Solutions Manual for Operations and Supply Chain Management 10th Edition by Roberta S. Russell and B by ACADEMIAMILL

1 - Introduction to Operations and Supply Chain Management Answers to Questions 1-1.

The operations function involves organizing work, selecting processes, arranging layouts, locating facilities, designing jobs, measuring performance, controlling quality, scheduling work, managing inventory, and planning production. Operations interacts with marketing in product development, forecasting, production planning, and customer service. Operations and finance interact in capital budgeting, cost analysis, production and inventory planning, and expansion and technology plans. Operations and human resources work together recruiting, training and evaluating workers, designing jobs and working with unions. IT and operations work together daily on e-commerce, enterprise resource planning and supply chain management systems.

1-2.

b. c.

Operations at a bank involves transferring funds, processing funds, providing checks, cashing checks, preparing monthly statements, reconciling statements, approving loans, loaning money, keeping track of loan payments, approving credit cards, and more. Operations at a retail store involves purchasing goods, stocking goods, selling goods, keeping track of inventory, scheduling workers, laying out the store, locating the store, forecasting demand, and more. Operations at a hospital involves preparing the rooms, scheduling doctors, nurses and other workers, processing paperwork, ordering supplies, caring for patients, maintaining the facility, laying out the facility, ensuring quality and more. Operations at a cable TV company involves taking orders, installing equipment, maintaining equipment, keeping the shows on the air, scheduling work, processing statements and payments, and more.

1-3. Inventions during the industrial revolution brought workers together under one roof in a factory setting where division of labor and interchangeable parts encouraged the formation of separate worker and management jobs. Ideas from the scientific management era made work more efficient. Human relations theorists emphasized the importance of the human element in operations management. The management science era saw many advances in quantitative techniques and their application. The quality revolution focused management on meeting customer expectations and emphasized quality over quantity. The Internet brought numerous opportunities to do work faster and better. It also opened doors to new markets worldwide. Today’s successful companies compete worldwide for both market access and production resources. 1-4. Productivity is the efficiency with which inputs are transformed into outputs. It is calculated by dividing units of output by units of input. Output can be represented by units or dollars of sales made, products produced, customers served or calls answered. The most common input is labor hours, although a variety of inputs can be used, such as labor cost, labor cost + machine cost + energy cost, and so forth. 1-5.

Student answers will vary.

1-6.

Student answers will vary.

1-7.

Students can begin this assignment by accessing Fortune’s homepage and referring to the Fortune 500 or Global 500 by industry. The leaders in each industry are listed and there is usually some discussion of industry concerns. Individual data on companies can be found at Hoover’s website (www.hoovers.com).

1-8.

Student answers will vary.

1-9.

Student answers will vary.

1-10. Student answers will vary. Access www.worldbank.com. 1-11. The WTO is an international organization that works to establish and enforce rules of trade between nations. WTO agreements are ratified by the governing bodies of the nations involved. WTO’s dispute settlement process interprets agreements and rules on violations, thereby avoiding political or military conflict. The group promotes free trade and more recently, has helped developing nations enter the trade arena on more equitable grounds. Currently, there are 147 member nations. Membership is achieved by meeting certain environmental, human rights, and trade criteria, agreeing to abide by the rules of the organization, and being approved by two-thirds of the existing membership. See www.wto.org 1-12. Student answers will vary. Access www.executiveplanet.com 1-13. Student answers will vary. Access www.transparency.org 1-14. Student answers will vary. Access http://www.usdoj.gov/criminal/fraud/fcpa.html for basic information. 1-15. Students will find a variety of answers for this question. In general, it is easy to find mission or vision statements, but more difficult to find evidence of the mission or vision being applied. 1-16. Strategy formulation consists of four basic steps: (1) Defining a primary task—what is the purpose of the firm? What is the firm in the business of doing? (2) Assessing core competencies—what does a firm do better than anyone else? (3) Determining order winners and order qualifiers—what wins orders in the marketplace? What qualifies a product or service to be considered for purchase? (4) Positioning the firm— what one or two important things should the firm choose to concentrate on? How should the firm compete in the marketplace? Student answers will vary. Most start-ups try too much too soon. It’s often difficult to stick with what you do best. 1-17. Core competencies are the essential capabilities that create a firm’s sustainable competitive advantage. They have usually been built up over time and cannot be easily imitated. For example, First National Bank, one of our local banks, is known as a risk taker. Its core competence is its ability to size up the potential of investment opportunities. Through its familiarity with local businesses and its experience in loan making, the bank has developed the ability to predict which loans are worth taking extra risks. Bonomo’s, a successful retail store, is known for having just the right item in stock for special occasions. The store stocks a variety of stylish women’s clothing, but not too much of each style. They specialize in knowing individual customers and even keep track of evening wear purchases so that no one else at a particular party will be wearing the same dress. Toyota emphasizes superior quality at a price below its competitors with its Lexus line of automobiles. To establish a special reputation for quality over the lifetime of the car, the company set up separate sales and service facilities. When it is time for servicing, Lexus owners can have their vehicle picked up and delivered to their home or place of business. The car returns the same day, washed and vacuumed, often with a gift certificate inside for a night on the town complements of the dealer. 1-18. While the answers to this question vary considerably, most students feel competent in the technical areas of their major, but uncomfortable with their communication skills (both oral and written) and their ability to

make decisions. This opens the way for more project-oriented assignments from the instructor. The question also helps students prepare for the inevitable interview question—what are your strengths and weaknesses? 1-19. Order qualifiers are characteristics of a product or service that qualify it to be considered for purchase by a customer. An order winner is the characteristic of a product or service that wins orders in the marketplace— the final factor in the purchasing decision. 1-20. a.

Most companies approach quality in a defensive or reactive mode; quality is confined to minimizing defect rates or conforming to design specifications. To compete on quality, companies must view quality as an opportunity to please the customer, not just as a way to avoid problems or to reduce rework costs. The manufacturer of Rolex watches competes on quality. b. Companies that compete on cost relentlessly pursue the elimination of all waste. The entire cost structure is examined for reduction potential, not just direct labor costs. High volume production and automation may or may not provide the most cost-effective alternative. Wal-Mart competes on cost. c. Flexibility includes the ability to produce a wide variety of products, to introduce new products and to modify existing products quickly, and, in general, to respond to customer needs. National Bicycle Industrial Company competes on flexibility. d. Competing on speed requires a new type of organization characterized by fast moves, fast adaptations, and tight linkages. Citicorp competes on speed. e. Competing on dependability requires a stable environment, adequate resources, high standards for performance, and tight control. Maytag competes on dependability. F. Competing on service requires closeness to the customer, availability of resources, attention to detail, and flexible operations. Ritz-Carlton competes on service.

