Introduction to Management Science, 12E Bernard W Taylor Solution Manual by Ace Test Banks

Instructor’s Solution Manual For

Introduction to Management Science Twelfth Edition

Bernard W. Taylor III Virginia Polytechnic Institute and State University

Contents Chapter 1 Management Science ..................................................................................1-1 Chapter 2 Linear Programming: Model Formulation and Graphical Solution ...........2-1 Chapter 3 Linear Programming: Computer Solution and Sensitivity Analysis ..........3-1 Chapter 4 Linear Programming: Modeling Examples.................................................4-1 Chapter 5 Integer Programming ..................................................................................5-1 Chapter 6 Transportation, Transshipment, and Assignment Problems .......................6-1 Chapter 7 Network Flow Models ................................................................................7-1 Chapter 8 Project Management ...................................................................................8-1 Chapter 9 Multicriteria Decision Making....................................................................9-1 Chapter 10 Nonlinear Programming ...........................................................................10-1 Chapter 11 Probability and Statistics...........................................................................11-1 Chapter 12 Decision Analysis .....................................................................................12-1 Chapter 13 Queuing Analysis......................................................................................13-1 Chapter 14 Simulation .................................................................................................14-1 Chapter 15 Forecasting................................................................................................15-1 Chapter 16 Inventory Management .............................................................................16-1 Module A: The Simplex Solution Method........................................................................A-1 Module B: Transportation and Assignment Solution Methods ....................................... B-1 Module C: Integer Programming: The Branch and Bound Method ............................... C-1 Module D: Nonlinear Programming Solution Techniques .............................................D-1 Module E: Game Theory ................................................................................................. E-1 Module F: Markov Analysis .............................................................................................F-1

Chapter One: Management Science PROBLEM SUMMARY

33.

Linear programming

34.

Linear programming

35.

Linear programming

Total cost, revenue, profit, and break-even

36.

Forecasting/statistics

37.

Linear programming

Total cost, revenue, profit, and break-even

38.

Waiting lines

39.

Shortest route

Break-even volume

Graphical analysis (1−2)

Graphical analysis (1−4)

Break-even sales volume

Break-even volume as a percentage of capacity (1−2)

TC = cf + vcv = $8,000 + (300)(65) = $27,500;

Break-even volume as a percentage of capacity (1−3)

10.

Break-even volume as a percentage of capacity (1−4)

TR = vp = (300)(180) = $54,000; Z = $54,000 − 27,500 = $26,500 per month

1. 2.

Total cost, revenue, profit, and break-even

PROBLEM SOLUTIONS 1. a)

v = 300, cf = $8,000, cv = $65 per table, p = $180;

b) v =

cf 8,000 = = 69.56 tables per month p − cv 180 − 65

11.

Effect of price change (1−2)

12.

Effect of price change (1−4)

13.

Effect of variable cost change (1−12)

14.

Effect of fixed cost change (1−13)

TC = cf + vcv

15.

Break-even analysis

16.

Effect of fixed cost change (1−7)

17.

Effect of variable cost change (1−7)

= 18, 000 + (12, 000)(0.90) = $28,800; TR = vp = (12, 000)($3.20) = $38, 400;

18.

Break-even analysis

Z = $38, 400 − 28,800 = $9, 600 per year

19.

Break-even analysis

20.

Break-even analysis

21.

Break-even analysis; volume and price analysis

22.

Break-even analysis; profit analysis

23.

Break-even analysis

24.

Break-even analysis; profit analysis

25.

Break-even analysis; price and volume analysis

26.

Break-even analysis; price and volume analysis

27.

Break-even analysis; profit analysis

28.

Break-even analysis; profit analysis

29.

Break-even analysis; profit analysis

30.

Decision analysis

31.

Expected value

32.

Linear programming

2. a)

v = 12, 000, cf = $18, 000, cv = $0.90, p = $3.20;

b) v = 3. a)

cf 18, 000 = = 7,826 p − cv 3.20 − 0.90

v = 18,000, cf = $21,000, cv = $.45, p = $1.30; TC = cf + vcv = $21,000 + (18,000)(.45) = $29,100; TR = vp = (18,000)(1.30) = $23, 400; Z = $23, 400 − 29,100 = −$5,700 (loss)

b) v = 4.

cf = $25,000, p = $.40, cv = $.15, v=

1-1 .

cf 21,000 = = 24,705.88 yd per month p − cv 1.30 − .45

cf 25,000 = = 100,000 lb per month p − cv .40 − .15

13.

cf 25,000 = = 65,789.47 lb p − cv .60 − .22

per month; it increases the break-even volume from 55,555.55 lb per month to 65,789.47 lb per month. v=

14.

cf 39,000 = = 102,613.57 lb p − cv .60 − .22

per month; it increases the break-even

volume from 65,789.47 lb per month to 102,631.57 lb per month.

Initial profit: Z = vp − cf − vcv = (9,000)(.75) −

15.

4,000 − (9,000)(.21) = 6,750 − 4,000 − 1,890 = $860 per month; increase in price: Z = vp − cf − vcv = (5,700)(.95) − 4,000 − (5,700)(.21) = 5, 415 − 4,000 − 1,197 = $218 per month; the dairy should not raise its price. cf

$25,000 = 1,250 dolls 30 − 10

Break-even volume as percentage of capacity =

10.

p − cv

16.

cf p − cv

35,000 = 1,750 30–10

The increase in fixed cost from $25,000 to $35,000 will increase the break-even point from 1,250 to 1,750 or 500 dolls; thus, he should not spend the extra $10,000 for advertising.

Break-even volume as percentage of capacity v 24,750.88 = = = .988 = 98.8% k 25,000

17.

Original break-even point (from problem 7) = 1,250 New break-even point: v=

Break-even volume as percentage of v 100,000 = = .833 = 83.3% k 120,000

cf 17,000 = = 1,062.5 p − cv 30 − 14

Reduces BE point by 187.5 dolls. 18. a) v =

cf 18, 000 v= = = 9, 729.7 cupcakes p − cv 2.75 − 0.90

It increases the break-even volume from 7,826 to 9, 729.7

cf $27,000 = = 5,192.30 pizzas p − cv 8.95 − 3.75

5,192.3 = 259.6 days 20

c) Revenue for the first 30 days = 30(pv − vcv)

per year.

12.

v 7,826 = = .652 = 65.2% k 12, 000

capacity =

11.

= 30[(8.95)(20) − (20)(3.75)]

cf 25,000 v= = = 55,555.55 lb p − cv .60 − .15

= $3,120

per month; it reduces the break-even

$27,000 − 3,120 = $23,880, portion of fixed cost not recouped after 30 days.

volume from 100,000 lb per month to 55,555.55 lb.

New v =

1-2 .

cf $23,880 = = 5,685.7 pizzas p − cv 7.95 − 3.75

Total break-even volume = 600 + 5,685.7 = 6,285.7 pizzas

22. a) cf = $350,000 cv = $12,000

5,685.7 Total time to break-even = 30 + 20 = 314.3 days

p = $18,000 v=

19. a) Cost of Regular plan = $55 + (.33)(260 minutes) =

= $140.80 Cost of Executive plan = $100 + (.25)(60 minutes)

350,000 18,000 − 12,000

= 58.33 or 59 students

= $115

b) Z = (75)(18,000) − 350,000 − (75)(12,000) = $100,000

Select Executive plan. b) 55 + (x − 1,000)(.33) = 100 + (x − 1,200)(.25)

c) Z = (35)(25,000) − 350,000 − (35)(12,000) = 105,000

− 275 + .33x = .25x − 200 x = 937.50 minutes per month or 15.63 hrs. 20. a) 14,000 =

cf p − cv

This is approximately the same as the profit for 75 students and a lower tuition in part (b).

7,500 p − .35

23.

p = $400 cf = $8,000

p = $0.89 to break even

cv = $75

b) If the team did not perform as well as expected the crowds could be smaller; bad weather could reduce crowds and/or affect what fans eat at the game; the price she charges could affect demand.

Z = $60,000

c) This will be a subjective answer, but $1.25 seems to be a reasonable price. Z = vp − cf − vcv

Z + cf p − cv

60,000 + 8,000 400 − 75

v = 209.23 teams

Z = (14,000)(1.25) − 7,500 − (14,000)(0.35)

24.

= 17,500 − 12,400

Fixed cost (cf) = 875,000 Variable cost (cv) = $200

= $5,100 cv = $12 per pupil

Price (p) = (225)(12) = $2,700 v = cf/(p – cv) = 875,000/(2,700 – 200) = 350

p = $75

With volume doubled to 700:

1,700 v= 75 − 12

Profit (Z) = (2,700)(700) – 875,000 – (700)(200) = $875,000 cf = $26,000 cv = $0.67

21. a) cf = $1,700

25.

= 26.98 or 27 pupils b) Z = vp − cf − vcv

p = $3.75

$5,000 = v(75) − $1,700 − v(12) 63v = 6,700

v = 106.3 pupils

26, 000 3.75 − 0.67

= 8,442 slices

c) Z = vp − cf − vcv

Forecasted annual demand = (540)(52) = 28,080

$5,000 = 60p − $1,700 − 60(12)

Z = $105,300 – 44,813.6 = $60,486.40

60p = 7,420 p = $123.67

26.

1-3 .

Fixed cost (cf) = 100,000 Variable cost (cv) = $(.50)(.35) + (.35)(.50) + (.15)(2.30) = $0.695

Price (p) = $6 Profit (Z) = (6)(45,000) – 100,000 – (45,000)(0.695) = $138,725

No, she would make less money than (b) 29. a) v =

This is not the financial profit goal of $150,000.

cf 700 = p − cv 35 − 3

v = 21.88 jobs

The price to achieve the goal of $150,000 is, p = (Z + cf + vcv)/v = (150,000 + 100,000 + (45,000)(.695))/45,000 = $6.25

b) (6 snows)(2 days/snow)(10 jobs/day) = 120 jobs Z = (120)(35) − 700 − (120)(3) Z = $3,140 c) (6 snows)(2 days/snow)(4 jobs/day) = 48 jobs Z = (48)(150) − 1800 − (48)(28) Z = $4,056

The volume to achieve the goal of $150,000 is, v = (Z + cf)/(p − cv) = (150,000 + 100,000)/(6 − .695) = 47,125

Yes, better than (b) d) Z = (120)(35) − 700 − (120)(18) Z = $1,340

27. a) Monthly fixed cost (cf) = cost of van/60 months + labor (driver)/month = (21,500/60) + (30.42 days/month)($8/hr) (5 hr/day) = 358.33 + 1,216.80 = $1,575.13

Yes, still a profit with one more person 30.

This is a decision analysis problem – the subject of Chapter 12. The payoff table is: Weather Conditions

Variable cost (cv) = $1.35 + 15.00 = $16.35 Price (p) = $34 v = cf/(p − vc) = (1,575.13)/(34 − 16.35) v = 89.24 orders/month

Decision Alternatives

Good

Bad

$3.25

$12,800

$8,450

$4.00

$14,400

$5,275

The student’s decision depends on the degree of risk they are willing to assume. Chapter 12 includes decision criteria for this problem.

b) 89.24/30.42 = 2.93 orders/day − Monday through Thursday

31.

Double for weekend = 5.86 orders/day − Friday through Sunday

This problem uses expected value for the decision alternatives in problem 30. Expected value ($3.25) = ($12,800)(0.60) + ($8,450)(0.40) = $11,060

Orders per month = approximately (18 days) (2.93 orders) + (12.4 days)(5.86 orders)

Expected value ($4.00) = ($14,400)(0.60) + ($5,275)(0.40) = $10,750

= 125.4 delivery orders per month

Although the decision to sell hotdogs for $3.25 results in the greatest expected value, the results are so close, Annie would likely be indifferent.

Profit = total revenue − total cost = vp – (cf + vcv) = (125.4)(34) − 1,575.13 – (125.4)(16.35) = 638.18 cf 500 = 28. a) v = p − cv 30 − 14

32.

There are two possible answers, or solution points: x = 25, y = 0 or x = 0, y = 50 Substituting these values in the objective function: Z = 15(25) + 10(0) = 375 Z = 15(0) + 10(50) = 500

v = 31.25 jobs

Thus, the solution is x = 0 and y = 50

b) (8 weeks)(6 days/week)(3 lawns/day) = 144 lawns

This is a simple linear programming model, the subject of the next several chapters. The student should recognize that there are only two possible solutions, which are the corner points of the feasible solution space, only one of which is optimal.

Z = (144)(30) − 500 − (144)(14) Z = $1,804 c) (8 weeks)(6 days/week)(4 lawns/day) = 192 lawns Z = (192)(25) − 500 − (192)(14) Z = $1,612

1-4 .

33.

The solution is computed by solving simultaneous equations,

It is the only, i.e., “optimal” solution because there is only one set of values for x and y that satisfy both constraints simultaneously.

x = 30, y = 10, Z = $1,400 34. Labor usage

Clay usage

Profit

Possible

# mugs

12x + 15y < = 60

9x+5y < = 30

300x + 250y

solution?

250

yes

300

yes

550

yes

500

yes

600

yes

800

yes

850

yes

1100

yes, best solution

750

yes

900

yes

1050

yes

1150

1350

1400

1650

1200

1000

yes

1300

1450

1600

1700

# bowls

1-5 .

37. Determine logical solutions: 35. Maximize Z = $30xAN + 70xAJ + 40xBN + 60xBJ subject to xAN + xAJ = 400 xBN + xBJ = 400

Cakes

Bread

Total Sales

$12

$22

3 1 $36 4. 4 0 $40 Each solution must be checked to see if it violates the constraints for baking time and flour. Some possible solutions can be logically discarded because they are obviously inferior. For example, 0 cakes and 1 loaf of bread is clearly inferior to 0 cakes and 2 loaves of bread. 0 cakes and 3 loaves of bread is not possible because there is not enough flour for 3 loaves of bread.

xAN + xBN = 500 xAJ + xBJ = 300 The solution is xAN = 400, xBN = 100, xBJ = 300, and Z = 34,000 This problem can be solved by allocating as much as possible to the lowest cost variable, xAN = 400, then repeating this step until all the demand has been met. This is a similar logic to the minimum cell cost method.

Using this logic, there are four possible solutions as shown. The best one, 4 cakes and 0 loaves of bread, results in the highest total sales of $40. This problem demonstrates the cost trade-off inherent in queuing analysis, the topic of Chapter 13. In this problem the cost of service, i.e., the cost of staffing registers, is added to the cost of customers waiting, i.e., the cost of lost sales and ill will, as shown in the following table.

36. This is virtually a straight linear relationship between time and site visits; thus, a simple linear graph would result in a forecast of approximately 34,500 site visits.

38.

Registers staffed

Waiting time (mins)

1.7

0.5

0.1

Cost of service ($)

120

180

240

300

360

420

480

Cost of waiting ($)

850

550

300

Total cost ($)

910

670

480

290

300

360

420

480

The total minimum cost of $290 occurs with 4 registers staffed

39.

from consideration. As a result, the route 1-3-5-9 is the shortest.

The shortest route problem is one of the topics of Chapter 7. At this point, the most logical “trial and error” way that most students will probably approach this problem is to identify all the feasible routes and compute the total distance for each, as follows:

An additional aspect to this problem could be to have the students look at these routes on a real map and indicate which they think might “practically” be the best route. In this case, 1-2-5-9 would likely be a better route, because even though it’s two miles farther it is Interstate highway the whole way, whereas 1-3-5-9 encompasses U.S. 4-lane highways and state roads.

1-2-6-9 = 228 1-2-5-9 = 213 1-3-5-9 = 211 1-3-8-9 = 276 1-4-7-8-9 = 275 Obviously inferior routes like 1-3-4-7-8-9 and 1-2-5-8-9 that include additional segments to the routes listed above can be logically eliminated

1-6 .

If demand is less than 833 rafts, alternative 2 should not be selected, and alternative 1 should be used if demand is expected to be between 375 and 833.33 rafts.

CASE SOLUTION: CLEAN CLOTHES CORNER LAUNDRY a) v =

If demand is greater than 833.33 rafts, which alternative is best? To determine the answer, equate the two cost functions.

cf 1,700 = = 2,000 items per month p − cv 1.10 − .25

b) Solution depends on number of months; 36 used here. $16,200 ÷ 36 = $450 per month, thus monthly fixed cost is $2,150

3,000 + 12v = 10,000 + 8v 4v = 7,000

v = 1,750 cf 2,150 = = 2,529.4 items per month This is referred to as the point of indifference between p − cv 1.10 − .25 the two alternatives. In general, for demand lower than this 529.4 additional items per month point (1,750) the alternative should be selected with the c) Z = vp − cf − vcv lowest fixed cost; for demand greater than this point the alternative with the lowest variable cost should = 4,300(1.10) − 2,150 − 4,300(.25) be selected. (This general relationship can be observed by = $1,505 per month graphing the two cost equations and seeing where they intersect.) After 3 years, Z = $1,955 per month v=

d) v =

Thus, for the Ocobee River Rafting Company, the following guidelines should be used:

cf 1,700 = = 2,297.3 p − cv .99 − .25

demand < 375, do not start business; 375 < demand < 1,750, select alternative 1; demand > 1,750, select alternative 2

Z = vp − cf − vcv = 3,800(.99) − 1,700 − 3,800(.25) = $1,112 per month

e) With both options:

Since Penny estimates demand will be approximately 1,000 rafts, alternative 1 should be selected.

Z = vp − cf − vcv

Z = vp − cf − vcv = (1,000)(20) − 3,000 − (1,000)(12)

= 4,700(.99) − 2,150 − 4,700(.25)

Z = $5,000

= $1,328 She should purchase the new equipment but not decrease prices.

CASE SOLUTION: CONSTRUCTING A DOWNTOWN PARKING LOT IN DRAPER

CASE SOLUTION: OCOBEE RIVER RAFTING COMPANY

a) The annual capital recovery payment for a capital expenditure of $4.5 million over 30 years at 8% is,

Alternative 1: cf = $3,000

p = $20

(4,500,000)[0.08(1 + .08)30] / (1 + .08)30 − 1

cv = $12 v1 =

= $399,723.45

cf 3,000 = = 375 rafts p − cv 20 − 12

This is part of the annual fixed cost. The other part of the fixed cost is the employee annual salaries of $140,000. Thus, total fixed costs are,

Alternative 2: cf = $10,000

$399,723.45 + 140,000 = $539,723.45 cf v= p − cv

p = $20 cv = $8 v2 =

539,723.45 3.20 − 0.60 = 207,585.94 parked cars per year =

cf 10,000 = = 833.37 p − cv 20 − 8

If demand is less than 375 rafts, the students should not start the business.

1-7 .

b) If 365 days per year are used, then the daily usage is,

(d) Decrease in trips: Annual revenue = 657,000 Total variable cost = 443,475 207,585.94 = 568.7 or approximately 569 cars First year loss = (986,475) 365 Break even year: (5.62) years per day Bus operating hrs/day = 13.5 Operating cost/hr. = 90 This seems like a reachable goal given the size of Days/year = 365 the town and the student population. Total annual variable cost = 443,475 (e) $1,200,000 Grant: Fixed Cost = 0 First Year Revenue = 56,940

CASE SOLUTION: A BUS SERVICE FOR DRAPER Fixed cost (3 buses) = 1,200,000 Total Variable Cost = 591,300 Annual Revenue = 648,240 Passengers/bus/trip = 37 Passenger fare = 4 Trips/bus/day = 4 Number of buses = 3 Days/year = 365 Total annual revenue = 648,240 = (37)(4)(4)(3)(365) Bus operating hrs/day = 18 Operating cost/hr = 90 Days/year = 365 Total annual variable cost = 591,300 = (18)(90)(365) (a) First year loss = (1,143,060.00) (b) Years to break even: Loss in year 1 = –1,143,060.0 Not possible to break even (c) 45 passengers per trip: Annual Revenue = 788,400 First year loss = (1,002,900) Not possible to break even 50 passengers per trip: Annual revenue = 876,000 First year loss = (915,300) Break even year: (3.215) years

1-8 .

Chapter Two: Linear Programming: Model Formulation and Graphical Solution 36. Maximization, graphical solution

PROBLEM SUMMARY

37. Sensitivity analysis (2–36)

1. Maximization (1–28 continuation), graphical solution

38. Maximization, graphical solution

2. Minimization, graphical solution

39. Sensitivity analysis (2–38)

3. Sensitivity analysis (2–2)

40. Maximization, graphical solution

4. Minimization, graphical solution

41. Sensitivity analysis (2–40)

5. Maximization, graphical solution

42. Minimization, graphical solution

6. Slack analysis (2–5), sensitivity analysis

43. Sensitivity analysis (2–42)

7. Maximization, graphical solution

44. Maximization, graphical solution

8. Slack analysis (2–7)

45. Sensitivity analysis (2–44)

9. Maximization, graphical solution

46. Maximization, graphical solution

10. Minimization, graphical solution

47. Sensitivity analysis (2–46)

11. Maximization, graphical solution

48. Maximization, graphical solution

12. Sensitivity analysis (2–11)

49. Minimization, graphical solution

13. Sensitivity analysis (2–11)

50. Sensitivity analysis (2–49)

14. Maximization, graphical solution

51. Minimization, graphical solution

15. Sensitivity analysis (2–14)

52. Sensitivity analysis (2–51)

16. Maximization, graphical solution

53. Maximization, graphical solution

17. Sensitivity analysis (2–16)

54. Minimization, graphical solution

18. Maximization, graphical solution

55. Sensitivity analysis (2–54)

19. Standard form (2–18)

56. Maximization, graphical solution

20. Maximization, graphical solution

57. Sensitivity analysis (2–56)

21. Constraint analysis (2–20)

58. Maximization, graphical solution

22. Minimization, graphical solution

59. Sensitivity analysis (2–58)

23. Sensitivity analysis (2–22)

60. Multiple optimal solutions

24. Sensitivity analysis (2–22)

61. Infeasible problem

25. Sensitivity analysis (2–22)

62. Unbounded problem

26. Minimization, graphical solution 27. Minimization, graphical solution 28. Sensitivity analysis (2–27) 29. Minimization, graphical solution 30. Maximization, graphical solution 31. Minimization, graphical solution 32. Maximization, graphical solution 33. Sensitivity analysis (2–32) 34. Minimization, graphical solution 35. Maximization, graphical solution

2-1 .

6x1 + 6x2 ≥ 36 (phosphate, oz) x2 ≥ 2 (potassium, oz) x1,x2 ≥ 0

PROBLEM SOLUTIONS 1. a) x1 = # cakes x2 = # loaves of bread maximize Z = $10x1 + 6x2 subject to 3x1 + 8x2 ≤ 20 cups of flour 45x1 + 30x2 ≤ 180 minutes x1,x2 ≥ 0

a) Maximize Z = 400x1 + 100x2 (profit, $) subject to 8x1 + 10x2 ≤ 80 (labor, hr) 2x1 + 6x2 ≤ 36 (wood) x1 ≤ 6 (demand, chairs) x1,x2 ≥ 0

a) Minimize Z = .05x1 + .03x2 (cost, $) subject to

8x1 + 6x2 ≥ 48 (vitamin A, mg) x1 + 2x2 ≥ 12 (vitamin B, mg) x1,x2 ≥ 0 b)

a) In order to solve this problem, you must substitute the optimal solution into the resource constraint for wood and the resource constraint for labor and determine how much of each resource is left over. Labor

The optimal solution point would change from point A to point B, thus resulting in the optimal solution

8x1 + 10x2 ≤ 80 hr 8(6) + 10(3.2) ≤ 80 48 + 32 ≤ 80 80 ≤ 80

x1 = 12/5 x2 = 24/5 Z = .408 4.

a) Minimize Z = 3x1 + 5x2 (cost, $) subject to

There is no labor left unused.

10x1 + 2x2 ≥ 20 (nitrogen, oz)

2-2 .

Sugar

Wood

2x1 + 4x2 ≤ 16 2(0) + 4(4) ≤ 16 16 ≤ 16

2x1 + 6x2 ≤ 36 2(6) + 6(3.2) ≤ 36 12 + 19.2 ≤ 36 31.2 ≤ 36 36 − 31.2 = 4.8

There is no sugar left unused. 9.

There is 4.8 lb of wood left unused. b) The new objective function, Z = 400x1 + 500x2, is parallel to the constraint for labor, which results in multiple optimal solutions. Points B (x1 = 30/7, x2 = 32/7) and C (x1 = 6, x2 = 3.2) are the alternate optimal solutions, each with a profit of $4,000. 7. a) Maximize Z = x1 + 5x2 (profit, $) subject to 5x1 + 5x2 ≤ 25 (flour, lb) 2x1 + 4x2 ≤ 16 (sugar, lb) x1 ≤ 5 (demand for cakes) x1,x2 ≥ 0

10. a) Minimize Z = 80x1 + 50x2 (cost, $) subject to 3x1 + x2 ≥ 6 (antibiotic 1, units) x1 + x2 ≥ 4 (antibiotic 2, units) 2x1 + 6x2 ≥ 12 (antibiotic 3, units) x1,x2 ≥ 0

In order to solve this problem, you must substitute the optimal solution into the resource constraints for flour and sugar and determine how much of each resource is left over.

11. a)

Maximize Z = 300x1 + 400x2 (profit, $) subject to 3x1 + 2x2 ≤ 18 (gold, oz) 2x1 + 4x2 ≤ 20 (platinum, oz) x2 ≤ 4 (demand, bracelets) x1,x2 ≥ 0

Flour 5x1 + 5x2 ≤ 25 lb 5(0) + 5(4) ≤ 25 20 ≤ 25 25 − 20 = 5 There are 5 lb of flour left unused.

2-3 .

The profit for a necklace would have to increase to $600 to result in a slope of −3/2:

400x2 = Z − 600x1 x2 = Z/400 − 3/2x1 However, this creates a situation where both points C and D are optimal, ie., multiple optimal solutions, as are all points on the line segment between C and D. 14. a) Maximize Z = 50x1 + 40x2 (profit, $) subject to

12.

3x1 + 5x2 ≤ 150 (wool, yd2) 10x1 + 4x2 ≤ 200 (labor, hr) x1,x2 ≥ 0

The new objective function, Z = 300x1 + 600x2, is parallel to the constraint line for platinum, which results in multiple optimal solutions. Points B (x1 = 2, x2 = 4) and C (x1 = 4, x2 = 3) are the alternate optimal solutions, each with a profit of $3,000.

The feasible solution space will change. The new constraint line, 3x1 + 4x2 = 20, is parallel to the existing objective function. Thus, multiple optimal solutions will also be present in this scenario. The alternate optimal solutions are at x1 = 1.33, x2 = 4 and x1 = 2.4, x2 = 3.2, each with a profit of $2,000. 13. a) Optimal solution: x1 = 4 necklaces, x2 = 3 bracelets. The maximum demand is not achieved by the amount of one bracelet.

15.

b) The solution point on the graph which corresponds to no bracelets being produced must be on the x1 axis where x2 = 0. This is point D on the graph. In order for point D to be optimal, the objective function “slope” must change such that it is equal to or greater than the slope of the constraint line, 3x1 + 2x2 = 18. Transforming this constraint into the form y = a + bx enables us to compute the slope:

The feasible solution space changes from the area 0ABC to 0AB'C', as shown on the following graph.

2x2 = 18 − 3x1 x2 = 9 − 3/2x1 From this equation the slope is −3/2. Thus, the slope of the objective function must be at least −3/2. Presently, the slope of the objective function is −3/4:

The extreme points to evaluate are now A, B', and C'. A:

400x2 = Z − 300x1 x2 = Z/400 − 3/4x1

*B':

2-4 .

x1 = 0 x2 = 30 Z = 1,200 x1 = 15.8 x2 = 20.5 Z = 1,610

C':

x1 = 24 x2 = 0 Z = 1,200

18.

Point B' is optimal 16. a) Maximize Z = 23x1 + 73x2 subject to x1 ≤ 40 x2 ≤ 25 x1 + 4x2 ≤ 120 x1,x2 ≥ 0

19.

Maximize Z = 5x1 + 8x2 + 0s1 + 0s3 + 0s4 subject to 3x1 + 5x2 + s1 = 50 2x1 + 4x2 + s2 = 40 x1 + s3 = 8 x2 + s4 = 10 x1,x2 ≥ 0 A: s1 = 0, s2 = 0, s3 = 8, s4 = 0 B: s1 = 0, s2 = 3.2, s3 = 0, s4 = 4.8 C: s1 = 26, s2 = 24, s3 = 0, s4 = 10

20.

17. a) No, not this winter, but they might after they recover equipment costs, which should be after the 2nd winter. b) x1 = 55 x2 = 16.25 Z = 1,851

21.

It changes the optimal solution to point A (x1 = 8, x2 = 6, Z = 112), and the constraint, x1 + x2 ≤ 15, is no longer part of the solution space boundary. 22. a) Minimize Z = 64x1 + 42x2 (labor cost, $) subject to 16x1 + 12x2 ≥ 450 (claims) x1 + x2 ≤ 40 (workstations) 0.5x1 + 1.4x2 ≤ 25 (defective claims) x1,x2 ≥ 0

No, profit will go down c)

x1 = 40 x2 = 25 Z = 2,435 Profit will increase slightly

d) x1 = 55 x2 = 27.72 Z = $2,073 Profit will go down from (c)

2-5 .

27.

23.

Changing the pay for a full-time claims solution to point A in the graphical solution where x1 = 28.125 and x2 = 0, i.e., there will be no part-time operators. Changing the pay for a part-time operator from $42 to $36 has no effect on the number of full-time and parttime operators hired, although the total cost will be reduced to $1,671.95.

24.

Eliminating the constraint for defective claims would result in a new solution, x1 = 0 and x2 = 37.5, where only part-time operators would be hired.

25.

The solution becomes infeasible; there are not enough workstations to handle the increase in the volume of claims.

28. 29.

30.

26.

2-6 .

The problem becomes infeasible.

31.

35. a) Maximize Z = 800x1 + 900x2 (profit, $) subject to

32. a) Maximize Z = $4.15x1 + 3.60x2 (profit, $) subject to

2x1 + 4x2 ≤ 30 (stamping, days) 4x1 + 2x2 ≤ 30 (coating, days) x1 + x2 ≥ 9 (lots) x1,x2 ≥ 0

x1 + x2 ≤ 115 (freezer space, gals.) 0.93 x1 + 0.75 x2 ≤ 90 (budget, $) x1 2 ≥ or x1 − 2 x2 ≥ 0 (demand) x2 1

x1 ,x2 ≥ 0

36. a) Maximize Z = 30x1 + 70x2 (profit, $) subject to 4x1 + 10x2 ≤ 80 (assembly, hr) 14x1 + 8x2 ≤ 112 (finishing, hr) x1 + x2 ≤ 10 (inventory, units) x1,x2 ≥ 0

b) 33.

No additional profit, freezer space is not a binding constraint.

34. a) Minimize Z = 200x1 + 160x2 (cost, $) subject to 6x1 + 2x2 ≥ 12 (high-grade ore, tons) 2x1 + 2x2 ≥ 8 (medium-grade ore, tons) 4x1 + 12x2 ≥ 24 (low-grade ore, tons) x1,x2 ≥ 0

2-7 .

37.

The slope of the original objective function is computed as follows: Z = 30x1 + 70x2 70x2 = Z − 30x1 x2 = Z/70 − 3/7x1 slope = −3/7

39. a) 15(4) + 8(6) ≤ 120 hr 60 + 48 ≤ 120 108 ≤ 120 120 − 108 = 12 hr left unused b) Points C and D would be eliminated and a new optimal solution point at x1 = 5.09, x2 = 5.45, and Z = 111.27 would result.

The slope of the new objective function is computed as follows: Z = 90x1 + 70x2 70x2 = Z − 90x1 x2 = Z/70 − 9/7x1 slope = −9/7

40. a) Maximize Z = .28x1 + .19x2 x1 + x2 ≤ 96 cans x2 ≥2 x1

The change in the objective function not only changes the Z values but also results in a new solution point, C. The slope of the new objective function is steeper and thus changes the solution point. A: x1 = 0 x2 = 8 Z = 560

C: x1 = 5.3 x2 = 4.7 Z = 806

D: x1 = 8 x2 = 0 Z = 720

x1 = 3.3 x2 = 6.7 Z = 766

x1 ,x2 ≥ 0

38. a) Maximize Z = 9x1 + 12x2 (profit, $1,000s) subject to 4x1 + 8x2 ≤ 64 (grapes, tons) 5x1 + 5x2 ≤ 50 (storage space, yd3) 15x1 + 8x2 ≤ 120 (processing time, hr) x1 ≤ 7 (demand, Nectar) x2 ≤ 7 (demand, Red) x1,x2 ≥ 0

2-8 .

41.

The model formulation would become, maximize Z = $0.23x1 + 0.19x2 subject to x1 + x2 ≤ 96 –1.5x1 + x2 ≥ 0 x1,x2 ≥ 0 The solution is x1 = 38.4, x2 = 57.6, and Z = $19.78 The discount would reduce profit.

42. a) Minimize Z = $0.46x1 + 0.35x2 subject to .91x1 + .82x2 = 3,500 x1 ≥ 1,000 x2 ≥ 1,000 .03x1 − .06x2 ≥ 0 x1,x2 ≥ 0 b) 477 − 445 = 32 fewer defective items

44. a) Maximize Z = $2.25x1 + 1.95x2 subject to 8x1 + 6x2 ≤ 1,920 3x1 + 6x2 ≤ 1,440 3x1 + 2x2 ≤ 720 x1 + x2 ≤ 288 x1,x2 ≥ 0 b)

43. a) Minimize Z = .09x1 + .18x2 subject to .46x1 + .35x2 ≤ 2,000 x1 ≥ 1,000 x2 ≥ 1,000 .91x1 − .82x2 = 3,500 x1,x2 ≥ 0

2-9 .

45.

A new constraint is added to the model in

47.

The feasible solution space changes if the fertilizer constraint changes to 20x1 + 20x2 ≤ 800 tons. The new solution space is A'B'C'D'. Two of the constraints now have no effect.

x1 ≥ 1.5 x2 The solution is x1 = 160, x2 = 106.67, Z = $568

The new optimal solution is point C': A': x1 = 0 x2 = 37 Z = 11,100 B': x1 = 3 x2 = 37 Z = 12,300

46. a) Maximize Z = 400x1 + 300x2 (profit, $) subject to x1 + x2 ≤ 50 (available land, acres) 10x1 + 3x2 ≤ 300 (labor, hr)

*C': x1 = 25.71 x2 = 14.29 Z = 14,571 D': x1 = 26 x2 = 0 Z = 10,400

48. a) Maximize Z = $7,600x1 + 22,500x2 subject to

8x1 + 20x2 ≤ 800 (fertilizer, tons)

x1 + x2 ≤ 3,500 x2/(x1 + x2) ≤ .40 .12x1 + .24x2 ≤ 600 x1,x2 ≥ 0

x1 ≤ 26 (shipping space, acres) x2 ≤ 37 (shipping space, acres) x1,x2 ≥ 0 b)

2-10 .

49. a) Minimize Z = $(.05)(8)x1 + (.10)(.75)x2 subject to

for wine but only for beer. This amount “logically” would be the waste from 266.67 bottles, or $20, and the amount from the additional 53 bottles, $3.98, for a total of $23.98.

5x1 + x2 ≥ 800 5 x1 = 1.5 x2

51. a) Minimize Z = 3700x1 + 5100x2 subject to

8x1 + .75x2 ≤ 1,200 x1, x2 ≥ 0 x1 = 96 x2 = 320 Z = $62.40

x1 + x2 = 45 (32x1 + 14x2) / (x1 + x2) ≤ 21 .10x1 + .04x2 ≤ 6

x1 ≥ .25 ( x1 + x2 )

x2 ≥ .25 ( x1 + x2 ) x1, x2 ≥ 0 b)

50.

52. a) No, the solution would not change

The new solution is

b) No, the solution would not change

x1 = 106.67 x2 = 266.67 Z = $62.67

If twice as many guests prefer wine to beer, then the Robinsons would be approximately 10 bottles of wine short and they would have approximately 53 more bottles of beer than they need. The waste is more difficult to compute. The model in problem 53 assumes that the Robinsons are ordering more wine and beer than they need, i.e., a buffer, and thus there logically would be some waste, i.e., 5% of the wine and 10% of the beer. However, if twice as many guests prefer wine, then there would logically be no waste

Yes, the solution would change to China (x1) = 22.5, Brazil (x2) = 22.5, and Z = $198,000.

53. a) x1 = $ invested in stocks x2 = $ invested in bonds maximize Z = $0.18x1 + 0.06x2 (average annual return) subject to x1 + x2 ≤ $720,000 (available funds) x1/(x1 + x2) ≤ .65 (% of stocks) .22x1 + .05x2 ≤ 100,000 (total possible loss)

x1,x2 ≥ 0

2-11 .

being used anyway so assigning him more time would not have an effect.

One more hour of Sarah’s time would reduce the number of regraded exams from 10 to 9.8, whereas increasing Brad by one hour would have no effect on the solution. This is actually the marginal (or dual) value of one additional hour of labor, for Sarah, which is 0.20 fewer regraded exams, whereas the marginal value of Brad’s is zero. 56. a) x1 = # cups of Pomona x2 = # cups of Coastal Maximize Z = $2.05x1 + 1.85x2 subject to 16x1 + 16x2 ≤ 3,840 oz or (30 gal. × 128 oz) (.20)(.0625)x1 + (.60)(.0625)x2 ≤ 6 lbs. Colombian (.35)(.0625)x1 + (.10)(.0625)x2 ≤ 6 lbs. Kenyan (.45)(.0625)x1 + (.30)(.0625)x2 ≤ 6 lbs. Indonesian x2/x1 = 3/2 x1,x2 ≥ 0 b) Solution: x1 = 87.3 cups x2 = 130.9 cups Z = $421.09

54.

x1 = exams assigned to Brad x2 = exams assigned to Sarah minimize Z = .10x1 + .06x2 subject to x1 + x2 = 120 x1 ≤ (720/7.2) or 100 x2 ≤ 50(600/12) x1,x2 ≥ 0

55.

If the constraint for Sarah’s time became x2 ≤ 55 with an additional hour then the solution point at A would move to x1 = 65, x2 = 55 and Z = 9.8. If the constraint for Brad’s time became x1 ≤ 108.33 with an additional hour then the solution point (A) would not change. All of Brad’s time is not

57. a) The only binding constraint is for Colombian; the constraints for Kenyan and Indonesian are nonbinding and there are already extra, or slack, pounds of these coffees available. Thus, only getting more Colombian would affect the solution.

2-12 .

One more pound of Colombian would increase sales from $421.09 to $463.20.

60.

Increasing the brewing capacity to 40 gallons would have no effect since there is already unused brewing capacity with the optimal solution. b) If the shop increased the demand ratio of Pomona to Coastal from 1.5 to 1 to 2 to 1 it would increase daily sales to $460.00, so the shop should spend extra on advertising to achieve this result. 58. a) x1 = 16 in. pizzas x2 = hot dogs Maximize Z = $22x1 + 2.35x2 Subject to $10x1 + 0.65x2 ≤ $1,000 324 in2 x1 + 16 in2 x2 ≤ 27,648 in2 x2 ≤ 1,000 x1, x2 ≥ 0 b)

Multiple optimal solutions; A and B alternate optimal 61.

62.

59. a) x1 = 35, x2 = 1,000, Z = $3,120 Profit would remain the same ($3,120) so the increase in the oven cost would decrease the season’s profit from $10,120 to $8,120. b) x1 = 35.95, x2 = 1,000, Z = $3,140 Profit would increase slightly from $10,120 to $10, 245.46. c) x1 = 55.7, x2 = 600, Z = $3,235.48 Profit per game would increase slightly.

2-13 .

The graphical solution is shown as follows.

CASE SOLUTION: METROPOLITAN POLICE PATROL The linear programming model for this case problem is Minimize Z = x/60 + y/45 subject to 2x + 2y ≥ 5 2x + 2y ≤ 12 y ≥ 1.5x x, y ≥ 0 The objective function coefficients are determined by dividing the distance traveled, i.e., x/3, by the travel speed, i.e., 20 mph. Thus, the x coefficient is x/3 ÷ 20, or x/60. In the first two constraints, 2x + 2y represents the formula for the perimeter of a rectangle.

Changing the objective function to Z = $16x1 + 16x2 would result in multiple optimal solutions, the end points being B and C. The profit in each case would be $960.

The graphical solution is displayed as follows.

Changing the constraint from .90x2 − .10x1 ≥ 0 to .80x2 −.20x1 ≥ 0 has no effect on the solution.

CASE SOLUTION: ANNABELLE INVESTS IN THE MARKET x1 = no. of shares of index fund x2 = no. of shares of internet stock fund Maximize Z = (.17)(175)x1 + (.28)(208)x2 = 29.75x1 + 58.24x2 subject to 175x1 + 208x2 = $120, 000 x1 ≥ .33 x2

The optimal solution is x = 1, y = 1.5, and Z = 0.05. This means that a patrol sector is 1.5 miles by 1 mile and the response time is 0.05 hr, or 3 min.

x2 ≤2 x1 x1, x2 > 0

CASE SOLUTION: “THE POSSIBILITY” RESTAURANT

x1 = 203 x2 = 406 Z = $29,691.37

The linear programming model formulation is

x2 ≥ .33 x1 will have no effect on the solution. Eliminating the constraint

Maximize = Z = $12x1 + 16x2 subject to

x1 ≤2 x2 will change the solution to x1 = 149, x2 = 451.55, Z = $30,731.52.

x1 + x2 ≤ 60 .25x1 + .50x2 ≤ 20 x1/x2 ≥ 3/2 or 2x1 − 3x2 ≥ 0 x2/(x1 + x2) ≥ .10 or .90x2 − .10x1 ≥ 0 x1x2 ≥ 0

Eliminating the constraint

2-14 .

Increasing the amount available to invest (i.e., $120,000 to $120,001) will increase profit from Z = $29,691.37 to Z = $29,691.62 or approximately $0.25. Increasing by another dollar will increase profit by another $0.25, and increasing the amount available by one more dollar will again increase profit by $0.25. This

indicates that for each extra dollar invested a return of $0.25 might be expected with this investment strategy. Thus, the marginal value of an extra dollar to invest is $0.25, which is also referred to as the “shadow” or “dual” price as described in Chapter 3.

2-15 .

Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64.

PROBLEM SUMMARY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

QM for Windows Excel (1–34) Excel Excel solution Excel Excel (1–35) Model formulation Graphical solution; sensitivity analysis (3–7) Sensitivity analysis (3–7) Model formulation Graphical solution; sensitivity analysis (3–10) Sensitivity analysis (3–10) Model formulation Graphical solution; sensitivity analysis (3–13) Computer solution; sensitivity analysis (3–13) Model formulation Graphical solution; sensitivity analysis (3–16) Computer solution; sensitivity analysis (3–16) Model formulation Graphical solution; sensitivity analysis (3–19) Computer solution; sensitivity analysis (3–19) Model formulation Graphical solution; sensitivity analysis (3–22) Computer solution; sensitivity analysis (3–22) Model formulation Graphical solution; sensitivity analysis (3–25) Computer solution; sensitivity analysis (3–25) Model formulation Graphical solution; sensitivity analysis (3–28) Computer solution; sensitivity analysis (3–28) Model formulation; graphical solution Computer solution; sensitivity analysis (3–31) Model formulation Graphical solution; computer solution; sensitivity analysis (3–33). Model formulation Graphical solution; sensitivity analysis (3–35) Computer solution; sensitivity analysis (3–35) Computer solution (2–58) Computer solution Model formulation; computer solution Model formulation; computer solution Computer solution; sensitivity analysis

Model formulation Graphical solution; sensitivity analysis (3–43) Computer solution; sensitivity analysis (3–43) Model formulation; computer solution Sensitivity analysis (3–46) Model formulation; computer solution Sensitivity analysis (3–48) Model formulation Computer solution; sensitivity analysis (3–50) Model formulation Computer solution; sensitivity analysis (3–52) Model formulation Computer solution; sensitivity analysis (3–54) Model formulation Computer solution; sensitivity analysis (3–56) Computer solution Model formulation Sensitivity analysis (3–59) Model formulation Computer solution; sensitivity analysis (3–61) Model formulation; graphical solution Computer solution; sensitivity analysis (3–63)

PROBLEM SOLUTIONS 1.

3-1 .

7. a)

x1 = # bowls x2 = # mugs Maximize Z = 300x1 + 250x2 Subject to 12x1 + 15x2 ≤ 60 9x1 + 5x2 ≤ 30 x1, x2 ≥ 0 x1 = 2, x2 = 2.4, Z = $1,200 This solution is different because it is not restricted to interger values. Set Target cell: B13 Changing cells: B10:B12 Profit: = B10 * C4 + B11 * D4 + B12 * E4 Constraints: B10:B12 ≥ 0 G6 ≤ F6 G7 ≤ F7 x1 = 0 x2 = 9 Z = $54 F6: = C6 * B12 + D6 * B13 F7: = C7 * B12 + D7 * B13 F8: = C8 * B12 + D8 * B13 F9: = C9 * B12 + D9 * B13 G6: = E6 − F6 G7: = E7 − F7 G8: = E8 − F8 G9: = E9 − F9 B14: = C4 * B12 + D4 * B13 x1 = 0 x2 = 5.2 Z = 81.6 xAN = # Atlanta to Nashville XAJ = # Atlanta to Jacksonville XBN = # Birmingham to Nashville XBJ = # Birmingham to Jacksonville Minimize Z = 30xAN + 70xAJ + 40xBN + 60xBJ Subject to XAN + xAJ = 400 XBN + xBJ = 400 XAN + xBN = 500 XAJ + xBJ = 300 XAN = 400, xAJ = 0, xBN = 100, xBJ = 300, Z = $34,000 x1 = no. of basketballs x2 = no. of footballs maximize Z = 12x1 + 16x2 subject to

b) maximize Z = 12x1 + 16x2 + 0s1 + 0s2 subject to 3x1 + 2x2 + s1 = 500 4x1 + 5x2 + s2 = 800 x1, x2, s1, s2 ≥ 0

A: 3(0) + 2(160) + s1 = 500 s1 = 180 4(0) + 5(160) + s2 = 800 s2 = 0 B: 3(128.5) + 2(57.2) + s1 = 500 s1 = 0 4(128.5) + 2(57.2) + s2 = 800 s2 = 0 C: 2(167) + 2(0) + s1 = 500 s1 = 0 4(167) + 5(0) + s2 = 800 s2 = 132 b) Z = 12x1 + 16x2 and, x2 = Z/16 − 12x1/16 The slope of the objective function, −12/16, would have to become steeper (i.e., greater) than the slope of the constraint line 4x1 + 5x2 = 800, for the solution to change. The profit, c1, for a basketball that would change the solution point is, −4/5 = −c1/16 5c1 = 64 c1 = 12.8 Since $13 > 12.8 the solution point would change to B where x1 = 128.5, x2 = 57.2. The new Z value is $2,585.70. For a football, −4/5 = −12/c2 4c2 = 60 c2 = 15 Thus, if the profit for a football decreased to $15 or less, point B will also be optimal (i.e., multiple optimal solutions). The solution at B is x1 = 128.5, x2 = 57.2, and Z = $2,400.

3x1 + 2x2 ≤ 500 4x1 + 5x2 ≤ 800 x1, x2 ≥ 0

3-2 .

c) If the constraint line for rubber changes to 3x1 + 2x2 = 1,000, it moves outward, eliminating points B and C. However, since A is the optimal point, it will not change and the optimal solution remains the same, x1 = 0, x2 = 160, and Z = 2,560. There will be an increase in slack, s1, to 680 lbs.

9. a)

The lower limit is 0 since that is the lowest point on the x2 axis the constraint line can decrease to. Summarizing, 320 ≤ q1 ≤ ∞ 0 ≤ q2 ≤ 1,250

If the constraint line for leather changes to 4x1 + 5x2 = 1,300, point A will move to a new location, x1 = 0, x2 = 250, Z = $4,000. For c1 the upper limit is computed as −4/5 = −c1/16 5c1 = 64 c1 = 12.8 and the lower limit is unlimited. For c2 the lower limit is, −4/5 = −12/c2 4c2 = 60 c2 = 15 and the upper limit is unlimited. Summarizing, ∞ ≤ c1 ≤ 12.8 15 ≤ c2 ≤ ∞ For q1 the upper limit is ∞ since no matter how much q1 increases the optimal solution point A will not change. The lower limit for q1 is at the point where the constraint line (3x1 + 2x2 = q1) intersects with point A where x1 = 0, x2 = 160, 3x1 + 2x2= q1 3(0) + 2(160) = q1 q1 = 320 For q2 the upper limit is at the point where the rubber constraint line (3x1 + 2x2 = 500) intersects with the leather constraint line (4x1 + 5x2 = 800) along the x2 axis, i.e., x1 = 0, x2 = 250, 4x1 + 5x2 = q2 4(0) + 5(250) = q2 q2 = 1,250

Z = 2,560.000 Variable

Reduced Cost

Value

0.00

0.800

160.000

0.000

Slack/Surplus

Shadow Price

180.00

0.00

3.20

Constraint

The shadow price for rubber is $0. Since there is slack rubber left over at the optimal point, extra rubber would have no marginal value. The shadow price for leather is $3.20. For each additional ft.2 of leather that the company can obtain profit would increase by $3.20, up to the upper limit of the sensitivity range for leather (i.e., 1,250 ft.2). 10. a) x1 = no. of units of A x2 = no. of units of B maximize Z = 9x1 + 7x2 subject to 12x1 + 4x2 ≤ 60 4x1 + 8x2 ≤ 40 x1, x2 ≥ 0 b) maximize Z = 9x1 + 7x2 + 0s1 + 0s2 subject to 12x1 + 4x2 + s1 = 60 4x1 + 8x2 + s2 = 40 x1, x2, s1, s2 ≥ 0

b) Objective Coefficient Ranges Variables

Lower Limit

Current Values

Upper Limit

Allowable Increase Allowable Decrease

No limit

12.000

12.800

0.800

No limit

15.000

16.000

No limit

1.000

Allowable Increase

Allowable Decrease

Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit c1

320.000

500.000

No limit

180.000

0.000

800.000

1,250.000

450.000

800.000

3-3 .

If the profit for product B is increased to $15 the optimal solution point will not change, although Z would change from $57 to $81.

11.

If the profit for product B is increased to $20 the solution point will change from B to A, x1 = 0, x2 = 5, Z = $100. 12. a) For c1 the upper limit is computed as, −c1/7 = −3 c1 = 21 and the lower limit is, −c1/7 = −1/2 c1 = 3.50 For c2 the upper limit is, −9/c2 = −1/2 c2 = 18 and the lower limit is, −9/c2 = −3 c2 = 3 Summarizing, 3.50 ≤ c1 ≤ 21 3 ≤ c2 ≤ 18 b)

a) A: 12(0) + 4(5) + s1 = 60 s1 = 40 4(0) + 8(5) + s2 = 40 s2 = 0 B: 12(4) + 4(3) = 60 s1 = 0 4(4) + 8(3) + s2 = 40 s2 = 0 C: 12(5) + 4(0) + s1 = 60 s1 = 0 4(5) + 8(0) + s2 = 40 s2 = 20 b) The constraint line 12x1 + 4x2 = 60 would move inward resulting in a new location for point B at x1 = 2, x2 = 4, which would still be optimal. c) In order for the optimal solution point to change from B to A the slope of the objective function must be at least as flat as the slope of the constraint line, 4x1 + 8x2 = 40, which is −1/2. Thus, the profit for product B would have to be, −9/c2 = −1/2 c2 = 18

Z = 57.000 Variable

Value

Reduced Cost

4.000

0.000

3.000

0.000

Slack/Surplus

Shadow Price

0.000

0.550

0.000

0.600

Constraint

Objective Coefficient Ranges Variables

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

3.500

9.000

21.000

12.000

5.500

3.000

7.000

18.000

11.000

4.000

Right Hand Side Ranges Constraints

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

20.000

60.000

120.000

60.000

40.000

20.000

40.000

120.000

80.000

20.000

3-4 .

c) The shadow price for line 1 time is $0.55 per hour, while the shadow price for line 2 time is $0.60 per hour. The company would prefer to obtain more line 2 time since it would result in the greater increase in profit. 13. a) x1 = no. of yards of denim x2 = no. of yards of corduroy maximize Z = $2.25x1 + 3.10x2 subject to 5.0x1 + 7.5x2 ≤ 6,500 3.0x1 + 3.2x2 ≤ 3,000 x2 ≤ 510 x1, x2 ≥ 0

b) In order for the optimal solution point to change from B to C the slope of the objective function must be at least as great as the slope of the constraint line, 3.0x1 + 3.2x2 = 3,000, which is −3/3.2. Thus, the profit for denim would have to be, −c1/3.0 = −3/3.2 c1 = 2.91 If the profit for denim is increased from $2.25 to $3.00 the optimal solution would change to point C, where x1 = 1,000, x2 = 0, Z = 3,000. Profit for corduroy has no upper limit that would change the optimal solution point. c) The constraint line for cotton would move inward as shown in the following graph where point C is optimal.

b) maximize Z = $2.25x1 + 3.10x2 + 0s1 + 0s2 + 0s3 subject to 5.0x1 + 7.5x2 + s1 = 6,500 3.0x1 + 3.2x2 + s2 = 3,000 x2 + s3 = 510 x1, x2, s1, s2, s3 ≥ 0 14.

15.

a) 5.0(456) + 7.5(510) + s1 = 6,500 s1 = 6,500 − 6,105 s1 = 395 lbs. 3.0(456) + 3.2(510) + s2 = 3,000 s2 = 0 hrs. 510 + s3 = 510 s3 = 0 Therefore demand for corduroy is met.

3-5 .

Z = 2,607.000 Variable

Value

Reduced Cost

456.000

0.000

510.000

0.000

Constraint

Slack/Surplus

Shadow Price

395.000

0.000

0.750

0.000

0.700

Objective Coefficient Ranges Variables

Lower Limit Current Values Upper Limit Allowable Increase

Allowable Decrease

0.000

2.250

2.906

0.656

2.250

2.400

3.100

No limit

0.700

Right Hand Side Ranges Constraints Lower Limit Current Values Upper Limit Allowable Increase

Allowable Decrease

6,015.000

6,500.000

No limit

395.000

1,632.000

3,000.000

3,237.000

237.000

1,368.000

0.000

510.000

692.308

182.308

510.000

a) The company should select 237 additional hours of processing time, with a shadow price of $0.75 per hour. Cotton has a shadow price of $0 because there is already extra (slack) cotton available and not being used, so any more would have no marginal value. b) 0 ≤ c1 ≤ 2.906 6,105 ≤ q1 ≤ ∞ 2.4 ≤ c2 ≤ ∞ 1,632 ≤ q2 ≤ 3,237 0 ≤ q3 ≤ 692.308 The demand for corduroy can decrease to zero or increase to 692.308 yds. without changing the current solution mix of denim and corduroy. If the demand increases beyond 692.308 yds., then denim would no longer be produced and only corduroy would be produced. 16. x1 = no. of days to operate Mill 1 x2 = no. of days to operate Mill 2 minimize Z = 6,000x1 + 7,000x2 subject to 6x1 + 2x2 ≥ 12 2x1 + 2x2 ≥ 8 4x1 + 10x2 ≥ 5 x1,x2 ≥ 0

6(4) + 2(0) − s1 = 12 s1 = 12 2(4) + 2(0) − s2 = 8 s2 = 0 4(4) + 10(0) − s3 = 5 s3 = 11 b) The slope of the objective function, −6,000/7,000, must become flatter (i.e., less) than the slope of the constraint line, 2x1 + 2x2 = 8, for the solution to change. The cost of operating Mill 1, c1, that would change the solution point is, −c1/7,000 = −1 c1 = 7,000 Since $7,500 > $7,000, the solution point will change to B, where x1 = 1, x2 = 3, Z = $28,500. c) If the constraint line for high-grade aluminum changes to 6x1 + 2x2 = 10, it moves inward but does not change the optimal variable mix. B remains optimal but moves to a new location, x1 = 0.5, x2 = 3.5, Z = $27,500. 18.

17.

3-6 .

Z = 24,000 Variable

Value

4.000

0.000

Constraint

Slack/Surplus

Shadow Price

12.000

0.000

−3,000.000

11.000

0.000

Objective Coefficient Ranges Variables

Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

0.000

6,000.000

7,000.000

1,000.000

6,000.000

7,000.000

No limit

1,000.000

Right Hand Side Ranges Constraints

Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

No limit

12.000

24.000

12.000

No limit

4.000

8.000

No limit

4.000

No limit

5.000

16.000

11.000

No limit

b) At point D only corn is planted. In order for point D to be optimal the slope of the objective function will have to be at least as great (i.e., steep) as the slope of the constraint line, x1 + x2 = 410, which is −1. Thus, the profit for corn is computed as, −c1/520 = −1 c1 = 520 The profit for corn must be greater than $520 for the Bradleys to plant only corn.

a) There is surplus high-grade and low-grade aluminum so the shadow price is $0 for both. The shadow price for medium-grade aluminum is $3,000, indicating that for every ton that this constraint could be reduced, cost will decrease by $3,000. b) 0 ≤ c1 ≤ 7,000 ∞ ≤ q1 ≤ 24 6,000 ≤ c2 ≤ ∞ 4 ≤ q2 ≤ ∞ ∞ ≤ q3 ≤ 16 c) There will be no change. 19. x1 = no. of acres of corn x2 = no. of acres of tobacco maximize Z = 300x1 + 520x2 subject to x1 + x2 ≤ 410 105x1 + 210x2 ≤ 52,500 x2 ≤ 100 x1,x2 ≥ 0 20.

If the constraint line changes from x1 + x2 = 410 to x1 + x2 = 510, it will move outward to a location which changes the solution to the point where 105x1 + 210x2 = 52,500 intersects with the axis. This new point is x1 = 500, x2 = 0, Z = $150,000.

d) If the constraint line changes from x1 + x2 = 410 to x1 + x2 = 360, it moves inward to a location which changes the solution point to the intersection of x1 + x2 = 360 and 105x1 + 210x2 = 52,500. At this point x1 = 260, x2 = 100, and Z = $130,000. 21.

Z = 142,800.000 Variable x1

320.000

90.000

Slack/Surplus

Shadow Price

0.000

80.000

0.000

2.095

10.000

0.000

Constraint a)

x1 = 320, x2 = 90 320 + 90 + s1 = 410 s1 = 0 acres uncultivated 90 + s3 = 100 s3 = 10 acres of tobacco allotment unused

3-7 .

Value

Objective Coefficient Ranges Variables

Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

260.000

300.000

520.000

220.000

40.000

300.000

520.000

600.000

80.000

220.000

Allowable Increase

Allowable Decrease

Right Hand Side Ranges Constraints

Lower Limit Current Values Upper Limit

400.000

410.000

500.000

90.000

10.000

43,050.000

52,500.000

53,550.000

1,050.000

9,450.000

90.000

100.000

No limit

10.000

a) No, the shadow price for land is $80 per acre indicating that profit will increase by no more than $80 for each additional acre obtained. The maximum price the Bradleys should pay is $80 and the most they should obtain is at the upper limit of the sensitivity range for land. This limit is 500 acres, or 90 additional acres. Beyond 90 acres the shadow price would change. b) The shadow price for the budget is $2.095. Thus, for every $1 borrowed they could expect a profit increase of $2.095. If they borrowed $1,000 it would change the amount of corn and tobacco they plant; x1 = 310.5 acres of corn and x2 = 99.5 acres of tobacco. 22. x1 = no. of sausage biscuits x2 = no. of ham biscuits maximize Z = .60x1 + .50x2 subject to .10x1 ≤ 30 .15x2 ≤ 30 .04x1 + .04x2 ≤ 16 .01x1 + .024x2 ≤ 6 x1, x2 ≥ 0 23.

x1 = 300, x2 = 100, Z = $230 .10(300) + s1 = 30 s1 = 0 leftover sausage .15(100) + s2 = 30

s2 = 15 lbs. leftover ham .01(300) + .024(100) + s4 = 6 s4 = 0.6 hr. b) The slope of the objective function, −6/5, must become flatter (i.e., less) than the slope of the constraint line, .04x1 + .04x2 = 16, for the solution to change. The profit for ham, c2, that would change the solution point is,

24.

−0.6/c2 = −1 c2 = .60 Thus, an increase in profit for ham of 0.60 will create a second optimal solution point at C where x1 = 257, x2 = 143, and Z = $240. (Point D would also continue to be optimal, i.e., multiple optimal solutions.) A change in the constraint line from .04x1 + .04x2 = 16 to .04x1 + .04x2 = 18 would move the line outward, eliminating both points C and D. The new solution point occurs at the intersection of 0.01x1 + .024x2 = 6 and .10x = 30. This point is x1 = 300, x2 = 125, and Z = $242.50. Z = 230.000 Variable x1

300.000

100.000

Constraint

3-8 .

Value

Slack/Surplus

Shadow Price

0.000

1.000

15.000

0.000

12.500

0.600

0.000

Objective Coefficient Ranges Variables

Lower Limit

Current Values Upper Limit Allowable Increase Allowable Decrease

0.500

0.600

No limit

0.100

0.000

0.500

0.600

0.100

0.500

Right Hand Side Ranges Constraints

Lower Limit

Current Values Upper Limit Allowable Increase Allowable Decrease

25.714

30.000

40.000

10.000

4.286

15.000

30.000

No limit

15.000

12.000

16.000

17.000

1.000

4.000

5.400

6.000

No limit

0.600

26.

a) The shadow price for sausage is $1. For every additional pound of sausage that can be obtained profit will increase by $1. The shadow price for flour is $12.50. For each additional pound of flour that can be obtained, profit will increase by this amount. There are extra ham and labor hours available, so their shadow prices are zero, indicating additional amounts of those resources would add nothing to profit. b) The constraint for flour, indicated by the high shadow price. c) .50 ≤ c1 ≤ ∞ 25.714 ≤ q1 ≤ 40 The sensitivity range for profit indicates that the optimal mix of sausage and ham biscuits will remain optimal as long as profit does not fall below $0.50. The sensitivity range for sausage indicates the dual value of $1 will be maintained as long as the available sausage is between 25.714 and 40 lbs. 25. x1 = no. of telephone interviewers x2 = no. of personal interviewers minimize Z = 50x1 + 70x2 subject to 80x1 + 40x2 ≥ 3,000 80x1 ≥ 1,000 40x2 ≥ 800 x1, x2 ≥ 0

a) The optimal point is at B where x1 = 27.5 and x2 = 20. The slope of the objective function, −50/70, must become greater (i.e., steeper) than the slope of the constraint line, 80x1 + 40x2 = 3,000, for the solution point to change from B to A. The cost of a telephone interviewer that would change the solution point is, −c1/70 = −2 c1 = 140 This is the upper limit of the sensitivity range for c1. The lower limit is 0 since as the slope of the objective function becomes flatter, the solution point will not change from B until the objective function is parallel with the constraint line. Thus, 0 ≤ c1 ≤ 140

3-9 .

Since the constraint line is vertical, it can increase as far as point B and decrease all the way to the x2 axis before the solution mix will change. At point B, 80(27.5) = q1 q1 = 2,200 At the axis, 80(0) = q1 q1 = 0 Summarizing, 0 ≤ q1 ≤ 2,200 b) At the optimal point, B, x1 = 27.5 and x2 = 20. 80(27.5) − s2 = 1,000 s2 = 1,200 extra telephone interviews 40(20) − s3 = 800 s3 = 0 c) A change in the constraint line from 40x2 = 800 to 40x2 = 1,200 moves the lineup, but it does not change the optimal mix. The new solution values are x1 = 22.5, x2 = 30, Z = $3,225. 27.

a) Reduce the personal interview requirement; it will reduce cost by $0.625 per interview, while a telephone interview will not reduce cost; i.e., it has a shadow price equal to $0. b) 25 ≤ c2 ≤ ∞ 1,800 ≤ q1 ≥ ∞ 28. x1 = no. of gallons of rye x2 = no. of gallons of bourbon maximize Z = 3x1 + 4x2 subject to x1 + x2 ≥ 400 x1 ≥ .4(x1 + x2) x2 ≤ 250 x1 = 2x2 x1 + x2 ≤ 500 x1, x2 ≥ 0 29.

Z = 2,775.000 Variable

Value

27.500

20.000

Constraint

Slack/Surplus

Shadow Price

0.000

−0.625

1,200.000

0.000

−1.125

Objective Coefficient Ranges Variables

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

0.000

50.000

140.000

90.000

50.000

25.000

70.000

No limit

45.000

Right Hand Side Ranges Constraints

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

1,800.000

3,000.000

No limit

1,200.000

No limit

1,000.000

2,200.000

1,200.000

No limit

0.000

800.000

2,000.000

1,200.000

800.000

3-10 .

a) Optimal solution at B: x1 = 333.3 and x2 = 166.7 (333.3) + (166.7) − s1 = 400 s1 = 100 extra gallons of blended whiskey produced .6(333.33) − .4(166.7) − s2 = 0 s2 = 133.3 extra gallons of rye in the blend (166.7) + s3 = 250 s3 = 83.3 fewer gallons of bourbon than the maximum (333.3) + (166.7) + s4 = 500 s4 = 100 gallons of blend production capacity left over b) Because the “solution space” is not really an area, but a line instead, the objective function coefficients can change to any positive value and the solution point will remain the same, i.e., point B. Observing the graph of this model, no matter how flatter or steeper the objective function becomes, point B will remain optimal. 30.

−2.0 ≤ c1 ≤ ∞ −6.0 ≤ c2 ≤ ∞ Because there is only one effective solution point, the objective function can take on any negative (downward) slope and the solution point will not change. Only “negative” coefficients that result in a positive slope will move the solution to point A; however, this would be unrealistic. b) The shadow price for production capacity is $3.33. Thus, for each gallon increase in capacity, profit will increase by $3.33. c) This new specification changes the constraint, x1 − 2x2 = 0, to x1 − 3x2 = 0. This change to a constraint coefficient cannot be evaluated with normal sensitivity analysis. Instead the model must be solved again on the computer, which results in the following solution output.

Z = 1,666.667 Variable

Value

333.333

166.667

Constraint

Slack/Surplus

Shadow Price

100.000

0.000

133.333

0.000

83.333

0.000

3.333

Objective Coefficient Ranges Variables

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

−2.000

3.000

No limit

5.000

−6.000

4.000

No limit

10.000

Right Hand Side Ranges Constraints

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

No limit

400.000

500.000

100.000

No limit

0.000

133.333

No limit

166.667

250.000

No limit

83.333

−250.000

0.000

500.000

250.000

400.000

500.000

750.000

250.000

100.000

3-11 .

Z = 1,625.000 Variable

Value

375.000

125.000

Constraint Slack/Surplus Shadow Price c1

100.000

0.000

175.000

0.000

125.000

0.000

3.250

Objective Coefficient Ranges Variables

Lower Limit Current Values Upper Limit Allowable Increase

Allowable Decrease

−1.333

3.000

No limit

4.333

−9.000

4.000

No limit

13.000

Right Hand Side Ranges Constraints

Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

No limit

400.000

500.000

100.000

No limit

0.000

175.000

No limit

125.000

250.000

No limit

125.000

−500.000

0.000

500.000

400.000

500.000

1,000.000

500.000

100.000

31. a) Maximize Z = $8.65x1 + 10.95x2 subject to x1 + x2 ≤ 250 x1 + x2 ≥ 120 x1 − 2x2 ≤ 0 x1 − 1.2x2 ≥ 0 x1 ≤ 150 x2 ≤ 110 x1, x2 ≥ 0 b)

32.

x1 = 140 x2 = 110 Z = $2,415.50 a) beef; higher dual value of $2.30/lb., compared to no dual value for pork; there is pork left over. b) profit increases to $2,423.86, however, they should not do this because profit will increase to this amount if they only order 3.64 additional lbs., the upper limit of the sensitivity range for beef supply. c) No; the upper limit ratio of pork BBQ to beef is a non-binding constraint.

33.

Minimize Z = 11x1 + 16x2 subject to x1 + x2 = 500 .07x1 +.02x2 ≤ 25 x1 ≥ .20 x1 + x2 x2 ≥ .20 x1 + x2

x1, x2 ≥ 0

3-12 .

34.

36.

a) The optimal solution point is at B where x1 = $70,833.33, and x2 = $24,166.67. The slope of the objective function, −1.2/1.3, must become flatter than the slope of the constraint line, .18x1 + .30x2 = 20,000, for the solution point to change to A (i.e., only cattle). The return on cattle that will change the solution point is −1.2/c2 = −.18/30 c2 = 2 a)

x1 = 400,

Thus, the return must be 100% before Alexis will invest only in cattle. b) Yes, there is no slack money left over at the optimal solution. c) Since her investment is $95,000, she could expect to earn $21,416.67.

x2 = 100, z = $6,000 b) Minimize Z = .07x1 + .02x2 subject to 11x1 + 16x2 ≤ 7,000 x1 + x2 = 500 x1 ≥ .20 x1 + x2

37.

Variable

x2 ≥ .20 x1 + x2

35.

Z = 116,416.667

x1, x2 ≥ 0 x1 = 200 x2 = 300 Z = 20 x1 = $ amount invested in land x2 = $ amount invested in cattle maximize Z =1.20x1 + 1.30x2 subject to x1 + x2 ≤ 95,000 .18x1 + .30x2 ≤ 20,000 x1, x2 ≥ 0

Value

70,833.333

24,166.667

Constraint Slack/Surplus Shadow Price

3-13 .

0.000

1.050

0.000

0.833

Objective Coefficient Ranges Variables

Lower Limit Current Values Upper Limit

Allowable Increase

Allowable Decrease

0.780

1.200

1.300

0.100

0.420

1.200

1.300

2.000

0.700

0.100

Allowable Increase

Allowable Decrease

Right Hand Side Ranges Constraints

Lower Limit Current Values Upper Limit

66,666.667

95,000.000

111,111.111

16,111.111

28,333.333

17,100.000

20,000.000

28,500.000

8,500.000

2,900.000

a) The shadow price for invested money is $1.05. Thus, for every dollar of her own money Alexis invested she could expect a return of $0.05 or 5%. The upper limit of the sensitivity range is $111,111.11; thus, Alexis could invest $16,111.11 of her own money before the shadow price would change. b) This would change the constraint, .l8x1 + .30x2 = 20,000, to .30x1 + .30x2 = 20,000. In order to assess the effect of this change the problem must be solved again using the computer, as follows. Z = 86,666.667 Variable x1 x2

Value 0.000 66,666.667

Constraint Slack/Surplus Shadow Price 28,333.33 0.000 c1 c2

0.000

4.333

Objective Coefficient Ranges Variables x1 x2

Lower Limit No limit

Current Values 1.200

Upper Limit 1.300

Allowable Increase 0.100

Allowable Decrease No limit

1.200

1.300

No limit

0.100

Lower Limit 66,666.667

Current Values 95,000.000

Upper Limit No limit

Allowable Increase No limit

Allowable Decrease 28,333.333

0.000

20,000.000

28,500.000

8,500.000

20,000.000

Right Hand Side Ranges Constraints c1 c2

3-14 .

38.

x1 = # pizzas x2 = # hot dogs Maximize Z = $22x1 + 2.35x2 Subject to 10x1 + 0.65x2 ≤ $1,000 324x1 + 16x2 ≤ 27,648 in2 x2 ≤ 1,000 (break-even) 22x1 + 2.35x3 = 1,433.33 x1 = 65.15, x2 = 0

40.

39. a) x1 = # impatiens x2 = # daisies Maximize Z = 6x1 + 4x2 Subject to 8x1 + 4x2 ≤ 1,080 mins. 216x1 + 324x2 ≤ 43,200 in2 x1, x2 ≥ 0 x1 = 102.5, x2 = 65, Z = $875 b) x1 = 147.5, x2 = 35, Z = 1,025 Increase of $150

Maximize Z = 140x1 + 205x2 + 190x3 + 0s1 + 0s2 + 0s3 + 0s4 subject to 10x1 + 15x2 + 8x3 + s1 = 610 x1 − 3x2 + s2 = 0 .6x1 − .4x2 − .4x3 − s3 = 0 x2 − x3 − s4 = 0 x1, x2, s1, s2, s3, s4 ≥ 0 Z = 9,765.596 Variable

Value

Reduced Cost

22.385

0.000

16.789

0.000

16.789

0.000

Slack/Surplus

Shadow Price

0.000 27.982 0.000 0.000

16.009 0.000 −33.486 −48.532

Constraint c1 c2 c3 c4

Objective Coefficient Ranges

Variables

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

−237.857

140.000

171.739

31.739

377.857

132.000

205.000

325.227

120.227

73.000

117.000

190.000

No limit

73.000

3-15 .

Right Hand Side Ranges

Constraints

Lower Limit

Current Values

Upper Limit

Allowable Increase

Allowable Decrease

0.000

610.000

No limit

610.000

−27.982

0.000

No limit

27.982

−21.217

0.000

11.509

21.217

−20.890

0.000

28.154

20.890

41. a) Minimize Z = $400x1 + 180x2 + 90x3 subject to

Z = 206,000.000 Variable x1 x2 x3

x1 ≥ 200 x2 ≥ 300 x3 ≥ 100 4x3 − x1 − x2 ≤ 0

Value 200.000 600.000 200.000

Constraint Slack/Surplus Shadow Price c1 0.000 –220.000 c2 500.000 0.000 c3 100.000 0.000 c4 0.000 18.000

x1 + x2 + x3 = 1,000 x1, x2, x3 ≥ 0

Objective Coefficient Ranges Lower Limit 180.00

Current Values 400.00

Upper Limit No limit

Allowable Increase No limit

Allowable Decrease 220.00

90.00

180.00

400.00

220.00

90.00

No limit

90.00

180.00

90.00

No limit

Variables x1

Right Hand Side Ranges Constraints c1

Lower Limit 0.00

Current Values 200.00

Upper Limit 700.00

Allowable Increase 500.00

Allowable Decrease 200.00

No limit

100.00

600.00

500.00

No limit

100.00

200.00

100.00

No limit

−500.00

0.00

2,500.00

500.00

1,000.00

No limit

500.00

3-16 .

45.

42. a) x1 = 36.71

x2 = 10,000

x2 = 58.64

Z = $11,500

x3 = 0

a) The dual value of rack space is 0.75, so an increase in rack space to accommodate an additional 500 copies would result in increased advertising revenue of $375. An increase in rack space to 20,000 copies would be outside the sensitivity range for this constraint and require the problem to be solved again. The new solution is x1 = 8,000, x2 = 10,560, and Z = $11,920

x4 = 63.57 Z = 9,177.85 b)

34.6871 ≤ c1 ≤ 61.53 43.0808 ≤ c2 ≤ 71.23 ∞ ≤ c3 ≤ 65.4686 55 ≤ c4 ≤ ∞

b) 7,000 is within the sensitivity range for the entertainment guide (6,250 ≤ q3 ≤ 10,000). The dual value is $0.25; thus for every unit the distribution requirement can be reduced, revenue will be increased by $0.25, or $250. Therefore, Z = $11,750

529.08 ≤ q1 ≤ 747.99 350.33 ≤ q2 ≤ ∞ 3,488.55 ≤ q3 ≤ ∞ 1,363.63 ≤ q4 ≤ 1,761.47 −20.13 ≤ q5 ≤ 64.46 c)

46. a) Maximize Z = 4.25x1 + 5.10x2 + 4.50x3 + 5.20x4 + 4.10x5 + 4.90x6 + 3.80x7 subject to:

process 1 time is the most valuable with a dual value of $7.9275

x1 + x2 + x3 + x4 + x5 + x6 + x7 +

d) Product 3(x3) is not produced; it would require a profit of $65.46 to be produced. 43.

x8 ≤ 140,000 xi ≥ 15,000, i = 1, 2, ...,7

Maximize Z = $0.50x1 + 0.75x2 subject to:

xi ≥ 15,000, i = 1, 2, …,7

$0.17x1 + 0.25x2 ≤ $4,000 (printing budget)

∑x

x1 + x2 ≤ 18,000 (total copies, rack space)

≤ .20, i = 1, 2,..., 7

x1 ≥ 8,000 (entertainment guide)

x8 = .10 x4 + x6 + x7

x2 ≥ 8,000 (real estate guide)

xi ≥ 0

x1 , x2 ≥ 0

b) x1 = 15,000 x2 = 26,863 x3 = 20,588.24 x4 = 26,862.75 x5 = 15,000 x6 = 15,000 x7 = 15,000 x8 = 5,686 Z = $625,083 47. a) A 20,000 ft2 increase in store size to 160,000 ft2 would increase annual profit to $718,316. This is a $93,233 increase in profit. Given the price of the land ($190,000) relative to the increase in profit, it would appear that the cost of the land would be offset in about 2 years; therefore the decision should be to purchase the land.

44.

b) The decrease in profit in all departments would result in a new solution with Z = $574,653. This is a reduction of $50,430 annually in profit from the original solution with a 140,000

a) c2 ≥ .50 b) s1 = $140 c)

x1 = 8,000

There would be no feasible solution.

3-17 .

ft2 store; thus, given these conditions, MegaMart should not purchase the land.

a) More hours to assemble; the dual value for budget and space is zero, while the dual value for assembly is $30/hour.

48. a) Maximize Z = $0.97x1 + 0.83x2 + 0.69x3 subject to:

b) The additional net sales would be $900. Since the cost of the labor is $300, the additional profit would be $600. c) It would have no effect on the original solution. $700 profit for a cross country bike is within the sensitivity range for the objective function coefficient for x2.

x1 + x2 + x3 ≤ 324 cartons x3 ≥ x1 + x2 x3 ≥3 x1

x2 ≤ 120

52.

b) x1 = 54, x2 = 108, x3 = 162, Z = $253.80

Maximize Z = $0.35x1 + 0.42x2 + 0.37x3 subject to:

49. a) The shadow price for shelf space is $0.78 per carton; however, this is only valid up to 360 cartons, the upper limit of the sensitivity range for shelf space.

0.45x1 + 0.41x2 + 0.50x3 ≤ 960 x1 + x2 + x3 ≤ 2,000 x1 ≥ 200

b) The shadow price for available local dairy cartons is $0 so it would not increase profit to increase the available amount of local dairy milk.

x2 ≥ 200

x1, x2, x3 ≥ 0

x3 ≥ 200 x1 ≥ x2 + x3

The discount would change the objective function to, 53.

maximize Z = 0.86x1 + 0.83x2 + 0.69x3

x1 = 1,000 x2 = 800

and the constraint for relative demand would change to

x3 = 200

x3 ≥ 1.5 x1

Z = $760 a) Increase vending capacity by 100 sandwiches. There is already excess assembly time available (82 minutes) and the dual value is zero, whereas the dual value of vending machine capacity is $0.38. $38 is additional profit. b) x1 = 1,000

the resulting optimal solution is, x1 = 108, x2 = 54, x3 = 162, Z = $249.48 Since the profit declines the discount should not be implemented. 50.

x1 = road racing bikes

x2 = 1,000

x2 = cross country bikes

Z = $770

x3 = mountain bikes

The original profit is $760 and the new solution is $770. It would seem that a $10 difference would not be worth the possible loss of customer goodwill due to the loss of variety in the number of sandwiches available.

maximize Z = 600x1 + 400x2 + 300x3 subject to 1,200x1 + 1,700x2 + 900x3 ≤ $12,000 x1 + x2 + x3 ≤ 20

8x1 + 12x2 + 16x3 ≤ 120 x3 ≥ 2(x1 + x2) x1, x2, x3 ≥ 0 51.

x1 = 3 x3 = 6 Z = 3,600

3-18 .

Profit would increase to $810 but the solution values would not change. If profit is increased to $0.45 the solution values change to x1 = 1,600, x2 = 200, x3 = 200.

54. a)

Maximize Z = 7,500x1 + 8,200x2 + 10,500x3 subject to:

Mine 2 = 530 tons Mine 3 = 610 tons

.21x1 + .24x2 + .18x3 ≤ 17

Mine 4 = 240 tons Multiple optimal solutions exist

x1 + x2 + x3 ≤ 80 12x1 + 14.5x2 + 16x3 ≤ 2,500

a) Mine 4 has 240 tons of “slack” capacity.

x3 ≤ (x1 + x2)/2

x1 = 20

b) The dual values for the 4 constraints representing the capacity at the 4 mines show that Mine 1 has the highest dual value of $61, so its capacity is the best one to increase.

x2 = 33.3334

x1, x2, x3 ≥ 0 55.

x3 = 26.6667 Z = $703,333.40 a) The sensitivity range for x2 is 7,500 ≤ c2 ≤ 8,774.999. Since $7,600 is within this range the values for x1, x2, and x3 would not change, but the profit would decline to $683,333.30 [i.e., less the difference in profit, ($600)(x2 = 33.3334)].

d) The effect of simultaneous changes in objective function coefficients and constraint quality values cannot be analyzed using the sensitivity ranges provided by the computer output. It is necessary to make both changes in the model and solve it again. Doing so results in a new solution with Z = $73,080, which is $4,830 less than the original solution, so Exeter should make these changes.

b) One ton of grapes; the dual value is $23,333.35 c)

Grapes: (0.5)($23,333.35) = $11,666.68 Casks: (4)($3,833.329) = $15,333.32 Production: $0 Select the casks.

58.

d) $6,799; slightly less than the lower band of the sensitivity range for cj. 56.

Minimize Z = 8.2x1 + 7.0x2 + 6.5x3 + 9.0x4 + 0s1 + 0s2 + 0s3 + 0s4 subject to 6x1 + 2x2 + 5x3 + 7x4 − s1 = 820 .7x1 − .3x2 − .3x3 − .3x4 − s2 = 0

Minimize Z = $37x11 + 37x12 + 37x13 + 46x21 + 46x22 + 46x23 + 50x31 + 50x32 + 50x33 + 42x41 + 42x42 + 42x43 subject to:

−.2x1 + x2 + x3 − .2x4 + s3 = 0 x3 − x1 − x4 − s4 = 0 ***** Input Data *****

.7x11 + .6x21 + .5x31 + .3x41 = 400 tons

57.

The sensitivity range for Mine 1 is 242.8571 ≤ c1 ≤ 414.2857, thus capacity could be increased by 64.2857 tons before the optimal solution point would change.

.7x12 + .6x22 + .5x32 + .3x42 = 250 tons

Max. Z = 8.2x1 + 7.0x2 + 6.5x3 + 9.0x4 subject to

.7x13 + .6x23 + .5x33 + .3x43 = 290 tons

6x1 + 2x2 + 5x3 + 7x4 ≥ 820

x11 + x12 + x13 ≤ 350 tons

.7x1 − .3x2 − .3x3 − .3x4 ≥ 0

x21 + x22 + x23 ≤ 530 tons

−.2x1 + 1x2 + 1x3 − .2x4 ≤ 0

x31 + x32 + x33 ≤ 610 tons

−1x1 + 1x3 − 1x4 ≥ 0

x41 + x42 + x43 ≤ 490 tons x13 = 350 tons

***** Program Output ***** Infeasible Solution because Artificial variables remain in the final tableau.

x21 = 158.333 tons x22 = 296.667 tons

59.

x23 = 75 tons

x1 = cakes x2 = breads

x31 = 610 tons

x3 = pies

x42 = 240 tons

x4 = cookies

Z = $77,910

maximize Z = $12x1 + 8x2 + 10x3 + 6x4 subject to

Mine 1 = 350 tons

3-19 .

per minute. They would be willing to pay up to $0.22/min. c) The profit with 500 more lbs. of pecans is $20,243.50 and the profit with 30 more hours of oven time is $19,046.60. The family would choose to have the additional pecans. d) The profit with the bigger oven would be $19,046.60, which is only an increase of $396.83. It would take over 7 years to recoup the new oven cost. 63. a) x1 = no. of residential lawns

flour: sugar: eggs: oven:

2x1 + 9x2 + 1.3x3 + 2.5x4 ≤ 74 cups 2x1 + .25x2 + 1x3 + 1x4 ≤ 24 cups 2x1 + 0x2 + 5x3 + 2x4 ≤ 36 eggs 45x1 + 35x2 + 50x3 + 16x4 ≤ 480 mins x1, x2, x3, x4 ≥ 0 Cakes = 2.6

60.

Breads = 3.4 Pies = 0 Cookies = 15.4 Total sales = $150.3

x2 = no. of commercial properties

a) 2.6 cups of sugar left over

maximize Z = $23x1 + 61x2 subject to 1.2x1 + 5x2 ≤ 96

b) 6 eggs; each additional egg would increase sales by $1.24, or a total of (6)(1.24) = $7.44 c)

Eggs: (6)(1.24) = $7.44

1.2x1 + 20x2 ≤ 700

Flour: (20)(.08) = $1.60

x1 ≤ 50

Oven time: (30)(.21) = $6.30

x2 ≤ 15 x1, x2 ≥ 0

Six more eggs b)

d) She could round the values and then check to see if any of the constraints are violated: Cakes = 3, Breads = 3, Cookies = 15 Total sales = $150 The ability to generate integer solutions with Solver can also be shown using the “Add Constraints” integer feature. The Solver solution with integer restrictions is the same as the rounded solution above. 61.

x1 = pies x2 = batches of one dozen cookies x3 = 1 lb. bags of shelled pecans

64. a) Hours and budget

x4 = 5 lb. bags of unshelled pecans

b) Increasing the budget increases profit to $1,589.20; increasing the working hours increases profit to $1,628.78. Increase working day to 9 hours.

Maximize Z = $5x1 + 3x2 + 7x3 + 16x4 subject to: 8x1 + 12x2 + 32x3 + 80x4 ≤ 80,000 ounces 55 x1 15 x2 + ≤ 7,200 mins 4 2 6x1 + 4x2 + 10x3 + 1x4 ≤ 18,000 mins

x1, x2, x3, x4 ≥ 0 62.

x1 = 523.6 pies x2 = 0 cookies x3 = 1,449 lb. bags of shelled pecans x4 = 368 five-lb. bags of unshelled pecans Z = $18,649.77 a) No slack resources available. b) Oven time has the highest dual value of $0.22

3-20 .

The model solution can be rounded, but not up, which would violate the constraints. Excel and QM for Windows have integer capabilities.

CASE SOLUTION: MOSSAIC TILES LTD. a)

Maximize Z = $190x1 + 240x2 subject to

The optimal solution is at point B. For point C to become optimal the profit for a large tile, x1, would have to become steeper than the constraint line for glazing, .16x1 + .20x2 = 40: −c1/240 = −.16/.20

.30x1 + .25x2 ≤ 60 hr.—molding

c1 = 192

.27x1 + .58x2 ≤ 105 hr.—baking

This is the upper limit for c1. The lower limit is at point A which requires an objective function slope flatter than the constraint line for baking,

.16x1 + .20x2 ≤ 40 hr.—glazing 32.8x1 + 20x2 ≤ 6,000 lb.—clay b)

−c1/240 = −.27/.58

x1, x2 ≥ 0 Maximize Z = $190x1 + 240x2 + 0s1 + 0s2 + 0s3 + 0s4 subject to

c1 = 111.72 Thus, 111.72 ≤ c1 ≤ 192 The same logic is used to compute the sensitivity range for c2. The lower limit is computed as,

.30x1 + .25x2 + s1 = 60 .27x1 + .58x2 + s2 = 105

−190/c2 = −.16/.20

.16x1 + .20x2 + s3 = 40

c2 = 237.5

32.8x1 + 20x2 + s4 = 6,000

The upper limit is,

x1, x2, s1, s2, s3, s4 ≥ 0

−190/c2 = −.27/.58

c2 = 408.15 The sensitivity ranges for the constraint quantity values are determined by observing the graph and seeing where the new location of the constraint lines must be to change the solution point. For the molding constraint, the lower limit of the range for q1 is where the constraint line intersects with point B, .30(56.7) + .25(154.64) = q1 q1 = 55.67 The upper limit is ∞ since it can be seen that this constraint can increase indefinitely without changing the solution point. d)

Thus,

x1 = 56.70, x2 = 154.64

55.67 ≤ q1 ≤ ∞

.30(56.7) + .25(154.64) + s1 = 60 s1 = 4.33 hr. of molding time .27(56.7) + .58(154.64) + s2 = 105 s2 = 0 hr. of baking time .16(56.7) + .20(154.64) + s3 = 40 s3 = 0 hr. of glazing time 32.8(56.7) + 20(154.64) + s4 = 6,000 s4 = 1,047.42 lbs. of clay

For the baking constraint the lower limit of the range for q2 is where point C becomes optimal, and the upper limit is where the baking constraint intersects with the x2 axis (x2 = 200). At C:

.27(100) + .58(120) = q2 q2 = 96.6

At x2 axis: .27(0) + .58(200) = q2 q2 = 116

3-21 .

Thus,

f) 96.6 ≤ q2 ≤ 116

For the glazing constraint the lower limit of the range for q3 is at point A, and the upper limit is where the glazing constraint line, .16x1 + .20x2 = 40, intersects with the baking and molding constraints (i.e., x1 = 80.28 and x2 = 143.68). At A:

−190/c2 = −.27/.58 c2 = $408.14 g) Problem Title: Case Problem: Mossaic Tiles, Ltd.

.16(0) + .20(181.03) = q3 q3 = 36.21

***** Input Data *****

At intersection of constraints:

Max. Z = 190x1 + 240x2 subject to

.16(80.28) + .20(143.68) = q3 q3 = 41.58 Thus, 36.21 ≤ q3 ≤ 41.58 For the clay constraint the upper limit is ∞ since the constraint can increase indefinitely. The lower limit is at the point where the constraint line intersects with point B: At B:

The slope of the objective must be flatter than the slope of the constraint that intersects with the x2 axis at point A, which is the baking constraint,

.30x1 + .25x2 ≤ 60

.27x1 + .58x2 ≤ 105

.16x1 + .20x2 ≤ 40

c4 32.8x1 + 20x2 ≤ 6000

32.8(56.7) + 20(154.64) = q4 q4 = 4,952.56

Thus, 4,952.56 ≤ q4 ≤ ∞

3-22 .

***** Program Output ***** Final Optimal Solution At Simplex Tableau : 2 Z = 47,886.598 Variable x1 x2

Value 56.701 154.639 Slack/Surplus 4.330

Shadow Price

0.000

10.309

0.000

1,170.103

1,047.423

0.000

Constraint c1

0.000

Objective Coefficient Ranges Variables x1 x2

Lower Limit 111.724 237.500

Current Values 190.000 240.000

Upper Limit 192.000 408.148

Allowable Increase 2.000 168.148

Allowable Decrease 78.276 2.500

Right Hand Side Ranges Constraints c1 c2 c3 c4

Lower Limit 55.670 96.600 36.207 4,952.577

Current Values 60.000 105.000 40.000 6,000.000

Upper Limit No limit 116.000 41.577 No limit

Additional clay will have no effect on the solution since there is already slack clay left. Thus, Mossaic should not agree to the offer of extra clay.

Although an additional hour of glazing has the highest shadow price of $1,170.103, the upper limit of the sensitivity range for glazing hours is 41.577. Thus, with an increase of only 1.577 hours the solution will change and a new shadow price will exist. In order to assess the full impact of a 20-hour increase in glazing hours, the problem should be solved again using the computer with this change. This new solution results in a profit of $49,732.39, an increase in profit of only $1,845.79. The reason for this small increase

Allowable Decrease 4.330 8.400 3.793 1,047.423

can be observed in the graphical solution; as the glazing constraint increases it quickly becomes a “non-binding” constraint with a new solution point.

h) Since there are already slack molding hours left over, reducing the time required to mold a batch of tiles will only create more slack molding time. Thus, the solution will not change. i)

Allowable Increase No limit 11.000 1.577 No limit

k) A reduction of 3 hours is within the sensitivity range for kiln hours. However, the shadow price for kiln hours is $1,170.103 per hour. Thus, a loss of 3 kiln hours will reduce profit by (3)(1,170.103) = $3,510.31.

CASE SOLUTION: “THE POSSIBILITY” RESTAURANT—CONTINUED The solution is, x1 = 40 x2 = 20 Z = $800 a) The question regarding a possible advertising expenditure of $30 per day requires that the

3-23 .

sensitivity range for q1 be computed.

mix and the shadow price, so the impact could not be totally ascertained from the optimal simplex tableau. Solving the model again with q2 = 15 results in the following new solution.

q1: s3: 20 + 7∆ ≥ 0

s4: 14 − 1.1∆ ≥ 0

7 ∆ ≥ −20

1.1∆ ≥ −14

∆ ≥ −2.86 s1: 40 − 2∆ ≥ 0

s1 = 5.45

∆ ≤ 12.72

s3 = 81.82 x1 = 49.09

s2: 20 −∆ ≥ 0

2∆ ≥ −40

∆ ≥ −20

x2 = 5.45

∆ ≥ −20

∆ ≤ 20

Z = $676.36 Notice that simply using the shadow price of $16 for staff time (hr) would have indicated a loss in profit of only (5hr)(16) = $80, or Z = $720. The actual reduction in profit to $676.36 is greater. The final question concerns an increase in the coefficient for c1 from $12 to $14. This requires the computation of the range for c1.

Summarizing, −20 ≤ −2.86 ≤ ∆ ≤ 12.72 ≤ 20 and, −2.86 ≤ ∆ ≤ 12.72 Since q1 = 60 + ∆, ∆ = q1 − 60. Therefore, −26 ≤ q1 − 60 ≤ 12.72

57.14 ≤ q1 ≤ 72.72 Thus, an increase of 10 meals does not affect the shadow price for mean demand, which is $800. An increase of 10 meals will result in increased profit of ($8)($10) = $80, which exceeds the advertising expenditure of $30. The ad should be purchased.

c1, basic:

b) The reduction in kitchen staff from 20 to 15 hours requires the computation of the sensitivity range for q2. q2: s3: 20 − 10∆ ≥ 0

−8 −2Δ ≤ 0

−16 + 4Δ ≤ 0

−2Δ ≤ 8

4Δ ≤ 16

Δ ≤ −4

Δ≤ 4

–4≤ Δ≤ 4 Since c1 = 12 + Δ, Δ = c1 − 12. Therefore, −4 ≤ c1 − 12 ≤ 4

s4: 14 + 4∆ ≥ 0

− 20∆ ≥ −20

4∆ ≥ −14

∆≥1

∆ ≤ −3.5

s1: 40 − 40∆ ≥ 0

The final question concerns an increase in the coefficient for c1 from $12 to $14. This requires the computation of the range for c1.

8 ≤ c1 ≤ 16 Since c1 = $14 is within this range the price increase could be implemented without affecting Pierre’s meal plans.

s2: 20 + 4∆ ≥ 0

−40∆ ≥ −40

4∆ ≥ −20

∆≥1

∆ ≤ −5

CASE SOLUTION: JULIA’S FOOD BOOTH

Summarizing,

a) x1 = pizza slices, x2 = hot dogs,

−5 ≤ −3.5 ≤ ∆ ≤ 1 ≤ 10 and,

x3 = barbeque sandwiches

−3.5 ≤ Δ ≤ 1

The model is for the first home game,

Since q2 = 20 + Δ, Δ = q2 − 20.

maximize Z = $0.75x1 + 1.05x2 + 1.35x3 subject to:

Therefore,

$0.75x1 + 0.45x2 + 0.90x3 ≤ 1,500

−3.5 ≤ q2 − 20 ≤ 1

24x1 + 16x2 + 25x3 ≤ 55,296 in2 of oven space.

