: 1 TOPICS: Decision making with probabilities 69. Transrail is bidding on a project that it figures will cost $400,000 to perform. Using a 25% markup, it will charge $500,000, netting a profit of $100,000. However, it has been learned that another company, Rail Freight, is also considering bidding on the project. If Rail Freight does submit a bid, it figures to be a bid of about $470,000. Transrail really wants this project and is considering a bid with only a 15% markup to $460,000 to ensure winning regardless of whether or not Rail Freight submits a bid. a. Prepare a profit payoff table from Transrail's point of view. b. What decision would be made if Transrail were conservative? c. If Rail Freight is known to submit bids on only 25% of the projects it considers, what Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis decision should Transrail make? Given the information in (c), how much would a corporate spy be worth to Transrail to find d. out if Rail Freight will bid? ANSWER: a. Rail Freight Transrail Bid $470,000 Doesn't Bid Bid $500,000 $0 $100,000 Bid $460,000 $60,000 $ 60,000 b. Bid $460,000 c. Bid $500,000 d. $15,000 POINTS: 1 TOPICS: Decision making with and without probabilities 70. The Super Cola Company must decide whether or not to introduce a new diet soft drink. Management feels that if it does introduce the diet soda it will yield a profit of $1 million if sales are around 100 million, a profit of $200,000 if sales are around 50 million, or it will lose $2 million if sales are only around 1 million bottles. If Super Cola does not market the new diet soda, it will suffer a loss of $400,000. a. Construct a payoff table for this problem. b. Construct a regret table for this problem. Should Super Cola introduce the soda if the company: (1) is conservative; (2) is optimistic; c. (3) wants to minimize its maximum disappointment? An internal marketing research study has found P(100 million in sales) = 1/3; P(50 million in d. sales) = 1/2; P(1 million in sales) = 1/6. Should Super Cola introduce the new diet soda? A consulting firm can perform a more thorough study for $275,000. Should management e. have this study performed? ANSWER: PAYOFF a. Sales ($millions) TABLE 100 50 1 Introduce $1,000,000 $200,000 −$2,000,000 Do Not −$400,000 −$400,000 −$400,000 Introduce b.

REGRET TABLE

Sales ($millions)

Introduce Do Not Introduce

100 $0

50 $0

1 $1,600,000

$1,400,000

$600,000

c. (1) do not introduce; (2) introduce; (3) do not introduce d. Yes e. No POINTS: 1 TOPICS: Decision making with and without probabilities 71. Super Cola is also considering the introduction of a root beer drink. The company feels that the probability that the product will be a success is .6. The payoff table is as follows: Success (s1) Cengage Learning Testing, Powered by Cognero

Failure (s2) Page 15

Chapter 13 - Decision Analysis Produce (d1) Do Not Produce (d2)

$250,000 −$ 50,000

−$300,000 −$ 20,000

The company has a choice of two research firms to obtain information for this product. Stanton Marketing has market indicators, I1 and I2 for which P(I1 | s1) = .7 and P(I1 | s2) = .4. New World Marketing has indicators J1 and J2 for which P(J1 | s1) = .6 and P(J1 | s2) = .3. a. What is the optimal decision if neither firm is used? Over what probability of success range is this decision optimal? b. What is the EVPI? c. Find the EVSIs and efficiencies for Stanton and New World. d. If both firms charge $5,000, which firm should be hired? e. If Stanton charges $10,000 and New World charges $4,000, which firm should Super Cola hire? Why? ANSWER: a. Introduce root beer; p ≤ .483 b. EVPI = $112,000 NOTE: The answers to (c)-(e) are very sensitive to roundoff error. Figures in parentheses are for two decimal places only. c. Stanton: EVSI = $13,200 ($11,862) Efficiency = .118 (.106) New World: EVSI = $6,400 ($6,424) Efficiency = .057 (.057) d. Hire Stanton (Stanton) e. Hire New World (Stanton) POINTS: 1 TOPICS: Computing branch probabilities 72. Dollar Department Stores has just acquired the chain of Wenthrope and Sons Custom Jewelers. Dollar has received an offer from Harris Diamonds to purchase the Wenthrope store on Grove Street for $120,000. Dollar has determined probability estimates of the store's future profitability, based on economic outcomes, as: P($80,000) = .2, P($100,000) = .3, P($120,000) = .1, and P($140,000) = .4. a. Should Dollar sell the store on Grove Street? b. What is the EVPI? Dollar can have an economic forecast performed, costing $10,000, that produces indicators c. I1 and I2, for which P(I1 | 80,000) = .1; P(I1 | 100,000) = .2; P(I1 | 120,000) = .6; P(I1 | 140,000) = .3. Should Dollar purchase the forecast? ANSWER: a. Yes, Dollar should sell store b. EVPI = $8,000 c. No; survey cost exceeds EVPI POINTS: 1 TOPICS: Computing branch probabilities 73. An appliance dealer must decide how many (if any) new microwave ovens to order for next month. The ovens cost $220 and sell for $300. Because the oven company is coming out with a new product line in two months, any ovens not sold next month will have to be sold at the dealer's half price clearance sale. Additionally, the appliance dealer feels he suffers a loss of $25 for every oven demanded when he is out of stock. On the basis of past months' sales data, the dealer estimates the probabilities of monthly demand (D) for 0, 1, 2, or 3 ovens to be .3, .4, .2, and .1, respectively. The dealer is considering conducting a telephone survey on the customers' attitudes towards microwave ovens. The results of the survey will either be favorable (F), unfavorable (U) or no opinion (N). The dealer's probability estimates for the Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis survey results based on the number of units demanded are: P(F | D = 0) = .1 P(F | D = 1) = .2

P(F | D = 2) = .3 P(F | D = 3) = .9

P(U | D = 0) = .8 P(U | D = 1) = .3

P(U | D = 2) = .1 P(U | D = 3) = .1

a. What is the dealer's optimal decision without conducting the survey? b. What is the EVPI? c. Based on the survey results what is the optimal decision strategy for the dealer? d. What is the maximum amount he should pay for this survey? ANSWER: Demand For Ovens Ovens Ordered 0 1 2 3 0 0 −25 −50 −75 1 −70 80 55 30 2 −140 10 160 135 3 −210 −60 90 240 a. Order one oven: EV = $25.00 b. EVPI = $63.00 c. Favorable: order 2; Unfavorable: order 0; No opinion: order 1 d. EVSI = $9.10 POINTS: 1 TOPICS: Computing branch probabilities 74. Lakewood Fashions must decide how many lots of assorted ski wear to order for its three stores. Information on pricing, sales, and inventory costs has led to the following payoff table, in thousands. Order Size 1 lot 2 lots 3 lots

Demand Medium 15 25 35

Low 12 9 6

High 15 35 60

a. What decision should be made by the optimist? b. What decision should be made by the conservative? c. What decision should be made using minimax regret? ANSWER: a. 3 lots b. 1 lot c. 3 lots Regret Table Order Size 1 lot 2 lots 3 lots

Low 0 3 6

Demand Medium 20 10 0

High 45 25 0

Maximum Regret 45 25 6

POINTS: 1 TOPICS: Decision making without probabilities 75. The table shows both prospective profits and losses for a company, depending on what decision is made and what state of nature occurs. Use the information to determine what the company should do. State of Nature Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis Decision d1 d2 d3 d4

s1 30 100 −80 20

s2 80 30 −10 20

s3 −30 −40 120 20

a. if an optimistic strategy is used. b. if a conservative strategy is used. c. if minimax regret is the strategy. ANSWER: a. d3 b. d1 c. d4 Regret Table: Decision d1 d2 d3 d4

s1 70 0 180 80

State of Nature s2 0 50 90 60

s3 150 160 0 100

Maximum Regret 150 160 180 100

POINTS: 1 TOPICS: Decision making without probabilities 76. A payoff table is given as Decision d1 d2 d3

s1 10 14 7

State of Nature s2 8 15 8

s3 6 2 9

a. b. c. d.

What decision should be made by the optimistic decision maker? What decision should be made by the conservative decision maker? What decision should be made under minimax regret? If the probabilities of s1, s2, and s3 are .2, .4, and .4, respectively, then what decision should be made under expected value? e. What is the EVPI? ANSWER: a. d2 b. d3 c. a three way tie d. EV(d1) = 7.6 EV(d2) = 9.6 (the best) EV(d3) = 8.2 e. EVPI = 12.4 − 9.6 = 2.8 POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Decision making with and without probabilities 77. A payoff table is given as Decision d1 d2 d3

s1 250 300 500

State of Nature s2 750 −250 500

s3 500 1200 600

a. b. c.

What choice should be made by the optimistic decision maker? What choice should be made by the conservative decision maker? What decision should be made under minimax regret? d. If the probabilities of d1, d2, and d3 are .2, .5, and .3, respectively, then what choice should be made under expected value? e. What is the EVPI? ANSWER: a. d2 b. d3 c. d1 d. EV(d1) = 695 (the best) EV(d2) = 385 EV(d 3) = 500 e. EVPI = 925 − 695 = 230 POINTS: 1 TOPICS: Decision making with and without probabilities 78. A decision maker has developed the following decision tree. How sensitive is the choice between N and P to the probabilities of states of nature U and V?

ANSWER: Choose N if p ≤ .78. POINTS: 1 TOPICS: Sensitivity analysis 79. If p is the probability of Event 1 and (1 − p) is the probability of Event 2, for what values of p would you choose A? B? C? Values in the table are payoffs. Choice/Event Event 1 Event 2 A 0 20 B 4 16 C 8 0 ANSWER: Choose A if p ≤ .5, choose B is .5 ≤ p ≤ .8, and choose C if p ≥ .8. POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Sensitivity analysis 80. Fold back the decision tree and state what strategy should be followed.

ANSWER:

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Chapter 13 - Decision Analysis

Strategy: Select A. If C happens, select H. If D happens, you are done. If E happens, select K. POINTS: 1 TOPICS: Expected value and decision trees 81. Fold back this decision tree. Clearly state the decision strategy you determine.

ANSWER:

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Chapter 13 - Decision Analysis

Choose A. If F happens, choose K. POINTS: 1 TOPICS: Expected value and decision trees 82. If sample information is obtained, the result of the sample information will be either positive or negative. No matter which result occurs, the choice to select option A or option B exists. And no matter which option is chosen, the eventual outcome will be good or poor. Complete the table. Sample Result Positive Negative ANSWER:

States of Nature good poor good poor

Prior Probabilities .7 .3 .7 .3

Sample Result Positive

States of Nature good poor good poor

Negative

Conditional Probabilities P(positive | good) = .8 P(positive | poor) = .1 P(negative | good) = P(negative | poor) =

Prior Probabilities .7 .3 .7 .3

Joint Probabilities

Conditional Probabilities P(positive | good) = .8 P(positive | poor) = .1 P(negative | good) = .2 P(negative | poor) = .9

Posterior Probabilities

Joint Probabilities .56 .03 .14 .27

Posterior Probabilities .9492 .0508 .3415 .6585

POINTS: 1 TOPICS: Posterior probabilities 83. Use graphical sensitivity analysis to determine the range of values of the probability of state of nature s1 over which each of the decision alternatives has its largest expected value. Decision

State of Nature s1 s2 8 10 4 16 10 0

d1 d2 d3 ANSWER: EV(d1) and EV(d2) intersect at p = .6. EV(d1) and EV(d3) intersect at p = .8333. Therefore when 0 ≤ p ≤ .6, Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis choose d2. When .6 ≤ p ≤ .8333, choose d1. When p ≥ .8333, choose d3.

POINTS: 1 TOPICS: Graphical sensitivity analysis 84. Dollar Department Stores has received an offer from Harris Diamonds to purchase Dollar's store on Grove Street for $120,000. Dollar has determined probability estimates of the store's future profitability, based on economic outcomes, as: P($80,000) = .2, P($100,000) = .3, P($120,000) = .1, and P($140,000) = .4. a. Should Dollar sell the store on Grove Street? b. What is the EVPI? Dollar can have an economic forecast performed, costing $10,000, that produces indicators c. I1 and I2, for which P(I1 | 80,000) = .1; P(I1 | 100,000) = .2; P(I1 | 120,000) = .6; P(I1 | 140,000) = .3. Should Dollar purchase the forecast? ANSWER: a. Yes, Dollar should sell store. b. EVPI = $8,000 c. No; survey cost exceeds EVPI. POINTS: 1 TOPICS: Posterior probabilities 85. Characterize each of the non-probabilistic approaches to decision making (i.e. - minimin, minimax, maximin, and maximax) in terms of it relating to a minimization or maximization problem and whether it is a pessimistic or optimistic approach. ANSWER: Answer not provided. POINTS: 1 TOPICS: Decision making without probabilities 86. A paint company has three sources for buying bright red pigment for their paints: Vietnam, Taiwan, or Thailand. Unfortunately, the pigment is made from a bush whose annual growth is heavily dependent upon the amount of rainfall during the growing season. The tables below show probabilities and prices for wet, dry and normal growing seasons: Probabilities Dry Normal .2 .3 .3 .1 .4 .2

Vietnam Taiwan Thailand

Wet .5 .6 .4

Vietnam

Price/Pound ($) Wet Dry Normal .95 1.10 1.00

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Chapter 13 - Decision Analysis Taiwan Thailand

.85 .90

1.20 1.15

.98 1.05

What country should the company select and what is the expected value (price) associated with it? ANSWER: EV (Vietnam) = .5(.95) + .2(l.10) + .3(1.00) = $.995 EV (Taiwan) = .6(.85) + .3(l.20) + .1(.98) = $.968 EV (Thailand) = .4(.90) + .4(1.15) + .2(l.05) = $1.03 Select Taiwan, because it has the lowest EV. POINTS: 1 TOPICS: Expected value and decision trees 87. A regional fast‑food restaurant is considering an expansion program. The major factor influencing the success of such a program is the future level of interest rates. It is estimated that there is a 20 percent chance that interest rates will increase by 2 percentage points, a 50 percent chance that they will remain the same, and a 30 percent chance that they will decrease by 2 percentage points. The alternatives they are considering and possible payoffs are shown in the table below. Which alternative is best, based on expected value? Rates up 2 percent ‑$200,000 ‑$115,000 ‑$70,000

Rates unchanged $50,000 $26,000 0

Rates down 2 percent $150,000 $80,000 $5,000

Build 50 restaurants Build 25 restaurants Do nothing ANSWER: EV (A‑‑50 restaurants) = .2(‑200,000) + .5(50,000) + .3(150,000) = $30,000 EV (B‑‑25 restaurants) = .2(‑115,000) + .5(26,000) + .3(80,000) = $14,000 EV (C‑‑do nothing) = .2(‑70,000) + .5(0) + .3(5,000) = ‑$12,500 The company should build 50 restaurants; EV of $30,000 POINTS: 1 TOPICS: Expected value and decision trees 88. A chemical company is trying to decide whether to build a pilot plant now for a new chemical process or to build the full plant now. If they build a pilot plant now, they could expand it later to a full plant or license the plant to another company. It would cost them $2 million to build the pilot plant and another $2 million later to expand it. If they build the full plant now it would cost $3.5 million to construct. The returns they expect to get from the full production plant depend upon the market. They estimate there is a 60% chance the market will be robust, a 30% chance it will remain stable, and a 10% chance it will become stagnate. The returns are estimated to be $5 million if it is robust, $3 million if it is stable, and $1 million if it is stagnate. Before they expand the pilot plant, they plan to conduct a comprehensive study. Based on past experience, they expect the study to report a 60% chance of favorable outcome for expansion and a 40% unfavorable chance. In either case they will have to decide whether to expand to a full plant or license the pilot plant. If the report is favorable and they license it, they expect to get $3 million. However, if the report is unfavorable and they license it, they will only get $1 million. Develop a decision tree for this problem and determine the optimal decision strategy. ANSWER:

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Chapter 13 - Decision Analysis

Company should build pilot plant. If report is unfavorable, company should expand. If report is favorable, company should license. EV of $600,000. POINTS: 1 TOPICS: Expected value and decision trees 89. A manufacturing company is considering expanding its production capacity to meet a growing demand for its product line of air fresheners. The alternatives are to build a new plant, expand the old plant, or do nothing. The marketing department estimates a 35 percent probability of a market upturn, a 40 percent probability of a stable market, and a 25 percent probability of a market downturn. Georgia Swain, the firm's capital appropriations analyst, estimates the following annual returns for these alternatives: Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis Market Upturn

Stable Market

Market Downturn

Build new plant

$690,000

$(130,000)

$(150,000)

Expand old plant

490,000

(45,000)

(65,000)

Do nothing

50,000

(20,000)

a. Use a decision tree analysis to analyze these decision alternatives. b. What should the company do? c. What returns will accrue to the company if your recommendation is followed? ANSWER: a.

b. Decision: Build the new plant c. Returns to accrue: $690,000; ($130,000); or ($150,000) POINTS: 1 TOPICS: Expected value and decision trees 90. The Sunshine Manufacturing Company has developed a unique new product and must now decide between two facility plans. The first alternative is to build a large new facility immediately. The second alternative is to build a small plant initially and to consider expanding it to a larger facility three years later if the market has proven favorable. Marketing has provided the following probability estimates for a ten-year plan: First 3-Year Demand

Next 7-Year Demand

Probability

Unfavorable Unfavorable Favorable Favorable

Unfavorable Favorable Favorable Unfavorable

.2 .0 .7 .1

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Chapter 13 - Decision Analysis If the small plant is expanded, the probability of demands over the remaining seven years is 7/8 for favorable and 1/8 for unfavorable. The accounting department has provided the payoff for each outcome: Demand Favorable, favorable Favorable, unfavorable Unfavorable, unfavorable Favorable, favorable Favorable, unfavorable Favorable, favorable Favorable, unfavorable Unfavorable, unfavorable

Facility Plan

Payoff

1 1 1 2--expanded 2--expanded 2--not expanded 2--not expanded 2--not expanded

$5,000,000 2,500,000 1,000,000 4,000,000 100,000 1,500,000 500,000 300,000

With these estimates, analyze Sunshine's facility decision and: a. Perform a complete decision tree analysis. b. Recommend a strategy to Sunshine. c. Determine what payoffs will result from your recommendation. ANSWER: a.

b. Recommended strategy: Build the large plant c. Possible payoffs that will result: $5,000,000; $2,500,000; or $1,000,000 POINTS: 1 TOPICS: Expected value and decision trees 91. A Pacific Northwest lumber company is considering the expansion of one of its mills. The question is whether to do it Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis now, or wait for one year and re-consider. If they expand now, the major factors of importance are the state of the economy and the level of interest rates. The combination of these two factors results in five possible situations. If they do not expand now, only the state of the economy is important and three conditions characterize the possibilities. The following table summarizes the situation: Probabilities

Revenues

.2 .2 .1 .3 .2

$80,000 $60,000 $20,000 -$20,000 -$30,000

.2 .5 .3

$50,000 $30,000 $10,000

Expand very favorable favorable neutral unfavorable very unfavorable Don't expand expansion steady contraction

a. Draw the decision tree for this problem. b. What is the expected value for expanding? c. What is the expected value for not expanding? d. Based on expected value, what should the company’s decision(s) be? ANSWER: a.

b. EV(Expanding) = 2(80,000) + .2(60,000) + .1(20,000) + .3(-20,000) + .2(-30,000) = $18,000 c. EV(Not Expanding) = 2(50,000) + .5(30,000) + .3(10,000) = $28,000 d. Do not expand POINTS: 1 TOPICS: Expected value and decision trees 92. For the payoff table below, the decision maker will use P(s1) = .15, P(s2) = .5, and P(s3) = .35. Decision

State of Nature s2

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s3 Page 28

Chapter 13 - Decision Analysis −5000 −15,000

d1 d2 a. b.

10,000 40,000

What alternative would be chosen according to expected value? For a lottery having a payoff of 40,000 with probability p and −15,000 with probability (1 − p), the decision maker expressed the following indifference probabilities. Payoff 10,000 1000 −2000 −5000

1000 −2000

Probability .85 .60 .53 .50

Let U(40,000) = 10 and U(−15,000) = 0 and find the utility value for each payoff. What alternative would be chosen according to expected utility?

ANSWER:

a. b.

EV(d1) = 3250 and EV(d2) = 10750, so choose d2. Payoff Probability Utility 10,000 .85 8.5 1000 .60 6.0 .53 5.3 −2000 .50 5.0 −5000

EU(d1) = 6.725 and EU(d2) = 6.15, so choose d1.

POINTS: 1 TOPICS: Expected utility approach 93. A decision maker who is considered to be a risk taker is faced with this set of probabilities and payoffs Decision d1 d2 d3 Probability

s1 5 −25 −50 .30

State of Nature s2 10 0 −10 .35

s3 20 50 80 .35

For the lottery p(80) + (1 − p)(−50), this decision maker has assessed the following indifference probabilities Payoff 50 20 10 5 0 −10 −25

Probability .60 .35 .25 .22 .20 .18 .10

Rank the decision alternatives on the basis of expected value and on the basis of expected utility. ANSWER: EV(d1) = 12 EV(d2) = 10 EV(d3) = 9.5 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis EU(d1) = 2.76

EU(d3) = 3.1

EU(d3) = 4.13

POINTS: 1 TOPICS: Expected utility approach 94. Three decision makers have assessed utilities for the problem whose payoff table appears below. Decision d1 d2 d3 Probability

State of Nature s2 100 150 200 .6

s3 −400 100 300 .2

Indifference Probability for Person A B C .95 .68 .45 .94 .64 .32 .91 .62 .28 .89 .60 .22 .75 .45 .10

Payoff 300 200 150 100 −100 a. b. c.

s1 500 200 −100 .2

Plot the utility function for each decision maker. Characterize each decision maker's attitude toward risk. Which decision will each person prefer?

ANSWER:

b. c.