1-21. Operations can play two important roles in corporate strategy: (1) it can provide support for the strategy of a firm (help with order qualifiers), and (2) it can serve as a firm’s distinctive competence (win orders). 1-22. Strategic decisions in operations and supply chain management involve products and services, processes and technology, capacity and facilities, human resources, quality, sourcing, and operating systems. 1-23. Policy deployment tries to focus everyone in an organization on common goals and priorities by translating corporate strategy into measurable objectives down through the various functions and levels of the organization. As a result, everyone in the organization should understand the strategic plan, be able to derive several goals from the plan, and determine how each goal ties into their own daily activities. 1-24. The balanced scorecard examines a firm’s performance in four critical areas – its finances, customers, processes and capacity for learning and growing. Although operational excellence is important in all four areas, the tools in operations are most closely associated with process. 1-25. Student answers will vary. 1-26. Student answers will vary. The balanced scorecard worksheet in Table 2.1 is helpful. Finances might refer to future income, customers to potential employers who are interested in both grades and experience, processes to how students will raise their grades and gain experience, and learning and growing to developing skills in several areas.

Solutions to Problems 1-1.

The Blacksburg store is the most productive. Store Sales volume Labor hours Productivity

1-2.

Blacksburg $12,000 60 $200

Charlottesville $60,000 500 $120

Danville $25,000 200 $125

Danville is the most productive.

Based on productivity, the Annandale store should be closed. Other factors to consider include total revenue, potential for growth, and options for reducing costs.

Sales volume Labor hours Labor cost/hr Rent Productivity 1-2.

Annandale $40,000 250 $160

Annandale $40,000 250 $15.00 $2800 $6

Blacksburg $12,000 60 $10.00 $1200 $7

Charlottesville $60,000 500 $9.00 $2000 $9

Danville $25,000 200 $8.50 $800 $10

a. By number, Jim was more productive last year.

Hours fishing Bass caught Bass/hr

Last yr 4 12 3

This yr 6 15 2.5

Last yr 4 12 20 60

This yr 6 15 25 62.5

b. By weight, Jim was more productive this year.

Hours fishing Bass caught Average weight Lbs/hr 1-4.

Productivity could be measured by total account dollars per hour worked, new account dollars per hour worked or existing account dollars per hour worked. Bates is the most productive based on total output. Albert and Duong have the newest accounts, and thus the greater potential returns in the future. However, Duong cannot work many more hours a week and Bates is only working half time. Bates has the potential to sell more if he works more hours. Agents New accounts Existing accounts Labor hours Total $/hr $ New accts/hr $ Existing accts/hr

Albert $100,000 $40,000 40 $3,500.00 $2,500.00 $1,000.00

Bates $40,000 $40,000 20 $4,000.00 $2,000.00 $2,000.00

Cressey $80,000 $150,000 60 $3,833.33 $1,333.33 $2,500.00

Duong $200,000 $100,000 80 $3,750.00 $2,500.00 $1,250.00

1-5.

The U.S. is the most productive. Labor Hours 89.5 83.6 72.7

U.S. Germany Japan 1-6.

Cinncinnati 10,000 1,000

Frankfurt 12,000 2,200

Guadalajara 5,000 3,000

Bejiing 8,000 6,000

Costs (in 000’s) Labor costs Material costs Energy costs Transportation costs Overhead costs

$3,500 $3,500 $1,000 $250 $1,200

$4,200 $3,000 $1,500 $2,500 $3,000

$2,500 $2,000 $1,200 $2,000 $2,500

$800 $2,500 $800 $5,000 $500

Labor productivity Total productivity

3.14 1.16

3.38 1.00

3.20 0.78

17.50 1.46

Henry is the most productive in terms of rushing yards and touchdowns per carry. However, Fournette has the highest number of rushing yards and touchdowns. Using “carries” as the input variable skews the results. Productivity can be measured in many different ways. Make sure it is expressed in a manner that is useful to you or your customer. Candidates Rushing yards # Carries # Touchdowns Yards/carry Touchdowns/carry

1-8.

Productivity 1.52 1.20 1.40

Omar should probably close the plant in Guadalajara because its multifactor productivity is the lowest, its labor productivity is the second lowest, and its output is the least of the four plants. Units (in 000’s) Finished goods Work-in-process

1-7.

Units of Output 136 100 102

Henry 2,110 105 15

McCaffrey 3,623 875 20

Fournette 6,925 1,186 70

20.10 0.14

4.14 0.02

5.84 0.06

1 1,225 4 3 102.08

2 1,435 3 5 95.67

3 2,500 5 6 83.33

Productivity decreases from week to week. Installation Square Yards # workers # hours Square yds/hr

1-9. Center Pieces processed Workers/hr Hourly wage rate Overhead/hr Multifactor productivity a. b. c.

1 1,000 10 $5.50 $10 15.38

2 2,000 5 $10 $25 26.67

3 3,000 2 $12 $50 40.54

3c. 5,000 2 $12 $80 48.08

Work center # 3 is the most productive. With a 10% raise in center 1, productivity goes down to 14.18 pieces per dollar spent. With new equipment in center 3, productivity goes up to 48 pieces. Install the new equipment.

1-10. Material productivity is stable over the 4 weeks. Labor productivity increases in week 2 and decreases in weeks 3 and 4. Week Units of output # workers Hours per week Labor cost per hour Material (lbs.) Material cost per lb.

1 2,000 4 40 $10 286 $4

2 4,000 4 48 $10 570 $4

3 5,000 5 56 $10 720 $4

4 7,000 6 70 $10 1,000 $4

Labor productivity Material productivity Multifactor productivity

1.25 1.75 0.73

2.08 1.75 0.95

1.79 1.74 0.88

1.67 1.75 0.85

1-11. John is the most productive.

# ads sold # hours spent Output/hr

Jake

Josh

Jennifer

John

100 40 2.50

50 15 3.33

200 85 2.35

35 10 3.50

1-12. Choose Cold Case.

Purchase cost

Alaskan Seal

Brr Frost

Cold Case

Deep Freeze

$3,270

$4,000

$4,452

$5,450

3.61

3.88

6.68

29.07

Daily energy consumption (kwh) Cost per kwh

$0.10

Daily energy cost

$0.36

$0.39

$0.67

$2.91

Daily purchase cost

$2.99

$3.65

$4.07

$4.98

Total cost

$3.35

$4.04

$4.73

$7.88

Volume (cu ft)

1-13.

Productivity (cu ft/$)

7.47

8.66

10.35

10.02

Cost/cu ft

$0.13

$0.12

$0.10

a. Productivity of current process = 100/10 = 10 b. Productivity of proposed process = 200/15 = 13.33 c. Multifactor productivity = Output/ ($ material + $ hours) = 100 / [(1000*6.80) + (10*25)] = 0.0142 lbs. per dollar d. Multifactor productivity = Output/ ($ material + $ hours) = 200 / [(1800*6.80) + (15*25)] = 0.0159 lbs. per dollar e. Choose the new process.

1-14.

a. Labor productivity = Output/Labor hrs = 60 jeans /(3 workers *8 hours) = 2.5 pairs of jeans per hr b. Manufacturing cost = (Labor cost + Material cost + Energy cost + Machine cost) per pair of jeans = [(3 workers * 8 hrs * $20) + (60 * $10) + (3 machines * 8 hrs*$1) + (3 machines *8 hrs*$10)]/60 = 1344/60 = $22.40 Since the problem provides us with the resources to produce 60 pairs of jeans, we calculate the cost of producing 60 pairs and divide the result by 60 to get the per unit manufacturing cost. c. Multifactor productivity = Output / (Labor cost + Material cost + Energy cost + Machine cost) = 60/ [(3 workers * 8 hrs * $20) + (60 * $10) + (3 machines * 8 hrs*$1) + (3 machines *8 hrs*$10)] = 60/1344 = 0.0446 pairs of jeans per $ input Multifactor productivity is the number of jeans that can be produced per dollar of resource used. Note, the reciprocal of multifactor productivity is the cost to manufacture one pair of jeans, $22.40.