16/5 ≤ q2 ≤ 21 A reduction of 5 hours to 15 hours would exceed the lower limit of the sensitivity range. This would result in a change in the solution

x1 ≥ x2 + x3

3-24 .

b) Yes, she would increase her profit; the dual value is $1.50 for each additional dollar. The upper limit of the sensitivity range for budget is $1,658.88, so she should only borrow approximately $158. Her additional profit would be $238.32 or a total profit of $2,488.32.

x2 ≥ 2.0 x3

x1, x2, x3 ≥ 0 *Note that the oven space required for a pizza slice was determined by dividing the total space required by a pizza, 14 × 14 = 196 in2, by 8, or approximately 24 in2 per slice. The total space available is the dimension of a shelf, 36 in. × 48 in. = 1,728 in2, multiplied by 16 shelves, 27,648 in2, which is multiplied

by 2, the times before kickoff and halftime the oven will be filled = 55,296 in2.

Yes, she should hire her friend. It appears impossible for her to prepare all of the food items given in the solution in such a short period of time. The additional profit she would get if she borrowed more money as indicated in part B would offset this additional expenditure.

d) The biggest uncertainty is the weather. If the weather is very hot or cold, fans might eat less. Also, if it is rainy weather for a game or games, the crowd might not be as large, even though the games are all sellouts. The model results show that Julia will reach her goal of $1,000 per game—if everything goes right. She has little slack in her profit margin; thus it seems unlikely that she will achieve $1,000 for each game.

Solution: x1 = 1,250 pizza slices x2 = 1,250 hot dogs x3 = 0 barbecue sandwiches Z = $2,250 Julia should receive a profit of $2,250 for the first game. Her lease is $1,000 per game so that leaves her with $1,250. Her cost of leasing a warming oven is $100 per game, thus she will make a little more than what she needs to (i.e., $1,000), for it to be worth her while to lease the booth. A “tricky” aspect of the model formulation is the $1,500 used to purchase the ingredients. Since the objective function reflects net profit, the $1,500 is recouped and can be used for the next home game to purchase food ingredients; thus, it’s not necessary for Julia to use any of her $1,150 profit to buy ingredients for the next game.

3-25 .

Chapter Four: Linear Programming: Modeling Examples PROBLEM SUMMARY

35. Transportation (minimization) 36. Scheduling (minimization)

1. “Product mix” example

37. Production line scheduling (maximization)

2. “Diet” example

38. College admissions (maximization)

3. “Investment” example

39. Network flow (minimization)

4. “Marketing” example

40. Blend (maximization)

5. “Transportation” example

41. Personal scheduling (maximization)

6. “Blend” example

42. Employee allocation (minimization)

7. Product mix (maximization)

43. Trim loss (minimization)

8. Sensitivity analysis (4–7)

44. Multiperiod investment (maximization)

9. Diet (minimization)

45. Multiperiod sales and inventory (maximization)

10. Product mix (minimization) 11. Product mix (maximization) 12. Ingredients mix (minimization)

46. Multiperiod production and inventory (minimization)

13. Transportation (maximization)

47. Employee assignment (maximization)

14. Product mix (maximization)

48. Data envelopment analysis

15. Ingredients mix blend (minimization)

49. Data envelopment analysis

16. Crop distribution (maximization)

50. Network flow (maximization)

17. Monetary allocation (maximization)

51. Multiperiod workforce planning (minimization)

18. Diet (minimization), sensitivity analysis

52. Integer solution (4–51)

19. Transportation (maximization)

53. Machine scheduling (maximization), sensitivity analysis

20. Transportation (minimization) 21. Warehouse scheduling (minimization)

54. Cargo storage (maximization)

22. School busing (minimization)

55. Broadcast scheduling (maximization)

23. Sensitivity analysis (4–22)

56. Product mix (maximization)

24. Ingredients mixture (minimization)

57. Product mix/advertising (maximization)

25. Interview scheduling (maximization)

58. Scheduling (minimization)

26. Multiperiod investments mixture (maximization)

59. Consultant project assignment (minimization) 60. Multiperiod workforce planning (minimization)

27. Bake sale mix (maximization) 28. Vegetable garden mix (maximization)

61. Multiperiod workforce (4–60)

29. Advertising mix (minimization), sensitivity analysis

62. Coal transportation (minimization) 63. Soccer field assignment (minimization)

30. Blend (maximization)

64. Data envelopment analysis

31. Multiperiod borrowing (minimization)

65. Airline crew scheduling (maximization)

32. Multiperiod production scheduling (minimization)

66. Product flow/scheduling (minimization)

33. Blend (maximization), sensitivity analysis

68. Assignment (minimization)

67. Transshipment (minimization)

34. Assignment (minimization), sensitivity analysis

4-1 .

Many different combinations of maximum servings of each of the 10 food items could be used. As an example, limiting the four hot and cold cereals, x1, x2, x3, and x4, to four cups, eggs to three, bacon to three slices, oranges to two, milk to two cups, orange juice to four cups, and wheat toast to four slices results in the following solution:

PROBLEM SOLUTIONS 1.

Since the profit values would change, the shadow prices would no longer be effective. Also, the sensitivity analysis provided in the computer output does not provide ranges for constraint parameter changes. Thus, the model would have to be resolved. The reformulated model would have unit costs increased by 10 percent. This same amount would be subtracted from unit profits. The individual processing times would be reduced by 10 percent. This would result in a new, lower solution of $43,310. Thus, the suggested alternative should not be implemented.

x3 = 2 cups of oatmeal x4 = 1.464 cups of oat bran x5 = .065 eggs x8 = 1.033 cups of milk x10 = 4 slices of wheat toast Z = $0.828

Probably not. The t-shirts are a variable cost and any additional t-shirts purchased by Quick-Screen would likely reduce unit profit, which would change the current shadow price for blank t-shirts. The shadow price is effective only if the profit is based on costs that would be incurred without regard to the acquisition of additional resources.

Further limiting the servings of the four hot and cold cereals to two cups, x1 + x2 + x3 + x4 ≤ 2, results in the following solution: x3 = 2 cups of oatmeal x6 = .750 slices of bacon x8 = 2 cups of milk x9 = .115 cups of orange juice

The new requirement is that

x10 = 4 slices of wheat toast

x1 = x2 = x3 = x4

Z = $0.925

This can be achieved within the model by creating three additional constraints,

x1 = x2 x1 = x3 x1 = x4 If x1 equals x2, x3 and x4 then x2, x3 and x4 must also equal each other. These constraints are changed to, x1 − x2 = 0 x1 − x3 = 0 x1 − x4 = 0 4.

The new solution is x1 = x2 = x3 = x4 = 112.5. 2.

With a minimum of 500 calories, the three food items remain the same; however, the amount of each and the total cost increases: x3 = 2.995 cups of oatmeal, x8 = 1.352 cups of milk, x10 = 1.005 slices of toast, and Z = $0.586. With a minimum of 600 calories, the food items change to x3 = 4.622 cups of oatmeal and 1.378 cups of milk and Z = $0.683. A change of variables would be expected given that 600 calories is greater than the upper limit of the sensitivity range for calories.

It would have no effect; the entire $70,000 would be invested anyway. Since the upper limit of the sensitivity range for the investment amount is “unlimited,” an increase of $10,000 will not affect the shadow price, which is $0.074. Thus, the total increase in return will be $740 (i.e., $10,000 × .074 = $740). No, the entire amount will not be invested in one alternative. The new solution is x1 = $22,363.636, x3 = $43,636.364, and x4 = $14,000. The shadow price is 1.00; thus for every $1 increase in budget, up to the sensitivity range upper limit of $14,000, audience exposure will increase by 1.00. The total audience increase for a $20,000 budget increase is 20,000. This new requirement results in two new model constraints, 20,000x1 = 12,000x2 20,000x1 = 9,000x3 or, 20,000x1 − 12,000x2 = 0 20,000x1 − 9,000x3 = 0

4-2 .

4,500 to 4,501 results in an increase in total cost to $76,820. 7. a) Maximize Z = $190x1 + 170x2 + 155x3 subject to

The new solution is x1 = 3.068, x2 = 5.114, x3 = 6.818 and Z = 184,090. This results in approximately 61,362 exposures per type of advertising (with some slight differences due to computer rounding). 5.

3.5x1 + 5.2x2 + 2.8x3 ≤ 500

The slack variables for the three ≤ warehouse constraints would be added to the constraints as follows:

1.2x1 + 0.8x2 + 1.5x3 ≤ 240 40x1 + 55x2 + 20x3 ≤ 6,500 x1,x2,x3 ≥ 0

x1A + x1B + x1C + s1 = 300

b) x1 = 41.27, x2 = 0, x3 = 126.98, Z = $27,523.81

x2A + x2B + x2C + s2 = 200 x3A + x3B + x3C + s3 = 200 These three slacks would then be added to the objective function with the storage cost coefficients of $9 for s1, $6 for s2, and $7 for s3. This change would not result in a new solution. The model must be reformulated with three new variables reflecting the shipments from the new warehouse at Memphis (4) to the three stores, x4A, x4B, and x4C. These variables must be included in the objective function with the cost coefficients of $18, $9, and $12 respectively. A new supply constraint must be added, x4A + x4B + x4C ≤ 200

s1 = s2 = 0, s3 = 2,309.52 8. a) It would not affect the model. The slack apples are multiplied by the revenue per apple of $.08 to determine the extra total revenue, i.e., (2,309.52)($.08) = $184.76. b) This change requires a new variable, x4, and that the constraint for apples be changed from ≤ to =. No, the Friendlys should not produce cider. The new solution would be x1 = 135, x2 = 0, x3 = 0, x4 = 18.33 and Z = $26,475. This reduction in profit occurs because the requirement that all 6,500 apples be used forces resources to be used for cider that would be more profitable to be used to produce the other products. If the final model constraint for apples is ≤ rather than =, the previous solution in 7(b) results.

The solution to this reformulated model is x1C = 200 x2B = 50

9. a)

x3A = 150

x2 = no. of bacon strips

x4B = 200

x3 = no. of cups of cereal

Z = 6,550

minimize Z = 4x1 + 3x2 + 2x3 subject to

Yes, the warehouse should be leased. The shadow price for the Atlanta warehouse shows the greatest decrease in cost, $6 for every additional set supplied from this source. However, the upper limit of the sensitivity range is 200, the same as the current supply value. Thus, if the supply is increased at Atlanta by even one television set the shadow price will change. 6.

x1 = no. of eggs

2x1 + 4x2 + x3 ≥ 16 3x1 + 2x2 + x3 ≥ 12 x1, x2, x3 ≥ 0 b)

x1 = 2 x2 = 3 Z = $0.17

This change would not affect the solution at all since there is no surplus with any of the three constraints.

10. a)

xi = number of boats of type i, i = 1 (bass boat), 2 (ski boat), 3 (speed boat) In order to break even total revenue must equal total cost:

Component 1 has the greatest dual price of $20. For each barrel of component 1 the company can acquire, profit will increase by $20, up to the limit of the sensitivity range which is an increase of 1,700 bbls. or 6,200 total bbls. of component 1. For example an increase of 1 bbl. of component 1 from

23,000x1 + 18,000x2 + 26,000x3 = 12,500x1 + 8,500x2 + 13,700x3 + 2,800,000 or, 10,500x1 + 9,500x2 + 12,300x3 = 2,800,000 minimize Z = 12,500x1 + 8,500x2 + 13,700x3

4-3 .

subject to

b) x2 = .1153 ton – ore 2

10,500x1 + 9,500x2 + 12,300x3 = 2,800,000

x3 = .8487 ton – ore 3

x1 ≥ 70

x4 = .0806 ton – ore 4

x2 ≥ 50

x5 = .4116 ton – ore 5

x3 ≥ 50

Z = $40.05 13. a)

x1 ≤ 120 x2 ≤ 120 x3 ≤ 120 x1,x2,x3 ≥ 0

x3 = 75.203

maximize Z = 1,800x1a + 2,100x1b + 1,600x1c+ 1,000x2a + 700x2b + 900x2c + 1,400x3a + 800x3b + 2,200x3c

Z = $2,925,284.553

subject to

b) x1 = 70.00 x2 = 120.00

11. a)

x1 = no. of gallons of Yodel

x1a + x1b + x1c = 30

x2 = no. of gallons of Shotz

x2a + x2b + x2c = 30

x3 = no. of gallons of Rainwater

x3a + x3b + x3c = 30

maximize Z = 1.50x1 + 1.60x2 + 1.25x3 subject to

x1a + x2a + x3a ≤ 40

x1 + x2 + x3 = 1,000

x1c + x2c + x3c ≤ 50

x1b + x2b + x3b ≤ 60

1.50x1 + .90x2 + .50x3 ≤ 2,000

xij ≥ 0

x1 ≤ 400

b) x1b = 30

x2 ≤ 500

x2a = 30

x3 ≤ 300

x3c = 30

x1, x2, x3 ≥ 0

Z = $159,000

b) x1 = 400

14. a)

x2 = 500

x1 = no. of sofas x2 = no. of tables

x3 = 100

x3 = no. of chairs

Z = $1,525.00 12. a)

xij = number of trucks assigned to route from warehouse i to terminal j, where i = 1 (Charlotte), 2 (Memphis), 3 (Louisville) and j = a (St. Louis), b (Atlanta), c (New York)

maximize Z = 400x1 + 275x2 + 190x3 subject to

xi = ore i (i = 1,2,3,4,5,6) minimize Z = 27x1 + 25x2 + 32x3 + 22x4 + 20x5 + 24x6 subject to

7x1 + 5x2 + 4x3 ≤ 2,250 12x1 + 7x3 ≤ 1,000 6x1 + 9x2 + 5x3 ≤ 240

.19x1 + .43x2 + .17x3 + .20x4 + .12x6 ≥ .21

x1 + x2 + x3 ≤ 650

.15x1 + .10x2 + .12x4 + .24x5 + .18x6 ≤ .12

x1, x2, x3 ≥ 0

.12x1 + .25x2 + .10x5 + .16x6 ≤ .07

b) x1 = 40

.14x1 + .07x2 + .53x3 + .18x4 + .31x5 + .25x6 ≥ .30

Z = $16,000

.14x1 + .07x2 + .53x3 + .18x4 + .31x5 + .25x6 ≤ .65

15. a) xij = lbs. of seed i used in mix j, where i = t (tall fescue), m (mustang fescue), b (bluegrass) and j = 1,2,3. minimize Z = 1.70 (xt1 + xt2 + xt3) + 2.80 (xm1 + xm2 + xm3) + 3.25 (xb1 + xb2 + xb3)

.60x1 + .85x2 + .70x3 + .50x4 + .65x5 + .71x6 = 1.00

4-4 .

subject to .50xt1 − .50xm1 − .50xb1 ≤ 0

700(xc1 + xp1 + xs1) − 500(xc3 + xp3 + xs3) = 0

−.20xt1 + .80xm1 − .20xb1 ≥ 0 −.30xt2 − .30xm2 + .70xb2 ≥ 0

xc2 = 100

−.30xt2 + .70xm2 − .30xb2 ≥ 0

xc3 = 300

.80xt2 − .20xm2 − .20xb2 ≤ 0

xp2 = 700

.50xt3 − .50xm3 − .50xb3 ≥ 0

xs3 = 400

.30xt3 − .70xm3 − .70xb3 ≤ 0

Z = $975,000

−.10xt3 − .10xm3 + .90xb3 ≥ 0

17. a)

xt1 + xm1 + xb1 ≥ 1,200

x1 = $ allocated to job training x2 = $ allocated to parks

xt2 + xm2 + xb2 ≥ 900

x3 = $ allocated to sanitation

xt3 + xm3 + xb3 ≥ 2,400

x4 = $ allocated to library

xij ≥ 0

xt3 = 1,680

maximize Z = .02x1 + .09x2 + .06x3 + .04x4 subject to x1 + x2 + x3 + x4 = 4,000,000 x1 ≤ 1,600,000

xm1 = 600

x2 ≤ 1,600,000

xm2 = 450

x3 ≤ 1,600,000

xm3 = 480

x4 ≤ 1,600,000

b) xt1 = 600 xt2 = 180

xb1 = 0

x2 − x3 − x4 ≤ 0

xb2 = 270

x1 − x3 ≥ 0

xb3 = 240

x1, x2, x3, x4 ≥ 0

Z = $10,123.50 16. a)

xij ≥ 0

b) xc1 = 500

b) x1 = 800,000

xij = acres of crop i planted on plot j, where i = c (corn), p (peas), s (soybeans) and j = 1,2,3

x2 = 1,600,000 x3 = 800,000

maximize Z = 600(xc1 + xc2 + xc3)

subject to

x4 = 800,000 Z = 240,000

+ 450(xp1 + xp2 + xp3) + 300(xs1 + xs2 + xs3)

18. a)

xc1 + xp1 + xs1 ≥ 300 xc1 + xp1 + xs1 ≤ 500 xc2 + xp2 + xs2 ≥ 480 xc2 + xp2 + xs2 ≤ 800 xc3 + xp3 + xs3 ≥ 420 xc3 + xp3 + xs3 ≤ 700 xc1 + xc2 + xc3 ≤ 900

Minimize Z = .80x1 + 3.70x2 + 2.30x3 +.90x4 + .75x5 + .40x6 + .83x7 subject to 520x1 + 500x2 + 860x3 + 600x4 + 50x5 + 460x6 + 240x7 ≥ 1,500 520x1 + 500x2 + 860x3 + 600x4 + 50x5 + 460x6 + 240x7 ≤ 2,000 4.4x1 + 3.3x2 + .3x3 + 3.4x4 +.5x5 + 2.2x6 + .2x7 ≥ 5 30x1 + 5x2 + 75x3 + 3x4 + 10x7 ≥ 20 30x1 + 5x2 + 75x3 + 3x4 +10x7 ≤ 60 17x1 + 85x2 + 82x3 + 10x4 + 6x5 + 10x6 + 16x7 ≥ 30

xp1 + xp2 + xp3 ≤ 700 xs1 + xs2 + xs3 ≤ 1,000 800(xc1 + xp1 + xs1) − 500(xc2 + xp2 + xs2) = 0

30x4 + 70x6 + 22x7 ≥ 40 180x1 + 90x2 + 350x3 + 20x7 ≤ 30

700(xc2 + xp2 + xs2) − 800(xc3 + xp3 + xs3) = 0

xi ≥ 0

4-5 .

b) x4 = 1.667

Ash: .03x11 − .01x21 − .02x31 ≤ 0

x6 = 0.304

.04x12 − .0x22 − .01x32 ≤ 0

x7 = 1.500

.04x13 − .0x23 − .01x33 ≤ 0

Z = $2.867

.03x14 − .01x24 − .02x34 ≤ 0

19. a)

The model becomes infeasible and cannot be solved. Limiting each food item to onehalf pound is too restrictive. In fact, experimentation with the model will show that one food item in particular, dried beans, is restrictive. All other food items can be limited except dried beans. xij = number of units of products i (i = 1,2,3) produced on machine j (j = 1,2,3,4) maximize Z = $7.8x11 + 7.8x12 + 8.2x13 + 7.9x14 + 6.7x21 + 8.9x22 + 9.2x23 + 6.3x24 + 8.4x31 + 8.1x32 + 9.0x33 + 5.8x34 subject to 35x11 + 40x21 + 38x31 ≤ 9,000

Sulfur: .01x11 −.01x21− .02x31 ≤ 0 .01x12 −.01x22−.02x32 ≤ 0 .01x13 −.03x23−.04x33 ≤ 0 .0x14 − .02x24−.03x34 ≤ 0 xij ≥ 0 b) x11 = 42 x13 = 18 x14 = 72 x21 = 10 x22 = 160 x31 = 58 x33 = 72 x34 = 108

41x12 + 36x22 + 37x32 ≤ 14,400

Z = $41,726

34x13 + 32x23 + 33x33 ≤ 12,000

21. a)

39x14 + 43x24 + 40x34 ≤ 15,000 x11 + x12 + x13 + x14 = 400

x23 = 170.00

minimize Z = 1.7x11 + 2(1.4)x12 + 3(1.2)x13 + 4(1.1)x14 + 5(1.05)x15 + 6(1.00)x16 +1.7x21 + 2(1.4)x22 + 3(1.2)x23 + 4(1.1)x24 + 5(1.05)x25 + 1.7x31 + 2(1.4)x32 + 3(1.2)x33 + 4(1.1)x34 + 1.7x41 + 2(1.4)x42 + 3(1.2)x43 + 1.7x51 + 2(1.4)x52 + 1.7x61

x31 = 121.212

subject to

x21 + x22 + x23 + x24 = 570 x31 + x32 + x33 + x34 = 320 xij ≥ 0 b) x11 = 15.385 x14 = 384.615 x22 = 400.00

x33 = 198.788

x11 + x12 + x13 + x14 + x15 + x16 = 47,000

Z = $11,089.73 20. a)

xij = space (ft2) rented in month i for j months, where i = 1,2,...,6, and j = 1,2,...,6

x12 + x13 + x14 + x15 + x16 + x21 + x22

Minimize Z = 69x11 + 71x12 + 72x13 +74x14 + 76x21 + 74x22 + 75x23 + 79x24 + 86x31 + 89x32 + 80x33 + 82x34

x13 + x14 + x15 + x16 + x22 + x23 + x24

subject to

x14 + x15 + x16 + x23 + x24 + x25 + x32

+ x23 + x24 + x25 = 35,000 + x25 + x31 + x32 + x33 + x34 = 52,000 + x33 + x34 + x41 + x42 + x43 = 27,000

x11 + x12 + x13 + x14 ≤ 220

x15 + x16 + x24 + x25 + x33 + x34 + x42

x21 + x22 + x23 + x24 ≤ 170 x31 + x32 + x33 + x34 ≤ 280

+ x43 + x51 + x52 = 19,000

x11 + x21 + x31 = 110

x16 + x25 + x34 + x43 + x52 + x61 = 15,000

x12 + x22 + x32 = 160 x13 + x23 + x33 = 90 x14 + x24 + x34 = 180

4-6 .

b) x11 = 12,000

xww = 300

x13 = 8,000

xws = 50

x14 = 8,000

xcc = 250

x15 = 4,000

xcw = 250

x16 = 15,000

Z = 20,400 b) Change the 3 demand constraints in the (a) formulation from ≤ 1,200 to = 1,000.

x31 = 17,000 Z = $224,300 c) 22. a)

xnc = 400

x16 = 52,000 Z = $312,000

xnw = 300

xij = no. of students bused from district i to school j, where i = n, s, e, w, c and j = c,w,s

xsw = 150

minimize Z = 8xnc + 11xnw + 14xns + 12xsc + 9xsw + 0xss + 9xec + 16xew + 10xes + 8xwc + 0xww + 9xws + 0xcc + 8xcw + 12xcs subject to

xes = 850

xss = 150 xec = 50 xwc = 300 xww = 300 xcc = 250

xnc + xnw + xns = 700

xcw = 250

xsc + xsw + xss = 300

Z = 21,200

xec + xew + xes = 900

24. a)

x1 = no. of lb of oats x2 = no. of lb of corn x3 = no. of lb of soybeans x4 = no. of lb of vitamin supplement minimize Z = .50x1 + 1.20x2 + .60x3 +2.00x4 subject to x1 ≤ 300 x2 ≤ 400 x3 ≤ 200 x4 ≤ 100 x3/(x1 + x2 + x3 + x4) ≥ .30 x4/(x1 + x2 + x3 + x4) ≥ .20 x2/x1 ≤ 2/1 x1 ≤ x3 x1 + x2 + x3 + x4 ≥ 500 x1, x2, x3, x4 ≥ 0 b) x1 = 200 x2 = 0 x3 = 200 x4 = 100.00 Z = $420 25. a) x1 = no. of day contacts by phone x2 = no. of day contacts in person x3 = no. of night contacts by phone x4 = no. of night contacts in person

xwc + xww + xws = 600 xcc + xcw + xsc = 500 xnc + xsc + xec + xwc + xcc ≤ 1,200 xnw + xsw + xew + xww + xcw ≤ 1,200 xns + xss + xes + xws + xcs ≤ 1,200 xij ≥ 0 b) xnc = 700 xss = 300 xes = 900 xww = 600 xcc = 500 Z = 14,600 23. a)

Add the following 3 constraints to the original formulation: xss ≤ 150 xww ≤ 300 xcc ≤ 250 xnc = 700 xnw = 0 xsw = 150 xss = 150 xes = 900 xwc = 250

4-7 .

maximize Z = 2x1 + 4x2 + 3x3 + 7x4 subject to x2 + x4 ≤ 300 6x1 + 15x2 ≤ 1,200 5x3 + 12x4 ≤ 2,400 x1, x2, x3, x4 ≥ 0 b) x1 = 200 x3 = 480 Z = 1,840 26. a) xij = dollar amount invested in alternative i in year j, where i = p (product research and development), m (manufacturing operations improvements), a (advertising and sales promotion) and j = 1,2,3,4 (denoting year):

6x1 + 2x2 +1x3 ≤ 60 cups butter 12x1 + 4x2 + 1x3 ≤ 72 cups sugar 18x1 + 6x2 + 2x3 ≤ 120 cups flour x1, x2, x3 ≥ 0 b) x1 = 0, x2 = 0 x3 = 52.5 batches of cookies = 3,150 Z = $6,300 c)

Marginal value of oven time = $15; up to one additional hour. Increasing oven time by one hour will increase sales by $900.

28. a) x1 = asparagus (acres) x2 = corn (acres) x3 = tomatoes (acres) x4 = green beans (acres) x5 = red peppers (acres)

sj = slack, or uninvested funds in year j maximize Z = s4 + 1.2xa4 + 1.3xm3 + 1.5xp3 subject to

Maximize Z = (1.90)(2,000)x1 + (0.10)(7,200)x2 +(3.25)(25,000)x3 + (3.40)(3,900)x4 + (3.45)(12,500)x5

xa1 ≥ 30,000

Subject to

xm1 ≥ 40,000

x1 + x2 + x3 + x4 + x5 = 1

xp1 ≥ 50,000 xa1 + xm1 + xp1 + s1 = 500,000

1,800x1 + 1,740x2 + 6,000x3 + 3,000x4 + 2,700x5 ≤ 5,000

xa2 + xm2 + xp2 + s2 = s1 + 1.2xa1

(2,000)x1 ≤ 1,200

xa3 + xm3 + xs3 = s2 + 1.2xa2 + 1.3xm1

(25,000)x3 ≤ 10,000

xa4 + xs4 = s3 + 1.2xa3 + 1.3xm2 + 1.5xp1 xij, sj ≥ 0 Note: Since it is assumed that any amount of funds can be invested in each alternative—i.e., there is no minimum investment required—and funds can always be invested in as short a period as one year yielding a positive return, it is apparent that the sj variables for uninvested funds will be driven to zero in every period. Thus, these variables could be omitted from the model formulation for this problem. b) xa1 = 410,000 xm1 = 40,000 xa2 = 492,000 xp1 = 50,000 xa3 = 642,400 Z = $1,015,056 xa4 = 845,880 27. a) x1 = oven batches of (12) cakes x2 = oven batches of (48) cupcakes x3 = oven batches of (60) cookies Maximize Z = $180x1 + 144x2 + 120x3 Subject to 40x1 + 25x2 + 8x3 ≤ 420 minutes 24x1 + 8x2 + 2x3 ≤ 120 eggs

(3,900)x4 ≤ 2,000 (12,500)x5 ≤ 5,000 x1, x2, x3, x4, x5 ≥ 0 b) x1, x2 = 0, x3 = 0.4, x4 = 0.2, x5 = 0.4; Z = $52,402 29. a) x1 = no. of television commercials x2 = no. of newspaper ads x3 = no. of radio commercials minimize Z = 15,000x1 + 4,000x2 + 6,000x3 subject to x3/x2 ≥ 2/1 25,000x1 + 10,000x2 + 15,000x3 ≥ 100,000 (15,000x1 + 3,000x2 + 12,000x3)/ (10,000x1 + 7,000x2 + 3,000x3) ≥ 2/1 (15,000x1 + 4,000x2 + 9,000x3)/ (25,000x1 + 10,000x2 + 15,000x3) ≥ .30 x2 ≤ 7 x1, x2, x3 ≥ 0

4-8 .

b) x2 = 2.5 x3 = 5.0 Z = 40,000 c) This reformulation of the model would result in a fourth variable x4, with model parameters inserted accordingly. However, it would have no effect on the solution. 30. a)

31. a)

x1 = $ amount borrowed for six months in July yi = $ amount borrowed in month i (i = 1, 2, ..., 6) for one month ci = $ amount carried over from month i to i+1 6

minimize Z = .11x1 + .05∑ yi

xij = lbs. of coffee i used in blend j per week, where i = b (Brazilian), o (Mocha), c (Colombian), m (mild) and j = s (special), d (dark), r (regular)

i =1

subject to July: x1 + y1 + 20,000 − c1 = 60,000 August:

maximize Z = 4.5xbs + 3.75xos + 3.60xcs + 4.8xms + 3.25xbd + 2.5xod + 2.35xcd + 3.55xmd + 1.75xbr + 1.00xor + 0.85xcr + 2.05xmr

c1 + y2 +30,000 − c2 = 60,000 + y1

September: c2 +y3+ 40,000 − c3 = 80,000 + y2 October:

c3+y4 + 50,000 − c4 = 30,000 + y3

November: c4 + y5 + 80,000 − c5 = 30,000 + y4

subject to

December: c5 + y6 + 100,000 − c6= 20,000 + y5 End: x1 + y6 ≤ c6

.6xcs − .4xbs − .4xos − .4xms ≥ 0 −.3xbs + .7xos − .3xcs − .3xms ≥ 0 .4xbd − .6xod − .6xcd − .6xmd ≥ 0

−.1 xbd − .1xod − .1xcd + .9xmd ≤ 0 −.6xbr − .6xor − .6xcr + .4xmr ≤ 0 .7xbr − .3xor − .3xcr − .3xmr ≥ 0 xbs + xbd + xbr ≤ 110 xos + xod + xor ≤ 70 xcs + xcd + xcr ≤ 80 xms + xmd + xmr ≤ 150 xij ≥ 0 b) xos = 60 Special: xos + xcs + xms = 200 lbs. xcs = 80 Dark: xbd + xmd = 72 lbs. xms = 60 Regular: xbr + xor + xmr = 138 lbs. xbd = 64.8 xmd = 7.2 xbr = 45.2 xor = 10 xmr = 82.8 Z = $1,296

x1, yi, ci ≥ 0 Solution x1 = 70,000 y3 = 40,000 y4 = 20,000 y1 = y2 = y5 = y6 = 0 c1 = 30,000 c5 = 30,000 c6 = 110,000 Z = $10,700 Changing the six-month interest rate to 9% results in the following new solution: x1 = 90,000 y3 = 20,000 c1 = 50,000 c2 = 20,000 c5 = 50,000 c6 = 130,000 Z = $9,100

32. a)

xij = production in month i to meet demand in month j, where i = 1,2,...7 and j = 4,5,6 and 7 yj = overtime production in month j where j = 4,5,6,7. minimize Z = 150x14 + 100x24 + 50x34 + 200x15 + 150x25 + 100x35 + 250x16 + 200x26 +150x36 + 300x17 + 250x27 + 200x37 + 400y4 + 400y5 + 400y6 + 400y7

4-9 .

subject to

Production for Month j

x14 + x24 + x34 + x44 + y4 = 60

Month i Capacity 4

x15 + x25 + x35 + x55 + y5 = 85 x16 + x26 + x36 + x66 + y6 = 100 x17 + x27 + x37 + x77 + y7 = 120

x14 + x15 + x16 + x17 ≤ 30

x24 + x25 + x26 + x27 ≤ 30

x34 + x35 + x36 + x37 ≤ 30

x44 ≤ 40

20 20

Capacity

40 60

60 90

90 50

x55 ≤ 60

Overtime —

x66 ≤ 90

Demand

85 100

120

x77 ≤ 50

y4 ≤ 20

x34 = 20 x44 = 40

y5 ≤ 20

x25 = 5

y6 ≤ 20

x35 = 20

y7 ≤ 20

x55 = 60

b) x14 = 20

y5 = 5

x44 = 40

y6 = 10

x35 = 20

y7 = 20

x55 = 60

Z = $31,500

x26 = 10 x66 = 90 x17 = 40 x27 = 25 x77 = 50

x66 = 90

y7 = 5

x17 = 10

33. a)

x27 = 30 x37 = 10 x77 = 50

4-10 .

Z = $26,000 xij = amount of ingredient i in wiener type j, where i = c, b, p, a represent chicken, beef, pork, and additives, and j = r,b,m represent regular, beef, and all-meat, respectively maximize Z = .7xcr + .6xbr + .4xpr + .85xar + 1.05 xcb + .95xbb + .75xpb + 1.20xab + 1.55xcm + 1.45xbm + 1.25xpm + 1.70xam subject to xcr + xcb + xcm ≤ 200

xbr + xbb + xbm ≤ 300

the press, lathe, and grinder. The model would be reformulated as, minimize Z = 22x1 + 18x2 + 35x3 + 41x4 + 30x5 + 28x6 + 25x7 + 36x8 + 18x9 + 20x10 + 20x11 + 20x12 subject to x1 + x2 + x3 ≤ 1 x4 + x5 + x6 ≤ 1 x7 + x8 + x9 ≤ 1 x10 + x11 + x12 ≤ 1 x1 + x4 + x7 + x10 = 1 x2 + x5 + x8 + x11 = 1 x3 + x6 + x9 + x12 = 1 xi ≥ 0 The new solution is x2 = 1, x9 = 1, x10 = 1, Z = 56. Joe should hire Kelly. 35. a) This is a transportation problem. x1 = no. of tons of carpet shipped from St. Louis to Chicago x2 = no. of tons of carpet shipped from St. Louis to Atlanta x3 = no. of tons of carpet shipped from Richmond to Chicago x4 = no. of tons of carpet shipped from Richmond to Atlanta minimize Z = 40x1 + 65x2 + 70x3 + 30x4 subject to x1 + x2 = 250 x3 + x4 = 400 x1 + x3 = 300 x2 + x4 = 350 x1, x2, x3, x4 ≥ 0 b) x1 = 250 x3 = 50 x4 = 350 Z = 24,000 36. a) xi = number of nurses that begin the 8-hour shift in period i, where i = 1,2,...12 and period 1 = 12 AM–2 AM, period 2 = 2AM–4AM, etc.

xpr + xpb + xpm ≤ 150 xar + xab + xam ≤ 400 .90xbr + .90xpr − .l0xcr − .l0xar ≤ 0 .80xcr −.20xbr − .20xpr − .20xar ≥ 0 .25xbb − .75xcb − .75xpb − .75xab ≥ 0 *xam = 0 .5xbm + .5xpm − .5xcm − .5xam ≤ 0 xij ≥ 0 *Also feasible to delete xam from the problem. b) xcr = 75 xar = 300 xcm = 125

xab = 100

xbb = 300 xpm = 125 34. a)

Z = $1,062.50 This is an assignment problem. x1 = operator 1 to drill press x2 = operator 1 to lathe x3 = operator 1 to grinder x4 = operator 2 to drill press x5 = operator 2 to lathe x6 = operator 2 to grinder x7 = operator 3 to drill press x8 = operator 3 to lathe x9 = operator 3 to grinder minimize Z = 22x1 + 18x2 + 35x3 + 41x4 + 30x5 + 28x6 + 25x7 + 36x8 + 18x9 subject to x1 + x2 + x3 = 1 x4 + x5 + x6 = 1 x7 + x8 + x9 = 1 x1 + x4 + x7 = 1 x2 + x5 + x8 = 1 x3 + x6 + x9 = 1 x1, x2, x3, x4, x5, x6, x7, x8, x9 ≥ 0

minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 subject to x10 + x11 + x12 + x1 ≥ 30 x11 + x12 + x1 + x2 ≥ 20 x12 + x1 + x2 +x3 ≥ 40

b) x1 = 1

x5 = 1 x9 = 1 Z = 70 c)

This would require the model to be reformulated with three new variables, x10, x11, x12, representing Kelly’s assignment to

x1 + x2 + x3 + x4 ≥ 50

4-11 .

( )

x2 + x3 + x4 + x5 ≥ 60 x4 + x5 + x6 + x7 ≥ 80

x1 + y1 ≤ 470

x5 + x6 + x7 + x8 ≥ 70

x2 + y2 ≤ 1,300

x6 + x7 + x8 + x9 ≥ 70

x3 + y3 ≤ 240

x7 + x8 + x9 + x10 ≥ 60

x4 + y4 ≤ 820

x8 + x9 + x10 + x11 ≥ 50

x5 + y5 ≤ 1,060

x9 + x10 + x11 + x12 ≥ 50

x6 + y6 ≤ 610

xij ≥ 0

∑ SATIi ( xi ) + ∑ SATOi ( yi ) ≥ 1,150 ∑ xi + ∑ yi

b) x1 = 30

x3 = 10

∑ yi ≤ .47 ∑ xi + ∑ yi

x4 = 10 x5 = 40

xi ≥ .30, i = 1, 2, …6 xi + yi

x6 = 20 x7 = 10

37. a)

x9 = 40

∑ xi + ∑ yi ≤ 4,500

x10 = 10

xi , yi ≥ 0

Z = 170

Solution:

x1 = no. of hours molding

x1 (in-state Architecture) = 470

x2 = no. of hours smoothing

x2 (in-state A&S) = 557

x3 = no. of hours painting

x3 (in-state Agriculture) = 103

maximize Z = 175(7x1) = 1,225x1 subject to 8 x1 + 5 x2 + 6.5 x3 ≤ 3,000

x4 (in-state Business) = 539 x5 (in-state Engineering) = 454 x6 (in-state Human Resource) = 261

x1 + x2 + x3 ≤ 120

y1 (out-of-state Architecture) = 0

90(7 x1 ) ≤ 10,000 7 x1 = 12 x 2 = 10 x 3

(The no. of units going through each process must be equal. More bathtubs cannot be smoothed than are first molded.)

y2 (out-of-state A&S) = 743 y3 (out-of-state Agriculture) = 137

⎧ 7 x1 – 12 x 2 = 0 ⎪ ⎪12 x 2 – 10 x 3 = 0 ⎨ ⎪ 7 x1 – 10 x 3 = 0 ⎪ x ,x ,x ≥0 ⎩ 1 2 3

y4 (out-of-state Business) = 281 y5 (out-of-state Engineering) = 606 y6 (out-of-state Human Resources) = 349 Z = $61,119,000 39. a)

b) x1 = 15.87 hours

x13 = tons shipped from 1 to 3 x14 = tons shipped from 1 to 4

x2 = 9.26 hours

x12 = tons shipped from 1 to 2

x3 = 11.11 hours 38.

( )

maximize Z = 8,600 ∑ x + 19,200 ∑ y i i subject to

x3 + x4 + x5 + x6 ≥ 80

x34 = tons shipped from 3 to 4

Z = 19,444.44 xi = In-state freshman in college i; i = 1,2,...,6 yi = Out-of-state freshman in college i; i = 1,2,...,6

x45 = tons shipped from 4 to 5 x25 = tons shipped from 2 to 5 minimize Z = 3x13 + 4x14 + 5x12 + 2x34 + 7x45 + 8x25

4-12 .

subject to

41. a)

x13 + x14 + x12 = 5 x45 + x25 = 5 x13 = x34

Maximize Z = 0.30x1 + 0.20x2 + 0.05x3 + 0.10x4 + 0.15x5 subject to 0.3x1 + 0.2x2 + 0.05x3 + 0.1x4 + 0.15x5 ≤ 4.0 x1 + x2 + x3 + x4 + x5 ≤ 24

x14 + x34 = x45

x1 ≤ 4

x12 = x25

x2 ≤ 8

x13, x14, x12, x34, x45, x25 ≥ 0

x3 ≤ 10

b) x14 = 5

x4 ≤ 3

x45 = 5

x4 ≥ 2

Z = $55,000 40. a) xij = barrels of component i used in gasoline grade j per day, where i = 1, 2, 3, 4 and j = R (regular), P (premium), and L (diesel)

x5 ≤ 10 x5 ≥ 3 xi ≥ 0

Regular: x1R + x2R + x3R + x4R

x1 = 4 hrs.

Premium: x1P + x2P + x3P + x4P

x2 = 8 hrs.

Low lead: x1L + x2L + x3L + x4L maximize Z = 3x1R + 5x2R + 0x3R + 6x4R

x3 = 5 hrs. x4 = 2 hrs.

+ 9x1P + 11x2P + 6x3P + 12x4P + x1L + 3x2L − 2x3L + 4x4L

x5 = 5 hrs. Z = 4.0 42. a) Minimize Z = 211x1 + 173x2 + 410x3 + 152x4 + 263x5 + 414x6 + 302x7 subject to x1 + x2 + x3 + x4 + x5 + x6 + x7 = 14 3400x1 + 3920x2 + 4760x3 + 3560x4 + 4980x5 + 4050x6 + 3240x7 ≤ 60,000 x1 ≤ 4 x2 ≤ 3 x3 ≤ 5 x4 ≤ 2 x5 ≤ 6 x6 ≤ 3 x7 ≤ 2 xi ≥ 0 b) x1 = 4, x2 = 3, x4 = 2, x5 = 5, Z = 2,982 miles c) x1 = 4, x2 = 3, x4 = 2, x6 = 3, x7 = 2, Z = $51,110 43. a) First, all possible patterns that contain the desired lengths must be determined. Pattern

subject to x1R + x2R + x3R + x4R ≥ 3,000 x1P + x2P + x3P + x4P ≥ 3,000 x1L + x2L + x3L + x4L ≥ 3,000 x1R + x1P + x1L ≤ 5,000 x2R + x2P + x2L ≤ 2,400 x3R + x3P + x3L ≤ 4,000 x4R + x4P + x4L ≤ 1,500 .6x1R – .4x2R – .4x3R – .4x4R ≥ 0 –.2x1R + .8x2R – .2x3R – .2x4R ≤ 0 –.3x1R – .3x2R + .7x3R – .3x4R ≥ 0 –.4x1P – .4x2P + .6x3P – .4x4P ≥ 0 –.5x1L + .5x2L – .5x3L – .5x4L ≤ 0 .9x1L – .1x2L – .1x3L – .1x4L ≥ 0 all xij ≥ 0 b) x1R = 2,100

x2R = 1,000 x3R = 900 x2P = 2,300

Length (ft)

x3P = 3,100

x4P = 1,500

x1L = 2,900

Z = $71,400

Total used (ft)

4-13 .

xi = no. of standard-length boards to cut using pattern i minimize Z = x1 + x2 + x3 + x4 + x5 + x6 subject to

Bi = amount of money invested in bonds at the beginning of year i, i = 1,2,3,4,5 C2 = amount of money invested in certificates of deposit in year 2 Ri = amount of money invested in real estate at the beginning of year i, i = 5,6 Ii = amount of money held idle and not invested at beginning of year i

3x1 + 2x2 + 2x3 + x4 = 700 x3 + 2x4 + x5 = 1,200 x2 + x5 + 2x6 = 300 xi ≥ 0 b)

maximize Z = 1.20S5 + 1.40B4 + 1.10R6 + I6

x2 = 50

subject to

x4 = 600

S1 + B1 + I1 = $1,000,000

x6 = 125

S2 + B2 + C2 − I1 + I2 = 0

Z = 775

−1.20S1 + S3 + B3 − I2 + I3 = 0

−1.20S2 + S4 − 1.40B1 + B4 − I3 + I4 = 0 Pattern

−1.20S3 + S5 − 1.40B2 + R5 − I4 + I5 = 0

Length (ft)

Trim loss (ft)

−1.20S4 − 1.40B3 − 1.80C2 − 1.10R5 + R 6 − I5 + I6 = 0 5

i =1

0.7∑ Si − 0.3∑ Bi − 0.3C2 − 6

0.3∑ Ri ≤ 0 i =5

−0.25∑ Si − 0.25∑ Bi +

xi = no. of standard-length boards to cut using pattern i; coefficients of objective function = trim loss using pattern i

i =1

0.75C2 − 0.25∑ Ri ≥ 0

minimize Z = 4x1 + x2 + 2x3 + 0x4 + 6x5 + 5x6

i =5

Si , Bi , Ci , Ri , I i ≥ 0

subject to 3x1 + 2x2 + 2x3 + x4 ≥ 700

I1 = 1,000,000

x3 + 2x4 + x5 ≥ 1,200

C2 = 1,000,000

x2 + x5 + 2x6 ≥ 300

R6 = 1,800,000

xi ≥ 0

Z = $1,980,000

x2 = 300

45. a)

xi = amount sold in month i

x4 = 600

yi = amount purchased in month i

Z = 300

si = amount held in storage in month i (until the beginning of the following month)

44. a)

s0 = amount available at the beginning of month 1 4

i =1

maximize Z = ∑ Pi xi − ∑ ci yi

subject to xi ≤ si − 1 (sales ≤ amount held in storage from previous month) si = si − 1 – xi + yi (ending storage = beginning storage – sales + purchases) Si = amount of money invested in stocks at the beginning of year i, i = 1,2,3,4,5

4-14 .

si ≤ 10,000 (amount held in storage ≤ storage capacity) or maximize Z = 4x1 + 8x2 + 6x3 + 7x4 – 5y1 – 6y2 – 7y3 subject to s0 = 2,000 x1 ≤ s0 s1 = s0 – x1 + y1 s1 ≤ 10,000 x2 ≤ s1 s2 = s1 – x2 + y2 s2 ≤ 10,000 x3 ≤ s2 s3 = s2 – x3 + y3 s3 ≤ 10,000 x4 ≤ s3 xi, yi, si ≥ 0 x1 = 0 y1 = 8,000 s0 = 2,000 b) x2 = 10,000 y2 = 10,000 s1 = 10,000 x4 = 10,000 s2 = 10,000 s3 = 10,000 Z = $50,000 46. a) xi = regular production in month i, where i = 1,2,3,4,5

x4 = 2,000

y4 = 600

x5 = 2,000

y5 = 600

w4 = 1,400

Z = $152,300 47. a) xij = employee i assigned to department j, where i = 1,2,3,4 and j = a (lamps), b (sporting goods), c (linen) maximize Z = 130x1a + l50x1b + 90x1c + 275x2a + 300x2b + 100x2c + 180x3a + 225x3b + 140x3c + 200x4a + 120x4b + 160x4c subject to

x1a + x1b + x1c = 1 x2a + x2b + x2c = 1 x3a + x3b + x3c = 1 x4a + x4b + x4c = 1 1≤ (x1a + x2a + x3a + x4a) ≤ 2 1≤ (x1b + x2b + x3b + x4b) ≤ 2 1 ≤ (x1c + x2c + x3c + x4c) ≤ 2 xij ≥ 0 b)

x1a = 1 x2b = 1 x3b = 1 x4c = 1 Z = 815

c) xij = employee i assigned to department j, where i = 1,2,3,4, and j = a (lamps), b (sporting goods), c (linen)

yi = overtime production in month i, where i = 1,2,3,4,5 wi = units in inventory at end of month i, where i = 1,2,3,4

maximize Z = 130x1a + 150x1b + 90x1c + 275x2a + 300x2b + 100x2c + 180x3a + 225x3b + 140x3c + 200x4a + 120x4b + 160x4c

minimize Z = $10(x1 + x2 + x3 + x4 + x5) + 15(y1 + y2 + y3 + y4 + y5) + 2(w1 + w2 + w3 + w4)

subject to

x1a + x1b + x1c ≤ 1

xi ≤ 2,000 (i = 1,2,3,4,5)

x2a + x2b + x2c ≤ 1

yi ≤ 600 (i = 1,2,3,4,5)

x3a + x3b + x3c ≤ 1

x1 + y1 – w1 = 1,200

x4a + x4b + x4c ≤ 1

x2 + y2 + w1 – w2 = 2,100

x1a + x2a + x3a + x4a = 1

x3 + y3 + w2 – w3 = 2,400

x1b + x2b + x3b + x4b = 1

x4 + y4 + w3 – w4 = 3,000

x1c + x2c + x3c + x4c = 1

x5 + y5 + w4 = 4,000

xi ≥ 0

xi, yi, wi ≤ 0

x2a = 1

b) x1 = 2,000

y1 = 300

w1 = 1,100

x3b = 1

x2 = 2,000

y2 = 600

w2 = 1,600

x4c = 1

x3 = 2,000

y3 = 600

w3 = 1,800

Z = 660

4-15 .

48.

Z values: A = 1.000, B = 1.000, C = .6700, D = 1.000. Bank C is inefficient

49.

Z values: A = 1.000, B = 1.000, C = 1.000. All 3 hospitals are efficient.

50. a)

xij = amount shipped from city i to city j maximize Z = xAB+ xAC or maximize

Z = x = 4.4545 lawyers y1 = 18.1818 y2 = 236.3636 y3 = 304.5454 y4 = 472.7273 y5 = 440.9091

Z = xEF + xCF

y6 = 459.09

subject to xAB ≤ 7 xAC ≤ 6 xBE ≤ 2 xBD ≤ 5 xBC ≤ 10 xCD ≤ 3 xCF ≤ 9 xDE ≤ 4 xEF ≤ 3 xAB = xBC + xBD + xBE xAC + xBC = xCD + xCF xBD + xCD = xDE xBE + xDE = xEF xij ≥ 0 b) xAB = 6 xAC = 6 xBE = 2 xBD = 1 xBC = 3 xCF = 9 xDE = 1 xEF = 3 Z = 12 51. a) Minimize Z = x subject to 150x = 650 + y1 150x + y1 = 450 + y2 150x + y2 = 600 + y3 150x + y3 = 500 + y4 150x + y4 = 700 + y5 150x + y5 = 650 + y6 150x + y6 = 750 + y7 150x + y7 = 900 + y8 150x + y8 = 800 + y9

y7 = 377.2727 y8 = 145.4545 y9 = 13.6364 y10 = 31.8182 y11 = 0 52.

The original optimal solution has x = 4.4545 lawyers being hired. To get an “optimal” integer solution you would have to round either down to x = 4 or up to x = 5. Adding a new constraint, x = 4, to the original model formulation results in an infeasible solution, so rounding down is not possible. Adding the constraint, x = 5, instead (i.e., rounding up) results in a feasible solution so (by logical deduction) it must be optimal. The original solution had 168.1818 surplus hours while the new integer solution has 1,150 surplus hours, a large difference in excess capacity. However, increasing the number of lawyers (to 6, for example) only increases the surplus so x = 5 has to be the optimal integer solution.

53. a)

y = quantity of assembled product produced per day x1 = quantity of component 1 produced per day x2 = quantity of component 2 produced per day x3 = quantity of component 3 produced per day maximize Z = y subject to (Component 1) y = x1 (Component 2) y = x2 y = x3 (Component 3) Note: The above constraints reflect the fact that the quantity of completely assembled product cannot exceed the quantities of components 1, 2, and 3 produced per day, i.e., no in-process inventory.

150x + y9 = 650 + y10 150x + y10 = 700 + y11 150x + y11 ≥ 500

4-16 .

x1 = 43.077 x2 = x3 = y = 29.31 The best alternative is to remove the machine balancing requirement. If both requirements are relaxed the production output is x1 = x2 = x3 = y = 32. 54. a) xij = amount of commodity i put into hold j, where i = b (beef), g (grain) and j = f (fore), a (aft)

The time available on one lathe is: 60 min/hr. × 8 hrs/day = 480 min. In order to reflect the availability of two lathes (with the work load shared evenly between them) the total minutes of lathe time available is: 480 min/day/machine × 2 machines = 960 min. Thus, the lathe use/availability constraint is: 10x1 + 8x2 + 6x3 ≤ 960 (Lathe Time) By the same rationale, the press use/ availability constraint (for three presses) is: 9x1 + 21x2 + 15x3 ≤ 1,440 (Press Time)

maximize Z = .35(xbf + xba) + .12(xgf + xga) subject to xbf + xgf ≤ 70,000

Assuming that the work loads are shared equally among the lathes, and similarly for presses, the individual machine utilizations are: (10x1 + 8x2+ 6x3)/2 = 5x1 + 4x2 + 3x3

xba + xga ≤ 90,000 .2xbf + .4xgf ≤ 30,000 .2xba + .4xga ≤ 40,000 xbf + xba ≤ 85,000

(9x1 + 21x2 +15x3)/3 = 3x1 + 7x2 + 5x3

xgf + xga ≤ 100,000

The balance condition is reflected by specifying that the absolute difference between individual machine time consumed (on lathes versus presses) must be less than or equal to 60 minutes.

xij ≤ 0 b) xba = 85,000

xgf = 70,000 xga = 5,000

|(5x1 + 4x2 + 3x3) – (3x1 + 7x2 + 5x3)| ≤ 60

Z = 38,750

or,

55.

|2x1 – 3x2 – 2x3| ≤ 60

b) c)

Absolute value constraint conditions can be reflected by 2x1 – 3x2 – 2x3 ≤ 60 –2x1 + 3x2 + 2x3 ≤ 60 The model is summarized as follows: maximize Z = y subject to y – x1 = 0 y – x2 = 0 y – x3 = 0 10x1 + 8x2 + 6x3 ≤ 960 9x1 + 21x2 + 15x3 ≤ 1,440 2x1 – 3x2 – 2x3 ≤ 60 –2x1 + 3x2 + 2x3 ≤ 60 x1, x2, x3, y ≥ 0 x1 = x2 = x3 = y = 20 If the machine balancing requirement is relaxed such that the constraints 2x1 – 3x2 – 2x3 ≤ 60 and –2x1 + 3x2 + 2x3 ≤ 60 are eliminated, production output would increase: x1 = x2 = x3 = y = 32 Alternatively, if the restriction that there be no excess component parts at the end of the day is relaxed such that y ≤ x1, y ≤ x2 and y ≤ x3, production output would increase:

xi = no. of commercial minutes during broadcast segment i, where i = l (local news), n (national news), s (sports), w (weather) yi = no. of minutes for broadcast segment i maximize Z = $850xl + 600xn + 750xS + 1,000xw subject to xl + xn + xs + xw = 18 xl + xn + xs + xw + yl + yn + ys + yw = 60 400yl + 100yn + 175ys + 90yw ≤ $9,000 10 ≤ yl ≤ 25 5 ≤ yn ≤ 10 5 ≤ ys ≤ 10 5 ≤ yw ≤ 10 xl ≤ 6 xn ≤ 6 xs ≤ 6 xw ≤ 6 xi,yi, ≥ 0

4-17 .

y2 = $ amount spent on advertising apple juice

Solution: xl = 6

yl = 14.44

xn = 0

yn = 10

xs = 6

ys = 7.56

xw = 6

yw = 10

maximize Z = .85x1 + .90x2 – y1 – y2 subject to x1 ≤ 5,000 + 3y1 x2 ≤ 4,000 + 5y2

Z = $15,600

.60x1 + .85x2 + y1 + y2 ≤ $16,000

Multiple optimal solutions exist. 56. a)

x1 ≥ .30(x1 + x2)

xij = amount (oz) of ingredient i in jar of j, where i = cabbage, tomato, onions and j = chow-chow, tomato

x1 ≤ .60(x1 + x2) x1, x2, y1, y2 ≥ 0

maximize Z

b) x1 = 5,458.128

⎛ 2.25 ⎞ ⎛ 1.95 ⎞ =⎜ ⎟ ( xcc + xtc + xoc ) + ⎜ ⎟ ( xct + xtt + xot ) ⎝ 16 ⎠ ⎝ 16 ⎠

x2 = 12,735.63 y1 = 152.709 y2 = 1,747.126

subject to

Z = $14,201.64

xcc+xtc+xoc+xct+xtt+xot ≤ (700)(16)

58. a)

xcc ≥ .60 xcc + xtc + xoc

xi = no. of employees assigned to time period i, where i = 1,2,...,6 (time period 1 = 12:00 midnight –4:00 A.M.;

xtt xoc ≥ .50, ≥ .05 xct + xtt + xot xcc + xtc + xoc

period 2 = 4:00–8:00 A.M.; etc.) minimize Z = x1 + x2 + x3 + x4 + x5 + x6

xoc xtc ≤ .10, ≥ .10 xcc + xtc + xoc xcc + xtc + xoc

subject to x6 + x1 ≥ 90

xot xot ≤ .10, ≥ .05 xct + xtt + xot xct + xtt + xot

x1 + x2 ≥ 215

xcc + xct ≤ (300)(16)

x3 + x4 ≥ 65

xtc + xtt ≤ (350)(16)

x4 + x5 ≥ 300

xoc + xot ≤ (30)(16)

x5+ x6 ≥ 125

x2 + x3 ≥ 250

xi ≥ 0

xcc + xtc + xoc xct ≥ 1.3, ≥ .10 xct + xtt + xot xot + xtt + xct

b) x1 = 90

x2 = 250

xij ≥ 0 b) chow-chow relish xcc = 4,608 oz

tomato relish xct = 192 oz

x3 = 0

xtc = 2,688 oz 384 oz xoc = 7,680 oz 480 jars

xtt = 1,632 oz 96 oz xot = 1,920 oz 120 jars

x5 = 125

x4 = 175 x6 = 0 Z = 640 59. a)

Z = $1,313.66 57. a) x1 = no. of jars of applesauce

xij = consultant i hours assigned to project j; i = A,...,F and j = 1,...,8

x2 = no. of bottles of apple juice

Minimize Z = ∑ ∑ (ranking) • xij

y1 = $ amount spent on advertising applesauce

i = A j =1

4-18 .

subject to

Note: Objective function coefficient of $4,880 for x is computed as $610/week (8 weeks).

∑ xij ≤ available hours, i = A,..., F j =1

360x + 270y6 = 29,800

b) Solution x = 53.6 full-time operators y1 = 0.76 part-time operators y2 = 6.31 part-time operators y3 = 23.35 part-time operators y4 = 29.27 part-time operators y5 = 52.24 part-time operators y6 = 38.90 part-time operators y7 = 28.53 part-time operators y8 = 43.35 part-time operators Z = $361,788 61. Minimize Z = 4,880x1 + 4,270x2 + 3,660x3 + 3,050x4 + 2,440x5+ 1,830x6 + 1,220x7 +610x8 + 450 ∑ yi subject to 360x1 + 270y1 = 19,500 360(x1 + x2) + 270y2 = 21,000 360(x1 + x2 + x3) + 270y3 = 25,600 360(x1 + x2 + x3 + x4) + 270y4 = 27,200 360(x1 + x2 + x3 + x4 + x5) + 270y5 = 33,400 360(x1 + x2 + x3 + x4 + x5 + x6) + 270y6 = 29,800 360(x1 + x2 + x3 + x4 + x5 + x6+ x7) + 270y7 = 27,000 360(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8) + 270y8 = 31,000 1.1x1 + 2.7y1 ≤ 200 1.1(x1 + x2) + 2.7y2 ≤ 200 1.1(x1 + x2 + x3) + 2.7y3 ≤ 200 1.1(x1 + x2 + x3 + x4) + 2.7y4 ≤ 200 1.1(x1 + x2 + x3 + x4 + x5) + 2.7y5 ≤ 200 1.1(x1 + x2 + x3 + x4 + x5 + x6) + 2.7y6 ≤ 200 1.1(x1 + x2 + x3 + x4 + x5 + x6 + x7) + 2.7y7 ≤ 200 1.1(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8) + 2.7y8 ≤ 200 xi, yi ≥ 0 Solution: x1 = 0 y1 = 72.22 x2 = 4 y2 = 72.44 Z = $360,196 x3 = 18.4 y3 = 64.95 x4 = 6.4 y4 = 62.34 x5 = 24.8 y5 = 52.24 y6 = 38.90

360x + 270y7 = 27,000

y7 = 28.53

360x + 270y8 = 31,000

y8 = 43.35

∑ xij ≤ project hours, j = 1,...,8 i =1 F

∑ (hourly rate) • xij ≤ Budget, j = 1,...,8

i=A

b) Solution: xA3 = 400

xD2 = 131.7

xA4 = 50

xD7 = 15.93

xB4 = 250

xE1 = 208.33

xB5 = 350

xE7 = 149.07

xC4 = 175

xF1 = 291.67

xC7 = 274.1

xF2 = 108.3

xC8 = 50.93

xF6 = 460

Z = $12,853.33 xA4 = 257.2 xE1 = 129.7 xA5 = 192.8

xE2 = 240

xB4 = 95

xE3 = 270

xC1 = 333.3

xE8 = 33.3

xC8 = 166.7

xF5 = 110

xD3 = 130

xF6 = 460

xD4 = 112.8

xF7 = 290

xD5 = 47.2 Z = $576,250 60. a)

x = full-time operators yi = part-time operators hired in week i, where i = 1,2,...8 8

Minimize Z = $4,880 x + $450∑ yi i =1

subject to 360x + 270y1 = 19,500 360x + 270y2 = 21,000 360x + 270y3 = 25,600 360x + 270y4 = 27,200 360x + 270y5 = 33,400

1.1x + 2.7yi ≤ 200, i = 1,2,...,8 x, yi ≥ 0

4-19 .

62. xij = tons of coal shipped from supplier i, where i = A,B,C,D,E,F, to power plant j, where j = 1,2,3,4.

subject to

minimize Z = $23 (xA1 + xA2 + xA3 + xA4) + 28 (xB1+ xB2 + xB3 + xB4) + 24 (xC1 + xC2 + xC3 + xC4) + 31 (xD1 + xD2 + xD3 + xD4) + 29 (xE1 + xE2 + xE3 + xE4) + 34 (xF1 + xF2 + xF3 + xF4) + 12.20xA1 + 14.25xA2 + 11.15xA3 + 15.00xA4 + 10.75xB1 + 13.70xB2 + 11.75xB3 + 14.45xB4 + 15.10xC1 + 16.65xC2 + 12.90xC3 + 12.00xC4 + 14.30xD1 + 11.90xD2 + 16.35xD3 + 11.65xD4 + 12.65xE1 + 9.35xE2 + 10.20xE3 + 9.55xE4 + 16.45xF1 + 14.75xF2 + 13.80xF3 + 14.90xF4 subject to xA1 + xA2 + xA3 + xA4 = 190,000 xB1 + xB2 + xB3 + xB4 = 305,000 xC1 + xC2 + xC3 + xC4 = 310,000 xD1 + xD2 + xD3 + xD4 ≤ 125,000 xE1 + xE2 + xE3 + xE4 ≤ 95,000 xF1 + xF2 + xF3 + xF4 ≤ 190,000 26.2xA1 + 27.1xB1 + 25.6xC1 + 21.4xD1 +19.2xE1 + 23.6xF1 ≥ 4,600,000 26.2xA2 + 27.1xB2 + 25.6xC2 + 21.4xD2 +19.2xE2 + 23.6xF2 ≥ 6,100,000 26.2xA3 + 27.1xB3 + 25.6xC3 + 21.4xD3 +19.2xE3 + 23.6xF3 ≥ 5,700,000 26.2xA4 + 27.1xB4 + 25.6xC4 + 21.4xD4 +19.2xE4 + 23.6xF4 ≥ 7,300,000 xij ≥ 0 Solution: xA3 = 190,000 xB1 = 169,741.70 xB2 = 132,084.87 xB3 = 3,173.43 xC3 = 24,843.75 xC4 = 285,156.25 xD2 = 32,546.73 xE2 = 95,000 Z = $34,921,937.85 63. xij = team i, where i = 1, 2, . . . 16, assigned to field j, where j = 1, 2, . . . 12. minimize Z = ∑ xij (priority ij )

i =1

∑ x = 1, where i = 1, 2,...,16 ij

∑ x ≤ available slots j, where j = 1,...,12 i =1

Solution: U11B: 3,3–5M U11G: 1,3–5T U12B: 1,3–5T U12G: 1,3–5M U13B: 2,3–5T U13G: 1,3–5M U14B: 1,5–7M U14G: 2,3–5M U15B: 1,5–7T U15G: 3,3–5M U16B: 3,5–7T U16G: 2,5–7T U17B: 2,5–7M U17G: 1,5–7T U18B: 3,3–5T U18G: 1,5–7M Z = 27 The U15G and U18B teams did not get one of their selected times/locations. By having the teams select “4” or more times/locations, the model might assign these teams a more desirable location. It may occur that the model solution can be used as is and a team simply assigned what seems to be a reasonable location and time. For example, the U15G team’s selections are all for the 5–7 time period; however, their assignment was for 3–5M, which is probably not feasible for them. Looking at the solution results it can be seen that field 3 has openings on both MW and TTh at 5–7 so they could be assigned one of these slots. Another way to avert this problem is to assign very high values (for example “10”) to location and times that are unacceptable for teams, i.e., assign a value of “10” to all the 3–5 time slots for the U15G team.

where priority ij = 1, 2, 3 or 4 (for a field not selected)

64. Z values: A = .9276, B = 1.000, C = 1.000, D = 1.000. Police station A is inefficient.

4-20 .

65. Minimize Z = 1x1D + 3x1E + 5x1F + 6x1G + 9x1H + 10x1I + 12x1J + 1x2E + 3x2F + 4x2G + 7x2H + 8x2I + 10x2J + 2x3F + 3x3G + 6x3H + 7x3I + 9x3J + 1x4I + 3x4J + 1x5J + 6xA4 + 8xA5 + 10xA6 + 11xA7 + 5xB4 + 7xB5 + 9xB6 + 10xB7 + 4xC4 + 6xC5 + 8xC6 + 9xC7 + 2xD4 + 4xD5 + 6xD6 + 7xD7 + 2xE5 + 4xE6 + 5xE7 + 2xF6 + 3xF7 + 1xG6 + 2xG7 subject to

Solution: 1 – D (1 hr) 2 – E (1 hr) 3 – F (2 hr) 3 – H (6 hr) 4 – I (1 hr) 5 – J (1 hr) A – 5 (8 hr) B – 4 (5 hr)

1 ≤ x1D + x1E + x1F + x1G + x1H + x1I + x1J ≤ 2 1 ≤ x2E + x2F + x2G + x2H + x2I + x2J ≤ 2 1 ≤ x3F + x3G + x3H + x3I + x3J ≤ 2 x4I + x4J ≤ 2 x5J = 1 1 ≤ xA4 + xA5 + xA6 + xA7 ≤ 2 1 ≤ xB4 + xB5 + xB6 + xB7 ≤ 2 1 ≤ xC4 + xC5 + xC6 + xC7 ≤ 2 1 ≤ xD4 + xD5 + xD6 + xD7 ≤ 2 xE5 + xE6 + xE7 ≤ 2 xF6 + xF7 ≤ 2 xG6 + xG7 ≤ 2 1 ≤ x1D + xD4 + xD5 + xD6 + xD7 ≤ 2 1 ≤ x1E + x2E + xE5 + xE6 + xE7 ≤ 2 1 ≤ x1F + x2F + x3F + xF6 + xF7 ≤ 2 1 ≤ x1G + x2G + x3G + xG6 + xG7 ≤ 2 1 ≤ xA4 + xB4 + xC4 + xD4 + x4I + x4J ≤ 2 1 ≤ xA5 + xB5 + xC5 + xD5 + xE5 + x5J ≤ 2 1 ≤ xA6 + xB6 + xC6 + xD6 + xE6 + xF6 + xG6 ≤ 2 1 ≤ xA7 + xB7 + xC7 + xD7 + xE7 + xF7 + xG7 ≤ 2 x1H + x2H + x3H ≥ 1 x1I + x2I + x3I + x4I ≥ 1 x1J + x2J + x3J + x4J + x5J ≥ 1 xA4 + xB4 + xC4 + xD4 ≤ 1 xA5 + xB5 + xC5 + xD5 + xE5 ≤ 1 xA6 + xB6 + xC6 + xD6 + xE6 + xF6 + xG6 ≤ 1 xA7 + xB7 + xC7 + xD7 + xE7 + xF7 + xG7 ≤ 1 x1E + x2E ≤ 2 x1F + x2F + x3F ≤ 1 x1G + x2G + x3G ≤ 1 x1H + x2H + x3H ≥ 1 x1L + x2L + x3L + x4L ≥ 1 x1J + x2J + x3J + x4J + x5J ≥ 1

C – 6 (8 hr) G – 7 (2 hr) Z = 35 hours (ground time) 6 crews originate at Pittsburgh; 4 in Orlando. One crew ferries on flight 3 from Pittsburgh and flies H back from Orlando. One crew flies A from Orlando and then flies 5 back to Orlando while a crew ferries on flight 5 to Orlando and then flies J back to Pittsburgh. A crew flies B from Orlando and then flies 4 back to Orlando while a crew ferries on flight 4 and flies I back to Pittsburgh. There are multiple optimal solutions. 66. xij = flow of product from node i to node j

maximize Z = .17(x26 +x27 + x28) + .20(x36 + x37 + x38) + .18(x46 + x47 + x48) + .16(x56 + x57 + x58) + .26(x69 + x610 + x611) + .29(x79 + x710 + x711) + .27(x89 + x810 + x811) + .12(x912) + .11(x1012) +.14(x1112) subject to x12 = x26 + x27 + x28 x13 = x36 + x37 + x38 x14 = x46 + x47 + x48 x15 = x56 + x57 + x58 .25x26 + .25x36 + .25x46 + .25x56 = x69 + x610 + x611 x26 = x36 x36 = x46 x46 = x56 .25x27 + .25x37 + .25x47 + .25x57 = x79 + x710 + x711 x27 = x37 x37 = x47

xij ≥ 0

x47 = x57 .25x28 + .25x38 + .25x48 + .25x58 = x89 + x810 + x811

4-21 .

x28 = x38

Solution:

x38 = x48

x12 = x13 = x14 = x15 = 37,000

x48 = x58

x26 = x36 = x46 = x56 = 12,000 x27 = x37 = x47 = x57 = 6,000

x69 + x79 + x89 = x912

x28 = x38 = x48 = x58 = 19,000

x610 + x710 + x810 = x1012

x69 = 5,000

x611 + x711 + x811 = x1112

x79 = 6,000

x912 + x1012 + x1112 = 37,000

x89 = 3,000

x12 ≤ 60,000

x810 = 16,000

x13 ≤ 50,000

x611 = 7,000

x14 ≤ 55,000

x912 = 14,000

x15 ≤ 60,000

x1012 = 16,000

x26 + x27 + x28 ≤ 50,000

x1112 = 7,000 Z = $40,680

x36 + x37 + x38 ≤ 40,000

67. xij = bushels shipped from farm i, where i = 1, 2, 3, to plant j, where j = 4, 5, and gallons shipped from plant i, where i = 4, 5, to distribution center j, where j = 6, 7, 8.

x46 + x47 + x48 ≤ 46,000 x56 + x57 + x58 ≤ 50,000 x69 + x610 + x611 ≤ 12,000

Minimize Z = .41x14 + .57x15 + .37x24 + .48x25 + .51x34 + .60x35 + .22x46

x79 + x710 + x711 ≤ 14,000

+ .10x47 + .20x48 + .15x56

x89 + x810 + x811 ≤ 19,000

+ .16x57 + .18x58

x912 ≤ 14,000

subject to

x1012 ≤ 16,000

x14 + x15 ≤ 24,000

x1112 ≤ 12,000

x24 + x25 ≤ 18,000

xij ≥ 0

x34 + x35 ≤ 32,000 x14 + x24 + x34 ≤ 48,000 bushels x15 + x25 + x35 ≤ 35,000 bushels (x14 + x24 + x34)/2 = x46 + x47 + x48 gals. (x15 + x25 + x35)/2 = x56 + x57 + x58 gals. x46 + x56 = 9,000 gals. x47 + x57 = 12,000 gals. x48 + x58 = 15,000 gals. xij ≥ 0 Solution: x14 = 24,000 x24 = 18,000 x34 = 6,000 x35 = 24,000 x47 = 12,000

4-22 .

x48 = 12,000

b) Minimize

x56 = 9,000 x58 = 3,000 Z = $39,450 68. a)

Z = 10∑ xil + 20∑ xi 2 + 35∑ xi 3

subject to 3

Minimize Z = ∑∑ xij pij i

∑ x = 1, for i = 1,2,...,15 ij

where xij = victim i (i = 1,2,...,15) transported to hospital j (j = 1,2,3); and pij = injury priority for victim i at hospital j.

∑x ≤8 il

∑ x ≤ 10 i2

subject to

∑x ≤ 7

∑ xij = 1, for i = 1,2,...,15

⎛ 15 3 ⎞ ⎜ ∑∑ xij pij ⎟ / 15 ≤ 1.50 ⎝ i j ⎠

∑x ≤8 il

∑ x ≤ 10

Solution:

∑x ≤7 i3

10∑ xil + 20∑ xi 2 + 35∑ xi 3 ≤ 22 xij = integer and ≥ 0

Solution: Victim

Hospital

Montgomery Regional

Lewis Galt

Montgomery Regional

Radford Memorial

Montgomery Regional

Lewis Galt

Radford Memorial

Montgomery Regional

Lewis Galt

Montgomery Regional

Lewis Galt

Victim

Hospital

Montgomery Regional

Lewis Galt

Radford Memorial

Montgomery Regional

Radford Memorial

Montgomery Regional

Lewis Galt

Z = avg. transport time = 16.67 mins. Average priority = 1.47

CASE SOLUTION: SUMMER SPORTS CAMP AT STATE UNIVERSITY Model Formulation: wi = new sheets purchased for week i (i = 1,2,...,8)

Average Priority =1.07

xi = sheets cleaned at laundry at end of week i

4-23 .

yi = sheets cleaned by Mary’s friends at end of week i

xi = overtime production of product i in stage 1 Ai = regular production of product i in stage 2

minimize Z = 10(w1 + w2 + w3 + w4 + w5 + w6 + w7 + w8) + 4(x1 + x2 + x3 + x4 + x5 + x6) + 2(y1 + y2 + y3 + y4 + y5)

yi = overtime production of product i in stage 2 minimize Z = $6R1 + 10R2 + 8R3 + 10R4 + 7.2S1 + 12S2 + 9.6S3 + 12S4 + 6.2x1 + 10.7x2 + 8.5x3 + 10.7x4 + 3A1 + 5A2 + 4A3 + 5A4 + 3.1y1 + 5.4y2 + 4.3y3 + 5.4y4

subject to w1 = 115 x1 + y1 = 115

subject to .04R1 + .17R2 + .06R3 + .12R4 ≤ 500 hrs. .04x1 + .17x2 + .06x3 + .12x4 ≤ 100 hrs. .05R1 + .14R2 + .14R4 ≤ 400 hrs. .05x1 + .14x2 + .14x4 ≤ 100 hrs. 1.2R1 + 1.6R2 + 2.1R3 + 2.4R4 + 1.2x1 + 1.6x2 + 2.1x3 + 2.4x4 ≤ 10,000 ft2 .06A1 + .13A2 + .05A3 + .10A4 ≤ 600 hrs. .06y1 + .13y2 + .05y3 + .10y4 ≤ 100 hrs. .05A1 + .21A2 + .02A3 + .10A4 ≤ 550 hrs. .05y1 + .21y2 + .02y3 + .10y4 ≤ 100 hrs. .03A1+.15A2 +.04A3 + .15A4 ≤ 500 hrs. .03y1 + .15y2 + .04y3 + .15y4 ≤ 100 hrs. R1 + S1 + x1 = A1 + y1 R2 + S2 + x2 = A2 + y2 R3 + S3 + x3 = A3 + y3 R4 + S4 + x4 = A4 + y4 y1 + A1 = 1,800 y2 + A2 = 1,400 y3 + A3 = 1,600 y4 + A4 = 1,800 Ri, Si, xi, Ai, yi ≥ 0 Solution: R1 = 1,691.954 s1 = s2 = s3 = 0 R2 = 1,319.54 s2 = 866.6667 R3 = 1,600 y1 = 0 Z = $85,472.60 R4 = 933.33 y2 = 315.1514 A1 = 1,800 y3 = 0 A2 = 1,084.849 y4 = 388.1822 A3 = 1,600 x1 = 108.0457 A4 = 1,461.818 y2 = 80.4599 x3 = x4 = 0

w2 = 210 x2 + y2 = 210 w3 + .8x1 = 250 x3 + y3 = 250 w4 + .8x2 + .8y1 = 230 x4 + y4 = 230 w5 + .8x3 + .8y2 = 260 x5+ y5 = 260 w6 + .8x4 + .8y3 = 300 x6 = 300 w7 + .8x5+ .8y4 = 250 w8 + .8x6 +.8y5 ≥ 190 wi, xi, yi ≥ 0 Model Solution: w1 = 115 w2 = 210 w3 = 250 w4 = 138 w5 = 50 w6 = 0 w7 = 0 w8 = 0 Z = $11,940

x1 = 0 x2 = 0 x3 = 52.5 x4 = 177.5 x5= 260 x6 = 300

y1 = 115 y2 = 210 y3 = 197.5 y4 = 52.5 y5 = 0

CASE SOLUTION: SPRING GARDEN TOOLS

CASE SOLUTION: SUSAN WONG’S PERSONAL BUDGETING MODEL

Model Formulation: Let i = 1 (trowel), 2 (hoe), 3 (rake), 4 (shovel) Ri = regular production of product i in stage 1

a) Let xij = amount invested for i months in month j where i = 1,3 and 7, and j = 1,2,...,12.

Si = subcontracted production of product i in stage 1

j =1

max Z = ∑ 0.005 x1j + ∑ 0.02 x3 j + ∑ 0.07 x7 j

4-24 .

subject to Jan: Feb: Mar: Apr: May: June: July: Aug: Sep: Oct: Nov: Dec:

Month (j) 1. January 2. February 3. March 4. April 5. May 6. June 7. July 8. August 9. September 10. October 11. November 12. December

–x11 – x31 – x71 + 2,450 + 3,800 = 2,750 x11 – x12 – x32 – x72 + 2,450 = 2,860 x12 – x13 – x33 – x73 + 2,450 = 2,335 x13 – x14 + x31 – x34 – x74 + 2,450 = 2,120 x14 – x15 + x32 – x35 – x75 + 2,450 = 1,205 x15 – x16 + x33 – x36 – x76 + 2,450 = 1,600 x16 – x17 + x34 – x37 – x77 + 2,450 = 3,050 x17 – x18 + x35 – x38 + x71 – x78 + 2,450 = 2,300 x18 – x19 + x36 – x39 + x72 – x79 + 2,450 = 1,975 x19 – x110 + x37 – x310 + x73 – x710 + 2,450 = 1,670 x110 – x111 + x38 – x311 + x74 – x711 + 2,450 = 2,710 x111 – x112 + x39 – x312 + x75 – x712 + 2,450 = 2,980 Investments (i = 1,3,7) 1-month 3-month x11 = 410 x31 = 390

7-month

x13 = 115 x74 = 3,535 x75 = 1,245 x16 = 600

x36 = 250

x18 = 150 x39 = 875 x110 = 780 x111 = 4,055 x712 = 5,645 Z = $844.60 earned in interest payments

Using sensitivity analysis for the January constraint, the lower range for the right hand side is –410. Thus, Susan needs $710 out of her original $3,800 to make the model feasible (i.e., avoid an infeasible solution).

subject to xnv + xnm + xnt + xni ≤ 1,400 xpv + xpm + xpt + xpi ≤ 1,100 xov + xom + xot + xoi ≤ 1,700 xnv + xpv + xov ≤ 1,200 xnm + xpm + xom ≤ 1,100 xnt + xpt + xot ≤ 1,400 xni + xpi + xoi ≤ 1,400 yjv + yjm + yjt + yji = 1,200 ykv + ykm + ykt + yki = 900 ylv + ylm + ylt + yli = 700 yjv + 2ykv + 1.5ylv = xnv + xpv + xov yjm + 2ykm + 1.5ylm = xnm + xpm + xom yjt + 2ykt + 1.5ylt = xnt + xpt + xot yji + 2yki + 1.5yli = xni + xpi + xoi xij, yij ≥ 0 Solution: xni = 1,400 yji = 1,200 xpm = 1,100 ykv = 75 xov = 150 ykm = 25 xot = 1,400 ykt = 700 yki = 100 ylm = 700 Z = $10,606,000

CASE SOLUTION: WALSH’S JUICE COMPANY xij = tons of unprocessed grape juice transported from vineyard i to plant j where i = n (New York), p (Pennsylvania), o (Ohio), and j = v (Virginia), m (Michigan), t (Tennessee), and i (Indiana). yij = tons of grape juice processed into product i at plant j, where i = j (juice), k (concentrate), and l (jelly) minimize Z = $850xnv + 720xnm + 910xnt + 750xni + 970xpv + 790xpm + 1,050xpt + 880xpi + 900xov + 830xom + 780xot + 820xoi + 2,100yjv + 2,350yjm + 2,200yjt + 1,900yji + 4,100ykv + 4,300ykm + 3,950ykt + 3,900yki + 2,600ylv + 2,300ylm + 2,500ylt + 2,800yli

4-25 .

Solution: y1 = 100 y2 = 171.5 y3 = 213.2 y4 = 417.6 y5 = 54.3 y6 = 408.4 y7 = 147.0 y8 = 284.9 y9 = 189.7 y10 = 222.1 y11 = 152.4 y12 = 302.4 y13 = 279.9 y14 = 73.1 y15 = 517.8 y16 = 0

CASE SOLUTION: KING’S LANDING AMUSEMENT PARK xi = experienced employees in week i yi = new employees in week i zi = pool of employees available in week i minimize z = Σyi subject to 10yi + 30xi ≥ weekly hours needed, where i = 1,...,16 30xi ≥ weekly hours needed, where i = 17, 18, 19, 20 x1 = 700 xi = .85xi–1 + yi, where i = 2,3,...,16 x17 = .25x16 xi = .90xi–1, where i = 18,19,...,10 yi ≤ zi, where i = 2,3,...,16 z1 = 1,500 z2 = 1,500 – y1 + 200 z3 = z2 – y2 + 200 z4 = z3 – y3 + 200 z5 = z4 – y4 + 200 z6 = z5 – y5 + 200 z7 = z6 – y6 + 200 z8 = z7 – y7 + 200 z9 = z8 – y8 + 100 z10 = z9 – y9 + 100 z11 = z10 – y10 + 100 z12 = z11 – y11 + 100 z13 = z12 – y12 + 100 z14 = z13 – y13 + 100 z15 = z14 – y14 + 100 z16 = z15 – y15 + 100

x1 = 700.0 x2 = 695.0 x3 = 762.3 x4 = 861.1 x5 = 1,148.6 x6 = 1,030.5 x7 = 1,284.3 x8 = 1,238.7 x9 = 1,336.8 x10 = 1,326.0 x11 = 1,349.2 x12 = 1,299.2 x13 = 1,406.7 x14 = 1,475.6 x15 = 1,327.4 x16 = 1,646.1 x17 = 411.5 x18 = 370.4 x19 = 333.3 x20 = 300.0

Z = 3,532 Note: This would logically be best solved as an integer programming model; however, the model is so large that it exceeds the capabilities of Excel to solve in a reasonable amount of time.

4-26 .

Chapter Five: Integer Programming PROBLEM SUMMARY

29. Sensitivity analysis (5–28) 30. 0–1 integer model, computer solution

1. Integer model

31. 0–1 integer model, computer solution

2. Integer model 3. Integer model

32. Set covering problem, 0–1 integer model and computer solution

4. Integer model

33. 0–1 integer model

5. Integer model

34. 0–1 integer model

6. Integer model

35. 0–1 integer model and computer solution

7. Integer model

36. Set covering problem, 0–1 integer model and computer solution

8. Mixed integer model

37. Facility location problem, 0–1 integer model and computer solution

9. 0–1 integer model 10. Integer model

38. 0–1 integer model and computer solution

11. Integer (fixed charge) model, formulation and computer solution

39. Scheduling, 0–1 integer model and computer solution

12. 0–1 integer model, formulation and computer solution

40. “Traveling salesman” problem, 0–1 integer model

13. Set covering problem, integer model, formulation and computer solution

41. Leasing selection; 0–1 integer model and computer solution

14. Integer model, formulation and computer solution

42. 0–1 integer model and computer solution 43. Assignment, 0–1 integer model and computer solution

15. Integer model, formulation and computer solution

44. Set covering problem 0–1 integer model and computer solution

16. Integer model, knapsack problem, formulation and computer solution

45. 0–1 integer model and computer solution

17. Integer model, formulation and computer solution

PROBLEM SOLUTIONS

18. Plant location problem, integer model, formulation and computer solution

x1 (cakes) = 3

19. Integer model, formulation and computer solution

x2 (breads) = 3

20. 0–1 integer-model, computer solution

x4 (cookies) = 15

21. Integer model, computer solution

Z (Total sales) = 150

x3 (pies) = 0

22. Integer model, formulation and computer solution

x1 (residential) = 45 x2 (commercial) = 8

23. Integer model with 0–1 restriction (5–22)

Z = $1,523

24. Integer model, formulation and computer solution

3. a) Maximize Z = 50x1 + 40x2 (profit) subject to

25. 0–1 integer model, computer solution

3x1 + 5x2 ≤ 150 yd2

26. Integer model, formulation and computer solution

10x1 + 4x2 ≤ 200 hr. x1,x2 ≥ 0 and integer

27. Mixed-integer model, formulation and computer solution (5–26) 28. 0–1 integer model, computer solution

b) Relaxed solution:

x1 = 2.73, x2 = 5.91, Z = 372.9

x1 = 10.5, x2 = 23.7, Z = 1,473 Rounded-down solution:

Rounded-down solution:

x1 = 10, x2 = 23, Z = 1,420

x1 = 2, x2 = 5, Z = 300

Integer solution:

x1 = 10, x2 = 24, Z = 1,460

x1 = 4, x2 = 4, Z = 360

The rounded-down solution is not optimal.

4. a) Maximize Z = $400x1 + 100x2 subject to

8. a) Maximize Z = $8,000x1 + 6,000x2 subject to

8x1 + 10x2 ≤ 80

70x1 + 30x2 ≤ 500

2x1 + 6x2 ≤ 36

x1 + 2x2 ≤ 14

x1 ≤ 6

x1 ≥ 0 and integer

x1,x2 ≥ 0 and integer

x2 ≥ 0

b) Relaxed solution:

b) x1 = 5, x2 = 4.5, Z = 67,000

x1 = 6, x2 = 3.2, Z = 2,720 Rounded-down solution:

x1 = 1, x2 = 0, x3 = 1, Z = 1,800

10.

Minimize Z = 81x1 + 50x2 subject to

x1 = 6, x2 = 3, Z = 2,700

76x1 + 53x2 ≥ 600

Integer solution:

x1 + x2 ≤ 10

x1 = 6, x2 = 3, Z = 2,700

1.3x1 + 4.1x2 ≤ 24

Integer solution same as rounded-down solution.

x1,x2 ≥ 0 and integer

5. a) Maximize Z = 50x1 + 10x2 subject to

Solution: x1 = 6

x1 + x2 ≤ 15

x2 = 3

4x1 + x2 ≤ 25

Z = $636

x1,x2 ≥ 0 and integer

11. a) Maximize Z = 85, 000 x1 + 60, 000 x2 –18, 000 y1 subject to

b) x1 = 6, x2 = 1, Z = 310 6. a) Maximize Z = 600x1 + 540x2 + 375x3 subject to

x1 + x2 ≤ 10 10,000x1 + 7,000x2 ≤ 72,000

x1 + x2 + x3 ≤ 12

x1 – 10y1 ≤ 0

x1 ≤ 5

x1,x2 ≥ 0 and integer

80x1 + 70x2 + 50x3 ≤ 750

y1 = 0 or 1

x1,x2,x3 ≥ 0 and integer

b) x1 = 0, x2 = 10, y1 = 0, Z = $600,000

b) x1 = 0, x2 = 10, x3 = 1, Z = 5,775

12.

7. a) Maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35

Maximize Z = $.36x1 + .82x2 + .29x3 + .16x4 + .56x5 + .61x6 + .48x7 + .41x8 subject to 60x1 + 110x2 + 53x3 + 47x4 +

3x1 + 2x2 ≤ 20

92x5 + 85x6 + 73x7 + 65x8 ≤ 300

x1,x2 ≥ 0 and integer

15. a) x1 = tv ads

7x1 + 9x2 + 8x3 + 4x4 + 7x5 +

x2 = newspaper ads

6x6 + 8x7 + 5x8 ≤ 40

x3 = radio ads

x2 – x5 ≤ 0

minimize Z = $25,000x1 + 7,000x2 + 9,000x3 subject to

xi = 0 or 1 Z = $1.99 million; x1 = 0, x2 = 1, x3 = 0, x4 = 0, 13.

x5 = 1, x6 = 1, x7 = 0

53, 000 x1 + 30, 000 x2 + 41, 000 x3 ≥ 200, 000

xi = no. of employees assigned to time period i, i = 1, 2, . . , 6 (time period 1 = 12:00 midnight – 4:00 A.M.; period 2 = 4:00–8:00 A.M.; etc.)

≥ 1.5 (21, 000 x1 + 10, 000 x2 + 23, 000 x3 )

32, 000 x1 + 20, 000 x2 + 18, 000 x3

34, 000 x1 + 12, 000 x2 + 24, 000 x3

≥ .60 (53, 000 x1 + 30, 000 x2 + 41, 000 x3 )

Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 subject to

x1 ,x2 ,x3 ≥ 0 and integer

x6 + x1 ≥ 90

Integer solution:

x1 + x2 ≥ 215

x1 = 4

x2 + x3 ≥ 250

x2 = 0

x3 + x4 ≥ 65

x3 = 0

x4 + x5 ≥ 300

Z = $99,999.99

x5 + x6 ≥ 125

b) Non-integer solution:

xi ≥ 0

14.

x1 = 2.9275

x1 = 90, x2 = 250, x3 = 0, x4 = 175, x5 = 125,

x2 = .9713

x6 = 0, Z = 640

x3 = .383

x1 = day contacts by phone

Z = $83,433.65

x2 = day contacts in person

16.

x3 = night contacts by phone

Maximize Z = 90x1 + 150x2 + 30x3 subject to 2x1 + 3x2 + x3 ≤ 5

x4 = night contacts in person

Solution: Z = $240, x1 = 1, x2 = 1, x3 = 0

Maximize Z = $16x1 + 33x2 + 17x3 + 37x4 subject to

17.

x2 + x4 ≤ 575 6x1 + 13x2 ≤ 1,320

x1 = no. of salespeople to East, x2 = no. of salespeople to Midwest, x3= no. of salespeople to West Maximize Z = 25,000x1 + 18,000x2 + 31,000x3 subject to

7x3 + 19x4 ≤ 2,580 x1,x2,x3,x4 ≥ 0 and integer

x1 + x2 + x3 = 100

Integer solution:

5, 000 x1 + 11, 000 x2 + 7, 000 x3 ≤ 700, 000

x1 = 220

x1 ≥ 10

x3 = 368

x2 ≥ 10

Z = $9,776

x3 ≥ 10

The non-integer solution is:

x1 , x2 , x3 ≥ 0 and integer Solution: x1 = 20, x2 = 10, x3 = 70, Z = 2,850,000

x1 = 220 x3 = 368.57 Z = $9,785.71 The rounded-down solution is only slightly less (i.e., $9.71).

18.

20. a) Minimize Z = 5x1 + 10x2 + 8x3 + 12x4 + 7x5

xij = vehicles [1,000s shipped from plant i (i = 1, 2, 3, 4, 5) to warehouse j (j = A, B, C, D)], yi = plant i (i = 1,2,3,4,5) = 0 or 1

+ 10x6 + 8x7

subject to

minimize Z = 2,100,00y1 + 850,000y2 + 1,800,000y3

9 x1 + 6 x2 + 6 x3 + 3 x4 + 6 x5 + 3 x6 + 9 x7 ≥ 2.00 3( x1 + x2 + x3 + x4 + x5 + x6 + x7 )

+ 1,100,000y4 + 900,000y5 + 56x1A + 21x1B + 32x1C + 65x1D

3( x1 + x2 + x3 + x4 + x5 + x6 + x7 ) ≥ 12

+ 18x2A + 46x2B + 70x2C

x2 + x3 + x4 + x6 ≤ 2

+ 35x2D + 12x3A + 71x3B

x1 + x2 + x6 + x7 ≥ 3

+ 41x3C + 52x3D + 30x4A

xi = 0 or 1

+ 24x4B + 61x4C + 28x4D + 45x5A

b) x1 = 1 (Management I)

+ 50x5B + 26x5C + 31x5D

x2 = 1 (Principles of Accounting) x5 = 1 (Marketing Management)

subject to

x7 = 1 (English Literature)

12, 000 y1 − x1A − x1B − x1C − x1D = 0

Z = 30 hours per week

18, 000 y2 − x2A − x2B − x2C − x2D = 0

Minimum grade point average = 2.50

14, 000 y3 − x3A − x3B − x3C − x3D = 0

21. a) Maximize Z = 1,650x1 + 850x2 + 790x3 subject to

10, 000 y4 − x4A − x4B − x4C − x4D = 0 16, 000 y5 − x5A − x5B − x5C − x5D = 0

6.3x1 + 3.9x2 + 3.1x3 ≤ 125

x1A + x2A + x3A + x4A + x5A = 6, 000

17x1 + 10x2 + 7x3 ≤ 320

x1B + x2B + x3B + x4B + x5B = 14, 000

x1,x2,x3 ≥ 0 and integer

x1C + x2C + x3C + x4C + x5C = 8, 000

b) x1 = 10

x1D + x2D + x3D + x4D + x5D = 10, 000

x3 = 20

xij , yi ≥ 0

Z = 32,300

yi = 0 or 1

The relaxed, noninteger solution is,

Solution: y1, y4, y5 = 1,

x1 = 13.61

x1B = 12,000, x4A = 6,000,

x3 = 12.67

x4B = 2,000, x4D = 2,000,

Z = 32,460.46

x5C = 8,000, x5D = 8,000

The rounded-down solution is x1 = 13,

Z = $5,092,000 19.

x3 = 12, and Z = 30,930, which is not optimal.

Maximize Z = 12,100x1 + 8,700x2 + 10,500x3 subject to

22.

360x1 + 375x2 + 410x3 ≤ 30,000

Maximize Z = 575x1 + 120x2 subject to 40x1 + 15x2 ≤ 600

x1 + x2 + x3 ≤ 67

30x1 + 18x2 ≤ 480

14x1 + 10x2 + 18x3 ≤ 2,200

4x1 – x2 ≤ 0

x1/x3 ≥ 2

x1,x2 ≥ 0 and integer

x2/x1 ≥ 1.5

Optimal solution:

x1,x2,x3 ≥ 0 and integer

x1 = 4

Integer solution:

x2 = 20

x1 = 22

Z = 4,700

x2 = 33 x3 = 11 Z = $668,800

23.