Person A is a risk avoider, Person B is fairly risk neutral, and Person C is a risk avoider. For person A, EU(d1) = .734 EU(d2) = .912 EU(d3) = .904

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Chapter 13 - Decision Analysis For person B, EU(d1) = .56 EU(d2) = .62 EU(d3) = .61 For person C, EU(d1) = .332 EU(d2) = .276 EU(d3) = .302 Decision 1 would be chosen by person C. Decision 2 would be chosen by persons A and B. POINTS: 1 TOPICS: Risk avoiders versus risk takers 95. A decision maker has the following utility function Payoff 200 150 50 0 −50

Indifference Probability 1.00 .95 .75 .60 0

What is the risk premium for the payoff of 50? ANSWER: EV = .75(200) + .25(−50) = 137.50 Risk premium is 137.50 − 50 = 87.50 POINTS: 1 TOPICS: Developing utilities for monetary payoffs 96. Determine decision strategies based on expected value and on expected utility for this decision tree. Use the utility function Payoff 500 350 300 180 100 40 20 0

Indifference Probability 1.00 .89 .84 .60 .43 .20 .13 0

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Chapter 13 - Decision Analysis

ANSWER:

Let U(500) = 1 and U(0) = 0. Then After branch A J K B C

Expected value 120 316 150 127.2 100

Expected utility .336 .680 .522 .381 .430

Based on expected value, the decision strategy is to select B. If G happens, select J. Based on expected utility, it is best to choose C. POINTS: 1 TOPICS: Expected utility approach 97. Burger Prince Restaurant is considering the purchase of a $100,000 fire insurance policy. The fire statistics indicate that in a given year the probability of property damage in a fire is as follows: Fire Damage Probability a. b. Loss Utility

$100,000 .006

$75,000 .002

$50,000 .004

$25,000 .003

$10,000 .005

$0 .980

If Burger Prince was risk neutral, how much would they be willing to pay for fire insurance? If Burger Prince has the utility values given below, approximately how much would they be willing to pay for fire insurance? $100,000 0

$75,000 30

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$50,000 60

$25,000 85

$10,000 95

$5,000 99

$0 100 Page 32

Chapter 13 - Decision Analysis ANSWER:

a. b.

$1,075 $5,000

POINTS: 1 TOPICS: Decision making using utility 98. Super Cola is considering the introduction of a new 8 oz. root beer. The probability that the root beer will be a success is believed to equal .6. The payoff table is as follows:

Produce Do Not Produce

Success (s1) $250,000 −$50,000

Failure (s2) −$300,000 −$20,000

Company management has determined the following utility values: Amount Utility a. b.

$250,000 100

−$20,000 60

−$50,000 55

−$300,000 0

Is the company a risk taker, risk averse, or risk neutral? What is Super Cola's optimal decision?

ANSWER:

a. b.

Risk averse Produce root beer as long as p  60/105 = .571

POINTS: 1 TOPICS: Decision making using utility 99. Chez Paul is contemplating either opening another restaurant or expanding its existing location. The payoff table for these two decisions is: Decision New Restaurant Expand

s1 −$80,000 −$40,000

State of Nature s2 $20,000 $20,000

s3 $160,000 $100,000

Paul has calculated the indifference probability for the lottery having a payoff of $160,000 with probability p and −$80,000 with probability (1−p) as follows: Amount −$40,000 $20,000 $100,000 a. b.

Indifference Probability (p) .4 .7 .9

Is Paul a risk avoider, a risk taker, or risk neutral? Suppose Paul has defined the utility of −$80,000 to be 0 and the utility of $160,000 to be 80. What would be the utility values for −$40,000, $20,000, and $100,000 based on the indifference probabilities? Suppose P(s1) = .4, P(s2) = .3, and P(s3) = .3. Which decision should Paul make? Compare with the decision using the expected value approach.

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Chapter 13 - Decision Analysis ANSWER:

a. b.

A risk avoider Amount −$40,000 $20,000 $100,000

Decision is d2; EV criterion decision would be d1

Utility 32 56 72

POINTS: 1 TOPICS: Decision making using utility 100. The Dollar Department Store chain has the opportunity of acquiring either 3, 5, or 10 leases from the bankrupt Granite Variety Store chain. Dollar estimates the profit potential of the leases depends on the state of the economy over the next five years. There are four possible states of the economy as modeled by Dollar Department Stores and its president estimates P(s1) = .4, P(s2) = .3, P(s3) = .1, and P(s4) = .2. The utility has also been estimated. Given the payoffs (in $1,000,000's) and utility values below, which decision should Dollar make? Payoff Table Decision d1 -- buy 10 leases d2 -- buy 5 leases d3 -- buy 3 leases d4 -- do not buy

s1 10 5 2 0

State Of The Economy Over The Next 5 Years s2 s3 5 0 0 −1 1 0 0 0

+10 +10

+5 +5

s4 −20 −10 −1 0

Utility Table Payoff (in $1,000,000's) Utility

+2 +2

0 0

−1 −1

−10 −20

−20 −50

ANSWER:

Buy 3 leases POINTS: 1 TOPICS: Decision making using utility 101. Consider the following problem with four states of nature, three decision alternatives, and the following payoff table (in $'s): The indifference probabilities for three individuals are: Payoff $ 2600 $ 400 $ 200 $0 -$ 200 -$1400

Person 1 1.00 .40 .35 .30 .25 0

Person 2 1.00 .45 .40 .35 .30 0

Person 3 1.00 .55 .50 .45 .40 0

a. Classify each person as a risk avoider, risk taker, or risk neutral. Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis b. For the payoff of $400, what is the premium the risk avoider will pay to avoid risk? What is the premium the risk taker will pay to have the opportunity of the high payoff? c. Suppose each state is equally likely. What are the optimal decisions for each of these three people? s1 s2 s3 s4 d1 200 2600 -1400 200 d2 0 200 - 200 200 d3 -200 400 0 200 ANSWER: a. Person 1 -- risk taker; Person 2 -- risk neutral; Person 3 -- risk avoider b. Risk avoider would pay $400; Risk taker would pay $200 c. Person 1 -- d1; Person 2 -- d1; Person 3 -- d1 POINTS: 1 TOPICS: Risk avoiders and risk takers 102. Metropolitan Cablevision has the choice of using one of three DVR systems. Profits are believed to be a function of customer acceptance. The payoff to Metropolitan for the three systems is: Acceptance Level High Medium Low

System I $150,000 $ 80,000 $ 20,000

II $200,000 $ 20,000 -$ 50,000

III $200,000 $ 80,000 -$100,000

The probabilities of customer acceptance for each system are: System Acceptance Level High Medium Low

I .4 .3 .3

II .3 .4 .3

III .3 .5 .2

The first vice president believes that the indifference probabilities for Metropolitan should be: Amount $150,000 $ 80,000 $ 20,000 -$ 50,000

Probability .90 .70 .50 .25

The second vice president believes Metropolitan should assign the following utility values: Amount $200,000 $150,000 $ 80,000 $ 20,000 -$ 50,000 -$100,000

Utility 125 95 55 30 10 0

a. Which vice president is a risk taker? Which one is risk averse? b. Which system will each vice president recommend? c. What system would a risk neutral vice president recommend? Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis ANSWER:

a. Risk Taker -- Second Vice President Risk Avoider -- First Vice President b. First Vice President -- System I Second Vice President -- System III c. Risk Neutral Vice President -- System I POINTS: 1 TOPICS: Risk avoiders and risk takers Essay 103. When and why should a utility approach be followed? ANSWER: Answer not provided. POINTS: 1 TOPICS: Expected value versus utility 104. Give two examples of situations where you have decided on a course of action that did not have the highest expected monetary value. ANSWER: Answer not provided. POINTS: 1 TOPICS: Expected value versus utility 105. Explain how utility could be used in a decision where performance is not measured by monetary value. ANSWER: Answer not provided. POINTS: 1 TOPICS: Expected value versus expected utility 106. Explain the relationship between expected utility, probability, payoff, and utility. ANSWER: Answer not provided. POINTS: 1 TOPICS: Expected value versus expected utility 107. Draw the utility curves for three types of decision makers, label carefully, and explain the concepts of increasing and decreasing marginal returns for money. ANSWER: Answer not provided. POINTS: 1 TOPICS: Risk avoiders versus risk takers 108. Explain why the decision maker might feel uncomfortable with the expected value approach, and decide to use a non-probabilistic approach even when probabilities are available. ANSWER: Answer not provided. POINTS: 1 TOPICS: Decision making with probabilities 109. Why perform sensitivity analysis? Of what use is sensitivity analysis where good probability estimates are difficult to obtain? ANSWER: Answer not provided. POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 13 - Decision Analysis TOPICS: Sensitivity analysis 110. How can a good decision maker "improve" luck? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 111. Use a diagram to compare EVwPI, EVwoPI, EVPI, EVwSI, EVwoSI, and EVSI. ANSWER: Answer not provided. POINTS: 1 TOPICS: Expected value of sample information 112. Show how you would design a spreadsheet to calculate revised probabilities for two states of nature and two indicators. ANSWER: Answer not provided. POINTS: 1 TOPICS: Decision analysis and spreadsheets

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Chapter 14 - Multicriteria Decisions True / False 1. Objectives in multicriteria problems seldom conflict. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 2. Target values will never be met precisely in a goal programming problem. a. True b. False ANSWER: False POINTS: 1 TOPICS: GP: graphical solution procedure 3. Goal equations consist of a function that defines goal achievement and deviation variables that measure the distance from the target. a. True b. False ANSWER: True POINTS: 1 TOPICS: GP: developing the constraints and the goal equations 4. There can only be one goal at each priority level. a. True b. False ANSWER: False POINTS: 1 TOPICS: GP: developing an objective function with preemptive priorities 5. To solve a goal programming problem with preemptive priorities, successive linear programming programs, with an adjustment to the objective function and an additional constraint, must be solved. a. True b. False ANSWER: True POINTS: 1 TOPICS: GP: developing an objective function with preemptive priorities 6. If a problem has multiple goals at different priority levels, then usually they can all be achieved. a. True b. False ANSWER: False POINTS: 1 TOPICS: GP: computer solution Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions 7. For a scoring model, the decision maker evaluates each decision alternative using equally weighted criteria. a. True b. False ANSWER: False POINTS: 1 TOPICS: Scoring models 8. If airline A is moderately preferred to airline B, at a value of 3, then airline B is compared to airline A at a value of −3. a. True b. False ANSWER: False POINTS: 1 TOPICS: AHP: pair-wise comparisons matrix 9. An item's priority reveals how it compares to its competitors on a specific criterion. a. True b. False ANSWER: True POINTS: 1 TOPICS: AHP: synthesis 10. The priority matrix shows the priority for each item on each criterion. a. True b. False ANSWER: True POINTS: 1 TOPICS: AHP: developing an overall priority ranking 11. One limitation of a scoring model is that it uses arbitrary weights that do not necessarily reflect the preferences of the individual decision maker. a. True b. False ANSWER: False POINTS: 1 TOPICS: Scoring models 12. A consistency ratio greater than 0.10 indicates inconsistency in the pair-wise comparisons. a. True b. False ANSWER: True POINTS: 1 TOPICS: AHP: Consistency 13. Calculating the priority of each criterion in terms of its contribution to the overall goal is known as developing the hierarchy. a. True Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions b. False ANSWER: False POINTS: 1 TOPICS: Synthesization 14. The goal programming approach can be used when an analyst is confronted with an infeasible solution to an ordinary linear program. a. True b. False ANSWER: True POINTS: 1 TOPICS: Goal programming 15. A problem involving only one priority level is not considered a goal programming problem. a. True b. False ANSWER: False POINTS: 1 TOPICS: Goal programming 16. AHP allows a decision maker to express personal preferences about the various aspects of a multicriteria problem. a. True b. False ANSWER: True POINTS: 1 TOPICS: Analytic hierarchy process Multiple Choice 17. A decision with more than one objective a. cannot have an optimal solution. b. requires the decision maker to place the objectives in some order of importance. c. depends on the probability of satisfying each objective. d. should be decomposed into a separate model for each objective. ANSWER: b POINTS: 1 TOPICS: Introduction 18. Variables that indicate the distance a target is from the level achieved are called a. goal variables. b. target variables. c. deviation variables. d. preemptive variables. ANSWER: c POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions TOPICS: GP: deviation variables 19. Preemptive priorities in goal programming a. show the target values for the problem. b. prevent sacrifice of a goal to satisfy a lower level one. c. force the problem to be a standard linear program. d. limit deviations to d− only. ANSWER: b POINTS: 1 TOPICS: GP: preemptive priorities 20. Deviation variables that occur in the objective function indicate a. the targets. b. the priorities. c. only the areas that are of concern. d. the difference between all actual and target values. ANSWER: c POINTS: 1 TOPICS: GP: objective function 21. The variable d- measures a. the amount over the target and is similar to a slack. b. the amount under the target and is similar to a slack. c. the amount over the target and is similar to a surplus. d. the amount under the target and is similar to a surplus. ANSWER: b POINTS: 1 TOPICS: GP: deviation variables 22. The constraint 5x1 + 3x2 ≤ 150 is modified to become a goal equation, and priority one is to avoid overutilization. Which of the following is appropriate? a. Min P1d1− ; 5x1 + 3x2 + d1− − d1+ = 150 b. Min P1d1+ ;

5x1 + 3x2 + d1− − d1+ = 150

c. Min P1d1+ ;

5x1 + 3x2 + d1+ = 150

d. Min P1d1+ ;

5x1 + 3x2 − d1+ = 150

ANSWER: b POINTS: 1 TOPICS: GP: developing the constraints and the goal equations 23. The goal programming problem with the objective function min P1(d1+) +P2(d2−) is initially solved by the computer and the objective function value is 0. What constraint should be added for the second problem? a. d1+ = 0 b. d1+ + d2− = 0 Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions c. −d1+ + d2− = 0 d. d1+ ≤ 0 ANSWER: a POINTS: 1 TOPICS: GP: developing the constraints and the goal equations 24. A required step in the analytic hierarchy process is to determine a. the goals to be satisfied. b. the expected value of the criteria. c. the relative importance of a set of features based on a criterion. d. how many hierarchies to use. ANSWER: c POINTS: 1 TOPICS: AHP: developing the hierarchy 25. Pair-wise comparisons are used to a. compare criteria in terms of the overall goal. b. compare choices on each criterion. c. both a and b are true. d. neither a nor b is true. ANSWER: c POINTS: 1 TOPICS: AHP: pair-wise comparisons 26. The overall priorities for decision alternatives a. are the sum of the products of the criterion priority times the priority of the decision alternative with respect to that criterion. b. sum to 1. c. indicate what choice is preferred, but do not force that choice to be made. d. each of the above is true. ANSWER: d POINTS: 1 TOPICS: AHP: developing an overall priority ranking 27. The steps of the scoring model include all of the following EXCEPT: a. list the decision-making criteria and assign a weight to each. b. develop a pair-wise comparison matrix for each criterion. c. rate how well each decision alternative satisfies each criterion. d. compute the total score for each decision alternative. ANSWER: b POINTS: 1 TOPICS: Scoring models 28. Goal programming with preemptive priorities never permits trade-offs between Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions a. goals with the same priority level and the same weights. b. goals with different priority levels. c. goals with the same priority level and different weights. d. any goals. ANSWER: b POINTS: 1 TOPICS: Goal programming 29. Inconsistency in the pair-wise judgments is indicated by a consistency ratio that is a. less than zero b. greater than 0.10 c. greater than 0.50 d. greater than 1.00 ANSWER: b POINTS: 1 TOPICS: AHP: Consistency 30. When using a linear programming approach to solving a goal programming problem, a linear program must be solved for each a. goal. b. pair of deviation variables. c. priority level. d. pair-wise comparison. ANSWER: c POINTS: 1 TOPICS: Goal programming 31. Computing the consistency ratio for a criterion's pair-wise comparison matrix is the next step after a. developing the criterion's pair-wise comparison matrix. b. converting the criterion's pair-wise comparison matrix to a normalized matrix. c. developing the criterion's priority vector. d. developing the overall priority vector. ANSWER: c POINTS: 1 TOPICS: AHP: Consistency Subjective Short Answer 32. Solve the following problem graphically: Min s.t.

P1(d1+) + P2(d2−) 3x1 + 5x2 ≤ 45 3x1 + 2x2 − d1+ + d1− = 24 x1 + x2 − d2+ + d2− = 10 x1, x2, d1−, d1+, d2−, d2+ ≥ 0

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Chapter 14 - Multicriteria Decisions ANSWER:

POINTS: 1 TOPICS: GP: formulation and graphical solution 33. The Lofton Company has developed the following linear programming problem Max s.t.

x1 + x2 2x1 + x2 ≤ 10 2x1 + 3x2 ≤ 24 3x1 + 4x2 ≥ 36

but finds it is infeasible. In revision, Lofton drops the original objective and establishes the three goals Goal 1: Don't exceed 10 in constraint 1. Goal 2: Don't fall short of 36 in constraint 3. Goal 3: Don't exceed 24 in constraint 2. Give the goal programming model and solve it graphically. ANSWER: Min P1 d1+ , P2 d3− , P3 d2+ s.t. 2x1 + x2 + d1− − d1+ = 10 2x1 + 3x2 + d2− − d2+ = 24 3x1 + 4x2 + d3− − d3+ = 36

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Chapter 14 - Multicriteria Decisions The small triangle designated by the arrow satisfies the first two goals. After examining the corner points, the vertex (.8, 8.4) minimizes d2+. POINTS: 1 TOPICS: GP: formulation and graphical solution 34. Durham Designs manufactures home furnishings for department stores. Planning is underway for the production of items in the "Wildflower" fabric pattern during the next production period. Fabric required (yds) Time required (hrs) Packaging material Profit

Bedspread 7 1.5 3 12

Curtains 4 2 2 10

Dust Ruffle 9 .5 1 8

Inventory of the Wildflower fabric is 3000 yards. Five hundred hours of production time have been scheduled. Four hundred units of packaging material are available. Each of these values can be adjusted through overtime or extra purchases. Durham would like to achieve a profit of $3200, avoid purchasing more fabric or packaging material, and use all of the hours scheduled. Give the goal programming model. ANSWER: Let B = the number of bedspreads to make C = the number of curtains to make D = the number of dust ruffles to make Min

d1− + d2+ + d3+ + d4−

s.t.

12B + 10C + 8D + d1− − d1+ = 3200 7B + 4C + 9D + d2− − d2+ = 3000 1.5B + 2C + .5D + d3− − d3+ = 500 3B + 2C + 1D + d4− − d4+ = 400 B, C, D, d1−, d1+, d2−, d2+, d3−, d3+, d4− , d4+ ≥ 0

POINTS: 1 TOPICS: GP: formulation 35. An ATM is to be located in a campus union building so that it minimizes the distance from the food court, the gift shop, and the theater. They are located at coordinates (2,2), (0,6) and (8,0). Develop a goal programming model to use to locate the best place for the ATM. ANSWER: Let x1 = the first coordinate of the ATM x2 = the second coordinate of the ATM Min

Σ (di+ + di− )

s.t.

x1 + d1− − d1+ = 2 x1 + d2− − d2+ = 2 x1 + d3− − d3+ = 0

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Chapter 14 - Multicriteria Decisions x2 + d4− − d4+ = 6 x2 + d5− − d5+ = 8 x2 + d6− − d6+ = 0 POINTS: 1 TOPICS: GP: formulation 36. As treasurer of the school PTA, you chair the committee to decide how the $20,000 raised by candy sales will be spent. Four kinds of projects have been proposed, and facts on each are shown below. Project Basketball goals Encyclopedia sets Field trips Computer Stations

Number Requested 12 5 6 20

Unit Cost $400 750 300 800

Volunteers Needed (Person-days, each) 2 0 3 .5

Develop a goal programming model that would represent these goals and priorities. Priority 1 Goal 1 Spend the entire $20,000. Goal 2 Do not use more than 50 person-days of volunteer time Priority 2 Goal 3 Provide at least as many encyclopedias and computers as requested. Goal 4 Provide at least as many field trips as requested. Goal 5 Do not provide any more basketball goals than requested. ANSWER: Let B = number of basketball goals E = number of encyclopedia sets F = number of field trips C = number of computer stations Min

P1 [d1− + d2+] + P2 [d3− + d4− + d5− +d6+]

s.t.

400B + 750E + 300F + 800C + d1− − d1+ = 20000 2B + 3F + .5C + d2− − d2+ = 50 E + d3− − d3+ = 5 C + d4− − d4+ = 20 F + d5− − d5+ = 6 B + d6− − d6+ = 12 B, E, F, C, and all di ≥ 0

Note: This formulation assumes expenditures could exceed $20000. Delete d1+ from the problem if this is not so. POINTS: 1 TOPICS: GP: formulation 37. The goal programming problem below was solved with the Management Scientist. Min

P1(d1−) + P2(d2+) + P3(d3−)

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Chapter 14 - Multicriteria Decisions 72x1 + 38x2 + 23x3 ≤ 20,000 .72x1 − .76x2 − .23x3 + d1− − d1+ = 0 x3 + d2− − d2+ = 150 38x2 + d3− − d3+ = 2000 x1, x2, x3, d1−, d 1+, d2−, d2+, d3−, d3+ ≥ 0 Partial output from three successive linear programming problems is given. For each problem, give the original objective function expression and its value, and list any constraints needed beyond those that were in the original problem. s.t.

Objective Function Value = Variable D1MINUS X1 X2 X3 D1PLUS D2MINUS D2PLUS D3MINUS D3PLUS

Reduced Cost 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Objective Function Value = Variable D2PLUS X1 X2 X3 D1MINUS D1PLUS D2MINUS D3MINUS D3PLUS

Value 0.000 52.632 0.000 150.000 3.395 0.000 0.000 0.000 0.000

Value 0.000 52.632 0.000 150.000 0.000 3.395 0.000 0.000 0.000

0.000

Reduced Cost 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Objective Function Value =

0.000

Variable Value Reduced Cost D3MINUS 0.000 1.000 X1 52.632 0.000 X2 0.000 0.000 X3 150.000 0.000 D1MINUS 0.000 0.000 D1PLUS 3.395 0.000 D2MINUS 0.000 0.000 D2PLUS 0.000 0.000 D3PLUS 0.000 0.000 ANSWER: a. Min d1minus, Z = 0 b. Min d2plus, Z = 0, the constraint d1minus = 0 was included. c. Min d3minus, Z = 0, the constraint d2plus = 0 was included. POINTS: 1 TOPICS: GP: computer solution Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions 38. Rosie's Ribs is in need of an office management software package. After considerable research, Rosie has narrowed her choice to one of three packages: N-able, VersaSuite, and SoftTrack. She has determined her decision-making criteria, assigned a weight to each criterion, and rated how well each alternative satisfies each criterion. Criterion Ease of use Report generation Functional integration On-line help Entry error-checking Price Support cost

Weight 4 3 5 3 2 4 3

N-Able 3 8 5 8 8 4 6

Decision Alternatives VersaSuite 5 7 8 6 3 7 5

SoftTrack 8 6 6 4 4 5 7

Using a scoring model, determine the recommended software package for Rosie's. ANSWER: N-Able = 135, VersaSuite = 148, SoftTrack = 141 POINTS: 1 TOPICS: Scoring models 39. A consumer group is using AHP to compare four used car models. Part of the pair-wise comparison matrix for "repair frequency" is shown below. a. b.