1-15.

The most productive countries are Taiwan, China, Poland, Czech Republic and Singapore. The least productive countries are Norway, Germany, Belgium, Sweden and Switzerland.

See full table below.

Productivity per hr labor*

Hourly compensation**

Turkey

6.09

6.90

Mexico

3.91

5.37

India

1.69

5.33

Taiwan

9.82

5.30

Philippines

2.06

4.85

Hungary

8.60

4.42

Czech Republic

10.71

3.73

China

4.11

3.65

Brazil

7.98

2.26

Norway

48.62

1.97

United Kingdom

28.41

1.90

United States

39.03

1.84

Japan

26.46

1.78

France

37.72

1.78

Italy

32.49

1.63

Germany

43.18

1.62

South Korea

22.98

1.61

Argentina

16.77

1.61

Sweden

41.68

1.56

Switzerland

60.36

1.08

Country

* from Figure 1.10 **from Figure 1.6

Productivity per $ labor

CASE SOLUTION 1.1: Visualize This 1.

It is difficult to follow the four steps of strategy formulation for this case. Students will be able to easily identify VT’s core competency but will struggle with its primary task, and without a product, it’s impossible to determine an order winner and order qualifiers. “Developing the next generation of visualization tools” is probably not a marketable task. Students will come up with a variety of ideas from their Internet search.

Student answers will vary depending on how the primary task is derived in question 1.

That’s the crux of the problem for this case. Isaac needs to find a way to keep his business going to obtain the capital to pursue his dream. Great for class discussion.

(1) and (3) are more in keeping with VT’s earlier projects but require more hardware and do not promise future business. (2) is the most time-consuming, least challenging, but most sustainable. (4) and (5) are the most lucrative but do not advance VT’s knowledge of the field.

The selection of projects should reinforce the strategy determined by the student. This case is based on an actual situation. The company chose projects (1) and (3). The museum job consumed so much time and resources that the company had to turn down the bank training job. Without a “product” and no immediate repeat business, the company folded and the owner went back to academe. A student took on project (5), became quite successful and now has operations in three states.

CASE SOLUTION 1.2: Whither an MBA at Strutledge? 1.

The board of Regents should look at the proposal carefully and identify first what they are trying to achieve with this new program. If the program fits within their mission, and if they have the resources to pursue it, they need to assess the likelihood of their success or failure. It doesn’t appear that the board has sufficient information or insight to make the decision. A lot of questions remain. The focus of the program (i.e., interdisciplinary, problem solving, etc.) doesn’t seem like much of a focus at all. The desire to “try anything” to get more students is troublesome. A new program that Strutledge can’t support would damage their reputation. Strutledge needs to gather more information before a decision can be made.

Strutledge should go through the process of identifying its primary task. This would include the type of students it wishes to serve and their future role in society (i.e., community, state, regional, national, global). A clear assessment of Strutledge’s core competence is also needed. What special resources does the college have? What is it best known for? How does it compare to other institutions of similar size and mission? After those issues have been settled, the college needs to find out what its customers (i.e., students) look for when deciding where to go to school. What are some basic requirements that Strutledge should meet (i.e., order qualifiers)? What factor prompts the final determination of which school to attend (i.e., order winner)? If, as is hinted in the case, the ability to find employment upon graduation is important to prospective students, then the college should gather information from potential employers about their needs. It may very well be that an MBA program is needed in the area, but this needs to be determined from data. Only after the determination has been made, that the area needs another MBA program, should Strutledge explore the possibility of providing it. If the college concludes that it has the skills and resources necessary to pursue the task, then it should try to position itself properly in the market and find a special niche for its particular MBA program.

CASE SOLUTION 1.3 – Weighing Options at the Weight Club A Balanced Scorecard for the Weight Club Key Performance Indicator

Goal

Revenue Growth

Attract new customers

% increase in customers

25%

Quality

Meet or exceed customer needs

% customers satisfied

100%

Retention

Build sustainable customer base

% membership renewals

75%

# exercise classes/week

Fitness

Incr. participation in exercise classes Increase use of personal trainers

# client hours/week

100

Processes

Finances

Objectives Generate revenue for firstclass facility

Customers

Dimension

Enhance client experience Client services

Learning & Growing

Facilitate use of services Equipment maintenance

Maintain equipment in top working condition

Program development

Develop professional staff

Facility development

Provide first-class facilities and equipment

Organizational development

Develop management and administrative skills

% increase in revenue

30%

% participation in customer orientation # massage appointments/week Time required for check-in Hours of child care/week % fully operational % on regular maintenance schedule % new classes # innovative suggestions % equipment new or updated Months until facility expanded/ renovated

75% 200 1 min 90% 95% 60% 25 30 100% 6

# persons on Board of Directors

# full-time managers

Supplement 1 Operational Decision-Making Tools: Decision Analysis S1-1. a. Minimin: South Korea 15.2 China 17.6 Taiwan 14.9 Poland 13.8 Mexico 12.5  minimum Select Mexico b. Minimax: South Korea 21.7 China 19.0  minimum Taiwan 19.2 Poland 22.5 Mexico 25.0 Select China c. Hurwicz ( = 0.40 ) : South Korea: 15.2 ( 0.40 ) + 21.7 ( 0.60 ) = 19.10 China: 17.6 ( 0.40 ) + 19.0 ( 0.60 ) = 18.44 Taiwan: 14.9 ( 0.40 ) + 19.2 ( 0.60 ) = 17.48  minimum Poland: 13.8 ( 0.40 ) + 22.5 ( 0.60 ) = 19.02 Mexico: 12.5 ( 0.40 ) + 25.0 ( 0.60 ) = 20.0 Select Taiwan d. Equal likelihood: South Korea: 21.7 ( 0.33) + 19.1( 0.33) + 15.2 ( 0.33) = 18.48 China: 19.0 ( 0.33) + 18.5 ( 0.33) + 17.6 ( 0.33) = 18.18 Taiwan: 19.2 ( 0.33) + 17.1( 0.33) + 14.9 ( 0.33) = 16.90  minimum Poland: 22.5 ( 0.33) + 16.8 ( 0.33) + 13.8 ( 0.33) = 17.52 Mexico: 25.0 ( 0.33) + 21.2 ( 0.33) + 12.5 ( 0.33) = 19.37 Select Taiwan S1-2.