17x1D + 19x2A + 15x2B + 22x2C + 18x2D + 20x3A + 20x3B + 17x3C + 19x3D + 24x4A + 21x4B + 16x4C + 23x4D + 22x5A + 19x5B + 21x5C + 21x5D) ≤ .04

Maximize Z = $575x1 + 120x2 + 65x3 subject to 40x1 + 15x2 + 4x3 ≤ 600 30x1 + 18x2 + 5x3 ≤ 480

x1A + x1B + x1C + x1D ≤ 1

4x1 – x2 ≤ 0

x2A + x2B + x2C + x2D ≤ 1

x3 = 20y1

x3A + x3B + x3C + x3D ≤ 1

x1,x2,x3 ≥ 0 and integer

x4A + x4B + x4C + x4D ≤ 1

y1 = 0 or 1

x5A + x5B + x5C + x5D ≤ 1

Or the last restriction that y1 = 0 or 1 can be included in the model as a constraint, y1 ≤ 1.

x1A + x2A + x3A + x4A + x5A = 1

Solution:

x1B + x2B + x3B + x4B + x5B = 1

x1 = 3

x1C + x2C + x3C + x4C + x5C = 1

x2 = 16

x1D + x2D + x3D + x4D + x5D = 1 xij = 0 or 1

x3 = 20 b) x1C = 1

y1 = 1

x3D = 1

Z = $4,945

x4B = 1

They should produce the batch of 20 stools since the profit is slightly greater ($4,945 vs. $4,700). 24.

x5A = 1 Z = 83 parts

x1 = bass boat

26.

x2 = ski boat x3 = speed boat

Minimize Z = 120x1 + 75x2 subject to 220x1 + 140x2 ≥ 6,300

Maximize Z = 20,500x1 + 12,000x2 + 22,300x3 subject to

x1 + x2 ≤ 32 .4x1 + .9x2 ≤ 15

1.3x1 + 1.0 x2 + 1.5 x3 ≤ 210

x1,x2 ≥ 0 and integer

x1 ≤2 ( x2 + x3 )

Non-integer solution: x1 = 25.1409

x1 + 2 x3 ≤ 160

x2 = 5.493

x1 , x2 , x3 ≥ 0 and integer

Z = $3,428.87

Solution:

Integer solution:

x1 = 110

x1 = 25

x2 = 31

x2 = 1

x3 = 24

Z = $3,435

Z = $3,162,200 25. a) Maximize Z = 18x1A + 20x1B +21x1C +17x1D

27.

+19 x2 A + 15 x2 B + 22 x2C + 18 x2 D + 20 x3 A +

Minimize Z = 120x1 + 75x2 + 4.50x3 subject to 220x1 + 140x2 + 12x3 ≥ 6,300

20 x3 B + 17 x3C + 19 x3 D + 24 x4 A + 21x4 B +

8x1 + 8x2 + x3 ≤ 256

16 x4C + 23x4 D + 22 x5 A + 19 x5 B + 21x5C +

.4x1 + .9x2 + .16x3 ≤ 15

21x5 D

x1,x2 ≥ 0 and integer

subject to (.3x1A + .9x1B + .6x1C + .4x1D + .8x2A + .5x2B + 1.1x2C + .7x2D + 1.1x3A + 1.3x3B + .6x3C + .8x3D + 1.2x4A + .8x4B + .6x4C + .9x4D + 1.0x5A + .9x5B + 1.0x5C + 1.0x5D)/(18x1A + 20x1B + 21x1C +

x3 ≥ 0

30.

Solution: x1 = 28

NE-D, NW-D, SE-C

x2 = 0 31.

x3 = 11.67 Z = $3,412.50 28.

The minimum number of command centers (Z) = 3 Minimize Z = ∑ mij xij where mij = mileage for teacher “i” to location “j”

Maximize Z = 3y1 + 2y2 + 1y3 +1y4 + 2y5

xij = teacher “i” assigned to course “j”

where yi = 1, if project i is selected

subject to

= 0, if project i is not selected

subject to

∑ xij = 1, for j = 1(MIT 125) to 10 (MIT 500)

$2,600,000x1 + 950,000x2 + 38,000x3 + 365,000x4 + 175,000x5 ≤ 30,000,000

1 ≤ ∑ xij ≤ 2, for i = 1(Abrahams ) to 8(Hampton)

17,500x1 + 8,600x2 + 25x3 + 1,700x4 + 900x5 ≥ 250,000

j =1

( ∑ sij xij ) / 10 ≥ 4.21

220,000x1 + 125,000x2 + 26,000x3 + 75,000x4 + 45,000x5 ≥ 4,000,000 400,000x1 + 150,000x2 + 34,000x3 + 1,200x4 + 55,000x5 ≥ 5,000,000 x1 ≤ 75 x2 ≤ 75 x3 ≤ 165 x4 ≤ 75 x5 ≤ 75

29.

where sij = evaluation scores for teacher “i” for course “j” xij = 0 or 1 The solution approach used here is to minimize mileage in the objective function and then iteratively increase the teaching evaluation score until the optimal mileage (667) increases. This results in a maximum average teaching score of 4.21. In this solution approach start with a “low” evaluation score (for example, “2.00”). This will result in the minimum mileage solution (of 667 miles) with a recomputed evaluation score of “4.01.” Next, “zero” out this solution and resolve with a new increased evaluation score of (perhaps) “4.05.” Continue to resolve the problem with small incremental increases in the evaluation score until the total mileage increases above the optimal value of 667 miles. This occurs when the evaluation score increases above “4.21,” hence this is the overall optimal solution.

x1 ≥ 1y1 x2 ≥ 3y2 x3 ≥ 10y3 x4 ≥ 2y4 x5 ≥ 6y5 xi = 0 and integer Solution: y1,y2,y3,y4,y5 = 1 x1 = 1 x2 = 26 x3 = 10 x4 = 2 x5 = 7 Z=9 Change objective function to:

Solution: x13 = 1 (Abrahams – MIT 250) x21 = 1 (Bray – MIT 125)

Maximize Z = 220,000x1 + 125,000x2 + 26,000x3 + 75,000x4 + 45,000x5

x32 = 1 (Clayton – MIT 225)

Delete cost savings constraint.

x49 = 1 (Dennis – MIT 450)

Solution:

x58 = 1 (Evans – MIT 425) x5,10 = 1 (Evans – MIT 500)

y2 = 1, y3 = 1

x66 = 1 (Farah – MIT 375)

x2 = 29, x3 = 64

x67 = 1 (Farah – MIT 400)

Z = $5,289,000

x74 = 1 (Gonzalez – MIT 300)

x85 = 1 (Hampton – MIT 325)

x69 + x79 = x9, 10 + x9, 12

Z (mileage) = 667

x7, 10 + x8, 10 + x9, 10 = x10, 12 + x10, 13 x5, 11 + x8, 11 = x11, 13

Avg. score = 4.21

x9, 12 + x10, 12 = x12, 13

32. a) Maximize Z = 80 x1 + 500 x2 + 420 x3

4x12 + 3x13 + 5(x14 + x24) + 5x25 + 7x36 + 6(x37 + x47) + 4(x48 + x58) + 8(x69 + x79) + 6(x7, 10 + x9, 10) + 6(x5, 11 + x8, 11) + 5(x9, 12 + x10, 12) ≥ 30

+ 300 x4 + 270 x5 + 210 x6 subject to x1 + x2 + x3 + x4 +x5 + x6 = 70 x6 + x1 ≥ 10

4x12 + 3x13 + 5(x14 + x24) + 5x25 + 7x36 + 6(x37 + x47) + 4(x48 + x58) + 8(x69 + x79) + 6(x7, 10 + x9, 10) + 6(x5, 11 + x8, 11) + 5(x9, 12 + x10, 12) ≤ 55

x1 + x2 ≥ 12 x2 + x3 ≥ 20 x3 + x4 ≥ 25

Solution:

x4 + x5 ≥ 32

x14 = 1, x47 = 1, x79 = 1, x9, 10 = 1, x10, 12 = 1, x12, 13 = 1

x5 + x6 ≥ 18 xi ≥ 0 and integer

Z = 373 miles

Where xi = the number of drivers who start their 8-hour shift in period i, i = 1, 2,…,6 (period 1 = midnight to 4:00 a.m.; period 2 = 4:00–8:00 a.m.; period 3 = 8:00 a.m.–noon; period 4 = noon–4:00 p.m.; period 5 = 4:00–8:00 p.m.; period 6 = 8:00 p.m.–midnight).

Solution without passenger restriction: x13 = 1, x37 = 1, x7, 10 = 1, x10, 13 = 1 Z = 323 miles 34.

Maximize Z = 4.6x1 + 3.7x2 + 5.9x3+ 9.1x4 + 7.2x5 + 4.3x6 + 2.7x7 + 5.2x8 + 8.1x9 + 7.5x10

Solution: x1 = 0, x2 = 27, x3 = 1; x4 = 24, x5 = 8, x6 = 10, Z = $25,380

Subject to 2.1x1 + 10x2 + 8.7x3 + 4.8x4 + 6.0x5 + 4.3x6 + 3.1x7 + 5.5x8 + 4.0x9 + 4.65x10 ≤ 34

b) Add the constraint, x6 + x1 + x2 ≤ 15

Solution: x1 = 7, x2 = 5, x3 = 23, x4 = 17, x5 = 15, x6 = 3, Z = 22,500

2.1x1 + 10.0x2 + 17.4x3 + 24.0x4 + 18.0x5 + 4.3x6 + 3.1x7 + 16.5x8 + 24.0x9 + 9.3x10 ≤ 55

Solution:

1x1 + 1x2 + 2x3 + 5x4 + 3x5 + 1x6 + 1x7 + 3x8 + 6x9 + 2x10

x1 = 7, x2 = 5, x3 = 20, x4 = 20, x5 = 15, x6 = 3, Z = 22,140 33.

≤ 2.5

∑x

xij = route from stop “i” to stop “j” (i = 1, 2, …., 12 and j = 2, 3, …., 13) Minimize Z =

x1, x2, …., x10 = 0 or 1 = player signed

mileage i → j) · xij

25x1 + 28x2 + 31x3 + 32x4 + 34x5 + 27x6 + 36x7 + 29x8 +28x9 + 31x10

Subject to:

∑x

x12 + x13 + x14 = 1

x12 = x24 + x25

x2 + x5 + x10 ≥ 1

x13 = x36 + x37

xn ≥ 0

x14 + x24 = x47 + x48

Solution:

x25 = x58 + x5,11

x1 (Fish) = 1

x36 = x69

x4 (Cabrera) = 1

x37 + x47 = x79 + x7,10

x6 (Longier) = 1

x48 + x58 = x8, 10 + x8, 11

≤ 31

x7 (Goldsmith) = 1

Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 subject to

36.

x10 (Fists, p) = 1 Z (WAR) = 28.2

Atlanta: x1 + x2 + x7 ≥ 1

Z average = 5.64 35.

Charlotte: x1 + x2 + x9 ≥ 1

Maximize Z = 14x11 + 18x12 + 23x13 + 9x21 + 11x22 + 15x23 + 18x31 + 23x32 + 27x33 + 16x41 + 21x42 + 25x43 + 12x51 + 16x52 + 22x53 + 21x61 + 23x62 + 28x63 subject to

Cincinnati: x3 + x4 + x5 + x6 + x7 + x8 ≥ 1 Cleveland: x3 + x4 + x5 + x8 ≥ 1 Indianapolis: x3 + x4 + x5 + x6 + x7 + x10 ≥ 1 Louisville: x3 + x5 + x6 + x7 + x10 ≥ 1

14x11 + 9x21 + 18x31 + 16x41 + 12x51 + 21x61 ≥ 20

Nashville: x1 + x3 + x5 + x6 + x7 + x10 ≥ 1

18x12 + 11x22 + 23x32 + 21x42 + 16x52 + 23x62 ≥ 25

Pittsburgh: x3 + x4 + x8 + x9 ≥ 1

23x13 + 15x23 + 27x33 + 25x43 + 22x53 + 28x63 ≥ 35

Richmond: x2 + x8 + x9 ≥ 1

x11 + x12 + x13 ≤ 1

St. Louis: x5 + x6 + x7 + x10 ≥ 1

x21 + x22 + x23 ≤ 1

276x1 + 253x2 + 394x3 + 408x4 + 282x5 + 365x6 + 268x7 + 323x8 + 385x9 + 298x10 ≤ 900

x31 + x32 + x33 ≤ 1 x41 + x42 + x43 ≤ 1

xi = 0 or 1

x51 + x52 + x53 ≤ 1

x61 + x62 + x63 ≤ 1

x1 (Atlanta) = 1 x7 (Nashville) = 1

xij = 0 or 1

x8 (Pittsburgh) = 1

Solution:

Z=3

x13 = 1

b) x4 (Cleveland) = 1

x22 = 1

x7 (Nashville) = 1

x32 = 1

x9 (Richmond) = 1

x43 = 1

Z=3

x53 = 1

Total cost of this solution without cost constraint is $1,061 million. Total cost of previous solution with cost restraint = $867,000; a $194,000 difference.

x61 = 1 Z = $125 million

37.

Maximize Z = 127x1 + 83x2 + 165x3 + 96x4 + 112x5 + 88x6 + 135x7 + 141x8 + 117x9 + 94x10 subject to

Solution: Arsenal – Bayside United – Roadside

x1 + x3 ≤ 1

Wildcats – Bates

x1 + x2 + x4 ≤ 1

Rage – Harriott

x4 + x5 + x6 ≤ 1

Rapids – Holiday

x6 + x7 + x8 ≤ 1

Storm – Harriott

x6 + x9 ≤ 1

Tigers – Bayside

x8 + x10 ≤ 1

Stars – Marks

x9 + x10 ≤ 1

Comets – Marks

xi = 0 or 1

Hurricanes – Bates

Solution:

Strikers – Holiday

x2 = 1

Bees – Harriott

x3 = 1

Z = 12

x5 = 1

b) Min Z = ∑ ∑ (Priority Level) i Xij

x8 = 1

i =1 j = A

x9 = 1 Z = $618,000

Add the constraint,

∑ ∑ Xij = 12

38. a) Max Z= ∑ ∑ X ij

i =1 j = A

subject to

Solution:

Holiday: 100( x1A + x2A + x3A + x4A + x7A + x8A + x9A

Arsenal – Bayside

+ x10A ) + 17x5A + 10 x6A + 20 x11A + 14x12A ≤ 41

United – Roadside

Roadside: 100( x1B + x3B + x7B + x10B ) + 18x2B +12x4B

Wildcats – Bates

+17x5B + 10 x6B + 18x8B +20 x9B +20 x11B +14x12B

Rage – Harriott

≤ 26 Bates: 15x1C + 18x2C + 20 x3C + 12x4C + 17x5C + 10 x6C

Rapids – Hampson Storm – Marks

+ 18x7C + 18x8C + 20 x9C + 16x10C + 20 x11C

Tigers – Bayside

+ 14x12C ≤ 38

Stars – Marks

Hampson: 100( x1D + x2D + x3D + x4D + x6D + x7D + x8D

Comets – Harriott

+ x9D + x10D + x11D ) + 17x5D + 14x12D ≤ 25

Hurricanes – Bates

Tilton: 100( x1E + x2E + x3E + x4E + x6E + x7E + x8E + x9E

Strikers – Holiday

+ x10E + x11E ) + 17x5E + 14x12E ≤ 26 Marks: 100( x1F + x2F + x3F + x7F + x10F ) + 12x4F + 17x5F

Bees – Tilton

+10 x6F + 18x8F + 20 x9F + 20 x11F + 14x12F ≤ 38

Z = 15

Bayside: 100( x3G + x10G ) + 15x1G + 18x2G + 12x4G + 17x5G

39.

+10 x6G + 18x7G + 18x8G + 20 x9G + 20 x11G + 14x12G ≤ 35 Harriott: 100( x1H + x2H + x3H + x7H + x10H ) + 12 x4H + 17 x5H + 10 x6H + 18 x8H + 20 x9H + 20 x11H + 14 x12H ≤ 52 H

The variables for this problem require the number of different 3-day sequences available for each company based on their earliest and latest start days. The different combinations with required daily hours are shown in the following table.

∑ Xij = 1 i = 1, 2,...,12

j =A

xij = 0 or 1

in order to ensure that the training sessions will occur on 3 consecutive days.

Day (j) Company

Schedule (i)

1 2

1 2 3 4 5 6 7 4 6 3

4 6 3

4 B

5 6 8

x12 − x23 = 0

3 2 5 3 2 5 7 1 1 7 1 1

1 3 6 1 3 6

15 16

19 20

8 5 2 5 5 5 5 5 5 5 5 5

5 5 5

6 3 3

26 H

A feasible solution/schedule is,

8 5 2

i =1

8 5 2

18 F

∑ x ≤ 20, where j = 1.

1 3 6

6 3 3

Company

Days

1,2,3

4,5,6

5,6,7

4,5,6

1,2,3

5,6,7

4,5,6

Z = 100 hours

In part (b) the objective function is changed to minimize the hours for days 6 and 7. This results in the following solution,

4 6 6

Finally, there are 7 constraints limiting the training hours per day to 20 man-hours. For day 1 this constraint would take the form,

1 3 6

i =1 j =1

7 1 1

∑∑ x = 3

7 1 1

11 12

There is also one constraint for the schedule for each company that insures 3 sessions will occur, resulting in 8 constraints. For company A, this constraint would take the form,

3 2 5

x11 − x12 = 0

4 6 3

7 C

This will result in 56 constraints. For example, referring to the table opposite, for company A, schedule 1, two constraints are required,

4 6 3

4 6 6

Company

Days

Thus, there are 196 (0–1) decision variables, xij, where i = 1,…,28 and j = 1,…,7.

1,2,3

The objective is not specified; however, if the objective is to minimize the total number of training hours, it will result in a solution that will provide a feasible schedule for the 84 training hours required, i.e., Z = 100 hours.

2,3,4

4,5,6

1,2,3

5,6,7

3,4,5

The constraints include corequisite constraints of the form, xij = xi,j + 1 = xi,j + 2

Z = 14 weekend hours.

40.

subject to

This is an example of the classic “traveling salesman” problem.

xA1 + xA2 = 1

Minimize Z = 10 x12 + 15 x13 + 20 x14 + 40 x15 + 10 x21 +

xB1 + xB2 = 1

12 x23 + 16 x24 + 24 x25 + 15 x31 +

xC1 + xC2 = 1

12 x32 + 10 x34 + 20 x35 + 20 x41 +

xD1 + xD2 = 1

16 x42 + 10 x43 + 16 x45 + 40 x51 +

.9xA1 + .5xA2 + .9xB1 + .7xB2 + .95xC1 + .4xC2 + .95xD1 + .6xD2 ≥ 3

24 x52 + 20 x53 + 16 x54

subject to

xij ≥ 0

x21 + x31 + x41 + x51 = 1

Solution:

x12 + x32 + x42 + x52 = 1

xA2 = 1

x13 + x23 + x43 + x53 = 1

xB2 = 1

x14 + x24 + x34 + x54 = 1

xC1 = 1

x15 + x25 + x35 + x45 = 1

xD1 = 1

x12 + x13 + x14 + x15 = 1 x21 + x23 + x24 + x25 = 1 x31 + x32 + x34 + x35 = 1

Z = $4,035 per month 42.

x41 + x42 + x43 + x45 = 1

yj = carrier “j”

x51 + x52 + x53 + x54 = 1

xij and yij = 0 or 1

u2 − u3 + 5 x23 ≤ 4

minimize Z = ∑ y j

u2 − u4 + 5 x24 ≤ 4

subject to

u2 − u5 + 5 x25 ≤ 4

∑ x = 1, for all i

u3 − u2 + 5 x32 ≤ 4

u3 − u4 + 5 x34 ≤ 4

∑ x ≤ 10, for all j

u3 − u5 + 5 x35 ≤ 4

u4 − u2 + 5 x42 ≤ 4

∑ x = 10 ij

u4 − u3 + 5 x43 ≤ 4

u4 − u5 + 5 x45 ≤ 4

∑ x i (papers delivered) ≤ (capacity) i y for all j ij

u5 − u2 + 5 x52 ≤ 4

u5 − u3 + 5 x53 ≤ 4

∑ x i (delivery time) ≤ 120 mins.

u5 − u4 + 5 x54 ≤ 4

Solution:

xij = 0 or 1

41.

xij = route “i” (where i = 1,2,...,10) assigned to carrier “j” (where j = A,B,C,D,E,F)

x1A = 1, x2E = 1, x3E = 1, x4A = 1,

Solution: 1 - 3 - 4 - 5 - 2 - 1

x5E = 1, x6E = 1, x7D = 1, x8D = 1,

Z = 75 miles

x9D = 1, x10A = 1, yA = 1, yD = 1, yE = 1

xij = current or new tenant where i = space A, B, C, D and j = 1 (current) and 2 (new) xij = 0 or 1

A papers delivered = 560 D papers delivered = 400 E papers delivered = 550

Maximize Z =.9(3,600)xA1 + .5(7,200)xA2 + .9(2,400)xB1 + .7(3,600)xB2 + .95(3,000)xC1 + .4(6,000)xC2 +

Z = 3 carriers 43.

Maximize Z = ∑ (profit)i i xij ij

.95(3,300)xD1 + .6(5,400)xD2

subject to

∑ (load-lbs i) i x ≤ 80, 000 lbs for j = A,B,C ij

∑ (load-ft i) i x ≤ 5,500 ft for j = A,B,C 3

45.

∑ (time i) i x ≤ 90 hrs. for all j = A,B,C

a) Maximize Z = ∑ (annual usage i ) i xi

∑ x ≤ 1, for i = 1,2,...,12

subject to

∑ (acreage i) i x ≤ 55

xij = 0 or 1 for customer shipment i (i = 1,

∑ (cost i) i x ≤ $550, 000

2,...,12) assigned to truck j (j = A,B,C)

Solution:

∑ (priority i) i x − ( ∑ x i1.75) ≤ 0 i

x1B = 1

x2A = 1

xi = 0 or 1

x3A = 1

Solution: football fields, playground, walking/running trails, and softball fields.

x4B = 1

Z = 123,500

x7C = 1

b) Minimize Z = ∑ xi (priority i )

x11C = 1

Z = $78, 000

44.

xi = 1 if facility “i” is selected, and “0” if it is not.

subject to

Minimize Z = ∑ xi

∑ (acreage i) i x ≤ 55 i

subject to

∑ (cost i) i x ≤ $550, 000 i

3rd Street: x1 + x5 ≥ 1

10th Street: x2 + x3 + x10 ≥ 1

∑ (expected usage i) i x ≥ 120, 000 i

South Street: x3 + x2 + x8 ≥ 1

xi = 0 or 1

Mulberry Avenue: x4 + x12 ≥ 1

Solution: soccer fields, playground, walking/running trails

Rose Street: x1 + x5 + x11 ≥ 1 Wisham Avenue: x6 + x9 + x10 ≥ 1

Z = 4.0 or 1.33 average priority

Richmond Road: x7 + x10 ≥ 1

c) Maximize Z = ∑ (acreage i ) i xi

Hill Street: x3 + x8 ≥ 1

23rd Avenue: x6 + x9 ≥ 1

subject to

Broad Street: x6 + x7 + x10 ≥ 1

∑ (acreage i) i x ≤ 55 i

Church Street: x5 + x11 + x12 ≥ 1

∑ (cost i) i x ≤ $550, 000

Beamer Boulevard: x4 + x11 + x12 ≥ 1

xi = 0 or 1

∑ (priority i) i x − ( ∑ x i1.75) ≤ 0

Solution:

x3 (South Street) = 1

xi = 0 or 1

x5 (Rose Street) = 1 x9 (23rd Avenue) = 1

Solution: rugby fields, soccer fields, walking/running trails

x10 (Broad Street) = 1

Z = 52 acres

x12 (Beamer Boulevard) = 1

Expected annual usage = 83,700

Z = 5 stores Total cost = $899

Solution without integer restrictions:

CASE SOLUTION: PM COMPUTER SERVICES Minimize Z = 1,280(n1 + n2 + n3 + n4 + n5 + n6) + 12(y1 + y2 + y3 + y4 + y5 + y6) + 200(h1 + h2 + h3 + h4 + h5 + h6) + 320(f1 + f2 + f3 + f4 + f5 + f6) + 15(I1 + I2 + I3 + I4 + I5 + I6) subject to x1 + y1 – I1 = 63

n1 = 5

x1 = 63.5

y1 = 3

n2 = 6.221

x2 = 79.016

y2 = 3.733

n3 = 6.221

x3 = 79.016

y3 = 3.733

n4 = 5.288

x4 = 67.16

y4 = 3.172

n5 = 5.288

x5 = 67.16

y5 = 3.172

n6 = 5.288

x6 = 67.16

y6 = 3.172

h1 = 1.22

f4 = .933

x2 + y2 + I1 – I2 = 74 h6 = 1

x3 + y3 + I2 – I3 = 95

I1 = 3.49

x4 + y4 + I3 – I4 = 57

I2 = 12.25

x5 + y5 + I4 – I5 = 68

I4 = 13.33

x6 + y6 + I5 – I6 = 86

I5 = 15.66 Z = 44,088

x1 ≤ 12.7n1 x2 ≤ 12.7n2

CASE SOLUTION: THE TENNESSEE PTERODACTYLS

x3 ≤ 12.7n3 x4 ≤ 12.7n4

xi = player i, i = 1,2,...,12

x5 ≤ 12.7n5

a) Minimize Z = $8.2x1 + 6.5x2 +5.2x3 + 16.4x4 + 14.3x5 + 23.5x6 + 4.7x7 + 7.1x8 + 15.8x9 +

x6 ≤ 12.7n6 y1 ≤ 0.6n1

26.4x10 + 19.5x11 + 8.6x12

y2 ≤ 0.6n2

subject to

y3 ≤ 0.6n3

x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 = 5

y4 ≤ 0.6n4 y5 ≤ 0.6n5

14.7x1 + 12.6x2 + 13.5x3 + 27.1x4 +18.1x5 + 22.8x6 + 9.3x7 + 10.2x8 + 16.9x9 + 28.5x10 + 24.8x11 + 11.3x12 ≥ 80

y6 ≤ 0.6n6 n1 – h1 + f1 = 5

4.4x1 + 10.6x2 + 8.7x3 + 7.1x4 + 7.5x5 + 9.5x6 + 12.2x7 + 12.6x8 + 2.5x9 + 6.5x10 + 8.6x11 + 12.5x12 ≥ 40

n2 – h2 + f2 – n1 = 0 n3 – h3 + f3 – n2 = 0 n4 – h4 + f4 – n3 = 0

9.3x1 + 2.1x2 + 1.7x3 + 4.5x4 + 5.1x5 + 2.4x6 + 3.5x7 + 1.8x8 + 11.4x9 + 1.3x10 + 6.9x11 + 3.2x12 ≥ 25

n5 – h5 + f5 – n4 = 0 n6 – h6 + f6 – n5 = 0

40.3x1 + 34.5x2 + 29.3x3 + 42.5x4 + 41.0x5 + 38.5x6 + 31.5x7 + 44.4x8 + 42.7x9 + 38.1x10 + 42.6x11 + 39.5x12 ≥ 190

xi,yi,ni,Ii,hi,fi ≥ 0 and integer Solution: n1 = 6 x1 = 74

y1 = 0 h1 = 1

f 4 = 1 I1 = 11

n2 = 6 x2 = 76

y2 = 3 h6 = 1 I 2 = 16

n3 = 6 x3 = 76

y3 = 3

I4 = 9

n4 = 6 x4 = 63

y4 = 3

I5 = 7

n6 = 6 x5 = 63

y5 = 3

x6 = 76

y6 = 3

x1 + x3 + x4 + x5 + x10 ≤ 3 x2 + x6 + x7 + x8 + x9 + x11 + x12 ≤ 2 xi = 0 or 1 Solution: x1 (Mack Madonna) = 1 Z = $52.2 million x4 (Ramon Dion) = 1

Z = 45, 065

x5 (Joe Eastcoast) = 1 x7 (Hiram Grant) = 1 x12 (Grant Hall) = 1

b) The team does not have enough cash ($50 million) to sign the projected players. Reformulating the model with an objective function of points per game and a constraint for salary with a constraint value of ≤$50 million results in an infeasible solution. The requirements must be reduced. Arbitrarily reducing the assists to 20 and minutes to 185 will achieve the following feasible solution:

y4 + y6 ≥ 1 y1 + y5 = 1 xij ≥ 0 and integer yi = 0 or 1 Solution: y1(Atlanta) = 1

x11 = 4

x26 = 1

x46 = 3

y2(Boston) = 1

x12 = 2

x37 = 12

x47 = 1

x1 (Mack Madonna) = 1

y6(St. Louis) = 1

x16 = 3

x41 = 5

x2 (Darrell Boards) = 1

y7(Washington) = 1

x17 = 5

x42 = 12

x3 (Silk Curry) = 1

Z = 18

x4 (Ramon Dion) = 1

Total cost = $13,618,000

x12 (Grant Hall) = 1

CASE SOLUTION: SCHEDULING TELEVISION ADVERTISING SLOTS AT THE UNITED BROADCAST NETWORK

Z = 79.2 points per game

CASE SOLUTION: NEW OFFICES AT ATLANTIC MANAGEMENT SYSTEMS

Because of the nature of the constraints, a maximization objective function is required; thus, the top ranked city, Washington, is given a value of “7,” etc.

Maximize Z = ∑∑ Psw ⋅ xsw ⋅ kw s =1 w =1

Where, Psw = performance score for show s, in week w

maximize Z = 5y1 + 4y2 + 3y3 + 1y4 + 6y5 + 2y6 + 7y7 subject to

xsw = ad slot for show s, in week w

19x11 + 32x12 + 27x13 + 14x14 + 23x15 + 14x16

k w = weighting factor for week w

+ 41x17 + 14x21 + 47x22 + 31x23 + 28x24 + 35x25

= 0 or 1 s = 1 (Bayside, 30-sec), 2 (Bayside,

+ 18x26 + 53x27 + 16x31 + 39x32 + 26x33 + 23x34

15-sec), 3 (Newsline, 30-sec),

+ 31x35 + 19x36 + 48x37 + 22x41 + 26x42 + 21x43

4 (Newsline, 15-sec),...,

+ 18x44 + 28x45 + 24x46 + 43x47 + 1,700y1 + 3,600y2 + 2,100y3 + 2,500y4 + 3,100y5 + 2,700y6 + 4,100y7 ≤ 14,000

14 (ER Doctor, 15-sec)

x11 + x21 + x31 + x41 = 9y1

w = week 1,2,...,6

subject to 6

∑ xsw ≤ available inventory, (s = 1,...,14)

x12 + x22 + x32 + x42 = 14y2

w=1 14

x13 + x23 + x33 + x43 = 8y3 x14 + x24 + x34 + x44 = 12y4

∑ xsw ≥ weekly minimum, (w = 1,..., 6)

x15 + x25 + x35 + x45 = 11y5

s =1 14

x16 + x26 + x36 + x46 = 7y6

∑ xsw ≤ weekly maximum, (w = 1,..., 6)

x17 + x27 + x37 + x47 = 18y7

s =1 14 6

x11 + x12 + x13 + x14 + x15 + x16 + x17 ≤ 24

∑ ∑ Csw ⋅ xsw ⋅ kw ≤ $600, 000

x21 + x22 + x23 + x24 + x25 + x26 + x27 ≤ 19

s =1 w=1

x31 + x32 + x33 + x34 + x35 + x36 + x37 ≤ 16

where Csw = cost of show “s,” during week “w”

x41 + x42 + x43 + x44 + x45 + x46 + x47 ≤ 21

x3w + x4 w + x5 w + x6 w + x11w + x12 w 14

∑ ∑ xsw s =1 w=1

≥ .50

xsw + xs +1, w ≤ 1 where s = 1, 3, 5, 7, 9, 11, 13; w = 1, 2, …6

subject to 36

∑x = 7

Solution:

j =1

Newsline, 30-sec, October week 4 = 1

∑x = 7

Newsline, 30-sec, November week 2 = 1

j =1

Newsline, 30-sec, November week 3 = 1

∑x =7

Newsline, 30-sec, November week 4 = 1

j =1

Newsline, 15-sec, November week 1 = 1

∑x = 7

Cops & Lawyers, 30-sec, October week 3 = 1

j =1

Cops & Lawyers, 30-sec, November week 1 = 1

Cops & Lawyers, 30-sec, November week 3 = 1 Cops & Lawyers, 30-sec, November week 4=1

6.0 ≤

∑x j =1

6.0 ≤

Friday Night Football, 15-sec, November week 2 = 1

∑x j =1

6.0 ≤

Total Performance Score = 2,082.40

6.0 ≤

CASE SOLUTION: THE DRAPERTON PARKS AND RECREATION DEPARTMENT’S FORMATION OF GIRLS’ BASKETBALL TEAMS

∑x j =1

j =1

≤7

i =1

xij ≥ 0 xij = 0 or 1

xij = player j, where j = 1,2,…,36, assigned to team i, where i = 1, 2, 3, or 4

j =1

≤7

∑ x ≤ 1.0, for j = 1,2,...,36

xij = 0 or 1

≤7

ER Doctor, 15-sec, November week 4 = 1

j =1

Friday Night Football, 15-sec, November week 4 = 1

≤7

Cops & Lawyers, 15-sec, October week 4=1

maximize Z = ∑ (score j ) x1 j + ∑ (score j ) x2 j + ∑ (score j ) x3 j + ∑ (score j ) x4 j

Solution: Team 2

Team 1 Player

Score

Player

Team 3

Score

Team 4

Player

Score

Player

Score

Average = 7

Average = 6.3

Average = 7

Z = 191

Average = 7

∑x iI ≤ 2 ij

The teams seem to be competitive on paper; they might be made more equal by narrowing the range for the average evaluation score for each team.

∑ x i FIN ≤ 2 ( j = A, B, C, D, E, F) ij

The players who were cut were 7, 8, 20, 21, 22, 29, 30, and 31. None had a score above “3,” and the lowest score of any player assigned to a team was “4.” Therefore, it does not appear that any player was unfairly cut.

(FINi = 0 if student i is not a Finance major, 1 if yes) 18

∑ x i MKTG ≤ 2 ij

∑ x i MGT ≤ 2 ij

CASE SOLUTION: DEVELOPING PROJECT TEAMS FOR MANAGEMENT 4394

∑ x i ACCT ≤ 2 ij

Maximize Z = ∑ (GPA) j / 6, j = team A, B, C, D, E, F j

18 ⎛ F ⎡ 18 ⎜ ∑ ⎢ ∑ (xiA i GPA iA ) / 3 + ∑ (xiB i GPA iB ) / 3 + i ⎝ j ⎣ i

subject to

∑ (x i GPA ) / 3 + ∑ ( x i GPA ) / 3 + iC

∑ xij = 1 (i = 1, 2, 3, ..., 18)

⎤⎞

∑ ( x i GPA ) / 3 + ∑ ( x i GPA ) / 3⎥⎦ ⎟ /6

∑ xij = 3 ( j = A, B, C, D, E, F)

i 18

= AVG GPA ≥ 2.80 (GPAij = grade point average for student i on team j)

xij ≥ 0 and binary

∑ xij i Fi ≥ 1 ( j = A, B, C, D, E, F)

Solution:

(Fi = 0 if student i is male, and 1 if student i is female)

Team

Student

Average GPA

∑ xij i Fi ≤ 2

2,3,6

2.85

12,17,18

2.86

4,14,15

2.82

7,8,11

2.96

1,5,10

3.12

9,13,16

2.93

∑ x i I ≥ 1 ( j = A, B, C, D, E, F) ij

(Ii = 0 if student i is not international, 1 if yes)

⎠

Overall Average GPA = 2.92

X 6 + X11 ≤ 1

The teams seem relatively equitable and diverse but this is an objective assessment by the student.

X 6 + X16 ≤ 1

CASE SOLUTION: SCHEDULING THE LEAD BALLOON’S SUMMER TOUR

X10 + X16 ≤ 1

X8 + X10 ≤ 1 X8 + X11 ≤ 1 X11 + X16 ≤ 1

Maximize Z = arena ij i capacityi

Every week requires a set of constraints like this. (See Excel Solution for all constraints).

where i = city i, i = 1 to 16

Solution:

j = week j, j = 1 to 12

Atlanta – July week 3

subject to

Boston – July week 2 Cincinnati – August week 1

∑ xij ≤ 1, i = 1 to 16

Charlotte – May week 4

j =1

Cleverland – June Week 1

∑ x ≤ 2, j = 1 to 12 i =1

Detroit – June week 4

D.C. – July week 1

∑ x ≥ 1, j = 1 to 12 i =1

Philadelphia – June week 2

∑∑ x ≤ 16

Dallas – June week 3

For each week, there must be a set of constraints that eliminates combinations of two cities that are greater than 500 miles apart. The student needs to determine these combinations using a map like a road atlas. As an example, consider the first possible tour week, the 3rd week in May. This week requires the following constraints (where the variables are xi1)

Orlando – May week 3

Miami – May week 3

Houston – June week 3 Indianapolis – June week 1 Nashville – May week 4 Chicago – August week 2 New York – July week 4 Total arena capacity = 314,500

X 3 + X8 ≤ 1 (for example, Cincinnati and

Philadelphia are > 500 mi.) CASE SOLUTION: DEVELOPING X 3 + X10 ≤ 1 (Cincinnati and Miami are > 500 mi.) KATHLEEN TAYLOR’S 401(K) PLAN

Maximize Z = ( ∑ ri i xi ) / 1800

X 3 + X11 ≤ 1 X 3 + X16 ≤ 1

where ri = average 5-year return for fund “i”

X4 + X6 ≤ 1

xi = 0 or 1, if fund “i” is selected

X 4 + X8 ≤ 1 X 4 + X10 ≤ 1 X 4 + X11 ≤ 1 X 4 + X16 ≤ 1 X 6 + X8 ≤ 1 X 6 + X10 ≤ 1

Z = Average 5-year return = 7.47 subject to

Average Evening Star rating = 3.8

∑x =5

Average fund size = $10,097 million

Average expense ratio = 1.04

∑ xi = 1, for international funds 1

Changing the objective function to maximize the Evening Star rating would achieve the same rating but decrease the 5-year return.

∑ xi = 1, for small -cap funds 6

∑ x = 1, for mid -cap funds i

CASE SOLUTION: THE BASEBALL BALLPARK TOUR

∑ x = 1, for large-cap funds i

This is a “travelling salesman” problem similar to problem 40.

∑ xi = 1, for bond funds 20

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

invested ∑ y = 1800, where y = inamount fund "i" i

yi ≤ 1800xi for all “i” funds 5

90 ≤ ∑ yi ≤ 630 1 9

90 ≤ ∑ yi ≤ 450 6

Cincinnati Pittsburgh New York Chicago Dallas Miami Minneapolis St. Louis

xij = city “i” to city “j” (i = 1, 2, … 8 and j = 1, 2, …, 8)

90 ≤ ∑ yi ≤ 630 10

ui = (i = 1, 2, …., 8)

Mniimize Z =

360 ≤ ∑ yi ≤ 900 24

90 ≤ ∑ yi ≤ 180

Subject to

∑ ESR i x ≥ 3.7

∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (j = 1, …, 8) ∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8)

where ESRi = Evening Star Rating for all “i” funds

( ∑ FS i x ) / 6 ≥ 10, 000 i

where FSi = “i” fund size

( ∑ ER i x ) / 6 ≥ 1.10 i

i, j

∑( mileage ij) (x )

where ERi = expense ratio for fund “i”

xi = 0 or 1

yi ≥ 0 Solution:

x4 (Admiral Foreign Investor) = 1; y4 = $630 x6 (Maxam Small Cap Return) = 1; y6 = $90

x11 (T. Row Price Growth) = 1; y11 = $90

x17 (Draper Strategic Growth) = 1; y17 = $900 x21 (Parham High Yield) = 1; y21 = $90

∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8) ∑x = 1 (i = 1, …, 8)

Analysis:

u2 – u3 + 8x23 ≤ 7

u5 – u6 + 8x56 ≤ 7

u2 – u4 + 8x24 ≤ 7

u5 – u7+ 8x57 ≤ 7

u2 – u5 + 8x25 ≤ 7

u5 – u8 + 8x58 ≤ 7

u2 – u6 + 8x26 ≤ 7

u6 – u2 + 8x62 ≤ 7

u2 – u7 + 8x27 ≤ 7

u6 – u3 + 8x63 ≤ 7

u2 – u8 + 8x28 ≤ 7

u6 – u4 + 8x64 ≤ 7

u3 – u2 + 8x32 ≤ 7

u6 – u5 + 8x65 ≤ 7

u3 – u4 + 8x34 ≤ 7

u6 – u7 + 8x67 ≤ 7

u3 – u5 + 8x35 ≤ 7

u6 – u8 + 8x68 ≤ 7

u3 – u6 + 8x36 ≤ 7

u7 – u2 + 8x72 ≤ 7

u3 – u7 + 8x37 ≤ 7

u7 – u3 + 8x73 ≤ 7

u3 – u8 + 8x38 ≤ 7

u7 – u4+ 8x74 ≤ 7

u4 – u2 + 8x42 ≤ 7

u7 – u5+ 8x75 ≤ 7

u4 – u3 + 8x43 ≤ 7

u7 – u6 + 8x76 ≤ 7

u4 – u5 + 8x45 ≤ 7

u7 – u8 + 8x78 ≤ 7

u4 – u6 + 8x46 ≤ 7

u8 – u2 + 8x82 ≤ 7

u4 – u7 + 8x47 ≤ 7

u8 – u3 + 8x83 ≤ 7

u4 – u8 + 8x48 ≤ 7

u8 – u4 + 8x84 ≤ 7

u5 – u2 + 8x52 ≤ 7

u8 – u5 + 8x85 ≤ 7

u5 – u3 + 8x53 ≤ 7

u8 – u6 + 8x86 ≤ 7

u5 – u4 + 8x54 ≤ 7

u8 – u7 + 8x87 ≤ 7

7. 8.

If they leave Cincinnati Saturday morning they can easily get to Chicago (4.5 hours) for a 7:00 PM game. Leaving Chicago early Sunday morning they can get to Minneapolis (6.5 hours) for a 7:00 PM game, even changing time zones. Leaving Minneapolis early Monday morning they can get to St. Louis (8.5 hours) in time for a 7:00 PM game. Leaving St. Louis early Tuesday morning they can get to Dallas (9.5 hours) for a 7:00 PM game. Leaving Dallas on Wednesday morning they cannot make it to Miami (14.0 hours) in time for the game, especially given that they lose an hour changing time zones, so a day of rest, and they go to the Thursday night game Leaving early Friday morning to drive to New York (18.5 hours) they cannot arrive in time for a 7:00 PM game, so they go to the Saturday night game. Leaving New York Sunday morning they can get to Pittsburgh (6.0 hours) in time for the 7:00 PM game. If they leave Pittsburgh after the game (around 11:00 PM) they can get back to Cincinnati around 4:00 AM, in time for Chuck to go to work Monday morning.

The trip is “feasible,” but barely.

Solution: x14 (Cincinnati to Chicago) = 1, 4.5 hours x47 (Chicago to Minneapolis) = 1, 6.5 hours x78 (Minneapolis to St. Louis) = 1, 8.5 hours x85 (St. Louis to Dallas) = 1, 9.5 hours x56 (Dallas to Miami) = 1, 14.0 hours x63 (Miami to New York) = 1, 18.5 hours x32 (New York to Pittsburgh) = 1, 6.0 hours x21 (Pittsburgh to Cincinnati) = 1, 4.5 hours Z = 5,135 miles

Chapter Six: Transportation, Transshipment, and Assignment Problems PROBLEM SUMMARY

37. Balanced transportation 38. Transshipment

1. Balanced transportation

39. Transshipment

2. Unbalanced transportation

40. Transshipment

3. Unbalanced transportation

41. Transshipment

4. Unbalanced transportation

42. Transshipment

5. Unbalanced transportation, multiple optimal

43. Transshipment

6. Sensitivity analysis (6–5)

44. Transshipment

7. Unbalanced transportation, multiple optimal

45. Transshipment (6–44)

8. Unbalanced transportation

46. Transshipment

9. Unbalanced transportation

47. Assignment/Transportation, LP formulation

10. Balanced transportation

48. Sensitivity analysis (6–47)

11. Balanced transportation

49. Transshipment

12. Sensitivity analysis (6–11)

50. Assignment

13. Unbalanced transportation, multiple optimal

51. Assignment, LP formulation

14. Sensitivity analysis (6–13)

52. Assignment

15. Shortage costs (6–13)

53. Unbalanced assignment, multiple optimal

16. Unbalanced transportation

54. Assignment, multiple optimal

17. Unbalanced transportation, multiple optimal

55. Assignment

18. Balanced transportation

56. Unbalanced assignment, multiple optimal

19. Unbalanced transportation, multiple optimal

57. Assignment or transportation

20. Sensitivity analysis (6–19)

58. Prohibited routes (6–55)

21. Unbalanced transportation

59. Unbalanced assignment

22. Sensitivity analysis (6–21)

60. Unbalanced assignment

23. Sensitivity analysis (6–21)

61. Unbalanced assignment

24. Unbalanced transportation

62. Unbalanced assignment (maximization)

25. Unbalanced transportation

63. Unbalanced assignment

26. Unbalanced transportation

64. Assignment, multiple optimal

27. Balanced transportation (6–26)

65. Unbalanced assignment

28. Balanced transportation (6–27)

66. Unbalanced assignment

29. Balanced transportation (6–28)

67. Assignment

30. Unbalanced transportation; production scheduling

68. Assignment 69. Assignment (6–68)

31. Unbalanced transportation (6–30)

70. Assignment

32. Unbalanced transportation

71. Assignment

33. Sensitivity analysis (6–32)

72. Sensitivity analysis (6–71)

34. Shortage costs 35. Multiperiod scheduling 36. Balanced transportation

6-1 .

PROBLEM SOLUTIONS 1.

subject to xA1 + xA2 + xA3 + xA4 ≤ 150

x13 = 2

xB1 + xB2 + xB3 + xB4 ≤ 210

x14 = 10

xC1 + xC2 + xC3 + xC4 ≤ 320

x22 = 9

xA1 + xB1 + xC1 = 130

x23 = 8

xA2 + xB2 + xC2 = 70

x31 = 10

xA3 + xB3 + xC3 = 180

x32 = 1

xA4 + xB4 + xC4 = 240

Z = 20,200 2.

xij ≥ 0

Minimize Z = 6xA1 + 9xA2 + 100xA3 + 12xB1

xA2 = 70

+ 3xB2 + 5xB3 + 4xC1 + 8xC2

xA4 = 80

+ 11xC3

xB1 = 50

subject to

xB4 = 160

xA1 + xA2 + xA3 ≤ 130

xC1 = 80

xB1 + xB2 + xB3 ≤ 70

xC3 = 180

xC1 + xC2 + xC3 ≤ 100

Z = $8,260

xA1 + xB1 + xC1 = 80

There is no effect. The Gary mill has 60 tons left over as surplus with the current solution to Problem 5. Reducing the capacity at Gary by 30 still leaves a surplus of 30 tons.

Minimize Z = 100xA1 + 10xA2 + 5xA3 + 12xB1

xA2 + xB2 + xC2 = 110 xA3 + xB3 + xC3 = 60 xij ≥ 0 xA2 = 80 xB2 = 10

+ 9xB2 + 4xB3 + 7xC1 + 3xC2

xB3 = 60

+ 11xC3 + 9xD1 + 5xD2 + 7xD3

xC1 = 80

subject to

xC2 = 20

xA1 + xA2 + xA3 = 90

Z = $1,530 3.

xB1 + xB2 + xB3 = 50

x11 = 70

xC1 + xC2 + xC3 = 80

x13 = 20

xD1 + xD2 + xD3 = 60

x22 = 10

xA1 + xB1 + xC1 + xD1 ≤ 120

x23 = 20

xA2 + xB2 + xC2 + xD2 ≤ 100

x32 = 100

xA3 + xB3 + xC3 + xD3 ≤ 110

Z = $1,240 4.

xij ≥ 0

xA2 = 20 xA3 = 60

xA3 = 90

xB2 = 70

xB1 = 30

xC1 = 80

xB3 = 20

xC2 = 20

xC2 = 80

Z = $1,290

xD1 = 40

Minimize Z = 14xA1 + 9xA2 + 16xA3 + 18xA4 + 11xB1 + 8xB2 + 100xB3 + 16xB4 + 16xC1 + 12xC2 + 10xC3 + 22xC4

xD2 = 20 Z = $1,590 multiple optimal solutions

6-2 .

x3B = 25

Minimize Z = 9xTN + 14xTP + 12xTC + 17xTB + 11xMN + 10xMP + 100xMC + 10xMB + 12xFN + 8xFP + 15xFC + 7xFB subject to

x3D = 25 Z = $13,200 11.

xA4 = 950

xTN + xTP + xTC + xTB ≤ 200

xA6 = 750

xMN + xMP + xMC + xMB ≤ 200 xFN + xFP + xFC + xFB ≤ 200

xB1 = 1,600 xB3 = 1,500

xTN + xMN + xFN = 130

xB5 = 1,250

xTP + xMP + xFP = 170

xB6 = 650

xTC + xMC + xFC = 100

Z = $3,292.50

xTB + xMB + xFB = 150 xij ≥ 0 Tampa - NY = 100

12.

No effect

13.

x1B = 250 x1D = 170

Tampa - Chicago = 100

x2A = 520

Miami - NY = 30

x2C = 90

Miami - Philadelphia = 120

x3C = 130

Fresno - Philadelphia = 50

x3D = 210

Fresno - Boston = 150

Z = $21,930

Z = $5,080 9.

14. (1) x1B = 250

Minimize Z = 7x1A + 8x1B + 5x1C + 6x2A + 100x2B + 6x2C + 10x3A + 4x3B + 5x3C + 3x4A + 9x4B + 100x4C subject to x1A + x1B + x1C ≤ 5

(2)

x1B = 250

x1D = 350

x1D = 50

x2A = 520

x2C = 90

x3C = 310

x3D = 30

x3D = 30 x4D = 300

x2A + x2B + x2C ≤ 25

Z = $29,130

x3A + x3B + x3C ≤ 20

Z = $24,930

Select alternative 2; add a warehouse at Charlotte

x4A + x4B + x4C ≤ 25 x1A + x2A + x3A + x4A = 10

15.

x1B = 250

x1B + x2B + x3B + x4B = 20

x1D = 170

x1C + x2C + x3C + x4C = 15

x2A = 520

xij ≥ 0

10.

xA2 = 1,800

x2C = 90

x1C = 5

x3C = 130

x2C = 10

x3D = 210

x3B = 20

x4C = 180

x4A = 10

Z = $26,430

Z = $195

Total transportation cost = $21,930

x1A = 70

Total shortage cost = $4,500

x2B = 25 x2C = 90 x3A = 10

6-3 .

16.

17.

21.

SC - 1 to VA - T = 7

GA - 1 to VA - SW = 10

FL - 1 to NC - E = 9

A - 4 = 18

GA - 2 to NC - SW = 6

FL - 1 to NC - W = 6

B - 3 = 12

GA - 2 to VA - C = 4

FL - 2 to VA - C = 5

B - 5 = 27

SC - 1 to NC - SW = 1

Z = $841,000

D-3=5

SC - 1 to NC - P = 6

D - 6 = 35

x1B = 60

E - 1 = 25

x2A = 45

E - 2 = 15

x2B = 25

E-3=4

x2C = 35

Z = $1,528 (multiple optimal) 22.

x3B = 5 Z = $1,605 18.

x11 = 30

x54 = 10

x12 = 5

x55 = 30

x14 = 2

x63 = 6

x22 = 20

x64 = 2

x33 = 14

x66 = 20

x44 = 26 Z = 364 miles 19.

North A = 250 South B = 300

If Easy Time purchased all the baby food demanded at each store from the distributor, total profit would be $1,246, which is less than buying it from the other locations as determined in problem 23. This profit is computed by multiplying the profit at each store by the demand. In order to determine if some of the demand should be met by the distributor, a new source (F) must be added to problem 23. This source represents the distributor and has an available supply of 150 cases, the total demand from all the stores. The new optimal solution is shown as follows. A-3=8

South C = 40

A - 4 = 18

East A = 150

B - 2 = 13

East C = 160

B - 5 = 27

West D = 210

D - 6 = 35

Central B = 100

E - 1 = 25

Central D = 190

E-2=2

Z = 20,700 min. 20.

A-3=8

GA - 1 to NC - W = 2

F - 3 = 22

North A = 250

Z = $1,545

South B = 200 South C = 140

23.

Solve the model as a linear programming model to obtain the shadow prices. Among the 5 purchase locations, the store at Albany has the highest shadow price of $3. The sensitivity range for supply at Albany is 25 ≤ q1 ≤ 43. Thus, as much as 17 additional cases can be purchased from Albany which would increase profit by $51 for a total of $1,579.

24.

Charlotte - Atlanta = 30

East A = 100 East C = 210 West D = 210 Central B = 150 Central D = 140 Z = 21,200 min. The overall travel time increased by 500 minutes, which divided by all 1,400 students is only an increase of .357 minutes per student. This does not seem to be a significantly large increase.

Memphis - St. Louis = 30 Louisville - NY = 30 Z = 159,000

6-4 .

25.

26.

30.

1-C=2 1-E=5

Jan - Feb = 30

2 - C = 10

Feb - Feb = 230

3-E=5

Feb - March = 50

4-D=8

Feb - June = 20

5-A=9

March - March = 290

6-B=6

March - May = 10

Z = 1,275

April - April = 210

Arkansas – Charleston = 12,000 bales

April - May = 90

Mississippi – Savannah = 19,000 bales

May - May = 300

Mississippi – New Orleans = 14,000 bales

June - June = 300

Texas – Houston = 26,000 bales

Z = $1,803,750 31.

Z = $1,197,000 27.

RF - Feb = 300 OF - Feb = 20

Savannah – Karachi = 10,000 bales

OF - March = 120

Savannah – Saigon = 1,000 bales

RM - March = 180

New Orleans – Saigon = 14,000 bales

RM - April = 120

Charleston – Karachi = 12,000 bales

OM - March = 200

Z = $1,748,000

RA - April = 300

Shanghai – China = 5,425,000 yards

OA - April = 200

Shanghai – Japan = 1,675,000 yards

RM - May = 300

Shanghai – Turkey = 3,950,000 yards

OM - May = 130

Karachi – India = 6,350,000 yards

RJ - June = 300

Karachi – Turkey = 800,000 yards

OJ - June = 80

Saigon – Japan = 1,775,000 yards

Z = $3,010,040

Saigon – Italy = 3,100,000 yards

32.

Z = $882,950 29.

RJ - Jan = 300 OJ - Jan = 110

Houston – Shanghai = 26,000 bales Savannah – Shanghai = 8,000 bales

28.

Jan - Jan = 180

Sacramento - St. Paul = 13 Sacramento - Topeka = 5

China – New York = 3,149,000

Bakersfield - Denver = 8

China – New Orleans -= 468,000

Bakersfield - St. Paul = 2

India – New York = 3,978,000

San Antonio - Topeka = 10

India – Marseilles = 255,000

Montgomery - Denver = 12

Japan – Bristol = 368,000

Jacksonville - Akron = 15

Japan – New Orleans = 1,932,000

Jacksonville - Topeka = 5

Turkey – Marseilles = 3,167,000

Ocala - Louisville = 15

Italy – Bristol = 2,067,000

Z = $278,000

Z = $6,850,280

It is cheaper for National Foods to continue to operate its own trucking firm.

6-5 .

33.

Increasing the supply at Sacramento, Jacksonville, and Ocala to 25 tons would have little effect, reducing the overall monthly shipping cost to $276,000, which is still higher than the $245,000 the company is currently spending with its own trucks.

34.

L.A. - Singapore = 150 L.A. - Taipei = 300 Savannah - Hong Kong = 400 Savannah - Taipei = 200 Galveston - Singapore = 350 Order shortage in Hong Kong = 200

Alternatively, increasing the supply at San Antonio and Montgomery to 25 tons per month reduces the monthly shipping cost to $242,500, which is less than the company’s cost with its own trucks.

Z = $723,500 Penalty cost = $160,000

35. Period of Production Period of Use Beginning Inventory Regular 1

300 8,700

Overtime

300 300

9,000

1,000

Subcontract 2

3,000

Regular

10,000

Overtime

700

10,000 800

1,500

200

3,000

Regular

12,000

Overtime

2,000

Subcontract

1,000

Subcontract 3

Capacity

2,000

3,000

Regular

12,000

Overtime

2,000

Subcontract

3,000

Demand

9,000

12,000

Z = $1,198,500 (multiple optimal solutions)

6-6 .

16,000

19,000

36.

Solution:

38.

Portland − Omaha = 724

x59 (NY - Chicago) = 50

Portland – Kansas City = 376

x26 (Marseilles - Savannah) = 63

Fresno – Omaha = 55

x35 (Liverpool - NY) = 37

Fresno – Topeka = 665

x48 (Norfolk - St. Louis) = 42

Long Beach – Omaha = 446

x15 (Hamburg - NY) = 13

Long Beach – Tucson = 60

x67 (Savannah - Dallas) = 60

Long Beach – Denver = 281

x68 (Savannah - St. Louis) = 3

Long Beach – Wichita = 663

Z = $77,362

Salt Lake City − Denver = 980

HND = 38 HNS = 17

El Paso − Tucson = 650

MSD = 22

Houston − Topeka = 1,025 St. Louis − Memphis = 479 St. Louis − Kansas City = 851 Chicago − Milwaukee = 974 Chicago − Minneapolis = 301 Z = $149,777 37.

x14 (Hamburg - Norfolk) = 42

Al - Eagles (2) and Bengals (5) Barbara - Saints (5) and Jets (1) Carol - Cowboys (1) and Packers (2) Dave - Redskins (1) and Cardinals (7) Z = 24 Multiple optimal solutions exist Carol seems to have received the best allocation but overall the allocation seems relatively fair.

6-7 .

39.

x16 (Mexico - Houston) = 18

40. a) x14 = 72

x46 = 75

x24 (Puerto Rico - Miami) = 11

x25 = 105

x47 = 80

x34 (Haiti - Miami) = 23

x34 = 83

x56 = 15

x47 (Miami - NY) = 20

x58 = 90

x48 (Miami - St. Louis) = 12

Z = $10,043,000

x49 (Miami - LA) = 2 x69 (Houston - LA) = 18 Z = $479 or $479,000

6-8 .

Adding a capacity constraint at plants in Indiana and Georgia x14 = 72

x46 = 40

x25 = 105

x47 = 80

x34 = 48

x56 = 50

x35 = 35

x58 = 90

Z = $10,043,000 41.

x1C = 70

xBA = 10

x2B = 80

xCB = 30

x3A = 50 Z = 1,490 or $14,900

6-9 .

42.

x37 (Italy - Texas) = 2.1

x510 + x610 + x710 = 1,100

x15 (Germany - Mexico) = 5.2

x611 + x711 = 1,500

x26 (Belgium - Panama) = 6.3

x25 + x35 + x45 ≤ x58 + x59 + x510 x16 + x26 + x36 + x46 ≤ x68 + x69 + x610 + x611

x59 (Mexico - Ohio) = 5.2 x68 (Panama - Virginia) = 3.7

x17 + x27 + x37 + x47 ≤ x78 + x79 + x710 + x711

x69 (Panama - Ohio) = 2.6

xij ≥ 0

Z = $27.12 million 43.

Solution:

xij = potatoes shipped (in bushels) from farm i (where i = 1, 2, 3, 4) to distribution center j (where j = 5, 6, 7) and from distribution center i (where i = 5, 6, 7) to plant j (where j = 8, 9, 10, 11) Minimize Z = 1.09x16 + 1.26x17 + .89x25 + 1.32x26 + 1.17x27 + .78x35

x16 = 1,600 x27 = 1,100 x35 = 1,400 x46 = 600 x59 = 900

+ 1.22x36 + 1.36x37 + 1.19x45

x510 = 500

+ 1.25x46 + 1.42x47 + 4.56x58

x68 = 1,200

+ 3.98x59 + 4.94x510 + 3.43x68

x610 = 600 x611 = 400

+ 5.74x69 + 4.65x610 + 5.01x611

x711 = 1,100

+ 5.39x78 + 6.35x79 + 5.70x710

Z = $25,192

+ 4.87x711

44.

subject to x16 + x17 ≤ 1,600 x25 + x26 + x27 ≤ 1,100 x35 + x36 + x37 ≤ 1,400 x45 + x46 + x47 ≤ 1,900

xij = containers shipped from European port i (where i = 1, 2, 3) to U.S. Port j (where j = 4, 5, 6, 7); from U.S. Port i (where i = 4, 5, 6, 7) to Inland Port j (where j = 8, 9, 10); from Inland Port i (where i = 8, 9, 10) to distribution center j (where j = 11, 12, 13, 14, 15) Minimize Z = 1,725x14 + 1,800x15 + 2,345x16 + 2,700x17 + 1,825x24 + 1,750x25 + 1,945x26 + 2,320x27 + 2,060x34 + 2,175x35 + 2,050x36 + 2,475x37 + 825x48 + 545x49 + 320x410 + 750x58 + 675x59 + 450x510 + 325x68 + 605x69 + 690x610 + 270x78 + 510x79 + 1,050x710 + 450x811 + 830x812 + 565x813 +

x25 + x35 + x45 ≤ 1,800 x16 + x26 + x36 + x46 ≤ 2,200 x17 + x27 + x37 + x47 ≤ 1,600 x58 + x68 + x78 = 1,200 x59 + x69 + x79 = 900

6-10 .

420x814 + 960x815 + 880x911 + 520x912 + 450x913 + 380x914 + 660x915 + 1,350x1011 + 390x1012 + 1,200x1013 + 450x1014 + 310x1015

45. a) The model formulation is the same as problem 44 except the available shipments from the European ports are 5 each:

subject to x14 + x15 + x16 + x17 ≤ 125 x24 + x25 + x26 + x27 ≤ 210 x34 + x35 + x36 + x37 ≤ 160 x48 + x49 + x410 ≤ 85 x58 + x59 + x510 ≤ 110 x68 + x69 + x610 ≤ 100 x78 + x79 + x710 ≤ 130 x48 + x58 + x68 + x78 ≤ 170 x49 + x59 + x69 + x79 ≤ 240 x 410 + x510 + x610 + x710 ≤ 140 x811 + x911 + x1011 = 85 x812 + x912 + x 1012 = 60 x813 + x913 + x 1013 = 105 x814 + x914 + x 1014 = 50 x815 + x915 + x 1015 = 120 x14 + x24 + x34 = x48 + x49 + x410 x15 + x25 + x35 = x58 + x59 + x510 x16 + x26 + x36 = x68 + x69 + x610 x17 + x27 + x37 = x78 + x79 + x710 x48 + x58 + x68 + x78 = x811 + x812 + x813 + x814 + x815 x49 + x59 + x69 + x79 = x911 + x912 + x913 + x914 + x915 x410 + x510 + x610 + x710 = x1011 + x1012 + x1013 + x1014 + x1015 xij ≥ 0 Solution: x14 = 85 x15 = 40 x25 = 70 x26 = 15 x27 = 125 x36 = 85 x410 = 85 x59 = 55 x510 = 55 x68 = 100 x78 = 70 x79 = 55 x811 = 85 x813 = 35

x14 + x15 + x16 + x17 ≤ 5 x24 + x25 + x26 + x27 ≤ 5 x34 + x35 + x36 + x37 ≤ 5 and the demand constraints for the U.S. distributors are x811 + x911 + x1011 = 1 x812 + x912 + x1012 = 1 x813 + x913 + x1013 = 1 x814 + x914 + x1014 = 1 x815 + x915 + x1015 = 1 Solution: x24 = 5 x49 = 2 x410 = 3 x911 = 1 x913 = 1 x1012 = 1 x1014 = 1 x1015 = 1 Z = 144 days Tucson: Hamburg – Norfolk – Kansas City (31) Denver: Hamburg – Norfolk – Front Royal (28) Pittsburgh: Hamburg – Norfolk – Kansas City (29) Nashville: Hamburg – Norfolk – Front Royal (27) Cleveland: Hamburg – Norfolk – Front Royal (29) b) x14 = 2 x24 = 3 x49 = 2 x410 = 3 x911 = 1 x913 = 1 x 1012 = 1

x814 = 50

x 1014 = 1

x912 = 40

x 1015 = 1

x913 = 70

Z = 154 days

x1012 = 20 x1015 = 120 Z = $1,179,400

6-11 .

46.

x19 = Brazil − Savannah = 5.2

∑ xiH = 520 yH

x1, 10 = Brazil − Jacksonville = 1.4

i= A

x28 = Colombia − New Orleans = 0.3

∑ xiI = 490 yI

x29 = Colombia − Savannah = 2.9

i= A

x38 = Indonesia − New Orleans = 4.1

x47 = Kenya − Galveston = 1.2

∑ xiJ = 310 yJ

x48 = Kenya − New Orleans = 4.6

i= A

x5, 10 = Cote d’Ivoire − Jacksonville = 1.7

∑ xiK = 410 yK

x6, 10 = Guatemala − Jacksonville = 3.6

i= A

x = $2,105,600 47.

∑ xiL = 605 yL

Minimize Z = ∑ cij xij + ∑ c jk x jk + ∑ vi y j

i= A

where cij = shipping cost from Asian port “i”(i = A, B, C, D, E) to distribution enter “j” (j = F, G, H, I, J, K, L)

∑ x jM = 440

j=F L

xij = containers shipped from Asian port “i” to distribution center “j”

∑ x jN = 305

j=F

cjk = shipping cost from distribution center “j” (j = F, G, H, I, J, K, L) to U.S. port “k” (k = M, N, O, P)

∑ x jO = 190

j=F

xjk = containers shipped from distribution center “j” to U.S. port “k”

xAH (Hong Kong – Antwerp) = 235

xHO (Antwerp – Miami) = 190

xBI (Shanghai – Bremen) = 170

xHP (Antwerp – New Orleans) = 330

xCJ (Busan – Valencia) = 165

xIM (Bremen – New York) = 165

∑ xij = 235

xDH (Mumbai – Antwerp) = 285

xIN (Bremen – Savannah) = 305

xDJ (Mumbai – Valencia) = 40

xJN (Valencia – New York) = 275

xEI (Kaoshiung – Bremen) = 300

xJP (Valencia – New Orleans) = 35

j= f

xEJ (Kaoshiung – Valencia) = 105

vj = cost of distribution center “j” yj = 0 if distribution center “j” is selected, 1 if not selected subject to L

j= f

∑ xBj = 170 j= f

∑ xcj = 165 L

Z = $47,986,050

j= f

Shipping cost = $6,215,050

∑ xDj = 325 L

∑ xEj = 405 j= f E

∑ xiF = 565 yF

i= A E

∑ xiG = 485 yG

i= A

6-12 .

48.

yF (Rotterdam) = 1

49.

Arkansas – Charleston = 12,000 bales

yH (Antwerp) = 1

Mississippi – Savannah = 19,000 bales

yK (Lisbon) = 1

Mississippi – New Orleans = 14,000 bales

xAH (Hong Kong – Antwerp) = 235

xFM (Rotterdam – New York) = 395

Texas – Houston = 26,000 bales

xBK (Shanghai – Lisbon) = 170

xFO (Rotterdam – Miami) = 70

Savannah – Shanghai = 8,000 bales

xCK (Busan – Lisbon) = 165

xHN (Antwerp – Savannah) = 305

Savannah – Saigon = 1,000 bales

xDF (Mumbai – Rotterdam) = 60

xHO (Antwerp – Miami) = 120

Charleston – Karachi = 12,000 bales

xDH (Mumbai – Antwerp) = 190

xKM (Lisbon – New York) = 45

xDK (Mumbai – Lisbon) = 75

xKP (Lisbon – New Orleans) = 365

Houston – Shanghai = 26,000 bales Savannah – Karachi = 10,000 bales New Orleans – Saigon = 14,000 bales Z = $2,945,000 50.

2-A 3-B

xEF (Kaoshiung – Rotterdam) = 405

4-D Z = 37 min.

Z = total shipping cost = $5,843,715 Distribution center cost = $44,302,000 Total cost: $50,145,715 Shipping cost is reduced

51. a) 1 - B

2-D 3-C 4-A

∑ x jP = 365

Z = $32

j=F E

1-C

b) Minimize Z = 12x1A + 11x1B + 8x1C + 14x1D

∑ xiF = ∑ xFk

8x2D

+ 10x2A + 9x2B + 10x2C +

i= A

k=M

i= A

k=M

∑ xiH = ∑ xHk

x1A + x1B + x1C + x1D = 1

i= A

k=M

x2A + x2B + x2C + x2D = 1

x3A + x3B + x3C + x3D = 1

i= A

k=M

i= A

k=M

x1C + x2C + x3C + x4C = 1

x1D + x2D + x3D + x4D = 1

i= A

k=M

xij ≥ 0

i= A

k=M

+ 14x3A + 8x3B + 7x3C + 11x3D

∑ xiG = ∑ xGk

+ 6x4A + 8x4B + 10x4C + 9x4D subject to

∑ xiI = ∑ xIk

x4A + x4B + x4C + x4D = 1 x1A + x2A + x3A + x4A = 1

∑ xiJ = ∑ xJk

x1B + x2B + x3B + x4B = 1

∑ xiK = ∑ xKk 52.

∑ xiL = ∑ xLk

1-B 2-D 3-A

Solution: yH (Antwerp) = 1 yI (Bremen) = 1 yJ (Valencia) = 1

4-C 5-E Z = 51 days

6-13 .

53.

1-B

1-E

60.

2-E

2-A

2-D

3-A

3-B

4-C

4-A

5-D

5-C

6-F

Average score = 94.5

Z = $36 54.

1-C

61.

Debbie - Breaststroke

2-A

Erin - Freestyle

3-B

Fay - Butterfly

4-D

4-C

Z = 10.61 minutes 62.

Employee 3 - Jewelry

2-F

Employee 4 - Appliances

3-E

Employee 6 - China 63.

5-D

58.

C-1

Z = 85 defects

D-1 B-2

E-3

B-2

C-5

F-3

D-1

D-3

G-2

E-5

E-1

H-1

F-4

Z = $1,070 (multiple optimal solutions)

A-3

64.

1-B

1, 4 and 7 - Columbia

2-A

2-C

2, 6 and 8 - Athens

3-F

3, 5 and 9 - Nashville

4-D

Z = 985 (multiple optimal solutions)

5-C

5-A

3, 6 and 7 - Athens

6-E

1, 2 and 8 - Columbia

Z = 36 nights 65.

4, 5 and 9 - Nashville 59.

Z = $1,930 A-2

6-B

Z = 14 miles 57.

Employee 2 - Home Furnishings

1-C

4-A

56.

Annie - Backstroke

1-D

Z = $26 55.

1 - “Grades exams”

A - NJ

Z = $1,220

B - PA

Solution Summary:

C - NY

Al’s – Parents’ Brunch

D - FL

Bon Apetít – Post game Party

E - GA

Bon Apetít – Lettermen’s Dinner

F - FL

Divine – Booster Club Luncheon

G - VA

Epicurean – Contributors’ Dinner

Z = 498

University – Alumni Brunch

Average success rate = 71.1%

Total Cost (Z) = $103,800

6-14 .

A preference table based on scores of “1” for most preferred sections, “2” for next preferred, etc. can be developed as follows. The “X’s” should be translated as large numbers relative to the preference scores.

66.

Time Course

11M

11T

12M

12T

14M

14T

Math

History

English

Biology

Spanish

Psychology

a) Math 12T

∑ xi2 = 4

History 11T

i= A

English 9T

∑ xi3 = 3

Biology 12M Spanish 11M

i= A

Psychology 14T

xij = 0 or 1 Solution:

Z = 10

xA2 = 1

b) Math 12 T

67.

History 14M

xB1 = 1

English 9T

xC1 = 1

Biology 12M

xD2 = 1

Spanish 8T

xE1 = 1

Psychology 14T

xF1 = 1

Z = 15

xG2 = 1

It’s not possible to develop a complete schedule (i.e., feasible solution) for either Monday and Wednesday or Tuesday and Thursday.

xH3 = 1 xI3 = 1 xJ2 = 1 xK1 = 1

Maximize Z = 2 (3xA1 + 2xA2 + 1xA3) + 5 (3xB1 + 1xB2 + 2xB3) + 7 (3xC1 + 2xC2 + 1xC3) + 1 (1xD1 + 3xD2 + 2xD3) + . . . etc.

xL3 = 1 Z = 104

subject to:

68.

Minimize Z = ∑ ∑ (mileageij ) xij i

∑ xij = 1, for i = A, B, C

where xij = truck “i” assigned to customer “j”

j =1

subject to:

∑ xi1 = 5

∑ xij = 1, for all j = 1, . . . , L

i= A

6-15 .

∑ xij ≤ 1, for all i = 1, . . . , 8

∑ xij = 1, for j = 1 to 12

i =1

xij ≥ 0

xij = 0 or 1

Solution:

x1G = 1

Umpire

x2B = 1

x4L = 1

Not assigned

Add the constraint:

∑∑ ((capacityij ) xij )/8 ≥ 85

x1G = 1

Not assigned

x2B = 1

x5K = 1

x6E = 1

Not assigned

x6F = 1 x7D = 1 x8A = 1 Total mileage = 4,260

Solution:

x3D = 1 x4L = 1

x7A = 1 x8C = 1 Total mileage = 4,420 Mileage not significantly different; only 160 additional miles. 70.

Match

x3H = 1 x5K = 1

69.

Z = 155 71.

Minimize Z = ∑ ri p j xij

Minimize Z = ∑ mij xij

where mij = travel distance for team “i” to site “j”

where xij = 1 if umpire “i” is assigned to match “j”; 0 if umpire is not assigned ri = rating of umpire “i”

xij = 1 if team “i” is assigned to site “j”, 0 otherwise

pj = priority of match “j”

subject to

(note priorities must be numerically reversed in this formulation such that the priority 1 match has a value of 12, priority 2 has a value of 11, priority 3 has a value of 10 and so on.)

∑ xij = 1, for i = 1, 2, ..., 16 j =1

∑ xij = 2, for j = 1, ..., 8 i =1

subject to

∑ ri xij = 17, for j = 1, ..., 8

∑ xij ≤ 1, for i = 1 to 15

j =1

6-16 .

where ri = rank for team “i”

CASE SOLUTION: THE DEPARTMENT OF MANAGEMENT SCIENCE AND INFORMATION TECHNOLOGY AT TECH

17 = combinations of ranks for all pairs, i.e., 1 + 16 = 17, 2 + 15 = 17, 3 + 14 = 17, etc. Solution:

72.

x11 (Jackets − 1) = 1

x44(Tigers − 4) =1

x37(Knights − 7) = 1

x16,1(Panthers − 1) = 1

x13,4(Bears − 4) =1

x14,7(Hawks − 7) = 1

x22 (Big Red − 2) = 1

x85(Blue Devils − 5) = 1

x68(Wasps − 8) = 1

x15, 2(Lions − 2) = 1

x95(Cavaliers − 5) = 1

x11,8(Eagles − 8) = 1

x53(Bulldogs − 3) = 1

x76(Blue Jays − 6) = 1

Z = 3,168 miles

x12,3(Beavers − 3) = 1

x10,6(Rams − 6) =1

The problem is formulated as a transportation model with the ten faculty as the source, each with a supply of 2 sections, and with the eight courses as the destinations with demand of either 1, 2, or 3 sections. The solution generated by QM for Windows results in the following teaching schedule (however, there are multiple optimal solutions).

In order to achieve ranking flexibility, we developed constraints for pair ranking sum within a range of 14 to 20 (instead of exactly equaling 17): 16

14 ≤ ∑ ri xij ≤ 20 i =1

Clayton

–

2 sections of 3444

Houck

–

3454 and 4434

Huang

–

3424 and 3444

Major

–

2 sections of 4434

Moore

–

3424 and 3434

Ragsdale

–

2 sections of 4444

Rakes

–

4444 and 4454

Rees

–

2 sections of 4454

Russell

–

2 sections of 3434

Sumichrast

–

2 sections of 4464

The total preference score (i.e., “Z”) is 23.0. Since there are 10 faculty this is an average score of 2.3 and the minimum possible preference score is 2.0, which means everyone is getting a preferable schedule.

However, the student might establish a different range. This changed the assignments as follows: x31(Knights − 1) = 1

x85(Blue Devils − 5) = 1

x13,1(Bears − 1) = 1

x12,5(Beavers − 5) = 1

x22(Big Red − 2) = 1

x76(Blue Jays − 6) = 1

x15,2(Lions − 2) = 1

x96(Cavaliers − 6) = 1

x53(Bulldogs − 3) = 1

x17(Jackets − 7) = 1

x10,3(Rams − 3) = 1

x14, 7(Hawks − 7) = 1

x44(Tigers − 4) = 1 x16,4(Panthers − 4) = 1

CASE SOLUTION: STATELINE SHIPPING AND TRANSPORT COMPANY The total cost of this solution is $2,630 per week. There are multiple optimal solutions. The solution is summarized as follows. 1. Kingsport

→ 2. Danville = 16 bbls

1. Kingsport

→ 3. Macon = 19 bbls

2. Danville

→ B. Los Canos = 80 bbls

x68(Wasps − 8) = 1

3. Macon

→ C. Duras = 78 bbls

x11,8(Eagles − 8) = 1

4. Selma

→ 3. Macon = 17 bbls

4. Selma

→ 5. Columbus = 36 bbls

5. Columbus

→ A. Whitewater = 65 bbls

6. Allentown

→ 2. Danville = 38 bbls

Z = 2,625 miles (reduced by 543 miles)

6-17 .

assignment problem must be solved with the five remaining applicants and the two available jobs. The tableau for this problem is 1. Carding

2. Spinning

Acuff

Ball

Davis

Gantry

Harper

The optimal solution offers the carding job to Davis and the spinning job to Gantry. If a third job was turned down then a new assignment problem with the five applicants and the three vacant jobs would be solved, and so on. The selection of the two best applicants of the remaining five to select for clerical positions is more of a common sense problem than an assignment problem. Since no test scores exist for the clerical positions and Brenda does not know (or has not told us) which areas the next supervisory jobs will be in we have no means of evaluation. The most logical approach for Brenda to take would simply be to total up all 5 test module scores for the remaining five candidates and keep the two applicants with the highest overall scores. The total scores for each applicant are,

CASE SOLUTION: BURLINGHAM TEXTILE COMPANY This assignment problem has ten sources and five destinations. Thus, if it is solved by hand it would require five dummy columns. It is also a maximization problem which would require all the scores to be subtracted from the highest score in the tableau (i.e. 102) and then minimized. The solution output for this assignment problem obtained using Excel is as follows. Applicant

Job

Test Score

Angela Coe

Carding

Fred Evans

Weaving

Bob Frank

Shipping

Mary Inchavelia

Spinning

102

Marilu Jones

Inspection

Applicant

Total Score

Acuff

379

Ball

384

Davis

442*

Gantry

433*

Harper

406

*The two best scores are for Davis and Gantry so Brenda should probably select these two people to keep.

476

CASE SOLUTION: THE GRAPHIC PALETTE

If one or more of the applicants do not accept the job offer, this new problem should be approached iteratively. For example, if the person selected for the supervisory position in carding, which according to the optimal solution is Angela Coe, turns the job down, then the remaining applicant with the best score for carding should be selected. This would be Maureen Davis with a score of 87. If both the applicants selected for the carding and spinning jobs turn them down, then a new

Minimize Z = 18x14 + 23x15 + 25x16 + 21x17 + 20x24 + 26x25 + 24x26 + 19x27 + 24x34 + 24x35 + 22x36 + 23x33 + 36x48 + 41x49 + 40x58 + 52x59 + 42x68 + 46x69 + 33x78 + 49x79

6-18 .

subject to x14 + x15 + x16 + x17 ≤ 750 x24 + x25 + x26 + x27 ≤ 900 x34 + x35 + x36 + x37 ≤ 670 x14 + x24 + x34 ≤ 530 x15 + x25 + x35 ≤ 320 x16 + x26 + x36 ≤ 450 x17 + x27 + x37 ≤ 250 x14 + x24 + x34 − x48 − x49 = 0 x15 + x25 + x35 − x58 − x59 = 0 x16 + x26 + x36 − x68 − x69 = 0 x17 + x27 + x37 − x78 − x79 = 0 x48 + x58 + x68 + x78 = 620 x49 + x59 + x69 + x79 = 750 xij ≥ 0 Solution: x14 = 530 x69 = 370 x15 = 220 x78 = 250 x27 = 250 x36 = 370 Z = $82,540 x48 = 150 x49 = 380 x58 = 220 Multiple optimal solutions exist.

6-19 .

CASE SOLUTION: SCHEDULING AT HAWK SYSTEMS, INC. a) Minimize Z = 110xSS + 122xSO + 136xSN + 152xSD + 170xSJ + 190xSF + 212xS,MR + 236xSA + 262xS,MY + 112xOO + 126xON + 142xOD + 160xOJ + 180xOF + 202xO,MR + 226xOA + 252xO,MY + 114xNN + 130xND + 148xNJ + 168xNF + 190xN,MR + 214xNA + 240xN,MY + 116xDD + 134xDJ + 154xDF + 176xD,MR + 200xDA + 226xD,MY + 108xJJ + 128xJF + 150xJ,MR + 174xJA + 200xJ,MY + 110xFF + 132xF,MR + 156xFA + 182xF,MY + 112xMR,MR + 136xMR,A + 162xMR,MY + 114xAA + 140xA,MY + 111xMY,MY

xNN = 420

xF,MR = 20

xND = 200

xFA = 280

xNF = 80

xMR,MR = 300 xAA = 300 xMY,MY = 120

Z = $497,000 (Multiple optimal solutions exist)

subject to xSS + xSO + xSN + xSD + xSJ + xSF + xS,MR + xSA+ xS,MY ≤ 700 xOO + xON + xOD + xOJ + xOF + xO,MR + xOA + xO,MY ≤ 700 xNN + xND + xNJ + xNF + xN,MR + xNA + xN,MY ≤ 700 xDD + xDJ + xDF + xD,MR + xDA + xD,MY ≤ 700 xJJ + xJF + xJ,MR + xJA + xJ,MY ≤ 300 xFF + xF,MR + xFA + xF,MY ≤ 300 xMR,MR + xMR,A + xMR,MY ≤ 300

If Hawk Systems meets all demand according to its schedule (in 1), it will sell 3,950 monitors at $180 apiece for revenues of $711,000. Assuming all costs are included in the model, it will be able to pay back the entire $200,000 and have a profit of $14,800 left over.

The schedule with the supply pattern of 500 per month results in a cost of $445,800. Therefore, the change costs Hawk Systems $51,200.

Miriam made $119,800 on her lease arrangement, so she did better by $19,800.

CASE SOLUTION: GLOBAL SHIPPING AT ERKEN APPAREL INTERNATIONAL European Port to U.S. Port

xAA + xA,MY ≤ 300

Lisbon - Jacksonville = 2,062.5 lb. goat

xMY,MY ≤ 300

Lisbon - Savannah = 2,000 lb. goat

xSS = 340

Lisbon - Jacksonville = 1,350 lb. lamb

xSO + xOO = 650

Marseilles - Savannah = 2,500 lb. goat

xSN + xON + xNN = 420

Marseilles - Savannah = 3,000 lb. lamb

xSD + xOD + xND + xDD = 200

Caracas - New Orleans = 4,200 lb. goat

xSJ + xOJ + xNJ + xDJ + xJJ = 660

Caracas - Jacksonville = 1,237.5 lb. goat

xSF + xOF + xNF + xDF + xJF + xFF = 550

Caracas - New Orleans = 2,375 lb. lamb

xS,MR + xO,MR + xN,MR + xD,MR + xJ,MR + xF,MR + xMR,MR = 390

Caracas - Jacksonville = 475 lb. lamb U.S. Port to Distribution Center

xSA + xOA + xNA + xDA + xJA + xFA + xMR,A + xAA = 580

New Orleans - NC = 4,200 lb. goat

xS,MY + xO,MY + xN,MY + xD,MY + xJ,MY + xF,MY + xMR,MY + xA,MY + xMY,MY = 120

New Orleans - NC = 2,375 lb. lamb

xij ≥ 0

Jacksonville - PA = 300 lb. goat

xSS = 340

xDJ = 310

xSF = 150

xDF = 320

xOO = 690

xD,MR = 70

xOJ = 50

xJJ = 300

Jacksonville - IN = 3,000 lb. goat Jacksonville - IN = 1,825 lb. lamb Savannah - PA = 4,500 lb. goat Savannah - IN = 125 lb. lamb

6-20 .

Z = 12 jobs assigned Total time = 37 hours

Savannah - PA = 2,875 lb. lamb Tanning Factory to Plant Mende - Limoges = 4,000 lb. goat Mende - Limoges = 4,400 lb. lamb Foggia- Naples = 3,000 lb. lamb Saragossa - Madrid = 6,500 lb. goat Saragossa - Madrid = 1,300 lb. lamb Feira - Sao Paulo = 5,100 lb. goat El Tigre - Caracas = 3,600 lb. goat El Tigre - Sao Paulo = 2,500 lb. lamb El Tigre - Caracas = 3,200 lb. lamb Plant to Port Madrid - Lisbon = 4,062.5 lb. goat Madrid - Lisbon = 650 lb. lamb Naples - Marseilles 1,500 lb. lamb Limoges - Marseilles = 2,500 lb. goat Limoges - Lisbon = 700 lb. lamb Limoges - Marseilles = 1,500 lb. lamb Sao Paulo - Caracas = 3,187.5 lb. goat Sao Paulo - Caracas = 1,250 lb. lamb Caracas - Caracas 2,250 lb. goat Caracas - Caracas 1,600 lb. lamb Total cost = $606,965.63

(b) Maximize Z = ∑ ( timeij) (xij) ij

Subject to 12

∑ (timej) (xij) ≤ 8, i = 1, 2, …., 6 j =1 6

∑ xij = 1, j = 1, 2, …., 12 i =1 12

∑ xij ≥ 1, i = 1, 2, …., 6 j =1

∑ xij = 12, i = 1, 2, …., 6 ij

Solution: x18 = 1, x1,10 = 1 x23, x2, 12 = 1 x32 = 1, x35 = 1 x46 = 1, x4, 11 = 1 x51 = 1, x54 = 1

CASE SOLUTION: “GIVE-BACK” WEEKEND

x69 = 1 Z = 32.5 hours

(a) xij = 0 or 1 = team “i” (i = 1, 2, …., 6) assigned to job “j” (j = 1, 2, …., 12) Maximize Z = ∑xij

Average hours per team = 5.08 hours The solution is improved. All jobs are assigned and the total time is less.

Subject to 12

∑ (timej)(xij) ≤ 8, i = 1, 2, …., 6 j =1

CASE SOLUTION: WEEMOW LAWN SERVICE

∑ xij = 1, j = 1, 2, …., 12 i =1

Minimize total job time: 3 N

Minimize Z = ∑∑ xij tij

∑ xij ≥ 1, i = 1, 2, …., 6

j =1

Xij ≥ 0

where xij = team i (i = 1, 2, 3) assigned to job j (where j = A, B, C, . . ., N)

Solution:

tij = time (minutes) for team i to complete job j

x16 = 1 x22 = 1, x28 = 1 x35 = 1, x3, 10 = 1, x3, 12 = 1 x47 = 1, x 4,11 = 1 x53 = 1, x54 = 1 x61 = 1, x69 = 1

subject to 3

∑ xij = 1, for j = A, B, C,...,N j

6-21 .

Minimize total cost:

∑ xij ⋅ tij ≤ 450 minutes, for i = 1, 2, 3 j

3 N

minimize Z = ∑∑ xij ⋅ kij

subject to

∑∑ xij ⋅ kij ≤ 1,000, where kij j

= cost for team i performing job j xij ≥ 0 and integer (binary)

∑ xij = 1, for j = A, B, C, . . ., N i

∑ xij ⋅ tij ≤ 450, for i= 1, 2, 3

Solution: Team

Jobs

A, D, G, H, J, K, N

E, F, L, M

B, C, I

Z = 1,153 minutes

Team 1 = 408 minutes

Cost = $1,000

Team 2 = 405 minutes

xij ≥ 0 and integer (binary) Solution:

Team 3 = 340 minutes

Team

Jobs

A, D, F, H, J, K, N

B, C, G, I

E, L, M

Z = $967 Total time = 1,148 minutes Team 1 = 391 minutes Team 2 = 370 minutes Team 3 = 387 minutes This solution provides for a more uniform distribution of minutes between the 3 teams. Whether the student believes this to be “better” or not depends on their reasoning.

6-22 .

Chapter Seven: Network Flow Models

PROBLEM SUMMARY 1.

Shortest route

10.

Shortest route

11.

Shortest route

12.

Shortest route

13.

Shortest route

14.

Shortest route

15.

Shortest route

16.

Shortest route

17.

Shortest route

18.

Shortest route (5–33)

19.

Minimal spanning tree

20.

Minimal spanning tree

21.

Minimal spanning tree

22.

Minimal spanning tree

23.

Minimal spanning tree

24.

Minimal spanning tree

25.

Minimal spanning tree

26.

Minimal spanning tree (7–9)

27.

Minimal spanning tree

28.

Minimal spanning tree

29.

Minimal spanning tree (7–16)

30.

Maximal flow

31.

Maximal flow

32.

Maximal flow

33.

Maximal flow

34.

Maximal flow

35.

Maximal flow

36.

Maximal flow

37.

Maximal flow

38.

Maximal flow (7–37)

39.

Maximal flow

40.

Maximal flow

41.

Maximal flow

42.

Maximal flow (6–44)

PROBLEM SOLUTIONS 1.

7-1 .

7-2 .

7-3 .

7-4 .

7-5 .

7-6 .

7-7 .

7-8 .

7-9 .

7-10 .

7-11 .

7-12 .

7-13 .

7-14 .

7-15 .

Solution Steps

Branch Added

Branch Distance

Total Distance from Origin

1–3

1–4

1–7

3–2

2–5

7–6

7–8

4–11

3–9

6–10

10–12

Shortest route path = 1–7–6–10–12

7-16 .

10. Step

Permanent Set

Branch Added

Distance

{1}

1–3

{1,3}

1–2

{1,2,3}

1–4

{1,2,3,4}

3–7

154

{1,2,3,4,7}

2–5

164

{1,2,3,4,5,7}

3–6

167

{1,2,3,4,5,6,7}

4–8

177

{1,2,3,4,5,6,7,8}

7–9

208

{1,2,3,4,5,6,7,8,9}

7–10

239

{1,2,3,4,5,6,7,8,9,10}

9–12

263

{1,2,3,4,5,6,7,8,9,10,12}

8–11

283

{1,2,3,4,5,6,7,8,9,10,11,12}

10–13

323

17. This is an example of the application of the shortest route method to solve the scheduled replacement problem. The branch costs are determined using the formula,

11. 1 – 4 – 7 – 8 = 42 miles 12. 1 – 4 – 7 – 10 – 12 – 16 – 17 = 14 days 13. 1 – 3 – 11 – 14 = 13 days

Nome: 1 – 2 – 8 – 12 – 13 = 9 hours

cij = maintenance cost for year i, i + 1,…, + cost of purchasing a new car at the beginning of year i – selling price of a used car at the beginning of year j.

Stebbins: 1 – 2 – 7 – 10 = 8.5 hours

c12 = 3 + 26 – 15 = 14

14. 1 – 3 – 5 – 12 – 16 – 20 – 22 = 114 15. Kotzebue: 1 – 2 – 8 – 11 – 15 = 10 hours

c13 = 3 + 4.5 + 26 – 12 = 21.5

16. (a) Shortest route solution from St. Louis.

c14 = 3 + 4.5 + 6 + 26 – 8 = 31.5

(1) St. Louis – (5) St. Joseph – (7) Ft. Kearney – (9) Ft. Laramie – (11) Ft. Bridger – (13) Ft. Hall – (15) Ft. Boise – (17) Ft. Vancouver = 186 days

c15 = 3 + 4.5 + 6 + 8 + 26 – 4 = 43.5 c16 = 3 + 4.5 + 6 + 8 + 11 + 26 – 2 = 56.5 c17 = 3 + 4.5 + 6 + 8 + 11 + 14 + 26 + 28.5 – 0 = 101

(b) Shortest route solution from Ft. Smith, Arkansas.

c23 = 3 + 26.5 – 15 = 14.5 c24 = 3 + 4.5 + 26.5 – 12 = 22

(2) Ft. Smith – (8) Ft. Vasquez – (11) Ft. Bridger – (13) Ft. Hall – (15) Ft. Boise – (17) Ft. Vancouver = 182 days

c25 = 3 + 4.5 + 6 + 26.5 – 8 = 32 c26 = 3 + 4.5 + 6 + 8 + 26.5 – 4 = 44 c27 = 3 + 4.5 + 6 + 8 + 11 + 26.5 – 2 = 57 c34 = 3 + 27 – 15 = 15 c35 = 3 + 4.5 + 27 – 12 = 22.5 c36 = 3 + 4.5 + 6 + 27 – 8 = 32.5

7-17 .

c45 = 3 + 27.5 – 15 = 15.5

c57 = 3 + 4.5 + 28 – 12 = 23.5

c46 = 3 + 4.5 + 27.5 – 12 = 23

c67 = 3 + 28.5 – 15 = 16.5

c47 = 3 + 4.5 + 6 + 27.5 – 8 = 33

Solution: 1 – 4 – 7 = $64.5 = $64,500

c56 = 3 + 28 – 15 = 16 A car should be sold at end of year 3 (beginning of year 4) and a new one purchased.

18.

xij = route from stop “i” to stop “j” (i = 1, 2, …., 12 and j = 2, 3, …., 13)

6(x7, 10 + x9, 10) + 6(x5, 11 + x8, 11) + 5(x9, 12 + x10, 12) ≤ 55

Minimize Z = ∑( mileage i → j) · xij

Solution: x14 = 1, x47 = 1, x79 = 1, x9, 10 = 1, x10, 12 = 1, x12, 13 = 1

i, j

Subject to:

Z = 373 miles

x12 + x13 + x14 = 1 x12 = x24 + x25

Solution without passenger restriction:

x13 = x36 + x37

x13 = 1, x37 = 1, x7, 10 = 1, x10, 13 = 1

x14 + x24 = x47 + x48

Z = 323 miles

x25 = x58 + x5,11

19.

x36 = x69 x37 + x47 = x79 + x7,10 x48 + x58 = x8, 10 + x8, 11 x69 + x79 = x9, 10 + x9, 12 x7, 10 + x8, 10 + x9, 10 = x10, 12 + x10, 13 x5, 11 + x8, 11 = x11, 13 x9, 12 + x10, 12 = x12, 13 4x12 + 3x13 + 5(x14 + x24) + 5x25 + 7x36 + 6(x37 + x47) + 4(x48 + x58) + 8(x69 + x79) + 6(x7, 10 + x9, 10) + 6(x5, 11 + x8, 11) + 5(x9, 12 + x10, 12) ≥ 30 4x12 + 3x13 + 5(x14 + x24) + 5x25 + 7x36 + 6(x37 + x47) + 4(x48 + x58) + 8(x69 + x79) +

7-18 .

1–3, 1–4, 2–3, 4–8, 5–6, 6–7, 7–8, 7–9, 9–10,

4.1 4.8 3.6 5.5 2.1 2.8 2.7 2.7 4.6 32.9 = 32,900 feet

20.

1–2 1–3 2–4 3–6 5–6 6–7 7–8 21.

1–3 2–3 3–4 4–6 5–6 5–8 6–7 22.

1–2 2–3 3–6 4–8 5–6 6–7 7–9 7–8 9–10

7-19 .

25.

23.

1–2 2–4

1–4

2–5

2–4

3–4

3–6

4–7

4–6

5–6

5–7

6–8

6–7

8–9

7–8

26.

24.

1–4 2–3 3–4 3–5 5–6 5–7 5–8

7-20 .

27.

Total sidewalk = 1,086 ft. 28.

1 – 2 = 48

29. 1 – 4 = 220

1 – 4 = 52

2 – 3 = 180

4 – 7 = 35

3 – 4 = 90

3 – 5 = 39

4 – 6 = 40

5 – 6 = 29

5 – 6 = 30

5 – 8 = 56

5 – 7 = 210

5 – 9 = 48

7 – 8 = 310

6 – 7 = 80 9 – 10 = 71

8 – 9 = 120

9 – 12 = 71

9 – 11 = 280

10 – 11 = 38

10 – 12 = 380

11 – 14 = 57

11 – 13 = 150

12 – 13 = 105

12 – 13 = 350

Total = 729

13 – 15 = 220 14 – 15 = 380 14 – 16 =160 15 – 17 = 430 Total miles = 3,550

30.