Complete the matrix. Does it seem to be consistent?

Repair Frequency Model A Model B Model C Model D Model A 1 5 1/6 Model B 1/5 1 1/2 Model C 1/4 1/3 1 Model D 1/4 1 ANSWER: The matrix is not consistent. A to B is 5, B to C is 3, C to D is 4, yet D to A is 6. POINTS: 1 TOPICS: AHP: Consistency 40. A computer company looking for a new location for a plant has determined three criteria to use to rate cities. Pair-wise comparisons are given.

Recreation opportunities Proximity to university Cost of living

Recreation Opportunities 1 3 5

Proximity to University 1/3 1 4

Cost of Living 1/5 1/4 1

Determine priorities for the three relative to the overall location goal. ANSWER: Normalized pair-wise comparison matrix Priorities .1111 .0625 .1379 .1038 .3333 .1875 .1724 .2311 .5556 .7500 .6897 .6651 POINTS: 1 TOPICS: AHP: synthesis Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions 41. In an AHP problem, the priorities for three criteria are Criterion 1 Criterion 2 Criterion 3

.1722 .1901 .6377

The priority matrix is Criterion 1 .425 .330 .245

Choice 1 Choice 2 Choice 3

Criterion 2 .292 .251 .457

Criterion 3 .850 .105 .045

Compute the overall priority for each choice. ANSWER: a. States 1 and 2 (recover and die) are the absorbing states.

The probability that a person with symptom 2 will recover is .633. POINTS: 1 TOPICS: AHP: developing overall priorities 42. Like many high school seniors, Anne has several universities to consider when making her final college choice. To assist in her decision, she has decided to use AHP to develop a ranking for school R, school P, and school M. The schools will be evaluated on five criteria, and Anne's pair-wise comparison matrix for the criteria is shown below. Distance 1 4 1/2 1/3 1/4

Distance Program Size Climate Cost

Program 1/4 1 1/4 1/5 1/6

Size 2 4 1 1/3 1/3

Campus Climate 3 5 3 1 1/2

Cost 4 6 3 2 1

The universities' pair-wise comparisons on the criteria are shown below. Distance R P M

R 1 1/2 1/3

P 2 1 2/3

M 3 3/2 1

Programs R P M

R 1 2 1

P 1/2 1 1/2

M 1 2 1

Size R P M

R 1 1/4 1/2

P 4 1 2

M 2 1/2 1

Climate R

R 1

P 1

M 3

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Chapter 14 - Multicriteria Decisions P M

1 1/3

3 1

Cost R P M

R 1 5 1

P 1/5 1 1/5

M 1 5 1

a. What is the overall ranking of the five criteria? b. What is the overall ranking of the three universities? ANSWER: The spreadsheet shows all calculations. a. Distance Program Size Climate Cost

The ranking of the criteria is .2098 .4980 .1548 .1548 .0553

R P M

The ranking of the schools is .3705 .4030 .2265

AHP COLLEGE

Distance Program Size Climate Cost sums

Distance 1.0000 4.0000 0.5000 0.3333 0.2500 6.0833

Program 0.2500 1.0000 0.2500 0.2000 0.1667 1.8667

Size 2.0000 4.0000 1.0000 0.3333 0.3333 7.6666

Climate 3.0000 5.0000 3.0000 1.0000 0.5000 12.5000

Cost 4.0000 6.0000 3.0000 2.0000 1.0000 16.0000

NPCM 0.1644 0.6575 0.0822 0.0548 0.0411

Distance

NPCM

1.0000

2.0000

3.0000

0.5000

1.0000

0.3333

0.6667

sums

1.8333

3.6667

5.5000

Program

NPCM

1.0000

0.5000

1.0000

2.0000

1.0000

sums

4.0000

Size R

0.2609 0.5217 0.1304 0.0435 0.0435

0.5455

0.5454

0.5455

1.5000

0.2727

1.0000

0.1818

0.2500

2.0000

0.5000

1.0000

0.2500

2.0000

4.0000

NPCM

1.0000

4.0000

2.0000

0.5714

0.2500

1.0000

0.5000

0.1429

0.5000

2.0000

1.0000

0.2857

sums

1.7500

7.0000

3.5000

Climate

NPCM

1.0000

3.0000

0.4286

1.0000

3.0000

0.4286

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0.2400 0.4000 0.2400 0.0800 0.0400

0.2500 0.3750 0.1875 0.1250 0.0625

Avg 0.2098 0.4980 0.1548 0.0821 0.0553

0.1339 0.5357 0.1339 0.1071 0.0893

Avg

Avg.

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Chapter 14 - Multicriteria Decisions M

0.3333

1.0000

sums

2.3333

7.0000

0.1428

0.1429

Cost

NPCM

1.0000

0.2000

1.0000

5.0000

1.0000

sums

7.0000

0.1429

5.0000

0.7143

0.2000

1.0000

0.1429

1.4000

7.0000

Priority Matrix

0.1428

Avg.

Criterion Priority Vector Distance Program Size

Climate

Cost

0.5455

0.2500

0.5714

0.4286

0.1429

distance program size

climate

cost

0.2727

0.5000

0.1429

0.4286

0.7143

0.2098

0.0821

0.0553

0.1818

0.2500

0.2857

0.1428

0.1429

0.4980

0.1548

Overall Rankings R

0.3705

0.4030

0.2265

POINTS: 1 TOPICS: AHP: developing overall priorities 43. John Harris is interested in purchasing a new Harley-Davidson motorcycle. He has narrowed his choice to

one of three models: Sportster Classic, Heritage Softtail, and Electra Glide. After much consideration, John has determined his decision-making criteria, assigned a weight to each criterion, and rated how well each decision alternative satisfies each criterion.

Criterion Wind protection Fuel tank capacity Passenger comfort Seat height Acceleration Vehicle weight Storage capacity

Weight 5 3 2 3 4 3 3

Decision Alternative Sportster Heritage Classic Softtail 3 6 5 7 5 6 8 5 8 5 8 6 4 5

Electra Glide 8 6 8 6 3 3 8

Using a scoring model, determine the recommended motorcycle model for John. ANSWER: S1 = 5(3)+3(5)+2(5)+3(8)+4(8)+3(8)+3(4) = 132

S2 = 5(6)+3(7)+2(6)+3(5)+4(5)+3(6)+3(5) = 131 S3 = 5(8)+3(6)+2(8)+3(6)+4(3)+3(3)+3(8) = 137 (Electra Glide recommended) POINTS: 1 TOPICS: Scoring models 44. The campaign headquarters of Jerry Black, a candidate for the Board of Supervisors, has 100 volunteers. With one week to go in the election, there are three major strategies remaining: media advertising, door-to-door canvassing, and telephone campaigning. It is estimated that each phone call will take approximately four minutes and each door-to-door personal contact will average seven minutes. These times include time between contacts for breaks, transportation, dialing, etc. Volunteers who work on advertising will not be able to handle any other duties. Each ad will utilize the Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions talents of three workers for the entire week. Volunteers are expected to work 12 hours per day during the final seven days of the campaign. At a minimum, Jerry Black feels he needs 30,000 phone contacts, 20,000 personal contacts, and three advertisements during the last week. However, he would like to see 50,000 phone contacts and 50,000 personal contacts made and five advertisements developed. It is felt that advertising is 50 times as important as personal contacts, which in turn is twice as important as phone contacts. Formulate this goal programming problem with a single weighted priority to determine how the work should be distributed during the final week of the campaign. ANSWER: Variables x1 = number of volunteers doing phone work during the week x2 = number of volunteers making personal contacts during week x3 = number of volunteers preparing advertising during the week Goals 1) 50,000 phone contacts: d1+ and d1- = the amount this quantity is overachieved and underachieved, respectively 2) 50,000 personal contacts: d2+ and d2- = the amount this quantity is overachieved and underachieved, respectively 3) 5 advertisements: d3+ and d3- = the amount this quantity is overachieved and underachieved, respectively Objective Function It would not hurt Jerry Black if his goals were exceeded, however since the importance of his goals are in the ratio 1:2:100, the goal programming objective function would be: Min d1- + 2d2- + 100d3LP Constraints 1) At least 30,000 phone contacts: 1260x1 > 30,000 2) At least 20,000 personal contacts: 720x2 > 20,000 3) At least 3 advertisements: (1/3)x3 > 3 4) 100 volunteers: x1 + x2 + x3 = 100 Goal Constraints 5) Make 50,000 phone contacts: 6) Make 50,000 personal contacts: 7) Design 5 advertisements: POINTS: 1 TOPICS: Goal programming: formulation

1260x1 - d1+ + d1- = 50,000 720x2 - d2+ + d2- = 50,000 (1/3)x3 - d3+ + d3- = 5

Essay 45. Why are multicriteria problems of special interest to quantitative analysts? ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 46. Explain the difference between hard and soft constraints in a goal programming problem. ANSWER: Answer not provided. POINTS: 1 TOPICS: GP: hard and soft constraints Cengage Learning Testing, Powered by Cognero

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Chapter 14 - Multicriteria Decisions 47. Explain why goal programming could be a good approach to use after a linear programming problem is found to be infeasible. ANSWER: Answer not provided. POINTS: 1 TOPICS: GP: notes and comments 48. How can you be sure your rankings in AHP are consistent? ANSWER: Answer not provided. POINTS: 1 TOPICS: AHP: estimating the consistency ratio 49. Should the decision maker always accept the alternatives with the highest AHP rating? Explain. ANSWER: Answer not provided. POINTS: 1 TOPICS: AHP: developing an overall priority ranking 50. Explain the structure of a hierarchy diagram used in the analytic hierarchy process as a graphical representation of the problem. ANSWER: Answer not provided. POINTS: 1 TOPICS: AHP: developing the hierarchy

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Chapter 15 - Time Series Analysis and Forecasting True / False 1. Time series methods base forecasts only on past values of the variables. a. True b. False ANSWER: True POINTS: 1 TOPICS: Introduction 2. Quantitative forecasting methods can be used when past information about the variable being forecast is unavailable. a. True b. False ANSWER: False POINTS: 1 TOPICS: Introduction 3. Trend in a time series must be linear. a. True b. False ANSWER: False POINTS: 1 TOPICS: Trend pattern 4. All quarterly time series contain seasonality. a. True b. False ANSWER: False POINTS: 1 TOPICS: Seasonal pattern 5. A four-period moving average forecast for period 10 would be found by averaging the values from periods 10, 9, 8, and 7. a. True b. False ANSWER: False POINTS: 1 TOPICS: Moving averages 6. If the random variability in a time series is great, a high α value should be used to exponentially smooth out the fluctuations. a. True b. False ANSWER: False POINTS: 1 TOPICS: Exponential smoothing Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 7. With fewer periods in a moving average, it will take longer to adjust to a new level of data values. a. True b. False ANSWER: False POINTS: 1 TOPICS: Moving averages 8. Qualitative forecasting techniques should be applied in situations where time series data exists, but where conditions are expected to change. a. True b. False ANSWER: True POINTS: 1 TOPICS: Qualitative approaches 9. A time series model with a seasonal pattern will always involve quarterly data. a. True b. False ANSWER: False POINTS: 1 TOPICS: Seasonal pattern 10. Any recurring sequence of points above and below the trend line lasting less than one year can be attributed to the cyclical component of the time series. a. True b. False ANSWER: False POINTS: 1 TOPICS: Cyclical pattern 11. Smoothing methods are more appropriate for a stable time series than when significant trend or seasonal patterns are present. a. True b. False ANSWER: True POINTS: 1 TOPICS: Moving averages and exponential smoothing 12. The exponential smoothing forecast for any period is a weighted average of all the previous actual values for the time series. a. True b. False ANSWER: True POINTS: 1 TOPICS: Exponential smoothing Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 13. The mean squared error is influenced much more by large forecast errors than is the mean absolute error. a. True b. False ANSWER: True POINTS: 1 TOPICS: Forecast accuracy 14. If a time series has a significant trend pattern, then one should not use a moving average to forecast. a. True b. False ANSWER: True POINTS: 1 TOPICS: Trend pattern 15. If the random variability in a time series is great and exponential smoothing is being used to forecast, then a high alpha (α) value should be used. a. True b. False ANSWER: False POINTS: 1 TOPICS: Exponential smoothing 16. An alpha (α) value of .2 will cause an exponential smoothing forecast to react more quickly to a sudden drop in demand than will an α equal to .4. a. True b. False ANSWER: False POINTS: 1 TOPICS: Exponential smoothing 17. Exponential smoothing with α = .2 and a moving average with n = 5 put the same weight on the actual value for the current period. a. True b. False ANSWER: True POINTS: 1 TOPICS: Moving averages and exponential smoothing 18. Time series data can exhibit seasonal patterns of less than one month in duration. a. True b. False ANSWER: True POINTS: 1 TOPICS: Seasonal pattern 19. When using a moving average of order k to forecast, a small value for k is preferred if only the most recent values of Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting the time series are considered relevant. a. True b. False ANSWER: True POINTS: 1 TOPICS: Moving averages 20. In situations where you need to compare forecasting methods for different time periods, relative measures such as mean absolute error (MAE) are preferred. a. True b. False ANSWER: False POINTS: 1 TOPICS: Forecast accuracy Multiple Choice 21. All of the following are true about time series methods except a. They discover a pattern in historical data and project it into the future. b. They involve the use of expert judgment to develop forecasts. c. They assume that the pattern of the past will continue into the future. d. Their forecasts are based solely on past values of the variable or past forecast errors. ANSWER: b POINTS: 1 TOPICS: Time series patterns 22. Gradual shifting of a time series to relatively higher or lower values over a long period of time is called a. periodicity. b. cycles. c. seasonality. d. trend. ANSWER: d POINTS: 1 TOPICS: Trend component 23. Seasonal patterns a. cannot be predicted. b. are regular repeated patterns. c. are multiyear runs of observations above or below the trend line. d. reflect a shift in the time series over time. ANSWER: b POINTS: 1 TOPICS: Seasonal pattern 24. The focus of smoothing methods is to smooth out Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting a. the random fluctuations. b. wide seasonal variations. c. significant trend effects. d. long range forecasts. ANSWER: a POINTS: 1 TOPICS: Moving averages and exponential smoothing 25. Forecast errors a. are the difference in successive values of a time series b. are the differences between actual and forecast values c. should all be nonnegative d. should be summed to judge the goodness of a forecasting model ANSWER: b POINTS: 1 TOPICS: Forecast accuracy 26. To select a value for α for exponential smoothing a. use a small α when the series varies substantially. b. use a large α when the series has little random variability. c. use a value between 0 and 1 d. All of the alternatives are true. ANSWER: d POINTS: 1 TOPICS: Exponential smoothing 27. Linear trend is calculated as a. 11.25 b. 28.50 c. 39.75 d. 44.25 ANSWER: c POINTS: 1 TOPICS: Trend projection

. The trend projection for period 15 is

28. All of the following are true about qualitative forecasting methods except a. They generally involve the use of expert judgment to develop forecasts. b. They assume the pattern of the past will continue into the future. c. They are appropriate when past data on the variable being forecast are not applicable. d. They are appropriate when past data on the variable being forecast are not available. ANSWER: b POINTS: 1 TOPICS: Qualitative forecasting methods Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 29. The trend pattern is easy to identify by using a. a moving average b. exponential smoothing c. regression analysis d. a weighted moving average ANSWER: c POINTS: 1 TOPICS: Trend pattern 30. The forecasting method that is appropriate when the time series has no significant trend, cyclical, or seasonal pattern is a. moving average b. mean squared error c. mean average error d. qualitative forecasting ANSWER: a POINTS: 1 TOPICS: Moving averages 31. If data for a time series analysis is collected on an annual basis only, which pattern does not need to be considered? a. trend b. seasonal c. cyclical d. horizontal ANSWER: b POINTS: 1 TOPICS: Seasonal pattern 32. One measure of the accuracy of a forecasting model is the a. smoothing constant b. linear trend c. mean absolute error d. seasonal index ANSWER: c POINTS: 1 TOPICS: Forecast accuracy 33. Using a naive forecasting method, the forecast for next week’s sales volume equals a. the most recent week’s sales volume b. the most recent week’s forecast c. the average of the last four weeks’ sales volumes d. next week’s production volume ANSWER: a POINTS: 1 TOPICS: Forecast accuracy Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 34. Which of the following forecasting methods puts the least weight on the most recent time series value? a. exponential smoothing with α = .3 b. exponential smoothing with α = .2 c. moving average using the most recent 4 periods d. moving average using the most recent 3 periods ANSWER: b POINTS: 1 TOPICS: Moving averages and exponential smoothing 35. Using exponential smoothing, the demand forecast for time period 10 equals the demand forecast for time period 9 plus a. α times (the demand forecast for time period 8) b. α times (the error in the demand forecast for time period 9) c. α times (the observed demand in time period 9) d. α times (the demand forecast for time period 9) ANSWER: b POINTS: 1 TOPICS: Exponential smoothing 36. Which of the following exponential smoothing constant values puts the same weight on the most recent time series value as does a 5-period moving average? a. α = .2 b. α = .25 c. α = .75 d. α = .8 ANSWER: a POINTS: 1 TOPICS: Moving averages and exponential smoothing 37. All of the following are true about a cyclical pattern except a. It is often due to multiyear business cycles. b. It is often combined with long-term trend patterns and called trend-cycle patterns. c. It usually is easier to forecast than a seasonal pattern due to less variability. d. It is an alternating sequence of data points above and below the trend line. ANSWER: c POINTS: 1 TOPICS: Cyclical pattern 38. All of the following are true about a stationary time series except a. Its statistical properties are independent of time. b. A plot of the series will always exhibit a horizontal pattern. c. The process generating the data has a constant mean d. There is no variability in the time series over time. ANSWER: d POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting TOPICS: Time series patterns: horizontal pattern 39. In situations where you need to compare forecasting methods for different time periods, the most appropriate accuracy measure is a. MSE b. MAPE c. MAE d. ME ANSWER: b POINTS: 1 TOPICS: Forecast accuracy 40. Whenever a categorical variable such as season has k levels, the number of dummy variables required a. k − 1 b. k c. k + 1 d. 2k ANSWER: a POINTS: 1 TOPICS: Seasonality Subjective Short Answer 41. The number of cans of soft drinks sold in a machine each week is recorded below. Develop forecasts using a threeperiod moving average. 338, 219, 278, 265, 314, 323, 299, 259, 287, 302 ANSWER: FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 3 TIME PERIODS Time Period 1 2 3 4 5 6 7 8 9 10

Actual Value

Forecast

Forecast Error

265 314 323 299 259 287 302

278.33 254.00 285.67 300.67 312.00 293.67 281.67

−13.33 60.00 37.33 −1.67 −53.00 −6.67 20.33

THE FORECAST FOR PERIOD 11 282.67 POINTS: 1 TOPICS: Moving averages 42. Use a four-period moving average to forecast attendance at baseball games. Historical records show Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 5346, 7812, 6513, 5783, 5982, 6519, 6283, 5577, 6712, 7345 ANSWER: FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 4 TIME PERIODS Time Period 1 2 3 4 5 6 7 8 9 10

Actual Value

Forecast

Forecast Error

5,982 6,519 6,283 5,577 6,712 7,345

6,363.50 6,522.50 6,199.25 6,141.75 6,090.25 6,272.75

−381.50 −3.50 83.75 −564.75 621.75 1,072.25

THE FORECAST FOR PERIOD 11 POINTS: 1 TOPICS: Moving averages

6479.25

43. A hospital records the number of floral deliveries its patients receive each day. For a two-week period, the records show 15, 27, 26, 24, 18, 21, 26, 19, 15, 28, 25, 26, 17, 23 Use exponential smoothing with a smoothing constant of .4 to forecast the number of deliveries. ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.4 Time Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Actual Value

Forecast

Forecast Error

27 26 24 18 21 26 19 15 28 25 26 17 23

15.00 19.80 22.28 22.97 20.98 20.99 22.99 21.40 18.84 22.50 23.50 24.50 21.50

12.00 6.20 1.72 −4.97 0.02 5.01 −3.99 −6.40 9.16 2.50 2.50 −7.50 1.50

THE FORECAST FOR PERIOD 15 POINTS: 1 TOPICS: Exponential smoothing

22.10

44. The number of girls who attend a summer basketball camp has been recorded for the seven years the camp has been offered. Use exponential smoothing with a smoothing constant of .8 to forecast attendance for the eighth year. Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 47, 68, 65, 92, 98, 121, 146 ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8 Time Period 1 2 3 4 5 6 7

Actual Value

Forecast

Forecast Error

68 65 92 98 121 146

47.00 63.80 64.76 86.55 95.71 115.94

21.00 1.20 27.24 11.45 25.29 30.06

THE FORECAST FOR PERIOD 8 POINTS: 1 TOPICS: Exponential smoothing

139.99

45. The number of pizzas ordered on Friday evenings between 5:30 and 6:30 at a pizza delivery location for the last 10 weeks is shown below. Use exponential smoothing with smoothing constants of .2 and .8 to forecast a value for week 11. Compare your forecasts using MSE. Which smoothing constant would you prefer? 58, 46, 55, 39, 42, 63, 54, 55, 61, 52 ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.2 Time Period 1 2 3 4 5 6 7 8 9 10

Actual Value

Forecast

Forecast Error

46 55 39 42 63 54 55 61 52

58.00 55.60 55.48 52.18 50.15 52.72 52.97 53.38 54.90

−12.00 −0.60 −16.48 −10.18 12.85 1.28 2.03 7.62 −2.90

THE MEAN SQUARE ERROR 84.12 THE FORECAST FOR PERIOD 11 54.32 FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8 Time Period 1 2 3 4 5

Actual Value

Forecast

Forecast Error

46 55 39 42

58.00 48.40 53.68 41.94

−12.00 6.60 −14.68 0.06

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Chapter 15 - Time Series Analysis and Forecasting 6 7 8 9 10

63 54 55 61 52

41.99 58.80 54.96 54.99 59.80

21.01 −4.80 0.04 6.01 −7.80

THE MEAN SQUARE ERROR 107.17 THE FORECAST FOR PERIOD 11 53.56 Based on MSE, smoothing with α = .2 provides a better model. POINTS: 1 TOPICS: Exponential smoothing 46. A trend line for the weekly attendance at a restaurant's Sunday brunch is given by