EV ( South Korea ) = 21.7 (.30 ) + 19.1(.40 ) + 15.2 (.30 ) = 18.71

EV ( China ) = 19.0 (.30 ) + 18.5 (.40 ) + 17.6 (.30 ) = 18.38 EV ( Taiwan ) = 19.2 (.30 ) + 17.1(.40 ) + 14.9 (.30 ) = 17.07  minimum EV ( Poland ) = 22.5 (.30 ) + 16.8 (.40 ) + 13.8 (.30 ) = 17.61

EV ( Mexico ) = 25.0 (.30 ) + 21.2 (.40 ) + 12.5 (.30 ) = 19.73 Select Taiwan

Expected value of perfect information = 19 (.30 ) + 16.8 (.40 ) + 12.5 (.30 ) = 16.17 EVPI = 16.17 − 17.07 = $ − 0.9 million

The EVPI is the maximum amount the cost of the facility could be reduced (.9 million) if perfect information can be obtained. S1-3. a. Maximax criteria: Office building 4.5  maximum Parking lot 2.4 Warehouse 1.7 Shopping mall 3.6 Condominiums 3.2 Select office building b. Maximin criteria: Office building 0.5 Parking lot 1.5  maximum Warehouse 1.0 Shopping mall 0.7 Condominiums 0.6 Select parking lot c. Equal likelihood: Office building: 0.5 ( 0.33) + 1.7 ( 0.33) + 4.5 ( 0.33) = 2.21  maximum Parking lot: 1.5 ( 0.33) + 1.9 ( 0.33) + 2.4 ( 0.33) = 1.91 Warehouse: 1.7 ( 0.33) + 1.4 ( 0.33) + 1.0 ( 0.33) = 1.35 Shopping mall: 0.7 ( 0.33) + 2.4 ( 0.33) + 3.6 ( 0.33) = 2.21  maximum Condominiums: 3.2 ( 0.33) + 1.5 ( 0.33) + 0.6 ( 0.33) = 1.75 Select office building or shopping mall d. Hurwicz criteria ( = 0.3) : Office building: 4.5 ( 0.3) + 0.5 ( 0.7 ) = 1.70 Parking lot: 2.4 ( 0.3) + 1.5 ( 0.7 ) = 1.77  maximum Warehouse: 1.7 ( 0.3) + 1.0 ( 0.7 ) = 1.21 Shopping mall: 3.6 ( 0.3) + 0.7 ( 0.7 ) = 1.57 Condominiums: 3.2 ( 0.3) + 0.6 ( 0.7 ) = 1.38 Select parking lot S1-4. a) EV ( Office building ) = .5 (.50 ) + 1.7 (.40 ) + 4.5 (.10 ) = 1.38

EV ( Parking lot ) = 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 mall ) = 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 = $0.75 million

S1-5. a. Maximax: Risk fund, maximax payoff = $167,000 b. Maximin: Savings bond maximin payoff = $30,000 c. Equal likelihood: Bond fund, maximum payoff = $35,000 S1-6. a. Best decision, given probabilities: Bond fund, maximum payoff = $35,000 b. Expected value given perfect information = (5*0.1) + (4*0.2) + (4.2*0.4) + (9.3*0.2) + (16.7*0.1) = $6.51 EVPI = $6.51- $3.50 = $3.01 or $30,100 S1-7. Since the payoff table includes “costs,” the decision criteria must be reversed. a. Minimin: Philippines, minimum cost = $170,000 b. Minimax: Brazil, minimum cost = $570,000 c. Equal likelihood: Philippines, minimum cost = $399,000 d. Minimax regret: Philippines, minimum regret = $70,000 S1- 8

a. EV (China) = 5.328 EV (India) = 5.375 EV (Philippines) = 5.218 EV (Brazil) = 5.178 Select EV (Mexico) = 5.202 b. EV given perfect information = $(1.7) (0.09) + (3.8) (0.27) + (5.4)(0.64) = $4.635 EVPI = $5.178 – 4.365 = $0.813 or $813,000

S1– 9. 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 = $8.91 million d. Minimax regret: Philippines, minimum regret = $2 million S1-10.

EV (China) = $10.91 EV (India) = 7.21 Select EV (Thailand) = 9.77 EV (Philippines) = 7.54

S1-11. a. Product Widget

Expected Value 160, 000 ( 0.2 ) + 90, 000 ( 0.5 ) − 50, 000 ( 0.3) = $62, 000

Hummer

70, 000 ( 0.2 ) + 40, 000 ( 0.5 ) + 20, 000 ( 0.3) = $40, 000

Nimnot

45, 000 ( 0.2 ) + 35, 000 ( 0.5 ) + 30, 000 ( 0.3) = $35,500

The best option is to introduce the widget. b. EV given perfect information: 160, 000 ( 0.2 ) + 90, 000 ( 0.5 ) + 30, 000 ( 0.3) = $86, 000. EV without perfect information: Widget at $62,000. Value of perfect information: $86,000 − $62,000 = $24,000 The company would consider this a maximum; since perfect information is rare, it would probably pay less than $24,000. c. Maximax: Introduce widget, maximax payoff = $160,000

Maximin: Introduce nimnot, maximin payoff = $30, 000. Minimax regret: Introduce widget, Minimax regret = $80,000 Equal likelihood: Introduce widget, maximum payoff = $66,000 S1-12. a. Maximax: Major physical revision, maximum payoff = $972,000 b. Maximin: Paperback, maximum payoff = $68,000 c. Equal likelihood: Major content revision, maximum payoff = $419,430 d. Hurwicz: Major content revision, maximum payoff = $273,900

S1-13. Publication Decision Paperback Similar revision Major content revision Major physical revision

Expected Value $216,290 386,340 468,780 405,970

Best decision = major content revision Overall “best” decision appears to be a “major content revision” EVPI = (.23) (68,000) + (.46)(515,000) + (.31)(972,000) − 468,780 = $85,080 This is the maximum amount Wiley would pay an “expert” for additional information about the future competitive market. S1-14. a. Maximax: Singapore, maximum payoff = $71 million b. Maximin: Kaohsiung, maximum payoff = -$15 million c. Equal likelihood: Kaohsiung, maximum payoff = $28.05 million d. Hurwicz: Singapore, maximum payoff = $37.8 million e. Minimax regret: Singapore, minimum regret = $9 million S1-15. Expected value Port Hong Kong Singapore Shanghai Busan Kaohsiung

Expected Value $22.99 34.52 24.54 28.30 33.66

a. Best decision = Singapore b. Singapore appears to be the best “overall” decision. S1-16. a. Expected value Lease Decision 1 – year 2 – year 3 – year 4 – year 5 – year

Expected Value $65,980 103,010 133,810 154,300 114,210

b. EVPI = (.17) (1,228,000) + (.34)(516,000) + (.49)(−551,000) − 154,300 = $237,740 This is the maximum amount the restaurant owner would pay an energy “expert” for additional information about future energy prices. S1-17. a. Maximax: Food court, maximum payoff = $87,000 b. Maximin: Child care center, maximum payoff = $17,000 c. Hurwicz: Lockers and showers, maximum payoff = $32,250 d. Equal likelihood: Lockers and showers, maximum payoff = $34,980 S1-18. Service Facility Child care center Swimming pool Lockers and showers Food court Spa

Expected Value $30,560 7,610 44,150 15,440 20,580

Best decision = Lockers and showers S1-19. a.