7-21 .

31.

7-22 .

32.

7-23 .

7-24 .

33.

7-25 .

7-26 .

34.

7-27 .

Maximal flow network:

35.

7-28 .

36. Allocation Branch in Which Total Capacity Is Used

Step

Path

Flow Amount

1–2–5–7–9

2–5

1–3–5–7–9

3–5

1–3–6–8–9

1–3, 3–6

1–4–6–8–9

4–6, 6–8, 8–9

Maximum flow = 13,000 cars

37.

7-29 .

solution

Allocation Flow Amount

x12 = 6

x45 = 8

1–2–5–7–10

x13 = 2

x57 = 13

1–2–5–8–10

x14 = 8

x58 = 4

1–3–5–9–10

x25 = 0

x59 = 1

1–3–5–2–6–8–10

x26 = 6

x68 = 2

1–3–5–2–6–9–10

x35 = 0

x69 = 6

1–3–6–9–10

x36 = 2

x7,10 = 3

Step

Path

x8,10 = 6

Maximum flow = 17,000 units 38.

x9,10 = 7

This problem is solved using Excel with the addition of a cost constraint to the normal linear programming formulation for a maximum flow problem, as follows,

Z = 16 = 16,000 units Total cost = 684 = $684,000 39.

Maximize Z = x10,1 subject to: x10,1 – x12 – x13 – x14 = 0 x12 – x25 – x26 = 0 x13 – x35 – x36 = 0 x14 – x45 – x46 = 0 x25 + x35 + x45 – x57 – x58 – x59 = 0 x26 + x36 – x68 – x69 = 0 x57 – x7,10 = 0 x58 + x68 – x8,10 = 0

Allocation

x59 + x69 – x9,10 = 0

Path

x7,10 + x8,10 + x9,10 – x10,1 = 0

1–2–5–8–10

1–4–7–10

1–3–6–9–10

Flow Amount

x 12 ≤ 7

x57 ≤ 3

x 13 ≤ 10

x58 ≤ 4

1–4–7–9–10

x 14 ≤ 8

x 59 ≤ 1

1–3–4–5–7–9–10

x 25 ≤ 9

x 68 ≤ 6

1–4–6–7–8–10

x 26 ≤ 6

x 69 ≤ 8

1–4–6–9–10

x 35 ≤ 7

x 7,10 ≤ 8

x 36 ≤ 5

x 8,10 ≤ 7

x 45 ≤ 10

x9,10 ≤ 7

Maximum flow = 32 Branch 1–2 1–3 1–4 2–4 2–5 3–4 3–6 4–5 4–6 4–7

x 10,1 ≤ 100 3(x12 + x13 + x14) + 5(x25 + x26) + 7(x35 + x36) + 4(x45) + 22(x57 + x58 + x59) + 19(x68 + x69) + 12x7,10 + 14x8,10 + 16x9,10 ≤ 700

7-30 .

Allocation 7 9 16 0 7 6 3 6 4 12

Branch 5–7 5–8 6–7 6–9 7–8 7–9 7–10 8–10 9–10

Allocation 6 7 3 4 3 13 5 10 17

40.

1–2–6–10–12–13–15 = 25

1–2 = 16

3–4 = 16

1–2–6–12–13–15 = 35

2–5 = 12

4–10 = 4

1–2–9–12–15 = 10

5–7 = 12

9–10 = 4

1–3–6–10–12–15 = 30

2–6 = 12

10–13 = 8

1–3–7–10–13–15 = 20

6–7 = 4

13–14 = 4

1–4–7–10–13–15 = 20

7–12 = 16

4–8 = 12

1–4–7–6–10–13–15 = 10

12–15 = 22

8–14 = 12

1–4–7–6–12–15 = 5

1–3 = 22

14–15 = 16

1–4–8–13–15 = 25

3–9 = 6

1–5–8–13–15 = 35

9–11 = 2

1–5–8–14–15 = 5

11–12 = 2

1–5–11–14–15 = 30

13–12 = 4

maximum flow = 250

Total traffic = 38,000 cars.

Branch 1–2 1–3 1–4 1–5 2–6 2–9 3–6 3–7 4–7 4–8 5–8 5–11 6–10 6–12 7–10

Allocation 70 50 60 70 25 45 30 20 35 25 40 30 65 5 40

Branch 7–6 8–13 8–14 9–12 10–12 10–13 11–14 12–13 12–15 13–15 14–15

41.

42.

Allocation 15 60 5 45 55 50 30 60 45 170 35

Gdansk – Galveston = 125 Hamburg – Jacksonville = 110 Hamburg – New Orleans = 95 Hamburg – Galveston = 5 Lisbon – Norfolk = 85 Norfolk – KC = 75 Norfolk – Dallas = 10 Jacksonville – FR = 70 Jacksonville – KC = 40 NO – FR = 95 Galveston – Dallas = 130 FR – Denver = 105 FR – Pittsburgh = 60 KC – Cleveland = 65 KC – Nashville = 50 Dallas – Tucson = 85 Dallas – Cleveland = 55

7-31 .

*CASE SOLUTION: THE PEARLSBURG RESCUE SQUAD The network for the Pearlsburg Rescue Squad follows.

The shortest route solution to each network node is as follows: 1–2 = 10 min l–4–6–8 = 29 min 1–3 = 15 min l–4–6–9 = 24 min 1–4 = 12 min 1–4–10 = 23 min 1–2–5 = 24 min l–4–6–8–11 = 35 min 1–4–6 = 19 min 1–4–6–9–12 = 35 min 1–2–7 = 30 min l–4–6–8–13 = 44 min

7-32 .

(10) Thionville – (9) Havange = 3

CASE SOLUTION: AROUND THE WORLD IN 80 DAYS

(11) Bouillon – (12) Florenville = 7 (11) Bouillon – (16) Paliseul = 16

Using QSB+, the following optimal route for Phileas Fogg was determined (which is approximately the same route he travelled in the book).

(12) Florenville – (13) Tintigny = 7

1(London) – 6(Paris) – 7(Barcelona) – 12(Naples) – 17(Athens) – 22(Cairo) – 23(Aden) – 28(Bombay) – 31(Calcutta) – 33(Singapore) – 35(Hong Kong) – 37(Shanghai) – 39(Yokohama) – 41(San Francisco) – 47(Denver) – 50(Chicago) – 53(New York) – 1(London) = 81 days

(15) Luxembourg – (19) Diekirch = 21

(13) Tintigny – (17) Neufchateau = 12 (14) Arlon – (18) Martelange = 8 (16) Paliseul – (20) Recogne = 16 (17) Neufchateau – (18) Martelange = 12 (18) Martelange – (21) Bastogne = 23 (19) Diekirch – (21) Martelange = 3 (19) Diekirch – (21) Bastogne = 18

Note that 3 additional “end” nodes were added to the network for computer solution – 55, 56 and 57. Node 55 replaced node 3 (Casablanca); node 56 replaced node 2 (Lisbon), and node 57 replaced node 1 (London) at the end of the network. Additional branches were added to connect Casablanca with Lisbon (56–57) and Lisbon with London (56–57).

(19) Diekirch – (18) Martelange = 3 (20) Recogne – (21) Bastogne = 16 Total flow = 57,000 troops

CASE SOLUTION: NUCLEAR WASTE DISPOSAL AT PAWV POWER AND LIGHT

Although it appears that Phileas Fogg lost his wager, recall that, as in the novel, he travelled toward the east and eventually crossed the international date line. This saved him one day and allowed him to win his wager once he realized (just in time) the error in his calculations.

This is a “modified” shortest route problem. Instead of the minimum time as the objective function the population traveled through should be minimized. The time (which would normally be the objective function) should be a constraint ≤ 42 hours.

CASE SOLUTION: BATTLE OF THE BULGE

Solution: (1) Pittsburgh - (2) Columbus

(1) Verdun – (2) Stenay = 8

(2) Columbus - (7) Cincinnati

(1) Verdun – (3) Montmedy = 23

(7) Cincinnati - (11) Indianapolis

(1) Verdun – (5) Etain = 26

(11) Indianapolis - (15) Springfield - (16) Davenport/Moline/Rock Island

(2) Stenay – (11) Bouillon = 8

(16) Davenport/Moline/Rock Island (19) Des Moines

(3) Montmedy – (6) Virton = 10 (3) Montmedy – (11) Bouillon = 15

(19) Des Moines - (23) Omaha

(4) Longuyen – (3) Montmedy = 2

(23) Omaha - (28) Cheyenne

(4) Longuyen – (7) Longwy = 5

(28) Cheyenne - (31) Salt Lake City

(5) Etain – (4) Longuyen = 7

(31) Salt Lake City - (33) Nevada Site

(5) Etain – (8) Briey = 10

Total time = 41.7 hours

(5) Etain – (9) Havange = 9

Total population (Z) = 8.23 million

(6) Virton – (13) Tintigny = 10 (7) Longwy – (14) Arlon = 9 (8) Briey – (10) Thionville = 10 (9) Havange – (15) Luxembourg = 8 (10) Thionville – (15) Luxembourg = 7

7-33 .

CASE SOLUTION: A DAY IN PARIS

CASE SOLUTION: SUNTREK GLOBAL CONTAINER NETWORK

There is no direct network technique (from this chapter) that will solve this problem and that will result in an optimal solution. It is more a “test” of the students’ ability to use logic and a systematic approach based on the network analysis concepts from this chapter to find a “good” solution. In other words, it is a network problem, but it does not fit into any of the somewhat narrow, straightforward solution techniques that are required in most homework problems, i.e., it requires the student to “think.” The problem is also intended to be a “fun” exercise enabling the student to use the Internet on an interesting topic. Students may apply elements of the shortest route technique and/or the minimal spanning tree technique, or some other logical approach or technique of their own derivation. (The problem almost fits into the classic “traveling salesman problem” model, except that Kathleen does not have to make an optimal closed loop back to her starting destination; in any event it would be an enormous traveling salesman problem).

Nodes 1. Sources/Farms 2. Houston 3. Savannah 4. New Orleans 5. Charleston 6. Shanghai 7. Karachi 8. Saigon 9. Taiwan 10. India 11. Japan 12. Turkey 13. Italy 14. NY

It is suggested that an interesting and possibly “fun” class assignment might be to have all the students in the class compete to find the best, i.e., most time efficient, route for the sites Kathleen wants to see. The first thing the student will need to do is download a copy of the Metro map from the Web site given in the problem. The map is very detailed, and color-coded, and shows all the stations. They will then (using the Internet) have to locate the sites on the Metro map so they’ll know which subway station corresponds to each site.

15. Liverpool 16. Marseilles 17. New Orleans 18. End/Return

One possible approach is to use the minimal spanning tree technique to determine a “first” possible solution and then logically adjust the tree to improve the solution, since all sites will be connected to a tree, thus resulting in some degree of “back tracking.” Another possible approach is to use the shortest route technique to find the shortest route to the closest site, or “a” site, and then once that is determined, use it to find the next shortest route and so on.

7-34 .

2. Houston

Port

7. Karachi

6. Shanghai

8. Saigon

7. Karachi

6. Shanghai

Destination

Shipped

Containers 6. Shaghai

Facility

Port/

10. India

9. Taiwan

13. Italy

12. Turkey

11. Japan

10. India

9. Taiwan

Shipped

Containers 9. Taiwan

Facility

Port/

New Orleans

Marseilles

Liverpool

17. New Orleans

16. Marseilles

15. Liverpool

14. NY

Marseilles

Liverpool

11. Japan

Containers

(60 containers)

8. Saigon 30 11. Japan

U.S.

3. Savannah

6. Shanghai 30

12. Turkey

10 0

New Orleans

Shipped

(40 containers)

7. Karachi 10

13. Italy 9. Taiwan

Destination

4. New Orleans 8. Saigon 20 8. Saigon

10. India

Destination

(70 containers) 6. Shanghai 10

10. India

5. Charleston 7. Karachi 0

7. Karachi

(30 Containers) 8. Saigon

12. Turkey 11. Japan

Liverpool

12. Turkey

New Orleans

10 NY

Marseilles

Liverpool

Marseilles

13. Italy

New Orleans

7-35

Chapter Eight: Project Management

PROBLEM SUMMARY

33. General linear programming model formulation (8–4)

1. Gantt chart construction and analysis 2. Gantt chart construction and analysis

34. General linear programming model formulation

3. Gantt chart and network construction and analysis

35. Project crashing, linear programming model formulation

4. Network analysis

36. Project crashing, computer (8–12)

5. Network, earliest and latest event times, slack (8–4)

37. Project crashing, computer (8–6)

6. Network, earliest and latest event times, slack

SOLUTIONS TO PROBLEMS

7. Network, earliest and latest event times, slack 1.

8. Network, earliest and latest event times, slack 9. Network, earliest and latest event times, slack 10. Network construction and analysis 11. Network analysis 12. Network analysis 13. Network analysis (8–6) 14. Network analysis, probability analysis 15. Network analysis, probability analysis 16. Network analysis (8–8)

17. Probability analysis (8–13) 18. Network analysis, probability analysis 19. Network analysis, probability analysis 20. Network analysis, probability analysis 21. Probability analysis (8–8 and 8–16) 22. Network analysis 23. Network analysis, probability analysis 24. Network construction, probability analysis 25. Network construction, probability analysis 26. Network construction and analysis

Activity

27. Network construction, probability analysis

28. Network construction, probability analysis

29. Network construction, probability analysis

30. Network construction, probability analysis

31. Project crashing, linear programming model formulation

32. Project crashing, linear programming model formulation

8-1 .

Slack (weeks)

Paths:

2→5→7 10 + 4 + 2 = 16

Activity

2→4→6→7

Slack (weeks)

10 + 5 + 3 + 2 = 20*

1→3→6→7

7 + 6 + 3 + 2 = 18

Activity

5. Time

Paths:

LF Slack

1→3→7 4 + 8 + 2 = 14 1→3→6→8 4 + 8 + 5 + 6 = 23* 1→4→8 4 + 3 + 6 = 13

Critical path activities have no slack

2→5→8

Critical path = 2 − 4 − 6 − 7 = 20

7 + 9 + 6 = 22 2→9 7 + 5 = 12

8-2 .

Critical path = 2 − 6 − 9 − 11 − 12 = 38 months 7.

Critical path = 1 − 3 − 8 = 34 weeks

8-3 .

Critical path = 1 − 3 − 7 − 8 − 10 − 12 = 15 days 9. Activity

Time

Slack

Critical path = 2 − 4 − 10 − 13 − 14 − 17 Project completion time = 78 wk.

8-4 .

10.

Critical path = a − b − f − h = 15 weeks 11. Slack

σ2

10.16

2.25

25.66

18.00

5.43

14.33

22.16

14.33

1.35

19.83

16.00

25.66

5.83

4.00

10.16

20.66

10.16

20.66

4.67

11.16

10.16

21.33

20.50

31.66

10.33

3.35

6.00

19.83

25.83

25.66

31.66

5.83

1.00

Activity

10.16

7.66

10.16

7.66

18.00

7.83

9.66

10.16

10.50

9.50

7.83

17.33

22.16

31.66

14.33

3.35

19.66

20.66

40.33

20.66

40.33

28.41

11.33

7.83

19.16

29.00

40.33

21.16

4.00

8.66

25.83

34.50

31.66

40.33

5.83

4.00

Expected project completion time = 40.33 wk σ = 5.95 Critical path = 1 − 5 − 9

8-5 .

12.

Critical path = a − d − g − k = 33 weeks σ = 3.87

13. Activity

Slack

σ2

8.00

3.66

11.66

3.66

1.77

10.16

2.25

9.33

8.33

17.66

8.33

4.00

6.00

8.00

14.00

11.66

17.66

3.66

1.00

5.00

8.00

13.00

22.33

27.33

14.33

4.00

7.50

10.16

17.66

10.16

17.66

6.25

7.66

9.33

17.00

22.33

30.00

13.00

2.76

15.16

17.66

32.83

21.66

36.83

4.00

4.67

12.33

17.66

30.00

17.66

30.00

7.08

18.66

13.00

31.66

27.33

46.00

14.33

9.00

6.83

30.00

36.83

30.00

36.83

1.35

9.16

36.83

46.00

36.83

46.00

8.01

8-6 .

e. Critical path = 2 − 6 − 9 − 11 − 12 f. Expected project completion time = 46 mo σ2 = 25 mo 14.

Earliest Start

Earliest Finish

Latest Start

Latest Finish

Slack

Variance

1.833

0.000

1.833

0.028

1.833

3.833

1.833

3.833

0.111

3.833

9.667

3.833

9.667

0.694

3.833

5.833

10.667

12.667

6.833

5.833

6.833

12.667

13.667

6.833

3.833

6.000

11.500

13.667

7.667

9.667

13.667

9.667

13.667

1.833

5.667

10.667

14.500

8.833

13.667

16.667

14.500

17.500

0.833

13.667

17.500

13.667

17.500

0.250

17.500

23.83333

17.5

23.83333

23.833

29.667

23.833

29.667

0.694

29.667 Mean

32.833 32.833

29.667

32.833

0.250

Variance

3.472

Std. dev

1.863

Activity

Critical path = a − b − c − g − j − k − l − m Note that 6 months equal 26 weeks.

x−μ

26 − 32.83 = −4.97 1.86

P(x ≤ 26) = 0 Z=

x−μ

52 − 32.83 = 10.31 1.86

P(x ≤ 52) = 1.00

8-7 .

0.444

15.

Activity

Earliest Start

Earliest Finish

Latest Start

Latest Finish

Slack

Variance

0.000

2.000

-1E-14

0.111

2.000

5.000

2E+00

0.111

5.000

7.000

3E+01

5.000

19.333

2E+01

36.333

5.000

12.000

5E+00

5.000

12.000

29.333

36.333

24.333

12.000

36.333

7.000

13.333

36.333

48.667

36.333

48.667

61.333

48.667

61.333

63.333

61.333

63.333

0.111

63.333

67.167

63.333

67.167

0.25

67.167

69.333

67.167

69.333

0.25

69.333

71.167

69.333

71.167

0.028

Mean

71.167

Variance

12.306

Std. dev

3.508

Critical path = a − b − e − g − i − j − k − l − m − n Z=

x−μ

52 − 71.167 = −5.46 3.508

P(x ≤ 52) = 0 Z=

78 − 71.167 = 1.95 3.508

P(x ≤ 78) = .5000 + .4744 = .9744

8-8 .

1 5.444

16. Activity

3 4

2.50

0.694

3.00

2.50

5.50

7.50

10.50

5.00

0.436

5.50

2.50

8.00

2.50

8.00

1.35

6.83

2.50

9.33

2.66

9.50

0.16

3.35

1.5

1.50

8.00

9.50

8.00

9.50

0.026

3.50

8.00

11.50

9.00

12.50

1.00

0.689

4.50

8.00

12.50

10.50

15.00

2.50

1.35

3.00

9.50

12.50

9.50

12.50

0.436

1.66

12.50

14.16

15.33

17.00

2.83

0.436

4.50

12.50

17.00

12.50

17.00

1.35

2.00

12.50

14.50

15.00

17.00

2.50

0.109

1.00

17.00

18.00

17.00

18.00

e. Critical path = 1 − 3 − 7 − 8 − 10 − 12 f. Expected project completion time = 18 Mo

σ = 1.97 Mo 17.

18.

8-9 .

Slack

σ2

Slack

Standard Deviation

Activity

Time

Project

0.667

3.167

7.667

10.833

7.667

0.5

4.167

6.667

10.833

6.667

0.833

2.833

5.833

0.5

5.833

10.833

5.8333

10.833

1.70

1.833

10.833

12.667

15.167

4.333

0.167

5.833

10.833

16.667

10.833

16.667

0.833

3.833

12.667

16.5

20.833

4.333

0.5

4.167

16.667

20.833

16.667

20.833

0.5

2.167

20.833

0.5

Critical path = a − d − e − g − i − j = 23 days What is the probability for the project to be completed in 21 days? Z=

x−μ

21 − 23 = −1.18 Z= 1.7

P(x ≤ 21) = .119 19.

8-10 .

Slack

Standard Deviation

Activity

Time

Project

160.833

24.833

45.5

70.333

45.5

2.167

22.833

47.5

70.333

47.5

2.5

40.167

5.167

30.833

40.167

4.167

24.833

45.833

70.333

91.3333

45.5

17.167

88.167

2.5

11.833

88.167

100

88.167

100

2.167

19.167

45.833

91.333

110.5

45.5

2.5

15.167

45.833

95.333

110.5

49.5

2.167

10.5

100

110.5

100

110.5

1.167

110.5

138.5

110.5

138.5

3.333

10.167

110.5

120.667

128.333

138.5

17.8333

1.5

138.5

145.5

148

155

9.5

14.333

138.5

152.833

146.5

160.833

1.667

8.54

14.5

138.5

153

138.5

153

2.167

4.167

138.5

142.667

156.667

160.833

18.1667

0.5

5.833

145.5

151.333

155

160.833

9.5

0.5

7.833

153

160.833

153

160.833

0.833

Critical path: c − d − f − g − j − k − o − r Project duration = 160.83 Z=

x−μ

180 − 160.83 = 2.24 8.54

P(x ≤ 180 minutes) = .5000 + .4875 = .9875 20.

8-11 .

Activity

Slack

σ2

2.00

7.33

9.33

7.33

1.09

5.00

1.00

3.00

8.66

11.66

8.66

0.436

11.50

2.00

13.50

9.33

20.83

7.33

12.25

7.16

2.00

9.16

13.66

20.83

11.66

2.25

15.83

5.00

20.83

5.00

20.83

6.25

9.16

3.00

12.16

11.66

20.83

8.66

2.25

3.50

13.50

17.00

22.66

26.16

9.16

0.689

3.83

20.83

24.66

22.33

26.16

1.50

.689

5.33

20.83

26.16

20.83

26.16

1.77

26.16

28.16

26.16

28.16

c. Critical path = b − f − j − k

21.

d. Expected project completion time = 28.17 weeks.

σ = 3.00 e.

8-12 .

22.

Activity

Earliest Start

Earliest Finish

Latest Start

Latest Finish

Activity Slack

Critical Path Variance

0.00

7.00

0.00

7.00

0.00

1.00

7.00

18.50

7.00

18.50

0.00

2.25

18.50

23.50

21.33

26.33

2.83

23.50

30.83

26.33

33.67

2.83

18.50

27.33

18.50

27.33

0.00

3.36

27.33

33.67

27.33

33.67

0.00

1.00

33.67

42.83

33.67

42.83

0.00

2.25

42.83

49.50

42.83

49.50

0.00

1.00

49.50

63.83

49.50

63.83

0.00

2.78

63.83

86.67

63.83

86.67

0.00

6.25

86.67

104.83

86.67

104.83

0.00

4.69

Critical Path = a – b – e – f – g – h –i – j - k Project Completion Time = 104.83 Project variance = 24.58 Project std. dev. = 4.96

8-13 .

23. Activity

Slack

σ2

3.00

7.50

10.50

7.50

0.436

6.33

14.66

21.00

14.66

1.00

35.00

25.00

7.33

3.00

10.33

10.50

17.83

7.50

1.77

3.16

10.33

13.50

17.83

21.00

7.50

0.25

13.50

27.00

23.33

36.83

9.83

8.01

15.83

13.50

29.33

21.00

36.83

7.50

3.35

9.33

13.50

22.83

27.50

36.83

14.00

2.76

8.66

10.33

19.00

41.16

49.83

30.83

1.77

1.83

35.00

36.83

35.00

36.83

0.029

8.16

35.00

43.16

49.16

57.83

14.16

1.36

10.00

36.83

46.83

44.00

54.00

7.16

2.77

7.00

36.83

43.83

36.83

43.83

1.00

7.50

19.00

26.50

49.83

57.33

30.83

1.36

10.16

43.83

54.00

43.83

54.00

6.25

3.33

54.00

57.33

54.00

57.33

1.00

Critical path = 3 − 9 − 13 − 15 − 16 Expected project completion time = 57.33 days σ2 = 33.279 σ = 5.77 Z=

x−μ

67 − 57.33 5.77

P(x ≤ 67) = 0.9535 24.

8-14 .

Activity

Slack

Variance

5.33

10.33

9.67

5.33

9.17

13.67

17.5

8.33

0.69

5.33

11.67

5.33

11.67

10.33

17.83

27.5

17.17

2.25

10.33

19.83

29.5

9.67

3.36

9.17

21.17

17.5

29.5

8.33

7.11

9.17

18.33

22.33

31.5

13.17

1.36

11.67

19.17

31.5

12.33

3.36

11.67

5.44

21.17

29.5

42.33

8.33

3.36

19.17

31.6

42.33

12.33

2.25

17.83

25.17

42.33

17.17

1.78

34.5

4.69

34.5

42.33

34.5

42.33

4.69

Critical path = a − d − j − n − o Expected project completion time = 42.3 weeks σ = 4.10 Since probability is 0.90, Z = 1.29 129 =

x − 42.3 4.10

x − 42.3 = 5.29 x = 47.59 To be 90 percent certain of delivering the part on time, RusTech should probably specify at least 47.59 or 48 weeks in the contract bid. 25.

8-15 .

Activity

Time

Early Start

Early Finish

Late Start

Late Finish

Slack

0.0

8.83

64.5

73.33

64.5

24.16

39.16

19.5

39.16

58.66

39.16

58.66

8.16

39.16

47.33

51.5

59.66

12.33

13.66

47.33

59.66

73.33

12.33

20.16

81.16

73.33

93.5

12.33

58.66

83.66

58.66

83.66

14.66

58.66

73.33

78.16

92.83

19.5

58.66

81.66

61.83

84.83

3.16

8.66

81.66

90.33

84.83

93.5

3.16

7.16

83.66

90.83

83.66

90.83

73.33

78.33

92.83

97.83

19.5

4.33

90.83

94.66

93.5

97.83

3.16

90.83

97.83

90.83

97.83

5.5

90.83

95.83

113.16

118.66

22.83

20.833

97.83

118.66

97.83

118.66

Activity

std dev

1.66

1.16

2.5

2.16

1.16

2.33

1.5

3.33

2.33

1.33

1.5

0.66

0.83

2.5 x−μ

120 − 118.167 = 0.227 σ 5.85 P(x ≤ 120) = .5000 + .091 = 0.591

Critical path = a − c − d − h − l − o − q μ = 118.67 σ2 = 34.33; σ = 5.85

8-16 .

26.

Paths:

a−b−d−f 3 + 3 + 4 + 2 = 12 a−c−d−f 3 − 5 + 4 + 2 = 14* a−c−e−f= 3 + 5 + 3 + 2 = 13

27.

8-17 .

Activity Number

Activity Time

Early Start

Early Finish

Late Start

Late Finish

Slack

Standard Deviation

11.17

0.00

11.17

4.00

15.17

4.00

2.83

34.83

0.00

34.83

0.00

34.83

0.00

6.17

17.83

11.17

29.00

15.17

33.00

4.00

4.50

9.00

11.17

20.17

19.83

28.83

8.67

3.00

9.83

34.83

44.67

34.83

44.67

0.00

0.83

11.67

29.00

40.67

33.00

44.67

4.00

1.67

15.83

20.17

36.00

28.83

44.67

8.67

3.50

26.00

34.83

60.83

37.33

63.33

2.50

3.67

6.33

44.67

51.00

44.67

51.00

0.00

1.00

10.17

20.17

30.33

38.83

49.00

18.67

1.83

8.83

51.00

59.83

51.00

59.83

0.00

1.17

10.83

30.33

41.17

49.00

59.83

18.67

3.17

13.67

60.83

74.50

63.33

77.00

2.50

0.67

15.33

59.83

75.17

61.67

77.00

1.83

1.33

17.17

59.83

77.00

59.83

77.00

0.00

1.83

3.17

77.0

80.17

77.00

80.17

0.00

0.50

Critical path = 2 − 5 − 9 −11 − 15 − 16 Project duration = 80.17 months Standard deviation = 6.69 months P(x < 96) = .8315 28.

8-18 .

Activity

Time

7.17

18.00

10.17

Slack

Variance

7.17

0.69

33.67

15.67

7.11

10.17

37.17

1.36

22.17

37.17

12.25

30.00

7.17

37.17

7.17

37.17

8.67

37.17

45.83

37.17

45.83

1.78

7.00

45.83

52.83

45.83

52.83

21.33

39.33

58.33

7.11

20.17

38.17

33.67

53.83

15.67

3.36

13.33

52.83

66.17

52.83

66.17

7.83

39.33

47.17

58.33

66.17

0.69

12.33

38.17

50.5

53.83

66.17

15.67

1.78

10.17

37.17

47.33

55.83

18.67

1.36

7.33

45.83

53.17

58.83

66.17

1.78

18.33

66.17

84.5

66.17

84.5

7.17

84.5

91.67

84.5

91.67

0.69

8.17

66.17

74.33

83.5

91.67

17.33

1.36

7.00

66.17

73.17

84.67

91.67

18.5

27.83

45.83

73.67

63.83

91.67

6.25

17.50

47.33

64.83

83.5

18.67

4.69

8.17

64.83

83.5

91.67

18.67

1.36

The critical path is: a − e − f − g − j − o − p

2.78

From January 20 to April 29 is 101 days.

Activity “n,” send out acceptance letters, has ES = 45.83 (March 6) and LF = 66.17 (March 26), so it appears the club would meet the deadline of March 30 to send out acceptance letters.

x−μ Z 101 − 91.667 = 3.3082 = 2.82

Activity “q,” send out schedules, has ES = 66.16 (March 26) and LS = 83.50 (April 14) and LF = 91.67, so it seems likely the club would meet the deadline of April 15 for sending out game schedules.

Project duration = 91.667 days σ = 3.3082 P ( x ≤ 101) =

P ( x ≤ 101) = .9976

8-19 .

29. Activity

Time

Slack

Variance

8.17

0.67

8.83

0.67

1.36

5.83

3.00

8.83

3.00

0.69

21.50

6.33

27.83

6.33

6.25

31.00

0.00

31.00

13.44

7.00

8.17

15.17

8.83

15.83

0.67

1.00

17.33

15.17

32.50

15.83

33.17

0.67

4.00

5.33

21.50

26.83

27.83

33.17

6.33

1.00

2.17

31.00

33.17

31.00

33.17

0.25

35.83

33.17

69.00

33.17

69.00

17.36

6.33

69.00

75.33

69.00

75.33

1.00

15.17

30.33

60.17

75.33

45.00

3.36

6.83

75.33

82.17

75.33

82.17

0.69

6.17

82.17

88.33

82.17

88.33

1.36

4.00

88.33

92.33

88.33

92.33

1.00

The “suggested” network is as follows:

Critical path = d − h − i − j − l − m − n Expected project completion time = 92.33 days σ = 5.93 days 90 − 92.33 5.93 = −.39

P ( x ≤ 90 ) =

P ( x ≤ 90 ) = .5000 − .1517 = .348

8-20 .

30. Activity

Time

Slack

Variance

9.33

5.44

4.83

9.33

14.17

120.83

125.67

111.50

1.36

5.00

9.33

14.33

96.67

101.67

87.33

0.44

14.50

14.33

26.83

101.67

116.17

87.33

3.36

e f

9.50

28.83

38.33

116.17

125.67

87.33

1.36

2.17

14.33

16.50

118.5

120.67

104.17

0.25

5.00

16.5

21.50

120.67

125.67

104.17

0.44

13.67

9.33

23.00

9.33

23.00

5.44

6.17

29.17

119.5

125.67

96.50

1.36

22.17

45.17

103.5

125.67

80.50

6.25

9.67

9.33

19.00

116

125.67

106.67

4.00

25.67

23.00

48.67

18.78

12.50

48.67

61.17

48.67

61.17

4.69

15.83

61.17

77.00

61.17

77.00

6.25

45.00

77.00

122.00

25.00

6.83

48.67

55.50

102.17

109.00

53.50

1.36

5.00

55.50

60.50

109.33

114.33

53.83

1.00

12.33

60.50

72.83

114.33

126.67

53.83

1.00

4.17

122.00

126.17

122

126.17

0.69

17.17

55.50

72.67

109

126.17

53.50

3.36

2.00

55.50

57.50

124.17

126.17

68.67

0.11

0.50

45.17

45.67

125.67

126.17

80.50

0.00

0.50

126.17

126.67

126.17

126.67

0.00

The “suggested” network is as follows (although the student’s version may vary).

8-21 .

Critical path = a − h − l − m − n − o − s − w Expected project duration (μ) = 126.67 days σ = 8.14 days P ( x ≤ 150 days ) : Z =

x−μ

150 − 126.67 8.14 Z = 2.87 Z=

P ( x ≤ 150 ) = .5000 − .4979 = .9979

31. a)

b) Present (normal) critical path = a −d Normal critical path time = 30 wk Crash critical path (all crash time) = a − d Maximum possible project crash time = 20 wk c)

Normal cost = 3,950 Crash project cost = 4,700 Activity

Normal time

Crash time

Normal cost

Crash cost

1,000

1,480

1,200

1,400

700

1,190

500

820

550

730

3,950

5,620

Activity

Crash cost per day

Crash by

Crashing cost

400

200

150 750

8-22 .

f) Minimize Z = 40y12 + 50y14 + 70y13 + 80y24 + 30y34 subject to

Crashing: 1. Crash a 5 weeks; c − e becomes critical; cost = $200 2. Crash a and e 1 week; cost = $70

y12 ≤ 12

3. Crash a, b and e 4 weeks; cost = $480

y14 ≤ 4

d) Critical paths: a − d, b and c − e = 20 weeks

y13 ≤ 7

Total crash cost = $750

y24 ≤ 4

e) Minimize Z = x4 subject to

y34 ≤ 6 x1 + 20 − y12 ≤ x2

x2 − x1 ≥ 20

x1 + 14 − y13 ≤ x3

x3 − x1 ≥ 14

x1 + 24 − y14 ≤ x4

x4 − x1 ≥ 24

x2 + 10 − y24 ≤ x4

x4 − x2 ≥ 10

x3 + 11 − y34 ≤ x4

x4 − x3 ≥ 11

x4 ≤ 20

xi , xj ≥ 0

xi,xj,yij ≥ 0

32. a)

b) Normal critical path = a − d − h

Normal critical path time = 36 wk Project completion time = 36 Normal time

Crash time

Normal cost

Crash cost

Crash cost per day

2,000

4,400

300

1,000

1,800

160

500

700

100

600

1,300

700

1,500

3,000

750

800

1,600

400

3,000

4,500

500

5,000

8,000

600

8-23 .

33.

Total crash cost = $3,200

Minimize Z = x6 subject to

1. Crash a 3 wks; $900

x2 − x1 ≥ 10 x3 − x1 ≥ 7

2. Crash a and b 5 wks; $2,300 Critical paths: a − d − h and b − e − h = 28 wks.

x4 − x2 ≥ 6

c) Minimize Z = x6 subject to

x4 − x3 ≥ 5

x2 − x1 ≥ 16

x5 − x4 ≥ 3

x3 − x1 ≥ 14

x6 − x5 ≥ 2

x4 − x2 ≥ 8

xi, xj ≥ 0

x5 − x3 ≥ 4

x5 − x2 ≥ 5

The solution is x1 = 0, x2 = 7, x3 = 10, x4 = 15, x5 = 18, x6 = 20.

x5 − x3 ≥ 4 x6 − x3 ≥ 6 x6 − x4 ≥ 10

Minimize Z = x9 subject to

x6 − x5 ≥ 15

x2 − x1 ≥ 8

xi, xj ≥ 0

x3 − x1 ≥ 6

34.

x4 − x1 ≥ 3

d) The minimum project duration is 22 weeks.

x5 − x2 ≥ 0

Minimize Z = 300y12 + 160y13 + 100y24 + 700y25 + 750y35 + 400y36 + 500y46 + 600y56

x6 − x2 ≥ 5 x5 − x3 ≥ 3

subject to

x5 − x4 ≥ 4

y12 ≤ 8

x7 − x5 ≥ 7

y13 ≤ 5

x7 − x8 ≥ 0

y24 ≤ 2

x8 − x5 ≥ 4

y25 ≤ 1

x8 − x4 ≥ 2

y35 ≤ 2

x9 − x6 ≥ 4

y36 ≤ 2

x9 − x7 ≥ 9

y46 ≤ 3

xi, xj ≥ 0

y56 ≤ 5

The solution is x1 = 0, x2 = 9, x3 = 6, x4 = 3, x5 = 9, x6 = 14, x7 = 16, x8 = 16, x9 = 25.

x1 + 16 − y12 ≤ x2 x1 + 14 − y13 ≤ x3

35. a) Minimize Z = x5

x2 + 8 − y24 ≤ x4 x2 + 5 − y25 ≤ x5

subject to x2 − x1 ≥ 8

x3 + 4 − y35 ≤ x5

x3 − x1 ≥ 10

x3 + 6 − y36 ≤ x6

x3 − x2 ≥ 5

x4 + 10 − y46 ≤ x6

x4 − x2 ≥ 3

x5 + 15 − y56 ≤ x6

x4 − x3 ≥ 6

x6 ≤ 22

x5 − x3 ≥ 3

xi,xj,yij ≥ 0

x5 − x4 ≥ 4 xi, xj ≥ 0 The solution is x1 = 0, x2 = 8, x3 = 13, x4 = 19, x5 = 23.

8-24 .

y34 ≤ 2

y35 ≤ 0

Activity

(i–j)

Total Allowable Crash Time (weeks)

1–2

100

1–3

2–3

200

2–4

100

3–4

3–5

4–5

200

Crash cost per week ($)

y45 ≤ 1 x1 + 8 − y12 ≤ x2 x1 + 10 − y13 ≤ x3 x2 + 5 − y23 ≤ x3 x2 + 3 − y24 ≤ x4 x3 + 6 − y34 ≤ x4 x3 + 3 − y35 ≤ x5 x4 + 4 − y45 ≤ x5 x5 ≤ 15 xi,xj,yij ≥ 0

c) Minimize Z = 100y12 + 50y13 + 200y23 + 100y24 + 75y34 + 0y35 + 200y45

subject to y12 ≤ 3 y13 ≤ 3 y23 ≤ 2 y24 ≤ 2 The solution is Z = $1,150, y12 = 3, y13 = 2, y23 = 2, y24 = 0, y34 = 2, y45 = 1, y35 = 0, x1 = 0, x2 = 5, x3 = 8, x4 = 12, x5 = 15.

8-25 .

36. Activity

Normal time

Crash time

Normal cost

Crash cost

Normal cost/day

Crash by

Crashing cost

4,800

6,300

750

1,500

9,100

15,500

3,200

3,000

4,000

500

3,600

5,000

700

1,400

1,500

2,000

250

1,800

2,000

200

k 8 6 Project completion time = 33

5,000

7,000

1,000

2,000

Normal cost = 28,800 Minimum project completion time = 26 Crash cost = 33,900 Critical path: a − d − g − k. Crashing cost = $5,100 Total network cost = $33,900 37.

1, 2, 5, 6, 8, 9, 11, 12 Cost of crashing = $2,633,333

CASE SOLUTION: BLOODLESS COUP CONCERT

8-26 .

Activity

4.16

Slack

Std. dev.

4.16

0.83

5.33

3.50

8.83

3.50

0.66

2.16

4.16

6.33

6.66

8.83

2.50

0.50

5.50

4.16

9.66

6.66

12.16

2.50

1.16

3.50

4.16

7.66

4.16

7.66

1.16

4.16

7.66

11.83

7.66

11.83

0.83

3.00

7.66

10.66

10.16

13.16

2.50

0.66

3.00

9.66

12.66

12.16

15.16

2.50

0.33

3.33

11.83

15.16

11.83

15.16

0.66

2.00

10.66

12.66

13.16

15.16

2.50

0.33

6.33

12.66

8.83

15.16

2.50

1.66

5.5

6.33

11.83

9.66

15.16

3.33

1.83

Critical path = 1 − 4 − 6 − 10 Expected project completion time = 15.17 days σ = 1.79 Z=

x−μ

18 − 15.17 = 1.58 1.79

P(x ≤ 18) = 0.9429

CASE SOLUTION: MOORE HOUSING CONTRACTORS

8-27 .

Activity

Time

Early Start

Early Finish

Late Start

Late Finish

Slack

Std. dev.

4.16

0.0

4.16

0.0

0.5

3.16

4.16

7.33

4.16

7.33

0.0

0.5

3.83

7.33

11.16

7.83

11.66

0.5

2.16

7.33

9.5

33.83

26.5

0.5

7.33

9.33

7.33

9.33

0.0

0.33

3.83

11.16

11.66

15.5

0.5

3.16

9.33

12.5

9.33

12.5

0.0

0.5

4.16

9.33

13.5

17.66

4.16

0.83

2.83

17.83

23.83

0.5

2.16

17.16

15.5

17.66

0.5

5.16

12.5

17.66

12.5

17.66

0.0

0.83

6.5

17.83

24.33

23.83

30.33

0.83

8.33

17.66

0.0

3.33

24.33

27.66

30.33

33.66

0.66

2.33

27.66

33.66

0.66

3.5

33.5

39.5

0.83

4.16

30.16

0.5

6.33

33.5

39.83

39.5

45.83

5.83

30.16

40.83

4.83

1.5

4.33

30.16

34.5

30.16

34.5

3.33

30.16

33.5

31.17

34.5

0.66

6.33

34.5

40.83

34.5

40.83

45.83

40.83

45.83

2.83

40.83

43.66

45.83

2.16

0.5

Project completion time = 45.83 Project standard deviation = 2.40 Critical Path for Moore Contractors Critical path: a − b − e − g − k − m − q − t − v − w Notice that the expected completion time is 45.83 days which is very close to the realtor’s due date for completion. The probability of finishing in 45 days is 0.3647 36.47% is not a very high probability that the contractor will complete a house within 45 days; thus the Moores should probably inflate their bid.

8-28 .

Chapter Nine: Multicriteria Decision Making PROBLEM SUMMARY

30. AHP, student selection 31. AHP, athletic facilities 32. AHP, vacation locations 33. Pairwise comparisons (9–32) 34. AHP, major options 35. AHP, basketball players 36. AHP, school facilities 37. Student’s pairwise comparisons 38. AHP, primary care provider 39. AHP, Port selection 40. AHP, baseball player selection 41. Consistency (9–40) 42. Student response (9–20) 43. AHP, class sections 44. AHP, textbook selection 45. AHP, pro football draft 46. AHP, Civil War generals 47. Pairwise comparisons (9–46) 48. Soccer team location 49. Consistency (9–48) 50. AHP, sustainable project selection 51. AHP, basketball player team selection 52. Consistency (9–51) 53. Scoring model, healthcare clinics 54. Scoring model, team location (9–44) 55. Student car selection 56. Scoring model, plant site selection 57. Scoring model, site selection 58. Scoring model, baseball player free-agent selection 59. Scoring model, campus student center 60. Scoring model, college selection 61. Scoring model, time-share condominium 62. Scoring model, condominium selection 63. Scoring model, soccer tournament 64. Scoring model, retail stores 65. Scoring model, facility location 66. Scoring model, student’s restaurant selection model

1. Model formulation, product mix 2. Model formulation, transportation, computer solution (6–15) 3. Model formulation, urban recreation facility allocation 4. Model formulation, crop determination, computer solution 5. Model formulation, product mix, computer solution 6. Model formulation, OSHA safety compliance, computer solution 7. Computer solution, graphical solution 8. Computer solution, graphical solution 9. Computer solution 10. Model formulation, product mix, computer solution 11. Model formulation, product mix, computer solution 12. Model formulation, product mix, computer solution 13. Model formulation, clinic personnel selection, computer solution 14. Model formulation, production scheduling, computer solution, sensitivity analysis 15. Model formulation, employee scheduling, computer solution 16. Model formulation, R&D project selection 17. Model formulation, sustainable project selection 18. AHP, company takeover 19. Pairwise comparison (9–18) 20. AHP, apartment selection 21. Pairwise comparisons (9–20) 22. AHP, mutual funds 23. AHP (9–22) 24. AHP, utility vehicles 25. AHP, anchor persons 26. AHP, hotel selection 27. AHP, college selection 28. AHP, dating service 29. AHP, R&D projects

9-1 .

PROBLEM SOLUTIONS Minimize P1d1−, P2d2−, P3d1+, P4d3+ subject to

a) Minimize P1d1−, P2d2−, P3d3+, 3P4d4− + 6P4d5− + P4d6− + 2P4d7− subject to 80,000x1 + 24,000x2 + 15,000x3 + 40,000x4 + d1− = 600,000

5x1 + 2x2 + 4x3 + d1− − d1+ = 240 3x1 + 5x2 + 2x3 + d2− − d2+ = 500 4x1 + 6x2 + 3x3 + d3− − d3+ = 400 Minimize Z = P1 (d1− + d2− + d3−), P2d5−, P3d8−, P4 (d9− + d10− + d11− − + +

1,500x1 + 3,000x2 + 500x3 + 1,000x4 + d2− − d2+ = 20,000 4x1 + 8x2 + 3x3 + 5x4 + d3− − d3+ = 50 x1 + d4− − d4+ = 7 x2 + d5− − d5+ = 10

+ d12 ), P5 d13 , P6 d14

x3 + d6− − d6+ = 8

subject to x1A + x1B + x1C + x1D + d1− = 420

x4 + d7− − d7+ = 12 where x1 = no. of gymnasiums, x2 = no. of athletic fields, x3 = no. of tennis courts, x4 = no. of pools

x2A + x2B + x2C + x2D + d2− = 610 x3A + x3B + x3C + x3D + d3− = 340 x1A + x2A + x3A + d4− = 520

b) x1 = 6, x2 = 5, x3 = x4 = 0,

x1B + x2B + x3B + d5− = 250 x1C + x2C + x3C + d6− = 400 x1D + x2D + x3D + d7− = 380 x3D + d8− − d8+ = 80 x1A + x2A + x3A + d9− − d9+ = 416 x1B + x2B + x3B + d10− − d10+ = 200 x1C + x2C + x3C + d11− − d11+ = 320 x1D + x2D + x3D + d12− − d12+ = 304

d2+ = 4,000, d3+ = 14, d4+ = 4, d5− = 1, d6− = 5, d7− = 8, d8− = 12 4.

7x1 + 10x2 + 8x3 + d1− − d1+ = 6,000 100x1 + 120x2 + 70x3 + d2− − d2+ = 80,000 30x1 + 40x2 + 20x3 + d3− − d3+ = 105,000 x1 + x2 + x3 + d4− = 1,000

22x1A + 17x1B + 30x1C + 18x1D + 15x2A

x1 + d5− − d5+ = 200

+ 35x2B + 20x2C + 25x2D + 28x3A

x2 + d6− − d6+ = 500

+ 21x3B + 16x3C + 14x3D + d13− − d13+ = $24,717 x2C − d14+ = 0

x3 + d7− − d7+ = 300 where x1 = acres of corn, x2 = acres of wheat, x3 = acres of soybeans

Note that the negative deviational variables must be at the highest-priority level to force all the supply to be used.

b) x2 = 666.67, d1+ = 666.667, d3− = 78,333.33, d4− = 333.333, d5− = 200.00, d6+ = 166.67, d7− = 300.00; priority one and two are achieved; P3 (78,333.33), P4 (666.67), P5 (333.33), and P6 (600.00) are not achieved.

b) x1B = 215, x1C = 205, x2A = 520, x2D = 90, x3C = 50, x3D = 290 A 1 2

215

205

520 50

3 Demand Achieved

a) Minimize P1d1−, P2d2+, P3d3−, P4d1+, P5d4−, 1.5P6d5− + 2P6d6− + P6d7− subject to

420

a) Minimize P1d2+, P2d7+, 4P3d4− + 2P3d5− + P3d6−, P4d1−, P5d3− subject to

610

x1 + x2 + x3 + d1− = 2,000

290

340

800x1 + 1,500x2 + 500x3 + d2− − d2+ = 20,000

10x1 + 12x2 + 18x3+ d3− − d3+ = 800 520

215

255

380

x1 + d4− − d4+ = 800

The noninteger solution values were rounded to integer values (making sure no rim requirements were violated).

x2 + d5− − d5+ = 900 x3 + d6− − d6+ = 1,100 d3+ + d7− − d7+ = 100

9-2 .

where xl = tons of Supergro, x2 = tons of Dynaplant, x3 = tons of Soilsaver b)

The goal constraint d3+ + d7− − d8− = 100

must be converted to 10x + 12x2 + 18x3 + d7− − d7+ = 900 for QM for Windows input. The solution is x1 = 25, d1− = 1,975, d3− = 550, d4− = 775, d5− = 900, d6− = 1,100, d7− = 650. P1 and P2 are achieved; P3 (6,000), P4 (1,975), and P5 (550) are not achieved. 6.

a) Minimize P1d3+, P2 (d4− + d5− + d6− + d7−), P3d1−, P4d2− subject to .18x1 + .11x2 + .17x3 + .21x4 + d1− − d1+ = 20 1.21x1 + .48x2 + .54x3 +1.04x4 + d2− − d2+ = 115

x1 = 500, x2 = 300, d3− = 500 satisfied

135x1 + 87x2 + 58x3 + 160x4 + d3− − d3+ = 52,000

x1 + d4− = 60

x1 = 15, d1+ = 12, d2+ = 10, d4− = 6 not satisfied

10. a) Minimize P1d1−, P2d2−, P3d3−, P4d4+ subject to

x2 + d5− = 28 x3 + d6− = 35

8x1 + 6x2 + d1− − d1+ = 480 (production capacity, hr)

x4 + d7− = 17 where x1 = percentage points compliance in hazardous materials, x2 = percentage points compliance in fire protection, x3 = percentage points compliance in hand-powered tools, x4 = percentage points compliance in machine guarding

yd) yd)

x1 + d2− = 40 (demand, 100 x2 + d3− = 50 (demand, 100

d1+ + d4− − d4+ = 20 (overtime, hr) xj, di−, di+ ≥ 0

b) x1 = 60, x2 = 28, x3 = 35, x4 = 17, d1+ = 3.4, d2+ = 7.62, d3− = 36,714; P1, P2, and P4; P3 (3.4) not achieved.

b) x1 = 40, x2 = 50, d1+ = 140, d4− = 120 satisfied 11. a) Minimize P1d1−, P2d4+, 3P3d2− + 2P3d3−, P4d1+ subject to

x1 + x2 + d1− − d1+ = 80 x1 + d2− = 60 x2 + d3− = 35 d1+ + d4− − d4+ = 10 b) x1 = 60, x2 = 30, d1+ = 10, d3− = 5 satisfied 12. a) Minimize P1d5+, P2d2−, 5P3d3− + 4P3d4−, P4d1− subject to 5x1 + 8x2 + d1− − d1+ = 4,800

x1 = 20, d2− = 10, d3− = 50

.20x1 + .25x2 + d2− − d2+ = 300

satisfied

x1 + d3− = 500 x2 + d4− = 400 + d1 + d5− − d5+ = 480

9-3 .

b) The solution is x1 = 500, x2 = 347.5, d1+ = 480, d2− = 113.125, d4− = 52.5.

Goal achievement: P1: achieved, demand met

13. a) Minimize P1d1−, P2d2+, P3d3−, P4d4− subject to

P2: not achieved; overtime required for all processes (d1+ = 340, d2+ = 296, d3+ = 214, d4+ = 317)

x1 + d1− − d1+ = 30 20x1 + 40x2 + 150x3 + d2− − d2+ = 1,200

P3: achieved, profit goal met

x2 + x3 + d3− − d3+ = 20 x3 + d4− − d4+ = 6

P4: not achieved; 8,370 ft2 (d10+ = 8,370) extra material required. 15. a) Minimize P1d1+, P2(d2−, d8−), P3(d7−), P4(d3−, d4−, d5−, d6−)

b) x1 = 30 x2 = 15 Goal achievement:

subject to

P1: achieved

x1 + x2 + x3 + x4 + x5 + x6 + x7 + d1− − d1+ = 60

P2: achieved P3: not achieved, d3− = 5

x1 + x4 + x5 + x6 + x7 + d2− − d2+ = 47

P4: not achieved, d4− = 6

x1 + x2 + x5 + x6 + x7 + d3− − d3+ = 22 x1 + x2 + x3 + x6 + x7 + d4− − d4+ = 28

14. a) Minimize P1(d1+, d2+, d3+, d4+), P2(d5−, d6−, d7−, d8−), P3d9−, P3d10+ subject to −

−

x1 + x2 + x3 + x4 + x7 + d5− − d5+ = 35 x1 + x2 + x3 + x4 + x5 + d6− − d6+ = 34

.06 x1 + .17 x2 + .10 x3 + .14 x4 + d1 − d1 = 700

x2 + x3 + x4 + x5 + x6 + d7− − d7+ = 43

.18x1 + .20 x2 + .14x4 + d2 − d2 = 700

x3 + x4 + x5 + x6 + x7 + d8− − d8+ = 53

.07x1 + .20 x2 + .08x3 + .12x4 + d3 − d3 = 800

xi, di−, di+ ≥ 0

.09x1 + .12x2 + .07x3 + .15x4 + d4 − d4 = 600

b) Employees beginning work on day i = xi

x1 + d5− − d5+ = 2,600

x1 = 0 − Sunday

x2 + d6− − d6+ = 1,800

x2 = 7 − Monday

x3 + d7− − d7+ = 4,100

x3 = 0 − Tuesday

x4 + d8− − d8+ = 1,200

x4 = 27 − Wednesday

90 x1 + 100 x2 + 80 x3 + 120 x4 + d9− − d9+ = 700,000 −

x5 = 0 − Thursday

2.6x1 + 1.4x2 + 2.5x3 + 3.2x4 + d10 − d10 = 15,000

x6 = 25 − Friday

b) x1 = 2,600 x2 = 0 x3 = 0 x4 = 1,657.14

x7 = 1 − Saturday Goal achievement: P1: achieved P2: achieved P3: achieved

Goal achievement:

P4: achieved

P1: achieved; no overtime

16. a) Minimize P1d1+, P2d2+, P3d3−, P4(d4−, d5−), P5d6+, P6d7−

P2: not achieved; demand not met for parts 2 (d6− = 1800) and 4 (d7− = 4100)

subject to

P3: not achieved; profit goal not met by

.675x1 + 1.05x2 + .725x3 + .430x4 + 1.24x5 +.890x6 + 1.620x7 + 1.20x8 + d1− − d1+ = 5.0 6x1 + 5x2+ 7x3 + 8x4 + 10x5 + 6x6 + 7x7 + 6x8 + d2− − d2+ = 27 .820x1 + 1.75x2 + 1.60x3 + 1.90x4 + .930x5 + 1.70x6 + 1.30x7 + 1.80x8 + d3− − d3+ = 6.5

$72,166.67 (d9− = 267,143) c) x1 = 2,600 x2 = 1,800 x3 = 4,100 x4 = 1,200

9-4 .

220,000x1 + 125,000x2+ 26,000x3 + 75,000x4 + 45,000x5 + d3− − d3+ = $4,000,000

x1 + x3 + x4 + x6 + d4− − d4+ = 2 x2 + x5 + x7 + x8 + d5− − d5+ = 2 x2 + x3 + x5 + x6 + x7 + d6− − d6+ = 3 x5 + x6 + d7− − d7+ = 2

400,000x1 + 150,000x2+ 34,000x3 + 1,200x4 + 55,000x5 + d4− − d4+ = 5,000,000

b) Solving sequentially using Excel: x2 = 1

d1− = 1.01

x4 = 1

d2− = 1

x6 = 1

d3+ = 0.15

x7 = 1

d7− = 2

x1 + d5− = 75 x2 + d6− = 75 x3 + d7− = 165 x4 + d8− = 75 x5 + d9− = 75

All goals are achieved except the priority 6 goal; d1− = 1, meaning only one of the two projects was selected.

xi ≥ 0 and integer b) x1 = 2

17. a) Minimize Z = P1d1+, P2d2−, P3d3−, P4d4− subject to

x2 = 24 x3 = 5

2,600,000x1 + 950,000x2+ 38,000x3 + 365,000x4 + 17,500x5 + d1− − d1+ = $30,000,000

x4 = 1 x5 = 8

17,500x1 + 8,600x2+ 25x3 + 900x4 + 600x5 + d2− − d2+ = 250,000 18.

P1, P2, P3, P4 achieved

Normalized matrices: Profitability A

Row Averages

.2414

.2308

.4118

.2946

.7241

.6923

.5294

.6486

.0345

.0769

.0588

.0567

Company

1.0000 Growth Potential A

Row Averages

.1250

.1111

.1304

.1222

.2500

.2222

.2174

.2299

.6250

.6667

.6522

.6479

Company

1.0000

Criteria

Profit

Growth

Row Averages

Profit

.2500

Growth

.7500

.7500 1.0000

9-5 .

Profit

Growth

A ⎡.2946 B ⎢⎢.6486 C ⎢⎣.0567

.1222 ⎤ .2299 ⎥⎥ .6479 ⎥⎦

Profit Growth

Company

Score

.1653

.3346

.5001

⎡.25⎤ ⎢.75 ⎥ = ⎣ ⎦

1.0000 19.

Profitability: (1)

(2)

(3)

⎡ 1 1/3 7 ⎤ ⎡.2946 ⎤ ⎡ .9077 ⎤ ⎢ 3 ⎥ ⎢ ⎥ 1 9 ⎥ × ⎢.6486 ⎥ = ⎢⎢ 2.0427 ⎥⎥ ⎢ ⎢⎣1/7 1/9 1⎥⎦ ⎢⎣.0567 ⎥⎦ ⎢⎣ .1709 ⎥⎦

(3)/(2) ⎡3.0811⎤ ⎢3.1494 ⎥ ⎢ ⎥ ⎢⎣3.0133⎥⎦ 9.2438

9.2438 3.081 − 3 = 3.081, CI = = .041 3 2 .041 CI/RI = = .07 .58

Since .07 < .10 this preference comparison is consistent. Growth: (1)

(2)

(3)

⎡ 1 1/2 1/5⎤ ⎡.1222 ⎤ ⎡ .3667 ⎤ ⎢2 ⎥ × ⎢.2299 ⎥ = ⎢ .6003⎥ 1 1/3 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 5 ⎢⎣.6479 ⎥⎦ ⎢⎣1.9486 ⎥⎦ 3 1⎥⎦

(3)/(2) ⎡ 3.0011⎤ ⎢3.0025⎥ ⎢ ⎥ ⎣⎢3.0076 ⎦⎥ 9.0111

9.0111 3.004 − 3 = 3.004, CI = = .0018 3 2 CI /RI = .0018/.58 = .003

Since .003 < .10 this preference comparison is consistent. Since there are only two preferences for the criteria matrix and they are exact reciprocals, this preference comparison is perfectly consistent. 20. Criteria Apartment

Cost

Condition

Location

Criteria

Terraces

.2299

.3667

.6667

.6479

Vistas

.1222

.2333

.2222

Foxfield

.6479

.4000

.1222

9-6 .

.1600 .5253

Apartment

Score

Terraces

.3147

Vistas

.1600

Foxfield

.5253 1.000

21.

Criteria: (1)

(2)

(3)

(3)/(2)

3 5⎤ ⎡ 1 ⎡.6479 ⎤ ⎡1.9486 ⎤ ⎢1/3 ⎥ ⎢ ⎥ 1 2 ⎥ × ⎢.2299 ⎥ = ⎢⎢ .6903 ⎥⎥ ⎢ ⎢⎣1/5 1/2 1⎥⎦ ⎢⎣.1222 ⎥⎦ ⎢⎣ .3667 ⎥⎦

⎡3.0076 ⎤ ⎢3.0025 ⎥ 9.0111 = 3.004, CI = 3.004 − 3 = .0018 ⎢ ⎥ 3 2 ⎢⎣3.0011 ⎥⎦ .0018 CI /RI = 9.0111 = .003 .58

Since .003 < .10 this preference comparison is consistent. 22.

Normalized matrices: Return Fund

Global

Blue Chip

Bond

Row Averages

Global

.1818

.1765

.2222

.1935

Blue Chip

.7273

.7059

.6667

.6999

Bond

.0909

.1176

.1111

.1066

Risk Fund

Global

Blue Chip

Bond

Row Averages

Global

.222

.2500

.23

0.23

Blue Chip

.111

.1250

.12

0.12

Bond

.667

.6250

.65

0.65

Load Fund

Global

Blue Chip

Bond

Row Averages

Global

.2000

Blue Chip

.2000

Bond

.6000

Criteria

Return

Risk

Load

Row Averages

Return

.6522

.6667

.6250

.6479

Risk

.2174

.2222

.2500

.2299

Load

.1304

.1111

.1250

.1222

9-7 .

Global Blue Chip Bond

Return Risk

Load

⎡.1935 ⎢.6999 ⎢ ⎢⎣.1066

.2000 ⎤ ⎡.6479 ⎤ ⎥ .2000 ⎥ × ⎢⎢.2299 ⎥⎥ = ⎢⎣.1222 ⎥⎦ .6000 ⎥⎦

.23 .12 .65

Criteria

Fund

Score

Global

.203

Blue Chip

.506

Bond

.291 1.0000

Megan should invest in the Blue Chip Fund. 23.

Global = 85,000(.203) = $17,225.50 Blue Chip = 85,000(.506) = $43,014.50 Bond = 85,000(.291) = $24,760.00 $85,000.00

24.

Normalized matrices: Consumer Digest Vehicle

Explorer

Trooper

Passport

Row Averages

Explorer

.6316

.5714

.6667

.6232

Trooper

.1579

.1429

.1111

.1373

Passport

.2105

.2857

.2222

.2395

Price Vehicle

Explorer

Trooper

Passport

Row Averages

Explorer

.0909

.1429

.0526

.0955

Trooper

.3636

.5714

.6316

.5222

Passport

.5455

.2857

.3158

.3823

Passport

Row Averages

Appearance Vehicle

Explorer

Trooper

Explorer

.6316

.5714

.6667

.6232

Trooper

.1579

.1429

.1111

.1373

Passport

.2105

.2857

.2222

.2395

Criteria

Consumer Digest

Price

Appearance

Row Averages

Consumer Digest

.5714

.6000

.5000

.5571

Price

.2857

.3000

.3750

.3202

Appearance

.1429

.1000

.1250

.1226

9-8 .

Consumer Digest Explorer Trooper Passport

⎡.6232 ⎢ ⎢.1373 ⎢⎣.2395

Price Appearance .0955 .5222 .3823

Criteria

.6232 ⎤ .1373 ⎥⎥ .2395⎥⎦

⎡.5571 ⎤ ⎢ ⎥ ⎢.3202 ⎥ = ⎢⎣.1226 ⎥⎦

Vehicle

Score

Explorer

.4542

Trooper

.2605

Passport

.2852 1.0000

Alex should purchase the Explorer. 25.

Normalized matrices: Appearance Anchor

Pawlie

Cooric

Brokenaw

Row Averages

Pawlie

.6087

.6250

.5385

.5907

Cooric

.3043

.3125

.3846

.3338

Brokenaw

.0870

.0625

.0769

.0755

Intelligence Anchor

Pawlie

Cooric

Brokenaw

Row Averages

Pawlie

.1250

.1000

.1429

.1226

Cooric

.3750

.3000

.2857

.3202

Brokenaw

.5000

.6000

.5714

.5571

Speech Anchor

Pawlie

Cooric

Brokenaw

Row Averages

Pawlie

.2222

Cooric

.6667

Brokenaw

.1111

Criteria

Appearance

Intelligence

Speech

Row Averages

Appearance

.6857

.5714

.7143

.6571

Intelligence

.0857

.0714

.0476

.0683

Speech

.2286

.3571

.2381

.2746

Appearance Intelligence Speech Pawlie Cooric Brokenaw

⎡.5907 ⎢ ⎢.3338 ⎢⎣.0755

.1226 .3202 .5571

.2222 ⎤ .6667 ⎥⎥ .1111 ⎥⎦

Criteria

Anchor

Score

⎡.6571 ⎤ × ⎢⎢.0683 ⎥⎥ = ⎢⎣.2746 ⎥⎦

Pawlie

.4575

Cooric

.4243

Brokenaw

.1182 1.0000

9-9 .

26.

Normalized matrices: Ambiance Hotel

Cheraton

Milton

Harriott

Row Averages

Cheraton

.1250

.1111

.1304

.1222

Milton

.2500

.2222

.2174

.2229

Harriott

.6250

.6667

.6522

.6479

Location Hotel

Cheraton

Milton

Harriott

Row Averages

Cheraton

.6522

.5000

.7059

.6194

Milton

.1304

.1000

.0588

.0964

Harriott

.2174

.4000

.2353

.2842

Cost Hotel

Cheraton

Milton

Harriott

Row Averages

Cheraton

.5882

.5714

.6250

.5949

Milton

.2941

.2857

.2500

.2766

Harriott

.1176

.1429

.1250

.1285

Criteria

Ambiance

Location

Cost

Row Averages

Ambiance

.5714

.6000

.5000

.5571

Location

.2857

.3000

.3750

.3202

Cost

.1429

.1000

.1250

.1226

Ambiance Location Cheraton Milton Harriott

⎡ .1222 ⎢.2299 ⎢ ⎢⎣.6479

.6194 .0964 .2842

Cost

Criteria

.5949 ⎤ ⎡.5571⎤ = ⎥ .2766 ⎥ × ⎢⎢.3202 ⎥⎥ ⎢⎣.1226 ⎥⎦ .1285 ⎥⎦

Hotel

Score

Cheraton

.3393

Milton

.1929

Harriott

.4677 1.0000

The Harriott Hotel should be selected for the meeting. 27.

Normalized matrices: Academic College

Row Averages

.3000

.2857

.3750

.3202

.6000

.5714

.5000

.5571

.1000

.1429

.1250

.1226

9-10 .

Location A

Row Averages

.4286

.4545

.3333

.4055

.4286

.4545

.5556

.4796

.1429

.0909

.1111

.1150

College

Cost College

Row Averages

.1429

.2857

.5714

Social A

Row Averages

.6667

.6923

.6000

.6530

.2222

.2308

.3000

.2510

.1111

.0769

.1000

.0960

College

Criteria

Academic

Location

Cost

Social

Row Averages

Academic

.2791

.3333

.2459

.4000

.3146

Location

.0698

.0833

.0984

.0667

.0795

Cost

.5581

.4167

.4918

.4000

.4667

Social

.0930

.1667

.1639

.1333

.1392

Academic Location A B C

⎡.3202 ⎢.5571 ⎢ ⎢⎣.1226

.4055 .4796 .1150

Cost

Social

Criteria

⎡.3146 ⎤ .1429 .6530 ⎤ ⎢.0795 ⎥ ⎥ = .2857 .2510 ⎥⎥ × ⎢ ⎢.4667 ⎥ .5714 .0960 ⎥⎦ ⎢ ⎥ ⎣.1392 ⎦

Aaron should select Barton (by a small margin). Consistency: (1) ⎡ 1 ⎢1/4 ⎢ ⎢ 2 ⎢ ⎣1/3

(2)

(3)

4 1/2 3 ⎤ ⎡.3146 ⎤ ⎡1.2836 ⎤ 1 1/5 1/2 ⎥⎥ ⎢⎢.0795 ⎥⎥ ⎢⎢ .3211 ⎥⎥ × = 5 1 3 ⎥ ⎢.4667 ⎥ ⎢1.9110 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 2 1/3 1 ⎦ ⎣ .1392 ⎦ ⎣ .5586 ⎦

9-11 .

College

Score

Arrington

.2906

Barton

.3817

Claiborne

.3277 1.0000

(3)/(2) ⎡ 4.0799 ⎤ ⎢ 4.0389 ⎥ ⎢ ⎥ 16.2267 = 4.0567, CI = 4.0567 − 4 = .019 ⎢ 4.0947 ⎥ 4 3 ⎢ ⎥ .019 4.40132 = .021 CI/RI = ⎣ ⎦ .90 16.2297

Since .021 < .10 this preference comparison for criteria is consistent. 28.

Normalized matrices: Attractiveness Client

Robin

Terry

Kelly

Row Averages

Robin

.6522

.6667

.6250

.6479

Terry

.2174

.2222

.2500

.2299

Kelly

.1304

.1111

.1250

.1222

Intelligence Client

Robin

Terry

Kelly

Row Averages

Robin

.2857

Terry

.1429

Kelly

.5714

Personality Client

Robin

Terry

Kelly

Row Averages

Robin

.2222

.4000

.1818

.2680

Terry

.1111

.2000

.2727

.1946

Kelly

.6667

.4000

.5455

.5374

Attractiveness

Intelligence

Personality

Row Averages

Attractiveness

.4286

.4000

.5000

.4429

Intelligence

.4286

.4000

.3333

.3873

Personality

.1429

.2000

.1667

.1698

Criteria

Attractiveness Intelligence Personality Robin Terry Kelly

⎡.6479 ⎢.2299 ⎢ ⎢⎣ .1222

.2857 .1429 .5714

.2680 ⎤ .1946 ⎥⎥ .5374 ⎥⎦

Criteria

Client

Score

⎡.4429 ⎤ ⎢.3873 ⎥ = ⎢ ⎥ ⎢⎣ .1698 ⎥⎦

Robin

.4431

Terry

.1902

Kelly

.3667 1.0000

The best match for Chris is Robin.

9-12 .

29.

Normalized matrices: Profit A

Row Averages

.7059

.7273

.6667

.6999

.1765

.1818

.2222

.1935

.1176

.0909

.1111

.1066

Project

P (Success) A

Row Averages

.2222

.1111

.6667

Project

Cost A

Row Averages

.1250

.1000

.1429

.1226

.3750

.3000

.2857

.3202

.5000

.6000

.5714

.5571

Criteria

Profit

P (Success)

Cost

Row Averages

Profit

.6000

.6154

.5455

.5869

P (Success)

.3000

.3077

.3636

.3238

Cost

.1000

.0769

.0909

.0893

Project

Profit A ⎡.6999 B ⎢⎢.1935 C ⎢⎣.1066

P (Success ) .2222 .1111 .6667

Cost

Project

Criteria

.1226 ⎤ ⎡.5869 ⎤ ⎥ .3202 ⎥ × ⎢⎢.3238⎥⎥ = ⎢⎣.0893⎥⎦ .5571⎥⎦

Score

.4937

.1781

.3282 1.0000

30.

The solution to this problem depends on how the students set up the preference comparisons for the three degree disciplines.

31.

Normalized matrices: Usage Facility

Gym

Field

Tennis

Pool

Row Averages

Gym

.2069

.1869

.2500

.2727

.2291

Field

.6207

.5607

.4167

.5455

.5359

Tennis

.0690

.1121

.0833

.0455

.0775

Pool

.1034

.1402

.2500

.1364

.1575

9-13 .

Cost Facility

Gym

Field

Tennis

Pool

Row Averages

Gym

.1364

.1448

.1053

.2000

.1466

Field

.5455

.5793

.6316

.4667

.5558

Tennis

.2727

.1931

.2105

.2667

.2358

Pool

.0455

.0828

.0526

.0667

.0619

Criteria

Usage

Cost

Row Averages

Usage

.8333

Cost

.1667

Usage

⎡.2291 Field ⎢⎢.5359 Tennis ⎢.0775 ⎢ Pool ⎣.1575 Gym

Cost

Facility

Score

.1466 ⎤ Criteria ⎥ .5558⎥ ⎡.8333 ⎤ = × ⎢ ⎥ .2358⎥ ⎣.1667 ⎦ ⎥ .0619 ⎦

Gym

.215

Field

.539

Tennis

.104

Pool

.142 1.0000

Rank: (1) Field; (2) Gym; (3) Pool; (4) Tennis Usage: (1)

(2)

(3)

2⎤ ⎡ 1 1/3 3 ⎡.2291⎤ ⎡ .9552 ⎤ ⎢ 3 ⎥ ⎢ ⎥ ⎢ 2.2407 ⎥ 1 5 4⎥ .5359 ⎥ ⎢ ⎥ × ⎢ = ⎢ ⎢1/3 1/5 1 1/3⎥ ⎢.0775⎥ ⎢ .3135⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣1/2 1/4 3 ⎣.1575⎦ ⎣ .6385⎦

(3)/(2) ⎡ 4.1695⎤ ⎢ 4.1812 ⎥ ⎢ ⎥ 16.4506 = 4.1127, CI = 4.1127 − 4 = .038 ⎢ 4.0458⎥ 4 3 ⎢ ⎥ .038 4.0541 ⎣ ⎦ CI/CR = = .042 .90 16.4506

Since .042 < .10 this pairwise comparison is consistent. Cost: (1) ⎡ 1 1/4 1/2 ⎢ 4 1 3 ⎢ ⎢ 2 1/3 1 ⎢ ⎣1/3 1/7 1/4

(2)

(3)

3⎤ ⎡.1446 ⎤ ⎡ .5872 ⎤ ⎥ ⎢ ⎥ ⎢ 2.2749 ⎥ 7⎥ .5558⎥ ⎥ × ⎢ = ⎢ ⎢.2358⎥ ⎢ .9579 ⎥ 4⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣.0619 ⎦ ⎣ .2485⎦

9-14 .

(3)/(2) ⎡ 4.0605⎤ ⎢ 4.0930 ⎥ ⎢ ⎥ 16.2295 = 4.0573, CI = 4.0573 − 4 = .019 ⎢ 4.0622 ⎥ 4 3 ⎢ ⎥ .019 4.0137 ⎣ ⎦ CI/CR = = .021 .90 16.2295

Since .021 < .10 this pairwise comparison is consistent. The pairwise comparison matrix for criteria is consistent since it includes only one comparison and the preference for one is an exact reciprocal of the other. 32.

Normalized matrices: Weather Location

Row Averages

.1429

.4286

Cost Location

Row Averages

.6522

.6667

.6250

.6479

.2174

.2222

.2500

.2299

.1304

.1111

.1250

.1222

Fun Location

Row Averages

.3125

.2729

.5556

.3803

.6250

.5455

.3333

.5013

.0625

.1818

.1111

.1185

Criteria

Weather

Cost

Fun

Row Averages

Weather

.1905

.4000

.1724

.2543

Cost

.0476

.1000

.1379

.0952

Fun

.7619

.5000

.6897

.6505

9-15 .

Weather

Cost

Fun

Criteria

Location

Score

⎡ .1429 ⎢.4286 ⎢ ⎢⎣.4286

.6479

.3803⎤ .5013⎥⎥ .1185 ⎥⎦

⎡.2543⎤ × ⎢⎢.0952 ⎥⎥ = ⎢⎣.6505 ⎥⎦

.3454

.4570

.1977

MB DB FL

.2299 .1222

1.0000 They will select Daytona Beach. 33.

Weather: (1)

(2)

(3)

⎡ 1 1/3 1/3⎤ ⎡ .1429 ⎤ ⎡ .4286 ⎤ ⎢3 ⎥ ⎢ ⎥ 1 1⎥ × ⎢.4286 ⎥ = ⎢⎢1.2859 ⎥⎥ ⎢ ⎢⎣3 ⎢⎣.4286 ⎥⎦ ⎢⎣1.2859 ⎥⎦ 1 1⎥⎦

(3)/(2) ⎡ 2.9995⎤ ⎢3.0002 ⎥ 9.0000/3 = 3.0000 ⎢ ⎥ ⎢⎣3.0002 ⎥⎦ 9.0000

This set of pairwise comparisons is perfectly consistent. Cost: (1)

(2)

(3)

3 5⎤ ⎡ 1 ⎡.6479 ⎤ ⎡1.9486 ⎤ ⎢1/3 ⎥ ⎢ ⎥ 1 2 ⎥ × ⎢.2299 ⎥ = ⎢⎢ .6903⎥⎥ ⎢ ⎢⎣1/5 1/2 1⎥⎦ ⎢⎣.1222 ⎥⎦ ⎢⎣ .3667 ⎥⎦

(3)/(2) ⎡3.0076 ⎤ ⎢3.0025⎥ ⎢ ⎥ ⎢⎣ 3.0011⎥⎦ 9.0111

9.0111 3.0037 − 3 = 3.0037, CI = = .0019 3 2 .0019 = .003 CI/RI = .58

Since .003 < .10 this pairwise comparison for cost is consistent. Fun: (1)

(2)

(3)

⎡ 1 1/2 5⎤ ⎡.3803⎤ ⎡1.2235⎤ ⎢ 2 ⎥ ⎢ ⎥ 1 3⎥ × ⎢.5013⎥ = ⎢⎢1.6174⎥⎥ ⎢ ⎢⎣1/5 1/3 1⎥⎦ ⎢⎣.1185 ⎥⎦ ⎢⎣ .3617 ⎥⎦

(3)/(2) ⎡ 3.2171⎤ ⎢3.2264 ⎥ ⎢ ⎥ ⎣⎢3.0520 ⎦⎥ 9.4955

9.4955 3.1652 − 3 = 3.1652, CI = = .0825 3 2 .0825 CI/RI = = .14 .58

9-16 .

Since .14 > .10 this pairwise comparison for fun is inconsistent and the comparisons should be made again. Criteria: (1)

(2)

(3)

⎡ 1 4 1/4 ⎤ ⎡.2543⎤ ⎡ .7977 ⎤ ⎢1/4 1 1/5⎥ × ⎢.0952 ⎥ = ⎢ .2889⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 4 5 ⎢⎣.6505 ⎥⎦ ⎢⎣ 2.1437 ⎥⎦ 1⎥⎦

(3)/(2) 3.1369 ⎡ ⎤ ⎢3.0344 ⎥ ⎢ ⎥ ⎢⎣3.2955⎥⎦ 9.4668

9.4668 3.1556 − 3 = 3.1556, CI = = .0778 3 2 CI/RI =

.0778 = .13 .58

Since .13 > .10 this pairwise comparison for the criteria also appears inconsistent. 34.

Normalized matrices: Aptitude Option

DSS

Row Averages

DSS

.75

.25

Row Averages

Jobs Option

DSS

.8000

.2000

Faculty Option

DSS

Row Averages

DSS

.1667

.8333

Criteria

Aptitude

Faculty

Jobs

Row Averages

Aptitude

.1429

.1111

.1579

.1373

Faculty

.2857

.2222

.2105

.2395

Jobs

.5714

.6667

.6316

.6232

Aptitude Faculty DSS OM

⎡.75 ⎢.25 ⎣

.1667 .8333

Jobs

Criteria

⎡ .1373 ⎤ .8000 ⎤ × ⎢⎢.2395⎥⎥ = .2000 ⎥⎦ ⎣⎢.6232 ⎦⎥

The student should select the DSS option.

9-17 .

Option

Score

DSS

.6415

.3585 1.0000

35.

The solution to this problem depends on how the student develops the pairwise comparisons for the individual criteria and between the criteria.

36.

Normalized matrices: PTA Option

Renovate

New

Row Averages

Renovate

.25

New

.75

Students Proposal

Renovate

New

Row Averages

Renovate

.6667

New

.3333

Teachers Option

Renovate

New

Row Averages

Renovate

.10

New

.90

Town Council Proposal

Renovate

New

Row Averages

Renovate

.8333

New

.1667

Group

PTA

Teachers

Students

Town Council

Row Averages

PTA

.1754

.2941

.2424

.1570

.2172

Teachers

.0351

.0588

.0303

.0897

.0535

Students

.0877

.2353

.1212

.1256

.1424

Town Council

.7018

.4118

.6061

.6278

.5868

Criteria

PTA Teachers Students Town Council Renovate New

⎡.25 ⎢.75 ⎣

.10

.6667

.90

.3333

.8333⎤ .1667 ⎥⎦

a) The old middle school should be renovated. b)

(1) ⎡ 1 ⎢1/5 ⎢ ⎢1/2 ⎢ ⎣ 4

(2)

(3)

2 1/4 ⎤ ⎡.2172 ⎤ ⎡ .9162 ⎤ ⎥ ⎢ ⎥ ⎢ .2164 ⎥ 1 1/4 1/7 ⎥ .0535⎥ ⎥ × ⎢ = ⎢ ⎢ .1424 ⎥ ⎢ .5824 ⎥ 4 1 1/5⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 7 5 1⎦ ⎣.5868⎦ ⎣ 2.5421⎦ 5

9-18 .

⎡.2172 ⎤ ⎢.0535 ⎥ ⎢ ⎥ ⎢ .1424 ⎥ ⎢ ⎥ ⎣.5868 ⎦

Proposal

Score

Renovate

.6436

New

.3564 1.0000

(3)/(2) ⎡ 4.2182 ⎤ ⎢ 4.0443⎥ ⎢ ⎥ 16.6843 = 4.1711, CI = 4.1711 − 4 = .057 ⎢ 4.0896 ⎥ 4 3 ⎢ ⎥ 4.3321 .057 ⎣ ⎦ CI/RI = = .063 16.6843 .90

Since .063 < .10 the pairwise comparison for the group is consistent. 37.

Answers depend on student preferences.

38.

Preference matrix: Criteria Hospital

Medical Care

Distance

Speed of Attention

Cost

County

.0960

.7043

.3202

.6999

Memorial

.6530

.1686

.5571

.1066

General

.2510

.1271

.1226

.1935

Normalized matrix: Criteria Medical Care

Distance

Speed of Attention

Cost

Row Averages

Medical Care

.6154

.4706

.5217

.6792

.5717

Distance

.0769

.0588

.0435

.0377

.0542

Speed of Attention

.1026

.1176

.0870

.0566

.0909

Cost

.2051

.3529

.3478

.2264

.2831

Criteria

1.000 Hospital

Score

Memorial

.4633

County

.3203

General

.2163

39. Criteria Port

Cost

Labor

Container/Shipping

Location

Criteria

Shanghai

.3339

.5321

.1226

.3093

.3086

Saigon

.5679

.3661

.3202

.3140

Karachi

.0982

.1018

.5571

.3767

9-19 .

.0665 .1246

Port

Score

Shanghai

.2383

Saigon

.3989

Karachi

.3627 1.000

40. Criteria Player

WAR

Contract

Injury History

Criteria

Aaron

.2145

.0820

.2207

.1279

Bass

.1096

.1593

.5828

Cabrera

.6323

.4129

.0624

Donald

.0436

.3459

.1340

Player

.3601 .5119

Score

Aaron

.1699

Bass

.3697

Cabrera

.2615

Donald

.1987 1.000

41. WAR:

CI = 0.0833 CI/RI = 0.0926

Contract:

CI = 0.0440 CI/RI = 0.0489

Injury History: CI = 0.1181 CI/RI = 0.1312 Not consistent Criteria:

CI = 0.0543 CI/RI = 0.0937

42.

The answer depends on student selection of apartments and criteria.

43.

Preference matrix: Criteria Section

Time/Day

Grading

Atmosphere

Homework

Jokes

.5681

.0800

.5287

.0613

.2103

.2410

.5017

.2198

.2677

.6173

.1333

.3548

.1231

.5698

.1145

.0576

.0635

.1284

.1012

.0579

9-20 .

Criteria

Time/Day

.4220

Grading

.3157

Atmosphere

.0552

Homework

.1723

Jokes

.0347

Section

Score

.3120

.3398

.2772

.0709 1.000

44. Coverage Textbook

Row Averages

.1000

.1304

.0588

.0964

.5000

.6522

.7059

.6194

.4000

.2174

.2353

.2842

Readability Textbook

Row Averages

.5455

.6000

.4286

.5247

.2727

.3000

.4286

.3338

.1818

.1000

.1429

.1416

Cost Textbook

Row Averages

.1250

.1111

.1304

.1222

.2500

.2222

.2174

.2299

.6250

.6667

.6522

.6479

Supplements Textbook

Row Averages

.7179

.7500

.6364

.7014

.1795

.1875

.2727

.2132

.1026

.0625

.0909

.0853

9-21 .

Criteria

Coverage

Readability

Cost

Supplements

Row Averages

Coverage

.1333

.1064

.1304

.1818

.1380

Readability

.2667

.2128

.1739

.4545

.2770

Cost

.5333

.6383

.5217

.2727

.4915

Supplements

.0667

.0426

.1739

.0909

.0935

Coverage Readability Cost Supplements Criteria

T __ex_t_b_o_o_k_S __co_r_e

.2843 ⎡.1380 ⎤ .7014 ⎤ ⎢ ⎥ .3109 .2770 ⎥ .3338 .2299 .2132 ⎥⎥ × ⎢ = .4049 ⎢.4915 ⎥ .1416 .6479 .0853 ⎥⎦ ⎢ ⎥ 1.0000 ⎣.0935 ⎦ The committee should select the Cook/Smith textbook. A ⎡.0964 B ⎢⎢.6194 C ⎢⎣.2842

.5247

.1222

Consistency: (1)

(2)

(3)

⎡ 1 1/2 1/4 2 ⎤ ⎡ .1380 ⎤ ⎡.5864 ⎤ ⎢ 2 1 1/3 5⎥⎥ ⎢⎢.2770 ⎥⎥ ⎢⎢.1843⎥⎥ ⎢ × = ⎢ 4 3 1 3⎥ ⎢.4915⎥ ⎢.1550 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣1/2 1/5 1/3 1⎦ ⎣.0935⎦ ⎣.3817 ⎦

(3)/(2) ⎡ 4.2491⎤ ⎢ 4.2756 ⎥ ⎢ ⎥ , 16.9919 = 4.2489, CI = 4.2489 − 4 = .0827 ⎢ 4.3845 ⎥ 4 3 ⎢ ⎥ 4.0827 ⎣ ⎦ 16.9919 .0827 CI/RI = = .0918 .90

Since .0918 < .10 this preference comparision for criteria is consistent, barely. 45. Salary Player

Row Averages

.6522

.6250

.6667

.6479

.1304

.1250

.1111

.1222

.2174

.2500

.2222

.2299

Speed Player

Row Averages

.1622

.1304

.3000

.2014

.8108

.6522

.6000

.7071

.0270

.2174

.1000

.0915

9-22 .

Position Player

Row Averages

.1250

.1111

.1304

.1222

.2500

.2222

.2174

.2299

.6250

.6667

.6522

.6479

Size Player

Row Averages

.5714

.5000

.6000

.5571

.1429

.1250

.1000

.1226

.2857

.3750

.3000

.3202

Criteria

Salary

Speed

Position

Size

Row Averages

Salary

.0909

.0984

.0690

.0909

.0873

Speed

.4545

.4918

.6207

.3636

.4827

Position

.2727

.1639

.2069

.3636

.2518

Size

.1818

.2459

.1034

.1818

.1782

Player

Salary

Speed

Position Size

A B C

⎡.6479 ⎢.1222 ⎢ ⎢⎣.2299

.2014 .7071 .0915

.1222 .2299 .6479

Player

Score

.2838

.4317

.2844

Row Averages

.5571⎤ ⎡.0873 ⎤ ⎢ ⎥ ⎥ .1226 ⎥ × ⎢.4827 ⎥ ⎢.2518 ⎥ .3202 ⎥⎦ ⎢ ⎥ ⎣.1782 ⎦

Select Bruce Kowslaski CI = .0359 CI/RI = .0399 These pairwise comparisons for criteria appear to be consistent. 46. Criteria

Criteria

General

Interpersonal Skills/ Leadership

Tactical Skills

Impact on Battles

Decision Making

Overall Success

Grant

0.1176

0.0975

0.0895

0.2944

0.6402

Lee

0.5175

0.2903

0.2370

0.4658

0.1798

Jackson

0.3038

0.5573

0.6111

0.1603

Sherman

0.0611

0.0549

0.0623

0.0795

9-23 .

Interpersonal Skills/Leadership

0.0402

Tactical Skills

0.0944

Impact on Battles

0.1223

0.1074

Decision Making

0.2018

0.0725

Overall Success

0.5413

General

Score

Grant

0.4308

Lee

0.2685

Jackson

0.2300

Sherman

0.0705

Select Grant

0.9999 47.

Solution will depend on how the student develops the pairwise comparisons.

48. Criteria

Criteria 0.0877

City Population Soccer Interest Media Market Competition Playing Facility A

0.5944

0.1875

0.5664

0.0980

0.0646

0.1267

0.0734

0.1057

0.0680

0.1083

0.2321

0.2201

0.2572

0.2270

0.2344

0.0651

0.0467

0.5191

0.0706

0.6070

0.5926

0.2087

Score =

City

Score

0.2502

0.0900

0.2303

0.4295

Select Durham

1.0000 (1)

49. ⎡ 1 ⎢ 5 ⎢ ⎢ 2 ⎢ ⎢1/2 ⎢⎣ 4

(2)

(3)

1/5 1/2 2 1/4 ⎤ ⎡.0877 ⎤ ⎡ .4440 ⎤ ⎥ ⎢ ⎥ ⎢ 2.6109 ⎥ 1 3 6 4⎥ ⎢.4841⎥ ⎢ ⎥ 1/3 1 3 1/2 ⎥ × ⎢.1543⎥ = ⎢ .7907 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 1/6 1/3 1 1/2 ⎥ ⎢.0651⎥ ⎢ .3454 ⎥ ⎢⎣.2087 ⎥⎦ ⎢⎣1.1193 ⎥⎦ 1/4 2 2 1⎥⎦

(3)/(2) ⎡5.0632 ⎤ ⎢ 5.3933 ⎥ ⎢ ⎥ ⎢ 5.1245 ⎥ ⎢ ⎥ ⎢5.3059 ⎥ ⎢⎣ 5.3633 ⎥⎦

26.2503 = 5.2501 5 5.2501 − 5 = .0625 CI = 4 .0625 = .0558 < .10 CI/RI = 1.12

26.2503

Since .0558 < .10 this preference comparison is consistent.

9-24 .

0.4841 0.1543

50. Criteria

Criteria

Media/Public Response

Cost ($)

GHG Reduction

Energy Savings

Media/Public Response

0.4821

Solar Power

0.5579

0.0576

0.4598

0.5125

Cost($)

0.27

Wind Power

0.2633

0.1337

0.2537

0.3010

× GHG Reduction

0.157

Vehicles

0.1219

0.2410

0.2009

0.1310

Energy Savings

0.088

Waste

0.0569

0.5681

0.0856

0.0555

Project

Score

Rank

Solar Power

0.4025

Wind Power

0.2299

Vehicles

0.1675

Waste

0.2002

Sum

1.000

51. Criteria

Criteria

City

Salary($)

Media Exposure/ Endorsements

Cleveland

0.5406

0.1250

0.2145

0.0670

Miami

0.0843

0.1250

0.5655

0.5727

New York

0.2230

0.5000

0.0831

0.1813

Chicago

0.1521

0.2500

0.1370

0.1790

City

Score

Cleveland

0.1529

Miami

0.4486

New York

0.2192

Chicago

0.1794

Sum

1.0000

City/ Lifestyle

Championship Potential

Select Miami

9-25 .

Salary ($)

0.0756

Media Exposure/ Endorsements

0.1905

City/Lifestyle

0.2644

Championship Potential

0.4695

52.

Salary: CI = .0392 CI/RI = .0436 Media Exposure/Endorsements: CI = .0000 CI/RI = .0000 City/Lifestyle:

59.

CI = .0328

S (West A) = 74.50

CI/RI = .0364

S (West B) = 67.25

Championship Potential: CI = .0759

S (East) = 76.75

CI/RI = .0843

Recommended the East campus site.

Criteria: CI = .0397

60.

CI/RI = .0441 53.

S (Western) = 76.65 S (Tech) = 71.62

Score (Crabtree) = 74.25

S (State) = 69.89

Score (Dowling) = 78.75

61.

Select Dowling.

S (D) = 74 S (B) = 70

S (Charlotte) = 76.4

S (A) = 69

S (Durham) = 74.3

62.

Recommend Charlotte.

S (Alexandria) = 80.55 S (Manassas) = 77.00

Solution depends on student selections and preferences.

S (Dupont) = 76.39

S (Abbeton) = 75.65

63.

S (Bayside) = 79.80 S (Cane Creek) = 74.15 S (Dunnville) = 77.85

S (Roanoke) = 84.74 ⎫ ⎪ S (Richmond) = 83.79 ⎬ Enter S (Greensboro) = 82.61⎪⎭ S (Charlotte) = 81.10 S (Bristol) = 79.02 S (Knoxville) = 76.93

Select Bayside. 57.

S (Tysons) = 85.27 S (Fairfax) = 80.94

Overall recommendation is student choice but should be either Charlotte or Durham.

56.

S (C) = 76 S (E) = 75

S (Atlanta) = 65.0 S (Birmingham) = 55.8

55.

S (Central) = 77.74 S (A&M) = 76.69

Score (Ashcroft) = 77.5 Score (Brainerd) = 75.5

54.

The ranking is slightly different from the AHP result; Cabrera is clearly first with the scoring model whereas he ranked 2nd with AHP. S (South) = 73.80

S (1) = 77.50 S (2) = 80.80

64.

Depends on student’s selection of weights and other factors.

65.

Depends on student’s selection of weights.

66.

Depends on the student’s selection of restaurants, decision criteria, and preferences.

S (3) = 82.05 Site 3 is recommended. 58. Player

Score

Rank

Aaron

75.50

Bass

76.70

Cabrera

82.00

Donald

74.75

9-26 .

yss + yes +yws + d15− − d15+ = 1,140

CASE SOLUTION: OAKDALE COUNTY SCHOOL BUSING

yse + yee +ywe + d16− − d16+ = 1,140

* 1.

xnw + xsw + xew + xww + ynw +

xne + xse + xee + xwe + yne +

xij = no. of white students bused from district i (i = n, s, e, w) to district j ( j = n, s, e, w); yij = no. of black students bused from district i (i = n, s, e, w) to district j ( j = n, s, e, w). minimize P1(d1−, d2−, d3−, d4−, d5−, d6−, d7−, d8−), P2(d9+, d10+ , d11+ , d12+ ), P3 d13+ , P4 (d14+ , d15+ , d16+ , d17+ ) subject to xnn + xns + xne + xnw + d1− = 1,000

ysw + yew +yww + d17− − d17+ = 1,360 The first 8 constraints at the first priority ensure that all students will be allocated to a school. The second 4 constraints reflect the first goal of achieving no more than 60% white students at each school. The 13th constraint reflects the goal for busing mileage, and the last 4 constraints are the goals for avoiding overcrowding. Notice that each school’s normal capacity has been increased by (approximately) 14%, the percentage by which the total number of students (5,000) exceeds the total school capacity (4,400).

ynn + yns + yne + ynw + d2− = 300 xsn + xss + xse + xsw + d3− = 450 ysn + yss + yse + ysw + d4− = 800 xen + xes + xee + xew + d5− = 1,050

yen + yes + yee + yew + d6− = 400

xnn = 645, xns = 355, xsw = 121, xss = 329, ynn = 300, yss = 456, yse = 300, ysw = 44,

xwn + xws + xwe + xww + d7− = 500

xen = 171, xec = 684, xew = 195, yen = 244,

ywn + yws + ywe + yww + d8− = 500

yee = 156, xww = 500, yww = 500,

.4xnn + .4xsn + .4xen + .4xwn − .6ynn − .6ysn −

School

.6yen − .6ywn + d9− − d9+ = 0

North

816

544

1,360

.4xns + .4xss + .4xes +

South

684

456

1,140

.4xws − .6yns − .6yss −

East

684

456

1,140

.6yes − .6yws + d10− − d10+ = 0

West

816

544

1,360

.4xne + .4xse + .4xee +

Total

3,000

2,000

5,000

.4xwe − .6yne − .6yse − .6yee − .6ywe + d11− − d11+ = 0 .4xnw + .4xsw + .4xew + .4xww − .6ynw − .6ysw − .6yew − .6yww + d12− − d12+ = 0 0xnn + 0ynn + 30xns + 30yns + 12xne + 12yne + 20xnw + 20ynw + 30xsn + 30ysn + 0xss + 0yss + 18xse + 18yse + 26xsw + 26ysw + 12xen + 12yen + 18xes + 18yes + 0xee 20ywn

+ 0yee + 24xew + 24yew + 20xwn + + 26xws + 26yws + 24xwe + 24ywe + 0xww + 0yww + d13− − d13+ = 30,000

xnn + xsn + xen + xwn + ynn + ysn + yen +ywn + d14− − d14+ = 1,360 xns + xss + xes + xws + yns +

9-27 .