How many guests would you expect in week 20? ANSWER: POINTS: 1 TOPICS: Trend projection 47. The number of new contributors to a public radio station's annual fund drive over the last ten years is 63, 58, 61, 72, 98, 103, 121, 147, 163, 198 Develop a trend equation for this information, and use it to predict next year's number of new contributors. ANSWER: FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION:

where

trend value of the time series in period t

Time Period Actual Value Forecast 1 63 39.35 2 58 54.69 3 61 70.04 4 72 85.38 5 98 100.73 6 103 116.07 7 121 131.42 8 147 146.76 9 163 162.11 10 198 177.45 MEAN SQUARE ERROR 154.11 FORECAST FOR PERIOD 11 192.80 POINTS: 1 TOPICS: Trend pattern Cengage Learning Testing, Powered by Cognero

Forecast Error 23.66 3.31 −9.03 −13.38 −2.73 −13.07 −10.42 0.24 0.89 20.55

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Chapter 15 - Time Series Analysis and Forecasting 48. The average SAT verbal score for students from one high school over the last ten exams is 508, 490, 502, 505, 493, 506, 492, 490, 503, 501 Do the scores support an increasing or a decreasing trend? ANSWER: FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION:

where

trend value of the time series in period t

The negative slope for the trend indicates that scores are decreasing. (However, we have not tested for the significance of this coefficient.) POINTS: 1 TOPICS: Trend component 49. The number of properties newly listed with a real estate agency in each quarter over the last four years is given below. Assume the time series has seasonality without trend. Year Quarter 1 2 3 4

1 73 89 123 92

2 81 87 115 95

3 76 91 108 87

4 77 88 120 97

a. Develop the optimization model that finds the estimated regression equation that minimize the sum of squared error. b. Solve for the estimated regression equation. c. Forecast the four quarters of Year 5. ANSWER: a. Min {(73 - Ȳ1)2 + (89 - Ȳ2)2 + (123 - Ȳ3)2 + . . . + (97 - Ȳ16)2} s.t. Ȳ1 = b0 + 1b1 + 0b2 + 0b3 Ȳ2 = b0 + 0b1 + 1b2 + 0b3 Ȳ3 = b0 + 0b1 + 0b2 + 1b3 Ȳ4 = b0 + 0b1 + 0b2 + 0b3 . . . Ȳ13 = b0 + 1b1 + 0b2 + 0b3 Ȳ14 = b0 + 0b1 + 1b2 + 0b3 Ȳ15 = b0 + 0b1 + 0b2 + 1b3 Ȳ16 = b0 + 0b1 + 0b2 + 0b3 b. Ȳt = 92.75 − 16Qtr1 − 4Qtr2 + 23.75Qtr3 c. Ȳ17 = 76.75, Ȳ18 = 88.75, Ȳ19 = 116.50, Ȳ20 = 92.75 POINTS: 1 TOPICS: Seasonal component Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting 50. Quarterly billing for water usage is shown below. Year Quarter Winter Spring Summer Fall

1 64 103 152 73

2 66 103 160 72

3 68 104 162 78

4 73 120 176 88

a. Solve for the forecast equation that minimizes the sum of squared error. b. Forecast the summer of year 5 and spring of year 6. ANSWER: a. Ȳt = 64.875 − 6.1375Qtr1t + 32.325Qtr2t + 86.0375Qtr3t + 1.2875t b. Summer, Year 5 = 175.375; Spring, Year 6 = 125.525 POINTS: 1 TOPICS: Seasonality with trend 51. A customer comment phone line is staffed from 8:00 a.m. to 4:30 p.m. five days a week. Records are available that show the number of calls received every day for the last five weeks. Week Day 1 M T W TH F 2 M T W TH F 3 M T W TH F

Number 28 12 16 15 23 25 10 14 14 26 32 15 15 13 21

Week Day 4 M T W TH F 5 M T W TH F

Number 27 13 16 18 24 26 11 18 17 25

Develop the optimization model that finds the estimated regression equation that minimize the sum of squared error. b. Solve for the estimated regression equation. c. Forecast the five days of week 6. ANSWER: a. Min {(28 - Ȳ1)2 + (12 - Ȳ2)2 + (16 - Ȳ3)2 + . . . + (25 - Ȳ25)2} s.t. Ȳ1 = b0 + 1b1 + 0b2 + 0b3 + 0b4 Ȳ2 = b0 + 0b1 + 1b2 + 0b3 + 0b4 Ȳ3 = b0 + 0b1 + 0b2 + 1b3 + 0b4 Ȳ4 = b0 + 0b1 + 0b2 + 0b3 + 1b4 Ȳ5 = b0 + 0b1 + 0b2 + 0b3 + 0b4 . . a.

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Chapter 15 - Time Series Analysis and Forecasting . Ȳ21 = b0 + 1b1 + 0b2 + 0b3 + 0b4 Ȳ22 = b0 + 0b1 + 1b2 + 0b3 + 0b4 Ȳ23 = b0 + 0b1 + 0b2 + 1b3 + 0b4 Ȳ24 = b0 + 0b1 + 0b2 + 0b3 + 1b4 Ȳ25 = b0 + 0b1 + 0b2 + 0b3 + 0b4 b. Ȳt = 23.8 + 3.8M − 11.6T − 8.0W − 8.4TH c. Ȳ26 = 27.6, Ȳ27 = 12.2, Ȳ28 = 15.8, Ȳ29 = 15.4, Ȳ30 = 23.8 POINTS: 1 TOPICS: Seasonal pattern 52. Monthly sales at a coffee shop have been analyzed. The seasonal index values are Month Index Jan 1.38 Feb 1.42 Mar 1.35 Apr 1.03 May .99 June .62 July .51 Aug .58 Sept .82 Oct .82 Nov .92 Dec 1.56 and the trend line is 74123 + 26.9(t). Assume there is no cyclical component and forecast sales for year 8 (months 97 108). ANSWER: SI Month Trend Forecast 1.38 97 76994.2 106252.0 1.42 98 77023.8 109373.8 1.35 99 77053.4 104022.1 1.03 100 77083.0 79395.5 0.99 101 77112.6 76341.5 0.62 102 77142.2 47828.2 0.51 103 77171.8 39357.6 0.58 104 77201.4 44776.8 0.82 105 77231.0 63329.4 0.82 106 77260.6 63353.7 0.92 107 77290.2 71107.0 1.56 108 77319.8 120618.9 POINTS: 1 TOPICS: Seasonal component 53. A 24-hour coffee/donut shop makes donuts every eight hours. The manager must forecast donut demand so that the bakers have the fresh ingredients they need. Listed below is the actual number of glazed donuts (in dozens) sold in each of the preceding 13 eight-hour shifts. Date

Shift

Demand(dozens)

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Chapter 15 - Time Series Analysis and Forecasting June 3

June 4

June 5

June 6

June 7

Day Evening Night Day Evening Night Day Evening Night Day Evening Night Day

59 47 40 64 43 39 62 46 42 60 45 40 58

a. Develop the optimization model that finds the estimated regression equation that minimize the sum of squared error. b. Solve for the estimated regression equation. c. Forecast the demand for glazed donuts for the Day, Evening, and Night shifts of June 8. ANSWER: a. Min {(59 - Ȳ1)2 + (47 - Ȳ2)2 + (40 - Ȳ3)2 + . . . + (58 - Ȳ13)2} s.t. Ȳ1 = b0 + 1b1 + 0b2 Ȳ2 = b0 + 0b1 + 1b2 Ȳ3 = b0 + 0b1 + 0b2 . . . Ȳ10 = b0 + 1b1 + 0b2 Ȳ11 = b0 + 0b1 + 1b2 Ȳ12 = b0 + 0b1 + 0b2 Ȳ13 = b0 + 1b1 + 0b2 b. Ȳt = 40.25 + 20.35Dt + 5.00Et c. Day = 60.60, Evening = 45.25, Night = 40.25 POINTS: 1 TOPICS: Seasonal pattern 54. The number of plumbing repair jobs performed by Auger's Plumbing Service in each of the last nine months are listed below. Month March April May

Jobs 353 387 342

Month June July August

Jobs 374 396 409

Month September October November

Jobs 399 412 408

Assuming a linear trend function, forecast the number of repair jobs Auger's will perform in December using the least squares method. What is your forecast for December using a three-period weighted moving average with b. weights of .6, .3, and .1? How does it compare with your forecast from part (a)? ANSWER: a. Ȳ10 = 349.667 + (10)(7.4) = 423.667 a.

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Chapter 15 - Time Series Analysis and Forecasting Ȳ10 = .1(399) + .3(412) + .6(408) = 408.3

POINTS: 1 TOPICS: Trend component 55. Quarterly revenues (in $1,000,000's) for a national restaurant chain for a five-year period were as follows: Quarter 1 2 3 4

Year 1 33 36 35 38

Year 2 42 40 42 47

Year 3 54 53 54 62

Year 4 70 67 70 77

Year 5 85 82 87 99

a. Solve for the forecast equation that minimizes the sum of squared error. b. Forecast the four quarters of year 6. ANSWER: a. Ȳt = 24.475 + 2.23125Qtr1t − 2.3125Qtr2t − 3.65625Qtr3t + 3.34375t b. Year 6: Q1 = 96.925, Q2 = 95.725, Q3 = 97.725, Q4 = 104.725 POINTS: 1 TOPICS: Seasonal with trend 56. Business at Terry's Tie Shop can be viewed as falling into three distinct seasons: (1) Christmas (NovemberDecember); (2) Father's Day (late May - mid-June); and (3) all other times. Average weekly sales (in $'s) during each of these three seasons during the past four years has been as follows: Season 1 2 3

Year 1 1856 2012 985

Year 2 1995 2168 1072

Year 3 2241 2306 1105

Year 4 2280 2408 1120

Determine a forecast for the average weekly sales in years 5 and 6 for each of the three seasons. ANSWER: Ȳt = 797 + 1095.433S1t + 1189.467S2t + 36.467t Forecasts Year 5: Seas.1 = 2366.5, Seas.2 = 2497.0, Seas.3 = 1344.0 Forecasts Year 6: Seas.1 = 2475.9, Seas.2 = 2606.4, Seas.3 = 1453.4 POINTS: 1 TOPICS: Seasonality with trend 57. Sales (in thousands) of the new Thorton Model 506 convection oven over the eight-week period since its introduction have been as follows: Week 1 2 3 4 5 6 7 8 a. b.

Sales 18.6 21.4 25.2 22.4 24.6 19.2 21.7 23.8

Which exponential smoothing model provides better forecasts, one using α = .6 or α = .2? Compare them using mean squared error. Using the two forecast models in part (a), what are the forecasts for week 9?

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Chapter 15 - Time Series Analysis and Forecasting ANSWER:

a. b.

For α = .6, MSE = 9.12; for α = .2, MSE = 10.82 For α = .6, Ȳ9 = 22.86; for α = .2, Ȳ9 = 21.72

POINTS: 1 TOPICS: Exponential smoothing and forecast accuracy 58. Coyote Cable has been experiencing an increase in cable service subscribers over the last few years due to increased advertising and an influx of new residents to the region. The numbers of subscribers (in 1000's) for the last 16 months are as follows: Month

Sales

Month

Sales

Month

Sales

12.8

20.6

23.8

14.6

18.5

25.1

15.2

19.9

24.7

16.1

23.6

26.5

15.8

24.2

28.9

17.2

Forecast the number of subscribers for months 17, 18, 19, and 20. ANSWER: Month 17: 29.0; Month 18: 30.0; Month 19: 31.0; Month 20: 32.0 POINTS: 1 TOPICS: Linear trend projection 59. Weekly sales of the Weber food processor for the past ten weeks have been: Week 1 2 3 4 5

Sales 980 1040 1120 1050 960

Week 6 7 8 9 10

Sales 990 1030 1260 1240 1100

a. Determine, on the basis of minimizing the mean square error, whether a three-period or four-period simple moving average model gives a better forecast for this problem. b. For each model, forecast sales for week 11. ANSWER: a. The 3 week moving average gives the better forecast (MSE = 16,337) (compared to a MSE = 17,911 for the 4 week moving average) b. 3 week moving average forecast for week 11 = 1,200 4 week moving average forecast for week 11 = 1,158 POINTS: 1 60. Below you are given information on John's Hair Salon profit for the past 7 years. Year

Profit (In Thousands)

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Chapter 15 - Time Series Analysis and Forecasting 1 2 3 4 5 6 7

15.0 16.2 17.1 18.1 18.8 19.2 20.5

a. Use regression analysis to obtain an expression for the linear trend projection. b. Forecast John's Hair Salon profit for the next 5 years. ANSWER: a. b. 21.3, 22.2, 23.0, 23.9, 24.8 POINTS: 1 TOPICS: Linear trend projection Essay 61. Explain what conditions make quantitative forecasting methods appropriate. ANSWER: Answer not provided. POINTS: 1 TOPICS: Introduction 62. What is a stable time series, and what forecasting methods are appropriate for one? ANSWER: Answer not provided. POINTS: 1 TOPICS: Smoothing methods 63. How can error measures be used to determine the number of periods to use in a moving average? What are you assuming about the future when you make this choice? ANSWER: Answer not provided. POINTS: 1 TOPICS: Moving averages 64. Discuss the effects of using a small smoothing constant value and when it is most appropriate to use. Then, do the same for a large smoothing constant value. ANSWER: Answer not provided. POINTS: 1 TOPICS: Exponential smoothing 65. Explain and contrast three measures of forecast accuracy. ANSWER: Answer not provided. POINTS: 1 TOPICS: Forecast accuracy 66. Describe a time series plot and discuss its purpose and when in the forecasting process it should be constructed. ANSWER: Answer not provided. POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 15 - Time Series Analysis and Forecasting TOPICS: Time series plot

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Chapter 16 - Markov Processes True / False 1. Markov processes use historical probabilities. a. True b. False ANSWER: True POINTS: 1 TOPICS: Market share analysis 2. All entries in a matrix of transition probabilities sum to 1. a. True b. False ANSWER: False POINTS: 1 TOPICS: Transition probabilities 3. All Markov chain transition matrices have the same number of rows as columns. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transition probabilities 4. A unique matrix of transition probabilities should be developed for each customer. a. True b. False ANSWER: False POINTS: 1 TOPICS: Transition probabilities 5. The probability that the system is in state 2 in the 5th period is π5(2). a. True b. False ANSWER: False POINTS: 1 TOPICS: Market share analysis 6. The fundamental matrix is used to calculate the probability of the process moving into each absorbing state. a. True b. False ANSWER: True POINTS: 1 TOPICS: Fundamental matrix, absorbing state 7. Steady state probabilities are independent of initial state. Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes a. True b. False ANSWER: True POINTS: 1 TOPICS: Steady-state probabilities 8. A Markov chain cannot consist of all absorbing states. a. True b. False ANSWER: False POINTS: 1 TOPICS: Absorbing states 9. If an absorbing state exists, then the probability that a unit will ultimately move into the absorbing state is given by the steady state probability. a. True b. False ANSWER: False POINTS: 1 TOPICS: NR matrix 10. All Markov chains have steady-state probabilities. a. True b. False ANSWER: False POINTS: 1 TOPICS: Steady-state probabilities 11. All entries in a row of a matrix of transition probabilities sum to 1. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transition probabilities 12. A state i is a transient state if there exists a state j that is reachable from i, but the state i is not reachable from state j. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transition probabilities 13. A state i is an absorbing state if pii = 0. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes ANSWER: False POINTS: 1 TOPICS: Absorbing states 14. When absorbing states are present, each row of the transition matrix corresponding to an absorbing state will have a single 1 and all other probabilities will be 0. a. True b. False ANSWER: True POINTS: 1 TOPICS: Absorbing states 15. For Markov processes having the memoryless property, the prior states of the system must be considered in order to predict the future behavior of the system. a. True b. False ANSWER: False POINTS: 1 TOPICS: First-order Markov processes 16. The sum of the probabilities in a transition matrix equals the number of rows in the matrix. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transition matrix 17. Transition probabilities are conditional probabilities. a. True b. False ANSWER: True POINTS: 1 TOPICS: Transition probabilities 18. A state, i, is an absorbing state if, when i = j, pij = 1. a. True b. False ANSWER: True POINTS: 1 TOPICS: Absorbing states 19. If a Markov chain has at least one absorbing state, steady-state probabilities cannot be calculated. a. True b. False ANSWER: True Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes POINTS: 1 TOPICS: Steady-state probabilities 20. State j is an absorbing state if pij = 1. a. True b. False ANSWER: False POINTS: 1 TOPICS: Absorbing states Multiple Choice 21. In Markov analysis, we are concerned with the probability that the a. state is part of a system. b. system is in a particular state at a given time. c. time has reached a steady state. d. transition will occur. ANSWER: b POINTS: 1 TOPICS: Introduction 22. For a situation with weekly dining at either an Italian or Mexican restaurant, a. the weekly visit is the trial and the restaurant is the state. b. the weekly visit is the state and the restaurant is the trial. c. the weekly visit is the trend and the restaurant is the transition. d. the weekly visit is the transition and the restaurant is the trend. ANSWER: a POINTS: 1 TOPICS: Market share analysis 23. A transition probability describes a. the probability of a success in repeated, independent trials. b. the probability a system in a particular state now will be in a specific state next period. c. the probability of reaching an absorbing state. d. None of the alternatives is correct. ANSWER: b POINTS: 1 TOPICS: Introduction 24. The probability of going from state 1 in period 2 to state 4 in period 3 is a. p12 b. p23 c. p14 d. p43 Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes ANSWER: c POINTS: 1 TOPICS: Market share analysis 25. The probability that a system is in a particular state after a large number of periods is a. independent of the beginning state of the system. b. dependent on the beginning state of the system. c. equal to one half. d. the same for every ending system. ANSWER: a POINTS: 1 TOPICS: Market share analysis 26. At steady state a. π1(n+1) > π1(n) b. π1 = π2 c. π1 + π2 ≥ 1 d. π1(n+1) = π1 ANSWER: d POINTS: 1 TOPICS: Market share analysis 27. Analysis of a Markov process a. describes future behavior of the system. b. optimizes the system. c. leads to higher order decision making. d. All of the alternatives are true. ANSWER: a POINTS: 1 TOPICS: Introduction 28. If the probability of making a transition from a state is 0, then that state is called a(n) a. steady state. b. final state. c. origin state. d. absorbing state. ANSWER: d POINTS: 1 TOPICS: Absorbing states 29. Absorbing state probabilities are the same as a. steady state probabilities. b. transition probabilities. c. fundamental probabilities. Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes d. None of the alternatives is true. ANSWER: d POINTS: 1 TOPICS: Fundamental matrix 30. The probability of reaching an absorbing state is given by the a. R matrix. b. NR matrix. c. Q matrix. d. (I − Q)−1 matrix ANSWER: b POINTS: 1 TOPICS: Fundamental matrix Subjective Short Answer 31. Calculate the steady state probabilities for this transition matrix.

ANSWER: π1 = 3/7, π2 = 4/7 POINTS: 1 TOPICS: Steady-state probabilities 32. Two airlines offer conveniently scheduled flights to the airport nearest your corporate headquarters. Historically, flights have been scheduled as reflected in this transition matrix. Current Flight Airline A Airline B

Next Flight Airline A Airline B .6 .4 .2 .8

a. If your last flight was on B, what is the probability your next flight will be on A? b. If your last flight was on B, what is the probability your second next flight will be on A? c. What are the steady state probabilities? ANSWER: a. .2 b. .28 c. 1/3, 2/3 POINTS: 1 TOPICS: State of the system, steady-state probabilities 33. The matrix of transition probabilities below deals with brand loyalty to Bark Bits and Canine Chow dog food. Current Purchase Bark Bits Canine Chow

Next Purchase Bark Bits Canine Chow .75 .25 .20 .80

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Chapter 16 - Markov Processes a.

What are the steady state probabilities? What is the probability that a customer will switch brands on the next purchase after a large b. number of periods? ANSWER: a.

(π1 π2) = (π1 π2)

π1 = 4/9, π2 = 5/9 P(switching) = (4/9)(1/4) + (5/9)(1/5) = 2/9

b. POINTS: 1 TOPICS: Market share analysis

34. Bark Bits Company is planning an advertising campaign to raise the brand loyalty of its customers to .80. a. The former transition matrix is

What is the new one? What are the new steady state probabilities? If each point of market share increases profit by $15000, what is the most you would pay for c. the advertising? ANSWER: b.

b. c.

π1 = .5, π2 = .5 The increase in market share is .5 − .444 = .056. (5.6 points)($15,000/point) = $84,000 value for the campaign.

POINTS: 1 TOPICS: Market share analysis 35. The daily price of a farm commodity is up, down, or unchanged from the day before. Analysts predict that if the last price was down, there is a .5 probability the next will be down, and a .4 probability the price will be unchanged. If the last price was unchanged, there is a .35 probability it will be down and a .35 probability it will be up. For prices whose last movement was up, the probabilities of down, unchanged, and up are .1, .3, and .6. a. Construct the matrix of transition probabilities. b. Calculate the steady state probabilities. ANSWER: a.

π1 = .305, π2 = .333, π3 = .365

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Chapter 16 - Markov Processes POINTS: 1 TOPICS: Steady-state probabilities 36. Appointments in a medical office are scheduled every 15 minutes. Throughout the day, appointments will be running on time or late, depending on the previous appointment only, according to the following matrix of transition probabilities: Next Appointment On Time Late .75 .25 .30 .70

Previous Appointment On Time Late

The day begins with the first appointment on time. What are the state probabilities for periods 1, 2, 3 and 4? b. What are the steady state probabilities? ANSWER: a. π(0) = (1, 0) π(1) = (.75, .25) π(2) = (.6375, .3625) π(3) = (.5869, .4131) π(4) = (.5641, .4359) b. π1 = .5455, π2 = .4545 POINTS: 1 TOPICS: State of the system, steady-state probabilities a.

37. A city is served by three cable TV companies: Xcellent Cable, Your Cable, and Zephyr Cable. A survey of 1000 cable subscribers shows this breakdown of customers from the beginning to the end of August. Company on August 1 Xcellent Your Zephyr

Company on August 31 Xcellent Your Zephyr 300 50 50 10 200 40 40 80 230

a. Construct the transition matrix. b. What was each company's share of the market at the beginning and the end of the month? c. If the current trend continues what will the market shares be? ANSWER: a.