Payoff table:

Stock (lb) 20 21 22 23 24

20 0.10 20.00 18.50 17.00 15.50 14.00

21 0.20 20.00 21.00 19.50 18.00 16.50

Demand 22 0.30 20.00 21.00 22.00 20.50 19.00

23 0.30 20.00 21.00 22.00 23.00 21.50

24 0.10 20.00 21.00 22.00 23.00 24.00

EV ( 20 ) = $20

EV ( 21) = 18.50 ( 0.1) + 21( 0.2 ) + 21( 0.3) + 21( 0.3) + 21( 0.1) = $20.75 EV ( 22 ) = 17 ( 0.1) + 19.50 ( 0.2 ) + 22 ( 0.3) + 22 ( 0.3) + 22 ( 0.1) = $21.00

EV ( 23) = 15.50 ( 0.1) + 18 ( 0.2 ) + 20.50 ( 0.3) + 23 ( 0.3) + 23 ( 0.1) = $20.50 EV ( 24 ) = 14 ( 0.1) + 16.50 ( 0.2 ) + 19 ( 0.3) + 21.50 ( 0.3) + 24 ( 0.1) = $19.25

Order 22 lb of apples for a profit of $21.00. b. Maximax: Stock 24 lb for a maximax profit of $24.00. Maximin: Stock 20 lb for a maximin profit of $20.00.

S1-20. a. Payoff table: Demand Stock (lb) (boxes) 25 26 27 28 29 30

25 0.10 50 49 48 47 46 45

26 0.15 50 52 51 50 49 48

27 0.30 50 52 54 53 52 51

28 0.20 50 52 54 56 55 54

29 0.15 50 52 54 56 58 57

30 0.10 50 52 54 56 58 60

EV ( 25 ) = 50 ( 0.1) + 50 ( 0.15 ) + 50 ( 0.3) + 50 ( 0.2 ) + 50 ( 0.15 ) + 50 ( 0.1) = $50.00 EV ( 26 ) = 49 ( 0.1) + 52 ( 0.15 ) + 52 ( 0.3) + 52 ( 0.2 ) + 52 ( 0.15 ) + 52 ( 0.1) = $51.70

EV ( 27 ) = 48 ( 0.1) + 51( 0.15 ) + 54 ( 0.3) + 54 ( 0.2 ) + 54 ( 0.15 ) + 54 ( 0.1) = $52.95 EV ( 28 ) = 47 ( 0.1) + 50 ( 0.15 ) + 53 ( 0.3) + 56 ( 0.2 ) + 56 ( 0.15 ) + 56 ( 0.1) = $53.30

EV ( 29 ) = 46 ( 0.1) + 49 ( 0.15 ) + 52 ( 0.3) + 55 ( 0.2 ) + 58 ( 0.15 ) + 58 ( 0.1) = $53.05 EV ( 30 ) = 45 ( 0.1) + 48 ( 0.15 ) + 51( 0.3) + 54 ( 0.2 ) + 57 ( 0.15 ) + 60 ( 0.1) = $52.35 Best decision: Stock 28 boxes, for a profit of $53.30. b. Expected value under uncertainty: EV = 500 ( 0.10 ) + 52 ( 0.15 ) + 54 ( 0.30 ) + 56 ( 0.20 ) + 58 ( 0.15 ) + 60 ( 0.10 ) = $54.90 EVPI = $54.90 − $53.30 = $1.60

S1-21. 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; minimum regret = $4. S1-22. 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 S1-23.

S1-24.

EV (Shanghai) = 5.328 EV (Mumbai) = 5.375 EV (Manila) = 5.218 EV (Santos) = 5.178 Select EV (Veracruz) = 5.202 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

S1-25. 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. S1-26.

S1-27. a. Maximax = Real Estate b. Maximin = Nursing c. Equal Likelihood: select Real Estate Graphic design = $170,000 Nursing = $187,500 Real Estate = $202,500 Medical Technology = $195,000 Culinary technology = $170,000 Computer information technology = $186,250 d. Hurwicz (alpha = 0.25): select Nursing Graphic design = $141,250 Nursing = $161,250 Real Estate = $158,750 Medical Technology = $157,500 Culinary technology = $136,250 Computer information technology = $158,750 S1-28. EV (Graphic design) = $164,250 EV (Nursing) = $183,500 EV (Real Estate) = $174,400 EV (Medical Technology) = $187,500 EV (Culinary technology) = $149,250 EV (Computer information technology) = $174,750

S1-29. a. Maximax = Hong Kong b. Maximin = Pusan c. Equal likelihood: Shanghai = $0.44 billion Singapore = $0.37 billion Pusan = $0.43 billion Kaoshiung = $0.41 billion Hong Kong = $0.47 billion d. Hurwicz (alpha = .55): Shanghai = $0.47 billion Singapore = $0.41 billion Pusan = $0.46 billion Kaoshiung = $0.54 billion Hong Kong = $0.77 billion S1-30. EV(Shanghai) = $0.608 billion EV(Singapore) = $0.606 billion EV(Pusan) = $0.502 billion EV(Kaoshiung) = $0.487 billion EV (Hong Kong) = $0.724 billion

S1-31. EV ( snow shoveler ) = $30 (.12 ) + 60 (.19 ) + 90 (.24 ) + 120 (.22 ) + 150 (.13) + 180 (.08 ) + 210 (.02 ) = $101.10 The cost of the snow blower ($575) is much more than the annual cost of the snow shoveler, thus on the basis of one year the snow shoveler should not be purchased. 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.

S1-32.

Since cost of installation ($900,000) is greater than expected value of not installing ($552,000), do not install an emergency power generator.

S1-33.

Select strategy 3; Change oil regularly; EV = $98.80

S1-34.

Select Strategy 4; Change oil and sample; EV = $716.40

S1-35. a.

.98 9.2 x + 1.5 (1 − x ) + (.02 )(1.5) = 3.810 .98  7.7 x + 1.5 + .030 = 3.810 7.546 x + 1.47 + .030 = 3.810 7.546 x = 2.31

x = .306 probability of winning in overtime

S1-36.

S1-37. The following table includes the medical costs for all the final nodes in the decision tree. Expense 100 500 1,500 3,000 5,000 10,000

E (1) = 954 E ( 2 ) = 976.5 E ( 3) = 810 Select plan 3

Plan 1 481 884 984 1,134 1,334 1,834

Plan 2 160 560 1,290 1,440 1,640 2,140

Plan 3 318 438 738 1,188 1,788 3,288

S1-38.

S1-39.

CASE S1.1: Whither an MBA at Strutledge? -Continued a. Maximax: IT, maximum payoff = $517,000 b. Maximin: Health Administration, maximum payoff = −$75,000 c. Equal likelihood: Nursing, maximum payoff = $114,500 d. Hurwicz: Nursing, maximum payoff = $86,000 e. They do not have sufficient insight into the probability of the future success of the programs to indicate either optimism or pessimism; or for “political” reasons they feel it is imprudent to express a “preference.” f. Best decision = Nursing Graduate Program MBA Computer Science Information Technology Nursing Health Administration

Expected Value −27,470 −45,000 10,790 126,760 124,250

g. Nursing appears to be the best overall decision. h. Depends on student answer.

CASE S1.2: Transformer Replacement at Mountain State Electric Service The decision tree solution for this problem is shown below. The decision should be to retain the existing transformer; the cost of replacement ($85,000) is greater than the cost of retention ($61,000).

CASE S1.3: Evaluating Projects at Nexcom Systems Project 1 2 3 4 5

EV 404,368 434,976 442,891 344,490 262,252

2 - Quality Management Answers to Questions 2-1.