White

Black

Total

CASE SOLUTION: CATAWBA VALLEY HIGHWAY PATROL

CASE SOLUTION: KATHERINE MILLER’S JOB SELECTION The solution depends on the student’s preferences and the criteria the student uses.

The goal programming model formulation is as follows. minimize P1d1+, P2d2−, P3d3−, P4d4−, P5d5− subject to

CASE SOLUTION: SELECTING NATIONAL BASEBALL HALL OF FAME MEMBERS

x1 + x2 + x3 + x4 + x5 + x6 = 23

The solution depends on the players the student selects and the criteria selected.

x1 ≥ 2 x2≥ 2

CASE SOLUTION: SUNTREK DISTRIBUTION CENTER

x3≥ 2 x4≥ 2

The solution depends on the grades the student assigns to each location for each criterion.

x5≥ 2 x6≥ 2 x1 ≤ 5 x2 ≤ 5 x3 ≤ 5 x4 ≤ 5 x5 ≤ 5 x6≤ 5 20x1 + 18x2 + 22x3 + 24x4 +17x5 + 19x6 + d1− − d1+ = 450 .27x1 + .21x2 + .28x3 + .19x4 + .23x5 + .33x6 + d2− − d2+ = 5 18x1 + 26x2 + 10x3 + 34x4 + 25x5 + 17x6 + d3− − d3+ = 350 1,700x1 + 900x2 + 650x3 + 230x4 + 1,600x5 + 520x6 + d4− − d4+ = 30,000 .32x1 + .65x2 + .43x3 + .87x4 + .55x5 + .49x6 + d5− − d5+ = 13 This model was solved iteratively using Excel. Following is the model formulation and solution output. The first 3 goals were achieved; however, d4− = 16,340 sight contacts and d5− = 0.96 minutes. (Thus, the actual response time is approximately 16 minutes.) x1 = 5, x2 = 5, x3 = 4, x4 = 2, x5 = 5, x6 = 2 d1− = 1, d2+ = 0.71, d3+ = 497, d4− = 4,900, d5− = 0.96

9-28 .

Chapter Ten: Nonlinear Programming PROBLEM SUMMARY

1. Profit analysis, model solution 2. Graphical analysis (10–1) 3. Profit analysis, model solution 4. Profit analysis, model solution 5. Profit analysis, model solution 6. Profit analysis, model solution 7. Profit analysis, model solution 8. Constrained optimization, model solution 9. Constrained optimization, model solution 10. Constrained optimization, model solution 11. Investment portfolio selection model 12. Constrained optimization model solution 13. Lagrange multiplier (10–12) 3. Z = 93,460p – 1,153.8p2 – 1,350,000

14. Model formulation and solution 15. Model formulation and solution

p = $40.50

16. Model formulation and solution

v = 28,269.60

17. Facility location model

Z = $542,641.95

18. Facility location model

4. Z = 18,983.1p – 5,666p2 – 13,950

19. Facility location model

p = $1.68

20. Portfolio selection model

v = 7,508.45 yd.

21. Portfolio selection model

Z = $1,950.02

22. Project selection model

5. Z = 242.75p – 4.75p2 – 4,300 p = $25.55

PROBLEM SOLUTIONS 1.

v = 78.64

Z = 448p – 1.2p2 – 23,500

Z = $1,198.56 (loss)

p = 186.67 v = 176 chairs Z = $18,313.33

10-1 .

6. Z = 4,800p – 80p2 – 65,000 p = 30 v = 1,600 Z = $7,000 7. p = $35.16 Z = $10,257.57 8. x1 = 12.08, x2 = 10.34, Z = $80.73 9. x1 = 10, x2 = 45, Z = $212.50 10. x1 = 121.5, x2 = 157, Z = $2,064.28 11. x1 = 0.09 x3 = 0.497 x4 = 0.413 Total return = 0.136 12. x1 = 15.4, x2 = 12.3, Z = $382 13. Lagrange multiplier = .27. The company would be willing to pay $0.27 for one additional hour of labor. 14. Maximize ⎛

solution: ph = $56.40 per kwh pl = $16.70 per kwh Z = $167.94 ⎛ .15 ⎞ ⎛ .21 ⎞ 16. Minimize Z = ⎜ .24 + ⎟ + ⎜ .37 + ⎟ x1 ⎠ ⎝ x2 ⎠ ⎝ ⎛ .12 ⎞ ⎛ .30 ⎞ + ⎜ .21 + ⎟ + ⎜ .48 + ⎟ x3 ⎠ ⎝ x4 ⎠ ⎝

subject to x1 + x2 + x3 + x4 ≤ 20

⎞

⎛

⎞

⎛

⎞

Z = ⎜⎜15,000 − 9,000 ⎟⎟ + ⎜⎜ 24,000 − 15,000 ⎟⎟ x1 ⎠ ⎝ x2 ⎠ ⎝ ⎛

+ ⎜⎜ 8,100 − 5,300 ⎟⎟ + ⎜⎜12,000 − 7,6000 ⎟⎟ x3 ⎠ ⎝ x4 ⎠ ⎝

11 ≤ 10 x1

8 ≤ 10 x2

10 ≤ 10 x3

9 ≤ 10 x4

solution: x1 = 5

⎛ ⎞ + ⎜⎜ 21,000 − 12,500 ⎟⎟ x 5 ⎝ ⎠

x2 = 5 x4 = 4

subject to

x5 = 6

x1 + x2 + x3 + x4 + x5 ≤ 15

Z = 1.45 crimes per 1,000 population

355x1 + 540x2 + 290x3 + 275x4 + 490x5 ≤ 6,500

17. x = 151.20

xi ≥ 0

y = 484.08

solution:

Z = 52,684.29

x1 = 3 18. Using Columbia, SC as the origin (0,0),

x2 = 4 x3 = 2

x = 56.81

x4 = 3

y = 115.97

x5 = 3

Z = 30,333.21 miles

Z = $64,000

Raeford, NC might be a good choice.

15. Maximize Z = (5.8 – .06ph + .005pl)ph + (3.0 – .11pl + .008ph)pl subject to 5.8 – .06ph + .005pl ≤ 2.5 3.0 –.11pl + .008ph ≤ 2.5 pl, ph ≥ 0

10-2 .

19.

x = 1, 665.4

x1 + x2 ≤ 1400 freshmen

y = 1,562.9

x1 + x2 ≥ 700 freshmen .55 x1 + .72 x2 ≤ 800 dorm rooms 960 x1 + 1,150 x2 ≥ 1,000 SAT score x1 + x2 x2 ≤ .55 of class, out-of-state x1 + x2

where: x1 = 21,000 – 12t1 x2 = 35,000 – 6t2 t1 = in-state tuition t2 = out-of-state tuition solution: t1 = $1,703.38

20. xA = .250

t2 = $5,719.37

xB = .750

Z = $4,863,622

Z (total variance) = .0382

x1 = 559.44

Total return = 0.11

x2 = 683.76

21. x1 = 0.25

CASE SOLUTION: SELECTING A EUROPEAN DISTRIBUTION CENTER SITE FOR AMERICAN INTERNATIONAL AUTOMOTIVE INDUSTRIES

x3 = .615 x4 = .359 Z (total variance) = .0361 Total return = 0.12 22. x1 = 5

Students will need to use a map of Europe to determine location coordinates. Bern, Switzerland was used as a reference point (0,0) to set up coordinates to plot the coordinates of the seven plant sites, as follows,

x2 = 3 x3 = 8 x4 = 3 x5 = 2 x6 = 5 x7 = 5 x8 = 4 Z = $4,193,450.37

CASE SOLUTION: ADMISSIONS AT STATE UNIVERSITY Maximize Z = x1t1 + x2t2 subject to

Plant sites (x, y)

Load

Vienna (300, 60)

160

Leipzig (180, 225)

100

Budapest (390, 50)

180

Prague (240, 160)

210

Krakow (400, 170)

Munich (150, 60)

120

Frankfurt (40, 160)

The solution is, x = 260 y = 127 Z = 100,195.59 total miles With this set of coordinates a good selection might be Prague.

10-3 .

Chapter Eleven: Probability and Statistics PROBLEM SUMMARY

35. Normal distribution 36. Normal (sampling) distribution

1. Probabilities

37. Normal (sampling) distribution

2. Probabilities

38. Normal (sampling) distribution

3. Determining event probabilities

39. Normal (sampling) distribution

4. Determining event probabilities

40. Chi-square test

5. Binomial distribution

41. Chi-square test

6. Binomial distribution

PROBLEM SOLUTIONS

7. Binomial distribution 8. Binomial distribution

9. Binomial distribution

a) Subjective b) Subjective

10. Conditional probability, Bayesian analysis

c) Subjective and objective

11. Probability tree, marginal and joint probability, Bayesian analysis

d) Subjective

12. Probability tree

e) Objective

13. Probability tree, marginal, conditional, and joint probability, Bayesian analysis

g) Objective

14. Probability tree, marginal, conditional, and joint probability

h) Objective 2.

Subjective

a) 12/52

15. Bayes’ rule

b) 4/52

16. Bayes’ rule

c) 13/52 = 1/4

17. Expected value

d) 1/52

18. Expected value

a) Snowfall

Probability

19. Expected value

0–19

.066

20. Decisions and expected value

20–29

.233

21. Decisions and expected value

30–39

.266

22. Expected value

40–49

.266

23. Expected value, cumulative distribution

50+

.166

24. Normal distribution

b) Yes; only one of the events can take place each winter.

25. Normal distribution 4.

26. Normal distribution

a) Employee

Probability

27. Normal distribution

Female and union

.12

28. Normal distribution (CPM/PERT)

Female and nonunion

.25

29. Normal distribution (inventory)

Male and union

.21

30. Normal distribution

Male and nonunion

.42 1.00

31. Normal distribution (break-even analysis)

b) Yes; two or more of the events cannot occur at the same time.

32. Normal distribution 33. Normal distribution

34. Normal distribution

11-1 .

.63

d) Yes; contains all the employees in the industry. These are all of the events that can possibly occur. 5.

Binomial distribution: P(r ≥ 5) = .0433, rejected; P(r < 5) = .9567, accepted

Binomial distribution: P(r = 3) = .0574

Binomial distribution: P(r = 4) = 0.0367

P(r ≥ 3) = 1 − P(r < 3)

= 1 − [ P(r = 0) + P(r = 1) + P(r = 2)] 7! 7! ⎡ = 1− ⎢ (.2)0 (0.8)7 + (.2)t (0.8)6 1!(7 − 1)! ⎣ 0!(7 − 2)! 7! ⎤ (.2)2 (.8)5 ⎥ + 2!(7 − 2)! ⎦ = 1 − (.2097 + .3668 + .27527) = 1 − .852 P(r ≥ 3) = .148 9.

P(r ≥ 4) = 1 − P(r ≤ 3) = 1 − [ P(r = 0) + P(r = 1) + P(r = 2) + P(r = 3)] 12! ⎡ 12! (.05)0 (.95)12 + (.05)1 (.95)11 =⎢ 1!(12 − 1)! ⎣ 0!(12 − 0)! 12! 12! ⎤ (.05)2 (.95)10 + (.05)3 (.95)9 ⎥ + 2!(12 − 2)! 3!(12 − 3)! ⎦ = 1 – (.5404 + .3413 + .0988 + .0173) = 1 – .9976 = .0024

10.

P(S) = .23

P(I|S) = .18

P(NS) = .77

P(I|NS) = .06

P(S|I ) = =

P(I|S)P(S) P(I|S)P(S) + P(I|NS)P(NS) (.18)(.23) (.18)(.23) + (.06)(.77)

= .47

11-2 .

11. a)

b) Outcome

Firm

Win

Lose

Marginal Probabilities

Abercrombie

.28

.12

.40

Olson

.36

.24

.60

Marginal Probabilities

.64

.36

1.00

P (O|W ) = =

P (W|O)P (O) P (W|O)P (O) + P(W|A)P (A)

(.60)(.60) = .563 (.60)(.60) + (.70)(.40)

12.

Probability of passing is P(PD) and P(PR) = .23 + .24 = .47. Using expected value, the expected number of votes needed to pass is E = .47(100) = 47; therefore the bill will not pass.

11-3 .

13. a)

b) Quality

Marginal Probabilities

Company

Good

Defective

.198

.002

.200

.392

.008

.400

.368

.032

.400

Marginal Probabilities

.958

.042

1.00

P(B|D) = .008/.042 = .19

14. a)

11-4 .

20.

b) Test

c) 15.

District

Pass

Fail

Marginal Probabilities

North

.225

.025

.250

South

.340

.060

.400

Central

.332

.018

.350

Marginal Probabilities

.897

.103

1.00

21.

P(fail) = P(FN) + P(FS) + P(FC) = .103 P(A) = .50, P(B) = .50, P(D|A) = .20, P(D|B) = .10 P (D|A)P (A) P (A|D) = P (D|A)P (A) + P (D|B)P (B) =

16.

(.20)(.50) .10 = = .667 (.20)(.50) + (.10)(.50) .15

P(E) = .35, P(W) = .65, P(F|E) = .10, P(F|W) = .25 P (E|F ) = =

17.

18.

x − E(x)

[x − E(x)]2

[x − E(x)]2P(x)

144

43.2

484

96.8

1,764

176.4

−28

784

235.2

−38

1,444

144.4 696.0

σ = 696 = 26,381 Bonds: E(bonds) = 30,000(.6) + 40,000(.4) = 18,000 + 16,000 = 34,000

(.10)(.35) .035 = .177 = (.10)(.35) + (.25)(.65) .1975

E(grade) = 4(.1) + 3(.3) + 2(.4) + 1(.1) + 0(.1) = .4 + .9 + .8 + .1 + 0 = 2.2 x x − E(x)

19.

P (F|E)P (E) P(F|E) P (E) + P(F|W)P (W)

E(barrels) = 6(.10) + 7(.20) + 8(.40) + 9(.25) + 10(.05) = .6 + 1.4 + 3.2 + 2.25 + .50 = 7.95 barrels

4 3 2 1 0

E(A) = .6(900,000) + .4(−8000,000) = 540,000 − 320,000 = $220,000 E(B) = .6(120,000) + .4(70,000) = 72,000 + 28,000 = 100,000 Alternative A results in the largest expected value; however, before making a decision, the investor should consider the large potential loss with A. This is a consideration discussed in Chapter 12. Office building: E(building) = 50,000(.3) + 60,000(.2) + 80,000(.1) + 10,000(.3) = $38,000

1.8 .8 −.2 −1.2 −2.2

[x − E(x)]2

[x − E(x)]2P(x)

3.24 .64 .04 1.44 4.84

.324 .192 .016 .144 .484 1.160

x − E(x)

[x − E(x)]2

[x − E(x)]2P(x)

9.6

14.4 24.0

σ = 24 = 4.899, or $4,899 “Building” has a greater expected value, but it varies to a much greater extent. Therefore, “bond” is probably a less risky investment. 22.

P(win) = 4/2,000 = .002 E(win) = 6,000(.002) + 0(.998) − $4 = $12 − 4 = $8 Hours

Probability

Cumulative Probability

.15

23.

σ 2 = 1.160 E(lead time) = 1(.2) + 2(.5) + 3(.2) + 4(.1) = .2 +1.0 + .6 + .4 = 2.2 days 2

.20

.35

[x − E(x)] P(x)

.40

.75

.25

1.00

x − E(x)

[x − E(x)]

−1.2

1.44

.288

−.2

.04

.020

.64

.128

1.8

3.24

.324 .760

1.00 E(time between breakdowns) = 1(.15) + 2(.20) + 3(.40) + 4(.25) = 2.75 hr between breakdowns

σ = .760 = .87

11-5 .

28.

24.

500 − 400 100 = = 1.25 80 80

19 − 14 5 = = 1.25; P ( x ≥ 19) = .1056; 4 4 this is the type of probabilistic analysis done in Chapter 8 on CPM/PERT. Z=

From Table A.1: Z = 1.25; P = .3944 + .5000; P(x ≥ 400) = .8944 25.

29.

3.5 − 2.6 .9 = = 1.50; P( x ≥ 3.5) = .0668 .6 .6

26. Z=

6,000 − 4,500 1,500 = = 1.67; P( x ≥ 6,000) = .0475; 900 900

this is the type of probabilistic analysis done in Chapter 16 on inventory control. 30.

55 − 50 5 = = .83; P (45 ≤ x ≤ 55) = .2967 6 6 + .2967 = .5934

27. x−μ

x − 180 σ 60 A probability of .4000 has a Z value of 1.28: Z=

x − 180 60 76.8 = x − 180 x = 256.8 1.28 =

900 − 700 = 1.00; P( x ≥ 900) = .1587 200

or approximately 257 recorders must be ordered to meet 90% of customer demand.

11-6 .

35.

31.

36. Z=

32.

x−μ

670 − 805 _ 135 = = −.65 207 207

μ = 125,000 σ = 40,000 x − μ 200,000 − 125,000 75,000 Z= = = = 1.88 σ 40,000 40,000 p(x ≥ 200,000) = .4699 + .5000 = .9699 x=

⎛ n ⎞ ⎜ ∑ xi ⎟ n 2 xi − ⎝ i =1 ⎠ ∑ 4,569 − 4,284.9 n s 2 = i =1 = 9 n −1 s 2 = 31.57

σ From Table A.1: Z = .65; P = .2422; P(x ≥ 670) = .2422 + 5,000 = .7422 μ = 63

σ = 10

s = s 2 = 31.57 = 5.61

xB − 63 (for an area equal to .20, Z =.52) 10

An “A” is a grade greater than or equal to 75.8

25 − 20.7 5.61 = .766 P(x ≥ 25) = .2206 The goal is P(x ≥ 25) = .10; thus the hotel is not providing the desired level of customer service. 366,200 = 30,516.67 x= 12 s = 3,997 x − μ 27,000 − 30,516.67 = Z= σ 3,997 = −.8798

A “B” is a grade between 68.2 and 75.8

P(x ≤ 27,000) = .5000 − .3106

.52 =

5.2 = x − 63 xB = 68.2 x A − 63 (for an area equal to .40, 10 Z = 1.28 12.8 = xA – 63

1.28 =

37.

xA = 75.8 xC = 63 – 5.2 = 57.8 xD = 63 – 12.8 = 50.2

= .1894

A “C” is a grade between 57.8 and 68.2 A “D” is a grade between 50.2 and 57.8 33.

38.

An “F” is a grade less than 50.2 μ = 1,050 σ = 120 1,200 − 1,050 Z= 120 = 1.25

1,669 = 83.45 20 s = 6.78 85 − 83.45 Z= = .23 6.78 x=

P(x ≥ 85) = .5000 − .0910

P(x ≥ 1,200) = .5000 − .3944 = .1056 .1056 × 620 = 65.5 34.

207 10

μ = 2.67 σ = 0.58

39.

P(x ≥ = ?) = .10 Z = 1.28

x − 2.67 = 1.28 0.58 x = 3.41 GPA

= .409 There are 62 days in July and August; thus, the expected number of days the high temperature will be 85° or greater is (62)(.409) = 25.36. x = 72.35 s = 16.68 75 − 72.35 2.65 = = .159 Z= 16.68 16.68 P(x ≥ 75) = .5000 − .0636 = .4364

11-7 .

40.

(fo−ft)2/ft

20.75

.870

40.75

.811

.255

63.75

.166

2.5−3.0

.255

63.75

.519

3.0−3.5

.163

40.75

.958

3.5−4.0

.083

20.75

.365

Ranges

Area

0−0.5

.003

.75

0.5−1.0

.016

4.0

1.0−1.5

.064

16.0

1.5−2.0

.163

2.0−2.5

⎫ ⎪ ⎬ ⎪ ⎭

3.689 2 χ.05,3 = 7.815 > 3.689

The distribution is normal. 41. Ranges

Area

Theoretical Frequency

0−5

.006

2.4

5−10

.117

46.8

10−15

.452

180.8

15−20

.363

145.2

20−25

.060

24.0

25+

.002

(fo−ft)2/ft

49.2

127

123.03

180.8

139

9.66

145.2

105

11.13

24.8

36.78

⎫ ⎪ ⎬ ⎪ ⎭

180.59 2 χ.05,1 = 3.841

Since 180.59 > 3.841, the distribution is normal.

CASE SOLUTION: VALLEY SWIM CLUB

Number of days attendance greater than 500 = 92 × .0351 = 3.22. This criteria is met.

The mean and standard deviation were obtained using Excel, as follows, for daily attendance. x = 314.3 s = 102.3 x = 432.27 s = 77.51

Evaluating the criteria: Z=

x−μ

The current average daily attendance is 314.3, which is less than 320, so this criteria is met.

The average daily weekend attendance is 432.27, which is less than 500, so this criteria is met. Valley Swim Club should issue new shares of stock. The current daily average is 314.3, and 400 – 314.3 = 85.7. Dividing 85.7 by 2 equals 42.85. Thus approximately 43 new shares should be issued. This will generate $43,000 from the share sale plus the annual dues of $175 per shareholder, $7,525, for a total increase in revenue of $50,525.

The statistics for weekend attendance are,

500 − 314.3 = 1.81 102.3

P(x ≥ 500) = .5000 – .4649 = .0351

11-8 .

Chapter Twelve: Decision Analysis 39. Expected value (12–38) 40. Decision-making criteria without probabilities 41. Expected value (12–40) 42. Decision tree (12−25) 43. Sequential decision tree 44. Sequential decision tree 45. Sequential decision tree 46. Bayesian analysis, EVSI (12–15) 47. Bayesian analysis, EVSI (12–18), decision tree 48. Bayesian analysis, EVSI (12–25) 49. Bayesian analysis 50. EVPI (12–10 and 12–21) 51. EVPI (12–33 and 12–34) 52. Sequential decision tree 53. Sequential decision tree 54. Sequential decision tree 55. Sequential decision tree 56. Sequential decision tree 57. EVSI, EVPI (12-56) 58. Sequential decision tree 59. Sequential decision tree 60. Sequential decision tree (12–59) 61. Sequential decision tree (12–14, 12–37) 62. Bayesian analysis, EVSI (12–13, 12–24) 63. Utility 64. Expected value, utility 65. Sequential decision tree

PROBLEM SUMMARY 1. Decision-making criteria without probabilities 2. Decision-making criteria without probabilities 3. Decision-making criteria without probabilities 4. Decision-making criteria without probabilities 5. Decision-making criteria without probabilities 6. Decision-making criteria without probabilities 7. Decision-making criteria without probabilities 8. Decision-making criteria without probabilities 9. Decision-making criteria without probabilities 10. Decision-making criteria without probabilities 11. Decision-making criteria without probabilities 12. Decision-making criteria without probabilities 13. Decision-making criteria without probabilities 14. Decision-making criteria without probabilities 15. Expected value 16. Expected value and opportunity loss 17. Expected value 18. Expected value and opportunity loss, EVPI 19. Expected value 20. Indifferent probability 21. Expected value (12–10) 22. Expected value (12–11) 23. Expected value, EVPI (12–12) 24. Expected value, EVPI (12–13) 25. Expected value 26. Payoff table, expected value 27. Payoff table, expected value, EVPI 28. Payoff table, expected value, EVPI 29. Payoff table, expected value, EVPI 30. Payoff table, decision making without probabilities (12–28) 31. Payoff table, decision making without probabilities, with costs 32. Expected value (12–31) 33. Payoff table, decision making without probabilities, with costs 34. Expected value (12–33) 35. Payoff table, decision making without probabilities, with costs 36. Expected value (12-35) 37. Payoff table, decision making without probabilities, with costs 38. Expected value (12-33)

PROBLEM SOLUTIONS 1. a) Lease land; maximum payoff = $90,000 b) Savings certificate; maximum of minimum payoffs = $10,000 2. a) Drive-up window; maximum payoff = $20,000 b) Breakfast; maximum of minimum payoffs = $4,000 3. a) Bellhop Management

Good

Recession

25,000

35,000

Choose bellhop job. b) Bellhop: 120,000(.4) + 60,000(.6) = $84,000; management: 85,000(.4) + 85,000(.6) = $85,000; select management job.

12-1 .

c) Bellhop: 120,000(.5) + 60,000(.5) = $90,000; management: 85,000(.5) + 85,000(.5) = 85,000; select bellhop job. 4. a) Course III, maximax payoff = A b) Course I, maximin payoff = D

8. a) Purchase motel; maximax payoff = $20,000 b) Purchase theater; maximin payoff = $5,000 c)

5. a) Plant corn; maximax payoff = $35,000 b) Plant soybeans; maximin payoff = $20,000 c)

Pass

Fail

Corn

6. a) b) 7. a) b)

12,000

Peanuts

17,000

8,000

Soybeans

13,000

Houses

Stable

Increase

35,000

10,000

20,000

65,000

Shopping center Lease

Stable

Surplus

Motel

14,000

Restaurant

4,000

7,000

14,000

9,000

15,000

Theater

Select either motel or restaurant (both have minimum regret values of $14,000). d) Motel: 20,000(.4) − 8,000(.6) = $3,200; restaurant: 8,000(.4) + 2,000(.6) = $4,400; theater: 6,000(.4) + 5,000(.6) = $5,400; select theater. e) Motel: − 8,000(.33) + 15,000(.33) + 20,000(.33) = $8,910; restaurant: 2,000(.33) + 8,000(.33) + 6,000(.33) = $5,280; theater: 6,000(.33) + 6,000(.33) + 5,000(.33) = $5,610; select motel. 9. a) Select Army vs. Navy; maximax payoff = 12.5 b) Select Alabama vs. Auburn; maximin payoff = 5.4 c) Equal likelihood criterion: EV(A|A) = 10.2(.33) + 7.3(.33) + 5.4(.33) = 7.6 EV(G|GT) = 9.6(.33) + 8.1(.33) + 4.8(.33) = 7.4 EV(A|N) = 12.5(.33) + 6.5(.33) + 3.2(.33) = 7.3 Select Alabama vs. Auburn. 10. a) Risk fund, maximax payoff = $147,000 b) Savings bonds maximin payoff = $30,000 c) Money market: 2(.2) + 3.1(.2) + 4(.2) + 4.3(.2) + 5(.2) = 36,000; stock growth: −3(.2) − 2(.2) + 2.5(.2) + 4(.2) + 6(.2) = 15,000; bond: 6(.2) + 5(.2) + 3(.2) + 3(.2) + 2(.2) = 38,000; government: 4(.2) + 3.6(.2) + 3.2(.2) + 3(.2) + 2.8(.2) = 33,200; risk: −9(.2) −4.5(.2) + 1.2(.2) + 8.3(.2) + 14.7(.2) = 21,400; savings bonds: 3(.2) + 3(.2) + 3.2(.2) + 3.4(.2) + 3.5(.2) = 32,200; select bond fund.

Plant corn; minimum regret = $12,000 Corn: $35,000(.3) + 8,000(.7) = $16,100; peanuts: 18,000(.3) + 12,000(.7) = $13,800; soybeans: 22,000(.3) + 20,000(.7) = $20,600; plant soybeans. Corn: 35,000(.5) + 8,000(.5) = $21,500; peanuts: 18,000(.5) + 12,000(.5) = $15,000; soybeans: 22,000(.5) + 20,000(.5) = $21,000; plant corn. Note that this payoff table is for costs. Product 3, minimin payoff = $3.00 Product 3, minimax payoff = $6.50 Build shopping center; maximax payoff = $105,000 Lease equipment; maximin payoff = $40,000

Shortage

Build shopping center. d) Houses: $70,000(.2) + 30,000(.8) = $38,000; shopping center: $105,000(.2) + 20,000(.8) = $37,000; lease: $40,000(.2) + 40,000(.8) = $40,000; lease equipment. e) Houses: 70,000(.5) + 30,000(.5) = $50,000; shopping center: 105,000(.5) + 20,000(.5) = $62,500; lease: 40,000(.5) + 40,000(.5) = $40,000; build shopping center.

12-2 .

11.

Wide 54

Tackle

Nickel

Blitz

Off tackle

−2

−1

Option

−1

−2

Toss sweep

−5

Draw

−2

−3

Pass

−7

−8

Screen

−5

−2

South Korea: 21.7(.33) + 19.1(.33) + 15.2(.33) =18.48 China: 19.0(.33) + 18.5(.33) + 17.6(.33) = 18.18 Taiwan: 19.2(.33) + 17.1(.33) + 14.9(.33) = 16.90 ← minimum Philippines: 22.5(.33) + 16.8(.33) + 13.8(.33) = 17.52 Mexico: 25.0(.33) + 21.2(.33) + 12.5(.33) = 19.37 Select Taiwan 13. a) Maximax criteria:

a) Pass, maximax payoff = 20 yd b) Either off tackle or option, maximin payoff = −2 yd c) Off tackle: 3(.2) −2(.2) + 9(.2) + 7(.2) −1(.2) = 3.2; option: −1(.2) + 8(.2) −2(.2) + 9(.2) + 12(.2) = 5.2; toss sweep: 6(.2) + 16(.2) −5(.2) + 3(.2) + 14(.2) = 6.8; draw: −2(.2) + 4(.2) + 3(.2) + 10(.2) −3(.2) = 2.4; pass: 8(.2) + 20(.2) + 12(.2) −7(.2) −8(.2) = 5.0; screen: −5(.2) −2(.2) + 8(.2) + 3(.2) + 16(.2) = 4.0; use toss sweep. 12. a) Minimin: South Korea 15.2 China 17.6 Taiwan 14.9 Philippines 13.8 Mexico 12.5 ← minimum Select Mexico b) Minimax: South Korea 21.7 China 19.0 ← minimum Taiwan 19.2 Philippines 22.5 Mexico 25.0 Select China c) Hurwicz (α = .40): South Korea: 15.2(.40) + 21.7(.60) = 19.10 China: 17.6(.40) + 19.0(.60) = 18.44 Taiwan: 14.9(.40) + 19.2(.60) = 17.48 ← minimum Philippines: 13.8(.40) + 22.5(.60) = 19.02 Mexico:12.5(.40) + 25.0(.60) = 20.0 Select Taiwan d) Equal likelihood:

Office park 4.5 ← maximum Office building 2.5 Warehouse 1.7 Shopping center 3.6 Condominiums 3.2 Select office park b) Maximin criteria: Office park 0.5 Office building 1.5 ← maximum Warehouse 1.0 Shopping center 0.7 Condominiums 0.6 Select office building c) Equal likelihood Office park: 0.5(.33) + 1.7(.33) + 4.5(.33) = 2.21 ← maximum Office building: 1.5(.33) + 1.9(.33) + 2.5(.33) = 1.95 Warehouse: 1.7(.33) + 1.4(.33) + 1.0(.33) = 1.35 Shopping center: 0.7(.33) + 2.4(.33) + 3.6(.33) = 2.21 ← maximum

12-3 .

Condominiums: 3.2(.33) + 1.5(.33) + 0.6(.33) = 1.75 Select office park or shopping center

17.

d) Hurwicz criteria (α = .3) Office park: 4.5(.3) + 0.5(.7) = 1.70 Office building: 2.5(.3) + 1.5(.7) =1.80 ← maximum Warehouse: 1.7(.3) + 1.0(.7) = 1.21 Shopping center: 3.6(.3) + 0.7(.7) = 1.57 Condominiums: 3.2(.3) + 0.6(.7) = 1.38 Select office building 14. a) Maximax = Gordan b) Maximin = Johnson c) Hurwicz (α = .60) Byrd = 4.4(.6) + (−3.2)(.4) = $1.36M O’Neil = 6.3(.6) + (−5.1)(.4) = $1.74M Johnson = 5.8(.6) + (−2.7)(.4) = $2.40M Gordan = 9.6(.6) + (−6.3)(.4) = $3.24M Select Gordan d) Equal likelihood Byrd = 4.4(.33) + (1.3)(.33) + (−3.2)(.33) = +$0.83M O’Neil = 6.3(.33) + (1.8)(.33) + (−5.1)(.33) = +$.99M Johnson = 5.8(.33) + (0.7)(.33) + (−2.7)(.33) = +$1.254M Gordan = 9.6(.33) + (−1.6)(.33) + (−6.3)(.33) = $.561M Select Johnson 15. EV(press) = 40,000(.4) − 8,000(.6) = $11,200; EV(lathe) = 20,000(.4) + 4,000(.6) = $10,400; EV(grinder) = 12,000(.4) + 10,000(.6) = $10,800; purchase press. 16. a) EV(sunvisors) = −500(.3) −200(.15) + 1500(.55) = $645; EV(umbrellas) = 2,000(.3) + 0(.15) − 900(.55) = $105; carry sunvisors. b) Opportunity loss table:

EV (snow shoveler) = $30(.13) + 60(.18) + 90(.26) + 120(.23) + 150(.10) + 180(.07) + 210(.03) = $99.60 The cost of the snow blower ($625) is much more than the annual cost of the

snow shoveler; thus on the basis of one year the snow shoveler should be used. However, the snow blower could be used for an extended period of time such that after approximately 6 years the cost of the snow blower would be recouped. Thus, the decision hinges on whether or not the decision maker thinks 6 years is too long to wait to recoup the cost of the snow blower. 18. a) EV(widget) = 120,000(.2) + 70,000(.7) − 30,000(.1) = $70,000; EV(hummer) = 60,000(.2) + 40,000(.7) + 20,000(.1) = $42,000; EV(nimnot) = 35,000(.2) + 30,000(.7) + 30,000(.1) = $31,000; introduce widget. b)

Rain

Overcast

Sunshine

Sunvisors

2,500

200

Umbrellas

2,400

Favorable

Stable

Unfavorable

Widget

60,000

Hummer

60,000

30,000

10,000

Nimnot

85,000

40,000

EOL(A) = 0 + 0 + 60,000(.1) = $6,000; EOL(B) = 60,000(.2) + 30,000(.7) + 10,000(.1) = $34,000; EOL(C) = 85,000(.2) + 40,000(.7) + 0 = $45,000; select A (widget). c) Expected value given perfect information = 120,000(.2) + 70,000(.7) + 30,00(.1) = 76,000; EVPI = 76,000 − EV(widget) = 76,000 − 70,000 = $6,000; the company would consider this a maximum, and since perfect information is rare, it would pay less than $6,000 probably. 19. EV(operate) = 120,000(.4) + 40,000(.2) + (−40,000)(.4) = $40,000; leasing = $40,000; if conservative, the firm should lease. Although the expected value for operating is the same as leasing, the lease agreement is not subject to uncertainty and thus does not contain the potential $40,000 loss. However, the risk taker might attempt the $120,000 gain. 20. To be indifferent, the expected value for the investments would equal each other: EV(stocks) = EV(bonds). Next, let the probability of good economic conditions equal p and the probability of bad conditions equal 1 − p:

EOL(sunvisors) = 2,500(.3) + 200(.15) + 0 = $780; EOL(umbrellas) = 0 + 0 + 2,400(.55) = $1,320; select sunvisors since it has the minimum expected regret.

12-4 .

EV(stocks) = 10,000( p) + −4,000(1 − p) EV(bonds) = 7,000(p) + 2,000(1 − p) EV(stocks) = EV(bonds) 10,000(p) + (−4,000) (1 − p) = 7,000(p) + 2,000(1 − p) 10,000p − 4,000 + 4,000p = 7,000p + 2,000 − 2,000p 9,000p = 6,000 p = .667 Therefore, probability of good conditions = p = .667; probability of bad conditions = 1 − p = .333. 21. EV(money market) = 2(.2) + 3.1(.3) + 4(.3) + 4.3(.1) + 5(.1) = 34,600; EV(stock growth) = −3(.2) − 2(.3) + 2.5(.3) + 4(.1) + 6(.1) = 5,500: EV(bond) = 6(.2) + 5(.3) + 3(.3) + 3(.1) + 2(.1) = 41,000; EV(government) = 4(.2) + 3.6(.3) 3.2(.3) + 3(.1) + 2.8(.1) = 34,200; EV(risk) = −9(.2) − 4.5(.3) + 1.2(.3) + 8.3(.1) + 14.7(.1) = −49,000; EV(savings bonds) = 3(.2) + 3(.3) 3.2(.3) + 3.4(.1) + 3.5(.1) = 31,500; purchase bond fund. 22. a) EV(off tackle) = 3(.4) − 2(.10) + 9(.20) + 7(.20) − 1(.10) = 4.10; EV(option) = −1(.4) + 8(.10) − 2(.20) + 9(.2) + 12(.10) = 3; EV(toss sweep) = 6(.4) + 16(.10) − 5(.20) + 3(.20) + 14(.10) = 5.0; EV(draw) = − 2(.4) + 4(.10) + 3(.20) + 10(.20) − 3(.10) = 1.9; EV(pass) = 8(.4) + 20(.10) + 12(.20) − 7(.20) − 8(.10) = 5.4; EV(screen) = − 5(.4) − 2(.10) + 8(.20) + 3(.20) + 16(.10) = 1.6; PASS is best, followed by toss sweep, off tackle, option, draw, and screen. b) EV(off tackle) = 3(.10) − 2(.10) + 9(.10) + 7(.10) − 1(.60) = 1.1; EV(option) = −1(.10) + 8(.10) − 2(.10) + 9(.10) + 12(.60) = 8.6; EV(toss sweep) = 6(.10) + 16(.10) − 5(.10) + 3(.10) + 14(.60) = 10.4; EV(draw) = − 2(.10) + 4(.10) + 3(.10) + 10(.10) − 3(.60) = −.3; EV(pass) = 8(.10) + 20(.10) + 12(.10) − 7(.10) − 8(.60) = −1.5; EV(screen) = − 5(.10) − 2(.10) +

8(.10) + 3(.10) + 16(.60) = 10.0; select toss sweep. Yes, it is likely Tech will make the first down. 23. EV (South Korea) = 21.7(.40) + 19.1(0.5) + 15.2(.10) = 19.75 EV (China) = 19.0(.40) + 18.5(.50) + 17.6(.10) = 18.61 EV (Taiwan) = 19.2(.40) + 17.1(.50) + 14.9(.10) = 17.72 ← minimum EV (Philippines) = 22.5(.40) + 16.8(.50) + 13.8(.10) = 18.78 EV (Mexico) = 25.0(.40) + 21.2(.50) + 12.5(.10) = 21.85 Select Taiwan Expected value of perfect information = 19(.40) + 16.8(.50) + 12.5(.10) = 17.25 EVPI = 17.25 − 17.72 = $ − 0.47 million The EVPI is the maximum amount the cost of the facility could be reduced ($0.47 million) if perfect information can be obtained. 24. a) EV (Office park) = .5(.50) + 1.7(.40) + 4.5(.10) = 1.38 EV (Office building) = 1.5(.50) + 1.9(.40) + 2.4(.10) = 1.75 EV (Warehouse) = 1.7(.50) + 1.4(.40) + 1.0(.10) = 1.51 EV (Shopping center) = 0.7(.50) + 2.4(.40) + 3.6(.10) = 1.67 EV (Condominiums) = 3.2(.50) + 1.5(.40) + .06(.10) = 2.26 ← maximum Select Condominium project b) EVPI = Expected value of perfect information − expected value without perfect information = 3.01 − 2.26 EVPI = $0.75 million 25. Using expected value; EV(compacts) = 300,000(.6) + 150,000(.4) = $240,000; EV(full-sized) = −100,000(.6) + 600,000(.4) = $180,000; EV(trucks) = 120,000(.6) + 170,000(.4) = $140,000; select the compact car dealership.

12-5 .

26.

Payoff matrix: 20

Demand 22

Stock (lb)

.10

.20

.30

.10

$20.00

18.50

21.00

17.00

19.50

22.00

15.50

18.00

20.50

23.00

14.00

16.50

19.00

21.50

24.00

EV(20) = $20.00; EV(21) = 18.50(.1) + 21.00(.2) + 21.00(.3) + 21.00(.3) + 21.00(.1) = $20.75; EV(22) = 17.00(.1) + 19.50(.2) + 22.00(.3) + 22.00(.3) + 22.00(.1) = $21.00; EV(23) = 15.50(.1) + 18.00(.2) + 20.50(.3) + 23.00(.3) + 23.00(.1) = $20.50; EV(24) = 14.00(.1) + 16.50(.2) + 19.00(.3) + 21.50(.3) + 24.00(.1) = $19.25; stock 22 lb. 27. Revenue and cost data: sales revenue = d) Expected value with perfect information = $12.00/case; cost = $10/case; salvage for $30(.2) + 32(.25) + 34(.4) + 36(.15) = unsold cases = $2/case; shortage cost = $33; EVPI = 33 − EV(16) = 33 − 27.20 = $4/case. $5.80. a) Payoff matrix: 28. a) Payoff matrix: .10

.15

Stock (boxes)

Demand .30 .20

$22

$18

Demand Stock (Milk Cases)

.20

.25

.40

$30

$26

.15

.10

b) EV(15) = 30(.2) + 26(.25) + 22(.4) + 18(.15) = $24.00; EV(16) = 22(.2) + 32(.25) + 28(.4) + 24(.15) = $27.20; EV(17) = 14(.2) + 24(.25) + 34(.4) + 30(.15) = $26.90; EV(18) = 6(.2) + 16(.25) + 26(.4) + 36(.15) = $21.00; stock 16 cases. c) Opportunity loss table: 15

b) EV(25) = 50(.10) + 50(.15) + 50(.30) + 50(.20) + 50(.15) + 50(.10) = 50.0; EV(26) = 49(.10) + 52(.15) + 52(.30) + 52(.20) + 52(.15) + 52(.10) = 51.7; EV(27) = 48(.10) + 51(.15) + 54(.30) + 54(.20) + 54(.15) + 54(.10) = 52.95; EV(28) = 47(.10) + 50(.15) + 53(.30) + 56(.20) + 56(.15) + 56(.10) = 53.3; EV(29) = 46(.10) + 49(.15) + 52(.30) + 55(.20) + 58(.15) + 58(.10) = 53.05; EV(30) = 45(.10) + 48(.15) + 51(.30) + 54(.20) + 57(.15) + 60(.10) = 52.35; since EV(28) = $53.30 is the maximum, 28 boxes of Christmas cards should be stocked. c) Compute expected value under certainty: EV= 50(.10) + 52(.15) + 54(.30) + 56(.20) + 58(.15) + 60(.10) = $54.90; EVPI = $54.90 − $53.30 = $1.60.

EOL(15) = 0(.2) + 6(.25) + 12(.4) + 18(.15) = $9.00; EOL(16) = 8(.2) + 0(.25) + 6(.4) + 12(.15) = $5.80; EOL(17) = 16(.2) + 8(.25) + 0(.4) + 6(.15) = $6.10; EOL(18) = 24(.2) + 16(.25) + 8(.4) + 0(.15) = $12.00; stock 16 cases.

12-6 .

29. a) Payoff matrix: Demand .05

.10

.25

.30

.20

.10

Stock (dozens)

20.00

18.00

16.00

14.00

12.00

10.00

17.50

22.00

20.00

18.00

16.00

14.00

15.00

19.50

24.00

22.00

20.00

18.00

12.50

17.00

21.50

26.00

24.00

22.00

10.00

14.50

19.00

23.50

28.00

26.00

7.50

12.00

16.50

21.00

25.50

30.00

b) EV(20) = 20.00(.05) + 18.00(.10) + 16.00(.25) + 14.00(.30) + 12.00(.20) + 10.00(.10) = $14.40; EV(22) = 17.50(.05) + 22.00(.10) + 20.00(.25) + 18.00(.30) + 16.00(.20) + 14.00(.10) = $18.08; EV(24) = 15.00(.05) + 19.50(.10) + 24.00(.25) + 22.00(.30) + 20.00(.20) + 18.00(.10) = $21.10; EV(26) = 12.50(.05) + 17.00(.10) + 21.50(.25) + 26.00(.30) + 24.00(.20) + 22.00(.10) = $22.50; EV(28) = 10.00(.05) + 14.50(.10) + 19.00(.25) + 23.50(.30) + 28.00(.20) + 26.00(.10) = $21.95; EV(30) = 7.50(.05) + 12.00(.10) + 16.50(.25) + 21.00(.30) + 25.50(.20) + 30.00(.10) = $20.10; since EV(26) = $22.50 is the maximum, the green house owner should grow 26 dozen carnations. c) Opportunity loss table:

Stock (dozens) 20 22 24 26 28 30

.05

.10

Demand .25 .30

22 4.00 0 2.50 5.00 7.50 10.00

24 8.00 4.00 0 2.50 5.00 7.50

0 2.50 5.00 7.50 10.00 12.50

26 12.00 8.00 4.00 0 2.50 5.00

.20

.10

28 16.00 12.00 8.00 4.00 0 2.50

30 20.00 16.00 12.00 8.00 4.00 0

EOL(20) = 0(.05) + 4.00(.10) + 8.00(.25) + 12.00(.30) + 16.00(.20) + 20.00(.10) = $11.20; EOL(22) = 2.50(.05) + 0(.10) + 4.00(.25) + 8.00(.30) + 12.00(.20) + 16.00(.10) = $7.53; EOL(24) = 5.00(.05) + 2.50(.10) + 0(.25) + 4.00(.30) + 8.00(.20) + 12.00(.10) = $4.50; EOL(26) = 7.50(.05) + 5.00(.10) + 2.50(.25) + 0(.30) + 4.00(.20) + 8.00(.10) = $3.10; EOL(28) = 10.00(.05) + 7.50(.10) + 5.00(.25) + 2.50(.30) + 0(.20) + 4.00(.10) = $3.65; EOL(30) = 12.50(.05) + 10.00(.10) + 7.50(.25) + 5.00(.30) + 2.50(.20) + 0(.10) = $5.50; since EOL(26) = $3.10 is the minimum, 26 dozen carnations should be grown. 31. Since this payoff table includes “losses,” the decision criteria must be reversed. a) Minimin: Thailand, minimum loss = $3 million b) Minimax: India, minimum loss = $14 million c) Equal likelihood: India, minimum loss = $9 million d) Minimax regret: Philippines, minimum regret = $2 million

d) The expected value under certainty: EV = $20.00(.05) + 22.00(.10) + 24.00(.25) + 26.00(.30) + 28.00(.20) + 30.00(.10) = $25.60; EVPI = $25.60 − 22.50 = $3.10 30. a) Stock 25, maximum of minimum payoffs = $50 b) Stock 30, maximum of maximum payoffs = $60 c) 25: 50(.4) + 50(.6) = 50; 26: 52(.4) + 49(.6) = 50.2; 27: 54(.4) + 48(.6) = 50.4; 28: 56(.4) + 47(.6) = 50.6; 29: 58(.4) + 46(.6) = 50.8; 30: 60(.4) + 45(.6) = 51; stock 30 boxes. d) Stock 28 or 29 boxes; maximum regret = $4.

12-7 .

38. a) Maximax: Graphic Design 260,000 Nursing 215,000 Real Estate 320,000 ← maximum Medical Technology 280,000 Culinary Technology 305,000 Computer Information Technology 250,000 b) Maximin: Graphic Design 145,000 Nursing 150,000 ← maximum Real Estate 115,000 Medical Technology 130,000 Culinary Technology 115,000 Computer Information Technology 125,000 c) Equal likelihood: Graphic Design 200,000 Nursing 187,500 Real Estate 205,000 ← maximum Medical Technology 200,000 Culinary Technology 200,000 Computer Information Technology 178,750 d) Hurwicz (α = .50) Graphic Design 202,500 Nursing 182,500 Real Estate 217,500 ← maximum Medical Technology 205,000 Culinary Technology 210,000 Computer Information Technology 187,500

32. EV (China) = $10.91 EV (India) = 7.21 Select EV (Thailand) = 9.77 EV (Philippines) = 7.54 33. Since the payoff table includes “costs,” the decision criteria must be reversed. a) Minimin: Manila, minimum cost = $170,000 b) Minimax: Veracruz, minimum cost = $570,000 c) Equal likelihood: Manila, minimum cost = $403,000 d) Minimax regret: Veracruz, minimum regret = $70,000 34.

EV (Shanghai) = 5.328 EV (Mumbai) = 5.375 EV (Manila) = 5.218 EV (Santos) = 5.178 Select EV (Veracruz) = 5.202

35. a) Maximax = Pusan = $.526 billion b) Maximin = Pusan = $.119 billion c) Equal likelihood: Shanghai = $.443 billion Singapore = $.370 billion Pusan = $.429 billion Kaoshiung = $.407 billion Hong Kong = $.469 billion ← Select d) Hurwicz (α = .55) Shanghai = $.34 billion ← Select Singapore = $.22 billion Pusan = $.20 billion Kaoshiung = $.17 billion Hong Kong = $.22 billion

39.

36. a) EV (Shanghai) = $.608 billion EV (Singapore) = $.606 billion EV (Pusan) = $.502 billion EV (Kaoshiung) = $.487 billion EV (Hong Kong) = $.724 billion ← Select 37.

EV (Byrd) = (−3.2)(.15) + (1.3)(.55) + (4.4)(.30) = $1.56M EV (O’Neil) = (−5.1)(.18) + (1.8)(.26) + (6.3)(.56) = $3.08M EV (Johnson) = (−2.7)(.21) + (0.7)(.32) + (5.8)(.47) = $2.38M EV (Gordan) = (−6.3)(.30) + (−1.6)(.25) + (9.6)(.45) = $2.03M Select O’Neil

12-8 .

EV (Graphic Design) = 191,000 EV (Nursing) = 185,000 EV (Real Estate) = 187,000 EV (Medical Technology) = 189,000 EV (Culinary Technology) = 182,000 EV (Computer Information Technology) = 167,000 Student’s decision on the best degree program to recommend will be based on their own risk-taking preferences.

40. a) Maximax: Jose Diaz 153 Jerry Damon 173 ← maximum Frank Thompson 133 Derek Rodriguez 105 Ken Griffin 127 b) Maximin: Jose Diaz 76 Jerry Damon 46 Frank Thompson 88 Derek Rodriguez 95 ← maximum Ken Griffin 75 c) Equal likelihood: Jose Diaz 115.83 Jerry Damon 116.82 ← maximum Frank Thompson 110.88 Derek Rodriguez 99.33 Ken Griffin 99.00

d) Hurwicz (α = .35) Jose Diaz 102.95 ← maximum Jerry Damon 90.45 Frank Thompson 103.75 Derek Rodriguez 98.50 Ken Griffin 93.20 41. a) EV (Jose Diaz) = 111.3 EV (Jerry Damon) = 112.1 EV (Frank Thompson) = 108.7 EV (Derek Rodriguez) = 99.6 EV (Ken Griffin) = 94.0 b) Depends on the student's judgment and analysis c) EV (Jose Diaz) = 119.80 EV (Jerry Damon) = 104.34 EV (Frank Thompson) = 117.70 EV (Derek Rodriguez) = 102.40 EV (Ken Grifffin) = 96.60 Selection depends on the student’s judgment and analysis.

12-9 .

42.

Select compact car.

43.

12-10 .

44.

Since cost of installation ($800,000) is greater than expected value of not installing ($495,000), do not install an emergency power generator. 45.

12-11 .

46. P(c) = probability of contract = .40; P(n) = Probability of no contract = .60; P(f|c) = .70; P(u|c) = .30; P(u|n) = .80; P(f|n) = .20

P (c|u) =

P(f|c) P(c) P (c|f ) = P(f|c) P (c) + P(f|n) P (n) (.70)(.40) = = .70 (.70)(.40) + (.20)(.60)

Decision strategy: If report is favorable, purchase a lathe. If report is unfavorable, purchase a grinder. EV (strategy) = $16,480; EVSI = EV with information − EV without information = $16,480 − 11,200 = $5,280 47.

P(f) = favorable market conditions = .2; P(s) = stable market conditions = .7; P(u) = unfavorable market conditions = .1; P(p|f) = .60; P(n|f) = .40; P(p|s) = .30; P(n|s) = .70; P(p|u) = .10; P(n|u) = .90

12-12 .

Posterior probability table for a positive report: (1)

(2)

(3)

(4)

(5)

Market Conditions

Prior Probabilities

Conditional Probabilities

(2) × (3)

Posterior Probabilities (4) ÷ Σ(4)

Favorable

P(f) = .2

P(p|f) = .60

.12

P(f|p) = .12/.34 = .353

Stable

P(s) = .7

P(p|s) = .30

.21

P(s|p) = .21/.34 = .618

Unfavorable

P(u) = .1

P(p|u) = .10

.01

P(u|p) = .01/.34 = .029

P(p) = .34 Posterior probability table for a negative report: (1)

(2)

(3)

(4)

(5)

Market Conditions

Prior Probabilities

Conditional Probabilities

(2) × (3)

Posterior Probabilities (4) ÷ Σ(4)

P(f) = .2

P(n|f) = .40

.08

P(f|n) = .08/.66 = .121

Favorable Stable

P(s) = .7

P(n|s) = .70

.49

P(s|n) = .49/.66 = .742

Unfavorable

P(u) = .1

P(n|u) = .90

.09

P(u|n) = .09/.66 = .137

P(n) = .66

Decision strategy: Produce the widget regardless of the report. EV (strategy) = $69,966; EVSI = EV with

information − EV without information = $69,966 − $70,000 ≈ 0. Additional information has no value, since the owner will

produce the widget in either case.

12-13 .

P(S−) = .66 = .818

48. a) Let s− = shortage; s+ = surplus; P(s−) = .6; P(s+) = .4. Let S− = report of shortage; S+ = report of surplus; P(S−|s−) = .90; P(S+|s−) = .10; P(S+|s+) = .70; P(S−|s+) = .30.

P(s+|S−) = 1 − .818 = .182 P (S+ | s + ) P (s + ) P(S+ | s + ) P (s + ) + P (S+ | s − ) P(s − ) (.70)(.40) = (.70)(.40) + (.10)(.60)

P (s + | S+ ) =

P (S− | s − ) P (s − ) P (S | s ) P (s − ) + P (S− | s + ) P(s + ) (.90)(.60) = (.90)(.60) + (.30)(.40)

P (s − | S− ) =

−

P(S+) = .34 = .824 P(s−|S+) = 1 − .824 = .176

Decision strategy: If the report indicates a gas shortage, stock compacts. If the report indicates a surplus, stock full-sized cars. EV(strategy) = $342,094; EVSI = EVwith information − EVwithout information = $342,094 − $240,000 = $102,094 Expected value given perfect information = 300,000(.6) + 600,00(4) = $420,000; EVPI = $420,000 − 240,000 = $180,000; EVSI = $102,094; efficiency = EVSI/EVPI = $102,094/$180,000 = .567 or 56.7%

12-14 .

49.

P(s) = .10 P(f) = .90 G = good review B = bad review P(G|s) = .70 P(B|s) = .30 P(G|f) = .20 P(B|f) = .80

P(G|s)P (s) P (G|s)P (s) + P(G|f)P(f) (.70)(.10) = = .28 (.70)(.10) + (.20)(.90) P(f|G) = .72 P(s|B) = .04 P(G) = .25 P(s|G) =

P(f|B) = .96

P(B) = .75

EVSI = EVwith information − EVw/o information Hire Roper; if good review produce, if bad review don’t produce. = $0.31M − (− 4.7M) = $5.01M 50. Best decision, given probabilities: Bond fund, maximum payoff = $41,000 Expected value given perfect information = (6*0.2) + (5*0.3) + (4*0.3) + (8.3*0.1) + (14.7*0.1) = $6.20 EVPI = $6.20- $4.10 = $2.10 or $21,000 51. EV given perfect information = $(1.7)(0.09) + (3.8)(0.27) + (5.4)(0.64) = $4.635 EVPI = $5.178 – 4.635 = $0.543 or $543,000

12-15 .

52. a)

.98[7.2x + 1.7(1 − x)] + (.02)(1.7) = 3.515 .98(5.5x + 1.7) + .034 = 3.515 5.39x + 1.666 + .034 = 3.515 5.39x = 1.815 x = .337 probability of winning in overtime

12-16 .

53.

54.

The following table includes the medical costs for all the final nodes in the decision tree. Expense

Plan 1

Plan 2

Plan 3

100

484

160

388

500

884

560

438

1,500

984

1,290

738

3,000

1,134

1,440

1,188

5,000

1,334

1,640

1,788

10,000

1,834

2,140

3,288

E(1) = 954 E(2) = 976.5 E(3) = 820.5 Select plan 3.

12-17 .

55.

12-18 .

56.

57.

EVSI = $150,000 + 84,000 = $234,000 EVPI = $800,000 + 450,000 = $350,000

The EV without the test market is $450,000, which is $84,000 less than the EV with the test market. Since the cost of the test market is $150,000,

12-19 .

58.

Ellie should invest in the index fund with an expected return of $65,200.

12-20 .

59.

Select strategy 3.

12-21 .

60.

12-22 .

61.

P(1) = .21 P(c) = .35 P(p) = .44

P(L|1) = .75 P(C|1) = .15 P(P|1) = .10

P(L|c) = .10 P(C|c) = .80 P(P|c) = .10

P(L|p) = .05 P(C|p) = .10 P(P|p) = .85

12-23 .

P (L|1)P (1) (.75)(.21) .1575 = = = .734 P (L|1)P(1) + P (L|c) P (c) + P(L|p) P(p) (.75)(.21) + (.10)(.35) + (.05)(.44) .2145 (.10)(.35) P (c|L) = = .163 .2145 (.05)(.44) P (p|L) = = .103 .2145 P(1|C) = .089 P(c|C) = .788 P(p|C) = .124 P(1|P) = .049 P(c|P) = .081 P(p|P) = .870 P(L) = P(L|1) P(1) + P(L|c) P(c) + P(L|p) P(p) = (.75)(.21) + (.10)(.35) + (.05)(.44) = .2145 P(C) = P(C|1)P(1) + P(C|c) P(c) + P(C|p) P(p) = (.15)(.21) + (.8)(.35) + (.10)(.44) = .3555 P(P) = .430 EVSI = EV with information − EV w/o information = $3.75M − 3.08M = $670,000 P(1|L) =

62.

P(d) = .50 P(s) = .40 P(i) = .10 P(D|d) = .80 P(S|s) = .70 P(I|i) = .90

EVSI = $.38 million EVSI efficiency = EVPI .464 = = .6187 .750 Place-Plus would be willing to pay the consulting firm up to $.464 million and the information is 61.9% as efficient as perfect information. 63. Because the couple had limited funds, the utility of money was greater than it might be for a wealthier person. The $5,000 had as much utility to them as perhaps $20,000 or $30,000. This makes the couple risk averters. 64. a) EV (money market) = $162,000(1) = $162,000 EV (oil investment) = 0(.5) + 300,000(.5) = 150,000 According to expected value Annie should invest in the money market. b) EV (money market) = .80(1.0) + .80(0) = .80 EV (oil investment) = 0(.5) + 1(.5) = .50 Thus, according to utility, the money market is the best investment.

P(S|d) = .10 P(I|d) = .10 P(D|s) = .10 P(I|s) = .20 P(D|i) = .02 P(S|i) = .08

Joint probabilities: P(Dd) = .40 P(Ds) = .04 P(Di) = .002

P(Sd) = .05 P(Id) = .05 P(Ss) = .28 P(Is) = .08 P(Si) = .008 P(Ii) = .09

Marginal probabilities: P(D) = .442

P(S) = .338

P(I) =.220

Posterior probabilities: P(d|D) = .905 P(d|S) = .148 P(d|I) = .227 P(s|D) = .090 P(s|S) = .828 P(s|I) = .364 P(i|D) = .005 P(i|S) = .024 P(i|I) = .409 EVwith information = 3.03(.442) + 2.18(.338) + 2.49(.220) = 2.64 EVSI = EVwith information − EVwithout information = 2.64 − 2.26

12-24 .

65.

12-25 .

*CASE SOLUTION: STEELEY ASSOCIATES VS. CONCORD FALLS a)

The decision tree for Steeley Associates is as follows.

12-26 .

Program: Decision Theory / Decision Tree Problem Title : Steeley Associates vs. Concord Falls ***** Input Data ***** Type of Problem : Profit Problem Nodes Type 1 −> 2 Decision 1 −> 3 Decision 1 −> 4 Decision 2 −> 5 Event 2 −> 6 Event 4 −> 7 Event 4 −> 8 Event 5 −> 9 Decision 5 −> 10 Decision 5 −> 11 Decision 9 −> 12 Event 9 −> 13 Event 9 −> 14 Event 11 −> 15 Event 11 −> 16 Event 14 −> 17 Decision 14 −> 18 Decision 17 −> 19 Event 17 −> 20 Event 18 −> 21 Event 18 −> 22 Event ***** Program Output ***** Nodes Type Probability 1−>2 Decision None 1−>3 Decision None 1−>4 Decision None 2−>5 Event 0.900 2−>6 Event 0.100 4−>7 Event 0.700 4−>8 Event 0.300 5−>9 Decision None 5 − > 10 Decision None 5 − > 11 Decision None 9 − > 12 Event 0.100 9 − > 13 Event 0.400 9 − > 14 Event 0.500 11 − > 15 Event 0.700 11 − > 16 Event 0.300 14 − > 17 Decision None 14 − > 18 Decision None 17 − > 19 Event 0.500 17 − > 20 Event 0.500 18 − > 21 Event 0.500 18 − > 22 Event 0.500 The conditional payoff of solution: 1767000.000 ***** End of Output******

12-27 .

Probability None None None 0.900 0.100 0.700 0.300 None None None 0.100 0.400 0.500 0.700 0.300 None None 0.500 0.500 0.500 0.500 Payoff 1767000.000 900000.000 970000.000 1467000.000 300000.000 910000.000 60000.000 1630000.000 700000.000 970000.000 −20000.000 1600000.000 350000.000 910000.000 60000.000 700000.000 650000.000 450000.000 250000.000 600000.000 50000.00

Payoff 0.000 900000.000 0.000 0.000 3000000.000 1300000.000 200000.000 −300000.000 700000.000 0.000 −200000.000 4000000.000 0.000 1300000.000 200000.000 0.000 0.000 900000.000 500000.000 1200000.000 100000.000 Decision < = choice

< = choice

This output indicates that the recommended decision is, “Request a permit for the apartment,” with an expected value of $1,767,000.

The expected value is twice as much as the certain gain for selling the property, so a decision maker might make the conservative decision to sell.

CASE SOLUTION: TRANSFORMER REPLACEMENT AT MOUNTAIN STATES ELECTRIC SERVICE The decision tree solution for this problem is as follows.

The decision should be to retain the existing transformer; the cost of replacement ($85,000) is greater than the cost of retention ($61,000).

12-28 .

The decision tree analysis is as follows. Program: Decision Theory / Decision Tree Problem Title : Mountain States Electric Service ***** Input Data ***** Type of Problem : Cost Problem Nodes

Type

Probability

Payoff

1−>2

Decision

None

85000.000

1−>3

Decision

None

0.000

3−>4

Event

0.500

0.000

3−>5

Event

0.500

0.000

4−>6

Event

0.004

0.000

4−>7

Event

0.996

0.000

5−>8

Event

0.001

0.000

5−>9

Event

0.999

0.000

6 − > 10

Event

0.200

90000000.000

6 − > 11

Event

0.800

8000000.000

7 − > 12

Event

1.000

0.000

8 − > 13

Event

0.200

90000000.000

8 − > 14

Event

0.800

8000000.000

9 − > 15

Event

1.000

0.000 Payoff

***** Program Output ***** Nodes

Type

Probability

1−>2

Decision

None

85000.000

1−>3

Decision

None

61000.000

3−>4

Event

0.500

48800.000

3−>5

Event

0.500

12200.000

4−>6

Event

0.004

97600.000

4−>7

Event

0.996

0.000

5−>8

Event

0.001

24400.000

5−>9

Event

0.999

0.000

6 − > 10

Event

0.200

18000000.000

6 − > 11

Event

0.800

6400000.000

7 − > 12

Event

1.000

0.000

8 − > 13

Event

0.200

18000000.000

8 − > 14

Event

0.800

6400000.000

9 − > 15

Event

1.000

0.000

< = choice

The conditional payoff of solution: 61000.000 ***** End of Output*****

CASE SOLUTION: THE CAROLINA COUGARS

CASE SOLUTION: EVALUATING R&D PROJECTS AT WESTCOM SYSTEMS PRODUCTS COMPANY Project

421,880

421,840

442,253

329,725

120,560

Chapter Thirteen: Queuing Analysis PROBLEM SUMMARY

36. Multiple-server, decision analysis 37. Multiple server, decision analysis

l. Discussion

38. Multiple-server model, decision analysis

2. Discussion

39. Multiple-server model, decision analysis

3. Discussion

40. Multiple-server model, decision analysis

4. Discussion

41. Single-server model analysis

5. Discussion 6. Discussion

42. Finite calling population, decision analysis (13–41)

7. Single-server model analysis

43. Single-server, finite queue, decision analysis

8. Single-server model analysis

44. Multiple-server model analysis

9. Single-server model analysis

45. Multiple-server model analysis

10. Single-server model analysis

46. Multiple-server model analysis

11. Single-server model analysis (13–10)

47. Single-server model, finite queue

12. Single-server model, decision analysis

48. Multiple-server model, decision analysis

13. Single-server model, decision analysis

49. Multiple-server model, decision analysis

14. Single-server model, decision analysis

50. Multiple-server model, decision analysis

15. Single-server model, decision analysis

51. Multiple-server model

16. Single-server model (13–15), Pn analysis

52. Finite calling population

17. Multiple-server model (13–14), decision analysis

53. Finite queue 54. Finite calling population

18. Multiple-server model analysis

55. Single-server model

19. Single-server, finite calling population

56. Multiple-server model

20. Single-server model analysis

57. Multiple-server model (13–56)

21. Single-server, constant service time

57. Multiple-server model (13–28)

22. Single-server, constant service time

58. Multiple-server model (13–55)

23. Single-server, constant service time

60. Multiple-server model

24. Single-server, constant service time

61. Constant service time model

25. Single-server, finite calling population 26. Single-server, finite calling population

PROBLEM SOLUTIONS

27. Single-server, general service time (13–14)

1. a) Hair salon: multiple-server; first-come, first-served or appointment; calling population can be finite (appointments only) or infinite (off-the-street business)

28. Single-server, general service time 29. Multiple-server model 30. Multiple-server model (13–18), decision analysis

b) Bank: multiple-server; first-come, first-served; infinite calling population

31. Multiple-server model, decision analysis 32. Single-server, decision analysis

c) Laundromat: multiple-server; first-come, first-served; infinite calling population

33. Single-server, finite calling population, decision analysis

d) Doctor’s office: single- (or multiple-) server; appointment (usually); finite calling population

34. Single-server, finite queue 35. Single-server, decision analysis

13-1 .

e) Advisor’s office: single-server; first-come, first-served or appointment; finite calling population f) Airport runway: single-server; first-come, first-served; finite calling population g) Service station: multiple-server; first-come, first-served; infinite calling population h) Copy center: single- or multiple-server; first-come, first-served; infinite calling population i) Team trainer: single-server; first-come, first-served or appointment; finite calling population j) Web site multiple-server; first-come, first-served (or priority level); infinite calling population 2. The addition of a new counter created two queues. The multiple-server model is for a single queue with more than one server. 3. a) False. The operating characteristic values may be higher or lower depending on the magnitude of the standard deviation compared to the mean of the exponentially distributed service time. b) True. Since there is no variability the operating characteristics would always be lower. 4. When arrivals are random, in the short run more customers may arrive than the serving system can accommodate. 5. When customers are served according to a prearranged schedule or alphabetically, or are picked at random. 6. 7.

10 = .41 hr (24.6 min); 12(2) λ 10 U= = = .833 μ 12

6 = .15 hr (9 min); if the arrival rate is 10(4) increased to 12 per hour, the arrival rate would exceed the service rate; thus, an infinite queue length would result.

10.

60 = 8 per hour; μ = 10 per hour; 7.5 λ2 (8)2 = = 3.2 parts; Lq = μ ( μ − λ ) 10(2) λ 8 = .80; I − U = I − .80 = .20 U= = μ 10

11.

12.

λ ; μ = 10 per hour; U = .90; μ λ therefore, .90 = , or λ = 9 per hour,

10 or 1 part every 6.67 min.

P0 = (.67)3 (.33) = .099;

(16) λ = = 1.33; μ ( μ − λ ) 24(8)

The arrival rate must be on an hourly basis;

λ=

Lq =

λ2 = μ (μ − λ )

λ 1 = .25 hr (15 min); Wq = = 4 μ (μ − λ )

⎛λ⎞ λ 16 P0 = 1 − = 1 − = .33; P3 = ⎜ ⎟ ; 24 μ ⎝μ⎠

λ = 6; μ = 10; Lq =

(6)2 1 = .9 car; W = = 10(4) μ −λ

λ = 16 per hour; μ = 24 per hour;

λ 16 = = 2.0; μ −λ 8

λ μ (μ − λ )

When μ = σ

λ = 10; μ = 12; Wq =

= .125 hr (7.5 min);

μ −λ 8 16 λ = = .083 hr (5 min); Wq = μ ( μ − λ ) 24(8) λ 16 = .67 U= = μ 24

λ = 4 per hour; μ =

60 = 5 per hour 12

a) Lq =

(4)2 λ2 = = 3.2 μ ( μ − λ ) 5(1)

b) Wq =

λ 4 = = .80 hr (48 min) μ ( μ − λ ) 5(1)

c) W =

1 = 1 hr (60 min) μ−λ

45 , or .75, 60 utilization factor. U cannot exceed .75.

d) 45 min per hour is

13-2 .

λ 4 = = .80 presently. Therefore, one μ 5

λ = 200 per day; μ = 220 per day

15. a)

more air traffic controller must be hired.

λ = 12 per hour;

13.

1 1 = = .05 day; μ − λ 20 8 hr/day × 60 min/hr = 480 min; .05 × 480 = 24 min; so W = 24 min;

One window:

λ 12 = = .26 hr (16 min). μ ( μ − λ ) 15(3) Two windows; μ = 15 per hour (does not Wq =

Wq =

change). However, the arrival rate for each window is now split. λ = 6 per hour; λ 6 = = .044 hr (2.67 min); Wq = μ ( μ − λ ) 15(9) 16 − 2.67 = 13.33 min, reduction in waiting time; 13.33 × $2,000 = $26,660; cost of window = $20,000; $26,660 > $20,000; therefore, the second drive-in window should be installed.

200 220(20) = .045 day (21.6 min)

b) 24 − 15 = 9 min; 9 min × $10,000 = $90,000 per year loss currently. With a new set of scales the arrival rate would be split. λ = 100 per day per scales; 1 1 W= = = .008 day (4 min). μ − λ 220 − 100 The entire $90,000 would be saved. Since the scales cost $50,000 per year, they should be installed.

60 = 30 per hour 2

L= Lq =

λ μ −λ

28 = 14; 2

16.

λ2 (28)2 = = 13.1; μ ( μ − λ ) 30(2)

1 1 = = .5 hr (30 min); μ −λ 2 28 λ = Wq = μ ( μ − λ ) 30(2) = .47 hr (28.2 min);

λ μ (μ − λ )

λ = 28 per hour; μ=

λ2 (200)2 = = 9.09 trucks; μ ( μ − λ ) 220(20) W=

60 μ= = 15 per hour. 4

14.

Lq =

Pn = the probability of n customers in the system; P4 = the probability of exactly 4 in the system. However, we need the probability of 4 or more. Therefore, compute P0, P1, P2, and P3 and subtract from 1.0. n

⎛λ⎞ Pn = ⎜ ⎟ ⋅ P0 ⎝μ⎠ P0 = 1 −

λ 28 = = .93 = 93% μ 30

λ 200 = 1− = .09 μ 220

P1 = (.91)1 ⋅ .09 = .082 P2 = (.91)2 ⋅ .09 = .075

1 ; let W = 10 min = .167 hr; μ −λ 1 .167 = , so ( μ − 28)(.167) = 1; μ − 28 μ − 28 = 6, so μ = 34 students 60 per hour. = 1.76 min required to 34 approve a schedule in order to meet the dean’s goal. Since each assistant will reduce the service time by .25 min, 1 more assistant is all that is needed (i.e., 2.00 min − .25 =1.75 min).

b) W =

P3 = (.91)3 ⋅ .09 = .068

P0 + P1 + P2 + P3 = .090 + .082 + .075 + .068 P(4 or more trucks) = 1 − .315 = .685

13-3 .

17.

λ = 28 per hour; μ = 30 per hour; c = 2 1

P0 =

⎡ 1 ⎛ 28 ⎞ ⎤ 1 ⎛ 28 ⎞2 ⎡ (2)(30) ⎤ ⎢∑ ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎥ ⎢⎣ 0 n! ⎝ 30 ⎠ ⎥⎦ 2! ⎝ 30 ⎠ ⎣ 60 − 28 ⎦ 1 = 0 ⎡ 1 ⎛ 28 ⎞ 1 ⎛ 28 ⎞1 ⎤ 1 ⎛ 28 ⎞2 ⎛ 60 ⎞ ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥+ ⎜ ⎟ ⎜ ⎟ ⎣⎢ 0! ⎝ 30 ⎠ 1! ⎝ 30 ⎠ ⎦⎥ 2 ⎝ 30 ⎠ ⎝ 32 ⎠

1 1 = = .36 (1 + .93) + (.436)(1.875) 2.75 c

⎛λ⎞ ⎟ λ ⎝μ⎠ L= ⋅ P0 + 2 μ (c − 1)! (cμ − λ )

λμ ⎜

(28)(30)(.93)2 726.5 (.36) + .93 = (.36) + .93 1024 1! (60 − 28)2 = 1.19

Lq = L −

λ = 1.19 − .93 = .26 μ

1.19 = .043 hr (2.55 min) λ 28 Lq .26 = = .009 hr (.56 min) Wq = λ 28 W=

18.

The student will probably recommend adding the advisor. λ = 100 per hour; μ = 60 per hour; c = 2 P0 =

1 ⎡ 1 1 ⎛ 100 ⎞ n ⎤ 1 ⎛ 100 ⎞2 ⎡ 2(60) ⎤ ⎢∑ ⎜ ⎟ ⎥+ ⎜ ⎟ ⎢ ⎥ ⎣⎢ 0 n! ⎝ 60 ⎠ ⎦⎥ 2! ⎝ 60 ⎠ ⎣120 − 100 ⎦ 1 1 = = .09 (1 + 1.67) + 8.34 11 c

⎛λ⎞ ⎟ λ ⎝μ⎠ L= ⋅ P0 + 2 μ (c − 1)! (cμ − λ )

λμ ⎜

(100)(60)(1.67)2 (.09) + 1.67 = 5.44 1! (20)2

Lq = L −

5.44 = .054 hr (3.26 min) 100 Lq 3.77 = = .038 hr (2.26 min) Wq = λ 100 W=

λ = 5.44 − 1.67 = 3.77 μ

13-4 .

No. of machines: 5

19.

Breakdowns: λ =

1 per day 4

Repairs: μ = 1 per day No. of repairmen: 1 a)

P0 =

1 5

⎡

n=0

⎢⎣

⎛1⎞ ⎤ n

∑ ⎢ (5 − n)! ⎜⎝ 4 ⎟⎠ ⎥ ⎥⎦

1 ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 + 5 ⎜ ⎟ + 20 ⎜ ⎟ + 60 ⎜ ⎟ + 120 ⎜ ⎟ + 120 ⎜ 1024 ⎟ 4 16 64 256 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 1 = = ≅ .2 (repairmen idle 20% of the time) 1 + 1.25 + 1.25 + .94 + .47 + .12 5.03 =

⎛ .25 + 1.0 ⎞ Lq = 5 − ⎜ ⎟ (1 − .2) ⎝ .25 ⎠ ⎛ 1.25 ⎞ = 5−⎜ ⎟ (.8) = 5 − 4 = 1 machine ⎝ .25 ⎠

(13.33)2 (21.43)(21.43 − 13.33) = 1.02 customers waiting =

Mean no. machines waiting for repairman = 1 machine Lq Wq = ( N − L )λ

= .077 hr or 4.61 minutes waiting in line

This is not very good service for a fast food drive-through window.

1 1 Wq = = = 1.25 days (5 − 1.8).25 .8

60 = 20.7 per hour 2.9 60 μ= = 22.22 per hour, constant service time 2.7 Lq = 6.3434 customers

21. λ =

Mean time waiting for repair = 1.25 days 3

P3 =

λ2 (13.33)2 = μ (μ − λ ) (21.43)(21.43 − 13.33)

Wq =

Where L = Lq + (1 − P0) = 1 + (1 − .2) = 1.8

λ2 μ (μ − λ )

Lq =

5! ⎛ 1 ⎞ (.2) (5 − 3)! ⎜⎝ 4 ⎟⎠

⎛ 1 ⎞ = (60) ⎜ ⎟ (.2) = .1825 ⎝ 64 ⎠

Wq = .3064 hr or 18.39 minutes

22.

18% chance of the three machines in the system (being repaired or waiting for repair).

λ = 10/hour μ = 13.33/hour Lq =

60 20. λ = = 13.33 customers per hour 4.5 60 μ= = 21.43 customers per hour 2.8

λ2 2μ (μ − λ )

(10)2 2(13.33)(13.33 − 10) Lq = 1.13drivers =

1.13 λ 10 = .112 hour or 6.76 minutes

Wq =

13-5 .

23.

λ = 60/hour or 1/minute μ = 3/minute Lq =

The patrol car is out of service an average of 23.6778 hours when being repaired. Whether or not this is adequate depends on how busy the police department is.

λ2 2μ (μ − λ )

26.

(1)2 = 2(3)(3 − 1) 1 = = .083 people in line 12 Lq .083 Wq = = 1 λ = .083 minutes waiting

24.

μ = 3.0/hour c=4 c μ = 12.0 P0 ≅ .07 1 − P0 = .93 c

⎛λ⎞ ⎟ λ ⎝μ⎠ L= P + 2 0 μ (c − 1)! (cμ − λ )

λμ ⎜

The appropriate model is Poisson arrivals with constant service times:

λ = 149 + 30 = 179 cars per 24-hour day 1

⎛ 7.5 ⎞ (7.5)(3.0) ⎜ ⎟ ⎝ 3.0 ⎠ = (.07) (4 − 1)![(4)(3.0) − 7.5]2 7.5 =+ = 3.03 3.0 λ 7.5 Lq = L − = 3.03 − = .533 μ 3.0 L 3.03 W= = = .404 hour = 24.24 minutes λ 7.5 Lq .533 Wq = = = .071 hour = 4.26 minutes λ 7.5

= 8 minutes

⎛1⎞ Thus, μ = ⎜ ⎟ (60)(24) = 180/day ⎝8⎠ 2

⎛ 179 ⎞ ⎛5⎞ ⎜ 180 ⎟ ⎜ ⎟ ⎠ = ⎝6⎠ Lq = ⎝ = 89.0003 cars 2 ⎛ 179 ⎞ ⎛ 5⎞ 2 ⎜1 − ⎟ 2 ⎜1 − ⎟ ⎝ 180 ⎠ ⎝ 6⎠ Wq =

25.

λ = 7.5/hour

89.003 = .4972 days 179 λ .4972(24) 11.93 hours =

27.

λ = 28 μ = 30 σ = 1/12 = .083

Yes, the port authority can assure the coal company that their cars will not have to wait longer than 12 hours each on the average.

Lq = 47.04

λ = .00208/hr

Wq = 1.68 hours

μ = .05555/hr

W = 1.71 hours

c=1

All statistics are greater than in problem 14(a), as would be expected with a general service time distribution.

L = 47.97

N=8 P0 = .7145

28.

1 – P0 = .2855 ⎛λ+μ⎞ Lq = N − ⎜ ⎟ 1 − P0 = 0.09 ⎝ λ ⎠

L = 2.4

L = Lq + (1 – P0) = .3755 Wq =

Lq ( N − L )λ

W = Wq +

λ =3 μ=4 σ = 1/6 = .167 Lq = 1.63 W = .79 hours = 47.54 minutes

= 5.676 hr

Wq = .54 hours = 32.54 minutes

= 23.6778 hr

13-6 .

λ = 40/hour μ = 15/hour

29.

30.

c=4 cμ = 60 P0 = .061

1 ⎡ 2 1 ⎛ 100 ⎞ n ⎤ 1 ⎛ 100 ⎞3 ⎛ 180 ⎞ ⎢∑ ⎜ ⎟ ⎥+ ⎜ ⎟ ⎜ ⎟ ⎢⎣ 0 n ! ⎝ 60 ⎠ ⎥⎦ 3! ⎝ 60 ⎠ ⎝ 180 − 100 ⎠ 1 = 1 (1 + 1.67 + 1.39) + (1.67)3 (2.25) 6 1 1 = = = .17 1 + 1.67 + 1.39 + 1.75 5.81

P0 =

1 − P0 = .9390 c

⎛λ⎞ ⎟ λ ⎝μ⎠ L= P + 2 0 μ (c − 1)! (c μ − λ )

λμ ⎜

⎛ 40 ⎞ (40)(15) ⎜ ⎟ 40 ⎝ 15 ⎠ = (.061) + = 3.42 15 (4 − 1)![(4)(15) − 40]2 40 λ Lq = L − = 3.42 − = .756 15 μ L 3.42 W= = = .086 hour = 5.1 minutes 40 λ Lq .756 Wq = = = .019 hour = 1.13 minutes λ 40

ρ=

λ = 100; μ = 60; c = 3

⎛λ⎞ ⎟ λ ⎝μ⎠ L= ⋅ P0 + 2 μ (c − 1)! (c μ − λ )

λμ ⎜

⎛ 100 ⎞ (100)(60) ⎜ ⎟ ⎝ 60 ⎠ (.17) + 100 = .37 + 1.67 = 60 (2)! (180 − 100)2 = 2.04 Lq = L −

λ 40 = = .67 c μ 60

With 2 presses: Lq = 3.77; with 3 presses: Lq = .37; Lq(2) − Lq(3) = 3.77 − 0.37 = 3.4 reduction; $50 × 3.4 = $170; therefore, the third press should be added, since it costs only $150.

idle time = 1 − .67 = .33; thus the postal workers are idle 33% of the time, which seems excessive. b)

Wq = 1.13 min. and Lq = .756 customers, neither which seems excessive.

A customer can expect to walk in and get served without waiting approximately 6% (i.e., P0 = .061) of the time.

λ = 2.04 − 1.67 = .37 μ

31.

λ = 500; μ = 200; c = 3 P0 =

Overall, the system seems very satisfactory from a customer service perspective; however, the post office might want to analyze the system with 3 stations instead of 4 because of the large percentage of idle time.

1 ⎡ 2 1 ⎛ 500 ⎞ n ⎤ 1 ⎛ 500 ⎞3 ⎛ 600 ⎞ ⎢∑ ⎜ ⎟ ⎥+ ⎜ ⎟ ⎜ ⎟ ⎣⎢ 0 n! ⎝ 200 ⎠ ⎦⎥ 3! ⎝ 200 ⎠ ⎝ 100 ⎠ 1

1 (1 + 2.5 + 3.13) + (15.6)(6) 6 = .045

1 22.23

⎛λ⎞ ⎟ λ ⎝μ⎠ L= ⋅ P0 + μ (c − 1)!(cμ − λ )2

λμ ⎜

(500)(200)(2.5)3 (.045) + 2.5 = 6.02 (2)!(100)2

Lq = L − Lq

λ = 6.02 − 2.5 = 3.52 μ

3.52 = .007 hr(.42 min ) λ 500 No, another wrapper is not needed. Wq =

13-7 .

35. 32.

33.

34.

Regular copier

High-speed copier

One drive-through window:

λ = 10 per hour

λ = 7/hr.

μ = 15 per hour

μ = 10/hr.

μ = 20/hr.

(a) No, Wq = 8 minutes

Wq = 14 minutes

Wq = 1.62 minutes

W = 20 minutes

W = 4.62 minutes

(b) Yes, μ = 24 and Wq = .11 minutes (c) Yes, λ = 20, μ = 24 and Wq = .53 minutes

Regular L = 2.33 Thus, hourly cost of people = 2.33(10) = 23.3/hour Cost of machine per hour = 8 Total cost with Regular per hour = 31.3 High Speed Copier L = 0.538 Thus, hourly cost of people = 0.538(10) = 5.38/hour Cost of machine per hour = 16 Total cost with Deluxe per hour = 21.38 Deluxe copier is cheaper to operate. Single server with finite calling population Current repair service. λ = 0.05/day μ = 1/day W = 2.6 days Downtime Cost = 2.6(24)($5) = $312 New repair service. λ = 0.05/day μ=2 W = .77 days Downtime Cost = .77(24)(5) = $92.4 + 120 = 212.27 Even at $10 per hour more, the new service is better. λ = 7.5 cabs/hr. μ = 7.05 cabs/hr. In this finite queue system the hotel guests are treated as the servers and the cabs are the customers. The guests have a service rate of 8.5 minutes (5 minutes to arrive and 3.5 minutes to load), thus 7.05 cabs per hour are “served.” The cabs arrive at the cab line at the rate of 7.5 per hour. The system size is 6 customers or cabs.

36.

λ = depends on the number of registers μ = 8.57 customers/hr./register QM for Windows is used to analyze a system with a number of single server queues. For example, if 2 registers are open then λ = 35 customers (per register). If 4 registers are open, λ = 17.5 and W = 7.66 minutes, which meets the service goal of W = 12 minutes. Thus 4 registers need to be opened.

37.

Multiple server model:

λ = 40/hr. μ = 15/hr. c=3 Wq = 9.57 minutes At $2 per minute, a 9.57 minute wait “costs” $19.14 per customer. The cost of an employee is only $0.40 per minute; thus the hotel should hire enough clerks so there is virtually no wait, i.e., 5 servers will result in almost no wait. 38.

λ = 6 per hour; μ = 4 per hour; c = 2 P0 =

1 ⎡ 1 1 ⎛ 6 ⎞ n ⎤ 1 ⎛ 6 ⎞2 ⎛ 8 ⎞ ⎢∑ ⎜ ⎟ ⎥ + ⎜ ⎟ ⎜ ⎟ ⎣⎢ 0 n! ⎝ 4 ⎠ ⎦⎥ 2! ⎝ 4 ⎠ ⎝ 2 ⎠ 1 1 = = .143 (1 + 1.5) + 4.5 7 c

⎛λ⎞ ⎟ λ ⎝μ⎠ L= ⋅ P0 + 2 μ (c − 1)! (c μ − λ )

λμ ⎜

(6)(4)(1.5)2 (.143) + 1.5 = 3.43 1! (2)2

Lq = L −

(a) Wq = the average time a cab must wait for a fare 22.81 minutes

Wq =

(b) Probability (x ≥ 6) = .171

13-8 .

λ = 3.43 − 1.5 = 1.93 μ =

1.93 = .32 hr (19.2 min ) 6

For c = 3:

41.

1 ⎡ 1 ⎛ 6 ⎞ ⎤ 1 ⎛ 6 ⎞3 ⎢∑ ⎜ ⎟ ⎥ + ⎜ ⎟ ( 2 ) ⎢⎣ 0 n! ⎝ 4 ⎠ ⎥⎦ 3! ⎝ 4 ⎠ 1 1 = = = .21 1 4.75 (1 + 1.5 + 1.13) + (3.38)(2) 6

P0 =

μ −λ

(6)(4)(1.5)3 (.21) + 1.5 = 1.74 2! (6)2

Lq = L − Lq

Average repair time =

λ = 1.74 − 1.5 = .24 μ

42.

.24 = .04 hr (2.4 min ) λ 6 Hire a third doctor. Wq =

39.

Pw =

1 ⎛ λ ⎞ ⎡ cμ ⎤ 1 ⎛ 10 ⎞ P0 = ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ c! ⎝ μ ⎠ ⎣ cμ − λ ⎦ 2! ⎝ 6 ⎠

43.

⎛λ⎞ ⎛6⎞ Pn ≥ 31 = ⎜ ⎟ = ⎜ ⎟ = .008 ≈ 0.84% ⎝7⎠ ⎝μ⎠

λ = 0.25 per hour per patient μ = 6 per hour

N = 15 patients

⎡ (2)(6) ⎤ ⎢ (2)(6) − 10 ⎥ × (.0909) = .75758. ⎣ ⎦

P0 =

Thus, 2 salespeople are not enough. For c = 3: P0 = .172662 and

1 N! ⎛ λ ⎞ ∑ ⎜ ⎟ n = 0 (N − n)! ⎝ μ ⎠ N

1 15

15!

⎛ .25 ⎞

∑ (15 − n)! ⎜⎝ 6 ⎟⎠

n =0

= .423

⎛λ+μ⎞ Lq = N − ⎜ ⎟ (1 − P0 ) ⎝ λ ⎠ ⎛ .25 + 6 ⎞ = 15 − ⎜ ⎟ (1 − .423) ⎝ .25 ⎠ = 0.592 patients

1 ⎛ 10 ⎞ ⎡ (3)(6) ⎤ (.172662) = .29976. 3! ⎜⎝ 6 ⎟⎠ ⎢⎣ (3)(6) − 10 ⎥⎦ Thus, 3 salespeople are sufficient to meet the company policy that a customer should have to wait no more than 30% of the time. Pw =

40.

8 hr ≈ 1.14 hr. 7

chance that the number of TVs will exceed shop capacity.

λ = 10; μ = 6. For c = 2: P0 = .090909 and c

1 =1 μ −6 μ = 1+ 6 = 7 1 1 = day μ 7

⎛λ⎞ λμ ⎜ ⎟ λ ⎝μ⎠ L= ⋅ P0 + μ (c − 1)! (c μ − λ )2 =

Determine the required value of 1/μ (average repair time) if λ, the average arrival rate, is 6 per day and if W, the average time in the system, is assumed to be 1 day.

λ = 40; μ = 15; c = 3; P0 = .028037; λμ (λ /μ )c λ L= P + 2 0 μ (c − 1)! (c μ − λ )

L = Lq + (1 − P0) = .592 + (1 − .423) = 1.168 patients

(40)(15)(40 /15)3 2![(3)(15) − 40]2 40 (.028037) + = 9.0467; 15 λ 40 Lq = L − = 9.0467 − = 6.38; μ 15 Lq 6.38 = = .1595 hr = 9.57 min. Wq = λ 40 Thus, 3 servers should be sufficient. =

Wq =

Lq (N − L )λ

0.592 (15 − 1.168)(.25)

= .171 hr or 10.27 minutes waiting W = Wq +

= .171 +

1 6

= .338 hr or 20.27 minutes in system Nurse Attaberry is correct. The waiting time for a patient who calls averages about 10 minutes. However, she is idle a little over 40 percent of the time; thus, the supervisor cannot achieve both objectives.

13-9 .

44.

in revenue for two weeks with the 4 teams paid $2,000 for two weeks or approximately $4,800 per week.

60 = 3.33 parties per hour 18 60 μ= = .705 parties per hour 85

λ=

47.

c=6 P0 = .007 L = 6.545 Lq = L −

λ 3.33 = 6.545 − .705 μ

= 1.821 Wq =

1.821 3.33

= .547 hr or 32.8 minutes waiting Wq =

45.

6.545 3.33

= 1.96 hr or 117.92 minutes at the restaurant A 32.8 minute waiting time may seem long, but actually restaurant customers sometimes perceive a waiting line and a reasonably long waiting time as an indicator of “quality.” Lq = 2.2469 manuscripts

Arrival rate: λ = 40 units per hour. Processing times: (1) without additional employees: 1/μ1 = 1.2 min per unit; thus, μ1 = 50 units per hour; (2) with additional employees: 1/μ2 = .9 min per unit; thus, μ2 = 66.67 units per hour. In-process inventory = number in process and waiting to be processed = L = number in the system = λ/(μ − λ); (1) without additional employees: L1 = 40/(50 − 40) = 40/10 = 4 units in system; (2) with additional employees: L2 = 40/(66.67 − 40) = 40/26.67 = 1.5 units in system. Decision analysis: Cost of in-process inventory: (1) without additional employees: (4) ($31) = $124.00/day; (2) with additional employees: (1.5) ($31) = $46.50/day. Difference = $124.00 − $46.50 = 77.50/day. Thus, the optimal decision is to add additional employees at a cost of $52.00 per day, yielding a net expected savings of $77.50 − $52.00 = $25.50/day.

48. a) λ = 5; μ = 2; c = 3 P0 =

L = 12.2469 manuscripts Wq = .321 weeks W = 1.7496 weeks

1 ⎡ 1 ⎛ 5 ⎞0 1 ⎛ 5 ⎞1 1 ⎛ 5 ⎞2 ⎤ 1 ⎛ 5 ⎞3 (3)(2) ⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥+ ⎜ ⎟ ⎣⎢ 0! ⎝ 2 ⎠ 1! ⎝ 2 ⎠ 2! ⎝ 2 ⎠ ⎦⎥ 3! ⎝ 2 ⎠ (3)(2) − 5

= .045

U = .8333 46. a) By testing several different numbers of servers (teams) for the multiple server model in QM for Windows, it is determined that at least 4 teams are required to be within the two-week waiting period. The operating characteristics are

(5)(2)(5 / 2)3 5 (.045) + = 6.0; 2 2 (3 − 1)![(3)(2) − 5]

L q = 6 − 5 = 3.5; W q = 3.5 = 0.70 hr (42 m in); 2 5 W = 6.0 = 1.20 hr (72 m in) 5

b) λ = 5; 1/ μ = 25 min; therefore, μ = 2.4; c = 3; P0 = .0982;

Lq = 1.128 jobs L = 3.98 jobs

(5)(2.4)(5 2.4)3 (3 − 1)![(3)(2.4) − 5]2 5 (.0982) + = 3.18; 2.4 5 Lq = 3.18; − = 1.1; 2.4 1.1 Wq = = .22 hr 5

Wq = .28 weeks

W = 1.00 week b) This can be determined in two ways. First, if the average number of jobs arriving each week is 4, at $1,700 apiece they will generate $6,800 in revenue per week whereas 4 painting teams will cost $2,000 per week for a difference of $4,800 per week. Alternatively, if there are approximately 4 jobs in the system (L = 3.98) over a two week period (W = 1.00) then that will result in $6,800

(13.2 min waiting time). The improvement in average waiting time per truck is 42 – 13.2 = 28.8 min. The

13-10 .

estimated value of this time saving is ($750)(28.8) = $21,600. Since the cost of achieving the improved service is only $18,000, the firm should implement the improved system, yielding an expected savings of $21,600 − $18,000 = $3,600.

50.

a) c = 4, P0 = .043, L = 4.197, Lq = 1.28, W = .839 hr = 50.34 min, Wq = .256 hr = 15.36 min

c) Alternative 1: Add a fourth loading/ unloading location at the dock, yielding four locations, where each location has a mean service rate of μ = 2 per hour. Alternative 2: Add extra employees and equipment at the existing three dock locations to reduce loading/unloading times from 30 min to 23 min per truck, yielding μ = 2.6 per hour. Decision analysis: The tendency of the student will be to compute the waiting time for both alternatives. However, this is not required, since the alternatives can be evaluated by using the concept of “effective service rate,” which is determined by multiplying the number of servers by the mean service rate. The purpose of this part of the problem is to introduce this concept; thus, the instructor may wish to give the student a hint before assigning this problem. Computing the effective service rate for each alternative gives, for alternative 1, (no. of servers) (mean service rate) = (4)(2) = 8 trucks per hour, and for alternative 2, (no. of servers) (mean service rate) = (3)(2.6) = 7.8 trucks per hour. Since the cost of each alternative is approximately equal, alternative 1, to add a fourth dock location, is superior because it increases the effective service rate to 8 trucks per day; whereas adding extra resources to the existing dock increases the effective service rate to only 7.8 trucks per day. 49.