Market shares Aug 1 X: .40, Y: .25, Z: .35 Market shares Aug 31 X: .35, Y: .33, Z: .32 Steady states: .2156, .48125, .30315

c. POINTS: 1 TOPICS: Market share analysis

38. A television ratings company surveys 100 viewers on March 1 and April 1 to find what was being watched at 6:00 p.m. -- the local NBC affiliate's local news, the CBS affiliate's local news, or "Other" which includes all other channels and not watching TV. The results show Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes March 1 Choice NBC CBS Other

Number 30 40 30

Record of Switches During March to NBC CBS Other -5 10 15 -5 5 5 --

a. What are the numbers in each choice for April 1? b. What is the transition matrix? c. What ratings percentages do you predict for May 1? ANSWER: a. NBC 35, CBS 30, Other 35

π3 =

POINTS: 1 TOPICS: Market share analysis 39. Accounts receivable have been grouped into the following states: State 1: Paid State 2: Bad debt State 3: 0-30 days old State 4: 31-60 days old Sixty percent of all new bills are paid before they are 30 days old. The remainder of these go to state 4. Seventy percent of all 30 day old bills are paid before they become 60 days old. If not paid, they are permanently classified as bad debts. a. Set up the one month Markov transition matrix. b. What is the probability that an account in state 3 will be paid? ANSWER: a. Paid Bad 0-30 31-60 Paid 1 0 0 0 Bad 0 1 0 0 0-30 .6 0 0 .4 31-60 .7 .3 0 0

NR =

So, the probability an account in state 3 will be paid is .88. POINTS: 1 TOPICS: Accounts receivable analysis Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes 40. The medical prognosis for a patient with a certain disease is to recover, to die, to exhibit symptom 1, or to exhibit symptom 2. The matrix of transition probabilities is Recover 1 0 1/4 1/4

Recover Die S1 S2

Die 0 1 1/4 1/8

S1 0 0 1/3 1/8

S2 0 0 1/6 1/2

a. What are the absorbing states? b. What is the probability that a patient with symptom 2 will recover? ANSWER: a. States 1 and 2 (recover and die) are the absorbing states. b.

NR =

The probability that a person with symptom 2 will recover is .633. POINTS: 1 TOPICS: Absorbing states 41. Rent-To-Keep rents household furnishings by the month. At the end of a rental month a customer can: a) rent the item for another month, b) buy the item, or c) return the item. The matrix below describes the month-to-month transition probabilities for 32-inch stereo televisions the shop stocks. This Month Rent Buy Return

Rent .72 0 0

Next Month Buy .10 1 0

Return .18 0 1

What is the probability that a customer who rented a TV this month will eventually buy it? ANSWER: P(Buy) = .357, P(Return) = .643 POINTS: 1 TOPICS: Absorbing states 42. A recent study done by an economist for the Small Business Administration investigated failures of small business. Failures were either classified as due to poor financing, poor management, or a poor product. The failure rates differed for new businesses (under one year old) versus established businesses (over one year old.) As the result of the economist's study, the following probabilities were determined. For new businesses the probability of failure due to financing was .15, due to management .20, and due to product .05. The corresponding probabilities for established businesses were .10, .06, and .03 respectively. a. Determine a five-state Markov Chain transition matrix with states for new, established, and each of the three failure states. Write it in the form of I, O, R, and Q submatrices. b. Determine the probability that a new business will survive during the next three years. c. What proportion of new businesses eventually fail due to: (1) poor financing? (2) poor management? (3) poor product? ANSWER: a. Fail.-Fail. Fail.-Mgt. Fail.-Prod. New Estab. Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes Failure-Failure Failure-Management Failure-Product New Established

1 0 0 .15 .10

0 1 0 .20 .06

0 0 1 .05 .03

0 0 0 0 0

0 0 0 .60 .81

b. .394 c. (1) .47; (2) .39; (3) .14 POINTS: 1 TOPICS: Probabilities at stage n 43. On any particular day an individual can take one of two routes to work. Route A has a 25% chance of being congested, whereas route B has a 40% chance of being congested. The probability of the individual taking a particular route depends on his previous day's experience. If one day he takes route A and it is not congested, he will take route A again the next day with probability .8. If it is congested, he will take route B the next day with probability .7. On the other hand, if on a day he takes route B and it is not congested, he will take route B again the next day with probability .9. Similarly if route B is congested, he will take route A the next day with probability .6. Construct the transition matrix for this problem. (HINT: There are 4 states corresponding to a. the route taken and the congestion. The transition probabilities are products of the independent probabilities of congestion and next day choice.) b. What is the long-run proportion of time that route A is taken? ANSWER: a. Next Day This Day RA-NC RA-C RB-NC RB-C Route A – Not Congested (RA.600 .200 .120 .080 NC) Route A – Congested (RA-C) .225 .075 .420 .280 Route B – Not Congested (RB-NC) .075 .025 .540 .360 Route B – Congested (RB-C) .450 .150 .240 .160 b. .36 + .12 = .48 POINTS: 1 TOPICS: Steady-state probabilities 44. Henry, a persistent salesman, calls North's Hardware Store once a week hoping to speak with the store's buying agent, Shirley. If Shirley does not accept Henry's call this week, the probability she will do the same next week is .35. On the other hand, if she accepts Henry's call this week, the probability she will not do so next week is .20. a. Construct the transition matrix for this problem. b. How many times per year can Henry expect to talk to Shirley? What is the probability Shirley will accept Henry's next two calls if she does not accept his c. call this week? What is the probability of Shirley accepting exactly one of Henry's next two calls if she d. accepts his call this week? ANSWER: a. The transition matrix is: Next Week's Call This Week's Call Refuses Accepts Refuses .35 .65 Accepts .20 .80 Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes b. 39.76 or about 40. c. .52. d. .13 + .16 = .29. POINTS: 1 TOPICS: State of the system, steady-state probabilities 45. Rent-To-Keep rents household furnishings by the month. At the end of a rental month a customer can: a) rent the item for another month, b) buy the item, or c) return the item. The matrix below describes the month-to-month transition probabilities for 32-inch stereo televisions the shop stocks.

This Month

Rent

Rent .72

Next Month Buy .10

Buy

Return .18 0

Return 0 0 1 What is the probability that a customer who rented a TV this month will eventually buy it? ANSWER: P(Buy) = .357; P(Return) = .643 POINTS: 1 TOPICS: State of the system, steady-state probabilities 46. Joe Ferris, a stock trader at the brokerage firm of Smith, Jones, Johnson, and Thomas, Inc. has noticed that price changes in the shares of Dollar Department Stores at each trade are dependent upon the previous trade's price change. His observations can be summarized by the following transition matrix. Current Price Change +1/8 0 -1/8

+1/8 .7 .3 .2

Next Price Change 0 .2 .4 .1

-1/8 .1 .3 .7

a. What is the long-run average change in the value of a share of Dollar Department Stores' stock per trade? b. If the shares of Dollar Department Stores are currently traded at $18 and the last trade was at 17 7/8, what is the probability the shares will sell at 18 in two trades? ANSWER: a. Let, p1 = the long-run probability of a +1/8 stock price change p2 = the long-run probability of a 0 stock price change p3 = the long-run probability of a -1/8 stock price change To determine the long-run (steady state) probabilities, solve:

[ p1 p2 p3 ]

.7 .3 .2

.2 .4 .1

.1 .4 .7

= [ p1 p2 p3 ]

This gives the following set of equations .7p1 + .3p2 + .2p3 = p1 (1) .2p1 + .4p2 + .1p3 = p2 (2) .1p1 + .3p2 + .7p3 = p3 (3) Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes p1 + p2 +

p3 = 1

(4)

Note, that any one of (1), (2) or (3) can be considered redundant. Delete (1) and solve (2), (3), and (4). From (4) substitute p1 = 1 − p2 − p3. This leaves the following two equations in two unknowns: .2(1 - p2 -p3) + .4p2 + .1p3 = p2 and .1(1 - p2 -p3) + .3p2 + .7p3 = p3 or .8p2 + .1p3 = .2 and -.2p2 + .4p3 = .1 Solving these two gives p2 = 7/34 and p3 = 12/34. Substituting into (4) gives p1 = 15/34. Hence, the long run average change per trade is: (15/34)(1/8) + (7/34)(0) + (12/34)(-1/8) = $.011. b. The fact that the last trade was 17 7/8 means that the last price change was +1/8. The tree diagram for the problem is:

The probability that the stock will sell at 18 after 2 trades is .07 + .08 + .02 = .17. POINTS: 1 TOPICS: Market share analysis 47. Joe Isley, the owner of Big I HiFi, believes that the store's inventory can be modeled as a Markov process. If items are either classified as in stock, out of stock, discontinued from stock or put on clearance sale, then the following transition matrix has been estimated: Next Month Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes This Month In Stock Out of Stock Discontinued Clearance Sale

In Stock .67 .48 0 0

Out of Stock .20 .42 0 0

Discontinued .05 .10 1 0

Clearance Sale .08 0 0 1

a. Rewrite the transition matrix for the problem in the form of I, O, R, and Q submatrices. b. Compute the fundamental matrix for this problem. c. What is the probability of an item currently in stock being out of stock in two months? d. What is the probability of an item currently out of stock eventually being discontinued from stock? ANSWER: a. Rearranging the states gives: This Month Discontinued Clearance Sale In Stock Out of Stock

Discontinued 1 0 .05 .10

Next Month Clearance Sale In Stock 0 0 1 0 .08 .67 0 .48

Out of Stock 0 0 .20 .42

b. N=

6.08 5.03

2.10 3.46

c. The tree diagram is:

Therefore, the probability of an item currently in stock being out of stock in two months is .13 + .08 = .21. d. The probability of eventually moving into each of the absorbing states from the nonabsorbing states is given by: Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes Discontinued Clearance In Stock .51 .49 Out of .60 .40 Stock

NR =

Hence, the probability of an item currently out of stock eventually being discontinued is .60. POINTS: 1 48. The evening television news broadcast that individuals view on one evening is influenced by which broadcast they viewed previously. An executive at the C network has determined the following transition probability matrix describing this phenomenon. Current Network News Watched A C N

A .80 .08 .08

Next Network News Watched C N .12 .08 .85 .07 .09 .83

a. Which network has the most loyal viewers? b. What are the three networks' long-run market shares? c. Suppose each of the three networks earns $1,250 in daily profit from advertising revenue for each 1,000,000 viewers it has. If on the average 40,000,000 people watch the evening television news, compute the long run average daily profit each network generates from its evening news broadcast. ANSWER: a. Network C b. (pA, pC, pN) = (.29, .41, .30) c. Network A: $14,300; Network C: $20,550; Network N: $15,200 POINTS: 1 TOPICS: State of the system, steady-state probabilities 49. Precision Craft, Inc. manufactures ornate pedestal sinks. On any day, the status of a given sink is either: a) somewhere in the normal manufacturing process, b) being reworked because of a detected flaw, c) finished successfully, or d) scrapped because a flaw could not be corrected. The transition matrix is: Today's Status In-Process Rework Finished Scrapped

In-Process .30 .40 0 0

Tomorrow's Status Rework Finished .15 .50 .10 .30 0 1 0 0

Scrapped .05 .20 0 1

a. What is the probability of a sink eventually being finished if it is currently in process? b. What is the probability of a sink eventually being scrapped if it is currently in rework? c. What is the probability that a sink currently in rework will have a "finished" status either tomorrow or the next day? (HINT: there are three ways this can happen.) ANSWER: a. .868 b. .281 c. .3 + .03 + .2 = .53 POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes TOPICS: State of the system, steady-state probabilities 50. Southside College has modeled its student loan program as a Markov process. Each year a student with a prior loan borrows again, defers repayment for a year, makes payments, pays the loan balance in full, or defaults on repayment. The transition matrix is as follows: This Year Borrowing Deferring Paying Paid-Off Defaulted

Borrowing .60 .15 0 0 0

Deferring .30 0 0 0 0

Next Year Paying 0 .65 .75 0 0

Paid-Off .10 .10 .15 1 0

Default 0 .10 .10 0 1

a. If currently a student is making payments on his/her loan, what is the probability the loan will be paid in full eventually? b. Is the probability of eventually defaulting greater for a student who is currently borrowing more or a student who is making payments? c. What is the probability a student who is borrowing this year will repay the loan balance in full in two years or less? ANSWER: a. .6 b.. .06 + .03 + .10 = .19 POINTS: 1 TOPICS: Probability at stage n Essay 51. Explain the concept of memorylessness. ANSWER: Answer not provided. POINTS: 1 TOPICS: Memoryless property 52. Where is a fundamental matrix, N, used? How is N computed? ANSWER: Answer not provided. POINTS: 1 TOPICS: Fundamental matrix 53. Why is a computer necessary for some Markov analyses? ANSWER: Answer not provided. POINTS: 1 54. What assumptions are necessary for a Markov process to have stationary transition probabilities? ANSWER: Answer not provided. POINTS: 1 TOPICS: Transition probabilities 55. Give two examples of how Markov analysis can aid decision making. ANSWER: POINTS: 1 Cengage Learning Testing, Powered by Cognero

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Chapter 16 - Markov Processes TOPICS: Introduction 56. Discuss three types of information provided by analysis of a Markov process. ANSWER: POINTS: 1 TOPICS: Fundamental matrix

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Chapter 17 - Linear Programming: Simplex Method True / False 1. When a system of simultaneous equations has more variables than equations, there is a unique solution. a. True b. False ANSWER: False 2. To determine a basic solution set of n−m, the variables equal to zero and solve the m linear constraint equations for the remaining m variables. a. True b. False ANSWER: True 3. A basic feasible solution satisfies the nonnegativity restriction. a. True b. False ANSWER: True 4. Every extreme point of the graph of a two variable linear programming problem is a basic feasible solution. a. True b. False ANSWER: True 5. In a simplex tableau, there is a variable associated with each column and both a constraint and a basic variable associated with each row. a. True b. False ANSWER: True 6. At each iteration of the simplex procedure, a new variable becomes basic and a currently basic variable becomes nonbasic, preserving the same number of basic variables and improving the value of the objective function. a. True b. False ANSWER: True 7. Coefficients in a nonbasic column in a simplex tableau indicate the amount of decrease in the current basic variables when the value of the nonbasic variable is increased from 0 to 1. a. True b. False ANSWER: True 8. If a variable is not in the basis, its value is 0. a. True b. False ANSWER: True 9. The purpose of row operations is to create a unit column for the entering variable while maintaining unit columns for Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method the remaining basic variables. a. True b. False ANSWER: True 10. We recognize infeasibility when one or more of the artificial variables do not remain in the solution at a positive value. a. True b. False ANSWER: False 11. Artificial variables are added for the purpose of obtaining an initial basic feasible solution. a. True b. False ANSWER: True 12. A solution is optimal when all values in the cj − zj row of the simplex tableau are either zero or positive. a. True b. False ANSWER: False 13. The variable to enter into the basis is the variable with the largest positive cj − zj value. a. True b. False ANSWER: True 14. The variable to remove from the current basis is the variable with the smallest positive cj − zj value. a. True b. False ANSWER: False 15. The coefficient of an artificial variable in the objective function is zero. a. True b. False ANSWER: False Multiple Choice 16. Algebraic methods such as the simplex method are used to solve a. nonlinear programming problems. b. any size linear programming problem. c. programming problems under uncertainty. d. graphical models. ANSWER: b 17. The basic solution to a problem with three equations and four variables would assign a value of 0 to Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method a. 0 variables. b. 1 variable. c. 3 variables. d. 7 variables. ANSWER: b 18. A basic solution and a basic feasible solution a. are the same thing. b. differ in the number of variables allowed to be zero. c. describe interior points and exterior points, respectively. d. differ in their inclusion of nonnegativity restrictions. ANSWER: d 19. When a set of simultaneous equations has more variables than constraints, a. it is a basic set. b. it is a feasible set. c. there is a unique solution. d. there are many solutions. ANSWER: d 20. Which is not required for a problem to be in tableau form? a. Each constraint must be written as an equation. b. Each of the original decision variables must have a coefficient of 1 in one equation and 0 in every other equation. c. There is exactly one basic variable in each constraint. d. The right-hand side of each constraint must be nonnegative. ANSWER: b 21. Unit columns are used to identify a. the tableau. b. the c row. c. the b column. d. the basic variables. ANSWER: d 22. The values in the cj − zj , or net evaluation, row indicate a. the value of the objective function. b. the decrease in value of the objective function that will result if one unit of the variable corresponding to the jth column of the A matrix is brought into the basis. c. the net change in the value of the objective function that will result if one unit of the variable corresponding to the jth column of the A matrix is brought into the basis. d. the values of the decision variables. ANSWER: c 23. The purpose of the tableau form is to provide Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method a. infeasible solution. b. optimal infeasible solution. c. initial basic feasible solution. d. degenerate solution. ANSWER: c 24. Which of the following is not a step that is necessary to prepare a linear programming problem for solution using the simplex method? a. formulate the problem. b. set up the standard form by adding slack and/or subtracting surplus variables. c. perform elementary row and column operations. d. set up the tableau form. ANSWER: c 25. A minimization problem with four decision variables, two greater-than-or-equal-to constraints, and one equality constraint will have a. 2 surplus variables, 3 artificial variables, and 3 variables in the basis. b. 4 surplus variables, 2 artificial variables, and 4 variables in the basis. c. 3 surplus variables, 3 artificial variables, and 4 variables in the basis. d. 2 surplus variables, 2 artificial variables, and 3 variables in the basis. ANSWER: a 26. In the simplex method, a tableau is optimal only if all the cj − zj values are a. zero or negative. b. zero. c. negative and nonzero. d. positive and nonzero. ANSWER: a 27. What coefficient is assigned to an artificial variable in the objective function? a. zero. b. one. c. a very large negative number. d. a very large positive number. ANSWER: c 28. When there is a tie between two or more variables for removal from the simplex tableau, a. post-optimality analysis is required. b. their dual prices will be equal. c. converting the pivot element will break the tie. d. a condition of degeneracy is present. ANSWER: d 29. An alternative optimal solution is indicated when in the simplex tableau a. a non-basic variable has a value of zero in the cj − zj row. Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method b. a basic variable has a positive value in the cj − zj row. c. a basic variable has a value of zero in the cj − zj row. d. a non-basic variable has a positive value in the cj − zj row. ANSWER: a 30. Infeasibility exists when one or more of the artificial variables a. remain in the final solution as a negative value. b. remain in the final solution as a positive value. c. have been removed from the basis. d. remain in the basis. ANSWER: b Subjective Short Answer 31. Write the following problem in tableau form. Which variables would be in the initial basic solution? Min Z = 3x1 + 8x2 s.t. x1 + x2 ≤ 200 x1 ≤ 80 x2 ≤ 60 ANSWER: Min Z = 3x1 + 8x2 + Ma1 + Ma2 s.t. x1 + x2 + a1 = 200 x1 + s1 = 80 x2 + s2 + a2 = 60 The initial basis includes a1, a2, and s2. 32. Given the following initial simplex tableau

Basis s1 s2 s3

cB 0 0 0 zj cj − zj

x1 5

x2 8

x3 12

s1 0

s2 0

s3 0

3 9 1

4 15 −1

5 20 2

1 0 0

0 1 0

0 0 0

80 250 20

0 5

0 8

0 12

0 0

a. b. c. d.

What variables form the basis? What are the current values of the decision variables? What is the current value of the objective function? Which variable will be made positive next, and what will its value be? Which variable that is currently positive will become 0? f. What value will the objective function have next? ANSWER: a. s1, s2 , s3 Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method b. c. d. e. f.

x1 = 0, x2 = 0, x3 = 0, s1 = 80, s2 = 250, s3 = 20 0 x3, 10 s3 z = 120

33. Given the following simplex tableau

Basis

x1 3 1/2 0

x2 4 1 0

x3 5 0 1

s1 0 1/2 −1/4

s2 0 −1/2 1

6 3

cj cj − zj a. What variables form the basis? b. What are the current values of the decision variables? c. What is the current value of the objective function? d. Which variable will be made positive next, and what will its value be? e. Which variable that is currently positive will become 0? f. What value will the objective function have next? ANSWER: a. x2, x3 b. x1 = 0, x2 = 6, x3 = 3 c. 39 d. x1, 12 e. x2 f. 51 34. A simplex tableau is shown below.

Basis s1 s2 x3

cB 0 0 8

x1 3

x2 5

x3 8

s1 0

s2 0

s3 0

3 −5/2 1/2

6 −1/2 1/2

0 0 1

1 0 0

0 1 0

−9 −9/2 1/2

126 45 18

zj cj − zj

4 −1

4 1

8 0

0 0

4 −4

144

a. Do one more iteration of the simplex procedure. b. What is the current complete solution? c. Is this solution optimal? Why or why not? ANSWER: a. x1 x2 x3 Basis 3 5 8 cB 5 1/2 1 0 x2 0 −9/4 0 0 s2 Cengage Learning Testing, Powered by Cognero

s1 0

s2 0

s3 0

1/6 1/12

0 1

−3/2 −21/4

21 111/2 Page 6

Chapter 17 - Linear Programming: Simplex Method 8

1/4

−1/12

−5/4

15/2

zj cj − zj

9/2 −3/2

5 0

8 0

1/6 −1/6

0 0

5/2 −5/2

165

b. x1 = 0, x2 = 21, x3 = 7.5, s1 = 0, s2 = 55.5, s3 = 0, Z = 165 c. The solution is optimal because all cj − zj ≤ 0. 35. A simplex table is shown below.

Basis s1 x3 s3

cB 0 8 0

x1 5

x2 4

x3 8

s1 0

s2 0

s3 0

2/5 4/5 4/5

−3/5 4/5 9/5

0 1 0

1 0 0

−2/5 1/5 1/5

0 0 1

4 8 10

zj cj − zj

32/5 −7/5

32/5 −12/5

8 0

0 0

8/5 −8/5

0 0

a. What is the current complete solution? b. The 32/5 for z1 is composed of 0 + 8(4/5) + 0. Explain the meaning of this number. c. Explain the meaning of the −12/5 value for c 2 − z2. ANSWER: a. x1 = 0, x2 = 0, x3 = 8, s1 = 4, s2 = 0, s3 = 10, Z = 64

If we make one unit of x 1 it will require .4 units of s1, at no cost, .8 units of x3 production will have to be foregone at a cost of 8 each, and .8 units of s3 will be needed at no cost. In order to make a unit of x1, profit of 32/5 will be given up.

The profit for a unit of x2 is 4. However, in order to make a unit of x2, profit of 32/5 would be given up as resources are diverted. The net improvement is a loss, shown by the −12/5. Therefore it makes no sense to produce x2.