Consumers perceive quality to be how well a product meets its intended use—that is, how well it does what it is supposed to do—whereas from the producers’s perspective, quality is how well the product conforms to its design during the production process.

2-2.

1. Performance: operating characteristics of a product 2. Features: extra items added to basic characteristics 3. Reliability: probability that a product will operate properly 4. Conformance: the degree to which a product meets standards 5. Durability: how long the product lasts 6. Serviceability: ease and speed of repair and courtesy of repair person 7. Aesthetics: how a product looks, feels, sounds, smells, or tastes 8. Safety: Assurance that the customer will not suffer harm; especially important for autos. 9. Other: subjective perceptions based on brand name or advertising

2-3.

Quality of design is the degree to which quality characteristics are designed into a product, whereas quality of conformance is how effectively the production process is able to conform to the specifications required by design.

2-4.

The cost of quality assurance is the cost of maintaining an effective quality program and includes prevention and appraisal costs. The cost of nonconformance, or poor quality, is the result of internal and external failures. These two costs react oppositely to each other; as the cost of quality assurance increases, the cost of poor quality decreases.

2-5.

Internal failure costs are incurred when poor quality is discovered before the product is delivered to the customer, whereas external failure costs are incurred after a customer receives a poor-quality product. Internal failure costs include scrap, rework, process failure, and downtime, whereas external failure costs include customer complaints, product returns, warranty claims, product liability, and lost sales.

2-6.

The contractor could be experiencing low productivity yields and have extensive internal failure costs, including scrap, rework, process failure, and downtime costs.

2-7.

a. From the consumer’s (e.g., student or parent) perspective, quality is probably determined by whether the college education provides the job opportunity expected and whether the graduate perceives he or she has acquired an anticipated level of knowledge that will enable the graduate to perform the job effectively. From the producer’s (e.g., university) perspective, quality is how effectively it is able to deliver knowledge (i.e., required courses) and provide the quality of life experience expected by the student. b. The education achieved by the student provides the job opportunities expected and a level of knowledge that enables the graduate effectively to perform the job achieved. c. Quality-assurance costs include the cost of hiring the best faculty, administrators, and support personnel, the cost of designing and redesigning courses and curriculum to meet changing needs, the cost of providing a good physical and mental environment (i.e., housing, food, entertainment, security, etc.), the cost of modern technical teaching equipment, the cost of information systems, and the cost of assessing alumni satisfaction with their education. Costs of poor quality include students who fail or drop out, reduced funding from the state or private donors, and fewer

enrollments. d. Quality circles could be developed within administrative and operational units and academic departments. Circles might include both faculty or administrators and classified employees. The normal quality circle stages of training, problem identification, analysis, solution, and presentation could be followed. 2-8.

Improving quality will increase product yield—that is, the number of acceptable units—thus increasing productivity.

2-9.

The cost of poor quality could include external failure costs for customer complaints, returned DVD players to be repaired under warranty, lost future sales, and liability costs if someone is hurt because of the problem. Costs of quality improvement might include improved design costs for the DVD player, process costs, and inspection costs for the final product and at various stages of the production process.

2-10. Cell phone: Visual attractiveness, size, weight, clarity of sound and resolution of screen, memory, battery life, etc. Pizza: Size, ingredients, taste, smell, service in delivery, temperature. Athletic shoes: Size, weight, comfort, visual attractiveness, durability. 2-11. The input is customer inquiries and the final product is responses that result in customer satisfaction. Associated quality costs might include prevention costs, such as designing a telephone system to ensure prompt connections without waiting and a properly designed computer system to provide accurate customer account information, and training costs to make certain service operators are courteous and knowledgeable. Appraisal costs might include the cost of monitoring service calls to ascertain response rates and operator courtesy. Poor quality might result in complaints from customers and lost accounts. A quality management program could incorporate a system to monitor calls to ensure prompt, courteous, and knowledgeable service. An employee-involvement program, wherein operators might identify problems, would be beneficial. 2-12. Prevention costs are directed at preventing poor quality products from reaching the customer, thus avoiding the various internal and external failure costs associated with poor quality. 2-13. It is important to have a means for assessing the impact of quality improvement programs on the organization’s profitability and productivity. 2-14. W.E. Deming: Introduced the Japanese to quality management principles and philosophy, embodied in his 14 points. Joseph Juran: A major contributor to the Japanese quality movement. Phillip Crosby: Changed general perceptions of cost of quality and promoted zero defects. Armand Feigenbaum: Introduced the concept of total quality control, a total company-wide approach to quality management. Kaoru Ishikawa: Introduced quality circles and cause and effect diagrams. Genichi Taguchi: Developed the Taguchi method for product and process design.

2-15. The Baldrige Award has had a pervasive impact on American companies, in general promoting quality improvement. Thousands of companies request award applications each year to use simply to establish quality management programs based on Baldrige Award criteria.

2-16. This should be a student project. The journal can be found in most libraries, and the articles are generally easy to read. 2-17. The student could provide many reasons for failure including lack of total commitment, ineffective planning, goals too easy or too difficult to achieve, improper measurement techniques, ineffective leadership, not enough employee training, etc. See G. Salegra and Farzaneh, “Obstacles to Implementing Quality,” Quality Progress, 33, no. 7 (July 2000): 53–57. 2-18. The dimensions of quality for a service company are located in the text. The student should identify these or similar ones for the company they select. 2-19. The two service companies should be in the community and the quality characteristics the students will tend to focus on will include courtesy and quickness of service. 2-20. Although students in this class might suggest that grades are a quality measurement a more realistic approach to evaluation are student evaluations of the class or surveys of students. Quality characteristics might include course organization, presentation of lectures, class environment, physical appearance of the classroom, schedule (i.e., are the lectures completed on time), the quality of supplementary material, physical appearance and demeanor of the instructor, including friendliness and courtesy, the accuracy and completeness of assignments, etc. 2-21. The answer depends on the company selected by the student. For example, there is a particular hotel that has never gotten a room reservation right for us, and, the instructions for ordering tickets at the web site for the 1996 Olympics in Atlanta were littered with pitfalls. Airlines are a favorite example of a poor quality service for students who travel. 2-22. A similar question to 26. Restaurants, retail stores and grocery stores are examples of local businesses that, in our experience, tend to vary in quality. We have never had a bad ordering experience with L.L. Bean although that’s not true of some other mail order operations we have dealt with. In most cases, if a service has been identified by the student it will be because of courteous, helpful employees, while if a manufacturing product has been identified, it will be because of superior physical traits, such as durability. 2-23. TQM tends to give some focus and structure to strategic planning. TQM provides identifiable goals, and many well-documented initiatives for quality improvement such as quality circles, employee training, empowerment, etc. TQM also provides a means for measuring success which is essential in a strategic planning process. 2-24. Many U.S. suppliers cannot do business with companies overseas unless they have ISO certification. In addition, many U.S. companies also desire or request their suppliers to comply with ISO 9000 standards. 2-25. Common characteristics that the students will discover include strong leadership at the top, total company commitment, employee training, involvement and empowerment, challenging goals for quality achievement, focus on customer satisfaction, and extensive use of statistical quality control techniques, among other things. 2-26. Some companies believe their quality is “good enough.” However, primary reasons for not implementing a TQM program are lack of time and the cost involved; some companies do not have the resources available to undertake a TQM program.