λ = 5, μ = 1.714

b) c = 5, P0 = .051, L = 3.220, Lq = .303, W = .644 hr = 38.64 min, Wq = .061 hr = 3.66 min. Although the customer waiting time is reduced from 15.36 to 3.66 min, 15 min does not seem excessive for a hairstylist; thus, the impact of adding a fifth stylist is probably not significant. 51.

52.

7:00 to 9:00 A.M.: λ = 10, μ = 2.5; c must equal at least 5 for the mean effective service rate to exceed the arrival rate; Wq = .222 min with 5 cashiers; therefore, 5 is sufficient. 9:00 A.M. to noon: λ = 4, μ = 2.5; c must equal at least 2; Wq = .711 min with 2 cashiers; therefore, 2 is sufficient. Noon to 2:00 P.M.: λ = 14, μ = 2.5; c must equal at least 6 cashiers; Wq = .823 min with 6 cashiers; therefore, 6 is sufficient. 2:00 to 5:00 P.M.: λ = 8, μ = 2.5; c must equal at least 4 cashiers; Wq = .298 min with 4 cashiers; therefore, 4 is sufficient.

13-11 .

40 ships per month; one terminal λ = 1.33 ships/day μ = 2 ships/day L = 2 ships Lq = 1.32 ships W = 35.82 hours Wq = 23.82 hours

60 ships per month λ =2 μ =2 c=2 Po = .3333 L = 1.33 ships Lq = .333 ship W = 16 hours Wq = 4 hours

80 ships per month λ = 2.67 ships/day μ = 2 ships/day c=2 L = 2.48 ships Lq = 1.15 ships W = 22.33 hours Wq = 10.334 hours

Finite calling population: .7 λ = = .04375 calls per room per day 16 μ = 2 calls/hour average utilization = 1 – P0 = 1 − .5776 = .4224 Wq = .31 hr. = 19.1 min. W = .81 hr. = 49.1 min. The system seems adequate. 19.1 minutes is somewhat long to wait for service; however the staff person is only busy

42.2% of the time as is. Thus adding another person seems excessive. Finite Queue Model; assumes 4 week 53. month 14 λ = = 1.17 customers/month 12 5 weeks = 1.25 months 1 μ= = .80 piece/month 1.25 W = 8.92 months Busy = 98.9% P(place an order) = 1 – Pm = .6765 Finite calling population 54. λ = 0.0625/hour µ = 0.67/hour N = 10 a) Lq = 1.177 athletes waiting Wq = 2.33 hours waiting W = 3.82 hours in the system Utilization = .7551 The system does not seem effective. An athlete must wait 2.33 hours which seems too long, and Judith does not have much time (P0 = .24) to work on her own studies. b) By reducing the number of athletes Judith is responsible for to 6 (i.e., N = 6) the waiting time is reduced to .89 hour (53.4 minutes) which seems more reasonable, and Judith has 51 percent of her time free (i.e., P0 = .5109). 55.

W = .12 day = .97 hours Wq = .02 day = .16 hours

The current staff meets the service guarantee b) λ = 45/day

μ = 6 calls/day 3 service trucks do not meet the cμ ≥ λ requirement and thus the waiting line (and time) will grow infinitely. 8 trucks are necessary for the cμ ≥ λ requirement, and are also sufficient to meet the service guarantee L = 22.11 customers Lq = 14.61 customers W = .49 day = 3.93 hours Wq = .32 day = 2.6 hours 57.

μ = 12 calls/day c = 4 trucks are necessary to meet the cμ ≥ λ requirement. 4 trucks also achieve the service guarantee – 1 truck must still be borrowed. L = 16.73 customers Lq = 12.98 customers W = .37 day = 5.95 hours Wq = .29 day = 4.61 hours

The long-term revenue from additional new customers will offset the short-term installation costs resulting from either borrowing trucks and crews from other offices and/or paying current staff overtime. However, there may be a tradeoff between borrowing trucks and crews and paying local staff overtime.

λ = 40 passengers per hour μ = 50 passengers per hour Lq = =

λ2 μ (μ − λ ) (40)2 (50)(50 − 40)

58.

λ=8 μ=3

= 3.2 passengers Wq =

λ2 μ (μ − λ )

c = 4 delivery people required to provide a “reasonable” level of service, as follows:

(40) (50)(50 − 40)

Lq = .757

L = 3.42 W = .43 hour = 25.68 minutes

= 4.8 minutes 56. a)

λ = 45/day

Wq = .09 hour = 5.68 minutes

λ = 15/day L = 1.81

A multiple server model with 3 security gates can accommodate this increased passenger traffic level.

Lq = .25

λ = 110 passengers per hour

59.

μ = 9.6 calls/day

13-12 .

μ = 50 passengers per hour

*CASE SOLUTION: THE COLLEGE OF BUSINESS COPY CENTER

c=3 P0 = .101 ⎡ λμ (λ /μ )c ⎤ λ L=⎢ P + 2 ⎥ 0 μ ⎣ (c − 1)!(c μ − λ ) ⎦

A multiple-server queuing model must be evaluated for a center with 2 copiers and 3 copiers for the normal academic year and the summer. Normal academic year: 2 copiers: λ = 7.5, μ = 5, c = 2; W = .475 hr = 27.42 min. In an 8-hr day there are 60 jobs (8 × 7.5 = 60). 60 jobs × .475 hr = 27.42 hr of total secretarial time in the college spent on copying jobs. $8.50/hr × 27.42 hr = $233.07 in secretarial wages spent daily for copying in the college. 177 days in the normal academic year × $233.07 = $41,253.39 per year. 3 copiers: λ = 7.5, μ = 5, c = 3; W = .232 hr = 13.92 min. In an 8-hr day there are 60 jobs (8 × 7.5 = 60). 60 jobs × 0.232 hr = 13.92 hr of total secretarial time in the college spent on copying jobs. $8.50/hr × 13.92 hr = $118.32 in secretarial wages spent daily for copying in the college. 177 days in the normal academic year × $118.32 = $20,942.64 per year. Summer month: 2 copiers: λ = 3.75, μ = 5, c = 2; W = .233 hr = 13.98 min. jobs/day = 30; 30 jobs × .233 hr = 6.99 hr; $8.50/hr × 6.99 = $59.42/day; 70 days × 59.42 = $4,159.05. 3 copiers: λ = 3.75, μ = 5, c = 3; W = .204 hr = 12.24 min. jobs/day = 30; 30 jobs × .204 hr = 6.12 hr; $8.50/hr × 6.12 = $52.02; 70 days × 52.02 = $3,641.40. Current copy center total wage copying costs = $41,253.39 + 4,159.05 = $45,412.44. Copy center with 3 copiers total wage copying cost = $20,942.64 + $3,641.40 = $24,584.04. Total annual copying wage savings by adding a third machine = $45,412.44 − $24,584.00 = $20,828.40 per year. Since a copying machine costs $36,000 and has maintenance costs of $8,000 per year, the total cost over the life of the copier will be $84,000 (with no present value discounting). Over the same 6-year period, adding a third copier would save $124,970.40 in copying wage costs, which outweighs the cost of a new copier. However, Dr. Moore may still not be able to convince Dr. Burris. The wage saving are not really savings to the college but a measure of secretarial time that could be reallocated to other tasks within the departments. The college would not save any money; it would simply incur the cost of the copier. The departments could argue that other tasks the secretaries might perform instead of copying would be a more efficient use of $20,828.40 in annual wages, but Dr. Burris would probably be a hard sell with this argument.

⎡ (110)(50)(110 / 50)3 ⎤ 110 =⎢ ⎥ .022 + 2 50 (2)!(120 110) − ⎣ ⎦

= 4.06 passengers Lq = L −

λ μ

= 4.06 − (110 / 50) = 1.857 passengers

Wq =

1.857 = 110 = 1.01 minutes

60.

A multiple server model with 4 cranes will accommodate the level of container traffic at the inland port. λ = 8 containers per hour μ = 2.14 containers per hour c = 4 cranes P0 = .007 L = 16.01 containers Lq = 12.27 containers W = 2 hours Wq = 1.53 hours

61.

Constant service time model μ = 30 passengers/6.8 minutes = 4.41 passengers / minute λ = 4 passengers / minute Lq =

λ2 2μ (μ − λ )

(4) 2 2(4.41)(4.41 − 4) = 4.425 groups of 30 passengers waiting = (4.425)(30) = 132.75 passengers =

13-13 .

different numbers of operators until the goal of one-half minute waiting time is achieved.

CASE SOLUTION: NORTHWOODS BACKPACKER

8 operators: Wq = 5.76 minutes

There are four system configurations to be considered, as follows.

9 operators: Wq = 1.20 minutes 10 operators: Wq = 0.42 minutes

1. 5-day, 8-hour per day service 2. 7-day, 8-hour per day service 3. 5-day, 16-hour per day service

(Probability of waiting = .18) Since 5 extra operators are required to reach the waiting time goal the cost of this alternative is $3,600 + (5)(3,800) = $22,600.

4. 7-day, 16-hour per day service In each case the first step is to determine the number of servers that are required to make the system feasible, i.e., cμ > λ. Remember, the current system has 5 operators (servers), and, μ = 60/3.6 = 16.67 customers per hour.

5-day, 16-hour service

λ = 87.5, μ = 16.67 Cost for 16 hour service = $11,500 Cost per extra operator = $4,700

5-day, 8-hour service: λ = 175, μ = 16.67; c > λ/μ or c > 175/16.67 = 10.49. Thus, at least 11 total operators are required for this (the current) system to be feasible. Since the current physical facility can only accommodate a maximum of 10 work stations, this alternative is eliminated.

At least 6 operators are required for this configuration to be feasible. 6 operators: Wq = 3.24 minutes 7 operators: Wq = 0.78 minutes 8 operators: Wq = 0.27 minutes (Probability of waiting = .21) Since 3 extra operators are required to reach the waiting time goal the cost of this alternative is $11,500 + (3)(4,700) = $25,600.

7-day, 8-hour service: λ = 125, μ = 16.67; c > λ/μ or c > 125/16.67 = 7.49. Thus, at least 8 operators are required for this system to be feasible.

7-day, 16-hour service

λ = 62.5, μ = 16.67 Cost for 16-hour service = $11,500 Cost for 7-day service = $7,200

5-day, 16-hour service: λ = 87.5; μ = 16.67; c > λ/μ or c > 87.5/16.67 = 5.24. Thus, at least 6 operators are required for this system to be feasible.

Cost per extra operator = $6,300 At least 4 operators are required for this configuration to be feasible; however, since 5 operator stations already exist, the starting point is 5 operators.

7-day, 16-hour service: λ = 62.5, μ = 16.67; c > λ/μ or c > 62.5/16.67 = 3.74. Thus, at least 4 operators are required for this system to be feasible.

5 operators: Wq = 1.32 minutes 6 operators: Wq = 0.36 minutes (Probability of waiting = .10) Since only 1 extra operator is required to reach the waiting time goal the cost of this alternative is $11,500 + 7,200 + (1)(6,300) = $25,000.

Therefore, only the first configuration (the current one) is not feasible and is eliminated. Next the costs of the remaining 3 alternatives are evaluated.

The 7-day, 8-hour service configuration has the lowest cost. However, all three alternatives are very close according to cost. All three also meet the goal of a customer getting immediate service at least 70 percent of the time. Thus, other factors may be taken into consideration. For example, both of the 16-hour service alternatives might be more convenient for customers who work during the day.

7-day, 8-hour service

λ = 125, μ = 16.67 Cost for 7 day service = $3,600 Cost per extra operator = $3,800 Recall that at least 8 operators are required for this configuration to be feasible. Thus, starting at this point we must compute the waiting times for

13-14 .

CASE SOLUTION: ANALYZING DISASTER SITUATIONS AT TECH

Wq = 14.41 minutes W = 26.41 minutes

a) Wq = 8.104 minutes

Wq (undefined) = 9.012 minutes b)

Lq = 1.695 victims

Lq = .481 victims Wq = 4.09 minutes

Lq = .198 victims

W = 25.07 minutes

Wq = 1.39 minutes

W = 21.39 minutes

Requires QM for Windows Lq = 15.97 victims

c) Requires QM for Windows

Wq = 67.03 minutes

Lq = 7.2 victims

W = 88 minutes

Wq = 18 minutes

g) Student analysis

W = .38 minutes

CASE SOLUTION: FORECASTING AIRPORT PASSENGER ARRIVALS Current (planned) Security System

Proposed (Required) System

4 A.M.–6 A.M.:

6 checkpoints

8 checkpoints

(λ = 2,463/hr)

6 A.M.–8 A.M.:

6 checkpoints

9 checkpoints

(λ = 2,764/hr)

8 A.M.–10 A.M.:

6 checkpoints

10 checkpoints

(λ = 2,892/hr)

10 A.M.–noon:

6 checkpoints

(λ = 1,624/hr)

Noon–2 P.M.:

6 checkpoints

7 checkpoints

(λ = 2,010/hr)

2 P.M.–4 P.M.:

6 checkpoints

7 checkpoints

(λ = 2,125/hr)

4 P.M.–6 P.M.:

6 checkpoints

7 checkpoints

(λ = 1,863/hr)

6 P.M.–8 P.M.:

3 checkpoints

(λ = 862/hr)

8 P.M.–10 P.M.:

2 checkpoints

1 checkpoint

(λ = 264/hr) μ = 310.3/hr

The planned system is not sufficient from 4 A.M. to 10 A.M., and from noon to 6 P.M. It is excessive from 8 P.M. to 10 P.M. The likely result during these times when the system is not sufficient is that the waiting lines will grow infinitely.

13-15 .

Chapter Fourteen: Simulation

35. Crystal Ball, hotel room rates 36. Crystal Ball, CPM/PERT network

PROBLEM SUMMARY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

Rescue squad emergency calls Car arrivals at a service station Machine breakdowns Income analysis (12–19) Decision analysis (12–16a) Machine repair time (14–3) Product demand and order receipt Bank drive-in window arrivals and service Loading dock arrivals and service Product demand and order receipt Football game Student advising, arrival and service Markov process Inventory analysis Rental car agency CPM/PERT network analysis Store robbery and getaway Stock price movement Hospital emergency room staffing Break-even analysis Rating dates Model analysis (14–21) Production capacity Baseball game Crystal Ball, Computer World example Crystal Ball (14–25) Crystal Ball, Bigelow Mfg. example Crystal Ball (14–27) Crystal Ball, maintenance cost Crystal Ball, B-E analysis (14–20) Crystal Ball, rating dates (14–21) Crystal Ball, production capacity (14–23) Crystal Ball, investment selection Crystal Ball, inventory management

PROBLEM SOLUTIONS 1.

Time Between Calls (hr) 1 2 3 4 5 6

a) r 65 71 20 15 48 89 18 83 08 90 05 89 18 08 26 47 94 06 72 40 62

Cumulative Probability .05 .15 .45 .75 .95 1.00

Time Between Calls 4 4 3 2 4 5 3 5 2 5 1 5 3 2 3 4 5 2 4 3 4

Random Numbers 01–05 06–15 16–45 46–75 76–95 96–99, 00

Cumulative Clock 4 8 11 13 17 22 25 30 32 37 38 43 46 48 51 55 60 62 66 69 73

= 3.48 hr between calls; EV = 1(.05) + b) μ = 73 21

2(.10) + 3(.30) + 4(.30) + 5(.20) + 6(.05) = 3.65. The results are different because there were not enough simulations to enable the simulated average to approach the analytical result.

*Note: Many of these solutions may be different from the solution the instructor or student derives depending on the stream of random numbers used. The Excel solutions (provided on the Companion Web site) will also be different because they use Excel-generated random numbers, whereas many of these solutions were developed using the random number table in the text.

14-1 .

21 calls; no, this is not the average number of calls per 3 days. In order to determine this average, this simulation would have to be repeated a number of times in order to get enough observations of calls per 3-day period to compute an average. Time Between Arrivals (min) 1

Cumulative Probability .15

Random Numbers 01–15

.45

16–45

.85

46–85

1.00

86–99, 00

a) Arrival

Time Between Arrivals (min)

Time Between Arrivals

Cumulative Clock

r 65

μ = 62/24 = 2.58 min between arrivals

The results are different because of the few simulations combined with the different random number streams. If there were enough simulations, the different random numbers would have little or no effect.

μ = 58/20 = 2.9 min between arrivals

14-2 .

Machine Breakdowns/ Week 0

Cumulative Probability .10

Random Numbers 01–10

.20

11–20

.40

21–40

.65

41–65

.95

66–95

1.00

96–99, 00

Use fifth column of random numbers in Table 14.3

a) Week

Breakdowns

Random Number 45

Return 40,000

−40,000

–40,000

120,000

–40,000

120,000

120,000 120,000

120,000

–40,000

120,000

–40,000 40,000

–40,000

–40,000 560,000

μ = 59/20 = 2.95 breakdowns per week

640, 000 = $32, 000 20 The expected value is $40,000, so the simulated average for 20 weeks is significantly lower indicating the need for more trials. Average return =

4. Snowfall Level (in) >40

Cumulative Probability 0.40

RN 01–40

Financial Return $120,000

20–40

0.60

41–60

40,000

<20

1.00

61–99, 00

–40,000

14-3 .

Weather Conditions Rain

Random Numbers 01–30

Overcast

31–45

Sunshine

46–99, 00

Sixth column in Table 14.3. r 19

Profit Sun Visors Umbrellas –500 2,000

1,500

–900

1,500

–900

–500

2,000

1,500

–900

1,500

–900

–200

1,500

–900

1,500

–900

1,500

–900

–200

–500

2,000

–500

2,000

–500

2,000

1,500

–900

1,500

–900

–500

2,000

–500

2,000

1,500

–900

–200

Sun visors: µ = 10,900/20 = $545; umbrellas: µ = 5,000/20 = $250. The best decision according to the simulation (for only 20 weeks) would be sun visors.

14-4 .

* 6.

The first three columns are from Problem 3. Select as many r2’s as there are breakdowns from a different random number stream.

Week

Breakdowns

Repair Time

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

1 2 5 2 0 3 3 5 3 2 4 4 4 4 3 4 4 3 3 0

58 47, 23 69, 35, 21, 41, 14 59, 28 — 13, 09, 20 73, 77, 29 72, 89, 81, 20, 85 59, 72, 88 11, 89 87, 59, 66, 53 45, 56, 22, 49 08, 82, 55, 27 49, 24, 83, 05 81, 07, 78 92, 36, 53, 04 95, 79, 61, 44 37, 45, 18 65, 37, 30 —

2 2+1=3 2+2+1+2+1=8 2+1=3 0 1+1+1=3 2+2+1=5 2 + 3 + 3 + 1 + 3 = 12 2+2+3=7 1+3=4 3+2+2+2=9 2+2+1+2=7 1+3+2+1=7 2+1+3+1=7 3+1+2=6 3+2+2+1=8 3+2+2+2=9 2+2+1=5 2+2+1=5 0

Total repair time = 110 hr; μ = 110/20 = 5.50 hr/week. It could bias the results. If a high random number is selected—for example, 98—this results in a high number of breakdowns (i.e., 5). If the same random number is used, it will result in a high repair time (i.e., 3 hr). Thus, a relationship will result wherein high number of breakdowns equals high repair times, and vice versa. The effect in this model will not be too bad since several repair time random numbers are selected for each breakdown. Average weekly breakdown cost: 110 hours × $50 = $5,500; μ = $5,500/20 = $275.00 per week.

d) Breakdowns per Week

Cumulative Probability

Random Numbers

.20

01–20

.50

21–50

.70

51–70

.85

71–85

.95

86–95

1.00

96–99, 00

14-5 .

Week

Breakdowns

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

0 1 5 1 0 2 1 5 2 1 3 3 3 3 1 3 4 1 2 0

— 58 47, 23, 69, 35, 21 41 — 14, 59 28 13, 09, 20, 73, 77 29, 72 89 81, 20, 85 59, 72, 88 11, 89, 87 59, 66, 53 45 56, 22, 49 08, 82, 55, 27 49 24, 83 —

Repair Time 0 2 2+1+2+2+1=8 2 0 1+2=3 1 1+1+1+2+2=7 1+2=3 3 3+1+3=7 2+2+3=7 1+3+3=7 2+2+2=6 2 2+1+2=5 1+3+2+1=7 2 1+3=4 0

Total repair time = 76 hr; average weekly breakdown cost: 76 × $50 = $3,800; μ = $3,800/20 = $190. Reduction in average weekly repair cost = $275 –190 = $85. Since the maintenance program costs $150 and only $85 would be saved, it should not be put into effect. However, the results should be applied with some hesitancy, since they were derived for only one actual simulation. This whole simulation process should be repeated a number of times. Note: In part d the same random number streams were used as in part a. This was done to replicate as much as possible the conditions of the first simulation since the results were to be compared. However, if many simulations were conducted, this would not have been necessary. 7. Demand/Month 0

Random Numbers 01–04

Lead Time 1

Random Numbers 01–60

05–12

61–90

13–40

91–99, 00

41–80

81–96

97–98

99, 00

14-6 .

Month

Demand

Order Placed

Order Received

Balance

Carrying Cost

—

Order Cost

Stockout Cost

Total

120

100

—

220

100

—

100

120

100

—

220

—

5+5

360

—

360

—

160

—

160

—

120

100

—

220

—

160

—

160

—

100

—

180

—

100

400

500

—

100

400

500

5+5

280

—

280

—

160

—

160

240

—

240

—

120

100

—

220

—

100

400

500

5+5

280

—

280

—

200

—

200

—

120

100

—

220

200

—

200

—

200

—

200

5 —

μ = $256

14-7 .

5,120

* 8. a)

One-teller system: Arrival Clock

Enter Facility Clock

Waiting Time

Length of Queue at Entry

Service Time

Departure Time

Time in System

Customer

Arrival Interval

—

μ = 13.6

μ = 2.2

μ = 17.6

Notice that statistics are misleading since the values keep increasing. In other words, a steady state has not been reached; thus, it is difficult to draw inferences. Two-teller system:

Arrival Clock

Enter Facility Clock

Waiting Time

Length of Queue

Service Time

Departure Time

Time in System

Customer

Arrival Interval

—

μ = .4

14-8 .

μ = 3.8

Arrival Clock

Enter Facility Clock

Waiting Time

Length of Queue

Service Time

Departure Time

Time in System

Customer

Arrival Interval

μ = 1.4

Ties for queue length: Line

Random Numbers

00–49

50–99

Customer

μ=5

Days to Fill and Prepare 3

Cumulative Probability .10

Random Numbers 1–10

.30

11–30

.70

31–70

1.00

71–99, 00

Observation No ship–1st ship

Random Number 43

Time Between Arrivals 4

1st ship–2nd ship

2nd ship–3rd ship

3rd ship–4th ship

4th ship–5th ship

5th ship–6th ship

6th ship–7th ship

7th ship–8th ship

b) It appears that the two-teller system is most appropriate; however, some trade-off analysis between the cost of the extra facility and the cost of customers waiting and possibly leaving is necessary. 9. Days Between Arrivals

Cumulative Probability

Random Numbers

8th ship–9th ship

9th ship–10th ship

.05

1–5

10th ship–11th ship

.15

6–15

11th ship–12th ship

.35

16–35

12th ship–13th ship

.65

36–65

13th ship–14th ship

.85

66–85

14th ship–15th ship

.95

86–95

15th ship–16th ship

1.00

96–99, 00

16th ship–17th ship

17th ship–18th ship

18th ship–19th ship

19th ship–20th ship

4 86

14-9 .

Ship No. 1

Arrival Day 4

Time to Arrival of Next Ship 7

Day Loading Begins 4

Loading Time 5

Departure Day 9

Waiting Time 0

No. of Ships Waiting 0

103

108

125

14-10 .

Average time between arrivals = 86/20 = 4.3 days; average waiting time to load = 125/20 = 6.25 days; average number of tankers waiting to unload = waiting time/simulation period = 125/108 = 1.16 tankers (Note: You must use waiting time instead of tankers waiting per arrival.) b) It should first be pointed out that it would be an error to make inferences about the true system based upon summary statistics (i.e., averages) computed from the simulation of this problem. Why? Because the simulated system does not reach steady state. That is, note that ship waiting times are continuously increasing, from zero up to 17 days, over the 20 ship arrivals. This is easily explained by the fact that the mean arrival rate exceeds the service rate; thus, no steady state can be reached. The relevant output data for analysis of this problem is ship waiting time for each ship, as opposed to any summary statistics. The management should conclude that additional manpower or equipment must be added so that ship servicing time could be reduced to less than 4 days (mean of arrival distribution). If ship service times were reduced to 3.5 days, the simulated system would then reach steady state, and summary statistics would be relevant for analysis of the problem. Another problem that should be considered when computing summary statistics is the time it takes the simulated system to reach steady state and the output generated as the sample is concluded (ending conditions). In both cases, the output may bias the computations of summary statistics. That is, note the reduction in number of ships waiting for the last three observations (ship arrivals 17, 18, 19, and 20), which is simply due to the ending of the simulation. The conclusion reached is that a portion of both the beginning and the end of the simulation output data should be discarded when computing summary statistics. 10. Months to Receive Cumulative Random an Order Probability Numbers 1 .50 01–50 2 .80 51–80 3 1.00 81–99, 00

Demand per Month 1 2 3 4

Cumulative Probability .10 .40 .80 1.00

Random Numbers 01–10 11–40 41–80 81–99, 00

Reorder Random Months Random Demand Number Number Lead Time Numbers per Month 1 21 1 70 3 2

70, 52

25, 17

04, 11

1, 2 = 3

76, 29

3, 2 = 5

70, 17, 48 3, 2, 3 = 8

87, 20, 16 4, 2, 2 = 8

65, 20

3, 2 = 5

97, 46

4, 3 = 7

92, 55, 35 4, 3, 2 = 9

57, 99, 20 3, 4, 2 = 9

52, 45

3, 3 = 6

01, 73

1, 3 = 4

58, 24

3, 2 = 5

84, 35

11, 42

3, 3 = 6 4 2, 2 = 4

1 2, 3 = 5

35, 94, 91 2, 4, 4 = 10

10, 98, 32 2, 4, 2 = 8 57

3 4, 2 = 6

143

14-11 .

Total demand = 143; average demand during lead time = 143/30 = 4.76; thus, should reorder at 5-car level. 11. State University: Cumulative Random Play Probability Numbers (r1) Sweep .10 01–10 Pass .30 11–30 Draw .50 31–50 Off tackle 1.00 51–99, 00

Play Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Tech: Cumulative Defense Probability Wide tackle .30 Oklahoma .80 Blitz 1.00

Random Numbers (r2) 01–30 31–80 81–99, 00

r1 39 73 72 75 37 02 87 98 10 47 93 21 95 97 69 41 91 80 67 59 63 78 87 47 56 22 19 16 78 03 04 61 23 15 58 93 78 61 42 77

Play

Defense

Yardage

Draw 65 Oklahoma Off tackle 71 Oklahoma Off tackle 20 Wide tackle Off tackle 17 Wide tackle Draw 48 Oklahoma Sweep 89 Blitz Off tackle 18 Wide tackle Off tackle 83 Blitz Sweep 08 Wide tackle Draw 90 Blitz Off tackle 05 Wide tackle Pass 89 Blitz Off tackle 18 Wide tackle Off tackle 08 Wide tackle Off tackle 26 Wide tackle Draw 47 Oklahoma Off tackle 94 Blitz Off tackle 06 Wide tackle Off tackle 72 Oklahoma Off tackle 40 Oklahoma Off tackle 62 Oklahoma Off tackle 47 Oklahoma Off tackle 68 Oklahoma Draw 60 Oklahoma Off tackle 88 Blitz Pass 17 Wide tackle Pass 36 Oklahoma Pass 77 Oklahoma Off tackle 43 Oklahoma Sweep 28 Wide tackle Sweep 31 Oklahoma Off tackle 06 Wide tackle Pass 68 Oklahoma Pass 39 Oklahoma Off tackle 71 Oklahoma Off tackle 22 Wide tackle Off tackle 76 Oklahoma Off tackle 81 Blitz Draw 88 Blitz Off tackle 94 Blitz

1 3 7 7 1 12 7 −3 −3 20 7 −10 77 77 77 11 −3–3 77 33 33 33 33 33 11 –3–3 1212 44 44 33 –3–3 55 77 44 44 33 77 33 –3–3 2020 –3–3

Total yardage = 155 yd; the sportswriter will predict that Tech will win.

14-12 .

* 12. Time Between Arrivals 4

Cumulative Probability .20

Random Numbers (r1) 1–20

.50

21–50

.90

51–90

1.00

91–99, 00

Schedule Approval 6 7 8

Cumulative Probability .30 .80 1.00

Random Numbers (r2) 1–30 31–80 81–99, 00

Time Between Arrivals

Arrival Clock

101

108

116

124

22 198

Approval Departure Waiting Time Clock Line

Waiting Time

times gives E (time between arrivals) = 4(.2) + 5(.3) + 6(.4) + 7(.l) = 5.4 min; E (approval time) = 6(.3) + 7(.5) + 8(.2) = 6.9 min. Therefore, students are arriving at a faster rate than their schedules are being approved. The queue will never reach a steady state but will increase infinitely.

Total waiting time = 198 min; 17 arrivals; average waiting time = 198/17 = 11.65 min; average waiting line = 36/17 = 2.12. The queue size and waiting time seem to be increasing and not approaching a steady state. Computing the expected time between arrivals and approval

14-13 .

13.

Tribune: Random Paper Numbers Tribune 01–65 Daily 66–99, 00 News

Daily News: Random Paper Numbers Tribune 01–45 Daily 46–99, 00 News

Demand Random Numbers 15 01–20 16 21–45 17 46–85 18 86–99, 00

Fifth column in Table 14.3; start with Tribune. r 45

Daily News

Tribune 1

Demand

Profit

1,788 μ = 1,788/20 = $89.40; the complete simulation would be achieved by performing the same simulation (above) for all four demand sizes as order sizes and selecting the order size with the maximum profit.

Simulation results: [Tribune Daily News] = [.50 .50]; the results of Markov analysis: [Tribune Daily News] = [.563 .437]. The results differ because the simulation had too few iterations (runs) to approach the actual steady-state results determined by Markov analysis.

15. Customers/ Random Day Numbers 0 01–20 1 21–40 2 41–90 3 91–99, 00

14. Order size = 16 cases. Order size ≥ demand: profit = DP − QC − (Q/2)Cc = 16D − 164. Demand > order size: profit = QP − QC − (Q/2)Cc − Cs(D − Q) = 108 − D.

14-14 .

Random Duration Numbers 1 01–10 2 11–40 3 41–80 4 81–90 5 91–99, 00

Available Day Cars 1

r1 62

Duration/ Car Day Customers r2 Car Available 2 19 2 3 66 3 4

27 43

2 3

4 5

92 22

5 2

9 6

91 46

5 3

10 8

—

09 81

1 4

11 14

Probability of not having a car available = customers not served/total customers = 4/17 = .235; since almost 24% of customers are not served, expansion would probably be warranted. A simulation model of this system necessary to make a decision would need to perform this simulation for a number of different fleet sizes (in addition to the four cars used in this experiment). The simulation should also include the daily cost of the car to the rental agency and the daily rental price plus some estimate of lost current and potential sales when a customer is turned away. The fleet size selected would be the one that maximized average daily profit. 16.

Activity 1−2: Random Day Numbers 5 01−60 9

61−99, 00

21−70

71−99, 00

Activity 1−3: Random Day Numbers 6 01−40

Activity 4−5: Random Day Numbers 3 01−50

41−99, 00

Activity 2−4: Random Day Numbers 2 01−20

Activity 5−6: Random Day Numbers 1 01−40

21−90

91−99, 00

41−99, 00

Activity 3−4:

Activity 3−5: Random Day Numbers 3 01−20 5

Customers Not Served

Day 1

Random Numbers 01−30

31−60

61−99, 00

Path through the network: A = 1−3−5−6; B = 1−3−4−5−6; C = 1−2−4−5−6.

51−99, 00

14-15 .

1−2

1−3

2−4

3−4

3−5

4−5

5−6

Run

x12

x13

x24

x34

x35

x45

x56

18*

19*

23*

21*

19*

16*

19*

20*

21*

18*

Average project completion time (i.e., mean critical path time) = 19.4 weeks; % of time A is critical = 0%, % of time B is critical = 40%, % of time C is critical = 30%; % of time B and C are critical together = 30%. These results indicate that since activity times are probabilistic, a single path may not always be critical; as shown, B and C were both critical. Since the PERT technique always assumes only one path is critical all of the time, the PERT critical path time will be overly optimistic, i.e., it does not recognize that a path might be longer than the critical path. 17. Direction East (x = +1)

Probability .25

Cumulative Probability .25

RN Ranges 01−25

West (x = −1)

.25

.50

26−50

North (y = +1)

.25

.75

51−75

South (y = −1)

.25

1.00

76−99,00

Monte Carlo Simulation (using 16th row of random numbers from Table 14.3) Considering the city as a grid with an x and y axis with the store at point (0, 0), each random number selected indicates a movement of 1 unit (block) in either an x or y direction.

14-16 .

Trial 1 r (x, y)

End of Block 1 58 2 47 3 23 4 69 5 35 6 21 7 41 8 14 9 59 10 28 Within 2 blocks?

(0,1) (−1,1) (0,1) (0,2) (−1,2) (0,2) (−1,2) (0,2) (0,3) (−1,3) no

Trial 2 r (x, y)

Trial 3 r (x, y)

Trial 4 r (x, y)

Trial 5 r (x, y)

68 13 09 20 73 77 29 72 89 81

20 85 59 72 88 11 89 87 59 66

53 45 56 22 49 08 82 55 27 49

24 83 05 81 07 78 92 36 53 04

(0,1) (1,1) (2,1) (3,1) (3,2) (3,1) (2,1) (2,2) (2,1) (2,0) yes

(1,0) (1,−1) (1,0) (1,1) (1,0) (2,0) (2,−1) (2,−2) (2,−1) (2,0) yes

(0,1) (−1,1) (−1,2) (0,2) (−1,2) (0,2) (0,1) (0,2) (−1,2) (−2,2) no

In 2 of the 5 trials the robber is within 2 blocks of the store. As an example, at the end of 10 blocks in trial 1, the robber is 1 block west and 3 blocks north. 18. Stock Price Movement Increase (+)

Probability .45

Cumulative Probability .45

RN Ranges 01−45

Same (0)

.30

.75

46−75

Decrease (−)

.25

1.00

76−99, 00

Stock Price 1/8

Probability Increase .40

Cumulative Probability .40

RN Ranges 01–40

Probability Decrease .12

Cumulative Probability .12

RN Ranges 01–12

1/4

.17

.57

41–57

.15

.27

13–27

3/8

.12

.69

58–69

.18

.45

28–45

1/2

.10

.79

70–79

.21

.66

46–66

5/8

.08

.87

80–87

.14

.80

67–80

3/4

.07

.94

88–94

.10

.90

81–90

7/8

.04

.98

95–98

.05

.95

91–95

.02

1.00

99,00

.05

1.00

96–99, 00

14-17 .

(1,0) (1,−1) (2,−1) (2,−2) (3,−2) (3,−3) (3,−4) (2,−4) (2,−3) (3,−3) no

Day

Stock Price Movement

Price Change (+, −)

Stock Price

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

76 47 25 08 76 56 31 94 88 14 77 06 62 92 68 27 76 17 35 96 58 31 30 70 33 88 70 70 07 03

− 0 + + − 0 + − − + − + 0 − 0 + − + + − 0 + + 0 + − 0 0 + +

1/4

79 15 58

1/2 1/8 1/2

11 31 10 54 24 23

1/8 3/8 1/8 1/4 1/4 1/8

3/4

23 28 98 25 53

1/8 3/8 7/8 1/8 1/2

07 45

1/8 1/4

69 16

3/8 1/4

37 47

1/8 1/4

61 3/4 61 3/4 62 1/4 62 3/8 61 7/8 61 7/8 62 61 5/8 61 1/2 61 3/4 61 1/2 61 5/8 61 5/8 60 7/8 60 7/8 61 60 5/8 61 1/2 61 5/8 61 1/8 61 1/8 61 1/4 61 1/2 61 1/2 61 7/8 61 5/8 61 5/8 61 5/8 61 3/4 62

In order to expand the model for practical purposes, the length of the simulation trial would be increased to one year. Then this simulation would need to be repeated for many trials, i.e., 1,000 trials.

14-18 .

19. Time Between Arrivals 5

Cumulative Probability .06

RN Ranges 01–06

Service Doctor (D)

Cumulative Probability .50

.16

07–16

Nurse (N)

.70

51–70

.39

17–39

Both (B)

1.00

71–99,000

.68

40–68

.86

69–86

1.00

87–99, 00

Time 10

Doctor Cumulative Probability

RN Ranges

Time

Nurse Cumulative Probability

RN Ranges

Time

RN Ranges 01–50

Both Cumulative Probability

RN Ranges

.22

01–22

.08

01–08

.07

01–07

.53

23–53

.32

09–32

.23

08–23

.78

54–78

.83

33–83

.44

24–44

.90

79–90

1.00

84–99,000

.72

45–72

1.00

91–99,000

.89

73–89

1.00

90–99,00

Arrival

Time of Service (mins.)

Patient 1

r –

Clock –

r 31

Service D

r 98

Doctor 30

115

135

150

190

170

205

185

215

200

230

210

265

230

300

240

320

260

335

275

345

300

310

320

330

335

380

355

420

14-19 .

Nurse

Wait?

Departure Clock D N 30 50 80

115

140

170

200

265

P(wait) =

15 = .75 20

440 mins. = 22 mins. 20 It would seem that this system is inadequate given that the probability of waiting is high and the average time is high. Also, observing the actual simulation, three customers had to wait an hour and two others had to wait 35 minutes, which seems excessive. Of course in order to make a fully informed decision this simulation experiment would need to be extended for more patients and then repeated several hundred times. Average waiting time =

20. Sales Volume 300

RN1 Range 1–12

Price $22

RN2 Range 1–7

Variable Cost $8

RN3 Range 1–17

400

13–30

8–23

18–49

500

31–50

24–47

50–78

600

51–73

48–72

79–92

700

74–90

73–90

93–99,00

800

91–99,00

91–99,00 cf = $9,000

Month

RN1

Sales Volume (V)

RN2

Price (p)

RN3

Variable Cost (Cv)

Z = Vp–9,000–VCv

600

$24

2 3 4 5 6 7 8

69 41 28 09 77 89 85

600 500 400 300 700 700 700

35 14 68 20 29 81 59

24 23 25 23 24 26 25

21 59 13 73 72 20 72

9 10 8 10 10 9 10

0 −2,500 −2,200 −5,100 +800 +2,900 +1,500

9 10 11 12 13 14 15 16 17

88 87 53 22 82 49 05 78 53

700 700 600 400 700 500 300 700 600

11 59 45 49 55 24 81 92 04

23 25 24 25 25 24 26 27 22

89 66 56 08 27 83 07 36 95

11 10 10 8 9 11 8 9 12

−600 +1,500 −600 −2,200 +2,200 −2,500 −3,600 +3,600 −3,000

700

+2,200

500

−1,500

600

Probability of at least breaking even =

10 = .50 20

Average monthly loss = − $455

14-20 .

21. Attractiveness

RN1 Range

Intelligence

RN2 Range

Personality

RN3 Range

1–27

1–10

1–15

28–62

11–26

16–45

63–76

27–71

46–78

77–85

72–88

79–85

86–99,00

89–99,00

86–99,00

Date

RN1

Attractiveness

RN2

Intelligence

RN3

Personality

Average Rating

3.67

2.33

4.33

2.33

3.00

3.33

2.00

2.67

2.00

4.00

1.67

3.00

3.67

2.00

2.67

4.00

2.00

Average overall rating of Salem dates = 2.92 Confidence limits can also be developed for the average rating. However, this is best/easiest done with Excel. 23. Capacity

22. There are several ways to assess the accuracy of the results. First the student can determine the expected value for each characteristic and average them to see if this results in a value close to the simulated result. EV (Attractiveness) = 2.50 EV (Intelligence) = 3.05 EV (Personality) = 2.77 Average rating =

2.50 + 3.05 + 2.77 = 2.77 3

F ( x) =

x2 360,000

x = 360,000r

This is relatively close to the simulated result of 2.92 which tends to verify that result.

14-21 .

Demand

RN Range

1–3

100

4–15

200

16–35

300

36–70

400

71–90

500

91–99,00

Week

RN1

Capacity

RN2

Demand

Capacity>Demand?

.16

240.0

200

yes

.93

578.6

300

yes

.13

216.3

400

.27

311.8

300

yes

.53

436.8

200

yes

.18

254.6

200

yes

.24

293.9

yes

.06

146.9

200

.27

311.8

200

yes

.81

540.0

400

yes

.16

240.0

400

.37

365.0

300

yes

.74

516.1

400

yes

.39

374.7

500

.67

491.1

200

yes

.12

207.8

100

yes

.88

562.8

200

yes

.07

158.7

300

.51

428.5

300

yes

.89

566.0

500

yes

15 = .75 20 Average capacity = 362.07 Average demand = 280

Probability capacity > demand =

14-22 .

24.

White Sox:

Yankees:

Play Designation No advance

Cumulative Probability .03

Random Numbers (r) 01–03

Play Designation No advance

Groundout

.42

04–42

Groundout

.42

05–42

Double play

.48

43–48

Double play

.46

43–46

Long fly

.57

49–57

Long fly

.56

47–56

Very long fly

.65

58–65

Very long fly

.62

57–62

Walk

.71

66–71

Walk

.69

63–69

Infield single

.73

72–73

Infield single

.73

70–73

Outfield single

.83

74–83

Outfield single

.83

74–83

Long single

.86

84–86

Long single

.87

84–87

Double

.90

87–90

Double

.92

88–92

Long double

.95

91–95

Long double

.95

93–95

Triple

.97

96–97

Triple

.96

Home run

1.00

98–99, 00

Home run

1.00

97–99, 00

14-23 .

Cumulative Random Probability Numbers (r) .04 01–04

Inning 1: Team White Sox

Inning 3: r

Play

Groundout

Infield single

Outfield single

Groundout

Outs

rbi

Team

White Sox

1 Yankees

Single

Home run

Groundout

Fly

Double

Groundout

1 1 1

04 Out 61 Very long fly 23 Groundout

1 1 1

Team

Play

White Sox

Groundout

58 93

Very long fly Long double

Outfield single

Very long fly Groundout Outfield single Walk Infield single

42 77 65 71

91 Long double

Groundout

80 Outfield single 67 Walk

Groundout

Play

Outs

41 Groundout

1 1 1

Yankees 1 1

Outs rbi

rbi

95 Double 97 Triple 69 Walk

1 1

59 Very long fly

63 Very long fly

Inning 5: 4

Yankees

19 Groundout 16 Groundout 78 Outfield single 03 Out

Inning 4:

Inning 2: White Sox

Outs rbi

Team

Play

1 1

Yankees

Team

Play

White Sox

17 48 89 18

Groundout Out Double Groundout

78 Outfield single 87 Long single 47 Long fly

56 Long fly

22 Groundout

Outs rbi 1 1 1 0

Yankees 1

83 08 90 05 89 18

Outfield single Groundout Double Groundout Double Groundout

1 1 1 1 1 2

14-24 .

Inning 6: Team White Sox

Inning 9: r

Play

08 Groundout 26 Groundout 47 Out

Outs

rbi

1 1 1

Team

Play

Outs

White Sox

25 79 08 15

Groundout Outfield single Groundout Groundout

Yankees

94 06 72 62 47

Long double Groundout Infield single Very long fly Long fly

Line score:

1 1 1

Inning 7: Play

White Sox

68 60 88 17 36

Walk Very long fly Double Groundout Groundout

Outs 1 1 1

77 Outfield single 43 Double play 28 Groundout

2 1 0

Inning 8: Team White Sox

Play

Outs

31 Groundout

06 Groundout 68 Walk

39 Groundout

rbi

Yankees

3 4 5 6 7 8 9 Total

White Sox 1

0 1 0 0 1 0 0

Yankees

0 1 2 1 0 5 X

71 Infield single 22 Groundout

76 Outfield single 81 Outfield single

88 Double

94 Long double

76 Outfield single

23 Groundout

47 Out

mean weekly demand = 1.52 laptops mean weekly revenue = $6,516.71 26. mean weekly demand = 2.51 laptops mean weekly revenue = $10,775.04 27. The maintenance program statistics are: mean annual number of breakdowns = 24.84 mean annual repair time = 54.2 days mean annual repair cost = $108,389 28. For the improved maintenance program simulation, the following statistics are generated: mean annual number of breakdowns = 17.27 mean annual repair time = 29.5 days mean annual repair cost = $58,918 Since the mean annual repair cost for the current maintenance program is $108,389, the savings (i.e., $108,389–$58,918 = $49,471) exceeds the $20,000 cost; so the program should be implemented. 29. average maintenance cost for the life of the car = $3,594.73 P(cost ≤ $3,000) = .435 30. average profit = $813.11 probability of breaking even = .569 31. average rating = 2.91 probability of rating better than 3.0 (i.e., P(x ≥ 3.0)) = .531 32. average surplus capacity = 63.67 probability of sufficient capacity = .618 33. a) In Crystal Ball include the parameters (μ, σ) for each fund return distribution in a spreadsheet cell. Select 6 other spreadsheet cells for the investment combination returns. For example, if the return parameters for investment 1 are in cell D6, the return parameters for investment 2 are in cell D7 and the total return for the investment combination (1,2) is in cell C12, the formula in C12 is “=50,000* (1 + D6)3 + 50,000(1 + D67)3”

rbi

Yankees

25. r

1 1 0

Team

rbi

14-25 .

The simulation results are as follows, Combination

Return

P (Return ≥ 120.000)

1,2

μ = 157,572, σ = 21,459

(.974)

1,3

μ = 166,739, σ = 28,599

(.959)

1,4

μ = 161,888, σ = 23,016

(.981)

2,3

μ = 148,692, σ = 22,693

(.911)

2,4

μ = 144,841, σ = 13,683

(.980)

3,4

μ = 129,429, σ = 20,886

(.653)

The critical path is B. b) Enter the critical path time of 11.98 in the lower left-hand corner of the frequency chart window and hit enter. This results in a value of 25.4% which is the percentage of time (or probability) that this path will exceed 11.98, the critical path time. In CPM/PERT analysis the critical path is determined analytically, which assumes only one path will always be critical. The simulation result shows that other paths might also be critical a percentage of the time. Since the analytical result reflects only a single critical path, the more accurate critical path results are provided with simulation.

b) The highest probability of exceeding $120,000 is virtually tied at .98 between (1,4) and (2,4). 34. Q: μ = 1,860.83, σ = 435.83 TC: μ = 1,465.2, σ = 326.85 35. a) Select an Excel spreadsheet cell to include the normal distribution parameters (μ = 800, σ = 270) for conference rooms, for example, cell D4. In another cell, for example D9, include the cost equation for the number of rooms actually reserved that is being tested. For example, if 600 rooms are being tested, the cost formula in D9 would be, “= 80*MAX(600-D4,0) + 40*MAX(D4600,0).” The simulation average value (for 1,000 trials) for 600 rooms is $12,283. The costs for each room reservation value are Rooms Reserved

Cost ($)

600

12,283

700

11,855

800

13,149

900

16,176

1,000

20,756

CASE SOLUTION: JET COPIES The probability function for time between repairs, f(x), is, x f ( x ) = ,0 ≤ x ≤ 6 18 The cumulative function, F(x), is, x

F ( x) = ∫ 0

x x2 dx = 18 36

and, x2 36 x2 = 36r r=

x=6 r The cumulative distribution and random number ranges for the distribution of repair times are

Repair Time y (days)

Thus, the minimum cost is with 700 rooms reserved. b) Testing different values between 600 and 800 shows that a more exact value is approximately 690 rooms. 36. In Crystal Ball define a cell for each activity’s triangular distribution parameters, then add the specific cells on each path to get the total path times. a) Path A: μ = 10.96, σ = 1.60 Path B: μ = 11.98, σ = 1.44 Path C: μ = 9.84, σ = 1.53 Path D: μ = 10.85, σ = 1.35

Cumulative P(y) Probability

RN Ranges

.20

01–20

.45

.65

21–65

.25

.90

66–90

.10

1.00

91–99,000

The probability function for daily demand is developed by determining the linear function for the uniform distribution, which is, f ( z) =

or,

14-26 .

1 b−a

f ( z) =

1 1 = b−a 6

14-27 .

This function is integrated to develop the cumulative probability distribution, z

z z z z ⎛1⎞ F ( z ) = ∫ dz = |2 = − ⎜ ⎟ 6 6 6 ⎝3⎠ 2 =

z 1 − 6 3

Letting F(z) = r. z 1 r= − 6 3 and solving for z, 1⎞ ⎛ z = 6⎜r + ⎟ 3⎠ ⎝ z = 6r + 2 Monte Carlo Simulation:

Time Between Breakdowns x weeks

∑x

Repair Time y days

.45 .90 .84 .17 .74 .94

4.03 5.69 5.50 2.47 5.16 5.82

4.03 9.72 15.22 17.69 22.85 28.67

61 11 19 58 30 95

2 1 1 2 2 4

.07 .15 .04 .31 .07 .99 .97 .73

1.59 2.32 1.20 3.34 1.59 5.97 5.91 5.12

30.29 32.58 33.78 37.12 38.71 44.68 50.59 55.71

38 24 68 41 35 18 71 59

2 2 3 2 2 1 3 2

They need to purchase the back-up copier.

14-28 .

Copies Lost z

.19, .65 .51 .17 .63, .85 .37, .89 .76, .71, .34, .11 .27, .10 .59, .87 .08, .08, .89 .42, .79 .79, .97 .26 .87, .39, .28 .97, .69

3.14 + 5.90 = 9.04 5.06 3.02 5.78 + 7.10 = 12.88 4.22 + 7.34 = 12.88 6.56 + 6.26 + 4.04 + 2.66 = 19.52 3.62 + 2.60 = 6.22 5.54 + 7.22 = 12.76 2.48 + 2.48 + 7.34 = 12.30 4.52 + 6.74 = 11.26 6.74 + 7.82 = 14.56 3.56 7.22 + 4.34 + 3.68 = 15.24 7.82 + 6.14 = 13.96

Revenue Lost

$904 506 302 1,288 1,156 1,952 622 1,276 1,230 1,126 1,456 356 1,524 1,396 $15,094

Average total time: Montgomery Regional = 1.78 hrs. Raeford Memorial = 4.32 hrs. County General = 4.18 hrs. Lewis Galt = 2.89 hrs. HGA Healthcare = 4.21 hrs.

CASE SOLUTION: BENEFIT–COST ANALYSIS OF THE SPRADLIN BLUFF RIVER PROJECT mean B/C = 1.278 σ = .145 P(B/C ≥ 1.0) = .985

Average total time: Montgomery Regional = 2.17 hrs. Raeford Memorial = 4.74 hrs. County General = 4.76 hrs. Lewis Galt = 3.61 hrs. HGA Healthcare = 4.27 hrs.

Student response

CASE SOLUTION: DISASTER PLANNING AT TECH a)

Average number of victims/hospital: Montgomery Regional = 21.4 Raeford Memorial = 25.7 County General = 12.8 Lewis Galt = 8.6 HGA Healthcare = 17.1

14-29 .

Chapter Fifteen: Forecasting PROBLEM SUMMARY

36. Coefficient of determination (15–35) 37. Linear regression, correlation, discussion

1. Moving average, MAD

38. Linear trend line (15–37)

2. Moving average, MAD

39. Discussion (15–35)

3. Moving average, cumulative error

40. Linear regression

4. Graphical analysis, discussion (15–3) 5. Exponential smoothed, and moving average, MAD

41. Linear regression

6. Exponential smoothed, adjusted, MAPD

42. Linear regression

7. Exponential smoothed, adjusted, linear trend line, MAD, cumulative error 8. Exponential smoothed, adjusted, linear trend line, MAD, average error 9. Moving average, weighted moving average, exponential smoothed, MAD

43. Forecast model selection

10. Forecast model selection 11. Adjusted exponential smoothed, E and E

48. Model selection

12. Seasonally adjusted linear trend line (15–11)

50. Linear regression

13. Seasonally adjusted linear trend line (15–3)

51. Linear trend line (15–50)

14. Seasonally adjusted forecast

52. Linear regression

15. Adjusted exponential smoothed, linear trend line, MAD

53. Linear trend line, linear regression

16. Forecast model comparison

55. Linear trend line

17. Forecast model comparison

56. Multiple regression, Excel, QM for Windows

18. Seasonally adjusted linear trend line, MAD

57. Multiple regression, Excel

19. Seasonally adjusted forecast, linear trend line

58. Multiple regression, Excel (15–44)

20. Seasonally adjusted forecast, linear trend line

59. Multiple regression, Excel

21. Adjusted exponential smoothed forecast (15–20)

60. Multiple regression, Excel

44. Model comparison 45. Exponential smoothing (15–44) 46. Seasonally adjusted forecast, linear regression 47. Adjusted exponential smoothing (15–46) 49. Model selection

54. Linear trend line

22. Seasonally adjusted forecast, linear trend line 23. MAD, MAPD and cumulative error 24. Discussion, forecast accuracy 25. Exponential smoothed (15–1) 26. MAD and cumulative error, linear trend line 27. Linear regression 28. Linear regression, correlation 29. Linear trend line, exponential smoothing, 3mo moving average 30. Linear trend line, 3-mo moving average 31. Linear regressions 32. Linear trend line, linear regression 33. Seasonally adjusted forecast 34. Seasonally adjusted forecast 35. Linear regression, correlation

15-1 .

2. a) and b)

PROBLEM SOLUTIONS 1. a) and b)

Weighted Moving Average

Month

Demand

8.00

—

12.00

—

Month Average

Sales

3-Month Moving Average

Jan

9.00

—

Feb

7.00

—

7.00

—

Mar

10.00

—

9.00

8.77

Apr

8.00

8.67

—

15.00

9.33

8.70

May

7.00

8.33

—

11.00

10.33

12.06

Jun

12.00

8.33

8.20

10.00

11.67

12.08

Jul

10.00

9.00

8.80

Aug

11.00

9.67

9.40

8 9

12.00 —

12.00 11.00

10.93 11.22

Sep

12.00

11.00

9.60

Oct

10.00

11.00

10.40

Nov

14.00

11.00

Dec

16.00

12.00

11.40

13.33

12.60

Jan c)

5-Month Moving Average

3-Month Moving Average

Three-month MAD = 1.89, 5-month MAD = 2.43. The dealer should use the 3-month forecast for January because the smaller MAD indicates a more accurate forecast.

15-2 .

Three-month MAD = 1.6; weighted 3-month MAD = 2.15. The 3-month moving average forecast appears to be more accurate.

a), b) and c)

Quarter

Demand

3-quarter Moving Average Forecast

105.00

—

150.00

—

93.00

—

121.00

116.00

5.00

—

113.85

7.15

140.00

121.33

18.67

—

116.69

23.31

170.00

118.00

52.00

121.80

48.20

125.74

44.26

105.00

143.67

−38.67

134.80

−29.80

151.77

−46.77

150.00

138.33

11.67

125.80

24.20

132.40

17.60

150.00

141.67

8.33

137.20

12.80

138.55

11.45

170.00

135.00

35.00

143.00

27.00

142.35

27.65

110.00

156.67

−46.67

149.00

−39.00

160.00

−50.00

130.00

143.33

−13.33

137.00

−7.00

136.69

−6.60

—

136.67

—

142.00

—

130.20

—

Error

5-quarter Moving Average

Cumulative Errors are: 3-quarter moving average, E = 32.0 5-quarter moving average, E = 36.4 Weighted 3-quarter moving average, E = 28.09 The weighted 3-quarter forecast appears to be the most accurate. All the forecasts exhibit a low bias.

4. There appears to be a slight upward trend in the demand data, and a pronounced seasonal pattern with a peak increase during the second quarter each year, followed by a substantial decrease in the third quarter.

15-3 .

Error

Weighted 3-quarter Moving Error

Error

5. a) and b) Enrollment

3-Semester Moving Average

Exponentially Smoothed Forecast

400

—

450

—

400.00

350

—

410.00

420

400.00

398.00

500

406.67

402.40

575

423.33

421.92

490

498.33

452.53

650

521.67

460.00

—

571.67

498.02

Semester

3-semester MAD = 80.33; Exponentially Smoothed MAD = 87.16; 3-semester moving average appears to be slightly more accurate.

6. a) and b)

Demand

Exponentially Smoothed Forecast (α = .30)

Adjusted Exponentially Smoothed Forecast (α = .30, β = .20)

Oct

800

800.00

—

Nov

725

800.00

Dec

630

777.50

773.00

Jan

500

733.25

720.80

Feb

645

663.27

639.32

Mar

690

657.79

637.53

Apr

730

667.45

653.18

May

810

686.21

678.55

Jun

1200

723.35

724.64

Jul

980

866.34

895.98

Aug

—

900.44

930.96

Month

Exponentially Smoothed MAPD = 1,282.86/7,710 = .166 = 16.63% Adjusted forecast MAPD = 1,264.59/7,710 = .1640 = 16.4% The adjusted exponentially smoothed forecast appears to be more accurate but only slightly.

15-4 .

Month

Price

Exponentially Smoothed Forecast (α = .40)

62.70

—

64.15

63.90

62.70

64.75

68.00

63.18

63.32

65.36

66.40

65.10

65.78

65.97

67.20

65.62

66.25

66.57

65.80

66.25

66.88

67.18

68.20

66.07

66.46

67.79

69.30

66.92

67.45

68.39

67.20

67.87

68.53

69.01

10 11

70.10 —

67.60 68.60

67.98 69.17

69.61 70.22

Adjusted Exponentially Smoothed Forecast (α = .40, β = .30)

Linear Trend Line y = 63.54 + .607x

Exponentially Smoothed

Adjusted Exponentially Smoothed

Linear Trend

Cumulative Error

14.75

10.73

—

MAD

1.89

1.72

1.09

The linear trend line forecast appears to be the most accurate. 8.

Year

Occupancy Rate

Exponentially Smoothed Forecast (α = .20)

Adjusted Exponentially Smoothed Forecast (α = .20, β = .20)

Linear Trend Line (y = .77 + 0.13x)

0.83

—

0.783

0.78

0.83

0.830

0.796

0.75

0.82

0.818

0.810

0.81

0.801

0.823

0.86

0.81

0.803

0.836

0.85

0.82

0.816

0.850

0.89

0.82

0.824

0.863

0.90

0.84

0.840

0.876

0.86

0.85

0.854

0.890

—

0.85

0.856

0.903

15-5 .

Exponentially Smoothed Forecast

Adjusted Forecast

Linear Trend Line Forecast

.0136

.0137

—

MAD

.0436

.0432

.0266

d) The lowest MAD values are with both the weighted 3-month moving average forecast and the exponentially smoothed forecast.

The linear trend line forecast appears to be the most accurate. 9. a)

3-month moving average forecast for month 21 = 74.67

10.

21 = 74.67 MAD = 3.12

Exponential smoothing forecasts (α = 0.3).

b) 3-month weighted moving average forecast for month 21 = 75.875 c)

Group data into 3 months to forecast periods 19, 20, and 21. Possible models include the following. F19,20,21 = 51.67, MAD = 18.93

MAD = 2.98

Linear trend line forecasts

Exponentially smoothed forecast (0.40) for month 21 = 74.60

F19,20,21 = 69.27, MAD = 1.10

MAD = 2.87 11. Adjusted Exponentially Smoothed Forecast (α = .50, β = .50)

Error

Quarter

Sales

Exponentially Smoothed Forecast (α = .50)

350

350.00

—

510

350.00

160.00

750

430.00

470.00

280.00

420

590.00

690.00

−270.00

370

505.00

512.50

−142.50

480

437.50

407.50

72.50

860

458.75

454.37

405.62

500

659.37

757.50

−257.50

450

579.69

588.91

−138.91

550

514.84

487.03

62.97

820

532.42

527.30

292.69

570

676.21

745.55

−175.55

—

623.11

E = 26.30 E = 289.336

The forecast seems to be biased low.

15-6 .

12.

Seasonal factors: Quarter 1: 1,170/6,630 = .18

14.

Quarter 2: 1,540/6,630 = .23 Quarter 3: 2,430/6,630 = .37 Quarter 4: 1,490/6,630 = .22 Forecast for 2006: y = 1,850 + 180x = 2,570 Seasonally adjusted forecasts: Quarter 1: 2,570(.18) = 453.53 Quarter 2: 2,570(.23) = 596.95 Quarter 3: 2,570(.37) = 941.95

Day

Daily Demand

1 2 3 4 5 6 7 8

212 182 215 201 158 176 212 188 ΣD = 1,544

Quarter 4: 2,570(.22) = 577.57

13.

D1 389 = = .25 ∑ D 1,544

The seasonality factor seems to provide a more accurate forecast.

S1 (10AM − 3PM) =

Seasonal factors: Quarter 1: 395/1,594 = .25

S 2 (3PM − 7PM) =

D2 567 = = .37 ∑ D 1,544

S3 (7PM − 11PM) =

D3 320 = = .21 ∑ D 1,544

Quarter 2: 490/1,594 = .31 Quarter 3: 308/1,594 = .19 Quarter 4: 401/1,594 = .25

S 4 (11PM − 12AM) =

D4 268 = = .17 ∑ D 1,544

Linear Trend Forecast for Day 9: y = 202.54 − 2.12x = 183.46 Day 9 Forecast for

Forecast for year 4: y = 440.33 + 45.5x = 622.33

10 AM–3 PM: 183.45(.25) = 45.87

Seasonally adjusted forecasts:

3 PM–7 PM: 183.46(.37) = 67.88

Quarter 1: 622.33(.25) = 155.6 Quarter 2: 622.33(.31) = 192.9

7 PM–11 PM: 183.46(.21) = 38.52

Quarter 3: 622.33(.19) = 118.2

11 PM–12 AM: 183.46(.17) = 31.18

Quarter 4: 622.33(.25) = 155.6 15. Exponentially Smoothed Forecast (αα = .30)

Adjusted Exponentially Smoothed Forecast (αα = .30, β = .20)

Linear Trend Line y = 4,690 − 211.67x

—

4,478.33

Year

Sales

4,260.00

0.00

4,510.00

4,260.00

4,266.67

4,050.00

4,335.00

4,350.00

4,055.00

3,720.00

4,249.50

4,244.40

3,843.33

3,900.00

4,090.65

4,054.80

3,631.67

3,470.00

4,033.45

3,993.34

3,420.00

2,890.00

3,864.42

3,798.52

3,208.33

3,100.00

3,572.09

3,460.91

2,996.67

—

3,313.19

2,785.00

15-7 .

Adjusted Exponentially Smoothed Forecast

Linear Trend Line

431.71

166.25

MAD

The linear trend line forecast appears to be the most accurate. 16.

18. a) Seasonally adjusted forecast January–March: D1 = 106.8

y = 354.35 + 30.195x

April–June: D2 = 135.6

Linear Trend Line:

July–September: D3 = 109.0

F(25) = 1109.23

October–December: D4 = 233.6

MAD = 119.83

ΣD = 585

Exponential Smoothing:

D1 106.8 = = .18 ∑D 585 D 135.6 S2 = 2 = = .23 ∑D 585 D 109.0 S3 = 3 = = .19 ∑D 585 D 233.6 S3 = 4 = = .40 585 ∑D S1 =

F(25) = 998.76 MAD = 164.02 3-Month Weighted Moving Average: F(25) = 1057.89 MAD = 109.18 17.

Linear trend line: y = 347.33 + 3.865x

Linear Trend Line Forecast for year 6: y = 96.33 + 6.89x = 137.67 January–March Forecast for year 6: SF1 = S1F6 = .18(137.67) = 24.78 April–June Forecast for year 6: SF2 = S2F6 = .23(137.67) = 31.66 July–September Forecast for year 6: SF3 = S3F6 = .19(137.67) = 26.16 October–December Forecast for year 6: SF4 = S4F6 = .40(137.67) = 55.07

F(37) = 347.33 MAD = 65.48 Exponentially smoothed model (α = 0.20): F(37) = 460.56 MAD = 74.92 5-month moving average: F(37) = 467.80 MAD = 65.75

b) Linear Trend Line Forecast for January–March year 6:

The 5-month moving average “seems” best. It reflects a recent trend upwards in sales but has a lower MAD than the exponentially smoothed forecast; however, the student might provide a different choice.

y = 16.29 + 1.69x = 26.43 Linear Trend Line Forecast for April–June year 6: y = 21.67 + 1.79x = 32.41 Linear Trend Line Forecast for July–September year 6: y = 18.35 + 1.15x = 25.25 Linear Trend Line Forecast for October–December year 6: y = 38.94 + 2.26x = 53.50

15-8 .

Year/Quarter

Orders

Seasonally Adjusted Forecast

*Dt − Ft*

Linear Trend Line Forecast

*Dt − Ft*

Jan–Mar

18.6

18.58

.02

17.98

.62

April–June

23.5

23.74

.24

23.46

.04

July–Sept

20.4

19.61

.79

19.50

.90

Oct–Dec

41.9

41.29

.61

42.20

.30

Jan–Mar

18.1

19.82

1.72

19.67

1.57

April–June

24.7

25.32

.62

25.25

.55

July–Sept

19.5

20.92

1.42

20.65

1.15

Oct–Dec

46.3

44.04

2.26

44.46

1.84

Jan–Mar

22.4

21.06

1.34

21.36

1.04

April–June

28.8

26.91

1.89

27.04

1.76

July–Sept

21.0

22.23

1.23

21.80

.80

Oct–Dec

45.5

46.80

1.30

46.72

1.22

Jan–Mar

23.2

22.30

.90

23.05

.15

April–June

27.6

28.49

.89

28.83

1.23

July–Sept

24.4

23.54

.86

22.95

1.45

Oct–Dec

47.1

49.56

2.50

48.98

1.88

Jan–Mar

24.5

23.54

.96

24.74

.24

April–June

31.0

30.08

.20

30.62

.38

July–Sept

23.7

24.85

1.15

24.10

.40

Oct–Dec

52.8

52.31

.49

51.24

1.56

Σ|Dt − Ft| = 21.39 Seasonally Adjusted Forecast MAD = ∑ | Dt − Ft | =

173.3 = .215 806.1 155.9 S (Winter) = = .193 806.1 203.5 S (Spring) = = .252 806.1 273.4 S (Summer) = = .339 806.1

19. S (Fall) =

21.39 = 1.07 20

Linear Trend Forecast for Seasons MAD = ∑ | D t − Ft | =

Σ|Dt − Ft| = 19.08

19.08 = .954 20

Although both forecasts seem to be relatively accurate, the linear trend line forecast for each season is slightly more accurate according to MAD.

y = 195.55 + 2.39(5) = 207.5 Forecasts for 2006: Fall:

(207.5)(.215) = 44.61

Winter:

(207.5)(.193) = 40.08

Spring:

(207.5)(.252) = 52.29

Summer: (207.5)(.339) = 70.34 Yes, there does appear to be a seasonal pattern.

15-9 .

20.

ΣD = 2,904

SF7(1 PM) = 43.39

Linear trend line forecast for year 7:

SF8(2 PM) = 34.95

y = 410.4 + 21.03x = 557.6

SF9(3 PM) = 27.84

SF1(7 AM) = 73.54

SF10(6 PM) = 47.04

SF2(8 AM) = 48.19

SF11(7 PM) = 43.78

SF3(9 AM) = 26.11

SF12(8 PM) = 22.85

SF4(10 AM) = 38.21

SF13(9 PM) = 15.55

SF5(11 AM) = 59.14

There does appear to be a seasonal pattern.

SF6(noon) = 77.00 21.

Error

Trend

Adjusted Exponentially Smoothed Forecast (α = .30, β = .20)

410.00

62.00

0.0000

410.00

62.00

473.5

428.60

56.40

3.7200

432.32

52.68

493

494.5

445.52

47.48

6.3600

451.88

41.12

513

515.6

459.76

53.24

7.9368

467.70

45.30

531

536.6

475.73

55.27

9.5436

485.28

45.72

557.6

492.31

10.951

503.27

Year

Pool Attendance

Linear Trend Line Forecast

Exponentially Smoothed Forecast (α = .30)

410

431.4

410.00

472

452.5

485

The linear trend line annual forecast appears more accurate. 22. ΣD = 2,300 1,347 = 0.586 2,300 953 S1 = = 0.414 2,300 S1 =

Linear trend line forecast for week 11: y = 203 + 4.91x = 257 Weekend forecast: 257(.586) = 150.61 Weekday forecast: 257(.414) = 106.40

15-10 .

Error

23. Actual Demand

Forecast Demand

Error

|Dt − Ft|

Running MAD

Cumulative Error

160

170

−10

10.00

−10

150

165

−15

12.50

−25

175

157

14.33

−7

200

166

19.25

Month

190

183

16.80

220

186

19.67

205

203

17.14

210

204

15.75

200

207

−7

14.78

220

203

15.00

150

Cumulative Error

86.00

8.60

MAD

15.00

MAPD

0.08

There is no way to determine if this is an accurate forecast method unless it is compared with some other method. 24. a) Month

Demand

Forecast

Error

|Dt − Ft|

Running MAD

Cumulative Error

Tracking Signal

Mar

120

—

Apr

110

120.0

−10

10.00

−10

−1.00

May

150

116.0

22.00

1.09

Jun

130

129.6

0.4

14.80

24.4

1.65

Jul

160

129.7

30.3

18.67

54.7

2.93

Aug

165

141.8

23.2

19.58

77.9

3.98

Sep

140

151.1

−11.1

11.1

18.17

66.8

3.67

Oct

155

146.7

8.3

16.76

75.1

4.48

150.0

75.10

117.30

Nov Bias

10.73

MAD

16.76

MAPD

0.1038

Cumulative Error

75.10

15-11 .

b. Month

Demand

3-Month Moving Forecast

Error

|Dt − Ft|

Mar

120

—

Apr

110

—

May

150

—

Jun

130

126.67

3.33

Jul

160

130.00

30.00

Aug

165

146.67

18.33

Sep

140

151.67

−11.67

11.67

Oct

155

155.00

0.00

153.33

39.99

63.33

Nov Bias 8.60 MAD 12.67 MAPD 0.08 MSE 276.64 Cumulative Error 39.99

The 3-month moving average seems to provide a better forecast. 25. Month

Demand

Forecast

|Dt – Ft|

Jan

9.00

—

Feb

9.00

2.00

Mar

8.60

1.40

Apr

8.88

0.88

May

8.70

1.70

Jun

8.36

3.64

Jul

9.09

0.91

Aug

9.27

1.73

Sep

9.62

2.38

Oct

10.09

0.09

Nov

10.07

3.92

Dec Jan

10.86 11.88

5.14

MAD

Moving Average Forecast (Prob. 1a)

Exponentially Smoothed Forecast (Prob. 23)

1.89

2.16

The 3-month average forecast appears to be more accurate.

15-12 .

26.

MAD = 1.79 Cumulative Error = 12.50 According to the above measures, the forecast appears to be fairly accurate. Year

Demand

Forecast

Error

Running MAD

Cumulative Error

16.8

—

14.1

16.8

−2.7

2.70

−2.7

15.3

15.7

−0.4

1.55

−3.1

12.7

15.5

−2.8

1.97

−5.9

11.9

14.4

−2.5

2.10

−8.4

12.3

13.4

−1.1

1.90

−9.5

11.5

12.9

−1.4

1.82

−10.9

10.8

12.4

−1.6

1.79

−12.5

Linear trend model: Year

Demand

Forecast

16.80

15.87

14.10

15.10

15.30

14.33

12.70

13.56

11.90

12.79

12.30

12.02

11.50

11.24

10.80

10.47

9.70

MAD = 0.688 The linear trend line forecast appears to be more accurate according to MAD.

15-13 .

27.

33.

y = 68.92 + 0.223x where y = occupancy rate x = wins

Forecast if the Blue Sox win 88 games; 68.92 + .223(88) = 88.58 occupancy rate; r = .4113 28. a) y = 2.36 + .267x where y = sales x = permits Forecast if 30 permits are filed: 2.36 + .268(30) = 10.40 b) Correlation coefficient = .699, indicating a moderately strong causal relationship 29. Linear trend line: y = 354.35 + 30.195x y (25) = 1109.23 MAD = 119.83 Exponential smoothing: month 25 forecast = 998.76 MAD = 164.02 3-month moving average: month 25 forecast = 1057.89 MAD = 109.18 3-month moving average appears to be most accurate. 30. Linear trend line: y = 68.955 + 4.253x y (13) = 124.25

Forecast i

3,338

0.15

1184.13

1,010

0.05

358.29

968

0.04

343.39

1,065

0.05

377.80

2,079

0.10

737.51

890

0.04

315.72

730

0.03

258.96

4,307

0.20

1527.87

3,010

0.14

1067.77

1,636

0.08

580.36

927

0.04

328.85

1,620

0.08

574.68

21,580

1.00

Si = monthly service calls/21,850 Forecast (i) = (Si)(7655.33) where, y = 6731.33 + 231x y(4) = 7655.33

10% growth = 136.68

34. Closely observing the data shows seasonal trend in July, August, September and October, likely due to sales as schools (college and high school) start in the early fall, with a smaller jump in demand before Christmas. However, demand is basically “flat” from year 1 to year 2, so a forecast model that reflects an upward trend or growth (like a linear trend line) is probably not appropriate. Alternative, the approach taken here is to simply sum demand for both years, develop seasonal factors for each month and then multiply these seasonal factors by the average of the previous two years total demand, or 29,330. This approach has a MAD value of only 57.33 whereas the MAD value of a seasonally adjusted linear trend line forecast (see Excel solution) has a MAD value of 699.45. This is not an approach described in the chapter so it requires some intuition on the part of the student, to recognize that there is seasonality but no trend.

3-month moving average: 113.65 10% growth = 125.02 Bernice should purchase the delivery truck 31.

Monthly Calls

y = 3513.72 – 13.83x y (115) = 1923.23 r = –0.952 A relatively strong relationship

32. a) Linear trend line: y = 25,392.86 + 2,523.81x y (9) = 48,107.14 b) Linear regression: y = 29,113.23 + 30.125x y (694) = 50,020.20 R = 0.95

15-14 .

Month

Year 1 Demand

Year 2 Demand

Sum

January

2,447

2,561

5,008

0.085

February

1,826

1,733

3,559

0.061

March

1,755

1,693

3,448

0.059

April

1,456

1,484

2,940

0.050

May

1,529

1,501

3,030

0.052

June

1,633

1,655

3,288

0.056

July

2,346

2,412

4,758

0.081

August

3,784

4,017

7,801

0.133

September

4,106

3,886

7,992

0.136

October

3,006

2,844

5,850

0.100

November

2,257

2,107

4,364

0.074

December

3,212

3,410

6,627

0.113

Total =

29,357

29,308

Month #

Month

Units Demanded

Forecast

Error

January

2,447

0.085

2,493

46.05

February

1,826

0.061

1,789

36.87

March

1,755

0.059

1,730

24.53

April

1,456

0.050

1,467

10.5

May

1,529

0.052

1,525

3.84

June

1,633

0.056

1,642

9.48

July

2,346

0.081

2,376

29.73

August

3,784

0.133

3,901

116.89

September

4,106

0.136

3,989

117.12

October

3,006

0.100

2,933

November

2,257

0.074

2,170

86.58

December

3,212

0.113

3,314

102.29

January

2,561

0.085

2,493

67.95

February

1,733

0.061

1,789

56.13

March

1,693

0.059

1,730

37.47

April

1,484

0.050

1,467

17.5

May

1,501

0.052

1,525

24.16

June

1,655

0.056

1,642

12.52

July

2,412

0.081

2,376

36.27

August

4,017

0.133

3,901

116.11

15-15 .

Month #

Month

Units Demanded

Forecast

Error

September

3,886

0.136

3,989

102.88

October

2,844

0.100

2,933

November

2,107

0.074

2,170

63.42

December

3,410

0.113

3,314

95.71

Year

Applications

Linear Trend Line Forecast

6,050

5,321.64

4,060

5,061.27

5,200

4,800.91

4,410

4,540.55

4,380

4,280.18

4,160

4,019.82

3,560

3,759.45

2,970

3,499.09

3,280

3,238.74

3,430

2,978.36

11 y = 5,582.00 − 260.36x

2,718.00

38.

35. a) y = −113.40 + 2.986x where y = gallons of ice cream x = temperature Forcast for temperature of 85°: −113.4 + 2.98(85) = 140.43 gallons b) Correlation coefficient = .929, indicating a strong causal relationship Coefficient of determination = (.929)2 = .863, indicating that 86.3% of the variation of ice cream sales can be attributed to the temperature.

36.

37. a) y = 6,541.66 − .449x where y = applications x = tuition If tuition is $9,000, forecast is 6,541.66 − .448(9,000) = 2,503.2 applications If tuition is $7,000, forecast is 6,541.66 − .448(7,000) = 3,400.6 applications b) Correlation coefficient = −.754, indicating a fairly strong linear relationship between tuition costs and number of applicants. c)

Correlation coefficient = −0.843 a) The linear regression forecast (from problem 37) has a MAD value of 466.9 while the MAD value for the linear trend line forecast in this problem is 372.1, indicating that the linear trend line forecast is somewhat better. The cumulative error (or bias) is zero (0) for both forecasts.

Number of class sections, number of dormitory rooms, number of persons per class, plus budgeting decisions.

b) The correlation coefficient is −.843 indicating a strong relationship between applications and time. 39. The slope, b = 2.986, indicates the rate of change, i.e., the number of gallons sold for each degree increase in temperature. 40. y = .284 + .407x where y = sales x = promotional expenditures correlation coefficient = .643 The correlation coefficient indicates a moderately strong linear relationship between

15-16 .

sales and promotional expenditures, thus linear regression model could probably be used.

y = 2.57% defects 42. y = 3.1968 + 0.0851x r = 0.45, which does not indicate a very strong relationship between hits and orders.

41. y = 13.476 − 0.4741x r = − 0.76

r2 = 0.198, which means only 19.8% of the variation in orders can be attributed to the number of web site hits.

r = 0.58 There seems to be a relatively strong relationship between production time and defects.

50,000 hits/month: y = 3.1968 + 0.0851(50)

Forecast for “normal” production time of 23 minutes:

y = 7.45 or 7,450 orders

y = 13.476 − 0.4741(23) 43. Month

Demand

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

8.20 7.50 8.10 9.30 9.10 9.50 10.40 9.70 10.20 10.60 8.20 9.90 10.30 10.50 11.70 9.80 10.80 11.30 12.60 11.50 10.80 11.70 12.50 12.80

Linear Trend Line Forecast

3-Month Moving Average

Adjusted Exponential Smoothing Forecast

— — — 7.93 8.30 8.83 9.30 9.67 9.87 10.10 10.17 9.67 9.57 9.47 10.23 10.83 10.67 10.77 10.63 11.57 11.80 11.63 11.33 11.67

8.20 8.20 7.99 8.02 8.40 8.61 8.88 9.34 9.44 9.67 9.95 9.42 9.57 9.79 10.00 10.51 10.30 10.45 10.70 11.27 11.34 11.18 11.33 11.68

8.24 8.42 8.59 8.77 8.95 9.13 9.31 9.49 9.67 9.84 10.02 10.20 10.38 10.56 10.74 10.92 11.09 11.27 11.45 11.63 11.81 11.99 12.17 12.34 12.52 Forecasting Alternatives

MAD

Linear Trend Line

.546

—

3-month moving average

.825

9.2

Adjusted Exponential smoothing

.817

9.47

15-17 .

306 = 0.22 1392 354 S2 = = 0.24 1392 404 S3 = = 0.29 1392 348 S4 = = 0.25 1392

All 3 of these methods appear to be relatively accurate. 44. a) y = 298.13 + 34.14x

S1 =

y(11) = 298.13 + 34.14(11) = 673.67 printers b) y = 74.77 + .34x y = 74.77 + (.34)(1,300)

Linear trend line forecast for year 6:

= 516.77 c)

y = 271.2 + 2.4x

MAD for the linear trend line forecast in (a) equals 60.25 while MAD for the linear regression forecast in (b) equals 85.09. In addition, the correlation coefficient for the linear trend line is r = 0.766 whereas the correlation coefficient for the linear regression is r = 0.599. This evidence seems to indicate the forecast model in (a) is best.

y(6) = 271.2 + 2.4(6) = 285.6 SF1 = (.22)(285.6) = 62.78 SF2 = (.26)(285.6) = 68.53 SF3 = (.29)(285.6) = 82.89 SF4 = (.25)(285.6) = 71.6 (b) Quarter 1: y = 52.69 + .3973x

45.

y(20) = 52.69 + .3973(20)

Year

Demand

Forecast

326

—

510

326.0

296

381.2

478

355.6

Quarter 3: y = 73.57 + .29x

305

392.3

y(25) = 73.57 + .29(25)

506

366.1

612

408.1

560

469.3

590

496.4

676

524.5

= 66.06 Quarter 2: y = 91.62 + .71x y(36) = 91.62 − (.71)(36) = 66.06

= 80.82 Quarter 4: y = 37.47 + 1.06x y(30) = 37.47 + 1.06(30) = 69.27 (c) This is an intuitive assessment, which managers must do on occasion. In general, the linear regression forecast provides a more conservative estimate.

569.9

The exponential smoothing model appears to be less accurate than the linear trend line forecast developed in 40(a).

47. The adjusted exponentially smoothed forecast (α = 0.4, β = 0.4) has a first quarter forecast for year 6 of 74.19 percent seat occupancy. It has a E (bias) value of 1.08 and a MAD value of 8.6, which seems low. Thus, this may be the best overall forecast model compared to the one developed in 46(a).

46. (a) Seasonally adjusted forecast. Quarter 1: D1 = 306 Quarter 2: D2 = 334 Quarter 3: D3 = 404 Quarter 4: D4 = 348

15-18 .

48. The following table shows several different forecast models developed using POM for Windows and selected measures of forecast accuracy. Forecast Method

Year 25 Forecast

MAD

Moving average (n = 3)

5.89

1.58

E (bias) −0.040

Linear trend line

8.22

1.86

0.000

Exponential smoothing (α = 0.3)

6.24

1.47

−0.330

Exponential smoothing (α = 0.5)

5.75

1.35

−0.410

Exponential smoothing (α = 0.3, β = 0.4)

6.24

1.33

−0.020

Exponential smoothing (α = 0.4, β = 0.5)

5.94 1.22 0.003 and thus would not seem to be particularly useful. In addition, while there is an obvious relationship between green episodes and sales (r = .756), the strength of this relationship is moderate (r2 = .571).

Although this selection of forecast models is not exhaustive, it does seem to indicate the exponential smoothing models are the most accurate, especially the adjusted model with α = 0.4 and β = 0.5. 49. Selected forecast models

51. The answer depends on the student’s approach to developing a solution. However, one approach would be to average the five estimates, which equals 40,200 units, then use this amount as the intercept for a linear trend line developed using the sales data in problem 50. This results in the following linear trend line equation:

5-day moving average forecasts for day 21: 11 − 12 = 80.8, MAD = 12.92 12 − 1 = 128.4, MAD = 28.32 1 − 2 = 93.0, MAD = 15.82 Exponentially smoothed (α = 0.3) forecasts for day 21:

y = 40,200 + 251.65x and to forecast for planning purposes in the 5th year, use 48 months (i.e., after 4 years) in the linear equation:

11 − 12 = 81.9, MAD = 12.92 12 − 1 = 129.7, MAD = 26.36

y(48) = 40,200 + 251.65x

1 − 2 = 99.61, MAD = 14.23

= 52,279

Linear trend line forecasts for day 21: 11 − 12 = 81.86, MAD = 11.25

52. y = 13.8399 + .00150x

12 − 1 = 132.42, MAD = 22.14

y(12,000) = 31.79

1 − 2 = 103.5, MAD = 12.44

r = 0.95 53. (a) y = 119.27 + 12.28x

The “best” forecast model depends on what models are selected for comparison. For the models tested above, they all seem to be relatively close, although the linear trend model consistently had the highest next period forecast and a slightly lower MAD value.

y(16) = 315.67 (b) y = −2,504.18 + .147x y(19,300) = 338.31 r2 = .966 y(20,000) = 444.41

50. The linear regression model is,

The club should use the linear regression model. The correlation coefficient shows that town population is a good predictor of the growth in the number of club players plus it provides a more favorable forecast for the club.

y = 28,923.02 + 3715.91x The forecast for “3” green episodes is, y(3) = 28,923.02 + 3715.91(3) = 40,070.74 units However, it is unclear how the company might use this forecast for planning purposes given that “green” episodes are unpredictable,

54. (a) y = 381.32 + 68.40x y(13) = 1,270.48

15-19 .

(c) y = 117.128 + .072(10,000) + 19.148(5) = $940.60 60. (a) y = 116.12 − 1.28x y(70) = 116.12 − 1.28(70) = 26 r2 = .537 (b) y = 116.9 − 1.24x1 + 0.14x2 y(70, 40) = 116.12 − 1.24(70.40) − 0.14(40) = 25 r2 = .538 Very little difference between the two forecasts. Annual budget appears to replicate endowment.

(b) r = .973 There appears to be a very strong linear relationship. 55. (a) Forecast of applicants: y = 13,803.07 + 470.55x y(11) = 18,979.12 applicants Forecast of % acceptances: y = 37.72 + .247x y(11) = 40.44% Estimated offers = 5,000/.4044 = 12,634 % offers = 12,364/18,979.12 = 65.15% (b) Forecast of % offers: y = 83 − 1.68x

CASE SOLUTION: FORECASTING AT STATE UNIVERSITY

y(11) = 64.54% (c) If the forecast of % acceptances is accurate then the number of applicants is not relevant; 12,634 offers will yield 5,000 acceptances.

Forecasting would be appropriate in a number of different areas. The university needs to be able to forecast future applications and enrollments both in the short and long term. A forecast of the college age population that will apply to State is very important for planning purposes. A multiple regression model that related applications to variables such as population, tuition levels, and entrance requirements would probably be most appropriate for this purpose.

56. (a) y = 43.09 + .0007x1 + 1.395x2 where y = SOL scores x1 = average teacher salary x2 = average tenure 2

(b) r = 0.696 Approximately 70% of the amount of variation in SOL scores can be attributed to teacher salaries and tenure. This is a moderately strong relationship indicating the superintendent is at least partially right. (c) y = 43.09 + .0007(30,000) + 1.397(9) = 76.65 No, the SOL score would only increase to 76.66. 57. (a) y = 745.91 − 2.226x1 + .163x2 (b) r2 = .992 (c) y = 7,186.91 58. (a) y = 608.795 + 0.215x1 − 0.985x2 (b) r2 = 0.766 (c) y = 608.795 + 0.215(1,500) − 0.985(300) = 636.65 59. (a) y = 117.128 + .072x1 + 19.148x2 (b) r2 = .822

Internal forecasts for classroom space, facilities, dormitory space, dining, etc. would enhance the planning process. Times series methods would probably be sufficient for this type of forecasting. The university might consider using a forecasting model to determine future funding from the state. Several models such as a multiple regression and perhaps a qualitative technique like the Delphi method might be combined. Forecasts for other sources of funding such as endowments and tuition increases could be forecast using more conventional methods such as regression or time series. The university’s TQM approach requires a forecast of what customers perceive educational quality to be in the future, i.e., a definition of quality according to students, parents, and legislators. Inhouse forecasting using key administrators, faculty, and students might be appropriate. Surveys and market research techniques of alumni, students, and parents might be useful in determining what quality factors will be important in the future.

15-20 .

CASE PROBLEM: THE UNIVERSITY BOOKSTORE STUDENT COMPUTER PURCHASE PROGRAM The following table shows several different forecast models developed using QM for Windows and selected measures of forecast accuracy. Forecast

Year 15 MAD

Method

E (bias)

Moving average (n = 3)

1,004.66

96.96

66.00

Linear trend line

1,020.07

73.24

0.00

Exponential smoothing (α = 0.3)

941.53

126.88

108.59

Exponential smoothing (α = 0.5)

1,066.11

109.49

46.96

Exponential smoothing (α = 0.3, β = 0.4)

983.22

109.58

62.19

Exponential smoothing (α = 0.4, β = 0.5)

1,031.09

105.13

61.31

bookstore could investigate which different majors and classes might be moving to more extensive computer usage in the future, thus driving up long run student demand. Additionally forecasts for other products would help the bookstore plan their inventory, warehouse usage, and distribution better.

Although this selection of different models is not exhaustive, it does seem to indicate that the linear trend line model is the best. Other forecast models that the bookstore might consider include forecasts of student enrollment and entering freshmen. Also for longer term forecasts the

CASE SOLUTION: VALLEY SWIM CLUB Attendance: Day M T W Th F Sa Su Total

1 139 273 172 275 337 402 487 2,085

2 198 310 347 393 421 595 497 2,761

3 341 291 380 367 359 463 578 2,779

4 287 247 356 322 419 516 478 2,625

5 303 223 315 258 193 378 461 2,131

Week 7 194 207 215 304 331 407 399 2,057

6 242 177 245 390 284 417 474 2,229

8 197 273 213 303 262 447 399 2,094

9 275 241 190 243 277 241 384 1,851

The seasonal factors for each weekday are as follows:

S6 (Saturday) =

S1 (Monday) =

3,139 = .110 28,539

S7 (Sunday) =

S2 (Tuesday) =

3,006 = .105 28,539

S3 (Wednesday) =

S4 (Thursday) = S5 (Friday) =

10 246 177 161 308 256 391 400 1,939

11 224 239 274 205 361 411 419 2,133

12 258 130 195 238 224 368 541 1,954

13 235 218 271 259 232 317 369 1,901

5,353 = .188 28,539

5,886 = .206 28,539

The linear trend line equation computed from the 13 weekly totals is, y = 2,598.2308 − 57.5604x

3,334 = .117 28,539

Using this forecast model to forecast weekly demand for each of the 13 weeks for the next summer and multiplying each weekly forecast by the daily seasonal factors will give the daily forecast for the next summer. For example, the daily forecast for week 1 is computed as,

3,865 = .135 28,539

3,956 = .139 28,539

y = 2,598.2308 − 57.5604(1) = 2,540.67

15-21 .

Total 3,139 3,006 3,334 3,865 3,956 5,353 5,886 28,539

Week 1 Forecasts

a) Seasonally adjusted forecast:

Monday = (.110)(2,540.67) = 279.5

4–6 AM: 4,926.50

Tuesday = (.105)(2,540.67) = 266.8

6–8 AM: 5,529.24

Wednesday = (.117)(2,540.67) = 297.3

8–10 AM: 5,783.28

Thursday = (.135)(2,540.67) = 343.0

10-noon: 3,247.80

Friday = (.139)(2,540.67) = 353.2

Noon-2 PM: 4,019.91

Saturday = (.188)(2,540.67) = 477.6

2–4 PM: 4,249.05

Sunday = (.206)(2,540.67) = 523.4

4–6 PM: 3,726.01

The remaining 12 weeks of daily forecasts would be developed similarly.

6–8 PM: 1,723.53 8–10 PM: 528.02

If the board of directors perceived that the pattern of weekly attendance totals would be closely followed next summer—i.e., low demand in the first week followed by high demand in weeks 2, 3 and 4 followed by gradually declining demand for the remaining 9 weeks—then a seasonally adjusted forecast could be used. That is, seasonal factors could be developed for all 13 weeks, and weekly forecasts could be computed by multiplying these weekly seasonal factors by the projected summer total attendance, rather than using the linear trend line forecast to compute forecasted weekly attendance.

b) Seasonally adjusted forecast: 4–6 AM: 4,455.06 6–8 AM: 5,000.12 8–10 AM: 5,229.85 10-noon: 2,937.01 Noon-2 PM: 3,635.22 2–4 PM: 3,842.43 4–6 PM: 3,369.45 6–8 PM: 1,558.6 8–10 PM: 477.59

CASE SOLUTION: FORECASTING AIRPORT PASSENGER ARRIVALS Seasonal Factors: 4–6 AM: 98,900/677,200 = .146 6–8 AM: 111,000/677,200 = .164 8–10 AM: 116,100/677,200 = .171 10–noon: 65,200/677,200 = .096 Noon–2 PM: 80,700/677,200 = .119 2–4 PM: 85,300/677,200 = .126 4–6 PM: 74,800/677,200 = .110 6–8 PM: 34,600/677,200 = .051 8–10 PM: 10,600/677,200 = .016 a) Linear trend line forecast for year 4 developed by averaging 10 sample days for each year (creating 3 data points): y = 11,413.3 + 5,580x y(4) = 33,733.3 b) Linear trend line forecast for year 4 developed using all 30 sample data points: y = 14,893 + 503.62x y(31) = 30,505.2

15-22 .

Chapter Sixteen: Inventory Management PROBLEM SUMMARY

38. Reorder point with variable demand and lead time

1. EOQ model

39. Quantity discount model (16–36)

2. EOQ cost analysis (16–1)

40. Service level

3. EOQ model

41. Service level (16–40)

4. EOQ model

42. Reorder point with variable lead time

5. EOQ model 6. Noninstantaneous receipt model

43. Safety stock and reorder point with variable demand and lead time

7. Shortage model

44. Reorder point (16–43) with variable demand

8. EOQ model and reorder point

45. Reorder point with variable demand

9. EOQ model and reorder point

46. Reorder point and service level with variable demand and lead time

10. Noninstantaneous receipt model

47. Reorder point (16–46) with variable demand

11. Noninstantaneous receipt model

48. Fixed period model, variable demand

12. EOQ model and reorder point

49. Fixed period model

13. EOQ model and reorder point (15–1)

50. Fixed period model

14. Noninstantaneous receipt model (15–2)

51. Reorder point with variable demand and lead time

15. Noninstantaneous receipt model (15–34) 16. Shortage model

PROBLEM SOLUTIONS

17. Noninstantaneous receipt model 18. Shortage model

19. Shortage model

D = 1,200 Co = $450

20. Shortage model

Cc = $170

21. Shortage model 22. EOQ model

TC = Co

23. Noninstantaneous receipt model 24. EOQ model

2Co D 2(450)(1,200) = = 79.7 Cc 170

25. EOQ model

D Q ⎛ 1,200 ⎞ + Cc = 450 ⎜ ⎟ + 170 2 Q ⎝ 79.7 ⎠

⎛ 79.7 ⎞ ⎜ 2 ⎟ = $13,550 ⎝ ⎠

26. “Practical” problem 27. EOQ model analysis 28. Noninstantaneous receipt model analysis

D 1,200 = = 15.05 orders Q 79.7

364 = 24.18 days 15.05

29. Carrying cost determination 30. Quantity discount model 31. Quantity discount model 32. Quantity discount model

2. Cases

33. Quantity discount model (16–32)

34. Quantity discount model

79.7

$12,195

35. Quantity discount model (16–34)

79.7

14,905

36. Quantity discount model

88.1

13,482

37. Quantity discount problem

72.1

13,482

16-1 .

D = 19,200

= 190,918.8 lbs.

Cc = $20 a)

2Co D = Cc

TC =

Co D Q ⎛ 19,200 ⎞ + Cc = 30 ⎜ ⎟ Q 2 ⎝ 240 ⎠

2(30)(19,200) = 240 20

b) TC = Co

D 19,200 = = 80 orders 240 Q

320 = 4 days 80

d) 4.