36. Solve the following problem by the simplex method. Max s.t.

14x1 + 14.5x2 + 18x3 x1 + 2x2 + 2.5x3 ≤ 50 x1 + x2 + 1.5x3 ≤ 30 x1 , x2 , x3 ≥ 0 ANSWER: Basis s1 s2

cB 0 0 zj cj − zj

x1 14

x2 14.5

x3 18

s1 0

s2 0

1 1

2 1

2.5 1.5

1 0

0 1

50 30

0 14

0 14.5

0 18

0 0

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Chapter 17 - Linear Programming: Simplex Method

Basis x3 s2

cB 18 0 zj cj − zj

Basis x3 x1

cB 18 14 zj cj − zj

Basis x2 x1

cB 14.5 14 zj cj − zj

x1 14

x2 14.5

x3 18

s1 0

s2 0

.4 .4

.8 −.2

1 0

.4 −.6

0 1

20 0

7.20 6.8

14.4 .1

18 0

7.2 −7.2

0 0

360

x1 14

x2 14.5

x3 18

s1 0

s2 0

0 1

1 −.5

1 0

1 −1.5

−1 2.5

20 0

14 0

11 3.5

18 0

−3 −3

17 −17

360

x1 14

x2 14.5

x3 18

s1 0

s2 0

0 1

1 0

1 .5

1 −1

−1 2

20 10

14 0

14.5 0

21.5 −3.5

.5 −.5

13.5 −13.5

430

37. Solve the following problem by the simplex method. Max s.t.

100x1 + 120x2 + 85x3 3x1 + 1x2 + 6x3 ≤ 120 5x1 + 8x2 + 2x3 ≤ 160 x1 , x2 , x3 ≥ 0 ANSWER: Basis s1 s2

x1 100

x2 120

x3 85

s1 0

s2 0

3 5

1 8

6 2

1 0

0 1

120 160

0 100

0 120

0 85

0 0

cB 0 120

x1 100

x2 120

x3 85

s1 0

s2 0

2.375 .625

0 1

5.75 .25

1 0

−.125 .125

100 20

zj cj − zj

75 25

120 0

30 55

0 0

15 −15

2400

x1 100

x2 120

x3 85

s1 0

s2 0

cB 0 0 zj cj − zj

Basis s1 x2

Basis

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Chapter 17 - Linear Programming: Simplex Method 85 120

.413 .522

0 1

1 0

.174 −.043

−.0217 .1304

17.391 15.652

zj cj − zj

97.745 2.283

120 0

85 0

9.63 −9.565

13.8035 −13.8035

3356.52

x1 100

x2 120

x3 85

s1 0

s2 0

0 1

−.792 1.917

1 0

.2083 −.0833

−.125 .250

5 30

100 0

124.38 −4.375

85 0

9.3755 −9.375

14.375 −14.375

3425

x3 x2

Basis

cB 85 100

x3 x1

zj cj − zj

38. Determine from a review of the following tableau whether the linear programming problem has multiple optimal solutions. Basis s3 x2 x1

cB 0 2 3

x1 0 0 1

x2 0 1 0

s1 1 0 0

s2 −1/5 1/5 1/5

s3 8/6 −3/5 2/5

zj cj −zj

3 0

2 0

0 0

0 −1

0 0

6 1 4 14

ANSWER: Since non-basic variable s3 has a relative profit of zero, this means that any increase in s3 will produce no change in the objective function value. Thus, since s3 can be made a basic variable, the resulting basic feasible solution will also have an optimum value of 14. An alternative optimal solution is indicated whenever there exists a non-basic variable whose relative profit (cj − z ) row coefficient is zero in the optimal solution. 39. Write the following problem in tableau form. Which variables would be in the initial basis? Max s.t.

x1 + 2x2 3x1 + 4x2 ≤ 100 2x1 + 3.5x2 ≥ 60 2x1 − 1x2 = 4 x1 , x2 ≥ 0 ANSWER: Max x1 + 2x2 − Ma1 − Ma2 s.t. 3x1 + 4x2 + s1 = 100 2x1 + 3.5x2 − s2 + a1 = 60 2x1 − 1x2 + a2 = 50 x1 , x2 , s1 , s2 , a1 , a2 ≥ 0 The initial basis would include s1, a1, and a2 . 40. Write the following problem in tableau form. Which variables would be in the initial basic solution? Min Z

= −3x1 + x2 + x3

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Chapter 17 - Linear Programming: Simplex Method x1 − 2x2 + x3 ≤ 11 −4 x1 + x2 + 2x3 ≥ 3 2x1 − x3 ≥ −1 ANSWER: Min Z = −3x1 + x2 + x3 + Ma1 + Ma2 s.t. x1 − 2x2 + x3 + s1 = 11 −4x1 + x2 + 2 x3 − s2 + a1 = 3 −2x1 + x3 + a2 = 1 The initial basis includes s1, a1, and a2. 2x1 + 3.5x2 − s2 + a1 = 60 2x1 − 1x2 + a2 = 50 s.t.

41. Comment on the solution shown in this simplex tableau.

Basis x3 s2

cB 9 0

x1 3

x2 5

x3 9

s1 0

s2 0

1/3 5/3

2/3 19/3

1 0

1/3 −1/3

0 1

40 100

0 0

360

3 6 9 3 zj 0 −1 0 −3 cj − zj ANSWER: The tableau indicates alternate optimal solutions. 42. Comment on the solution shown in this simplex tableau.

Basis a1 x3

cB −M 5

x1 1

x2 2

x3 5

s1 0

s2 0

a1 −M

−3 1

−1 1/2

0 1

−1 0

−2 1/2

1 0

4 4

2.5+2M −2.5−2M

−M 0

−4M+20

5+3M 2.5+M 5 M zj 0 −M cj − zj −4−3M −.5−M ANSWER: The tableau indicates an infeasible solution. 43. Comment on the solution shown in this simplex tableau.

Basis s1 x2

cB 0 6

x1 4

x2 6

x3 5

s1 0

s2 0

3 1

0 1

−1 1

1 0

−1 −1

26 10

−6 6

6 6 6 0 zj −2 0 −1 0 cj − zj ANSWER: The tableau indicates an unbounded solution

44. Joe Forrester, an operations analyst for a manufacturing company, developed the following LP formulation and wants to create an initial simplex tableau. Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method Max 40xl + 30x2 + 50x3 s.t. 2xl + 3x2 + 4x3 < 200 x1 + 2x2 + 2x3 < 300 3xl + x2 + 5x3 < 500 ANSWER: Initial tableau

Basis

cB 0 0 0

s1 s2 s3

zj cj - zj

x1 40

x2 30

x3 50

s1 0

s2 0

s3 0

3 9 1

4 15 -1

5 20 2

1 0 0

0 1 0

0 0 1

200 300 500

0 40

0 30

0 50

0 0

45. Jackie Quinn developed the following LP formulation for a problem she is working on and now needs to create an initial simplex tableau. Min 75xl + 45x2 s.t. 3xl + 2x2 > 10 xl + 6x2 > 15 ANSWER:

Initial tableau

Basis

x1 -75

x2 -45

a1 -M

a2 -M

s1 0

s2 0

3 1

2 6

1 0

0 1

-1 0

0 -1

10 15

-4M -75+4M

-8M -45+8M

-M 50

-M 0

M -M

-25M

cB 0 0

a1 a2

zj cj - zj

46. A student in a Management Science class developed this initial tableau for a maximization problem and now wants to perform row operations to obtain the next tableau and check for an optimal solution.

Basis s1 s2 s3

cB 0 0 0 zj cj - zj

x1 550

x2 350

s1 0

s2 0

s3 0

2 2 1

2 1 2

1 0 0

0 1 0

0 0 1

2000 1500 3000

0 550

0 350

0 0

ANSWER: Second tableau Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method

Basis s1 x1 s3

cB 0 550 0 zj cj - z j

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 1/2 3/2

1 0 0

-1 1/2 -1/2

0 0 1

500 750 2250

550 0

275 75

0 0

275 -275

0 0

412500

Solution is not optimal. There is a positive cj - zj value. 47. A student in a Management Science class wants to perform row operations on this second tableau and complete the third tableau to see if it is optimal. If it is optimal, what is the optimal answer?

Basis s1 x1 s3

cB 0 550 0 zj cj - zj

ANSWER:

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 1/2 3/2

1 0 0

-1 1/2 -1/2

0 0 1

500 750 2250

550 0

275 75

0 0

412500

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 0 0

1 -1/2 -3/2

-1 1 1

0 0 1

500 500 1500

550 0

350 0

75 -75

200 -200

0 0

450000

275 -275

Third tableau Basis x2 x1 s3

cB 350 550 0 zj cj - z j

Solution is optimal. There are no positive cj - zj values. x1 = 550, x2 = 350, and objective function value is 450,000. 48. An operations research analyst for a communications company has the following LP problem and wants to solve it using the simplex method. Max 50x1 + 20x2 s.t. 2x1 + x2 < 200 x1 + x2 < 350 xl + 2x2 < 275 ANSWER: Initial tableau Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method

Basis s1 s2 s3

cB 0 0 0 zj cj - zj

x1 50

x2 20

s1 0

s2 0

s3 0

2 1 1

1 1 2

1 0 0

0 1 0

0 0 1

200 350 275

0 50

0 20

0 0

Second tableau

Basis

cB 50 0 0

x1 s2 s3

zj cj - z j

x1 50

x2 20

s1 0

s2 0

s3 0

1 0 0

1/2 1/2 3/2

1/2 -1/2 -1/2

0 1 0

0 0 1

100 250 175

50 0

25 -5

25 -25

0 0

5000

Tableau is optimal. Optimal solution is x1 = 100 and the objective function = 5000. 49. The operations research analyst for a big manufacturing firm in Oregon developed the following variable definitions for a LP maximization problem she was working on. The company was trying to determine how many consoles of each model to produce next week given each console had to go through three production departments. She obtained the following optimal simplex tableau for the problem and wanted to interpret its meaning. Variable definitions: xl = number of model 1 consoles produced x2 = number of model 2 consoles produced s1 = unused personnel hours in department 1 s2 = unused personnel hours in department 2 s3 = unused personnel hours in department 3 objective function = total profit on model 1 and model 2 consoles produced in the coming week Optimal tableau Basis x1 s2 x2

ANSWER:

cB 180 0 350

180

350

1 0 0

0 0 1

1 -1 0

0 1 0

-1 1 1

15 10 20

zj cj - zj

180 0

350 0

180 -180

0 0

170 -170

9700

Optimal solution:

produce 15 model 1 consoles, produce 20 model 2 consoles, all personnel hours will be used in departments 1 and 3, 10 personnel hours will be unused in department 2, total profit next week will be $9,700.

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Chapter 17 - Linear Programming: Simplex Method 50. A manager for a food company is putting together a buffet and she is trying to determine the best mix of crab and steak to be served. Below are variable definitions she developed including vitamin, mineral and protein requirements. Also included below is an optimal simplex tableau she obtained from her computations. She is interested in interpreting what it means. Variable definitions: xl = amount of crab (oz) to be served per buffet batch x2 = amount of steak (oz) to be served per buffet batch s1 = vitamin A units provided in excess of requirements s2 = mineral units provided in excess of requirements s3 = protein units provided in excess of requirements Optimal tableau Basis s2 x2 x1

cB 0 -4 -7 zj cj - zj

ANSWER:

x1 -7

x2 -4

a1 -M

a2 -M

a3 -M

s1 0

s2 0

s3 0

0 0 1

0 1 0

4 1/50 -1/100

-2 0 0

-1 -1/100 1/100

-4 -1/50 1/100

1 0 0

2 1/100 -1/100

2500 20 15

-7 0

-4 0

1/100 -M

0 -M

-3/100 -M

1/100 -1/100

0 0

170 -3/100

-185

Optimal solution:

15 ounces of crab in each batch, 20 ounces of steak in each batch, 0 excess units of vitamin A provided per batch, 2500 excess units of minerals provided per batch, 0 excess units of protein provided per batch, total cost per batch = $185.

Essay 51. What is the criterion for entering a new variable into the basis? ANSWER: Answer not provided. 52. A portion of a simplex tableau is

Basis x2 s2

cB 25 0 zj cj − zj

x1 20 .2 3 5 15

Give a complete explanation of the meaning of the z1 = 5 value as it relates to x2 and s2. ANSWER: Answer not provided. 53. What is an artificial variable? Why is it necessary? Cengage Learning Testing, Powered by Cognero

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Chapter 17 - Linear Programming: Simplex Method ANSWER: Answer not provided. 54. Describe and illustrate graphically the special cases that can occur in a linear programming solution. What clues for these cases does the simplex procedure supply? ANSWER: Answer not provided. 55. For each of the special cases of infeasibility, unboundedness, and alternate optimal solutions, tell what you would do next with your linear programming model if the case occurred. ANSWER: Answer not provided. 56. List the steps to get a problem formulation to tableau form. ANSWER: Answer not provided. 57. What is degeneracy and what can be done in the simplex procedure to overcome the problem? ANSWER: Answer not provided.

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality True / False 1. The range of optimality is useful only for basic variables. a. True b. False ANSWER: False 2. The range of optimality is calculated by considering changes in the cj − zj value of the variable in question. a. True b. False ANSWER: False 3. As long as the objective function coefficient remains within the range of optimality, the variable values will not change although the value of the objective function could. a. True b. False ANSWER: True 4. If the simplex tableau is from a maximization converted from a minimization, the signs and directions of the inequalities that give the objective function ranges will need to be adjusted to apply to the original coefficients. a. True b. False ANSWER: True 5. The ranges for which the right-hand side values are valid are the same as the ranges over which the dual prices are valid. a. True b. False ANSWER: True 6. There is a dual price associated with each decision variable. a. True b. False ANSWER: False 7. The dual price for an equality constraint is the zj value for its artificial variable. a. True b. False ANSWER: True 8. The entries in the associated slack column of the final tableau indicate the changes in the values of the current basic variables corresponding to a one-unit increase in the right-hand side. a. True b. False ANSWER: True 9. The range of optimality for a basic variable defines the objective function coefficient values for which the variable will Cengage Learning Testing, Powered by Cognero

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality remain part of the current optimal basic feasible solution. a. True b. False ANSWER: True 10. The dual price is the improvement in value of the optimal solution per unit increase in the value of the right-hand side associated with a linear programming problem. a. True b. False ANSWER: True Multiple Choice 11. Dual prices and ranges for objective function coefficients and right-hand side values are found by considering a. dual analysis. b. optimality analysis. c. ranging analysis. d. sensitivity analysis. ANSWER: d 12. For the basic feasible solution to remain optimal a. all cj − zj values must remain ≤ 0. b. no objective function coefficients are allowed to change. c. the value of the objective function must not change. d. each of the above is true. ANSWER: a 13. A one-sided range of optimality a. always occurs for non-basic variables. b. always occurs for basic variables. c. indicates changes in more than one coefficient. d. indicates changes in a slack variable's coefficient. ANSWER: a 14. A linear programming problem with the objective function 3x1 + 8x2 has the optimal solution x1 = 5, x2 = 6. If c2 decreases by 2 and the range of optimality shows 5 ≤ c2 ≤ 12, the value of Z a. will decrease by 12. b. will decrease by 2. c. will not change. d. cannot be determined from this information. ANSWER: a 15. The improvement in the value of the optimal solution per-unit increase in a constraint's right-hand side is a. the slack value. b. the dual price. Cengage Learning Testing, Powered by Cognero

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality c. never negative. d. the 100% rule. ANSWER: b 16. The dual variable represents a. the marginal value of the constraint b. the right-hand side value of the constraint c. the artificial variable d. the technical coefficient of the constraint ANSWER: a 17. The range of feasibility indicates right-hand side values for which a. the value of the objective function will not change. b. the values of the decision variables will not change. c. those variables which are in the basis will not change. d. more simplex iterations must be performed. ANSWER: c 18. If the dual price for b1 is 2.7, the range of feasibility is 20 ≤ b1 ≤ 50, and the original value of b1 was 30, which of the following is true? a. There currently is no slack in the first constraint. b. We would be willing to pay up to $2.70 per unit for up to 20 more units of resource 1. c. If only 25 units of resource 1 were available, profit would drop by $13.50. d. Each of the above is true. ANSWER: d 19. The number of constraints to the dual of the following problem is: Max Z s.t.

= 3x1 + 2x2 + 6x3 4x1 + 2x2 + 3x3 ≥ 100 2x1 + x2 − 2x3 ≤ 200 4x2 + x3 ≥ 200

a. 1. b. 2. c. 3. d. 4. ANSWER: c 20. Given the simplex tableau for the optimal primal solution a. the values of the dual variables can be found from the cj − zj values of the slack/surplus variable columns. b. the values of the dual surplus variables can be found from the cj − zj values of the primal decision variable columns. c. the value of the dual objective function will be the same as the objective function value for the primal problem. d. each of the above is true. Cengage Learning Testing, Powered by Cognero

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality ANSWER: d Subjective Short Answer 21. Given the following linear programming problem Max Z s.t.

0.5x1 + 6x2 + 5x3 4x1 + 6x2 + 3x3 ≤ 24 1x1 + 1.5x2 + 3x3 ≤ 12 3x1 + x2 ≤ 12

and the final tableau is

Basis x2 x3 s3

cB 6 5 0

x1 .5

x2 6

x3 5

s1 0

s2 0

s3 0

1 0 2.33

1 0 0

0 1 0

.22 .11 −.22

−.22 −.44 .22

0 0 1

2.67 2.67 9.33

zj cj − zj

4 .5

6 0

5 0

.77 −.77

.88 −.88

0 0

29.33

a. Find the range of optimality for c1, c2, c3, c4, c5, and c6. b. Find the range of feasibility for b1, b2, and b3. ANSWER: a. c1 ≤ 4 2.5 < c2 ≤ 10 3 < c3 ≤ 12 c4 ≤ .77 c5 ≤ .88 −1.5 ≤ c6 ≤ 3.5 b.

12 ≤ b1 ≤ 48 6 ≤ b2 ≤ 24 b3 ≥ 2.67

22. For the following linear programming problem Max Z s.t.

−2x1 + x2 − x3 2x1 + x2 ≤ 7 1x1 + x2 + x3 ≥ 4

the final tableau is

Basis s2 x2

cB 0 1

x1 −2

x2 1

x3 −1

s1 0

s2 0

a2 −M

1 2

0 1

−1 0

1 1

1 0

−1 0

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3 7 Page 4

Chapter 18 - Simplex-Based Sensitivity Analysis and Duality 2 −4

zj cj − zj

1 0

0 1

1 −1

0 0

0 −M

a. Find the range of optimality for c1, c2 , c3. c4, c5 , and c6. b. Find the range of feasibility for b1, and b2. ANSWER: a. c1 ≤ 2 0 ≤ c2 c3 ≤ 0 c4 ≤ 1 −1 ≤ c5 ≤ 1 c6 ≤ 0 b.

4 ≤ b1 b2 ≤ 7

23. Write the dual of the following problem = 2x1 − 3x2 + 5x3 −3x1 + 2x2 + 5x3 ≥ 7 2x1 − x3 ≥ 5 4x 2 + 3x3 ≥ 8. ANSWER: Max Z2 = 7y1 + 5y2 + 8y3 s.t. −3y1 +2y2 ≤ 2 2y1 + 4y3 ≤ −3 5y1 − y2 + 3y3 ≤ 5 Min Z s.t.

24. Given the following linear programming problem Max s.t.

10x1 + 12x2 1x1 + 2x2 ≥ 40 5x1 + 8x2 ≤ 160 1x1 + 1x2 ≤ 40 x1, x2 ≥ 0

the final tableau is

Basis x2 x1 s3

cB 12 10 0 zj cj − zj

x1 10

x2 12

s1 0

s2 0

s3 0

0 1 0

1 0 0

−2.5 4 −1.5

−.5 1 −.5

0 0 1

20 0 20

10 0

12 0

10 −10

4 −4

0 0

240

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality a. Find the range of optimality for c1 and c2. b. Find the range of feasibility for b1, b2, and b3. c. Find the dual prices. ANSWER: a. 7.5 ≤ c1 < ∞ −∞ < c2 ≤ 16 b.

32 ≤ b1 < 0 160 ≤ b2 ≤ 200 20 ≤ b3 < ∞

u1 = 10 u2 = 4 u3 = 0

25. For this optimal simplex tableau the original right-hand sides were 100 and 90. The problem was a maximization.

Basis x3 x1

cB 8 2 zj cj − zj

x1 2

x2 4

x3 8

s1 0

s2 0

0 1

.48 .2

1 0

.12 −.2

−.04 .4

8.4 16

2 0

4.24 −.24

8 0

.56 −.56

.48 −.48

99.2

What would the new solution be if there had been 150 units available in the first constraint? What would the new solution be if there had been 70 units available in the second b. constraint? ANSWER: a. x3 = 14.4, x1 = 6, Z = 127.2 b. x3 = 9.2, x1 = 8, Z = 89.6 26. For this optimal simplex tableau, the right-hand sides for the two original ≥ constraints were 300 and 250. The problem was a minimization.

Basis x3 x1

cB −95 −100 zj cj − zj

x1 −100

x2 −110

x3 −95

s1 0

s2 0

0 1

−1.5 2

1 0

1 −1

−1.5 1

75 50

−100 0

−57.5 −52.5

−95 0

5 −5

42.5 −42.5

−12125

What would the new solution be if the right-hand side value in the first constraint had been 325? What would the new solution be if the right-hand side value for the second constraint had b. been 220? ANSWER: a. x3 = 50, x1 = 75, Z = 12250 a.

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality b.

x3 = 30, x1 = 80, Z = 10850

27. Creative Kitchen Tools manufactures a wide line of gourmet cooking tools from stainless steel. For the coming production period, there is demand of 1200 for 8 quart stock pots, and unlimited demand for 3 quart mixing bowls and large slotted spoons. In the following model, the three variables measure the number of pots, bowls, and spoons to make. The objective function measures profit. Constraint 1 measures steel, constraint 2 measures manufacturing time, constraint 3 measures finishing time, and constraint 4 measures the stock pot demand. Max s.t.

5x1 + 3x2 + 6x3 3x1 + 1x2 + 2x3 ≤ 15000 4x1 + 4x2 + 5x3 ≤ 18000 2x1 + 1x2 + 2x3 ≤ 10000 x1 ≤ 1200 x1, x2, x3 ≥ 0

The final tableau is:

Basis s1 s4 s3 x1

cB 0 0 0 5 zj cj − zj

x1 5

x2 3

x3 6

s1 0

s2 0

s3 0

s4 0

0 0 0 1

−2 1 −1 1

−1.75 1.25 −.5 1.25

1 0 0 0

−.75 .25 −.5 .25

0 0 1 0

0 1 0 0

1500 3300 1000 4500

5 0

5 −2

6.25 −.25

0 0

1.25 −1.25

0 0

22500

a. b. c.