2-27.

This will depend on the web site the student accesses. In general, they should adapt the attributes described for services.

2-28.

This depends on the airline the student selects. Example defects they might mention are flight delays or cancellations, lost luggage or luggage mishandling, discourteous employees, wrong or misleading flight information, uncomfortable seats, etc.

2-29.

This will depend on which websites the student selects.

2-30.

Categories of possible quality problems might be related to the ordering process, pizza construction, pizza ingredients, packaging/boxing, time to receive order, order accuracy/correctness, and pricing.

2-31.

If someone purchases a residence, then the dwelling is more of a product. However, renting a dormitory room or an apartment tends to fall into the service category because it is part of an ongoing process or interaction between the owner and renter. In other words, the owner retains responsibility for the product, i.e., the dwelling. As such, the quality of the living accommodation should be assessed according to the dimensions of quality for a service.

2-32.

Unfulfilled, late and erroneous orders, and defective items, would obviously have a negative effect on Amazon’s reputation. Amazon asks customers to rate sellers on its “open market” according to a number of criteria, and its assumed that sellers would be banned if they had continued poor performance. In their answer, students could reference the Amazon seller rating system.

2-33. The calculation of the “Power Circle Ratings” can be found on the J.D. Power Website http://www.jdpower.com/about-us/jdpower-ratings. To calculate Power Circle Ratings, J.D. Power begins with the syndicated study index scores or a specific standard of measurement which can be found, in most cases, in the associated press release. An example of an index score is found in the J.D. Power U.S. Retail Banking Satisfaction Study, where companies are ranked according to overall index scores based upon weighted responses to several survey factors. An example of a specific standard of measurement is found in the J.D. Power Initial Quality Study, where vehicles are ranked according to reported Problems per 100 vehicles (PP100). Using these measurements, Power Circle Ratings are calculated based on the range between the product or service with the highest score and the product or service with the lowest score. J.D. Power generates a Power Circle Rating of five, four, three, or two. 2-34. The answer will depend on the product and service the student selects. 2-35. The K&N Management Website at http://knmanagement.com/our-story/ has extensive information about the company, including several videos, and another source of information is the company profile on the Baldrige Award Website for award recipients at http://patapsco.nist.gov/Award_Recipients/index.cfm.

2-36.

The answer will depend on the service provider the student selects.

2-37.

They all focus on all aspects of the dining experience including the overall customer experience, including waiting time, service, food and price. As such, in many ways they are clones of each other, and must compete on all quality attributes. The student might want to reference the Darden Restaurant Group (Red Lobster, Olive Garden) and Bloomin’ Brands (Outback, Bonefish Grill, Carrabba’s).

2-38.

According to the Baldrige website, there is increasing pressure in the U.S. to overhaul our

health care system, and health care organizations around the country are looking for ways to improve safety and outcomes, while reducing cost. The history of the Baldrige Program shows that health care organizations of any size and type and in any location can benefit from using their health care criteria. It is likely that healthcare organizations have made more applications for the Baldrige Award, in part to effectively compete in an industry under close public scrutiny, and partly because health care organizations have the resources to prepare the extensive Baldrige application materials. 2-39. Sustainability, in its broadest sense, focuses on improving the “quality” of the workplace (i.e., human resources), which is one of the seven major categories of criteria for the Baldrige Award criteria. 2-40.

The quality management systems used by Amazon and Target can be applied to any number of “different” industries, including airlines; the fact that they have not are due more to the commitment of the company. In the case of airlines, their lack of competition and near monopolies in certain markets contributes to their unwillingness to adopt quality management principles. Further airlines seem to have concluded that they must compete on the basis of price and not quality, especially in the customer service areas.

2-41.

The obvious reason that companies outsource their customer service activities is labor cost; it’s cheaper. This is the same reason that manufacturing and retail companies have outsourced their supply to places like China and Mexico. Many quality-conscious companies like P&G and Wal-Mart seek to incorporate their overseas suppliers into their own quality management systems, whereas some companies have not.

2-42.

Apple’s overall product quality, customer service and technical innovations create demand that overcomes potential “glitches” in its initial product designs.

2-43.

Companies can insure quality across its supply chain by requiring its suppliers to adopt its own quality management system, and by closely monitoring its effectiveness. Companies that have failed in their attempts to insure quality among their overseas suppliers in countries like China, have allowed quality to be monitored locally or by a third-party firm, while successful companies have used their own offices and employees to monitor quality overseas, i.e., a physical presence is generally required combined with the willingness to change suppliers if quality is not forthcoming.

2-44. In the workplace, labor, materials, manufacturing processes, sourcing and transportation. 2-45.

Common characteristics include a commitment to product and process innovation, emulating “best practices” of quality leaders, and supply chain efficiency. These companies are typically involved in ISO certification processes, employ Six Sigma, and often adopt Baldrige Award criteria.

2-46.

There are several companies identified in the chapter that use Six Sigma, and the student can also search “Six Sigma” on the Internet.

2-47. Categories of quality problems for flight delays could relate to employees (not enough to check in, problems with the check-in process, deplaning problems, insufficient maintenance personnel to accomplish plane turn-around, etc.), mechanical problems, luggage problems, maintenance problems, weather, flight controller problems, over booking, over scheduling, etc.

2-48. This depends on the business the student selects. If for example, they selected a restaurant they eat at frequently they would need to identify the categories of quality problems which might include employees, food quality, restaurant environment, waiting time, price, service, menu, etc. 2-49. It should be obvious to the student that the most important defects are the engine problems and faulty brakes. The priority of the quality problems is almost the reverse of the frequencies; faulty brakes are clearly the most significant category of defects. This points out that when applying Pareto analysis the degree of importance must be the same for all defect categories. If not then the categories should be weighted according to their importance in order to adjust the chart. 2-50. Marketing has direct contact with the customer. Marketing is typically responsible for the consumer research that determines the quality characteristics that customers want and need, and the price they are willing to pay for it. Marketing also informs the consumer about the quality characteristics of a product through advertising and promotion. Sales provides feedback information through its interaction with the customer, which is a determinant of product design. Research and development will explore new ideas for products and be actively involved in product innovation. Engineering translates the product quality characteristics determined by marketing and top management into a product design, including technical specifications, material and parts requirements, equipment requirements, workplace and job design, and operator training and skills. Overdesigning the product is a drain on the company’s resources and can erode profits, whereas underdesigned products will generally not meet the customer’s quality expectations. Genichi Taguchi, the Japanese quality expert, estimates that poor product design is the cause of as much as 80 percent of all defective items. It is much cheaper and easier to make changes at the design stage than at the production stage, so companies need to focus on quality at all stages of the design process. Purchasing must make sure that the parts and materials required by the product design are of high quality. Quality of the final product will be only as good as the quality of the materials used to make it. Purchasing must select vendors who share the company’s commitment to quality and who maintain their own quality management program for providing high-quality service, materials, and parts. Human resources is responsible for hiring employees that have the required abilities and skills, and training them for specific job tasks. Employees not well trained in their tasks will probably contribute to poor quality or service. Personnel also have responsibility for educating employees about quality and ways to achieve quality in their tasks. TQM requires that all employees throughout the organization be responsible for quality. Employees, collectively and individually, must not only perform their tasks according to design specifications but also be responsible for identifying poor quality or problems that may lead to poor quality and taking action to correct these problems. Performance appraisal under TQM focuses more on quality improvement and group and company achievement than on individual job performance. Distribution makes sure that high-quality products are delivered on-time and undamaged to the customer. Packaging methods and materials, storage facilities and procedures, and shipping modes must ensure that final products are protected and that customers receive them on time. 2-51.