D 1,215,000 = = 6.36 orders Q 190,918.8

365 = 57.4 days 6.36

D = 5,000 d = 19.23 per day

D = 35,000

p = 64 per day

Co = $500

Cc = $0.35

Cc = $5

2Co D Q= Cc =

a) Q =

2(500)(35,000) .35

= 10,000 yards TC = Cc

10,000 ⎛ 35,000 ⎞ + (500) ⎜ ⎟ 2 ⎝ 10,000 ⎠

D 5,000 = = 4.18 production runs Q 1,195.6

260 = 62.2 working days 4.18

Q 1,195.6 = = 18.68 working days p 64

= $3,500 D Q 35,000 = 10,000

Number of orders =

= 3.5 per year Time between orders =

D Q⎛ d⎞ + Cc ⎜ 1 − ⎟ Q 2⎝ p⎠

⎛ 5,000 ⎞ ⎛ 1,195.6 ⎞ ⎛ 19.23 ⎞ = 500 ⎜ ⎟ ⎜ 1 − 64 ⎟ ⎟ + 5⎜ ⎠ ⎝ 1,195.6 ⎠ ⎝ 2 ⎠ ⎝ = $4,181.86

= 1,750 + 1,750

365 3.5

= 104.3 days 5.

2Co D 2(5000)(500) = = 1,195.6 ⎛ d⎞ ⎛ 19.23 ⎞ 5 ⎜1 − Cc ⎜ 1 − ⎟ ⎟ 64 ⎠ p⎠ ⎝ ⎝

b) TC = Co

Q D + Co 2 Q

= (.35)

D Q + Cc 2 Q

⎛ 1,215,000 ⎞ ⎛ 190,918.8 ⎞ = 1,200 ⎜ + .08 ⎜ ⎟ ⎟ 2 ⎝ ⎠ ⎝ 190,918.8 ⎠ = $15,273.51

⎛ 240 ⎞ + 20 ⎜ ⎟ = $4,800 ⎝ 2 ⎠

2Co D (1,200)(1,215,000)(2) = Cc .08

a) Q =

Co = $30

D = 1,215,000 Co = $1,200 Cc = $.08

16-2 .

D = (25)(52) = 1,300

10. d = 180 lbs./day

Cc = ($300)(.25) = $75

p = 250 lbs./day

Co = $100

D = 65,700

Cs = $250

Co = $125

Cc = $12

2Co D ⎛ Cs + Cc ⎞ ⎜ ⎟ Cc ⎝ C s ⎠

2Co D ⎛ d⎞ Cc ⎜ 1 − ⎟ p⎠ ⎝

2(100)(1,300) ⎛ 250 + 75 ⎞ ⎜ 250 ⎟ = 67.13 75 ⎝ ⎠

⎛ Cc ⎞ ⎛ 75 ⎞ S = Q⎜ ⎟ = 67.13 ⎜ ⎟ = 15.49 ⎝ 75 + 250 ⎠ ⎝ Cc + C s ⎠

2(125)(65,700) ⎛ 180 ⎞ 12 ⎜ 1 − ⎟ ⎝ 250 ⎠

S 2 Cc ( Q − S ) D TC = Cs + + Co 2Q 2Q Q 2

250 (15.49 )

2 ( 67.13 ) = $3,872.98

75 ( 67.13 − 15.49 ) 2 ( 67.13 )

= 2,211 2

⎛ 1,300 ⎞ + 100 ⎜ ⎟ ⎝ 67.13 ⎠

TC = Co

⎛ 65,700 ⎞ ⎛ 2,211 ⎞ = (125) ⎜ + (12) ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 2,211 ⎠

D = 2,000 Co = $2,600

⎛ 180 ⎞ * ⎜1 − ⎟ ⎝ 250 ⎠

Cc = $50 L = 10

= 3,714 + 3,714

2Co D 2(2,600)(2,000) Q= = = 456.07 gallons Cc 50

= $7,428 11. d = 1,500

D Q ⎛ 2,000 ⎞ + Cc = 2,600 ⎜ TC = Co ⎟ 2 Q ⎝ 456.07 ⎠

p = 2,000 D = 547,500

⎛ 456.07 ⎞ + 50 ⎜ ⎟ = $22,803.51 ⎝ 2 ⎠

Co = $6,500 Cc = $50

⎛ 2,000 ⎞ R = dL = ⎜ ⎟ (10 ) = 64.5 gallons ⎝ 310 ⎠

D = 4,000 Co = $60 =

Cc = $0.80 L=4 Q=

D Q⎛ d ⎞ + Cc ⎜ 1 − ⎟ 2⎝ Q p⎠

2(60) ( 4,000 ) 2Co D = = 774.6 boxes Cc .80

2Co D ⎛ d⎞ Cc ⎜ 1 − ⎟ p⎠ ⎝ 2(6,500)(547,500) ⎛ 1,500 ⎞ 50 ⎜ 1 − ⎟ ⎝ 2,000 ⎠

= 23,862 TC = Co

D Q ⎛ 4,000 ⎞ TC = Co + Cc = 60 ⎜ ⎝ 774.6 ⎟⎠ 2 Q

D Q⎛ d ⎞ + Cc ⎜ 1 − ⎟ Q 2⎝ p⎠

⎛ 547,500 ⎞ ⎛ 23,862 ⎞ = (6,500) ⎜ ⎟ + (50) ⎜ 2 ⎟ 23,862 ⎝ ⎠ ⎝ ⎠ 1,500 ⎛ ⎞ * ⎜1 − ⎟ ⎝ 2,000 ⎠

⎛ 774.6 ⎞ + .80 ⎜ ⎟ = $619.68 ⎝ 2 ⎠ ⎛ 4,000 ⎞ R = dL = ⎜ ⎟ (4) = 43.84 boxes ⎝ 365 ⎠

≅ 149,138 + 149,138

= $298,276

16-3 .

12.

Operates 360 days/year 12 converters

From problem 15-34, annual demand = 29,330 units

5 tons coal/day/converter

Co = $6,500

D = (5 tons)(12 converters)(360 days)

Cc = $115.75/unit

15.

p = 200 units/day

= 21,600 tons/year Co = $80

d = 80.36 units/day

Cc = 20% of average inventory level

Q = 2,346.59

Cc = (.20)($12) = $2.40

TC = $162,484.30

a) Q =

2Co D = Cc

Maximum inventory level = 2,346.03

2(80)(21,600) 2.4

R = 2,008.90 16.

1, 440,000 = 1,200 tons

b) TC = Cc

D = 715 Co = $6,000 Cc = $265

Q D + Co 2 Q

Cs = $14,000

⎛ 1,200 ⎞ ⎛ (80)(21,600) ⎞ = (2.4) ⎜ ⎟+⎜ ⎟ 1,200 ⎝ 2 ⎠ ⎝ ⎠

= 1,440 + 1,440 =

= $2,880 c) 13.

R=L

5(21,600) D = = 300 tons 360 360

2 ( 6000 )( 715 ) ⎛ 14000 + 265 ⎞ ⎜ 14000 ⎟ 265 ⎝ ⎠

= 181.6 ⎛ CC ⎞ S = Q⎜ ⎟ ⎝ CC + CS ⎠

From problem 15-1, annual demand = (13.33 motorcycles/month)(12) = 160 motorcycles/year

265 ⎛ ⎞ = 181.6 ⎜ ⎟ ⎝ 265 + 14000 ⎠

Co = $3200 Cc = $375

= 3.37

Q = 52.3 motorcycles

TC =

TC = $19,595.92 Orders = 3.06

CS S 2 CC (Q − S ) 2 Co D + + Q 2Q 2Q

(14000)(3.37)2 (265)(181.6 − 3.37) 2 715 + + (6000) 2(181.6) 2(181.6) 181.6 = $47,238.35 =

Time between orders = 119.2 days R = 13.33 14.

2Co D ⎛ CS + CC ⎞ ⎜ ⎟ Co ⎝ C S ⎠

From problem 15-2, annual demand = (11,000 yds/month)(12) = 132,000 yds/year

17.

D = 10,000 logs/year

Co = $425

T = 250 days/year

Cc = $0.63/yd

R = 60(250) = 15,000/year

p = 1200 yds/day

Co = $1,600

d = 132,000/360 = 366.67 yds/day

Cc = $15

Q = 16,014.31 yards a) Q =

TC = $7,006.23 R = 2,566.67 yards

2 (1,600 )(10,000 ) 2Co D = D 10,000 ⎞ ⎛ ⎞ Cc ⎜ 1 − ⎟ (15 ) ⎛⎜ 1 − ⎟ R⎠ ⎝ ⎝ 15,000 ⎠

= 6, 400, 000 = 2,529.8 logs

16-4 .

TC = Co

19. D = 270,000

D Q⎛ D⎞ + Cc ⎜ 1 − ⎟ Q 2⎝ R⎠

Co = $105 Cc = $.25

⎛ 10, 000 ⎞ = (1, 600) ⎜ ⎟ ⎝ 2,529.8 ⎠

Cs = $.70

⎡ ⎛ 2,529.8 ⎞ ⎛ 10, 000 ⎞ ⎤ + ⎢ (15) ⎜ ⎟ ⎜1 − ⎟⎥ ⎝ 2 ⎠ ⎝ 15, 000 ⎠ ⎦ ⎣

= 6,324.5 + 6,324.5 D 10,000 = = 3.95 orders/year Q 2,529.8 T 250 Tb = = = 63.3 days N 3.59

= 17,544

⎛ Cc ⎞ S = Q⎜ ⎟ ⎝ Cc + C s ⎠ ⎛ .25 ⎞ = (17,544) ⎜ ⎟ ⎝ .25 + .70 ⎠ = 4,616.84

d) Q = 2,529.8, R = 60

The number of operating days to receive the entire order is

TC =

18. D = 3,700

Co = $420 CS = $4 2Co D ⎛ Cs + Cc ⎞ ⎜ ⎟ Cc ⎝ C s ⎠

= $3,231.84

20. D = 400

2(420)(3, 700) ⎛ 1.75 + 4 ⎞ ⎜ ⎟ 1.75 ⎝ 4 ⎠

Co = $650 Cc = $45

= 1,598 tires

Cs = $60

⎛ Cc ⎞ S = Q⎜ ⎟ ⎝ Cc + C s ⎠ ⎛ 1.75 ⎞ = (1,598) ⎜ ⎟ ⎝ 1.75 + 4 ⎠

Q= =

= 486.3 C S 2 C (Q − S )2 D TC = s + c + Co 2Q 2Q Q (4)(486.3)2 (1.75) (1,598 − 486.3 ) = + 2(1,598) 2(1,598)

(.70)(4,562.1)2 (.25)(17,544 − 4,562.1)2 + 2(17,544) 2(17,544)

⎛ 270,000 ⎞ + (105) ⎜ ⎟ ⎝ 17,544 ⎠ = 425.24 + 1,190.66 + 1,615.94

CC = $1.75

C s S 2 Cc ( Q − S ) D + + Co 2Q 2Q Q 2

Q 2,529.8 = = 42.16 days R 60

2 (105 )( 270,000 ) ⎛ .70 + .25 ⎞ ⎜ .70 ⎟ .25 ⎝ ⎠

= $12,649 c)

2Co D ⎛ Cs + Cc ⎞ ⎜ ⎟ Cc ⎝ C s ⎠

2Co D ⎛ Cs + Cs ⎞ ⎜ ⎟ Cc ⎝ C s ⎠ 2(650)(400) ⎛ 60 + 45 ⎞ ⎜ 60 ⎟ = 142.21 sets 45 ⎝ ⎠

⎛ Cc ⎞ ⎛ 45 ⎞ S = Q⎜ ⎟ = 142.21⎜ ⎟ + C C ⎝ 45 + 60 ⎠ s ⎠ ⎝ c

= 60.95 sets

⎛ 3,700 ⎞ + (420) ⎜ ⎟ ⎝ 1,598 ⎠

TC =

Cs S 2 Cc (Q − S )2 Co D 60(60.95)2 + + = 2Q 2Q Q 2(142.21) +

= 295.98 + 676.72 + 972.47

45(142.21 − 60.95)2 ⎛ 400 ⎞ + 650 ⎜ ⎟ 2(142.21) ⎝ 142.21 ⎠

= $3,656.70

= $1,945.17

16-5 .

21. D = 1,200/yr.

23. Co = $1,700

Co = $350

Cc = $1.25

Cc = $2.75/book

D = 18,000/yr.

Cs = $45/book

P = 30,000/yr.

2(350)(1,200) ⎛ 45 + 2.75 ⎞ ⎜ 2.75 45 ⎟⎠ ⎝

= 569.32

= 11,062.62

⎛ 2.75 ⎞ S = 569.32 ⎜ ⎟ ⎝ 45 + 2.75 ⎠

(1,700)(18,000) ⎛ 11,062 ⎞ ⎛ 1 − 18,000 ⎞ + 1.25 ⎜ ⎟⎜ ⎟ 11,062.62 ⎝ 2 ⎠ ⎝ 30,000 ⎠ = $5,532.14

TC =

= 32.79 TC = 45

2(1,700)(18,000) ⎛ 18,000 ⎞ 1.25 ⎜ 1 − ⎟ ⎝ 30,000 ⎠

(32.79)2 (569.32 − 32.79)2 + 2.75 2(569.32) 2(569.32)

Number of production runs =

⎛ 1,200 ⎞ + (350) ⎜ ⎟ ⎝ 569.32 ⎠ = $1,475.45

⎛ d⎞ Maximum inventory level = Q ⎜ 1 − ⎟ p⎠ ⎝

22. D = 200/day

⎛ 18,000 ⎞ = 11,062.62 ⎜ 1 − ⎟ ⎝ 30,000 ⎠ = 4,425.7

Co = $20 CC = $0.20/min. = $120/day a)

2(20)(200) = 8.16 orders or 8 120

2,500 = Q(.4)

200 = 25deliveries per day 8 10 = 0.4 hour = every 24 minutes a delivery 25 truck goes out to deliver orders.

Q = 6,250 ⎛ (1,700)(18,000) ⎞ ⎛ 6,250 ⎞ + 1.25 ⎜ TC = ⎜ ⎟ (1 − 60) ⎟ 6,250 ⎝ 2 ⎠ ⎝ ⎠ = $6,458.50

(20)(200) (120)(8) + 8 2 = 500 + 480 = $980

24. D = 160,000

TC =

Co = $7,000 Cc = $0.80/ft.3

Q=6

2(7,000)(160,000) 0.80 = 52,915 ft.3

200 = 33.33 or 33 deliveries per day. 6 10 = .30 hr. = every 18 minutes a delivery truck 33 is sent out. TC =

Maximum inventory level = 2,500 ⎛ 18,000 ⎞ 2,500 = Q ⎜ 1 − ⎝ 30,000 ⎟⎠

The truck should carry approximately 8 orders each time it makes deliveries.

18,000 D = = 1.63 Q 11,062.62

(7,000)(160,000) 0.80(52,915) + 52,915 2 = $42,332

TC =

(20)(200) (120)(6) + 6 2

160,000 52,915 = 3.02

Number of orders =

= 666.67 + 360 = $1,026.67

16-6 .

25. Q = 661,800.9 bales TC = $10,158,664 Shipments = 1.88 Time between shipments = 194 days.

b) Co = $1,900

Cc = $4.50

26. This is more of a “practical” problem than an EOQ model, and thus requires some logic.

The annual shipping cost is fixed at $936,000 – 48,000 tons (rounded from problem 15-32) will be shipped during the year at $19.50 per ton, regardless of how many shipments there are and the amount of each shipment.

TC =

Co D Cc Q (1,900)(17, 400) + = Q 2 3,833.19

Select the new location. 28.

a) D = 270,000

CC = $0.12/lb. Co = $620 P = 305,000 Q=

2Co D 2(620)(270,000) = D ⎛ ⎞ ⎛ 270 ⎞ Cc ⎜ 1 − ⎟ (0.12) ⎜ 1 − ⎟ P⎠ ⎝ ⎝ 305 ⎠

= 155,925.81 ⎛ D⎞ maximum inventory level = Q ⎜ 1 − ⎟ P⎠ ⎝ = (155,925.81)(.1148 )

The average inventory on hand during a day will be 1,315 tons (the 10-day supply) plus 920 tons (the 7-day supply) divided by two, or approximately 1,120 tons per day. The handling cost per ton per day is $47 per ton divided by 365, or approximately $0.13/ton/day. Since an average of 1,120 tons will be on hand each day, the daily handing cost will be approximately $145 (1,120 tons times $0.13/ton), and thus, the total annual handling cost at the plant will be $145 multiplied by 365, or $52,925 per year.

= 17,900.3 Co D Cc Q ⎛ D ⎞ (620)(270,000) + 1− ⎟ = TC = 2 ⎜⎝ 155,925.81 Q P⎠ (0.12)(155,925.81) + (.1148) 2 = $2,147.60

b) P = 360,000 2(620)(270,000) = 105,640.90 ⎛ 270 ⎞ (0.12) ⎜ 1 − ⎟ ⎝ 306 ⎠ (620)(270,000) (0.12)(105,640.90) + TC = (0.25) 105,640.90 2 = $3,169.23 Q=

Thus the total annual inventory cost will be the shipping cost of $936,000 plus the handling cost of $52,935 or $988,925. 27. a) D = 17,400 CC = $3.75

No, should not increase production

Co = $2,600

TC =

2Co D 2(1,900)(17, 400) = = 3,833.19 Cc 4.50

= $17,249.36

If a 7- to 10-day supply is kept on hand then the relevant carrying cost is the cost/day/ton, multiplied by the average daily inventory, multiplied by 365 days. Using a forecasted annual coal consumption of approximately 48,000 tons (from problem 32 in chapter 15), the average coal consumption per day is 131.5 tons. A 10-day supply is 1,315 tons. If it takes 3 days to receive a shipment then the usage during the 3 days is 395 tons and at the end of the 3 days there are 920 tons available, which is a 7-day supply. As such, a shipment of 395 tons will reestablish the 10-day supply.

2Co D 2(2,600)(17, 400) = = 4,912.03 3.75 Cc

29.

D = 900 Co = $7,600

Co D Cc Q (2,600)(17, 400) + = Q 2 4,912.03 (3.75)(4,912.03) + 2 = $18,420.11

Cc = ? Q = 120 Q=

16-7 .

2Co D Cc

2(7, 600)(900) Cc

120 =

TC = Co

⎛ 900 ⎞ ⎛ 108.2 ⎞ = (160) ⎜ ⎟ + (.12)(205) ⎜ 2 ⎟ ⎝ 108.2 ⎠ ⎝ ⎠

2(7, 600)(900) Cc

(120 ) = 2

+ (205)(900)

CC = $950 30.

D = 10,000

= 1,330.86 + 1,330.86 + 184,500

Co = $150 CC = $0.75

= $187,161.72 With discount:

Order Size

0–4,999

$8.00

5,000 +

$6.50

Q = 300 P = 190 TC = Co

Without discount: Q=

2Co D 2(150)(10,000) = = 2,000 0.75 Cc

TC =

Co D Q ⎛ 10,000 ⎞ + Co + PD = 150 ⎜ ⎟ 2 Q ⎝ 2,000 ⎠

Take the discount, Q = 300 32.

D = 1,700 Co = 120 Cc = (.25)($38) = $9.50 Q=

Q = 5,000 TC = Co

D Q + Cc P + PD Q 2

⎛ 900 ⎞ ⎛ 300 ⎞ = (160) ⎜ ⎟ + (.12)(190) ⎜ ⎟ + (190)(900) 300 ⎝ ⎠ ⎝ 2 ⎠ = $174,900

⎛ 2,000 ⎞ + .75 ⎜ ⎟ + (10,000)(8) = $81,500 ⎝ 2 ⎠ With discount: D Q ⎛ 10,000 ⎞ + Cc + PD = (150) ⎜ ⎟ Q 2 ⎝ 5,000 ⎠

2Co D 2(120)(1,700) = = 207 Cc P (9.5)

⎛ 1,700 ⎞ ⎛ 207 ⎞ TC = 120 ⎜ + (9.50) ⎜ ⎟ ⎟ + (38) ⎝ 207 ⎠ ⎝ 2 ⎠

⎛ 5,000 ⎞ + .75 ⎜ ⎟ + 10,000(6.50) ⎝ 2 ⎠ = $67,175

(1,700) = $66,568.76 Q = 300: ⎛ 1,700 ⎞ ⎛ 300 ⎞ TC = 120 ⎜ ⎟ + 9.31⎜ 2 ⎟ 300 ⎝ ⎠ ⎝ ⎠

Select discount, Q = 5,000 31.

D Q + Co P + PD Q 2

D = 900

+ 37.24 (1,700) = $65,384.80

P = $205 CC = 12% (P)

Q = 500:

Co = $160

⎛ 1,700 ⎞ ⎛ 500 ⎞ TC = 120 ⎜ + 9.12 ⎜ ⎟ ⎟ ⎝ 500 ⎠ ⎝ 2 ⎠

Without discount: Q=

+ 36.48 (1,700) = $64,704

2Co D Cc P

Q = 800: ⎛ 1,700 ⎞ ⎛ 800 ⎞ TC = 120 ⎜ + 9.025 ⎜ ⎟ ⎟ ⎝ 800 ⎠ ⎝ 2 ⎠

⎛ 2(160)(900) ⎞ = ⎜ ⎝ (.12)(205) ⎟⎠

+ 36.10 (1,700) = $65,235

= 108.2

Select Q = 500, TC = $64,704

16-8 .

Select Q = 6,000, TC = $87,030.33

2(120)(1,700) = 226 8 ⎛ 1,700 ⎞ ⎛ 226 ⎞ TC = 120 ⎜ ⎟ + 8⎜ ⎟ ⎝ 226 ⎠ ⎝ 2 ⎠

33. Q =

2(28)(6,500) = 337.26 ≈ 337 boxes 3.20 ⎛ 6,500 ⎞ ⎛ 337 ⎞ TC = 28 ⎜ ⎟ + 3⎜ ⎟ ⎝ 337 ⎠ ⎝ 2 ⎠

35. Q =

+ 38 (1,700) = $66,406.65

+ 16 (6,500) = $105,079.20

Q = 300:

Q = 1,000:

⎛ 1,700 ⎞ ⎛ 300 ⎞ TC = 120 ⎜ ⎟ + 8⎜ ⎟ ⎝ 300 ⎠ ⎝ 2 ⎠

⎛ 6,500 ⎞ ⎛ 1,000 ⎞ + 2.80 ⎜ TC = 28 ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 1,000 ⎠

+ 37.24 (1,700) = $65,188

Q = 500:

+ 14 (6,500) = $92,582

⎛ 1,700 ⎞ ⎛ 500 ⎞ TC = 120 ⎜ ⎟ + 8⎜ ⎟ ⎝ 500 ⎠ ⎝ 2 ⎠

Q = 3,000: ⎛ 6,500 ⎞ ⎛ 3,000 ⎞ TC = 28 ⎜ ⎟ + 2.60 ⎜ 2 ⎟ ⎝ ⎠ ⎝ 3,000 ⎠

+ 36.48 (1,700) = $64,424

Q = 800:

+ 13 (6,500) = $88,460.67

⎛ 1,700 ⎞ ⎛ 800 ⎞ TC = 120 ⎜ ⎟ + 8⎜ ⎟ ⎝ 800 ⎠ ⎝ 2 ⎠

Q = 6,000: ⎛ 6,500 ⎞ ⎛ 6,000 ⎞ TC = 28 ⎜ ⎟ + 2.40 ⎜ 2 ⎟ 6,000 ⎝ ⎠ ⎝ ⎠

+ 36.10 (1,700) = $64,825

Select Q = 500, TC = $64,424

+ 12 (6,500) = $85,230.33

34. D = 6,500

Select Q = 6,000, TC = $85,230.33

Co = $28

36.

Cc = $3

D = 2,300,000/100 = 23,000 boxes Co = $320

2Co D 2(28)(6,500) Q= = = 348.32 ≈ 348 Cc 3

Cc = $1.90

⎛ 6,500 ⎞ ⎛ 348 ⎞ TC = 28 ⎜ ⎟ + 3⎜ ⎟ ⎝ 348 ⎠ ⎝ 2 ⎠

2Co D 2(320)(23,000) = = 2,783.4 Cc 1.90

≈ 2,784 boxes

+ 16 (6,500) = $105,045

⎛ 23,000 ⎞ ⎛ 2,784 ⎞ TC = 320 ⎜ ⎟ + 1.9 ⎜ 2 ⎟ 2,784 ⎝ ⎠ ⎝ ⎠

Q = 1,000: ⎛ 6,500 ⎞ ⎛ 1,000 ⎞ TC = 28 ⎜ ⎟ ⎟ + 3⎜ ⎝ 1,000 ⎠ ⎝ 2 ⎠

+ 47 (23,000) = $1,086,288.5

Q = 7,000:

+ 14 (6,500) = $92,682

⎛ 23,000 ⎞ ⎛ 7,000 ⎞ TC = 320 ⎜ + 1.9 ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 7,000 ⎠

Q = 3,000: ⎛ 6,500 ⎞ ⎛ 3,000 ⎞ TC = 28 ⎜ ⎟ ⎟ + 3⎜ ⎝ 3,000 ⎠ ⎝ 2 ⎠

+ 43 (23,000) = $996,701.43

Q = 12,000:

+13 (6,500) = $89,060.67

⎛ 23,000 ⎞ ⎛ 12,000 ⎞ + 1.9 ⎜ TC = 320 ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 12,000 ⎠

Q = 6,000: ⎛ 6,500 ⎞ ⎛ 6,000 ⎞ TC = 28 ⎜ ⎟ ⎟ + 3⎜ ⎝ 6,000 ⎠ ⎝ 2 ⎠

+ 41 (23,000) = $955,013.33

+ 12 (6,500) = $87,030.33

16-9 .

+ 43 (23,000) = $997,576.43

Q = 20,000:

Q = 12,000:

⎛ 23,000 ⎞ ⎛ 20,000 ⎞ + 1.9 ⎜ TC = 320 ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 20,000 ⎠

⎛ 23,000 ⎞ ⎛ 12,000 ⎞ TC = 320 ⎜ ⎟ + 2.05 ⎜ 2 ⎟ 12,000 ⎝ ⎠ ⎝ ⎠ + 47 (23,000) = $955,913.33 Q = 20,000:

+ 38 (23,000) = $893,368

Select Q = 20,000 boxes, TC = $893,368 37. D = 40,000

⎛ 23,000 ⎞ ⎛ 20,000 ⎞ TC = 320 ⎜ ⎟ + 1.90 ⎜ 2 ⎟ 20,000 ⎝ ⎠ ⎝ ⎠

Co = $180 Cc = $0.18

+ 38 (23,000) = $893,368

Without discount: Q=

2(180)(40,000) = 8,944.3 .18

TC =

Co D Q + Cc + PD 2 Q

Select Q = 20,000 boxes, TC = $893,368 40. d = 3, 000 L=6

σ d = 600 R = dL + Z σ d L

(180)(40,000) (.18)(8,944.3) + 8,944.3 2

R = 3, 000(6) + 1.64(600) 6 = 20, 410.3

+ (.34)(40,000)

Safety stock = 2,410 yards

= 804.98 + 804.98 + 13,600

41. If safety stock = 2,000

= 15,209.96

Z (600)( 6) = 2,000

With discount of Q = 30,000:

Z = 1.3608, which corresponds to a 91% service level

TC = $14,140 Take discount for Q = 30,000

42. d = 8,000

38. d = 2 packages/day

L= 7

σd = 0.8 packages/day

σ L = 1.6

L = 2 days

R = dL + Zdσ L

σL = 0.5 days

= 8,000(7) + 2.06 (8,000)(1.6)

Z = 2.33

R = 82,368 lbs.

R = dL + Z σ L + σ d 2 d

2 L

43. d = 18

σd = 4

= (2)(2) + 2.33 (.8)2 2 + (.5)2 (2)2

L =3 σ L = .8

= 7.52 packages of paper 2(320)(23,000) = 2,502.76 2.35 ≈ 2,503 boxes

39. Q =

R = dL + Z σ d2 L + σ L2 d

R = (18)(3) + 1.29 (4)2 (3) + (.8)2 (18)2

⎛ 23,000 ⎞ ⎛ 2,503 ⎞ TC = 320 ⎜ ⎟ + 2.35 ⎜ 2 ⎟ 2,503 ⎝ ⎠ ⎝ ⎠ + 47 (23,000) = $1,086,881.50

Safety stock = 20.61 gallons

Q = 7,000:

R = (18)(3) + 1.65 (4)2 (3) + (.8)2 (18)2

= 74.61 gallons

⎛ 23,000 ⎞ ⎛ 7,000 ⎞ TC = 320 ⎜ + 2.15 ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 7,000 ⎠

= 80.37 gallons Safety stock would increase to 26.37 gallons

16-10 .

50. d = 18 tb = 30

44. R = dL + Zσ d L

R = 18(3) + 1.28(4)( 3) = 62.94

L=2 σd = 4

R drops to 62.94 gallons; less safety stock is necessary

I = 25

45. R = dL + Zσ d L

Q = d (t b + L ) + Z σ d t b + L − I

R = 26(9) + .68(10)( 9) = 254.4 gallons for a 95% service level,

Q = 18(30 + 2) + 2.05(4) 30 + 2 − 25

= 597.4 bottles

Safety stock = Zσ d L = 1.65(10)( 9) = 49.5 The reorder point increases to 283.5 gallons

46. R = dL + Z σ L + σ d 2 d

2 L

51. d = 6/hr.

σ d = 2.5/hr.

L = 0.5/hr.

R = 2.5(25) + 1.28 (1.2)2 (25) + (10)2 (2.5)2 = 95.67 monitors Safety stock = 33.17 monitors

σ L = .133 hr. Z=?

47. R = dL + Zσ d L

R = 2.5(8) + 1.28(1.2)( 8) = 24.38

R = dL + Z σ d2 L + σ L2 d

1 = (6)(0.5) + Z (2.5)2 (0.5) + (.133)2 (6)2 1 = 3 + Z(1.94) –2 = Z(1.94) Z = −1.03 Service level = .5000 − .3485 = .1515 = 15.15% b) R = 3 + (2.05)(1.94) R = 6.977 pizzas

Decisions would be based on inventory holding cost, desire for low inventory, importance of reliable delivery, cost of the monitors from each source, etc. 48. d = 200 tb = 30

L=4 σ d = 80

CASE SOLUTION: THE NORTHWOODS GENERAL STORE

I = 60 Q = d (t b + L ) + Z σ d t b + L − I Q = 200(30 + 4) + 1.65(80) 30 + 4 − 60 = 7,509.69 oz.

This case requires a combination of classic EOQ analysis and common sense. Assuming that the demand of 7,500 gallons per year is constant, then monthly demand is 675 gallons. During the 4-month maple syrup season the total demand is 2,500 gallons which leaves demand of 5,000 gallons for the remaining 8 months of the year.

49. d = 8 tb = 10

L=3 σ d = 2.5

First, determine an optimal order quantity for the maple syrup season:

I=0 Q = d (t b + L ) + Z σ d t b + L − I

Q = 8(10 + 3) + 2.33(2.5) 10 + 3 − 0

= 125 pizzas 125 − 5 = 120 pizzas

2(450)(2,500) ≅ 435 gallons 15(1 − .205)

where d/p = 7,500/(365)(100) = .205

16-11 .

Q or 4.35 days and p 65.75 runs are needed to meet demand during the 4-month period. This also means a run needs to start about every 21 days. Using this information the following schedule can be developed:

The length of a production run is

CASE SOLUTION: THE TEXANS STADIUM STORE The objective of this case problem is to determine the reorder point with variable demand and lead time. The first step is to complete the average demand and standard deviation, and average lead time and standard deviation from the data provided in the problem. This is a good opportunity to allow students to use a statistical software package (if they have access to one) to compute these statistics.

February 1 – start run February 5 – end run February 26 – start run March 2 – end run

d = 42.57 caps per week

March 23 – start run March 27 – end run

σ d = 10.41 caps per week

April 17 – start run

L = 18.50 days

April 21 – end run

σ L = 4.67 days

May 12 – start run

Note that since demand is in terms of caps per week, lead time must be changed to weeks also,

May 16 – end run

L = 2.64 weeks

This schedule, producing approximately 435 gallons each run, will meet demand in the 4-month season.

σ L = .67 weeks

Next we will assume that for the remaining 5,000 gallons, it is desired to produce them as close to the end of the season as possible in order to minimize storage costs. Working from May 31, backwards, the following schedule can be developed producing the maximum 100 gallons per day until 5,000 gallons are produced.

The first question is, if R = 150, what level of service does this correspond to. Thus, we are seeking Z as follows, 150 = dL + Z σ d2 L + σ L2 d 2

150 = (42.57)(2.64) + Z (10.41)2 (2.64) + (.67)2 (42.57)2

May 21 to 31, produce 1,000 gallons May 17 to 21, a normal production run so only 65 extra gallons (500 − 435) are produced.

150 = 112.38 + Z(33.16) Z = 1.13 This Z value corresponds to a normal probability value of .7416, thus, the service level is approximately 74.2 percent. The desired service level is 99 percent (Z = 2.58). The reorder point and safety stock for this service level is determined as follows.

May 1 to May 17, produce 1,700 gallons April 27 to May 1, a normal production run so only 65 extra gallons are produced. April 10 to 27, produce 1,700 gallons. April 6 to 10, a normal production run so only 65 extra gallons are produced.

R = dL + Z σ L2 L + σ L2 d 2

April 3 to April 6, produce 400 gallons.

R = (42.57)(2.64) + 2.58

The total production from April 3 to May 31 results in approximately the extra 5,000 gallons plus the normal production to meet demand during that period. Therefore, the store should start producing syrup on a full time basis at about the first of the new year.

(10.41)2 (2.64) + (.67)2 (42.57)2

R = 112.38 + 85.55 R = 197.9 or 198 caps Ms. Jones could determine the order size with EOQ analysis by using the average demand, d, as D in the EOQ formula. However, she would also need the ordering and carrying costs. It is likely that the ordering cost is relatively high as compared to carrying cost since the caps are shipped from Jamaica while it would probably not be very expensive to store caps (given their small size and weight).

16-12 .

CASE SOLUTION: THE A-TO-Z OFFICE SUPPLY COMPANY

D 5,185,000 = = 13.944 loans/year Q 371,842 ≅ 14 loans / year for about

D = $17,000/day = $5,185,000/year (305 day year) Cc = $.09/dollar/year = $.09 Co = $1,200/loan + .0225 Q L = 15 days Optimal loan amount per loan: Q=

$371,000 per loan

R = (15)(17,000) = $255,000 reorder point. When cash balance gets down to $255,000 initiate another loan. Quantity Discount Analysis If Q > $500,000, points = 2% Since Q is unaffected by points, and Q was $371,842, we know we must set Q = $500,000 for this alternate option. D Q TC = Co + Cc + .02 D 2 Q

2Co D (2)(1,200)(5,185,000) = .09 Cc

= $371,842.26 loan amount per loan Memo: The .0225 Q cost per loan is not included in the calculation of Q since it is paid on the entire dollar amount of the loan regardless of loan size, and thus it is simply an annual cost, i.e., .0225 × D. TC = Co

⎛ 5,185,000 ⎞ ⎛ 500,000 ⎞ = (1,200) ⎜ + (.09) ⎜ ⎟⎠ ⎝ ⎝ 500,000 ⎟⎠ 2

D Q + Cc + .0225 D Q 2

+ (.02)(5,185,000) = 12,444 + 22,500 + 103,700 = $138,644

⎛ 5,185,000 ⎞ ⎛ 371,842 ⎞ = (1,200) ⎜ ⎟ ⎟ + (.09) ⎜ 371,842 2 ⎝ ⎠ ⎝ ⎠ + (.0225)(5,185,000) = $150,128.30 total cost of borrowing

Since this option yields a lower TC of $11,484 (150,128 – 138,644), it should be accepted.

CASE SOLUTION: DIAMANT FOODS COMPANY This problem requires the development of a forecast for product demand in year 4 (see Chapter 15). A seasonal forecast was developed, as follows. Year

Jan-March

April-May

June-Aug

Sept

Oct

Nov-Dec

Total

607

488

479

256

342

524

2,696

651

487

660

263

370

537

2,968

685

539

672

302

411

572

3,181

Total

1,943

1,514

1,811

821

1,123

1,633

8,845

S1 = 0.220 S2 = 0.171

SF2 = 587.68

S3 = 0.205

SF4 = 318.68

S4 = 0.093

SF5 = 435.91

SF3 = 702.97

S5 = 0.127 S6 = 0.185 Linear trend line forecast: y = 2,463.3 + 242.5x

SF6 = 633.88 Total 3,433.33 Q=

y(4) = 3,433.33 cases

2(4,700)(3, 433.33) = 527.5 116

Comparing monthly forecasts (with seasonal pattern) with order size, Q, using order frequency of 2 months:

SF1 = 754.21

16-13 .

D 3, 433.3 = = 6.5 orders Q 527.5

Co = $5,700

52 weeks = 8 weeks (2 months) per order 6.5 orders

Q = 52,340

No. of orders =

Cc = $0.45/lb. TC(1) = $353,380, Q = 52,340 TC(2) = $337,159, Q = 52,340

792 528 528 528 528 528

Monthly Forecast

Balance

TC(3) = 326,048, Q = 100,000

January

251

541

TC(4) = $319,023, Q = 150,000 * Optimal

February

251

289

March

251

566

Nuts

April

294

272

D = 77,242 lb.

May

294

506

Co = $6,300

June

234

272

Cc = $0.63/lb.

July

234

565

Q = 39,304 lbs.

August

234

331

TC(1) = $526,834, Q = 39,304

September

318

541

TC(2) = $507,524, Q = 39,304

October

436

105

TC(3) = $488,591, Q = 70,000 * Optimal

November

317

316

December

317

−1

Total

Filling

3,433

D = 61,794 lb.

Note that the “.5” order was added to the first month. The order size (Q = 528) seems to be adequate to offset seasonal patterns.

Co = $4,500 Cc = $0.55 Q = 31,799

Ingredient orders:

TC(1) = $110,180, Q = 31,799

Chocolate

TC(2) = $101,373, Q = 40,000 * Optimal

D = 108,140 lbs. (3,433.3 cases × 60 bags/case = 205,980 bags; 205,980 bags × 12 bars/bag = 2,471,760 bars; Demand = 2,471,760 bars × 0.70 oz chocolate/bar = 1,730,252 oz = 108,140 lbs.)

TC(3) = $102,718, Q = 80,000 Diamant Foods might experience quality problems with its large orders for ingredients that take advantage of price discounts; ingredients may be in storage for long periods. Also, the demand forecast is treated with certainty; if significant variation occurs it could create shortages and the need for safety stocks. Lead times are considered negligible, which could also create problems along the supply chain if they are significant in reality.

16-14 .

Module A: The Simplex Solution Method PROBLEM SUMMARY

42.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

43.

41.

Simplex short answer Simplex short answer Simplex short answer Simplex short answer Simplex short answer Simplex short answer 4 tableaus 2 tableaus 3 tableaus 3 tableaus 2 tableaus 5 tableaus 5 tableaus 5 tableaus 6 tableaus 4 tableaus 3 tableaus 3 tableaus 3 tableaus Simplex short answer 3 tableaus 4 tableaus Graphical analysis (A−22) 2 tableaus 2 tableaus 6 tableaus 5 tableaus Graphical analysis (A−27) Mixed constraint model transformation Mixed constraint model transformation 5 tableaus 3 tableaus 3 tableaus 4 tableaus 3 tableaus, multiple optimal 4 tableaus 3 tableaus, multiple optimal Graphical solution, 2 tableaus, infeasible Graphical solution, 2 tableaus, unbounded 4 tableaus, pivot row and column ties, multiple optimal Infeasible problem

44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54.

Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Sensitivity analysis, cj and qi Sensitivity analysis with duality Sensitivity analysis with duality Sensitivity analysis with duality

PROBLEM SOLUTIONS 1.

A-1 .

a) x1 = 10, x2 = 40, s3 = 30, z = 420 b) Yes; all cj − zj row values are zero or negative. c) x3 = 0; s2 = 0 d) Maximize Z = 10x1 + 2x2 + 6x3 e) 3 f) Since there are three decision variables, a three-dimensional graph is required. a) minimization; because zj − cj is being calculated on the bottom row and not cj − zj b) x1 = 20, x3 = 10, s1 = 10, z = 240 c) Yes; all zj − cj values are negative or zero. d) Minimize Z = 6x1 + 20x2 + 12x3 e) 3 constraints f) Yes, one; because there are 3 constraints but only 2 surplus variables. Since both ≤ and ≥ constraints have slack or surplus variables, and an equality does not, then one of the three constraints was an =. g) x2 = 0

a) b) c) d)

maximization; because of “cj − zj” x1 = 12, x2 = 0, x3 = 0, x4 = 15 s1 = 20, s2 = 0, s3 = 0, s4 = 45 If x2 is selected as the entering basic variable, Z will increase by 20 for every unit of x2 entered into the solution. This solution is not optimal because there are positive values in the cj − zj row.

Basic Variables

Quantity

60 x1

50 x2

45 x3

50 x4

0 s1

0 s2

0 s3

0 s4

1/2

1/10

45/8

−3/4

1/8

13,785/8

65/4

45/8

−65/4

−6

−45/8

cj − zj cj

Basic Variables

Quantity

60 x1

50 x2

45 x3

50 x4

0 s1

0 s2

0 s3

0 s4

−1/2

1/10

45/8

−3/4

1/8

2,123.125

65/4

45/8

−65/4

−6

−45/8

cj − zj 4.

Optimal a) Minimization; because “zj − cj” and a positive “M” in the objective function. b) x3 = 0 c) “M/6 − 5/3” has no real meaning since it includes “M”; it simply represents a large net decrease in cost if s2 is entered into the solution. d) two; since there are two artificial variables remaining in the tableau, it will take at least two more tableaus to eliminate them, and they must be eliminated to insure a feasible solution. e) no; because there are positive values in the zj − cj row.

A-2 .

Basic Variables

Quantity

8 x1

10 x2

4 x3

0 s1

0 s2

M A3

2/3

−1

1/6

1/3

−1/6

−2/3

−1/6

−1

7M + 220

−2M/3 + 18/3

M−4

−M/6−1

−M

−2M/3−2

M−4

−M/6−1

−M

zj − cj

Basic Variables

Quantity

8 x1

10 x2

4 x3

0 s1

0 s2

100

−1

1/3

−1/6

−2/3

−1/6

−1

500

10/3

−10/6

−4

−14/3

−10/6

−4

zj − cj 5.

Optimal a) Minimization; because zj − cj and “M” are positive in the objective function. b) x3 = 4 x2 = 6 c) no; because both constraints included a slack or surplus variable (s1 or s2) and an equation would have added only an artificial variable. d) s2 = 0 e) no; because there are positive values in the zj − cj row. cj

Basic Variables

Quantity

4 x1

6 x2

0 s1

0 s2

−2

−1

−4

−2

−4

zj − cj 6.

Optimal a) Maximization; because cj − zj b) x2 = 0 c) no; at this iteration no optimal solution exists. d) the cj − zj value of “5” means that if s1 was selected as the entering variable, Z would increase by 5 for every unit of s1 entered into the solution. e) no; there are positive values in the cj − zj row.

A-3 .

Basic Variables

Quantity

10 x1

5 x2

0 s1

0 s2

−1/2

1/2

−5

Quantity

10 x1

5 x2

0 s1

0 s2

cj − zj Basic Variables

cj 10

−10

cj − zj Optimal Multiple optimal solutions do not exist. 7.

Minimize Z = .05x1 + .03x2 (cost, $) subject to 8x1 + 6x2 ≥ 48 (vitamin A, mg) x1 + 2x2 ≥ 12 (vitamin B, mg) x1, x2 ≥ 0 cj

Basic Variables

Quantity

.05 x1

.03 x2

0 s1

0 s2

M A1

M A2

−1

60M

−M

9M − .05

8M − .03

−M

zj − cj cj

Basic Variables

Quantity

.05 x1

.03 x2

0 s1

0 s2

M A2

.05

3/4

−1/8

5/4

1/8

−1

6M + .30

.05

5M/4 + .38

M/8 − .006

−M

5M/4 + .38

M/8 − .006

−M

zj − cj

A-4 .

Basic Variables

Quantity

.05 x1

.03 x2

0 s1

0 s2

.05

12/5

−1/5

3/5

.03

24/5

1/10

−4/5

.264

.05

.03

−.007

.006

−.007

.006

Quantity

.05 x1

.03 x2

0 s1

0 s2

zj − cj cj

Basic Variables

5/3

−1/3

.03

4/3

−1/6

.24

.04

.03

−.005

−.01

−.005

Quantity

8 x1

2 x2

0 s1

0 s2

zj − cj Optimal 8. cj

Basic Variables

Quantity

8 x1

2 x2

0 s1

0 s2

cj − zj cj

Basic Variables

5/4

1/4

7/2

−1/2

−8

−20

Quantity

6 x1

4 x2

0 s1

0 s2

cj − zj

Optimal Maximize Z = 6x1 + 4x2 (profit, $) subject to 10x1 + 10x2 ≤ 100 (line 1, hr) 7x1 + 3x2 ≤ 42 (line 2, hr) x1, x2 ≥ 0 cj

Basic Variables

100

cj − zj

A-5 .

Basic Variables

Quantity

6 x1

4 x2

0 s1

0 s2

40/7

−10/7

3/7

1/7

18/7

6/7

10/7

−6/7

Quantity

6 x1

4 x2

0 s1

0 s2

cj − zj

Basic Variables

7/40

−1/4

−3/40

1/4

1/2

−1/4

−1/2

cj − zj Optimal 10. Maximize Z = 400x1 + 100x2 (profit, $) subject to 8x1 + 10x2 ≤ 80 (labor, hr) 2x1 + 6x2 ≤ 36 (wood, lb) x1 ≤ 6 (demand, chairs) x1, x2 ≥ 0 cj

Basic Variables

Quantity

400 x1

100 x2

0 s1

0 s2

0 s3

400

100

Quantity

400 x1

100 x2

0 s1

0 s2

0 s3

cj − zj

Basic Variables

−8

−2

400

2,400

400

100

−400

cj − zj

A-6 .

Basic Variables

Quantity

400 x1

100 x2

0 s1

0 s2

0 s3

100

3.2

1/10

−4/5

4.8

−3/5

14/5

400

2,720

400

100

320

−10

−320

Quantity

.1 x1

5 x2

0 s1

0 s2

0 s3

cj − zj Optimal 11. Maximize Z = x1 + 5x2 (profit, $) subject to 5x1 + 5x2 ≤ 25 (flour, lb) 2x1 + 4x2 ≤ 16 (sugar, lb) x1 ≤ 8 (demand for cakes) x1, x2 ≥ 0 cj

Basic Variables

Quantity

1 x1

5 x2

0 s1

0 s2

0 s3

cj − zj cj

Basic Variables

5/2

−5/4

1/2

1/4

5/2

5/4

−3/2

−5/4

cj − zj Optimal 12. Minimize Z = 3x1 + 5x2 (cost, $) subject to 10x1 + 2x2 ≥ 20 (nitrogen, oz) 6x1 + 6x2 ≥ 36 (phosphate, oz) x2 ≥ 2 (potassium, oz) x1, x2 ≥ 0

A-7 .

Basic Variables

Quantity

3 x1

5 x2

0 s1

0 s2

0 s3

M A1

M A2

M A3

−1

58M

16M

−M

16M − 3

9M − 5

−M

zj − cj cj

Basic Variables

Quantity

3 x1

5 x2

0 s1

0 s2

0 s3

M A2

M A3

1/5

−1/10

24/5

3/5

−1

26M + 6

29M/5 + 3/5

3M/5 − 3/10

−M

29M/5 − 22/5

3M/5 − 3/10

−M

zj − cj

Basic Variables

Quantity

3 x1

5 x2

0 s1

0 s2

0 s3

M A2

8/5

−1/10

1/5

72/5

3/5

−1

24/5

−1

72M + 74/5

3M/5 − 3/10

−M

24M/5 − 22/5

3M/5 − 3/10

−M

24M/5 − 22/5

zj − cj

Basic Variables

Quantity

3 x1

5 x2

0 s1

0 s2

0 s3

−1/8

1/24

1/8

−5/24

1/8

−5/24

1/4

−22/24

1/4

−22/24

zj − cj

A-8 .

Basic Variables

Quantity

3 x1

5 x2

0 s1

0 s2

0 s3

−1/6

−5/3

−1

−1/2

−2

−1/2

−2

zj − cj Optimal 13. cj

Basic Variables

Quantity

.06 x1

.10 x2

0 s1

0 s2

0 s3

M A1

M A2

M A3

−1

34M

12M

11M

−M

12M − .06

11M − .10

−M

zj − cj

Basic Variables

Quantity

.06 x1

.10 x2

0 s1

0 s2

0 s3

M A1

M A2

7/5

−1

4/5

24/5

−1

3/5

.06

2/5

−1/5

10M + .12

.06

31M/5 + .02

−M

7M/5 − .01

31M/5 − .08

−M

7M/5 − .01

zj − cj

Basic Variables

Quantity

.06 x1

.10 x2

0 s1

0 s2

0 s3

M A1

9/4

−1

7/24

5/8

.10

5/4

−5/24

1/8

.06

3/2

1/12

−1/4

9M/4 + .22

.06

.10

−M

7M/24 − .01

5M/8 − .002

−M

7M/24 − .01

5M/8 − .002

zj − cj

A-9 .

Basic Variables

Quantity

.06 x1

.10 x2

0 s1

0 s2

0 s3

7.71

−24/7

15/7

.10

2.88

−5/7

4/7

.06

.85

2/7

−3/7

.34

.06

.10

−.05

.035

−.05

.035

Quantity

.06 x1

.10 x2

0 s1

0 s2

0 s3

zj − cj

Basic Variables

3.6

−8/5

7/15

.10

1/5

−4/15

.06

2.4

−2/5

1/5

.22

.06

.10

−.004

−.014

−.004

−.014

zj − cj Optimal 14. Minimize Z = 200x1 + 160x2 (cost, $) subject to 6x1 + 2x2 ≥ 12 (high-grade ore, tons) 2x1 + 2x2 ≥ 8 (medium-grade ore, tons) 4x1 + 12x2 ≥ 24 (low-grade ore, tons) x1, x2 ≥ 0 cj

Basic Variables

Quantity

200 x1

160 x2

0 s1

0 s2

0 s3

M A1

M A2

M A3

−1

44M

12M

16M

−M

12M − 200

16M − 160

−M

zj − cj

Basic Variables

Quantity

200 x1

160 x2

0 s1

0 s2

0 s3

M A1

M A2

16/3

−1

1/6

4/3

−1

1/6

160

1/3

−1/12

12M + 320

20M/3 + 160/3

160

−M

M/3 − 40/3

20M/3 − 440/3

−M

M/3 − 40/3

zj − cj

A-10 .

Basic Variables

Quantity

200 x1

160 x2

0 s1

0 s2

0 s3

M A2

200

3/2

−3/6

1/32

1/4

1/8

160

3/2

1/16

−3/32

2M + 540

200

160

M/4 − 55/2

−M

M/8 − 70/8

M/4 − 55/2

−M

M/8 − 70/8

zj − cj

Basic Variables

Quantity

200 x1

160 x2

0 s1

0 s2

0 s3

200

−3/4

1/8

−4

1/2

160

1/4

−1/8

760

200

160

−110

Quantity

200 x1

160 x2

0 s1

0 s2

0 s3

zj − cj

Basic Variables

200

−1/4

1/4

−8

160

1/4

−3/4

680

200

160

−10

−70

−10

−70

zj − cj Optimal 15. cj

Basic Variables

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

M A1

M A2

M A3

400

−1

100

−1

1,200

−1

1,700M

11M

−M

11M − 2

M− 3

9M − 5

6M − 7

−M

zj − cj

A-11 .

Basic Variables

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

M A2

M A3

2/5

−1/10

100

−1

1,160

3/5

1/10

−1

1,260M + 80

23M/5 + 4/5

M/10 − 1/5

−M

M−3

−23M/5 − 21/5

6M − 7

M/10 − 1/5

−M

zj − cj

Basic Variables

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

M A3

2/5

−1/10

4/5

−1/5

1,140

−1/5

1/10

1/5

−1

1,240M + 220

−M/5 + 32/5

M/10 − 1/5

M/5 − 7/5

−M

M− 3

−M/5 + 7/5

M/10 − 1/5

M/5 − 7/5

−M

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

zj − cj Basic Variables

cj 2

2/5

−1/10

4/5

−1/5

1,140

−1/5

1/10

1/5

−1

3,640

29/5

1/10

−4/5

−3

4/5

−0

1/10

−4/5

−3

zj − cj

Basic Variables

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

−1/2

−1/10

1/10

5/4

−1/4

1,145

1/4

1/10

3/20

−1

3,620

1/10

−3/5

−3

−1

1/10

−3/5

−3

zj − cj

A-12 .

Basic Variables

Quantity

2 x1

3 x2

5 x3

7 x4

0 s1

0 s2

0 s3

1,175

−1/4

1/4

−1

5/4

−1/4

11,450

5/2

3/2

−10

2,475

23/4

−3/4

−2

−1

−5/4

−3/4

−2

Quantity

300 x1

400 x2

0 s1

0 s2

0 s3

zj − cj Optimal 16. cj

Basic Variables

300

400

Quantity

300 x1

400 x2

0 s1

0 s2

0 s3

cj − zj

Basic Variables

−2

−4

400

1,600

400

300

−400

Quantity

300 x1

400 x2

0 s1

0 s2

0 s3

cj − zj

Basic Variables

−3/2

300

1/2

−2

400

2,200

300

400

150

−200

−150

200

cj − zj

A-13 .

Basic Variables

Quantity

300 x1

400 x2

0 s1

0 s2

0 s3

1/4

−3/8

300

1/2

−1/4

400

−1/4

3/8

2,400

300

400

−50

−75

cj − zj Optimal 17. cj

Basic Variables

Quantity

5 x1

4 x2

0 s1

0 s2

150

3/10

1/2

2,000

Quantity

5 x1

4 x2

0 s1

0 s2

cj − zj

Basic Variables

19/50

−3/100

200

2/5

1/10

1,000

1/2

−1/2

cj − zj

Basic Variables

Quantity

5 x1

4 x2

0 s1

0 s2

4,500/19

50/19

−3/38

2,000/19

−20/19

5/38

28,000/19

100/19

13/28

−100/19

−13/38

cj − zj Optimal 18. cj

Basic Variables

Quantity

100 x1

150 x2

0 s1

0 s2

0 s3

160

300

100

150

cj − zj

A-14 .

Basic Variables

Quantity

100 x1

150 x2

0 s1

0 s2

0 s3

100

−1/5

1/2

−1/20

150

1/2

1/20

2,250

150

15/2

−15/2

Quantity

100 x1

150 x2

0 s1

0 s2

0 s3

cj − zj cj

Basic Variables

−16

3/5

100

−1/10

150

−1

1/10

2,500

100

150

−50

−5

cj − zj Optimal 19. cj

Basic Variables

Quantity

100 x1

20 x2

60 x3

0 s1

0 s2

0 s3

100

Quantity

100 x1

20 x2

60 x3

0 s1

0 s2

0 s3

cj − zj

Basic Variables

100

5/3

1/3

−4/3

−2/3

2,000

100

500/3

100/3

−440/3

−100/3

cj − zj

A-15 .

Basic Variables

Quantity

100 x1

20 x2

60 x3

0 s1

0 s2

0 s3

100

5/3

1/3

−2/3

−1/3

1/2

2/3

1/3

−1/2

3,800

100

380/3

40/3

−320/3

−40/3

−30

cj − zj

Optimal 20. a. Maximization, because cj − zj b. x2 = 10, s2 = 20, x1 = 10, Z = 30 c. maximize Z = x1 + 2x2 − x3 d. 3 e. No, because there are three constraints and three “slack” variables f. s1 = 0 g. yes, because cj −zj = 0 for s1 h. x2 = 3 1/3, s2 = 26 2/3, x1 = 23 1/3, Z = 30 21. cj

Basic Variables

Quantity

120 x1

40 x2

240 x3

0 s1

0 s2

M A1

M A2

−1

54M

−M

6M − 120

7M − 40

6M − 240

−M

zj − cj

Basic Variables

Quantity

120 x1

40 x2

240 x3

0 s1

0 s2

M A1

11/3

5/2

−1

1/6

1/3

1/2

−1/6

19M + 200

11M/3 + 40/3

5M/2 + 20

−M

M/6 − 20/3

11M/3 − 320/3

5M/2 − 220

−M

M/6 − 20/3

zj − cj

Basic Variables

Quantity

120 x1

40 x2

240 x3

0 s1

0 s2

120

15/22

−3/11

1/22

6/22

1/11

−2/11

840

120

1,020/11

−320/11

−20/11

−1,620/11

−320/11

−20/11

cj − zj Optimal

A-16 .

22. cj

Basic Variables

Quantity

.05 x1

.10 x2

0 s1

0 s2

M A1

M A2

−1

86M

11M

−M

11M − .05

7M − .10

−M

Quantity

.05 x1

.10 x2

0 s1

0 s2

M A2

zj − cj cj

Basic Variables

.05

1/3

−1/6

10/3

5/6

−1

20M + .3

.05

10M/3 + .02

5M/6 − .01

−M

10M/3 − .08

5M/6 − .01

−M

zj − cj cj

Basic Variables

Quantity

.05 x1

.10 x2

0 s1

0 s2

.05

−1/4

1/10

.10

1/4

−3/10

.80

.05

.10

.0125

−.025

.0125

−.025

Quantity

.05 x1

.10 x2

0 s1

0 s2

zj − cj

Basic Variables

.05

−1/5

−6/5

.50

.05

−.01

−.05

−.01

zj − cj Optimal

A-17 .

23.

24. cj

Basic Variables

Quantity

10 x1

12 x2

7 x3

0 s1

0 s2

0 s3

300

120

Quantity

10 x1

12 x2

7 x3

0 s1

0 s2

0 s3

cj − zj

Basic Variables

4/3

2/3

1/15

10/3

−10/3

−1/3

240

4/5

−6

−1

−4/5

cj − zj Optimal 25. cj

Basic Variables

Quantity

6 x1

2 x2

12 x3

0 s1

0 s2

cj − zj

A-18 .

Basic Variables

Quantity

6 x1

2 x2

12 x3

0 s1

0 s2

4/3

1/3

−2

−1

−10

−2

−4

cj − zj Optimal 26. cj

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

2,000

200

250

2,200

100

200

100

cj − zj cj

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

−3.75

−.015

100

1.25

.005

1,200

−1.25

200

−.50

1,000

100

1.25

.50

−35

−.50

cj − zj

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

−3.75

−.015

4.6

.002

−.005

100

1.25

.005

−.625

−.002

.005

1,570

100

65.6

.26

.475

24.4

−.26

−.475

cj − zj

A-19 .

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

−1.875

.50

−.007

4.6

.002

−.005

100

1.25

.005

−.625

−.002

.005

1,945

100

−75

37.5

−.30

.475

165

−37.5

.30

−.475

cj − zj

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

12.7

.405

−.006

−.002

4.1

.216

.001

−.001

100

4.9

−.270

.004

.001

8.6

.135

−.002

.004

2,623

100

37.5

35.7

−.211

.297

−37.5

−35.7

.211

−.297

cj − zj

Basic Variables

Quantity

100 x1

75 x2

90 x3

95 x4

0 s1

0 s2

0 s3

0 s4

1.5

.50

3.5

−.125

.25

−.001

1,125

231.25

−62.5

.313

.50

.005

2,860

148.75

37.5

22.5

.36

−48.75

−37.5

−22.5

−.36

cj − zj Optimal 27. cj

Basic Variables

Quantity

20 x1

16 x2

0 s1

0 s2

0 s3

M A1

M A2

M A3

−1

22M

−M

6M − 20

8M − 16

−M

zj − cj

A-20 .

Basic Variables

Quantity

20 x1

16 x2

0 s1

0 s2

0 s3

M A1

M A2

8/3

−1

1/6

2/3

−1

1/6

1/3

−1/6

32 + 6M

10M/3 + 16/3

−M

M/3 − 8/3

10M/3 − 44/3

−M

M/3 − 8/3

zj − cj

Basic Variables

Quantity

20 x1

16 x2

0 s1

0 s2

0 s3

M A2

3/2

−3/8

1/6

1/4

−1

1/8

3/2

1/8

−3/16

M + 27

M/4 − 53/6

−M + 16/3

M/8 − 25/12

M/4 − 53/6

−M + 16/3

M/8 − 25/12

zj − cj

Basic Variables

Quantity

20 x1

16 x2

0 s1

0 s2

0 s3

−3/2

1/4

−4

1/2

−1/4

−22

zj − cj

Basic Variables

Quantity

20 x1

16 x2

0 s1

0 s2

0 s3

−1/2

1/2

−8

1/2

−3/2

−2

−14

−2

−14

zj − cj Optimal

A-21 .

28.

29. Minimize Z = 8x1 + 2x2 + 7x3 + 0s1 + 0s2 + 0s3 −MA1 − MA2 − MA3 subject to 2x1 + 6x2 + x3 + A1 = 30 3x2 + 4x3 − s1 + A2 = 60 4x1 + x2 + 2x3 + s2 = 50 x1 + 2x2 − s3 + A3 = 20 x1, x2, x3 ≥ 0 30. Minimize Z = 40x1 + 55x2 + 30x3 + 0s1 + 0s2 + 0s3 + MA1 + MA2 + MA3 subject to x1 + 2x2 + 3x3 + s1 = 60 2x1 + x2 + x3 + A1 = 40 x1 + 3x2 + x3 − s2 + A2 = 50 5x2 −3x3 − s3 + A3 = 100 x1, x2, x3 ≥ 0 31. cj

Basic Variables

Quantity

40 x1

60 x2

0 s1

0 s2

0 s3

0 s4

−M A1

−M A2

−M

−1

−M

−1

−17M

−M

M + 40

M + 60

−M

cj − zj

A-22 .

Basic Variables

Quantity

40 x1

60 x2

0 s1

0 s2

0 s3

0 s4

−M A1

−M

−1

−5M + 720

−M

−60

−M

M + 40

−M

cj − zj cj

Basic Variables

Quantity

40 x1

60 x2

0 s1

0 s2

0 s3

0 s4

−1

920

−40

−60

cj − zj

Basic Variables

Quantity

40 x1

60 x2

0 s1

0 s2

0 s3

0 s4

1/2

−2

−1

25/2

1/2

950

−10

−30

cj − zj

Tie

Basic Variables

Quantity

40 x1

60 x2

0 s1

0 s2

0 s3

0 s4

−1/4

−4

1/2

−1

1/2

−1/4

960

−20

−5

cj − zj Optimal

A-23 .

32. cj

Basic Variables

Quantity

1 x1

5 x2

0 s1

0 s2

0 s3

−M A1

−M

−1

−25M

−5M

−M

5M + 1

5M + 5

−M

cj − zj

Basic Variables

Quantity

1 x1

5 x2

0 s1

0 s2

0 s3

−M A1

−M

5/2

−1

−5/4

1/2

1/4

−5M + 20

−5M/2 + 5/2

−5M/4 + 5/4

−M

5M/2 − 3/2

−M

−5M/4 − 5/4

cj − zj

Basic Variables

Quantity

1 x1

5 x2

0 s1

0 s2

0 s3

−2/5

−1/2

1/5

1/2

2/5

1/2

3/5

−3/5

−2

cj − zj Optimal 33. cj

Basic Variables

Quantity

3 x1

6 x2

0 s1

0 s2

0 s3

0 s4

M A1

−1

−M

M−3

M−6

−M

zj − cj

A-24 .

Basic Variables

Quantity

3 x1

6 x2

0 s1

0 s2

0 s3

0 s4

M A1

−3

−1

M + 12

−M

−M + 3

M−6

−M

−M +3

zj − cj

Basic Variables

Quantity

3 x1

6 x2

0 s1

0 s2

0 s3

0 s4

−1

−6

−3

−6

−3

Quantity

10 x1

5 x2

0 s1

0 s2

−M A1

−M A2

zj − cj Optimal 34. cj

Basic Variables

−M

−1

−M

−14M

−2M

−M

2M + 10

2M + 5

−M

cj − zj

Quantity

10 x1

5 x2

0 s1

0 s2

−M A2

1/2

−1/2

−M

7/2

1/2

−4M + 50

−M + 5

−5

−M

Basic Variables

cj − zj

A-25 .

Basic Variables

Quantity

10 x1

5 x2

0 s1

0 s2

−1/2

1/2

−5

Quantity

10 x1

5 x2

0 s1

0 s2 1

cj − zj

Basic Variables

−10

cj − zj Optimal 35. cj

Basic Variables

Quantity

1 x1

2 x2

−1 x3

0 s1

0 s2

0 s3

−1

cj − zj

Basic Variables

Quantity

1 x1

2 x2

−1 x3

0 s1

0 s2

0 s3

1/4

−1

1/2

−3/2

−1/2

cj − zj

A-26 .

Basic Variables

Quantity

1 x1

2 x2

−1 x3

0 s1

0 s2

0 s3

1/4

−3/4

3/4

−1/2

1/2

3/2

1/2

−5/2

−1/2

Quantity

1 x1

2 x2

−1 x3

0 s1

0 s2

0 s3

cj − zj Multiple optimum solution Alternate solution: Basic Variables

cj 2

10/3

1/2

−1/3

1/6

80/3

−1

4/3

−2/3

70/3

1/2

2/3

1/6

3/2

1/2

−5/2

−1/2

cj − zj 36. cj

Basic Variables

Quantity

1 x1

2 x2

2 x3

0 s1

0 s2

−M A1

−M A2

−M

−1

−28M

−3M

−2M

−4M

−M

1 + 3M

2 + 2M

2 + 4M

−M

cj − zj cj

Basic Variables

Quantity

1 x1

2 x2

2 x3

0 s1

0 s2

−M A2

1/5

3/5

2/5

1/5

−M

7/5

6/5

−1

−12M + 8

−4/5 − 7M/5

2/5 −6M/5

−M

1/5 + 7M/5

8/5 + 6M/5

−M

cj − zj

A-27 .

Basic Variables

Quantity

1 x1

2 x2

2 x3

0 s1

0 s2

16/7

3/7

1/7

4/7

−1/7

2/7

60/7

6/7

−5/7

68/7

4/7

−1/7

10/7

1/7

cj − zj

Basic Variables

Quantity

1 x1

2 x2

2 x3

0 s1

0 s2

16/3

7/3

1/3

4/3

1/3

−2

−1

52/3

10/3

1/3

−10/3

−1/3

cj − zj Optimal 37. cj

Basic Variables

Quantity

400 x1

350 x2

450 x3

0 s1

0 s2

0 s3

0 s4

120

160

100

400

350

450

cj − zj

Basic Variables

Quantity

400 x1

350 x2

450 x3

0 s1

0 s2

0 s3

0 s4

1/2

−1/2

135

13/4

5/2

−1/4

450

3/4

1/2

1/4

−1/4

1/2

−1/4

11,250

1,350/4

450/2

450

450/4

250/4

250/2

−450/4

cj − zj

A-28 .

Basic Variables

Quantity

400 x1

350 x2

450 x3

0 s1

0 s2

0 s3

0 s4

−1/2

1/2

−4

−5

450

1/2

−1

350

1/2

−1/2

15,000

400

350

450

250

−50

−250

cj − zj Multiple optimum solution at x1 Alternate solution: cj

Basic Variables

Quantity

400 x1

350 x2

450 x3

0 s1

0 s2

0 s3

0 s4

−5

−4

−1

400

−2

350

−1

15,000

400

350

450

250

−50

−250

cj − zj 38. (a)

(b) cj

Basic Variables

Quantity

3 x1

2 x2

0 s1

0 s2

−M A1

−M

−1

−2M

−M

M+3

M+2

−M

cj − zj

A-29 .

Basic Variables

Quantity

3 x1

2 x2

0 s1

0 s2

−M A1

−M

−1

3−M

−M

−1

−M

cj − zj Infeasible solution 39. (a)

(b) cj

Basic Variables

Quantity

1 x1

1 x2

0 s1

0 s2

−1

cj − zj

Tie for entering variable; if x1 is chosen, the solution is unbounded. Select x2 arbitrarily. Quantity

1 x1

1 x2

0 s1

0 s2

−1

Basic Variables

1 0

cj − zj

A-30 .

Basic Variables

Quantity

1 x1

1 x2

0 s1

0 s2

−1

−2

cj − zj Unbounded; no pivot row available 40. cj

Basic Variables

Quantity

7 x1

5 x2

5 x3

0 s1

0 s2

0 s3

0 s4

cj − zj

Basic Variables

Quantity

7 x1

5 x2

5 x3

0 s1

0 s2

0 s3

0 s4

1/2

−1/2

1/2

−1/2

140

7/2

cj − zj

3/2

−7/2

Basic Variables

Quantity

7 x1

5 x2

5 x3

0 s1

0 s2

0 s3

0 s4

−1

155

−3

−2

Tie

cj − zj Multiple optimum

A-31 .

Alternate Solution: cj

Basic Variables

Quantity

7 x1

5 x2

5 x3

0 s1

0 s2

0 s3

0 s4

−1

155

−3

−2

Quantity

15 x1

25 x2

0 s1

0 s2

0 s3

M A1

M A2

cj − zj 41. cj

Basic Variables

−1

18M

−M

5M − 15

5M − 25

−M

Quantity

15 x1

25 x2

0 s1

0 s2

0 s3

M A1

zj − cj

Basic Variables

5/2

−1

1/2

3M + 45

5M/2 + 15/2

−M

3M/2 − 15/2

5M/2 − 35/2

−M

3M/2 − 15/2

zj − cj

Basic Variables

Quantity

15 x1

25 x2

0 s1

0 s2

0 s3

M A1

−1

−6

−5

−2

−1

3M + 45

−M

−6M +45

−5M +45

−M

−6M +45

−5M +45

zj − cj Infeasible solution

A-32 .

42. a) minimize Zd = 90y1 + 60y2 subject to y1 + 2y2 ≥ 6 4y1 + 2y2 ≥ 10 y1, y2 ≥ 0 b) y1 = the marginal value of one additional lb of brass = $1.33 y2 = the marginal value of one additional hr of labor = $2.33 c)

c1, basic: Quantity

6+Δ x1

10 x2

0 s1

0 s2

1/3

−1/6

−1/3

2/3

260 + 10Δ

6+Δ

4/3 − Δ/3

7/3 + 2Δ/3

−4/3 + Δ/3

−7/3 − 2Δ/3

6 x1

10 + Δ x2

0 s1

0 s2

0 1 6 0

1 0

1/3 −1/3 4/3 + Δ/3 −4/3 − Δ/3

−1/6 2/3

Basic Variables

10 6+Δ

cj − zj Solving for the cj − zj inequalities: −4/3 + Δ/3 ≤ 0 Δ/3 ≤ 4/3 Δ≤4 Since c1 = 6 + Δ; Δ = c1 − 6. Thus c1 − 6 ≤ 4 c1 ≤ 10 −7/3 − 2Δ/3 ≤ 0 −2Δ/3 ≤ 7/3 −2Δ ≤ 7 Δ ≥ −7/2 Since c1 = 6 + Δ; Δ = c1 − 6. Thus c1 − 6 ≥ −7/2 c1 ≥ 5/2 Summarizing, 5/2 ≤ c1 ≤ 10. c2, basic: cj 10 + Δ 6

Basic Variables x2 x1 zj cj − zj

Quantity 20 10 280 + 20Δ

10 + Δ 0

A-33 .

7/3 − Δ/6 −7/3 + Δ/6

Solving for the cj − zj inequalities:

50y1 + 20y2 ≥ 300 y1, y2 ≥ 0

−4/3 − Δ/3 ≤ 0

y1 = $4.13 = the marginal value of one additional lb of chili beans; y2 = $4.67 = the marginal value of one additional lb of ground beef.

−Δ/3 ≤ 4/3 Δ ≥ −4 Since c2 = 10 + Δ; Δ = c2 − 10. Thus b)

c2 − 10 ≥ −4 c2 ≥ 6 −7/3 + Δ/6 ≤ 0 Δ/6 ≤ 7/3 Δ ≤ 14 Since c2 = 10 + Δ; Δ = c2 − 10. Thus c2 − 10 ≤ 14 c2 ≤ 24 Summarizing, 6 ≤ c2 ≤ 24. d) q1: x2:

20 + Δ/3 ≥ 0

x1:

Point C must become the optimal solution for x2 = 0; therefore the slope of the objective function must be greater than the slope of the constraint for ground beef, −34/20. Solving the following for the profit, p, of Razorback chili yields −p/300 = −34/20 p = $510 Thus, if the profit of Razorback chili is greater than $510, no Longhorn chili will be produced. The new optimal solution will be x1 = 23.5 and x2 = 0.

10 − Δ/3 ≥ 0

Δ/3 ≥ −20

−Δ/3 ≥ −10

Δ ≥ −60

Δ ≤ 30

Therefore, −60 ≤ Δ ≤ 30. Since q1 = 90 + Δ Δ = q1 − 90 −60 ≤ q1 − 90 ≤ 30 30 ≤ q1 ≤ 120 q2: x2: 20 − Δ/6 ≥ 0

x1:

10 + 2Δ/3 ≥ 0

− Δ/6 ≥ −20

2Δ/3 ≥ −10

Δ ≤ 120

Δ ≥ −15

Therefore, −15 ≤ Δ ≤ 120. Since q2 = 60 + Δ Δ = q2 − 60 −15 ≤ q2 − 60 ≤ 120 45 ≤ q2 ≤ 180 e)

The marginal value of 1 hr of labor is $2.33. From part d, the sensitivity range for q2, labor, is 45 ≤ q2 ≤ 180. Thus, the company would purchase up to 180 hr at the marginal value price.

The constraint line for chili beans rotates, creating a new, smaller solution space, and the optimal solution shifts from point B to point B´ where x1 = 21.43 and x2 = 3.57.

43. a) Minimize Zd = 500y1 + 800y2 subject to 10y1 + 34y2 ≥ 200

A-34 .

d) c1, basic: cj

Basic Variables

Quantity

200 + Δ x1

300 x2

0 s1

0 s2

300

17/750

−1/150

200 + Δ

−1/75

1/30

5,800 + 20Δ

200 + Δ

300

310/75 − Δ/75

70/15 + Δ/30

−310/75 + Δ/75

−70/15 − Δ/30

cj − zj Solving for the cj − zj inequalities: −310/75 + Δ/75 ≤ 0 Δ/75 ≤ 3310/75 Δ ≤ 310 Since c1 = 200 + Δ; Δ = c1 − 200. Thus c1 − 200 ≤ 310 c1 ≤ 510 −70/15 − Δ/30 ≤ 0 −Δ/30 ≤ 70/15 Δ ≥ −140 Since c1 = 200 + Δ; Δ = c1 − 200. Thus c1 − 200 ≥ −140 c1 ≥ 60 Summarizing, 60 ≤ c1 ≤ 510.

A-35 .

c2, basic: Quantity

300 + Δ x2

Basic

200

Variables

300 + Δ

17/750

−1/150

200

−1/75

1/30

5,800 + 6Δ

200

300 + Δ

310/75 + 17Δ/750

70/15 − Δ/150

−310/75 − 17Δ/750

−70/15 + Δ/150

cj − zj

q 2: Solving for the cj − zj inequalities:

x2: 6 − Δ/150 ≥ 0

−310/75 − 17Δ/750 ≤ 0

− Δ/150 ≥ −6

Δ/30 ≥ −20

Δ ≤ 900

Δ ≥ −600

−17Δ/750 ≤ 310/75 Δ ≥ −182.35

x1: 20 + Δ/30 ≥ 0

Therefore, −600 ≤ Δ ≤ 900. Since

Since c2 = 300 + Δ; Δ = c2 − 300. Thus

q2 = 800 + Δ

c2 − 300 ≥ −182.35

Δ = q2 − 800 −600 ≤ q2 − 800 ≤ 900

c2 ≥ 117.65

200 ≤ q2 ≤ 1,700 −70/15 + Δ/150 ≤ 0

Δ/150 ≤ 70/15 Δ ≤ 700 Since c2 = 300 + Δ; Δ = c2 − 300. Thus

g) Ground beef

c2 − 300 ≤ 700

h) No effect

c2 ≤ 1,000

44. a)

Minimize Zd = 60y1 + 40y2 subject to 12y1 + 4y2 ≥ 9 4y1 + 8y2 ≥ 7 y1, y2 ≥ 0 b) y1 = the marginal value of one additional hr of process 1; y2 = the marginal value of one additional hr of process 2 For the s1 column, the cj − zj value of $.55 is the marginal value of 1 hr of process 1 production time. For the s2 column, the cj − zj value of $0.60 is the marginal value of 1 hr of process 2 production time.

Summarizing, 117.65 ≤ c2 ≤ 1,000. e)

q1: x2: 6 + 17Δ/750 ≥ 0 x1: 20 − Δ/75 ≥ 0 17Δ/750 ≥ −6

−Δ/75 ≥ −20

Δ ≥ −264.7

The marginal value of 1 lb of chili beans is $4.13. The sensitivity range for q1, chili beans, is 235.3 ≤ q1q ≤ 2,000. Thus, the company would purchase up to 2,000 lb at the marginal value price.

Δ ≤ 1,500

Therefore, −264.7 ≤ Δ ≤1,500. Since q1 = 500 + Δ Δ = q1 − 500 −264.7 ≤ q1 − 500 ≤ 1,500 235.3 ≤ q1 ≤ 2,000

A-36 .

cj, basic: 9+Δ

Quantity

Basic cj

Variables

9+Δ

1/10

−1/20

3/20

57 + 4Δ

9+ Δ

11/20 − Δ/10

12/20 + Δ/20

−11/20 − Δ/10

−12/20 + Δ/20

7+ Δ

Quantity

cj − zj Solving for the cj − zj inequalities: −11/20 − Δ/10 ≤ 0 −Δ/10 ≤ 11/20 −Δ ≤ 11/2 Δ ≥ −11/2 Since c1 = 9 + Δ, Δ = c1 − 9. Thus c1 − 9 ≥ 11/2 c1 ≥ 7/2 −12/20 + Δ/20 ≤ 0 Δ/20 ≤ 12/20 Δ ≤ 12 Since c1 = 9 + Δ, Δ = c1 − 9. Thus c1 − 9 ≤ 12 c1 ≤ 12 Summarizing, 7/2 ≤ c1 ≤ 12. c2, basic: Basic cj

Variables

1/10

−1/20

7+Δ

−1/20

3/20

57 + 3Δ

7+Δ

11/20 − Δ/20

12/20 + 3Δ/20

−11/20 + Δ/20

−12/20 − 3Δ/20

cj − zj

A-37 .

Solving for the cj − zj inequalities: −11/20 + Δ/20 ≤ 0 Δ/20 ≤ 11/20 Δ ≤ 11 Since c2 = 7 + Δ; Δ = c2 − 7. Thus c2 − 7 ≤ 11 c2 ≤ 18 −12/20 − 3Δ/20 ≤ 0 −3Δ/20 ≤ 12/20 Δ ≥ −4 Since c2 = 7 + Δ; Δ = c2 − 7. Thus c2 − 7 ≥ −4 c2 ≥ 3 Summarizing, 3 ≤ c2 ≤ 18. d) q1: x1: 4 + Δ/10 ≥ 0 x2: 3 − Δ/20 ≥ 0 Δ/10 ≥ −4 −Δ/20 ≥ −3 −Δ ≥ −80 −Δ ≥ −60 Δ ≥ −40 Δ ≤ 60 Therefore, −40 ≤ Δ ≤60. Since q1 = 60 + Δ Δ = q1 − 60 −40 ≤ q1 − 60 ≤ 60 20 ≤ q1 ≤ 120 q2: x2: 3 +3Δ/20 ≥ 0 x1: 4 − Δ/20 ≥ 0 −Δ/20 ≥ −4 3Δ/20 ≥ −3 −Δ ≥ −80 Δ ≥ −20 Δ ≤ 80 Therefore, −20 ≤ Δ ≤ 80. Since q2 = 40 + Δ Δ = q2 − 40 −20 ≤ q2 − 40 ≤ 80 20 ≤ q2 ≤ 120 e) The marginal value of 1 hr of process 1 production time is $.55. The sensitivity range for q1, production hours, is 20 ≤ q1 ≤ 120. Thus, the company would purchase up to 120 hr at the marginal value price. 45. a) Minimize Zd = 180y1 + 135y2 subject to 2y1 + 3y2 ≥ 200 5y1 + 3y2 ≥ 300 y1, y2 ≥ 0

b) y1 = $33.33 = the marginal value of an additional hr of labor; y2 = $44.44 = the marginal value of an additional bd. ft. of wood c)

Point A must become the optimal solution for x1 = 0; therefore the slope of the objective function must be less than the slope of the constraint for labor, −2/5. Solving the following for the profit, p, of coffee tables gives −200/p = −2/5 p = $500 Thus, if the profit for coffee tables is greater than $500, no end tables will be produced. The new optimal solution will be x1 = 0 and x2 = 36. d)

The constraint line for wood moves outward, creating a new solution space, and the optimal solution point shifts from point B to point B´ where x1 = 31.67 and x2 = 23.33.

A-38 .

c1, basic: 200 + Δ

300

Quantity

Basic cj

Variables

300

1/3

−2/9

200 + Δ

−1/3

5/9

12,000 + 15Δ

200 + Δ

300

100/3 − Δ/3

400/9 + 5Δ/9

−100/3 + Δ/3

−400/9 − 5Δ/9

cj − zj Solving for the cj − zj inequalities: −100/3 + Δ/3 ≤ 0 Δ/3 ≤ 100/3 Δ ≤ 100 Since c1 = 200 + Δ; Δ = c1 − 200. Thus c1 − 200 ≤ 100 c1 ≤ 300 −400/9 − 5Δ/9 ≤ 0 −5Δ/9 ≤ 400/9 −Δ ≤ 80 Δ ≥ −80 = 200 + Δ; Δ = c1 − 200. Thus Since c1 c1 − 200 ≥ −80 c1 ≥ 120 Summarizing, 120 ≤ c1 ≤ 300.

A-39 .

c2, basic: 200

300 + Δ

Quantity

Basic cj

Variables

300 + Δ

1/3

−2/9

200

−1/3

5/9

12,000 + 300Δ

200

300 + Δ

100/3 + Δ/3

400/9 − 2Δ/9

−100/3 − Δ/3

−400/9 + 2Δ/9

cj − zj

Solving for the cj − zj inequalities: −100/3 − Δ/3 ≤ 0 −Δ/3 ≤ 100/3 −Δ ≤ 100 Δ ≥ −100 Since c2 = 300 + Δ; Δ = c2 − 300. Thus c2 − 300 ≥ −100 c2 ≥ 200 −400/9 + 2Δ/9 ≤ 0 2Δ/9 ≤ 400/9 Δ ≤ 200 Since c2 = 300 + Δ; Δ = c2 − 300. Thus c2 − 300 ≤ 200 c2 ≤ 500 Summarizing, 200 ≤ c2 ≤ 500. q1: x1: 15 − Δ/3 ≥ 0 x2: 30 + Δ/3 ≥ 0 Δ/3 ≥ −30 −Δ/3 ≥ −15 Δ ≥ −90 −Δ ≤ −45 Δ ≤ 45 Therefore, −90 ≤ Δ ≤ 45. Since q1 = 180 + Δ Δ = q1 − 180 −90 ≤ q1 − 180 ≤ 45 90 ≤ q1 ≤ 225

q 2: x1: 15 + 5Δ/9 ≥ 0 x2: 30 − 2Δ/9 ≥ 0 −2Δ/9 ≥ −30 5Δ/9 ≥ −15 −Δ ≥ −135 Δ ≥ −27 Δ ≤ 135 Therefore, −27 ≤ Δ ≤ 135. Since q2 = 135 + Δ Δ = q2 − 135 −27 ≤ q2 − 135 ≤ 135 108 ≤ q2 ≤ 270 g) The marginal value of 1 lb of wood is $44.44. From part f, the sensitivity range for q2, wood, is 108 ≤ q2 ≤ 270. Thus, the company would purchase up to 270 bd. ft. of wood at the marginal value price. h) The marginal value of labor is $33.33 and the marginal value of wood is $44.44; thus, wood should be purchased. 46. a) Minimize Zd = 19y1 + 14y2 + 20y3 subject to 2y1 + y2 + y3 ≥ 70 y1 + y2 + 2y3 ≥ 80 y1, y2, y3 ≥ 0 b) y1 =$20 = the marginal value of an additional hr of production time; y2 =$0 = the marginal value of an additional lb of steel; y3 = $30 = the marginal value of an additional ft of wire

A-40 .

c1, basic: 70 + Δ

Quantity

Basic cj

Variables

70 + Δ

2/3

−1/3

2/3

980 + 6Δ

70 + Δ

20 + 2Δ/3

30 − Δ/3

−20 − 2Δ/3

−30 + Δ/3

80 + Δ

Quantity

cj − zj Solving for the cj − zj inequalities: −20 − 2Δ/3 ≤ 0 −2Δ/3 ≤ 20 −Δ ≤ 30 Δ ≥ −30 Since c1 = 70 + Δ; Δ = c1 − 70. Thus c1 − 70 ≥ −30 c1 ≥ 40 −30 + Δ/3 ≤ 0 Δ/3 ≤ 30 Δ ≤ 90 Since c1 = 70 + Δ; Δ = c1 − 70. Thus c1 − 70 ≤ 90 c1 ≤ 160 Summarizing, 40 ≤ c1 ≤ 160. c2, basic: Basic cj

Variables

2/3

−1/3

80 + Δ

−1/3

2/3

980 + 7Δ

80 + Δ

20 − Δ/3

30 + 2Δ/3

−20 + Δ/3

−30 − 2Δ/3

cj − zj Solving for the cj − zj inequalities: −20 + Δ/3 ≤ 0 Δ/3 ≤ 20 Δ ≥ 60

A-41 .

Since c2 = 80 + Δ; Δ = c2 − 80. Thus c2 − 80 ≤ 60 c2 ≤ 140 −30 − 2Δ/3 ≤ 0 −2Δ/3 ≤ 30 −Δ ≤ 45 Δ ≥ −45 Since c2 = 80 + Δ; Δ = c2 − 80. Thus c2 − 80 ≥ −45 c2 ≥ 35 Summarizing, 35 ≤ c2 ≤ 140. d) q1: x1:

6 + 2Δ/3 ≥ 0 2Δ/3 ≥ −6 Δ ≥ −9

x2:

7 − Δ/3 ≥ 0 −Δ/3 ≥ −7 −Δ ≥ −21 Δ ≤ 21

s2:

1 − Δ/3 ≥ 0 −Δ/3 ≥ −1 −Δ ≥ −3 Δ≤3

Therefore, −9 ≤ Δ ≤ 3. Since q1 = 19 + Δ Δ = q1 − 19 −9 ≤ q1 − 19 ≤ 3 10 ≤ q1 ≤ 22 q2 x1:

6 + 0Δ ≥ 0

x2:

7 + 0Δ ≥ 0

s2:

1+Δ≥0 Δ ≥ −1

x2:

7 + 2Δ/3 ≥ 0 2Δ/3 ≥ −7 Δ ≥ −21/2

s2:

1 − Δ/3 ≥ 0 −Δ/3 ≥ −1 −Δ ≥ −3 Δ≤3

Therefore, Δ ≥ −1. Since q2 = 14 + Δ Δ = q2 − 14 q2 − 14 ≥ −1 q2 ≥ 13 q3: x1:

6 − Δ/3 ≥ 0 −Δ/3 ≥ −6 −Δ ≥ −18 Δ ≤ 18

Therefore, −21/2 ≤ Δ ≤ 3. Since q3 = 20 + Δ −21/2 ≤ q3 − 20 ≤ 3 Δ = q3 − 20 19/2 ≤ q3 ≤ 23 The sensitivity range for production hours is 10 ≤ q1 ≤ 22. Since 25 hr exceeds the upper limit of the range, it would change the optimal solution.

47. a)

A-42 .

Minimize Zd = 64y1 + 50y2 + 120y3 + 7y4 + 7y5 subject to 4y1 + 5y2 + 15y3 + y4 ≥ 9 8y1 + 5y2 + 8y3 + y5 ≥ 12 y1, y2, y3, y4, y5 ≥ 0

b) y1 = $.75 = the marginal value of one additional hr of labor for process 1; y2 = $1.20 = the marginal value of one additional hr of labor for process 2; y3, y4, y5 = $0; these resources have no value since there were units available which were not used. c) c1, basic: Solving for the cj − zj inequalities: −3/4 + Δ/4 ≤ 0 Δ/4 ≤ 3/4 Δ≤3

−6/5 − 2Δ/5 ≤ 0 −2Δ/5 ≤ 6/5 Δ ≥ −3

Since c1 = 9 + Δ; Δ = c1 − 9. Thus −3 ≤ Δ ≤ 3 −3 ≤ c1 − 9 ≤ 3 6 ≤ c1 ≤ 12 c2, basic: Solving for the cj − zj inequalities: −3/4 + Δ/4 ≤ 0 −Δ/4 ≤ 3/4 Δ ≥ −3

−6/5 + Δ/5 ≤ 0 Δ/5 ≤ 6/5 Δ≤6

Since c2 = 12 + Δ; Δ = c1 − 12. Thus −3 ≤ c2 − 12 ≤ 6 9 ≤ c2 ≤ 18 d) q1: x1:

4 − Δ/4 ≥ 0 −Δ/4 ≥ −4 Δ ≤ 16

s5:

1 − Δ/4 ≥ 0 −Δ/4 ≥ −1 Δ≤4

s 4:

3 + Δ/4 ≥ 0 Δ/4 ≥ −3 Δ ≥ −12

x2:

6 + Δ/4 ≥ 0 Δ/4 ≥ −6 Δ ≥ −24

s3:

Summarizing, −24 < −12 < −6.86 ≤ Δ ≤ 4 ≤ 16 and, therefore, −6.86 ≤ Δ ≤ 4 Since q1 = 64 + Δ; Δ = q1 − 64. Therefore, −6.86 ≤ Δ q1 − 64 ≤ 4 57.14 ≤ q1 ≤ 68

A-43 .

12 + 7Δ/4 ≥ 0 7Δ/4 ≥ −12 Δ ≥ −6.86

q3: x1:

4 + 0Δ ≥ 0 0Δ ≥ −4 Δ≤∞

s 5:

1 + 0Δ ≥ 0 Δ≤∞

s 4:

3 + 0Δ ≥ 0 Δ≤∞

x2:

6 + 0Δ ≥ 0 Δ≤∞

s3:

12 + Δ ≥ 0 Δ ≥ −12

−12 ≤ Δ ≤ ∞ Since q3 = 120 + Δ; Δ = q3 − 120. Therefore, −12 ≤ Δ ≤ ∞ −12 ≤ q3 − 120 ≤ ∞ 108 ≤ q3 ≤ ∞ Since 100 pounds is less than the lower limit of the range, the optimal solution mix will change. s2 enters the solution and s3 leaves. The new solution is, x1 = 3.27 s2 = 1.82 s5 = 0.64 s4 = 3.73 x2 = 6.36 Z = 105.82 48. a) Minimize Zd = 120y1 + 160y2 + 100y3 + 40y4 subject to 2y1 + 4y2 + 3y3 + y4 ≥ 40 3y1 + 3y2 + 2y3 + y4 ≥ 35 2y1 + y2 + 4y3 + y4 ≥ 45 y1, y2, y3, y4 ≥ 0 b) y1, y2 = 0; y3 = $5 = the marginal value of 1 hr of operation 3 time; y4 = $25 = the marginal value of 1 ft2 of storage space c) It does not have an effect. In the alternate solution the dual values remain the same, i.e., y3 = $5 and y4 = $25. d) c2, basic: Basic

35 + Δ

Variables

Quantity

−1/2

1/2

−4

−5

1/2

−1

35 + Δ

1/2

−1/2

1,500 + 35Δ

40 + Δ/2

35 + Δ

5 − Δ/2

25 + 2Δ

−Δ/2

−5 + Δ/2

−25 − 2Δ

cj − zj

A-44 .

Solving for the cj − zj inequalities: −Δ/2 ≤ 0 Δ≥0 Since c2 = 35 + Δ; Δ = c2 − 35. Thus c2 − 35 ≥ 0 c2 ≥ 35 −5 + Δ/2 ≤ 0 Δ/2 ≤ 5 Δ ≤ 10 Since c2 = 35 + Δ; Δ = c2 − 35. Thus c2 − 35 ≤ 10 c2 ≤ 45 −25 − 2Δ ≤ 0 −2Δ ≤ 25 Δ ≥ −12.5 Since c2 = 35 + Δ; Δ = c2 − 35. Thus c2 − 35 ≥ −12.5 c2 ≥ 22.5 Summarizing, 35 ≤ c2 ≤ 45. e)

x3:

10 − 4Δ ≥ 0 −4Δ ≥ −10 Δ ≤ 5/2

s2:

30 + 2Δ ≥ 0 2Δ ≥ −30 Δ ≤ −15

Therefore, −15 ≤ Δ ≤ 5/2. Since q4 = 40 + Δ Δ = q4 − 40 −15 ≤ q4 − 40 ≤ 5/2 25 ≤ q4 ≤ 42.5 f) The marginal value of 1 ft2 of storage is $25. From part e, the sensitivity range for q4 is 25 ≤ q4 ≤ 42.5. Thus, the company would purchase up to 42.5 ft2 of storage space at the marginal value price. 49. a) Maximize Zd = 20y1 + 30y2 + 12y3 subject to 4y1 + 12y2 + 3y3 ≤ .03 5y1 + 3y2 + 2y3 ≤ .02 y1, y2, y3 ≥ 0 b) y1 = $0 = marginal value of 1 mg of protein; y2 = $0 = marginal value of 1 mg of iron; y3 = $.01 = marginal value of 1 mg of carbohydrate (i.e., if one less mg of carbohydrate was required, it would be worth $.01 to the dietitian)

q 4: s 1:

10 − Δ ≥ 10 − Δ ≥ −10 Δ ≤ 10

60 − 5Δ ≥ 0 −5Δ ≥ −60 Δ ≤ 12

A-45 .

c) c1, basic: .03+Δ

.02

Quantity

Basic cj

Variables

.02

3.6

.20

−.80

.03 + Δ

1.6

−.13

.20

4.4

.47

−3.2

.12 + 1.6Δ

.03 + Δ

.02

0 − .133Δ

−.01 + .2Δ

0 − .133Δ

−.01 + .2Δ

.03

.02 + Δ

zj − cj Solving for the zj − cj inequalities: 0 − .133Δ ≤ 0 −.133Δ ≤ 0 −Δ ≤ 0 Δ≥0 Since c1 = .03 + Δ; Δ = c1 − .03. Thus c1 − .03 ≥ 0 c1 ≥ .03 −.01 + .2Δ ≤ 0 .2Δ ≤ .01 Δ ≤ .05 Since c1 = .03 + Δ; Δ = c1 − .03. Thus c1 − .03 ≤ .05 c1 ≤ .08 Summarizing, .03 ≤ c1 ≤ .08. d) When determining sensitivity ranges for qi values in a minimization problem, since artificial variables are eliminated, the surplus variable column coefficients must be used. This corresponds to a qi − Δ change. c2, basic: Basic cj

Variables

Quantity

.02 + Δ

3.6

.20

−.80

.03

1.6

−.13

.20

4.4

.47

−3.2

.12 + 3.6Δ

.03

.02 + Δ

0 + .2Δ

−.01 + .8Δ

0 + .2Δ

−.01 + .8Δ

zj − cj

A-46 .