Calculate the range of optimality for c1, c2, and c3. Calculate the range of feasibility for b1, b2, b3, and b4. Suppose that the inventory records were incorrect and the company really has only 14000 units of steel. What effect will this have on your solution? d. Suppose that a cost increase will change the profit on the pots to $4.62. What effect will this have on your solution? e. Assume that the cost of time in production and finishing is relevant. Would you be willing to pay a $1.00 premium over the normal cost for 1000 more hours in the production department? What would this do to your solution? ANSWER: a. 4.80 ≤ c1 < ∞ −∞ < c2 ≤ 5 −∞ < c3 ≤ 6.25 b.

13500 ≤ b1 < ∞ 480 ≤ b2 ≤ 20000 9000 ≤ b3 < ∞ −∞ < b4 ≤ 4500

This would affect only the amount of slack, decreasing it from 1500 to 500.

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality d.

This change is out of the range of optimality so the basis would change.

An increase of 1000 hours is within the range of feasibility and the cost is less than the dual price, so it makes sense to do this. The new solution would be s1 = 1500 − .75(1000) = 750 s4 = 3300 + .25(1000) = 3550 s3 = 1000 − .5(1000) = 500 x1 = 4500 + .25(1000) = 4750 Z = 22500 + (1.25 − 1)(1000) = 22750

28. Stelle Office Supplies must fill an order for 2000 modular office dividers. Each divider consists of a frame, a set of legs, and a panel. SOS has limited production and finishing time available and is considering the purchase of some of the components. Let x1, x2, and x3 be the number of frames, leg sets, and panels to make, and x4, x5, and x6 be the number of each to buy. The model reflects the costs to be minimized, the amount of production time, the amount of assembly time, and the need for 2000 of each component. Min s.t.

20x1 + 14x2 + 15x3 + 28x4 + 20x5 + 25x6 30x1 + 40x2 + 25x3 ≤ 180000 15x1 + 10x2 + 30x3 ≤ 90000 x1 + x4 = 2000 x2 + x5 = 2000 x3 + x6 = 2000 all xi ≥ 0

The final tableau is

Basis x6 s1 x1 x2 x3

cB −25 0 −20 −14 −15 zj cj − zj

x1 −20

x2 −14

x3 −15

x4 −28

x5 −20

x6 −25

s1 0

s2 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

.5 −17.5 1 0 −.5

.333 −31.67 0 1 −.333

1 0 0 0 0

0 1 0 0 0

−.033 −.833 0 0 .033

666.67 6666.67 2000 2000 1333.33

−20 −14 0 0

−15 0

−25 −3

−17.33 −2.67

−25 0

0 0

.33 −.33

68667

a. b. c.

Calculate the range of optimality for all of the objective function coefficients. Calculate the range of feasibility for the first two right-hand sides. How much less expensive would it have to be to buy frames before you would consider it? How much more expensive would legs have to be to make before you would change your d. solution? e. What would the total cost be if the cost to make a panel increased by $3.00? f. What would you be willing to pay for more production time? What would happen to the total cost if the amount of assembly time decreased by 2000 g. hours? ANSWER: a. −∞ < c1 ≤ 23 −∞ < c2 ≤ 16.67 Cengage Learning Testing, Powered by Cognero

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality 9 ≤ c3 ≤ 25 25 ≤ c4 < ∞ 17.33 ≤ c5 < ∞ 15 ≤ c6 ≤ 31 b.

173333.33 ≤ b1 < ∞ 50000 ≤ b2 ≤ 98000 6666.67 ≤ b3 ≤ 2380.95 0 ≤ b4 ≤ 2210.53 1333.33 ≤ b5 < ∞

$3.00 per frame.

$2.67 additional

Cost would increase by 1333.33(3) = 4000

Nothing

.33(2000) = 666.67 more cost

29. Write the dual to the following problem. Min s.t.

12x1 + 15x2 + 20x3 + 18x4 x1 + x2 + x3 + x4 ≥ 50 3x1 + 4x3 ≥ 60 2x2 + x3 − 2x4 ≤ 10 x1, x2, x3, x4 ≥ 0 ANSWER: Max 50u1 + 60u2 − 10u3 s.t. u1 + 3u2 ≤ 12 u1 − 2u3 ≤ 15 u1 + 4u2 − u3 ≤ 20 u1 + 2u3 ≤ 18 u1, u2, u3 ≥ 0 30. The primal problem is Min s.t.

2x1 + 5x2 + 4x3 x1 + 3x2 + 3x3 ≥ 30 3x1 + 7x2 + 5x3 ≥ 70 x1, x2, x3 ≥ 0

The final tableau for its dual problem is u1

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s3 Page 9

Chapter 18 - Simplex-Based Sensitivity Analysis and Duality Basis u2 s2 u1

cB 70 0 30

0 0 1

1 0 0

3/4 −3/2 −5/4

0 1 0

−1/4 −1/2 3/4

1/2 0 1/2

zj cj − zj

30 0

70 0

15 −15

0 0

5 −5

Give the complete solution to the primal problem. ANSWER: The complete primal solution is x1 = 15, x2 = 0, x3 = 5, s1 = 0, s2 = 0, Z = 50 31. The linear programming problem: Max s.t.

6x1 + 2x2 + 3x3 + 4x4 x1 + x2 + x3 + x4 ≤ 100 4x1 + x2 + x3 + x4 ≤ 160 3x1 + x2 + 2x3 + 3x4 ≤ 240 x1, x2, x 3, x4 ≥ 0

has the final tableau:

Basis x2 x1 x4

cB 2 6 4 zj cj − zj

x1 6

x2 2

x3 3

x4 4

s1 0

s2 0

s3 0

0 1 0

1 0 0

1/2 0 1/2

0 0 1

3/2 0 −1/3 1/3 −1/6 −1/3

−1/2 0 1/2

30 20 50

6 0

2 0

3 0

4 0

2.33 −2.33

1 −1

380

.67 −.67

Fill in the table below to show what you would have found if you had used The Management Scientist to solve this problem. LINEAR PROGRAMMING PROBLEM MAX

6X1+2X2+3X3+4X4

S.T. 1) 1X1 + 1X2 + 1X3 + 1X4 < 100 2) 4X1 + 1X2 + 1X3 + 1X4 < 160 3) 3X1 + 1X2 + 2X3 + 3X4 < 240 OPTIMAL SOLUTION Objective Function Value = Variable X1 X2 X3 X4

Value ______ ______ ______ ______

Reduced Cost ______ ______ ______ ______

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality Constraint 1 2 3

Slack/Surplus ______ ______ ______

Dual Price ______ ______ ______

OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value X1 ______ ______ X2 ______ ______ X3 ______ ______ X4 ______ ______

Upper Limit ______ ______ ______ ______

RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 ______ ______ ______ 2 ______ ______ ______ 3 ______ ______ ______ ANSWER: LINEAR PROGRAMMING PROBLEM MAX

6X1 + 2X2 + 3X3 + 4X4

S.T. 1) 1X1 + 1X2 + 1X3 + 1X4 < 100 2) 4X1 + 1X2 + 1X3 + 1X4 < 160 3) 3X1 + 1X2 + 2X3 + 3X4 < 240 OPTIMAL SOLUTION Objective Function Value = 380.000 Variable X1 X2 X3 X4

Value 20.000 30.000 0.000 50.000

Reduced Cost 0.000 0.000 0.000 0.000

Constraint Slack/Surplus Dual Price 1 0.000 0.333 2 0.000 0.667 3 0.000 1.000 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit X1 4.000 6.000 7.000 X2 2.000 2.000 4.000 X3 No Lower Limit 3.000 3.000 X4 4.000 4.000 6.000 RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 80.000 100.000 160.000 2 100.000 160.000 310.000 3 140.000 240.000 300.000 Essay 32. For an objective function coefficient change outside the range of optimality, explain how to calculate the new optimal solution. Must you return to the (revised) initial tableau? Cengage Learning Testing, Powered by Cognero

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Chapter 18 - Simplex-Based Sensitivity Analysis and Duality ANSWER: Answer not provided. 33. When sensitivity calculations yield several potential upper bounds and several lower bounds, how is the range determined? ANSWER: Answer not provided. 34. Explain why the zj value for a slack variable is the dual price. ANSWER: Answer not provided. 35. Explain how to put an equality constraint into canonical form and how to calculate its dual variable value. ANSWER: Answer not provided. 36. Explain the simplex tableau location of the dual constraint for each type of constraint. ANSWER: Answer not provided.

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems True / False 1. The transportation simplex method can be used to solve the assignment problem. a. True b. False ANSWER: True 2. The transportation simplex method is limited to minimization problems. a. True b. False ANSWER: False 3. For an assignment problem with 3 agents and 4 tasks, the assignment matrix will have 3 rows and 4 columns. a. True b. False ANSWER: False 4. If a transportation problem has four origins and five destinations, one of the destinations will not be fully supplied. a. True b. False ANSWER: False 5. When an assignment problem involves an unacceptable assignment, a dummy agent or task must be introduced. a. True b. False ANSWER: False 6. In assignment problems, dummy agents or tasks are created when the number of agents and tasks is not equal. a. True b. False ANSWER: True 7. The transportation simplex method is more efficient than general-purpose linear programming for solving large-sized transportation problems. a. True b. False ANSWER: True 8. A dummy origin in a transportation problem is used when supply exceeds demand. a. True b. False ANSWER: False 9. The net evaluation index for occupied cells in the transportation simplex method is 0. a. True b. False ANSWER: True Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems 10. Optimal assignments are made in the Hungarian method to cells in the reduced matrix that contain a 0. a. True b. False ANSWER: True 11. Using the Hungarian method, the optimal solution to an assignment problem is found when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of agents. a. True b. False ANSWER: True 12. To handle unacceptable routes in a transportation problem where cost is to be minimized, infeasible arcs must be assigned negative cost values. a. True b. False ANSWER: False Multiple Choice 13. A solution to a transportation problem that has less than m + n − 1 cells with positive allocations in the transportation tableau is a. an optimal solution. b. an initial feasible solution. c. a minimum-cost solution. d. a degenerate solution. ANSWER: d 14. The optimal solution is found in an assignment matrix when the minimum number of straight lines needed to cover all the zeros equals a. (the number of agents) − 1. b. (the number of agents). c. (the number of agents) + 1. d. (the number of agents) + (the number of tasks). ANSWER: b 15. The stepping-stone method requires that one or more artificially occupied cells with a flow of zero be created in the transportation tableau when the number of occupied cells is fewer than a. m + n − 2 b. m + n − 1 c. m + n d. m + n + 1 ANSWER: b 16. The per-unit change in the objective function associated with assigning flow to an unused arc in the transportation simplex method is called the a. net evaluation index. Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems b. degenerate value. c. opportunity loss. d. simplex multiplier. ANSWER: a 17. The difference between the transportation and assignment problems is that a. total supply must equal total demand in the transportation problem b. the number of origins must equal the number of destinations in the transportation problem c. each supply and demand value is 1 in the assignment problem d. there are many differences between the transportation and assignment problems ANSWER: c 18. An example of a heuristic is the a. minimum-cost method. b. stepping-stone method. c. Hungarian method. d. MODI method. ANSWER: a 19. Using the transportation simplex method, the optimal solution to the transportation problem has been found when a. there is a shipment in every cell. b. more than one stepping-stone path is available. c. there is a tie for outgoing cell. d. the net evaluation index for each unoccupied cell is ≥ 0. ANSWER: d 20. Identifying the outgoing arc in Phase II of the transportation simplex method is performed using the a. minimum cost method. b. MODI method. c. stepping-stone method. d. matrix reduction method. ANSWER: c 21. The MODI method is used to a. identify an outgoing arc. b. identify an incoming arc. c. identify unoccupied cells. d. identify an initial feasible solution. ANSWER: b 22. To use the transportation simplex method, a transportation problem that is unbalanced requires the use of a. artificial variables. b. one or more transshipment nodes. c. a dummy origin or destination. d. matrix reduction. Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems ANSWER: c 23. To use the Hungarian method, a profit-maximization assignment problem requires a. converting all profits to opportunity losses. b. a dummy agent or task. c. matrix expansion. d. finding the maximum number of lines to cover all the zeros in the reduced matrix. ANSWER: a 24. To use the transportation simplex method, a. there can be no unacceptable routes. b. the initial feasible solution cannot be degenerate. c. a minimization objective function must be the case. d. total supply must equal total demand. ANSWER: d Subjective Short Answer 25. Develop the transportation tableau for this transportation problem.

ANSWER: Cengage Learning Testing, Powered by Cognero

Page 4

Chapter 19 - Solution Procedures for Transportation and Assignment Problems Destination Origin

1 2 3 4 Demand

Supply

250

100 200 150 50

250

26. Solve the following transportation problem using the transportation simplex method. State the minimum total shipping cost. Origin A B

Supply 500 400

Destination X Y Z

Demand 300 300 300

Shipping costs are: Source A B ANSWER:

Destination Y 3 12

X 2 9

Origin

200

100

Demand

Z 5 10

Destination 2 3 300

300

12 300

Supply 5

300

500 400

300

Total shipping cost = $5,200. 27. Canning Transport is to move goods from three factories (origins) to three distribution centers (destinations). Information about the move is given below. Solve the problem using the transportation simplex method and compute the total shipping cost. Origin A B C

Supply 200 100 150

Destination X Y Z

Demand 50 125 125

Shipping costs are: Origin

Destination Y

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Z Page 5

Chapter 19 - Solution Procedures for Transportation and Assignment Problems A B C

3 2 5 9 10 -5 6 4 (Source B cannot ship to destination Z)

ANSWER: Origin

X 3

100

Demand

Destination Y 2 125

250

Supply 5

200

999

125

100

150

125

Total shipping cost = $2,000. 28. The following table shows the unit shipping cost between cities, the supply at each origin city, and the demand at each destination city. Solve this minimization problem using the transportation simplex method and compute the optimal total cost. Origin St. Louis Evansville Bloomington Demand ANSWER:

Terre Haute 8 5 3 150

Origin St. Louis

Terre Haute 8

Evansville

100

Bloomington

Demand

Destination Indianapolis Ft. Wayne 6 12 5 10 2 9 60 45

Destination Indianapolis Ft. Wayne 6 12 10 45

5 3 150

South Bend 9 8 10 45

50 60

Supply 100 100 100

South Bend 9 45

Supply 100 100 100

Ship 10 from St. Louis to Indianapolis, 45 from St. Louis to Ft. Wayne, 45 from St. Louis to South Bend, 100 from Evansville to Terre Haute, 50 from Bloomington to Terre Haute, and 50 from Bloomington to Indianapolis. The total cost is 1755. 29. After some special presentations, the employees of the AV Center have to move overhead projectors back to classrooms. The table below indicates the buildings where the projectors are now (the origins), where they need to go (the destinations), and a measure of the distance between sites. Determine the transport arrangement that minimizes the total transport distance. Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems

Origin Baker Hall Tirey Hall Arena Demand ANSWER:

Business 10 12 15 12 Origin Baker Hall Baker Hall Baker Hall Tirey Hall Arena Arena

Destination Parsons Hall 5 1 7 10

Education 9 11 14 20

Destination Business Education Holmstedt Hall Parsons Hall Holmstedt Hall (none)

Units 12 20 3 10 7 13 Total

Holmstedt Hall 2 6 6 10

Supply 35 10 20

Distance 120 180 6 10 42 0 358

30. Solve the following assignment problem using the Hungarian method. No agent can be assigned to more than one task. Total cost is to be minimized. Task Agent 1 2 3 ANSWER:

A 9 12 11 Agent 1 2 3 Unassigned

B 5 6 6

C 4 3 5

D 2 5 7

Task D C B A Total Cost

Cost 2 3 6 0 11

31. Use the Hungarian method to obtain the optimal solution to the following assignment problem in which total cost is to be minimized. All tasks must be assigned and no agent can be assigned to more than one task. Task Agent 1 2 3 4 ANSWER:

A 10 11 18 15

B 12 14 21 20

Agent 1 2 3 4

Task C B D A Total Cost

C 15 19 23 26

D 25 32 29 28

Cost 15 14 29 15 73

32. A professor has been contacted by four not-for-profit agencies that are willing to work with student consulting teams. The agencies need help with such things as budgeting, information systems, coordinating volunteers, and forecasting. Although each of the four student teams could work with any of the agencies, the professor feels that there is a difference in the amount of time it would take each group to solve each problem. The professor's estimate of the time, in days, is Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems given in the table below. Use the Hungarian method to determine which team works with which project. All projects must be assigned and no team can be assigned to more than one project. Project Team Budgeting Information Volunteers Forecasting A 32 35 15 27 B 38 40 18 35 C 41 42 25 38 D 45 45 30 42 ANSWER: Team A works with the forecast, Team B works with volunteers, Team C works with budgeting, and Team D works with information. The total time is 131. 33. A manufacturer of electrical consumer products, with its headquarters in Burlington, Iowa, produces electric irons at Manufacturing Plants 1, 2, and 3. The irons are shipped to Warehouses A, B, C, and D. The shipping cost per iron, the monthly warehouse requirements, and the monthly plant production levels are:

Plant 1 Plant 2 Plant 3 Monthly Warehouse Requirements (units)

A $.20 .15 .15

Warehouse B C $.25 $.15 .30 .20 .20 .20

D $.20 .15 .25

12,000

8,000

5,000

15,000

Monthly Plant Production (units) 10,000 20,000 10,000

How many electric irons should be shipped per month from each plant to each warehouse to minimize monthly shipping costs? a. Use the minimum cost method to find an initial feasible solution. b. Can the initial solution be improved? c. Compute the optimal total shipping cost per month. ANSWER: a.The minimum cost method (and breaking ties by choosing the cell corresponding to the arc over which the most units can be shipped) found the solution shown below. Origin Plant 1 Plant 2

Warehouse A .20

12,000

Demand

.15 .15

Plant 3 12,000

Destination Warehouse B Warehouse C .25 .15 10,000 .30

8,000 8,000

.20

3,000 2,000 15,000

.20

Warehouse D .20

5,000

.20

.15 .25

Supply 10,000 20,000 10,000

5,000

b. The solution cannot be improved. It is optimal. c. Total monthly shipping cost = $6,650. 34. Al Bergman, staff traffic analyst at the corporate headquarters of Computer Products Corporation (CPC), is developing a monthly shipping plan for the El Paso and Atlanta manufacturing plants to follow next year. These plants manufacture specialized computer workstations that are shipped to five regional warehouses. Al has developed these estimated requirements and costs: Warehouse Cengage Learning Testing, Powered by Cognero

Monthly Plant Page 8

Chapter 19 - Solution Procedures for Transportation and Assignment Problems Plant Chicago Dallas Denver New York San Jose Production (units) Atlanta $35 $40 $60 $45 $90 200 El Paso 50 30 35 95 40 300 Monthly Warehouse 75 100 25 150 150 Requirements (units) Determine how many workstations should be shipped per month from each plant to each warehouse to minimize monthly shipping costs, and compute the total shipping cost. a. Use the minimum cost method to find an initial feasible solution. b. Use the transportation simplex method to find an optimal solution. c. Compute the optimal total shipping cost. ANSWER: a. Initial feasible solution found using the minimum cost method is below. Total cost = $20,500. Origin Atlanta

Chicago 35 75 50

El Paso Demand

Dallas 40

100

Destination Denver 60 35

100

New York 45 125 95

150

San Jose 90

150

Supply 200 300

150

b. Optimal solution found using MODI and stepping-stone methods is below. Origin Atlanta

Chicago 35 50

El Paso

Demand

50 75

Dallas 40

100

Destination Denver 60

New York 45 150

150

San Jose 90

150

Supply 200 300

150

c. Total monthly shipping cost = $19,625. 35. Consider the transportation problem below. Origin A B C Demand

1 $ .50 .80 .90 300

Destination 2 $ .90 1.00 .70 800

3 $ .50 .40 .80 400

Supply 100 500 900

a. Use the minimum cost method to find an initial feasible solution. b. Can the initial solution be improved? c. Compute the optimal total shipping cost. ANSWER: a. The minimum cost method provided the solution shown below. Origin A

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.50

Destination 2 .90

3 .50

Supply 100 Page 9

Chapter 19 - Solution Procedures for Transportation and Assignment Problems 100 B

100

Demand b. c.

.80 .90

300

1.00

400

.70

800 800

.40 .80

500 900

400

The solution cannot be improved. It is optimal. Total shipping cost = $940.

36. Five customers needing their tax returns prepared must be assigned to five tax accountants. The estimated profits for all possible assignments are shown below. Only one accountant can be assigned to a customer, and all customers' tax returns must be prepared. What should the customer-accountant assignments be so that estimated total profit is maximized? What is the resulting total profit? Customer A B C D E ANSWER:

1 $500 625 825 590 450 Customer A B C D E

2 $525 575 650 650 750

Accountant 3 $550 700 450 525 660

Accountant 5 3 1 4 2 Total Profit

4 $600 550 750 690 390

5 $700 800 775 750 550

Profit $ 700 700 825 690 750 $3,665

37. Four jobs must be assigned to four work centers. Only one job can be assigned to each work center, and all jobs must be processed. The cost of processing each job at each work center is shown below. Determine which jobs should be assigned to which work center to minimize total processing cost. Compute the total processing cost.