Solving for the zj − cj inequalities: 0 + .2Δ ≤ 0 .2Δ ≤ 0 Δ≤0 Since c2 = .02 + Δ; Δ = c1 − .02. Thus c2 − .02 ≤ 0 c2 ≤ .02 −.01 + .8Δ ≤ 0 .8Δ ≤ .01 Δ ≤ .0125 Since c2 = .02 + Δ; Δ = c1 − .02. Thus c2 − .02 ≤ .0125 c2 ≥ .0075 Summarizing, c2 ≤ .0375. q1: x2:

3.6 + 0Δ ≥ 0

x1:

s1:

1.6 + 0Δ ≥ 0

4.4 +Δ ≥ 0 Δ ≥ −4.4

Therefore, Δ ≥ −4.4. Since q1 = 20 − Δ Δ = 20 − q1 20 − q1 ≥ −4.4 q1 ≤ 24.4 q2: x2:

3.6 + .2Δ ≥ 0 .2Δ ≥ −3.6 Δ ≥ −18

s 1:

4.4 + .47Δ ≥ 0 .47Δ ≥ −4.4 Δ ≥ −9.36

1.6 − .133Δ ≥ 0 .133Δ ≥ −1.6 −Δ ≥ −12 Δ ≤ 12

x1:

Therefore, −9.36 ≤ Δ ≤ 12. Since q2 = 30 − Δ Δ = 30 − q2 −9.36 ≤ 30 − q2 ≤ 12 18 ≤ q2 ≤ 39.36 q3: x2:

3.6 − .8Δ ≥ 0 .8Δ ≥ − 3.6 −Δ ≥ − 4.5 Δ ≥ 4.5

s 1:

4.4 − 3.2Δ ≥ 0 − 3.2Δ ≥ −4.4 − Δ ≥ −1.375 Δ ≥ −1.375

x1:

1.6 + .2Δ ≥ 0 .2Δ ≥ 1.6 Δ ≥ −8

A-47 .

Therefore, −8 ≤ Δ ≤ 1.375. Since q3 = 12 − Δ Δ = 12 − q3 −8 ≤ 12 − q3 ≤ 1.375 10.625 ≤ q3 ≤ 20 e) The marginal value of 1 mg of carbohydrates is $.01. From part d, the sensitivity range for q3, carbohydrates, is 10.625 ≤ q3 ≤ 20. Thus, the dietitian could lower the requirements for carbohydrates to 10.625 at the marginal value without the solution becoming infeasible. 50. a) Minimize Zd = 1,200y1 + 500y3 subject to .50y1 + y2 ≥ 1.25 1.2y1 − y2 + y3 ≥ 2.00 .80y1 + y2 ≥ 1.75 y1, y2, y3 ≥ 0 y1 = the marginal value of an additional hour of production time = $0 y2 = the marginal value of increasing the combined demand for cheese sandwiches by one sandwich = $1.75 y3 = the marginal value of producing an additional ham salad sandwich = $3.75 b) c1, non-basic: −5 + Δ ≤ 0 Δ ≤ .50 Since c1 = 1.25 + Δ; Δ = c1 − 1.25. Therefore, c1 − 1.25 ≤ .50 c1 ≤ 1.75 c2, basic: −3.75 − Δ ≤ 0 −Δ ≤ 3.75 Δ ≥ −3.75 Since c2 = 2 + Δ; Δ = c2 − 2. Therefore, c2 −2 ≥ −3.75 c2 ≥ −1.75 c3, basic: −.5 − Δ ≤ 0 −3.75 − Δ ≤ 0 −Δ ≤ .5 −Δ ≤ 3.75 Δ ≥ −.5 Δ ≥ −3.75 Summarizing, −3.75 ≤ −.5 ≤ Δ

and, therefore, −.5 ≤ Δ Since c3 = 1.75 + Δ; Δ = c3 − 1.75. Therefore, −.5 ≤ Δ −.5 ≤ c3 − 1.75 1.25 ≤ c3 q 3:

s1: 200 − 2Δ ≥ 0 − 2Δ ≥ −200

x3 : 500 + Δ ≥ 0

x2 : 500 + Δ ≥ 0 Δ ≥ −500

Δ ≥ −500

Δ ≤ 100

Summarizing, −500 ≤ Δ ≤ 100. Since q3 = 500 + Δ, Δ = q3 − 500. Therefore, −500 ≤ q3 − 500 ≤ 100 0 ≤ q3 ≤ 600 d) The marginal value for demand for cheese sandwiches is $1.75. The range for q2 is computed as follows. s1: 200 − .8Δ ≥ 0 − .8Δ ≥ −200 Δ ≤ 250

x3 : 500 + Δ ≥ 0

x2 : 500 + 0Δ ≥ 0

Δ ≥ −500

0Δ ≥ −500 Δ≤∞

Summarizing, −500 ≤ Δ ≤ 250 Since q2 = 0 + Δ, Δ = q2. Therefore, −500 ≤ q2 ≤ 250 Thus, the demand for cheese sandwiches can be increased up to a maximum of 250 sandwiches. The additional profit for 200 more cheese sandwiches would be, ($1.75) (200) = $350 Since the cost of advertising is $100, a $250 profit would result; therefore the company should advertise. 51. a) c3, nonbasic: −13 − Δ ≤ 0 −Δ ≤ 13 Δ ≥ −13 Since c3 = 2 + Δ, Δ = c3 − 2. And c3 − 2 ≥ −13 c3 ≥ −11

A-48 .

c1, basic: 3+Δ

Quantity

−4

−3/2

−1/2

3+Δ

−2

−1/2

1/2

−1

−1/2

165 + 5Δ

3+Δ

−11 − 2Δ

−4 − Δ/2

−1 + Δ/2

−13 − 2Δ

−4 − Δ/2

−1 + Δ/2

Basic Variables

zj − cj Solving for the zj − cj inequalities: −13 − 2Δ ≤ 0 −2Δ ≤ 13 Δ ≥ −13/2 Since c1 = 3 + Δ, Δ = c1 − 3. Thus c1 − 3 ≥ −13/2 c1 ≥ −7/2 −4 − 2Δ ≤ 0 −2Δ ≤ 4 Δ ≥ −2 Since c1 = 3 + Δ, Δ = c1 − 3. Thus c1 − 3 ≥ −2 c1 ≥ 1 −1 + Δ/2 ≤ 0 Δ/2 ≤ 1 Δ≤2 Since c1 = 3 + Δ, Δ = c1 − 3. Thus c1 − 3 ≤ 2 c1 ≤ 5 Summarizing, 1 ≤ c1 ≤ 5. c2: basic: Basic Variables

5+Δ

Quantity

−4

−3/2

−1/2

−2

−1/2

1/2

5+Δ

−1

−1/2

165 + 30Δ

5+Δ

−11 − 2Δ

−4 − Δ/2

−1 − Δ/2

−13 − 2Δ

−4 − Δ/2

−1 − Δ/2

zj − cj Solving for the zj − cj inequalities: −13 − 2Δ ≤ 0 −2Δ ≤ 13 Δ ≥ −13/2

A-49 .

Since c2 = 5 + Δ, Δ = c2 − 5. Thus c2 − 5 ≥ −13/2 c2 ≥ −3/2 −4 − Δ/2 ≤ 0 −Δ/2 ≤ 4 Δ ≥ −8 Since c2 = 5 + Δ, Δ = c2 − 5. Thus c2 − 3 ≥ −8 c2 ≥ −3 −1 − Δ/2 ≤ 0 −Δ/2 ≤ 1 Δ ≥ −2 Since c2 = 5 + Δ, Δ = c2 − 5. Thus c2 − 5 ≥ −2 c2 ≥ 3 Summarizing, c2 > 3. b) When determining sensitivity ranges for qi values, since artificial values are eliminated, the surplus variable column coefficients must be used. This corresponds to a qi − Δ change. q 1: s2:

15 − 3Δ/2 ≥ 0 −3Δ/2 ≥ −15

x1 :

Δ ≤ 10

5 − Δ/2 ≥ 0

x2 : 30 − Δ/2 ≥ 0

− Δ/2 ≥ −5 Δ ≤ 10

− Δ/2 ≥ −30 Δ ≤ 60

5 − 0Δ ≥ 0

x2 : 30 − 0Δ ≥ 0

5 + Δ/2 ≥ 0

x2 : 30 − Δ/2 ≥ 0

Therefore, Δ ≤ 10. Since q1 = 35 − Δ Δ = 35 − q1 35 − q1 ≤ 10 q1 ≥ 25 q 2: s 2:

15 + Δ ≥ 0 Δ ≥ −15

x1 :

Therefore, Δ ≤ −15. Since q2 = 50 − Δ Δ = 50 − q2 50 − q2 ≤ −15 q2 ≤ 65 q 3: s 2:

15 − Δ/2 ≥ 0 −Δ/2 ≥ −15 Δ ≤ 30

x1 :

Δ/2 ≥ −5 Δ ≥ −10

Therefore, −10 ≤ Δ ≤ 30. Since q3 = 25 − Δ and Δ = 25 − q3 −10 ≤ 25 − q3 ≤ 30 −5 ≤ q3 ≤ 35

A-50 .

− Δ/2 ≥ −30 Δ ≤ 60

52. a) y1 = 7/15 = $.467 Range for q1: x1 : 8 + Δ /15 ≥ 0 Δ /15 ≥ −8 Δ ≥ −120 x3 : 3 − Δ /40 ≥ 0

Since q5 = 40 + Δ, Δ = q5 − 40 q5 − 40 ≥ −8 q5 ≥ 32 No, increasing q5 from 40 to 50 will have no effect on the optimal solution. e) Since y1 = 7/15 = $.467, pears should be secured. 53. a) y2 = $0, spruce has no marginal value q 2: s2: 70 + Δ ≥ 0 Δ ≥ −70 Since q2 = 160 + Δ, Δ = q2 − 160 q2 − 160 ≥ −70 q2 ≥ 90 b) y3 = $2, marginal value of cutting hours q 3: s2: 70 + Δ/3 ≥ 0 s1: 80 − 4Δ/3 ≥ 0 −4Δ/3 ≥ −80 Δ/3 ≥ −70 Δ ≤ 60 Δ ≥ −210 x3: 20 + 2Δ/3 ≥ 0 x2: 10 − Δ/3 ≥ 0 2Δ/3 ≥ −20 −Δ/3 ≥ −10 Δ ≥ −30 Δ ≤ 30 Therefore, −30 ≤ Δ ≤ 30. Since q3 = 50 + Δ, Δ = q3 − 50 20 ≤ q3 ≤ 80 c) y3 = $2, cutting hours; y4 = $2, pressing hours. Since they both have the same marginal value, management could choose either. d) From part a, q2 ≥ 90; thus, a decrease from 160 to 100 lb of spruce will not affect the solution. e) Compute the range for c1, a nonbasic cj value. c1 = 4 + Δ −2 + Δ ≤ 0 Δ≤2 c1 − 4 ≤ 2 c1 ≤ 6 The unit profit from Western paneling would have to be $6 or more before it would be produced.

x2 : 16 − Δ /30 ≥ 0 − Δ /30 ≥ −16 Δ ≤ 480

Δ ≥ 120 s4 : 36 − Δ /30 ≥ 0 − Δ /30 ≥ −36 Δ ≤ 1,080

s5 : 8 − Δ /10 ≥ 0 − Δ /10 ≥ −8 Δ ≤ 80 − 120 ≤ Δ ≤ 80

Since q1 = 320 + Δ, Δ = q1 − 320 −120 ≥ q1 − 320 ≤ 80 200 ≤ q1 ≤ 400 As many as 400 pears can be purchased. b) y2 = 1/15 = $.067 Range for q2: x1: 8 + Δ/30 ≥ 0 x2: 16 + Δ/15 ≥ 0 Δ/30 ≥ −8 Δ/15 ≥ −16 Δ ≥ −240 Δ ≥ −240 s4: 36 − Δ/30 ≥ 0 x3: 3 − Δ/10 ≥ 0 −Δ/10 ≥ −3 −Δ/30 ≥ −36 Δ ≤ 30 Δ ≤ 1,080 −240 ≤ Δ ≤ 30 Since q2 = 400 + Δ, Δ = q2 − 400 −240 ≤ q2 − 400 ≤ 30 160 ≤ q2 ≤ 430 Range over which the value of peaches is valid c) Range for q3: s3 : 3 + Δ ≥ 0 Δ ≥ −3 Since q3 = 43 + Δ, Δ = q3 − 43 q3 − 43 ≥ −3 q3 ≥ 40 No, increasing q3 from 43 to 60 will not affect the optimal solution. d) Range for q5: s5: 8 + Δ ≥ 0 Δ ≥ −8

A-51 .

Compute the range for c3. c3 = 8 + Δ −2 − Δ/3 ≤ 0 −2 − 2Δ/3 ≤ 0 −2 +Δ/6 ≤ 0 −Δ/3 ≤ 2 −2Δ/3 ≤ 2 Δ/6 ≤ 2 Δ ≥ −6 Δ ≥ −3 Δ ≤ 12 Since Δ = c3 − 8, c3 − 8 ≥ −6 c3 − 8 ≥ −3 c3 − 8 ≤ 12 c3 ≥ 5 c3 ≤ c3 ≥ 2 20 Summarizing, 5 ≤ c3 ≤ 20. If the unit profit of Colonial paneling is increased to $13, the percent solution would not be affected.

54. y1 = $1.33 q 1: x4: 80 + 2Δ/3 ≥ 0 x2: 40 − Δ/3 ≥ 0 2Δ/3 ≥ −80 −Δ/3 ≥ −40 Δ ≥ −120 Δ ≤ 120 −120 ≤ Δ ≤ 120 Since q1 = 200 + Δ, Δ = q1 − 200 −120 ≤ q1 − 200 ≤ 120 80 ≤ q1 ≤ 320

A-52 .

Module B: Transportation and Assignment Solution Methods 32. Unbalanced transportation 33. Sensitivity analysis (B–32) 34. Shortage costs 35. Multiperiod scheduling 36. Unbalanced assignment, LP formulation 37. Assignment 38. Assignment 39. Assignment 40. Unbalanced assignment, multiple optimal 41. Assignment, multiple optimal 42. Assignment 43. Unbalanced assignment, multiple optimal 44. Balanced assignment 45. Balanced assignment (B–9) 46. Unbalanced assignment 47. Unbalanced assignment 48. Unbalanced assignment 49. Unbalanced assignment, prohibited assignment 50. Unbalanced assignment, multiple optimal 51. Assignment 52. Unbalanced assignment (maximization)

PROBLEM SUMMARY 1. Balanced transportation 2. Balanced transportation 3. Short answer, discussion 4. Unbalanced transportation 5. Balanced transportation 6. Unbalanced transportation 7. Unbalanced transportation 8. Unbalanced transportation 9. Unbalanced transportation, multiple optimal 10. Sensitivity analysis (B–9) 11. Unbalanced transportation, multiple optimal 12. Unbalanced transportation, degenerate 13. Unbalanced transportation, degenerate 14. Balanced transportation 15. Balanced transportation 16. Sensitivity analysis (B–15) 17. Unbalanced transportation, multiple optimal 18. Sensitivity analysis (B–17) 19. Shortage costs (B–17) 20. Unbalanced transportation 21. Unbalanced transportation, multiple optimal 22. Balanced transportation 23. Unbalanced transportation, multiple optimal 24. Sensitivity analysis (B–23) 25. Unbalanced transportation 26. Sensitivity analysis (B–25) 27. Sensitivity analysis (B–25) 28. Unbalanced transportation 29. Unbalanced transportation, degenerate 30. Unbalanced transportation 31. Unbalanced transportation, production scheduling (B–30)

PROBLEM SOLUTIONS 1.

B-1 .

Using the VAM initial solutions:

B-2 .

b) MODI Solution:

cij = ui + vj (occupied cells)

cij − ui − vj = k (empty cells)

u1 = 0 x12: u1 + v2 = 750, v2 = 750

x11: = 500 − 0 − 450 = +50

x14: u1 + v4 = 450, v4 = 450

x13: = 300 − 0 − 350 = −50*

x22: u2 + v2 = 800, u2 = 50

x21: = 650 − 50 − 450 = +150

x23: u2 + v3 = 400, v3 = 350

x24: = 600 − 50 − 450 = +100

x31: u3 + v1 = 400, v1 = 450

x33: = 500 − (−50) − 350 = +200

x32: u3 + v2 = 700, u3 = −50

x34: = 550 − (−50) − 450 = +150

Allocate 2 units to cell x13.

cij = ui + vj (occupied cells)

cij − ui − vj = k (empty cells)

u1 = 0 x13: u1 + v3 = 300, v3 = 300

x11: = 500 − 0 − 400 = +100

x14: u1 + v4 = 450, v4 = 450

x12: = 750 − 0 − 700 = +50

x22: u2 + v2 = 800, v2 = 700

x21: = 650 − 100 − 400 = +150

x23: u2 + v3 = 400, u2 = 100

x24: = 600 − 100 − 450 = +50

x31: u3 + v1 = 400, v1 = 400

x33: = 500 − 0 − 300 = +200

x32: u3 + v2 = 700, u3 = 0

x34: = 550 − 0 − 450 = +100

All positive; therefore optimal solution; TC = $20,200

B-3 .

3. a) Unbalanced; demand exceeds supply by 200 units, as indicated by the dummy row. b) Yes; there should be m + n − 1 = 7 occupied cells, and there are only 6 occupied cells. Place a dummy 0 in cell x4A (as one possible choice). c) Yes, cell x3C d) $13,700 e) x2B = 0 4. Initial solution using VAM:

VAM; TC = $2,480; this solution is not optimal. However, there were several penalty cost ties; thus, a different VAM solution could have occurred.

B-4 .

6. a)

b) Minimize Z = 6xA1 + 9xA2 + MxA3 + 12xB1 + 3xB2 + 5xB3 + 4xC1 + 8xC2 + 11xC3 subject to xA1 + xA2 + xA3 ≤ 130 xB1 + xB2 + xB3 ≤ 70 xC1 + xC2 + xC3 ≤ 100 xA1 + xB1 + xC1 = 80 xA2 + xB2 + xC2 = 110 xA3 + xB3 + xC3 = 60 xij ≥ 0 7.

B-5 .

8. a) Minimum cell cost; TC = $1,390

b) Solutions achieved using minimum cost cell method are optimal 9. a) The initial solution should be found using VAM, since it is the most efficient.

b) MODI solution:

B-6 .

cij = ui + vj

cij − ui − vj = k

uA = 0 xA2: uA + v2 = 9, v2 = 9

xA1: 14 − 0 − 13 = +1

xA4: uA + v4 = 18, v4 = 18

xA3: 16 − 0 − 7 = +9

xB1: uB + v1 = 11, v1 = 13

xA5: 0 − 0 − (−3) = +3

xB4: uB + v4 = 16, uB = −2

xB2: 8 − (−2) − 9 = +1

xC1: uC + v1 = 16, uC = 3

xB3: M − (−2) − 7 = M − 5

xC3: uC + v3 = 10, v3 = 7

xB5: 0 − (−2) − (−3) = +5

xC5: uC + v5 = 0, v5 = −3

xC2: 12 − 3 − 9 = 0 xC4: 22 − 3 − 18 = +1

This solution is optimal; TC = $8,260. c) Yes, there is a multiple optimum solution at cell xC2, since k = 0 for this cell. Allocate 70 units to cell xC2 for alternate; TC = $8,260.

d) Minimize Z = 14xA1 + 9xA2 + 16xA3 + 18xA4 + 11xB1 + 8xB2 + MxB3 + 16xB4 + 16xC1 + 12xC2 + 10xC3 + 22xC4 subject to xA1 + xA2 + xA3 + xA4 ≤ 150 xB1 + xB2 + xB3 + xB4 ≤ 210 xC1 + xC2 + xC3 + xC4 ≤ 320 xA1 + xB1 + xC1 = 130 xA2 + xB2 + xC2 = 70 xA3 + xB3 + xC3 = 180 xA4 + xB4 + xC4 = 240 xij ≥ 0 10.

There is no effect. The Gary mill has 60 tons left over as surplus with the current solution to Problem 11. Reducing the capacity at Gary by 30 still leaves a surplus of 30 tons.

B-7 .

11. a)

b) MODI solution:

cij = ui + vj

cij − ui − vj = k

uA = 0 xA3: uA + v3 = 5, v3 = 5

xA1: M − 0 − 13 = M − 13

xB1: uB + v1 = 12, v1 = 13

xA2: 10 − 0 − 9 = +1

xB3: uB + v3 = 4, uB = −1

xB2: 9 − (−1) −9 = +1

xC2: uC + v2 = 3, uC = −6

xC1: 7 − (−6) − 13 = 0

xD1: uD + v1 = 9, uD = −4

xC3: 11 − (−6) − 5 = +12

xD2: uD + v2 = 5, v2 = 9

xD3: 7 − (−4) − 5 = +6

xE1: uE + v1 = 0, uE = −13

xE2: 0 − (−13) − 9 = +4 xE3: 0 − (−13) − 5 = +8

Solution is optimal; TC = $1,590.

B-8 .

c) Yes, there is a multiple optimum solution. Cell xC1 has k = 0. The alternate solution follows; TC = $1,590

d) Minimize Z = MxA1 + 10xA2 + 5xA3 + 12xB1 + 9xB2 + 4xB3 + 7xC1 + 3xC2 + 11xC3 + 9xD1 + 5xD2 + 7xD3 subject to xA1 + xA2 + xA3 = 90 xB1 + xB2 + xB3 = 50 xC1 + xC2 + xC3 = 80 xD1 + xD2 + xD3 = 60 xA1 + xB1 + xC1 + xD1 ≥ 120 xA2 + xB2 + xC2 + xD2 ≥ 100 xA3 + xB3 + xC3 + xD3 ≥ 110 xij ≥ 0 12. a)

B-9 .

b) The initial solution is degenerate. A 0 is arbitrarily allocated to x33.

cij = ui + vj

cij − ui − vj = k

u1 = 0 x11: u1 + v1 = 9, v1 = 9

x12: 14 − 0 − (−M + 22) = M − 8

x13: u1 + v3 = 12, v3 = 12

x14: 17 − 0 − 4 = +13

x22: u2 + v2 = 10, v2 = −M + 22

x15: 0 − 0 − (−3) = +3

x23: u2 + v3 = M, u2 = M − 12

x21: 11 − (M − 12) − 9 = −M + 14

x33: u3 + v3 = 15, u3 = 3

x24: 10 − (M − 12) − 4 = −M + 18

x34: u3 + v4 = 7, v4 = 4

x25: 0 − (M − 12) − (−3) = −M + 15

x35: u3 + v5 = 0, v5 = −3

x31: 12 − 3 − 9 = 0 x32: 8 − 3 − (−M + 22) = M − 17

Not optimal; allocate 30 units to x21.

B-10 .

cij = ui + vj

cij − ui − vj = k

u1 = 0 x11: u1 + v1 = 9, v1 = 9

x12: 14 − 0 − 9 = +5

x13: u1 + v3 = 12, v3 = 12

x14: 17 − 0 − 4 = +13

x21: u2 + v1 = 11, u2 = 2

x15: 0 − 0 − (−3) = +3

x22: u2 + v2 = 10, v2 = 8

x23: M − 2 − 12 = M − 14

x33: u3 + v3 = 15, u3 = 3

x24: 10 − 2 − 4 = +4

x34: u3 + v4 = 7, v4 = 4

x25: 0 − 2 − (−3) = +1

x35: u3 + v5 = 0, v5 = −3

x31: 12 − 3 − 9 = 0 x32: 8 − 3 − 8 = −3

Not optimal; allocate 0 to x32.

cij = ui + vj u1 = 0 x11: u1 + v1 = 9, v1 = 9 x13: u1 + v3 = 12, v3 = 12 x21: u2 + v1 = 11, u2 = 2 x22: u2 + v2 = 10, v2 = 8 x32: u3 + v2 = 8, u3 = 0 x34: u3 + v4 = 7, v4 = 7 x35: u3 + v5 = 0, v5 = 0

cij − ui − vj = k x12: 14 − 0 − 8 = +6 x14: 17 − 0 − 7 = +10 x15: 0 − 0 − 0 = 0 x23: M − 2 − 12 = M − 14 x24: 10 − 2 − 7 = +1 x25: 0 − 2 − 0 = −2 x31: 12 − 0 − 9 = +3 x33: 15 − 0 − 12 = +3

Not optimal; allocate 50 units to x25.

B-11 .

cij = ui + vj u1 = 0 x11: u1 + v1 = 9, v1 = 9 x13: u1 + v3 = 12, v3 = 12 x21: u2 + v1 = 11, u2 = 2 x22: u2 + v2 = 10, v2 = 8 x25: u2 + v5 = 0, v5 = −2 x32: u3 + v2 = 8, u3 = 0 x34: u3 + v4 = 7, v4 = 7

cij − ui − vj = k x12: 14 − 0 − 8 = +6 x14: 17 − 0 − 7 = +10 x15: 0 − 0 − (−2) = +2 x23: M − 2 − 12 = M − 14 x24: 10 − 2 − 7 = +1 x31: 12 − 0 − 9 = +3 x33: 15 − 0 − 12 = +3 x35: 0 − 0 − (−2) = +2

Optimal; TC = $5,080 c) No multiple optimum solutions, since none of the k values equal 0 d) Minimize Z = 9xTN + 14xTP + 12xTC + 17xTB +11xMN + 10xMP + MxMC + 10xMB + 12xFN + 8xFP + 15xFC + 7xFB subject to xTN + xTP + xTC + xTB ≤ 200 xMN + xMP + xMC + xMB ≤ 200 xFN + xFP + xFC + xFB ≤ 200 xTN + xMN + xFN = 130 xTP + xMP + xFP = 170 xTC + xMC + xFC = 100 xTB + xMB + xFB = 150 xij ≥ 0 13. a)

B-12 .

b) The initial solution is degenerate. 0 is allocated arbitrarily to x4D.

c) Minimize Z = 7x1A + 8x1B + 5x1C + 6x2A + Mx2B + 6x2C + 10x3A + 4x3B + 5x3C + 3x4A + 9x4B + Mx4C

B-13 .

subject to x1A + x1B + x1C ≤ 5 x2A + x2B + x2C ≤ 25 x3A + x3B + x3C ≤ 20 x4A + x4B + x4C ≤ 25 x1A + x2A + x3A + x4A = 10 x1B + x2B + x3B + x4B = 20 x1C + x2C + x3C + x4C = 15 xij ≥ 0 14. a) Minimize cell cost method

b) Stepping-stone method

15. a)

B-14 .

b) Stepping-stone method

16.

VAM; TC = $3,292.50; optimal; this is the same total cost as determined in Problem 15. Therefore, the reduction in shipping cost from Tampa to Kentucky had no effect. (This probably could have been deduced by comparing shipping costs to Kentucky from both Tampa and St. Louis. The shipping cost from St. Louis, $.20, is the lowest in the model and significantly lower than $.65.) 17. a)

B-15 .

cij = ui + vj

cij − ui − vj = k

cij = ui + vj

u1 = 0

cij − ui − vj = k

u1 = 0

x12: u1 + v2 = 17, v2 = 17

x11: 22 − 0 −15 = +7

x12: u1 + v2 = 17, v2 = 17

x11: 22 − 0 −15 = +7

x14: u1 + v4 = 18, v4 = 18

x13: 30 − 0 −20 = +10

x14: u1 + v4 = 18, v4 = 18

x13: 30 − 0 −20 = +10

x21: u2 + v1 = 15, v1 = 15

x22: 35 − 0 −17 = +18

x21: u2 + v1 = 15, v1 = 15

x22: 35 − 0 −17 = +18

x23: u2 + v3 = 20, u2 = 0

x24: 25 − 0 −18 = +7

x23: u2 + v3 = 20, u2 = 0

x24: 25 − 0 −18 = +7

x33: u3 + v3 = 16, v3 =20

x31: 28 − (−4) − 15 = +17

x33: u3 + v3 = 16, v3 = 20

x31: 28 − (−4) − 15 = +17

x34: u3 + v4 = 14, u3 = −4

x32: 21 − (−4) − 17 = +8

x34: u3 + v4 = 14, u3 = −4

x32: 21 − (−4) − 17 = +8

x42: u4 + v2 = 0, u4 = −17

x41: 0 − (−17) − 15 = +2

x42: u4 + v2 = 0, u4 = −20

x41: 0 − (−20) − 15 = +5

x43: 0 − (−17) − 20 = −3

x42: 0 − (−20) − 17 = +3

x44: 0 − (−17) − 18 = −1

x44: 0 − (−20) − 18 = +2

Not optimal; allocate 180 units to x43.

Optimal.

B-16 .

18.

VAM; TC = $24,930; optimal; select alternative 2: add a warehouse at Charlotte.

B-17 .

19.

Optimal, TC = $26,430 Transportation cost = $21,930 Shortage cost = $ 4,500 $26, 430

20.

TC = $823,000 21.

B-18 .

cij = ui + vj

cij − ui – vj = k

u1 = 0 x1B: u1 + vB = 8, vB = 8

x1A: 5 – 0 − 9 = −4

x2A: u2 + vA = 10, vA = 9

x1C: 6 − 0 – 11 = −5

x2B: u2 + vB = 9, u2 = 1

x1D: 0 − 0 – 2 = −2

x2C: u2 + vC = 12, vC = 11

x2D: 0 − 1 – 2 = −3

x3B: u3 + vB = 6, u3 = −2

x3A: 7 – (−2) – 9 = 0

x3D: u3 + vD = 0, vD = 2

x3C: 8 – (−2) – 11 = −1

Optimal (with multiple optimal at x3A); TC = $1,605 22.

B-19 .

23.

VAM initial solution:

Allocate 40 students to cell (South, C):

Optimal (with multiple optimal solutions) Total travel time = 20,700 minutes

B-20 .

24.

VAM initial solution:

Allocate 40 students to cell (South, C):

Allocate 100 students to cell (East, A):

B-21 .

Optimal. Total travel time = 21,200 minutes. The overall travel time increased by 500 minutes, which divided by all 1,400 students is only an increase of .357 minutes per student. This does not seem to be a significantly large increase. 25.

The initial solution is determined using VAM, as follows.

Optimal, Total Profit = $1,528 with multiple optimal solutions at (B, 2), (E, 4) and (B, Dummy) Alternate, allocate 13 cases to (B, 2)

Alternate, allocate 17 cases to (E, 4)

B-22 .

26.

If Easy Time purchased all the baby food demanded at each store from the distributor total profit would be $1,246, which is less than buying it from the other locations as determined in problem 25. This profit is computed by multiplying the profit at each store by the demand. In order to determine if some of the demand should be met by the distributor a new source (F) must be added to the transportation tableau from problem 25. This source represents the distributor and has an available supply of 150 cases, the total demand from all the stores. The tableau and optimal solution are shown as follows.

Optimal, Total profit = $1,545 with multiple optimal solutions This new solution results in a greater profit than the solution in problem 25 ($1,545 > $1,528). Thus, some

B-23 .

27.

of the demand (specifically at store 3) should be met from the distributors. Solve the model as a linear programming model to obtain the shadow prices. Among the 5 purchase locations, the store at Albany has the highest shadow price of $3. The sensitivity range for supply at Albany is 25 ≤ q1 ≤ 43. Thus, as much as 17 additional cases can be purchased from Albany which would increase profit by $51 for a total of $1,579.

28.

29.

VAM initial solution (degenerate):

B-24 .

Allocate “0” to cell (4,Dummy):

Allocate 3 units to cell (3,Dummy):

B-25 .

Allocate “0” to cell (2,B):

30.

Initial transportation tableau (cost = $100s) with VAM initial solution

B-26 .

31.

Optimal transportation tableau (costs = $100s). Multiple optimal solutions exist.

32.

Cost = $1,000s

Total cost, Z = $278,000 It is cheaper for National Foods to continue to operate its own trucking firm. 33. Increasing the supply at Sacramento, Jacksonville, and Ocala to 25 tons would have little effect, reducing the overall monthly shipping cost to $276,000, which is still higher than the $245,000 the company is currently spending with its own trucks.

B-27 .

Alternatively, increasing the supply at San Antonio and Montgomery to 25 tons per month reduces the monthly shipping cost to $242,500 which is less than the company’s cost with their own trucks. 34.

Z = $723,500 Penalty costs = 200 × $800 = $160,000 35.

B-28 .

*36. This is an assignment problem.

Optimal Solution: 1-C

2-A

3-B

4-D

9 37 Minutes

Solution: 1-1

2-4

3-2

4-5

5-3

38. a)

78 37.

B-29 .

39.

Optimal Solution: 1-B

$11

2-D

Optimal Solution:

3-C

1-B

4-A

2-D

$32

3-A

4-C

5-E

b) Minimize Z = 12x1A + 11x1B + 8x1C + 14x1D + 10x2A + 9x2B + 10x2C + 8x2D + 14x3A + Mx3B + 7x3C + 11x3D + 6x4A + 8x4B + 10x4C + 9x4D

51 days 40.

subject to 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 + x2B + x3B + x4B = 1 x1C + x2C + x3C + x4C = 1 x1D + x2D + x3D + x4D = 1 xij ≥ 0

B-30 .

Opportunity cost table:

Solutions:

Solutions: 1-B

1-E

2-E

2-A

3-A

3-B

4-C

5-D

6-F

0 $36

6-F

0 $36

1-C

1-D

2-A

3-B

4-D

2 $26

4-C

4 $26

42.

41.

B-31 .

Solutions:

Solution:

A-3

A-6

B-2

C-6

C-5

4-A

D-1

D-3

5-D

E-5

E-1

6-B

14 85 defects

F-4

3 14 miles

F-4

3 14 miles

1-C

2-F

3-E

43.

44. a) It actually can be solved by either method—transportation or assignment. However, if the assignment method is used there will be 9 destinations, with each city repeated 3 times. b) The solution with the transportation method.

B-32 .

Total mileage, Z = 985; multiple optimal solutions exist. 45. This changes the solution and increases the total mileage. The new assignments are: Officials 3, 6 and 7 – Athens Officials 1, 2 and 8 – Columbia Officials 4, 5 and 9 – Nashville Total mileage, Z = 1,220 46. Solution Summary: Al’s – Parent’s Brunch Bon Apetít – Post-game Party Bon Apetít – Lettermen’s Dinner Divine – Booster Club Luncheon Epicurean – Contributor’s Dinner University – Alumni Brunch

Solution:

Total Cost (Z) = $103,600 47. Subtract all values from 100 and minimize the difference.

1-E

2-D

3-B

4-A

5-C

Average = 94.5

B-33

48.

Annie – 1. Backstroke Beth – dummy Carla – dummy Debbie – 2. Breaststroke Erin – 4. Freestyle Fay – 3. Butterfly Total time = 10.61 minutes 49.

Subtract all sales values from highest sales, $630.

B-34 .

Employee 1 – dummy Employee 2 – Home furnishing Employee 3 – Jewelry Employee 4 – Appliance Employee 5 – dummy Employee 6 – China

B-35 .

0 560 420 630 0 320 $1,930 total sales

50.

B-36 .

51.

Opportunity cost table:

Multiple optimal solutions: 1-B

1-B

2-A

2-C

3-F

4-D

5-C

5-A

6-E

9 36 nights

6-E

9 36 nights

52.

First subtract all the %’s from 100, and make 2 Florida columns:

Opportunity cost table:

B-37 .

A–NJ B–PA C–NY D–FL E–GA F–FL G–VA Z = 498 Average success rate =

498 = 71.1% 7

B-38 .

Module C: Integer Programming: The Branch and Bound Method PROBLEM SUMMARY

Relaxed solution: x2 = 0.25, x1 = 3, Z = 16 Addition of x1 ≥ 4 constraint:

1. Integer model, branch and bound solution, 5 nodes

maximize Z = 5x1 + 4x2 subject to

2. Integer model, branch and bound solution, 3 nodes

3x1 + 4x2 ≥ 10

3. Integer model, branch and bound solution, 3 nodes

x1, x2 ≥ 0

x1 ≥ 4 Relaxed solution: Infeasible

4. Integer model, branch and bound solution, 3 nodes 5. Integer model, branch and bound solution, 3 nodes 6. Integer model branch and bound solution, 5 nodes 7. Integer model, branch and bound solution, 7 nodes 8. Mixed integer model branch and bound solution, 3 nodes 9. 0–1 integer model, branch and bound solution, 3 nodes

Branch on x2 from node 2: x2 = 0 (not necessary to attempt ≤ 0 since negative values not possible), x1 ≥ 1

10. Integer model, branch and bound solution, 3 nodes

Addition of x2 = 0 constraint:

11. 0–1 integer model, implicit enumeration solutions (C–9)

maximize Z = 5x1 + 4x2 subject to

12. 0–1 integer model, implicit enumeration solution, discussion

3x1 + 4x2 ≤ 10 x1 ≤ 3 x1, x2 ≥ 0 Relaxed solution: s1 = 1, x1 = 3, x2 = 0, Z = 15 Addition of x2 ≥ 1 constraint:

PROBLEM SOLUTIONS 1. a) Relaxed solution: x1 = 3.33 UB = 16.65 (x1 = 3.33, x2 = 0)

maximize Z = 5x1 + 4x2 subject to

LB = 15 (x1 = 3, x2 = 0)

3x1 + 4x2 ≤ 10

Branch on x1: x1 ≤ 3, x1 ≥ 4 Addition of x1 ≤ 3 constraint:

x1 ≤ 3

maximize Z = 5x1 + 4x2 subject to

x2 ≥ 1 x1, x2 ≥ 0

3x1 + 4x2 ≤ 10

Relaxed solution: x1 = 2, s2 = 1, x2 = 1, Z = 14

x1 ≤ 3 x1, x2 ≥ 0

C-1 .

b) Original model:

Relaxed solution: x1 = 5.71, Z = 17.13 LB = 17.13 (x1 = 5.71, x2 = 0) UB = 18 (x1 = 6, x2 = 0) Branch on x1: x1 ≤ 5, x1 ≥ 6 Addition of x1 ≤ 5 constraint: minimize Z = 3x1 + 6x2 subject to 7x1 + 3x2 ≥ 40 x1 ≤ 5 x1, x2 ≥ 0 Relaxed solution x1 = 5, x2 = 1.67, Z = 25 Addition of x1 ≥ 6 constraint: minimize Z = 3x1 + 6x2 subject to 7x1 + 3x2 ≥ 40 x1 ≥ 6 x1, x2 ≥ 0 Relaxed solution x1 = 6, s1 = 2, Z = 18

C-2 .

3. a) Maximize Z = 50x1 + 40x2 (profit, $) subject to

The rounded down solution would not be optimal.

3x1 + 5x2 ≤ 150 (wool, yd2)

4. a) Maximize Z = 400x1 + 100x2 (profit, $) subject to

10x1 + 4x2 ≤ 200 (labor, hr) x1, x2 ≥ 0 and integer

8x1 + 10x2 ≤ 80

b) Relaxed solution: x1 = 10.5, x2 = 23.7, Z = 1,473

2x1 + 6x2 ≤ 36 x1 ≤ 6

UB = 1,473 (x1 = 10.5, x2 = 23.7)

x1, x2 ≥ 0 and integer

LB = 1,420 (x1 = 10, x2 = 23)

b) Relaxed solution: x1 = 6, x2 = 3.2, s2 = 4.8, Z = 2,720

Branch on x2: x2 ≤ 23, x2 ≥ 24 Addition of x2 ≤ 23 constraint:

UB = 2,720 (x1 = 6, x2 = 3.2)

maximize Z = 50x1 + 40x2 subject to

LB = 2,700 (x1 = 6, x2 = 3) Branch on x2: x2 ≤ 3, x2 ≥ 4

3x1 + 5x2 ≤ 150

Addition of x2 ≤ 3 constraint:

10x1 + 4x2 ≤ 200

maximize Z = 400x1 + 100x2 subject to

x2 ≤ 23 x1, x2 ≥ 0

8x1 + 10x2 ≤ 80

Relaxed solution: x1 = 10.8, x2 = 23, s1 = 2.6

2x1 + 6x2 ≤ 36

Addition of x2 ≥ 24 constraint:

x1 ≤ 6

maximize Z = 50x1 + 40x2 subject to

x2 ≤ 3 x1, x2 ≥ 0

3x1 + 5x2 ≤ 150 10x1 + 4x2 ≤ 200 x2 ≥ 24 x1, x2 ≥ 0 Relaxed solution: x1 = 10, x2 = 24.5, s2 = 4

C-3 .

UB = 312.5 (x1 = 6.25, x2 = 0) LB = 300 (x1 = 6, x2 = 0) Branch on x1: x1 ≤ 6, x1 ≥ 7 Addition of x1 ≤ 6 constraint: maximize Z = 50x1 + 10x2 subject to x1 + x2 ≤ 15 4x1 + x2 ≤ 25 x1 ≤ 6 x1, x2 ≥ 0 Relaxed solution: x1 = 6, x2 = 1, s1 = 8, Z = 310

Relaxed solution: x1 = 6, x2 = 3, s1 = 2, s2 = 6, Z = 2,700

Addition of x1 ≥ 7 constraint: maximize Z = 50x1 + 10x2 subject to

Addition of x2 ≥ 4 constraint: maximize Z = 400x1 + 100x2 subject to

x1 + x2 ≤ 15 4x1 + x2 ≤ 25

8x1 + 10x2 ≤ 80

x1 ≥ 7

2x1 + 6x2 ≤ 36

x1, x2 ≥ 0

x1 ≤ 6

Relaxed solution: Infeasible

x2 ≥ 4

c) Relaxed solution:

x1, x2 ≥ 0 Relaxed solution: x1 = 5, x2 = 4, s2 = 2, s3 = 1, Z = 2,400 5. a) Maximize Z = 50x1 + 10x2 subject to x1 + x2 ≤ 15 4x1 + x2 ≤ 25 x1, x2 ≥ 0 and integer

C-4 .

6. a) Maximize Z = 600x1 + 540x2 + 375x3 subject to x1 + x2 + x3 ≤ 12 x1 ≤ 5 80x1 + 70x2 + 50x3 ≤ 750 x1, x2, x3 ≥ 0 and integer b) Relaxed solution: x2 = 10.71, s2 = 1.28, s2 = 5, Z = 5,785.7 UB = 5,785.7 (x1 = 0, x2 = 10.71, x3 = 0) LB = 5,400 (x1 = 0, x2 = 10, x3 = 0)

Branch from node 2 on x1: x1 = 0, x1 ≥ 1

Branch on x2: x2 ≤ 10, x2 ≥ 11

Addition of x1 = 0 constraint:

Addition of x2 ≤ 10 constraint:

maximize Z = 600x1 + 540x2 + 375x3 subject to

x1 + x2 + x3 ≤ 12

x1 ≤ 5

80x1 + 70x2 + 50x3 ≤ 750

x2 ≤ 10

x1 = 0

x1, x2, x3 ≥ 0

Relaxed solution: x1 = 0.63, x2 = 10, s1 = 1.39, s2 = 4.37, Z = 5,778

Relaxed solution: x1 = 0, x2 = 10, x3 = 1, s1 = 1, s2 = 5, Z = 5,775

Addition of x2 ≥ 11 constraint:

Addition of x1 ≥ 1 constraint:

maximize Z = 600x1 + 540x2 + 375x3 subject to

x1 + x2 + x3 ≤ 12

x1 ≤ 5

80x1 + 70x2 + 50x3 ≤ 750

x2 ≥ 11

x2 ≤ 10

x1, x2, x3 ≥ 0

x1 ≥ 1

Relaxed solution is infeasible. (This can be determined by inspection or by working out the simplex model.)

x1, x2, x3 ≥ 0 Relaxed solution: x1 = 1, x2 = 9.57, s1 = 1.43, s2 = 4, s4 = .43, Z = 5,767.8

C-5 .

7. a) Maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x1, x2 ≥ 0 and integer b)

UB = 372.9 (x1 = 2.73, x2 = 5.91) LB = 300.0 (x1 = 2, x2 = 5) Branch on x2: x2 ≤ 5, x2 ≥ 6 Addition of x2 ≤ 5 constraint: maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x2 ≤ 5 x1, x2 ≥ 0 Relaxed solution: x1 = 3.33, x2 = 5, s1 = 3.33, Z = 366.5 Addition of x2 ≥ 6 constraint: maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x2 ≥ 6 x1, x2 ≥ 0 Relaxed solution: x1 = 2.5, x2 = 6, s2 = 0.5, Z = 365

Branch from node 2 on x1: x1 ≤ 3, x1 ≥ 4 Addition of x1 ≤ 3 constraint: maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x2 ≤ 5 x1 ≤ 3 x1, x2 ≥ 0 Relaxed solution: x1 = 3, x2 = 5, s1 = 4, s2 = 1, Z = 350 Addition of x1 ≥ 4 constraint: maximize Z = 50x1 + 40x2 subject to 2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x2 ≤ 5 x1 ≥ 4 x1, x2 ≥ 0 Relaxed solution: x1 = 4, x2 = 4, s1 = 7, s3 = 1, Z = 360

C-6 .

Addition of x1 ≥ 3 constraint:

Branch from node 3 (since it now has greatest UB) on x1: x1 ≤ 2, x1 ≥ 3

maximize Z = 50x1 + 40x2 subject to

Addition of x1 ≤ 2 constraint: maximize Z = 50x1 + 40x2 subject to

2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20

2x1 + 5x2 ≤ 35 3x1 + 2x2 ≤ 20 x2 ≥ 6 x1 ≤ 2 x1, x2 ≥ 0

x2 ≥ 6 x1 ≥ 3 x1, x2 ≥ 0 Relaxed solution is infeasible (which can be determined by inspection or simplex solution).

Relaxed solution: x1 = 2, x2 = 6.25, s2 = 0.5, s3 = 0.25, Z = 350

C-7 .

The rounded down solution would not be optimal. 8. a) maximize Z = 8,000x1 + 6,000x2 (annual return, $) subject to 70x1 + 30x2 ≤ 500 (capital outlay, $1,000s) x1 + 2x2 ≤ 14 (annual maintenance budget, $1,000s)

Relaxed solution: x1 = 1, x2 = 0.17, x3 = 1, s3 = 0.83, Z = 1,916.67 UB = 1,916.67 (x1 = 1, x2 = .17, x3 = 1) 1)

LB = 1,800 (x1 = 1, x2 = 0, x3 =

Branch on x2: x2 = 0, x2 = 1 Addition of x2 = 0 constraint:

x1 ≥ 0 and integer x2 ≥ 0

maximize Z = 1,000x1 + 700x2 + 800x3 subject to

b) x1 = 5.27, x2 = 4.37, Z = 68,380 UB = 68,380 (x1 = 5.27, x2 = 4.37) LB = 66,220 (x1 = 5, x2 = 4.37) Branch on x1 (since x1 is the only variable restricted to an integer value): x1 ≤ 5, x1 ≥ 6 Addition of x1 ≤ 5 constraint: maximize Z = 8,000x1 + 6,000x2 subject to 70x1 + 30x2 ≤ 500 x1 + x2 ≤ 14 x1 ≤ 5 x1 ≥ 0 and integer x2 ≥ 0 Relaxed solution: x1 = 5, x2 = 4.5, s1 = 15, Z = 67,000 Addition of x1 ≥ 6 constraint: maximize Z = 8,000x1 + 6,000x2 subject to 70x1 + 30x2 ≤ 500 x1 + 2x2 ≤ 14 x1 ≥ 6 x1 ≥ 0 and integer x2 ≥ 0 Relaxed solution: x1 = 6, x2 = 2.66, s2 = 2.68, Z = 63,960

5,000x1 + 6,000x2 + 4,000x3 ≤ 10,000 x1 ≤ 1 x2 = 0 (not necessary to also have x2 ≤ 1) x3 ≤ 1 x1, x2, x3 = 0 or 1 Relaxed solution: x1 = 1, x2 = 0, x3 = 1, s1 = 1,000, Z = 1,800 Addition of x2 = 1 constraint: maximize Z = 1,000x1 + 700x2 + 800x3 subject to 5,000x1 + 6,000x2 + 4,000x3 ≤ 10,000 x1 ≤ 1 x2 = 1 x3 ≤ 1 x1, x2, x3 = 0 or 1 Relaxed solution: x1 = 0.8, x2 = 1, s2 = 0.2, s3 = 1, Z = 1,500

C-8 .

10.

Relaxed solution: x1 = 2, x2 = 3.33, Z = 30

Relaxed solution: x1 = 2.2, x2 = 3, s2 = 0.8, Z = 29

UB = 30 (x1 = 2, x2 = 3.33, x3 = 0)

Addition of x2 ≥ 4 constraint:

LB = 28 (x1 = 2, x2 = 3, x3 = 0)

maximize Z = 5x1 + 6x2 + 4x3 subject to

Branch on x2: x2 ≤ 3, x2 ≥ 4 Addition of x2 ≤ 3 constraint:

5x1 + 3x2 + 6x3 ≤ 20

maximize Z = 5x1 + 6x2 + 4x3 subject to

x1 + 3x2 ≤ 12 x2 ≥ 4

5x1 + 3x2 + 6x3 ≤ 20

x1, x3 ≥ 0

x1 + 3x2 ≤ 12

x2 ≥ 0 and integer

x2 ≤ 3

Relaxed solution: x1 = 0, x2 = 4, x3 = 1.33, Z = 29.32

x1, x3 ≥ 0 x2 ≥ 0 and integer

C-9 .

11.

Solution

Feasibility

Feasible

1,000

Feasible

700

Feasible

800

Infeasible

∞

Feasible

1,800*

Feasible

1,500

Infeasible

∞

Feasibility

Feasible

Infeasible

∞

Feasible

60*

Infeasible

∞

Infeasible

∞

Feasible

Infeasible

∞

Infeasible

∞

Infeasible

∞

Infeasible

∞

Infeasible

∞

12. a) Solution

b) As the number of variables increases, the number of possible solutions increases dramatically. It is very difficult to manually identify all the possible 0–1 combinations for 5 variables and all but impossible for a model with greater than 5 variables. For n variables the number of 0–1 combinations (i.e., alternative solutions) is 2n. Thus, for 5 variables there are 25 = 32 solutions, for 6 variables, 26 = 64 solutions; etc. If the number of constraints is also increased, then it becomes more difficult to evaluate all possibilities. At such a point, computerized solution is necessary.

C-10 .

Module D: Nonlinear Programming Solution Techniques PROBLEM SUMMARY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Profit analysis Graphical analysis (D–1) Profit analysis Profit analysis Profit analysis Profit analysis Substitution method Substitution method Substitution method Substitution method Lagrange multiplier method (D–8) Lagrange multiplier method (D–9) Lagrange multiplier method (D–10) Lagrange multiplier method λ , discussion (D–14)

Z = vp − cf − vcv = (75,000 − 1,153.8 p) p − cf − (75,000 − 1,153.8 p)cv

PROBLEM SOLUTIONS 1.

v = 75,000 − 1,153.8p, cf = $150,000, cv = $16,

= 75,000 p − 1,153.8 p 2 − cf − 75,000cv + 1,153.8 pcv

v = 400 − 1.2 p; Z = vp − cf − vcv

= 75,000 p − 1,153.8 p 2 − 150,000 − 75,000(16) + 1,153.8 p(16)

= (400 − 1.2 p) p − cf − (400 − 1.2 p)cv = 400 p − 1.2 p 2 − cf − 400cv + 1.2 pcv ;

= 93, 460 p − 1,153.8 p2 − 1,350,000

substituting cf = $7,500 and cv = $40

∂Z = 93, 460 − 2,307.6 p ∂p

gives Z = 400 p − 1.2 p2 − 7,500 − 400(40) + 1.2 p(40) = 448 p − 1.2 p2 − 23,500. Then

0 = 93, 460 − 2,307.6 p

∂Z = 448 − 2.4 p ∂p 0 = 448 − 2.4 p 2.4 p = 448 p = $186.67 v = 400 − 1.2 p = 400 − 1.2(186.67) = 176 chairs

2,307.6 p = 93, 460 p = $40.50 v = 75,000 − 1,153.8 p = 75,000 − 1,153.8(40.5) = 28,271.1 Z = 93, 460 p − 1,153.8 p2 − 1,350,000 = 93, 460(40.5) − 1,153.8(40.5)2 − 1,350,000

Z = 448 p − 1.2 p − 23,500 2

= $3,785,130 − 3,242,520 = $542,610

= 448(186.67) − 1.2(186.67)2 − 23,500 = 83,628.16 − 65,314.83 = $18,313.33

D-1 .

v = 17,000 − 5,666p, cf = 8,000, cv = .35,

= (4,000 − 80 p) p − 25,000 − (4,000 − 80 p)10

Z = vp − cf − vcv

= 4,000 p − 80 p2 − 25,000 − 40,000 + 800 p

= (17,000 − 5,666 p) p − cf − (17,000 − 5,666 p)cv

= 4,800 p − 80 p 2 − 65,000 ∂Z = 4,800 − 160 p = 0 ∂p

= 17,000 p − 5,666 p 2 − cf − 17,000cv + 5,666 pcv

160 p = 4,800

= 18,983 p − 5,666 p 2

p = 30

− 13,950

v = 4,000 − 80(30)

∂Z = 18,983 − 11,332 p ∂p

= 1,600 Z = 4,800(30) − 80(30)2 − 65,000 = $7,000

0 = 18,983 − 11,332 p 11,332 p = 18,983 p = $1.68 v = 17,000 − 5,666 p = 17,000 − 5,666(1.68)

Z = vp − 12,000 − 17v; v = 800 − 15 p; substituting, Z = (800 − 15 p) p − 12,000 − 17(800 − 15 p) = 800 p − 15 p 2 − 12,000 − 13,600 + 255 p

= 7, 481 yd

= 1,055 p − 15 p 2 − 25,600

Z = 18,983 p − 5,666 p 2 − 13,950

∂Z = 0 = 1,055 − 30 p ∂p

= 18,983(1.68) − 5,666(1.68)2 − 13,950 = 31,891.44 − 15,991.72 − 13,950 = $1,949.72

Z = vp − cf − vcv

30 p = 1,055 p = $35.16

cf = $2,500, cv = $9, v = 200 − 4.75p,

Z = 1,055 p − 15 p 2 − 25,600

Z = vp − cf − vcv

= 1,055(35.16) − 15(35.16)2 − 25,600

= (200 − 4.75 p) p − cf − (200 − 4.75 p)cv

= 37,093.80 − 26,836.23 = $10,257.57

= 200 p − 4.75 p 2 − 2,500 − 200(9) + 4.75 p(9)

= 242.75 p − 4.75 p 2 − 4,300 ∂Z = 242.75 − 9.5 p ∂p 0 = 242.75 − 9.5 p 9.5 p = 242.75 p = $26.97

Maximize Z = 7x1 − .3x12 + 8x2 − .4x22 subject to 4x1 + 5x2 = 100 Solve the constraint for x1: 4 x1 = 100 − 5 x2 x1 = 25 − 1.25 x2

Substituting,

v = 200 − 4.75 p = 200 − 4.75(26.97) = 72

Z = 7(25 − 1.25 x2 ) − .3(25 − 1.25 x2 )2

Z = 242.75 p − 4.75 p − 4,300 2

+ 8 x2 − .4 x2 2

= 242.75(26.97) − 4.75(26.97) − 4,300 = $6,546.97 − 3, 455.06 − 4,300 2

= 175 − 8.75 x2 − .3(625 − 62.5 x2 + 1.56 x2 2 ) + 8 x2 − .4 x2 2

= −$1,208.09 (loss)

= 175 − 8.75 x2 − 187.5 + 18.75 x2 − .468 x2 2 + 8 x2 − .4 x2 2 = 18 x2 − 12.5 − .868 x2 2

D-2 .

10. Maximize Z = 10x1 − .02x12 + 12x2 − .03x22 subject to .2 x1 + .1x2 = 40

∂Z = 18 − 1.74 x2 ∂x 2 0 = 18 − 1.74 x2

Let .2 x1 = 40 − .1x2 and x1 = 200 − .5 x2 .

1.74 x2 = 18 x2 = 10.34

Z = 10(200 − .5 x2 ) − .02(200 − .5 x2 )2

4 x1 + 5 x2 = 100

+ 12 x2 − .03 x2 2

4 x1 + 5(10.34) = 100

= 2,000 − 5 x2 − .02(40,000 − 200 x2

4 x1 = 48.3

+ .25 x2 2 ) + 12 x2 − .03 x2 2

x1 = 12.08

= 2,000 − 5 x2 − 800 + 4 x2 − .005 x2 2

Z = 7(12.08) − .3(12.08) + 8(10.34) 2

+ 12 x2 − .03 x2 2

− .4(10.34) = 84.56 − 43.78 + 82.72 − 42.77 = 167.28 − 86.55 = $80.73 9. Maximize Z = 30x1 − 2x12 + 25x2 − .5x22 subject to 3x1 + 6x2 = 300 Let 3x1 = 300 − 6x2 and x1 = 100 − 2x2. Substituting, 2

= 1,200 + 11x2 + .035 x2 2 ∂Z = 0 = 11 − .07 x2 ∂x2 .07 x2 = 11 x2 = 157 .2 x1 + .1x2 = 40 .2 x1 + .1(157) = 40

Z = 30(100 − 2 x2 ) − 2(100 − 2 x2 )2

.2 x1 = 24.3

+ 25 x2 − .5 x2 2

x1 = 121.5

= 3,000 − 60 x2 − 2(10,000 − 400 x2

Z = 10 x1 − .02 x1 2 + 12 x2 − .03 x2 2

+ 4 x2 2 ) + 25 x2 − .5 x2 2

= 10(121.5) − .02(121.5)2 + 12(157)

= 3,000 − 60 x2 − 20,000 + 800 x2 − 8 x2 + 25 x2 − .5 x2 2

− .03(157)2 = 1,215 − 295.25 + 1,884 − 739.47 = 3,099 − 1,034.72 = $2,064.28

= 765 x2 − 8.5 x2 2 − 17,000

17 x2 = 765

11. Minimize Z = 7x1 − .3x12 + 8x2 − .4x22 subject to 4 x1 + 5 x2 = 100

x2 = 45

Then, 4 x1 + 5 x2 − 100 = 0 and

∂Z = 0 = 765 − 17 x2 ∂ x2

3 x1 + 6 x2 = 300

L = 7 x1 − .3 x1 2 + 8 x2 − .4 x2 2 −λ (4 x1 +5x2 − 100).

3 x1 + 6(45) = 300 3 x1 = 30 x1 = 10

∂L = 7 − .6 x1 − 4λ = 0 ∂x1

(1)

Z = 30(10) − 2(10)2 + 25(45) − .5(45)2 = 300 − 200 + 1,125 − 1,012.5

∂L = 8 − .8 x2 − 5λ = 0 ∂x2

(2)

∂L (3) = −4 x1 − 5 x2 + 100 = 0 ∂λ Divide equation (2) by 1.25 and subtract from equation (1): 7 − .6 x1 − 4λ = 0

= $212.5

−6.4

+ .64 x2 + 4λ = 0

.6 − .6 x1 + .64 x2

D-3 .

13. Maximize Z = 10x1 − .02x12 + 12x2 − .03x22 subject to .2 x1 + .1x2 = 40

Next, multiply equation (3) by .15 and subtract (4) from it: −.6 x1 − .75 x2 = −15 +.6 x1 − .64 x2 = +

Then .2 x1 + .1x2 − 40 = 0

− 1.39 x2 = −14.4

and L = 10 x1 − .02 x1 2 + 12 x2

x2 = 10.35

−.03 x2 2 − λ (.2 x1 + .1x2 − 40).

4 x1 + 5(10.35) − 100 = 0 4 x1 = 48.25 x1 = 12.06 Z = $80.73 12. Maximize Z = 30x1 − 2xl2 + 25x2 − .5x22 subject to 3 x1 + 6 x2 = 300

∂L = 10 − .04 x1 − .2λ = 0 ∂x1

(1)

∂L = 12 − .06 x2 − .1λ = 0 ∂x2

(2)

Then 3 x1 + 6 x2 − 300 = 0

∂L = −.2 x1 − .1x2 + 40 = 0 (3) ∂λ Multiply (2) by 2 and subtract from (1): 10 − .04 x1 − .2λ = 0

and L = 30 x1 − 2 x1 2

−24

+ 25 x2 − .5 x2 2 − λ (3 x1 + 6 x2 − 300).

−14 − .04 x1 + .12 x2

∂L = 30 − 4 x1 − 3λ = 0 ∂x1

(1)

∂L = 25 − 1x2 − 6λ = 0 ∂x2

(2)

∂L = −3 x1 − 6 x2 + 300 = 0 ∂λ Multiply (1) by 2 and subtract (2) from it: 60 − 8 x1 − 6λ = 0 −25 + 1x 2 + 6 λ = 0 35 − 8 x1 + 1x2

=0 =0

(4)

Multiply (3) by .2 and subtract (4): −.04 x1 − .02 x2 = −8 .04 x1 − .12 x2 = −14 − .14 x2 = −22 x2 = 157

(3)

.2 x1 + .1(157) = 40 .2 x1 = 24.3 x1 = 121.5 Z = $2,064.28

(4)

14. Maximize Z = $25x1 − .8x12 + 30x2 − 1.2x22 subject to x1 + 2 x2 = 40

Multiply (4) by 6 and add (3): −3 x1 − 6 x2 = −300 −48 x1 + 6 x2 = −210 −51x1

+ .12 x2 + .2λ

Let x1 + 2 x2 − 40 = 0

= −515

and L = 25 x1 − .8 x1 2 + 30 x2

x1 = 10

−1.2 x2 2 − λ ( x1 + 2 x2 − 40).

3 x1 + 6 x2 − 300 = 0 3(10) + 6 x2 = 300 6 x2 = 270 x2 = 45 Z = $212.50

∂L = 25 − 1.6 x1 − λ = 0 ∂x1

(1)

∂L = 30 − 2.4 x2 − 2λ = 0 ∂x2

(2)

∂L = − x1 − 2 x2 + 40 = 0 ∂λ Multiply (1) by 2 and subtract (2): 50 − 3.2 x1 − 2λ = 0 −30

+ 2.4 x2 + 2λ = 0

20 − 3.2 x1 + 2.4 x2

D-4 .

(3)

(4)

Multiply (3) by 3.2 and subtract (4): −3.2 x1 − 6.4 x2 = −128 3.2 x1 − 2.4 x2 =

15. Using equation (1) in Problem 14: ∂L = 25 − 1.6 x1 − λ = 0 ∂x1

25 − 1.6(15.454) − λ = 0 λ = .27

− 8.8 x2 = −108 x2 = 12.3

Therefore, we would be willing to pay $.27 for one additional hour or labor.

x1 + 2 x2 = 40 x1 + 2(12.3) = 40 x1 = 15.4 Z = 25(15.4) − .8(15.4)2 + 30(12.3) − 1.2(12.3)2 = 385 − 190 + 369 − 182 = $382

D-5 .

Module E: Game Theory 3.

PROBLEM SUMMARY

Washington

1. Pure strategy game Truman

3. Mixed strategy game, expected gain and loss method

4. Pure strategy game, dominance

2. Mixed strategy game

5. Mixed strategy game, dominance, expected gain and loss method

No pure strategy; thus, must use expected gain and loss method.

6. Mixed strategy game, dominance, expected gain and loss method

Truman: 7p + 6(1 − p) = 6 + p 3p + 10(1 − p) = 10 − 7p

PROBLEM SOLUTIONS

Strategy A: Strategy B:

where p = probability strategy 1 will be used and 1 − p = probability strategy 2 will be used.

Red Blue

1,800

2,000

1,700

2,300

900

1,600

6 + p = 10 − 7p 8p = 4 p = .5

1 − p = .5 Strategy C dominates A; thus, A can be eliminated. Blue: strategy 1, Red: strategy C; troop losses for Red Division = 1,700.

EG(Truman) = 7(.5) + 6(.5) = 6.5 percentage points gain Washington:

Strategy 1: Strategy 2:

Owners Players

7p + 3(1 − p) = 3 + 4p 6p + 10(1 − p) = 10 − 4p

3 + 4p = 10 − 4p 8p = 7 p = .875

1 − p = .125 EL(Washington) = 7(.875) + 3(.125) = 6.5 percentage points loss

a) Player/agent: strategy 1; owner/agent: strategy C b) A mixed strategy, since an equilibrium point was not determined; both game players will subsequently change strategies, which will result in a loop. Thus, mixed strategies, which will result in a loop. Thus, mixed strategies must be determined.

E-1 .

6p + 4 = 5 − 2p

8p = 1

Publisher Novelist

p = .125

1 − p = .875

80,000

120,000

90,000

130,000

90,000

80,000

110,000

140,000

100,000

EG(Smoothie) = 10(.125) + 4(.875) = 4.75 percent gain in market share Cooler Cola: Strategy 1: 10p + 3(1 − p) = 3 + 7p

Strategy C dominates B, and strategy 3 dominates strategy 1, leaving a 2 × 2 payoff table.

Strategy 2: 4p + 5(1 − p) = 5 − p 5 − p = 3 + 7p 8p = 2

b) A pure strategy exists; novelist: strategy 3; publisher: strategy C; equilibrium point = $100,000 gain for novelist and loss for publisher.

p = .25

1 − p = .75 EL(Cooler Cola) = 4(.25) + 5(.75) = 4.75 percent loss in market share

5. Cooler Cola

Smoothie

Tech

Zone

Man

Combination

Shuffle

−4

Overload

State

Strategy C dominates B, and strategy 1 dominates 3. Reduced payoff table: A

Combination defense dominated by zone defense Reduced payoff table: State

No pure strategy; therefore, mixed strategies must be determined.

Tech

Zone

Man

Shuffle

Overload

Smoothie: Strategy A: 10p + 4(1 − p) = 4 + 6p Strategy C:

No pure strategy; therefore, mixed strategies must be determined.

3p + 5(1 − p) = 5 − 2p

where p = probability strategy 1 will be used and 1 − p = probability strategy 2 will be used.

E-2 .

Tech:

State:

zone:

72p + 58(1 − p) = 58 + 14p

shuffle:

man:

60p + 91(1 − p) = 91 − 31p

overload: 58p + 91(1 − p) = 91 − 33p

where p = percentage of time Tech would use a shuffle and 1 − p = percentage of time Tech would use an overload.

72p + 60(1 − p) = 60 + 12p

60 + 12p = 91 − 33p 45p = 31 p = .69

91 − 31p = 58 + 14p

1 − p = .31

45p = 33

EL(State) = 72(.69) + 60(.31) = 68.4 points given by state

p = .73

1 − p = .27

State would employ a zone defense 69% of the time and man-to-man 31% of the time.

EG(Tech) = 60(.73) + 91(.27) = 68.4 points Tech would employ a shuffle 73% of the game time and an overload 27% of the time.

E-3 .

Module F: Markov Analysis 14. Absorbing state, debt example (F–13)

PROBLEM SUMMARY l. Decision trees, computing future state probabilities

15. Steady-state determination, 3 × 3 matrix

2. Markov properties, discussion (F–1)

17. Steady-state determination, 3 × 3 matrix

3. Decision trees, computing future state probabilities

18. Steady-state determination, 3 × 3 matrix

4. Matrix multiplication (F–1)

20. Steady-state analysis (F–19)

5. Matrix multiplication (F–3)

21. Steady-state analysis, 3 × 3 matrix

6. Matrix multiplication, steady-state determination

22. Steady-state analysis, 3 × 3 matrix (F–21)

16. Steady-state determination, 3 × 3 matrix

19. Steady-state determination, 3 × 3 matrix

23. Steady-state determination, 3 × 3 matrix

7. Steady-state determination

24. Steady-state determination (F–23)

8. Steady-state determination

25. Absorbing state, maintenance example

9. Steady-state analysis

26. Steady-state determination, 3 × 3 matrix (F–1)

10. Steady-state analysis (F–3)

27. Absorbing state, 4 × 4 matrix

11. Steady-state analysis (F–6)

28. Steady-state determination, 4 × 4 matrix

12. Steady-state analysis (F–7)

29. Steady-state determination, 4 × 4 matrix

13. Absorbing state, debt example

PROBLEM SOLUTIONS 1.

F-1 .

Probability in month 3: Starting State National 2. a)

Petroco

National

Gascorp

Sum

.14

.54

.32

1.00

The transition probabilities for a given beginning state of the system sum to 1.0.

The probabilities apply to all participants in the system.

The transition probabilities are constant over time.

P(Cheesedale) = .147 + .126 + .036 + .108 = .417

⎡.5 .3 .2 ⎤ ⎡⎣ Pp (3) N p (3) G p (3) ⎤⎦ = [.5 .3 .2 ] ⎢.1 .7 .2 ⎥ ⎢ ⎥ ⎢⎣.1 .1 .8 ⎥⎦ = [.30 .38 .32 ]

Notation: Let A = Creamwood and B = Cheesedale. ⎡.7 .3 ⎤ ⎥ = [.61 .39] ⎣.4 .6 ⎦ ⎡.7 .3 ⎤ [ Aa (4) Ba (4)] = [.61 .39] ⎢ ⎥ = [.583 .417] ⎣.4 .6 ⎦

[ Aa (3) Ba (3)] = [.7 .3] ⎢

Ba (4) = probability of Cheesedale at month 4 = .417

F-2 .

These are the probabilities of a customer purchasing a Tribune and Daily News in the long-run future regardless of which paper he or she purchases in week 1.

⎡.4 .6 ⎤ a) week 2: [ Oo (2) Bo (2)] = [.4 .6 ] ⎢ ⎥ ⎣.8 .2 ⎦ = [.64 .36 ]

⎡.4 .6 ⎤ week 3: [ Oo (3) Bo (3)] = [.64 .36 ] ⎢ ⎥ ⎣.8 .2 ⎦ = [.544 .456 ]

8. a)

⎡.9 .1 ⎤ month 3: [ Pn (3) N n (3)] = [.8 .2 ] ⎢ ⎥ ⎣.8 .2 ⎦

⎡.4 .6 ⎤ week 4: [ Oo (4) Bo (4)] = [.544 .456 ] ⎢ ⎥ ⎣.8 .2 ⎦ = [.582 .418]

= [.88 .12 ] ⎡.9 .1 ⎤ month 4: [ Pn (4) N n (4)] = [.8 .2 ] ⎢ ⎥ ⎣.8 .2 ⎦ = [.89 .11]

⎡.4 .6 ⎤ week 5: [ Oo (5) Bo (5)] = [.582 .418] ⎢ ⎥ ⎣.8 .2 ⎦ = [.567 .433]

⎡.4 .6 ⎤ week 6: [ Oo (6) Bo (6)] = [.567 .433] ⎢ ⎥ ⎣.8 .2 ⎦ = [.573 .427]

⎡.4 .6 ⎤ Bo ] ⎢ ⎥ ⎣.8 .2 ⎦ Oo = .4Oo + .8Bo; Bo = .6Oo + .2Bo; Oo + Bo = 1.0; substituting Bo = 1.0 − Oo into the first equation above gives Oo = .4Oo + .8(1.0 − Oo ) = .4O o + .8 − .8Oo

[ Oo

Bo ] = [ Oo

N n ] = [ Pn

Pn + N n = 1.0 .89 + N n = 1.0

and Oo + Bo = 1.0

9. a)

.571 + Bo = 1.0 Bo = .429

[ O B] = [.571 .429] ⎡.65 .35⎤ Dt ] ⎢ ⎥ ⎣.45 .55⎦ Tt = .65Tt + .45Dt, Dt = .35Tt + .55Dt, and Tt + Dt = 1.0; substituting Dt = 1.0 − Tt into the first equation above gives Tt = .65Tt + .45(1.0 − Tt ) D t ] = [ Tt

= .65Tt + .45 − .45Tt .8Tt = .45

N n = .11

N n ] = [.89 .11] ⎡.75 .25⎤

[ M W ] = [ M W ] ⎢.45 .55⎥

⎣ ⎦ M = .75M + .45W, W = .25M + .55W, M + W = 1.0. (Note: It is not necessary to use subscripts for steady-state computations.) Substituting W = 1.0 − M into the first equation above gives M = .75M + .45(1.0 − M) = .75M + .45 − .45M .7M = .45 M = .45/.70 = .642 M + W = 1.0 .642 + W = 1.0

W = .358

[ M W ] = [.642 .358]

Tt = .45 / .80 = .563 Tt + D t = 1.0

.563 + D t = 1.0

[ Tt

Pn = .8 / .9 = .89

Oo = .8 /1.4 = .571

[ Tt

⎡.9 .1 ⎤ Nn ] ⎢ ⎥ ⎣.8 .2 ⎦ Pn = .9Pn + .8Nn, Nn = .1Pn + .2Nn, and Pn + Nn = 1.0; substituting Nn = 1.0 − Pn into the first equation above gives Pn = .9Pn + .8(1.0 − Pn ) = .9Pn + .8 − .8Pn

.9Pn = .8

1.4Oo = .8

month 2: [ Pn (2) N n (2)] = [.8 .2 ]

D t = .437

D t ] = [.563 .437]

F-3 .

Tuned to movie: 1,200 × .642 = 770; tuned to western: 1,200 × .358 = 430 The movie, because it has more sets tuned to it

10. a)

Let A = Creamwood and B = Cheesedale. ⎡.7 .3 ⎤ [ A B] = [ A B] ⎢.4 .6 ⎥ ⎣ ⎦ A = .7A + .4B, B = .3A + .6B, A + B = 1.0, and B = 1.0 − A. A = .7A + .4 (1.0 − A ) = .7A + .4 − .4A

Operate (weeks) = 52 × .75 = 39 weeks; 39 – 30 = 9 weeks increase in operation = $9,000 in increased profit. Since preventive maintenance costs $8,000, it should be undertaken. 12. a)

Tribune: 20,000 × .563 = 11,260 copies; Daily News: 20,000 × .437 = 8,740 copies

.7A = .4 A = .4/.7 = .571 A + B = 1.0 .571 + B = 1.0 B = .429

[ A B] = [.571 .429] Creamwood: 600 × .571 = 343; Cheesedale: 600 × .429 = 257 ⎡.6 .4 ⎤ b) [ A B] = [ A B] ⎢ ⎥ ⎣.2 .8 ⎦ A = .6A + .2B, B = .4A + .8B, A + B = 1.0, and B = 1.0 − A. A = .6A + .2 (1.0 − A ) = .6A + .2 − .2A

⎡.5 .5 ⎤

[ T D] = [ T D] ⎢.3 .7⎥

⎣ ⎦ T = .5T + .3D, D = .5T + .7D, T + D = 1.0, and D = 1.0 − T. T = .5T + .3(1.0 − T) = .5T + .3 − .3T .8T = .3 T = .3/.8 = .375 T + D = 1.0 .375 + D = 1.0 D = .625

Increase in sales of Daily News: .625 × 20,000 = 12,500 copies, 12,500 – 8,740 = 3,760 copies increase; profit = 3,760 × .05 = $188. Since the promotional campaign costs $150 per week and it will earn $188 in profit, it should be conducted.

.6A = .2 A = .2/.6 = .33 A + B = 1.0 .33 + B = 1.0 B = .67

[ A B] = [.33 .67]

11.

[ T D] = [.563 .437] from Problem 7.

13.

Initial sales for Cheesedale: 257 gal; sales with promotion: 600 × .67 = 402 gal; increase in profit: 402 – 257 = 145 gal,145 × $1.00 = $145.00. Since $500 is greater than $145, the dairy should not institute the promotional campaign. Present situation (from Problem 6b): probability of operating = .571, Probability of breakdown = .429. Weeks per year: operate = .571 × 52 ≈ 30 weeks, breakdown = .429 × 52 ≈ 22 weeks. ⎡.7 .3⎤ [ O B] = [ O B] ⎢.9 .1⎥ ⎣ ⎦

p b p⎡1 0 ⎢ b 0 1 T= ⎢ 1 ⎢.80 0 ⎢ 2 ⎣.40 .60

1 2 0 0 ⎤ 0 0 ⎥⎥ , 0 .20 ⎥ ⎥ 0 0 ⎦

1 ⎡.80 0 ⎤ R= , 2 ⎢⎣.40 .60 ⎥⎦ 1 Q=

1 ⎡ 0 .20 ⎤ 2 ⎢⎣ 0 0 ⎥⎦

F = ( I − Q)

O = .7(O) + .9B, B = .3(O) + .1B, O + B = 1.0, and B = 1.0 − O. O = .7(O) + .9(1.0 − O) = .7(O) + .9 − .9(O) 1.2(O) = .9 O = .9/1.2 = .75 O + B = 1.0 .75 + B = 1.0 B = .25

−1

⎛ ⎡ 1 0 ⎤ ⎡0 .20 ⎤ ⎞ = ⎜⎢ ⎥−⎢ ⎥⎟ ⎝ ⎣0 1 ⎦ ⎣0 0 ⎦ ⎠ 1 −1

1 ⎡1 .20 ⎤ ⎡ 1 −.20 ⎤ =⎢ = ⎥ 1 ⎦ 2 ⎢⎣ 0 1 ⎥⎦ ⎣0

F-4 .

−1

1 FiR=

1 ⎡1 .20 ⎤ i 2 ⎢⎣ 0 1 ⎥⎦ p b

accounts receivable = 195,000 195,000 ] for the new credit plan [

1 ⎡.80 2 ⎢⎣.40

0 ⎤ .60 ⎥⎦

p ×

1 ⎡.88 .12 ⎤ 2 ⎢⎣.40 .60 ⎥⎦

Recall the accounts receivable for the existing plan (Problem 13): p b

1 ⎡.88 .12 ⎤ 2 ⎢⎣.40 .60 ⎥⎦ p b

[ 268,800 151,200]

The cost of the existing plan is (1) 20% loss of bad accounts = ($151,200)(.20) = $30,240; (2) $10 per dollar cost on remaining overdue accounts = (151,200 − 30,240)(.10) = $12,096; total cost = $42,336. The cost of the new credit plan is (1) the lost sales = 420,000 − 390,000 = $30,000; (2) 20% loss of bad accounts = (64,350)(.20) = $12,870; (3) $10 per dollar cost on remaining accounts = (64,350 − 12,870) × (.10) = $5,148; total cost = $48,018. The existing credit plan is cheaper ($42,336 < $48,018); thus the new plan should not be instituted.

= [ 268,800 151,200 ]

14.

First, compute the payment pattern with the new credit plan. p b 1 2 p⎡1 0 0 0 ⎤ b ⎢⎢ 0 1 0 0 ⎥⎥ = , I 1 ⎢.90 0 0 .10 ⎥ ⎢ ⎥ 2 ⎣.70 .30 0 0 ⎦ p

1 ⎡.90 0 ⎤ R= , 2 ⎢⎣.70 .30 ⎥⎦ 1 Q=

15. a)

P = .3P + .4C + .4G

⎛ ⎡ 1 0 ⎤ ⎡0 .10 ⎤ ⎞ F = ( I − Q ) −1 = ⎜ ⎢ ⎥−⎢ ⎥⎟ ⎝ ⎣0 1 ⎦ ⎣0 0 ⎦ ⎠ 1 2

−1

1 ⎡1 .10 ⎤ ⎡ 1 −.10 ⎤ =⎢ ⎥ = 2 ⎢0 1 ⎥ 0 1 ⎣ ⎦ ⎣ ⎦

FiR=

1 ⎡1 .10 ⎤ i 2 ⎢⎣ 0 1 ⎥⎦ p b

⎡.4 .3 .3 ⎤ [ P C G ] = [ P C G ] ⎢⎢.5 .1 .4 ⎥⎥ ⎢⎣.4 .2 .4 ⎥⎦ P = .4P + .5C + .4G ← eliminate P = .3P + .1C + .2G

1 ⎡ 0 .10 ⎤ 2 ⎢⎣ 0 0 ⎥⎦

1 ⎡.97 .03 ⎤ 2 ⎢⎣.70 .30 ⎥⎦

p b = [325,650 64,350 ]

1 2 accounts receivable = [ 210,000 210,000 ] p

p 1 ⎡.90 2 ⎢⎣.70

b 0 ⎤ .30 ⎥⎦

1 ⎡.97 .03 ⎤ 2 ⎢⎣.70 .30 ⎥⎦

F-5 .

P + C + G = 1.0 .3P − .9C + .2G = 0

(1)

.3P + .4C − .6G = 0

(2)

P + C + G = 1.0

(3)

(1)

.3P − .9C + .2G = 0

(2)

.3P + .4C − .6G = 0 →

−.3P − .4C + .6G = 0

(3)

P + C + G = 1.0

− 1.3C + .8G = 0 .3P + .3C + .3G = .3

(2)

.3P + .4C − .6G = 0 →

−.3P − .4C + .6G = 0 − .1C + .9G = .3

(A)

− 1.3C + .8G = 0

− 1.3C +

(B)

− .1C + .9G = .3 →

+1.3C − 11.7G = −3.9

(A)

(B)

.8G = 0

− 10.9G = −3.9 G = .358 (A)

− 1.3C + .8(.358) = 0 −1.3C = −.286 C = .220

(3)

P + C + G = 1.0

P + .220 + .358 = 1.0 P = .422

Therefore, [ P C G ] = [.422 .220 .358] . b) Multiply the steady-state probabilities by 9,000: P = 3,798 − Plant Plus, C = 1,980 − Crop Extra, G = 3,222 − Gro-fast. Plan Plus will gain 798 customers, Crop Extra will lose 2,020 customers, and Gro-fast will gain 1,222 customers. 16.

⎡.30 .38 .32 ⎤ ⎡.5 .3 .2 ⎤ P 3 = P i P 2 = ⎢⎢.15 .54 .31⎥⎥ ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎣.14 .18 .68 ⎥⎦ ⎢⎣.1 .1 .8 ⎥⎦ ⎡.220 .388 .392 ⎤ = ⎢⎢.156 .452 .392 ⎥⎥ ⎢⎣.156 .236 .608 ⎥⎦

⎡.8 .1 .1⎤ [ A B C] = [ A B C] ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎣.1 .3 .6 ⎥⎦ which yields A = .8A + .1B + .1C, B = .1A + .7B + .3C, C = .1A + .2B + .6C, and A + B + C = 1.0. Eliminating one redundant equation yields −.2A + .1B + .1C = 0, .1A − .3B + .3C = 0, and 1.0A + 1.0B + 1.0C = 1.0. Solving these equations simultaneously yields A = .333, B = .389, and C = .278; or [ A B C] = [.333 .389 .278]; and

⎡.220 .388 .392 ⎤ [3,000 5,000 2,000] ⎢⎢.156 .452 .392 ⎥⎥ ⎢⎣.156 .236 .608 ⎥⎦ = [1,752 3,896 4,352 ]

A = 2,332 customers, B = 2,723 customers, and C = 1,945 customers.

17. a)

⎡.5 .3 .2 ⎤ ⎡.5 .3 .2 ⎤ P i P = P 2 = ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎣.1 .1 .8 ⎥⎦ ⎢⎣.1 .1 .8 ⎥⎦

⎡.5 .3 .2 ⎤ E A B = E A B [ ] [ ] ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎣.1 .1 .8 ⎥⎦ E = .5E + .1A + .1B, A = .3E + .7A + .1B, B = .2E + .2A + .8B (eliminate), and E + A + B = 1.0. −.5E + .1A + .1B = 0 (1) .3E − .3A + .1B = 0 E + A + B = 1.0

⎡.30 .38 .32 ⎤ = ⎢⎢.15 .54 .31⎥⎥ ⎢⎣.14 .18 .68 ⎥⎦

F-6 .

(2) (3)

(1) (2)

− .5E + .1A + .1B = 0

.3E − .3A + .1B = 0 →

(2)

E + A + B = 1.0 .3E + .3A + .1B = 0 →

(A)

− .8E + .4A = 0

(B)

−.2E + .4A = .1 →

(A)

− .8(.166) + .4A = 0

(3)

−.3E + .3A − .1B = 0 −.8E + .4A =0 .1E + .1A + .1B = .1 −.3E + .3A − .1B = 0 −.2E + .4A = .1

(A)

(B)

−.8E + .4A = 0 +.2E − .4A = −.1 −.6E = −.1 E = 1 6 = .166

.4A = .133 (3)

A = .333 E + A + B = 1.0

.167 + .333 + B = 1.0

18.

B = .500 Therefore, [ E A B] = [.166 .333 .500 ] . To determine steady-state enrollments in each college, multiply each probability by 10,000: E = 1,666 engineering students, A = 3,333 liberal arts students, and B = 5,000 business students. ⎡.3 .5 .2 ⎤ = V N M V N M [ ] [ ] ⎢⎢.6 .2 .2 ⎥⎥ ⎢⎣.4 .1 .5 ⎥⎦

V = .3V + .6N + .4M, N = .5V + .2N + .1M, M = .2V + .2N + .5M (eliminate), and V + N + M = 1.0. −.7V + .6N + .4M = 0 (1) .5V − .8N + .1M = 0 (2) V + N + M = 1.0 (3) −.7V + .6N + .4M = 0 (1) − .7V + .6N + .4M = 0 (2) .5V − .8N + .1M = 0 → −2.0V + 3.2N − .4M = 0 −2.7V + 3.8N = 0 (A) (3) V + N + M = 1.0 .1V + .1N + .1M = 0 (2) .5V − .8N + .1M = 0 → −.5V + .8N − .1M = 0 −.4V + .9N = .1 (B) − 2.7V + 3.8N = 0 (A) − 2.7V + 3.8N = 0 − .4V + .9N = .1 → +2.7V − 6.08N = −.675 (B) − 2.28N = −.675 Ν = .296 (B) − .4V + .9(.296) = .1 −.4V = −.166

(3)

V = .415 V + N + M = 1.0 .415 + .296 + M = 1.0

M = .289 Therefore, [V N M] = [.415 .296 .289]. To determine the number of trucks in each state: V = (700) × (.415) = 291 trucks in Virginia; N = (700)(.296) = 207 trucks in North Carolina; M = (700)(.289) = 202 trucks in Maryland.

F-7 .

19. a) b)

Steady-state probabilities: [C O S] = [.772 .215 .013]

b) Number of viewers:

Ofrah = 12,085

Annual hours “in control” = (2,000)(.772) = 1,544 Annual hours “out of control” = (2,000)(.215) = 430 Annual hours “shut down” = (2,000)(.013) = 26

Josie = 8,907 Barney Fife = 6,010 24.

Next semester: State Eagle

B&T

= [ 9,000

3,000 ] •

5,000

State

Eagle

[ 7,620 5,090

Since $40,723.20 is greater than the cost of the ad campaign ($25,000) the station should undertake the campaign.

State Eagle

⎡.42 ⎢.57 ⎢ ⎢⎣.33

.34 .25 .26

.24 ⎤ .18 ⎥⎥ .41 ⎥⎦

B&T

4,290 ]

25.

26.

b) From problem 21, State could expect an average of 7,480 students per semester. At $175 per student this results in textbook sales revenue of $1,309,000. If a 10% price cut is implemented the number of students buying from the State Bookstore will increase to 8,833.4 per semester. At $157.50 per student (which is 10% less than $175) the total sales revenue would be $1,391,260 or an increase of $82,260.

Josie

Barney

[.4476 .3299

.2226 ]

Compute the steady-state conditions from the following equations: p1 = p4, p2 = .8p1 + .6p2, p3 = .1p1 + .2p2 + .5p3, p4 = .1p1 + .2p2 + .5p3, p1 + p2 + p3 + p4 = 1.0. Solving simultaneously results in the following steady-state probabilities: [ p1 p2 p3 p4 ] = [.2 .4 .2 .2]. The expected costs are EC = $0p1 + 100p2 + 400p3 + 800p4 = 0(.2) + 100(.4) + 400(.2) + 800(.2) = $280 per day.

Steady-state = [.520 .258 .222] The expected market share for State is 52% which slightly exceeds the University administration’s mandate to Don.

23. a) Ofrah

.1308]

4-month commercial revenue = (80 weekdays)($0.12)(4,242) = $40,723.20

Long run: Steady-state = [.440 .292 .268] State: (.440)(17,000) = 7,480 Eagle: (.292)(17,000) = 4,964 B&T: (.268)(17,000) = 4,556 22. a)

[.3821 .4870

State Eagle B&T

B&T

Barney

New Josie viewers = 13,149 – 8,907 = 4,242 per day

Annual hours “out of control” = (2,000)(.136) = 272 Annual hours “shut down” = (2,000)(.007) = 14 “Out of control” hours saved = 430 − 272 = 158 hr. “Shut down” hours saved = 26 − 14 = 12 hr. 21.

Josie

Number of Josie viewers = (27,000)(.4870) = 13,149

Steady-state probabilities: [ C O S] = [.857 .136 .007]

20.

Ofrah

⎡.5 .3 .2 ⎤ [ P N G ] = [ P N G ] ⎢⎢.1 .7 .2 ⎥⎥ ⎢⎣ .1 .1 .8 ⎥⎦ P = .5P + .1N + .1G (1) N = .3P + .7N + .1G (2) G = .2P + .2N + .8G ← eliminate P + N + G = 1.0

(3)

One of the first three equations is redundant. If P is known and N is known, then G must be known from P + N + G = 1.0. Thus, we will eliminate the third equation above and solve the remaining three simultaneously.

F-8 .

(1)

− .5P + .1N + .1G = 0

−.5P + .1N + .1G = 0

(2)

.3P − .3N + .1G = 0 →

−.3P + .3N − .1G = 0

(3)

P + N + G = 1.0

.1P + .1N + .1G = .1

(2)

.3P − .3N + .1G = 0 →

−.3P + .3N − .1G = 0

−.8P + .4N

−.2P + .4N (A)

− .8P + .4N = 0

−.8P + .4N = 0

(B)

−.2P + .4N = .1 →

.2P − .4N = −.1 −.6P = −.1

(A)

= .1

(B)

p = 1 6 = .167 (A) − .8(.167) + .4N = 0 .4N = .134 N = .335 P + N + G = 1.0

(B)

.167+.335 + G = 1.0 G = .498

Therefore, [ P N G ] = [.167 .335 .498] 27.

F So J Sr D G 0 .20 0 ⎤ ⎡.10 .70 0 So ⎢⎢ 0 .10 .80 0 .10 0 ⎥⎥ J ⎢ 0 0 .15 .75 .10 0 ⎥ ⎢ ⎥ Sr ⎢ 0 0 0 .15 .05 .80 ⎥ D ⎢0 0 0 0 1 0 ⎥ ⎢ ⎥ G ⎢⎣ 0 0 0 0 0 1 ⎥⎦

⎡1.11 .87 .81 .72⎤ ⎢ 0 1.11 1.05 .92⎥ ⎥, F = ( I − Q )− t = ⎢ ⎢ 0 0 1.18 1.04 ⎥ ⎢ ⎥ 0 0 1.18 ⎦ ⎣ 0

F Q = So J Sr

⎡.10 ⎢0 ⎢ ⎢0 ⎢ ⎣0

.70 .10 0 0

0 .80 .15 0

0 ⎤ .0 ⎥⎥ .75⎥ ⎥ .15 ⎦

⎡.43 F i R = So ⎢⎢.26 J ⎢.17 ⎢ Sr ⎣.06

.57 ⎤ .73 ⎥⎥ .83 ⎥ ⎥ .94 ⎦

a) 57% of the entering freshmen will eventually graduate, which is somewhat lower than two-thirds, or 67%. b) The probability that a freshman will eventually drop out is .43.

⎡.20 0 ⎤ ⎢.10 0 ⎥ ⎢ ⎥ ⎢.10 0 ⎥ ⎢ ⎥ Sr ⎣.05 .80 ⎦ 0 0 ⎤ ⎡.90 −.70 ⎢0 .90 −.80 0 ⎥⎥ I −Q = ⎢ , ⎢0 0 .85 −.75⎥ ⎢ ⎥ 0 0 .85 ⎦ ⎣0 F R = So J

28. a)

From the fundamental matrix, F, the time that all entering freshmen will remain at Tech is 1.11 + .87 + .81 + .72 = 3.51 years before dropping out or graduating.

Dry Wall Trim

[.3438

.2521

b) Number of days: Dry wall = 85.95 Trim = 63.03 Framing = 72.95 Roofing = 28.10

F-9 .

Framing Roofing .2918

.1124 ]

29. a) steady-state 63 Nickel Blitz ⎤ ⎡ 54 ⎢.2976 .4233 .1756 .1036 ⎥ ⎣ ⎦

54 = 34.22

F = (I − Q)−1

63 = 48.68

⎛ ⎡1 0 0 ⎤ ⎡0 .14 0 ⎤⎞ F = ⎜ ⎢⎢0 1 0 ⎥⎥ − ⎢⎢0 .22 .46 ⎥⎥⎟ ⎜ ⎟ ⎜⎝ ⎢0 0 0 ⎥ ⎢0 .54 .16 ⎥⎟⎠ ⎣ ⎦ ⎣ ⎦

Nickel = 20.19 Blitz = 11.91 (57.5)(.1036) = 5.96 or approximately 6 plays

C −1

0 ⎤ C ⎡ 1 −.14 = ⎢ ⎥ F = ⎢ 0 .78 −.46 ⎥ 1 ⎢⎣ 0 −.54 .84 ⎥⎦ 2

CASE SOLUTION: THE FRIENDLY CAR FARM The steady-state probabilities computed using AB:QM are: 0

C ⎡.86 0 ⎤ C ⎡ 0 .14 0 ⎤ ⎢ ⎥ R = 1 ⎢.32 0 ⎥ Q = 1 ⎢⎢ 0 .22 .46 ⎥⎥ 2 ⎢⎣.18 .12 ⎥⎦ 2 ⎢⎣ 0 .54 .16 ⎥⎦

b) Number of defensive plays:

Expected number of cars in stock = .284(0) + .284(1) + .265(2) + .167(3) = 1.315 cars

P ⎡1 0 ⎢ B ⎢0 1 C ⎢.86 0 ⎢ 1 ⎢.32 0 2 ⎢⎣.18 .12

| 0

⎡1 .289 0.158 ⎤ ⎢ 0 2.065 1.131 ⎥ ⎢ ⎥ ⎢⎣ 0 1.327 1.917 ⎥⎦ P

⎡.98 .02 ⎤ [750,000 400,000 200,000] ⎢⎢.86 .14 ⎥⎥ ⎣⎢.77 .23 ⎥⎦ = [1,235,457 114,542 ]

CASE SOLUTION: DAVIDSON’S DEPARTMENT STORE C

C ⎡.98 .02 ⎤ F • R = 1 ⎢.86 .14 ⎥ ⎢ ⎥ 2 ⎣⎢.77 .23 ⎦⎥

Average inventory holding cost = .284(75) + .265(175) + .167(310) = $119.64

C ⎡ 0.86 0⎤ ⎡1 .289 0.158 ⎤ ⎢ ⎥ ⎢ • F • R = ⎢0 2.065 1.131 ⎥ 1 ⎢ 0.32 0 ⎥⎥ ⎢⎣0 1.327 1.917 ⎥⎦ 2 ⎢⎣ 0.18 0.12 ⎥⎦

[.284 .284 .265 .167]

−1

Of the average outstanding balance of $1,350,000, each month, $1,235,457 will be paid and $114,542 will become bad debts. Thus, a $60,000 cash reserve each month is not sufficient.

0 ⎤ | 0 0 0 ⎥⎥ | 0 .14 0 ⎥ ⎥ | 0 .22 .46 ⎥ | 0 .54 .16 ⎥⎦ 0

F-10 .