Job A B C E ANSWER:

1 $50 25 65 55 Job A B C D

Work Center 2 3 $45 $50 40 35 60 55 65 75

4 $65 20 65 85

Work Center 2 4 3 1 Total Cost

Cost $ 45 20 55 55 $175

38. Four employees must be assigned to four projects. Only one employee can be assigned to each project, and all projects must be completed. The cost of each employee completing each project is shown below. Determine which employee Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems should be assigned to which project to minimize total project completion cost. Be sure to compute the total project completion cost. Project Employee Al Ben Cal Dan ANSWER:

1 $300 400 350 400

2 $325 525 400 350

Employee Al Ben Cal Dan

3 $500 575 600 450

Project 4 1 2 3 Total Cost

4 $350 600 500 450 Cost $ 350 400 400 450 $1,600

39. A large screen printer is faced with six jobs due on Tuesday. The plan is to do the jobs on Monday so they will be ready on time. The shop has six worker-machine pairs that can work on any of the six jobs. Because of differing experience levels and machine capabilities, processing times differ. The processing times presented in the table below are in minutes. What is the optimal assignment of jobs to worker-machine pairs that minimizes total processing time? Job A B C D E F

1 250 350 410 380 395 250

Worker-Machine Pair 3 4 175 425 375 410 325 275 350 375 280 390 410 385

2 375 310 450 245 250 285

5 225 275 315 210 410 300

6 350 225 275 350 375 295

ANSWER: Job A B C D E F

Worker-Machine Pair Time 3 175 6 225 4 275 5 210 2 250 1 250 Total Time = 1385

40. A company ships products from four factories to four warehouses. The factory capacities, warehouse requirements, and per-unit shipping costs are shown below: Warehouse Factory A Factory B

Monthly Factory Capacity (units)

$11 12

$13 10

$9 7

$6 9

5,000 10,000

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems Factory C Factory D Monthly Warehouse Min. Requirement (units)

19 7

16 6

15 4

21 9

3,000

8,000

12,000

5,000

10,000 5,000

How many products should the company ship from each factory to each warehouse to minimize monthly shipping costs? What will the monthly shipping cost be if the shipping plan is followed? (Use the minimum cost method to find an initial feasible solution and the transportation simplex method to find an optimal solution.) ANSWER: Factory A to Warehouse 4 = 5,000 units Factory B to Warehouse 3 = 10,000 units Factory C to Warehouse 2 = 8,000 units Factory D to Warehouse 1 = 3,000 units Factory D to Warehouse 3 = 2,000 units Total Cost = $257,000 41. The Des Moines plant of Tri-B Corp. has three fabrication departments with each producing a single unique product with equipment that is dedicated solely to its product. The three products are moved to four assembly departments where they are assembled. Although any of the three products can be processed in any of the assembly departments, the materials-handling and assembly costs are different because of the varying distances between departments and because of different equipment. Each fabrication and assembly department has a different monthly capacity, and it is desirable that each department operate at capacity. The variable costs and capacity for each department is shown below.

Fabrication Dept. A Fabrication Dept. B Fabrication Dept. C Monthly Assembly Dept. Capacity (units)

1 $1.20 0.70 0.50

Assembly Department 2 3 $0.70 $0.50 0.50 0.50 0.70 0.80

4 $0.60 0.60 1.20

3,000

10,000

12,000

15,000

Monthly Fabrication Dept. Capacity (units) 9,000 17,000 14,000

How many units of each product should be moved from each fabrication department to each assembly department to minimize total monthly costs? (Use the minimum cost method to find an initial feasible solution and the transportation simplex method to find an optimal solution.) Compute the optimal total monthly cost? ANSWER: Dept. A to Dept. 4 = 9,000 units Dept. B to Dept. 3 = 14,000 units Dept. B to Dept. 4 = 3,000 units Dept. C to Dept. 1 = 3,000 units Dept. C to Dept. 2 = 10,000 units Dept. C to Dept. 3 = 1,000 units Total shipping cost = $23,500.00 Essay 42. For an assignment problem where the number of agents does not equal the number of tasks, what adjustments must be made to allow the problem to be solved using the Hungarian method? ANSWER: No Answer provided. 43. Explain how the transportation simplex method can be used to solve a transportation problem that has a maximization Cengage Learning Testing, Powered by Cognero

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Chapter 19 - Solution Procedures for Transportation and Assignment Problems objective. ANSWER: No answer provided. 44. Explain what adjustments are made to the transportation tableau when there are unacceptable routes. ANSWER: No answer provided. 45. Explain what adjustments are made to the transportation tableau when total supply and total demand are not equal. ANSWER: No answer provided. 46. Explain how the Hungarian method can be used to solve an assignment problem that has a maximization objective. ANSWER: No answer provided.

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Chapter 20 - Minimal Spanning Tree True / False 1. The arcs in a minimal spanning tree problem can be measured in terms of criteria other than distance. a. True b. False ANSWER: True 2. Cases in which a greedy algorithm provides the optimal solution are rare. a. True b. False ANSWER: True 3. The minimum spanning tree allows a person to visit every node without backtracking. a. True b. False ANSWER: False 4. In the minimal spanning tree algorithm, you must consider the unconnected nodes that can be reached from any of the connected nodes, rather than arbitrarily considering only one of the connected nodes. a. True b. False ANSWER: True 5. The minimum spanning tree algorithm is considered a heuristic. a. True b. False ANSWER: False 6. The minimal spanning tree algorithm will lead to an optimal solution regardless of which node is chosen at the start of the algorithm. a. True b. False ANSWER: True 7. It is possible for minimal spanning tree problems to have alternative optimal solutions. a. True b. False ANSWER: True Multiple Choice 8. Consider a minimal spanning tree problem in which pipe must be laid to connect sprinklers on a golf course. When represented with a network, a. the pipes are the arcs and the sprinklers are the nodes. b. the pipes are the nodes and the sprinklers are the arcs. c. the pipes and the sprinklers are the tree. d. each sprinkler must be connected to every other sprinkler. Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Minimal Spanning Tree ANSWER: a 9. The minimal spanning tree algorithm is considered to be: a. a greedy algorithm. b. an arc algorithm. c. a non-optimal algorithm. d. a non-feasible algorithm. ANSWER: a 10. The minimal spanning tree algorithm has connected nodes 8 and 9. Node 8 could be connected to nodes 11 (distance 6) and 12 (distance 5) and node 9 could be connected to node 12 (distance 3) and node 13 (distance 2). Which will you do next? a. connect 8 to 11 b. connect 8 to 12 c. connect 9 to 12 d. connect 9 to 13 ANSWER: d 11. For a network consisting of N nodes, a minimal spanning tree will consist of: a. N − 2 arcs. b. N − 1 arcs. c. N arcs. d. N + 1 arcs. ANSWER: b 12. The minimal spanning tree algorithm will: a. sometimes fail to produce a feasible solution. b. always produce a feasible, but not necessarily optimal, solution. c. always produce an optimal solution. d. always produce an optimal, but not necessarily feasible, solution. ANSWER: c Subjective Short Answer 13. For the following eight cities with the given distances, find the minimal spanning tree path. To City From City 1 1 -2 5 3 2 4 6 5 -6 -7 -ANSWER: 1-3-4-2-5-6-7

2 5 --4 6 ---

3 2 --3 -12 --

4 6 4 3 -9 8 --

5 -6 -9 -0 14

6 --12 8 0 -4

7 ----14 4 --

14. Find the minimal spanning tree for this network. Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Minimal Spanning Tree

ANSWER:

15. Find the minimal spanning tree for this network.

ANSWER:

16. The numbers on this network represent times to distribute a message. Use the minimal spanning tree algorithm to determine how messages should be passed in order to minimize total time. Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Minimal Spanning Tree

ANSWER:

17. Ethernet cable costs $25 per foot to install. The table below gives the approximate direct routing distance in feet between various pairs of machines. WS1 WS2 LP1 LP2 CC

WS1 0

WS2 46 0

LP1 16 81 0

LP2 23 44 45 0

CC 100 67 57 38 0

FM 29 98 36 73 52

a. Use the minimal spanning tree algorithm to determine the least cost method of connecting all machines. b. How much cheaper would it be if the company decided not to hook up the second printer to this system? ANSWER: a. WS1-LP1, WS1-LP2, WS1-FM, LP2-WS2, LP2-CC; 150 feet; Total cost = $3,750 b. WS1-LP1, WS1-WS2, WS1-FM, FM-CC; 143 feet; Total cost = $3,575; $175 cheaper 18. Wondercamp is planning a new resort for the urban professional who wishes to "get back to nature". For a hefty fee, it plans to transport clients up the Crocodile River by canoe and then have their clients camp at one of eight designated campsites. Cengage Learning Testing, Powered by Cognero

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Chapter 20 - Minimal Spanning Tree They must build a 1000-yard trail from the river to the first campsite as there are no feasible alternatives. It then wishes to build a sequence of trails (of minimum total length) so that every campsite can be reached from any other campsite. A study of the terrain of the area has yielded the following possibilities for trails between campsites. Which trails should be built?

ANSWER:

There are two alternative optimal solutions: 1. (River-1),(1-4),(4-6),(4-5),(6-7),(4-3),(3-8),(3-2). 2. (River-1),(1-4),(4-6),(4-5),(6-7),(4-3),(3-8),(4-2). Total yards = 2050.

19. Griffith's Cherry Preserve is a combination wild animal habitat and amusement park. Besides their phenomenally successful wild animal safari tour, there are eight different theme areas in the amusement park. One problem encountered by management is to develop a method by which people can efficiently travel between each area of the park. Management has learned that a people mover can be constructed at a cost of $50 per foot. If the following network represents the distances (in feet) between each area of the park for which a people mover is possible, determine the minimum cost for such a system.

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Chapter 20 - Minimal Spanning Tree

ANSWER:

Optimal tree = (1-3),(1-4),(3-2),(3-6),(6-7),(6-8),(8-5),(8-9) Total distance = 9,110 feet. Total cost = $455,500.

Essay 20. What is a greedy algorithm? What is an example of a greedy algorithm? ANSWER: Answer not provided. 21. What are some other criteria (measures), besides distance, that might be minimized in a minimal spanning tree problem? Provide an example situation for each criterion. ANSWER: Answer not provided. 22. Describe three examples of minimal spanning tree problems. ANSWER: Answer not provided.

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Chapter 21 - Dynamic Programming True / False 1. Dynamic programming requires that its subproblems be independent of one another. a. True b. False ANSWER: False 2. Dynamic programming, when used for the shortest route problem, requires complete enumeration of paths from the beginning to ending node. a. True b. False ANSWER: False 3. The solution of stage k of a dynamic programming problem is dependent upon the solution of stage k−1. a. True b. False ANSWER: True 4. The output of stage k is the input for stage k−1. a. True b. False ANSWER: True 5. State variables are a function of a state variable and a decision. a. True b. False ANSWER: True 6. The return function for a shortest route problem refers to two directional arcs between nodes. a. True b. False ANSWER: False 7. In solving a shortest route problem using dynamic programming the stages represent how many arcs you are from the terminal node. a. True b. False ANSWER: True 8. Dynamic programming is a general approach rather than a specific technique. a. True b. False ANSWER: True 9. Dynamic programming must only involve a finite number of decision alternatives and a finite number of stages. a. True b. False Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming ANSWER: False 10. Dynamic programming is a general approach with stage decision problems differing substantially from application to application. a. True b. False ANSWER: True 11. In a production and inventory control problem, the states can correspond to the amount of inventory on hand at the beginning of each period. a. True b. False ANSWER: True 12. In a knapsack problem, if one adds another item, one must completely resolve the problem in order to find a new optimal solution. a. True b. False ANSWER: False 13. The subscripts used in dynamic programming notation refer to states. a. True b. False ANSWER: False 14. In order to use dynamic programming, one must be able to view the problem as a multistage decision problem. a. True b. False ANSWER: True 15. Finding the optimal solution to each stage of a dynamic programming problem will always lead to an optimal solution to the total problem. a. True b. False ANSWER: True 16. The stage transformation function identifies which state one reaches at the next stage for a given decision. a. True b. False ANSWER: True Multiple Choice 17. Stages of a dynamic programming solution procedure a. represent parts of a large mathematical model. b. often represent a sequence of decisions made over time. c. are usually not independent of each other. Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming d. All of the alternatives are true. ANSWER: d 18. State variables in a shortest route problem represent a. decisions. b. locations in the network. c. the minimum distance between nodes. d. None of the alternatives is true. ANSWER: b 19. The stage transformation function a. transforms the input into the output. b. transforms a stage into a state. c. is a different function for each stage. d. None of the alternatives is true. ANSWER: a 20. Stage transformation functions a. are linear. b. calculate the return. c. determine the output of the stage. d. All of the alternatives are true. ANSWER: c 21. A return function is a value such as profit or loss associated with making decision dn at: a. stage n for specific value of output variable xn. b. stage n for a specific value of input variable xn. c. stage n for a specific value of stage m. d. input n for a specific value of output variable xn. ANSWER: b 22. If x3 = t4 (x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4 the state variable is a. t b. x c. r d. d ANSWER: b 23. If x3 = t4(x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4, the stage transformation function is a. t b. x c. r d. d ANSWER: a Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming 24. If x3 = t4(x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4, the subscripts refer to a. state. b. stage. c. transformation. d. return. ANSWER: b 25. The knapsack problem is to determine how many units of each item to place in the knapsack to: a. minimize total value. b. maximize total value. c. minimize the number of items in the knapsack. d. maximize the number of items in the knapsack. ANSWER: b 26. Solutions in dynamic programming a. are not optimal. b. are unique. c. represent each stage. d. All of the alternatives are true. ANSWER: c Subjective Short Answer 27. Find the shortest path through the following network using dynamic programming.

ANSWER:

STAGE 1 Input node 8 9 STAGE 2 Input node 5 6 7 STAGE 3 Input node

Decision arc 8−10 9−10

Shortest distance 4 3

Decision arc 5−9 6−8 7−8

Shortest distance 6 10 11

Decision arc

Shortest distance

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Chapter 21 - Dynamic Programming 2 3 4 STAGE 4 Input node 1

2−5 3−5 4−6

16 13 17

Decision arc 1−4

Shortest distance 19

28. Audio Disks will be opening outlets in the greater Phoenix area. The estimated sales at each store are dependent not only on the store location, but on the number of sales personnel, as presented in the table below ($000/year). Each store requires at least 2 sales people, and a pool of 9 salespeople is available. 2 60 105 120

Store 1 Store 2 Store 3

Staff Size 3 85 120 145

4 90 130 160

5 100 150 175

a. b. c.

What would the states be in the dynamic programming formulation? Draw the network that represents the dynamic programming formulation. Given the above network, solve the sales personnel allocation problem by finding the longest path. ANSWER: a. The states represent the number of salesmen available at each stage (store).

c. Store Salespeople 1 3 2 3 3 3 Total Sales

Alternate optimal solutions: Sales Salespeople 85 3 120 2 145 4 350

Sales 85 105 160 350

29. Consider the following integer linear program Max s.t.

5x1 + 7x2 + 9x3 2x1 + 3x2 + 4x3 ≤ 8 x1 ≤ 3 x2 ≤ 2 x1, x2, x3 ≥ 0, integer

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Chapter 21 - Dynamic Programming a. Set up the network that represents the dynamic programming formulation. b. Solve the problem using dynamic programming. ANSWER:

x1 = 1, x2 = 2, x3 = 0

30. A driver wants to make a trip from city 1 to city 7. The road mileage between cities is given below. Find the shortest route. From City 1 2 3 1 − 2 2 2 − − − 3 − − − 4 − − − 5 − − − 6 − − − 7 − − − ANSWER: 1-3-6-7; 13 miles

To City: 4 5 4 − − 4 − 12 − 10 − − − − − −

6 − 10 6 − − − −

7 − − − − 14 2 −

31. The owner of a small construction firm is excavating at three sites. He wishes to assign his 5 additional trucks in such a way as to minimize his total costs. Each site can use 0 to 3 additional trucks; no site can use more than 3 trucks efficiently. The following site total costs are known. Number of Trucks 0 1 2 3

Site 1 $10000 10000 9200 8500

Cost of Excavating Site 2 $15000 14000 13250 12750

Site 3 $20000 18000 17500 17250

Use dynamic programming to find the assignment of the additional trucks that minimizes total cost. b. If the owner had only 4 trucks to assign, what would be the optimal assignment and total cost? ANSWER: a. The minimal cost truck assignments are: 0 trucks to site 1 3 trucks to site 2 a.

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Chapter 21 - Dynamic Programming 2 trucks to site 3 at a cost of $40,250. b.

There are two optimal assignments: Site 1 0 trucks 0 trucks Site 2 3 trucks or 2 trucks Site 3 1 truck 2 trucks at a cost of $40,750.

32. We have a number of types of items to be shipped as cargo. The total available weight in the truck is ten tons. We wish to determine the number of units of items to be shipped to maximize profit. Item Weight (Tons) A 3 B 4 C 2 ANSWER: Ship 0 tons of item A. Ship 2 tons of item B. Ship 1 ton of item C. Total profit 17.

Profit (1000s dollars) 5 7 3

33. A cargo company has a set of delivery patterns for its goods from its locations at city1 to a series of cities 2,3,4,5, and 6. The delivery times between cities are given I hours below. Find the shortest route. To City: From City: 1 2 1 − 6 2 6 − 3 10 4 4 7 − 5 − 5 6 − − 7 − − ANSWER: 1-2-5-7; 18 hours

3 10 4 − 5 3 4 −

4 7 − 4 − − 8 −

5 − 5 2 − − − 7

6 − − 4 8 − − 5

7 − − − − 7 5 −

34. Ajax Sound is in the business of fabricating printer connection cables. They purchase 30 foot spools of wire from National Electric at $3.00 per spool and cut the wire into various lengths. Each length of wire is fitted with printer connector jacks at both ends and then packaged. The printer connector jacks cost $.40 per pair (one pair is used in each cable package). Packaging and labor together cost $.20 per package. Because of Ajax's superior marketing skills they are in the enviable position of being able to sell all the printer connection cables they produce. They are currently contemplating offering four different sized cable packages: Size 5' 8' 12' 16'

Wholesale Package Price $1.30 $1.80 $2.50 $3.30

If unused wire from a spool can be sold for scrap at $.03 per foot, how many packages of each size cable should Ajax make from a 30-foot spool? ANSWER: Produce one package of 16 ft. wire. Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming Purchase one package of 12 ft. wire. 2 feet of wire will be sold for scrap. Total Profit = $1.66. 35. Unidyde Corporation is currently planning the production of red dye number 56 for the next four months. Production and handling costs, as well as production and storage capacity, vary from month to month. This data is given in the table below. Production and holding costs are in ($1,000's per batch) and production levels and storage capacities are in batches. Holding costs are based on inventory on hand at the end of the month. The number of orders for batches the sales department has received over the four-month period are also given. Month February March April May

Production Cost 11 15 16 9

Maximum Production 3 4 3 2

Holding Cost 3 2 2 1

Storage Capacity 4 3 5 2

Orders Received 2 4 2 3

Unidyne does not wish to have any inventory of the dye at the end of May. Its current inventory is 2 batches. Determine a production schedule for the next four months. ANSWER:

Produce 3 batches in February. Produce 1 batch in March Produce 3 batches in April. Produce 2 batches in May. Total Cost = $125,000.

36. Marvelous Marvin is planning his annual "Almost Everything Must Go" inventory clearance sale. Marvin has decided to allocate 18 shelf feet to the cooking section. He is considering offering up to five items for sale in this category: Item A B C D E

Shelf Feet Required 1 2 3 5 7

Expected Profit $ 10 $ 25 $ 40 $ 70 $100

If Marvin wants at least one item A and one item B on sale, what stock should he have on sale and what is the total expected profit? ANSWER: Three solutions: (1) 2-A, 2-E, 1-B (2) 1-A, 1-B, 3-D (3) 1-A, 1-B, 1-C, 1-D, 1-E Total Profit = $245. 37. Franklin Plate Company is in the business of manufacturing commemorative plates for holidays. The company is currently planning its production schedule for this year's Thanksgiving plate. Discussions with the sales manager have indicated that sales of the plate will run from August through November. The production manager has determined the manufacturing cost per plate for each of these months as well as the sales demand (in 1000's) and manufacturing capacity (in 1000's). The data is as follows: Manufacturing Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming Month August September October November

Demand 2 4 6 3

Cost Per Plate 2 3 3 4

Production 6 6 4 3

It is also known that the maximum inventory level possible at the end of any month is 5,000 plates. The storage cost per plate for each plate in inventory at the end of the month is $.50. Franklin has no inventory on hand at the beginning of August and wishes to have no inventory left after November. Use dynamic programming to determine an optimal production schedule. ANSWER: There are four alternative optimal solutions giving a total of $45,000: Month Solution 1 Solution 2 Solution 3 Solution 4 August 6 6 6 6 September 2 3 4 5 October 4 4 4 4 November 3 2 1 0 38. Walt's Custom Boats manufactures luxury yachts. Walt is phasing out production of his 42-foot Sportfisher that he plans to replace with a 44-foot Sportfisher. Walt currently has orders for the next four months for the 42-foot model after which he will cease production. The following data (with costs in $1,000's) is given for the next four months:

Month May June July August

Production Cost Per Boat 25 27 30 29

Maximum Production Level 4 3 2 3

Holding Cost Per Boat in Inventory at End of Month 1 1 2 1

Maximum Storage Capacity 6 5 3 4

Number of Boats to be Delivered in Month 1 3 3 4

If Walt's current inventory of 42-foot Sportfishers is 1 boat, how should Walt schedule production of the 42-foot Sportfisher over the next four months to minimize the total production and inventory holding costs? ANSWER: Produce 4 boats in May. Produce 3 boats in June. Produce 0 boats in July. Produce 3 boats in August. Total Cost = $278,000. 39. Mission Bay Development Corporation is engaged in developing an exclusive 25 acre parcel of property. They have received bids from five builders to purchase construction lots. Because of the different nature of the builders, they desire different sized lots. The data are as follows: Builder 1 2 3 4 5

Lot Size Requested 3 acre 1 acre 6 acre 20 acre 5 acre

Offered Price Per Lot $ 50,000 $ 15,000 $105,000 $350,000 $ 91,000

Maximum Number Of Lots Desired 6 5 8 1 2

How should Mission Bay sell their land to maximize total sales revenue? Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming ANSWER: Builder 1: 1 lot; Builder 3: 2 lots; Builder 5: 2lots; Total revenue = $442,000 40. Given the following shortest path problem in which travel is only permitted from left to right. Use dynamic programming to determine the shortest path from node 1 to: a. Node 15. b. Node 14.

ANSWER:

a. 1-2-7-12-15: Distance = 36 b. 1-4-9-13-14: Distance = 37

Essay 41. What is the Principle of Optimality, and what is its relationship to dynamic programming? ANSWER: Answer not provided. 42. Define the following terms as they relate to dynamic programming. a. Stage b. State variable c. Stage transformation function ANSWER: Answer not provided. 43. A stage in a dynamic programming problem is defined when 2 variables and 2 functions related to that stage are defined. Identify and define the 2 variables and 2 functions and illustrate them with an example of your choice. ANSWER: Answer not provided. Cengage Learning Testing, Powered by Cognero

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Chapter 21 - Dynamic Programming 44. Explain the divide-and-conquer solution strategy of dynamic programming. ANSWER: Answer not provided.

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