This answer depends on the award the student selects.

2-52. This answer depends on the company the student selects. 2-53. This answer depends on the article and company the student selects. 2-54. This question was adapted from: L. Fredendall, J. Patterson, C. Lenhartz and B. Mitchell,” What Should Be Changed?”, Quality Progress 35 (1; January 2002): p. 50–59. This is an excellent article about the use of cause and effect diagrams students can be referred to.

2-55.

Black Belt—the leader of a quality improvement project, which is a full-time position Green Belt—a project team member, which is a part-time position Master Black Belt—a teacher and mentor for Black Belts which, is also a full-time position. A Black Belt would have to have led several successful projects before being certified as a Master Black belt.

2-56. Breakthrough Strategy: 1. Define the process including who the customers are and what their problems are. 2. Measure the process and collect data. 3. Analyze the data in order to develop information that provides insight into the process, including causes of defects. 4. Improve the process by making changes and measuring the results. 5. Control the improved process by monitoring it and making sure the desired performance level is sustained. 2-57. This answer depends on the project the student selects. 2-58. In general, the ACSI model is a set of causal equations that link customer expectations, perceived quality, and perceived value to customer satisfaction (ACSI). In turn, satisfaction is linked to consequences as defined by customer complaints and customer loyalty—measured by price tolerance and customer retention. There are two menu items on the ACSI website that describe, in general terms, how the ACSI is determined—“What it measures,” and “Methodology.” The student should refer to these descriptions. As an example, the student could select two fast food restaurant

chains in the fast food industry and compare the company with the highest score with the lowest scoring company and explain the reasons for the difference in scores. 2-59.

Answer depends on the personal health improvement project the student selects.

2-60.

Answer depends on the personal improvement project the student selects.

2-61.

Answer depends on the infirmary process the student selects.

2-62.

Answer depends on the registration process at the student’s university.

2-63.

In general, the Japanese recognized that even though high quality might cost more in the “short run,” in the long run it would help them gain market share, which would increase long term profits. This is something American companies did not recognize. The Japanese economic climate and business and management culture was also more conducive to quality management programs than American companies.

2-64.

The student should go to the ISO website at www.iso.org to determine these steps.

2-65.

Answer depends on the store the student selects.

2-66.

5S described in Table 16.1 is a process for identifying defects, establishing goals for improvement, and then eliminating or correcting defects to achieve the goals, must like Deming’s PDCA cycle, or the Six Sigma Breakthrough Strategy DMAIC 5 step improvement process.

2-67.

The student should refer to the ISO website at www.iso.org and the Baldrige Award website at www.nist.gov to answer this question.

2-68.

The answer should include references to some form of customer feedback, such as a “voice of the customer (VoC)” process and surveys. Since assessing customer satisfaction is a critical part of the Baldrige Award criteria, the summaries of Baldrige Award winning companies at www.nist.gov are a good source for students to learn how companies and organizations assess customer satisfaction.

2-69.

In general, this will be a subjective answer on the part of the student. A good source for this question is an article by David Goldhill titled “How American Health Care Killed My Father,” that appeared in the September 2009 online edition of The Atlantic magazine (www.theatlantic.com).

Solutions to Problems 2-1.

a. Failure costs as percentage of quality costs: 2015: 157.7 / 187.2 = 0.8424, or 84.24% 161.8 = 0.8022, or 80.22% 201.7 153.6 = 0.7288, or 72.88% 2017: 212.5 127.2 = 0.6560, or 65.6% 2018: 193.9 97.3 = 0.5830, or 58.3% 2019: 166.9

2016:

The failure costs decrease as a percentage of total quality costs. This may be attributed to an increase in product monitoring and inspection. Fewer defective products are reaching the consumer, as evidenced by the sharp decline in external failure costs. b. Prevention costs as % of quality costs: 3.2 = 0.0171, or 1.71%; 187.2 10.7 = 0.0530, or 5.3%; 2016: 201.7 28.3 = 0.1332, or 13.32%; 2017: 212.5 42.6 = 0.2197, or 21.97%; 2018: 193.9 50 = 0.2996, or 29.96%; 2019: 166.9

Appraisal costs as % of quality costs:

2015:

0 1404 or 14 04% 26 3 = . , . 187. 2 0 1448 or 14 48% 29 2. = . , . . 201 7 0 144 or 14 4% 30. 6 = . , . . 212 5 24 1 . 0 1243 or 12 43% = . , . 193. 9 0 1174 or 11 74% 19 6. = . , . 166. 9 The increase in prevention costs as a percentage of total quality costs indicates that Backwoods . American is placing more emphasis on prevention of defects rather than correction of them. Perhaps they are spending more in the areas of quality planning, product design, process, training, and information. This is contributing to a decline in the need for inspection and testing, equipment testing, and operators to test quality; thus appraisal costs decline, both absolutely and as a percentage of total costs. Prevention also contributes to the decline in external and internal failures, because fewer defective products are produced to begin with. Increases in prevention expenditures will result in a decrease in all other quality costs.

2015 2016 2017 2018 2019

Quality Sales Index 6.93 7.50 7.85 6.90 5.79

Quality-Cost Index 44.48 47.64 50.04 44.46 38.32

These index values do not provide much information regarding the effectiveness of the quality assurance program. They are, however, useful in making comparisons from one period to the next and in showing trends in product quality over time. d. Examples of quality-related costs: • Prevention: Market research, that is, producing what consumers want; purchasing only highquality down and other materials, designing an efficient and effective manufacturing process; training employees in making quality products. • Appraisal: Inspection of raw materials, work-in-process, and finished product; equipment testing (pattern cutter, sewing machines, etc.), inspection. • Internal failure: Wasted materials and labor, defective products discovered during inspection, use of inefficient processes, equipment downtime, poorly trained employees. • External failure: Defective products, customer complaints, warranty costs, lost sales, loss of good will.

2-2.

a. Product yield 2015: 20,000 (0.83) + 20,000 (1-0.83) (0.20) = 16,600 + 680 = 17,280 parkas 2016: 20,000 (0.85) + 20,000 (0.15) (0.20) = 17,000 + 600 = 17,600 parkas 2017: 20,000 (0.87) + 20,000 (0.13) (0.20) = 17,400 + 520 = 17,920 parkas 2018: 20,000 (0.89) + 20,000 (0.11) (0.20) = 17,800 + 440 = 18,240 parkas 2019: 20,000 (0.91) + 20,000 (0.09) (0.20) = 18,200 + 360 = 18,560 parkas

b. Manufacturing cost per good parka: 420,900 + 12(680) 429,060 = = $24.83 17,280 17,280 423,400 + 12(600) 430,600 2016: = = $24.47 17,600 17,600 424,700 + 12(520) 430,940 2017: = = $24.05 17,920 17,920 436,100 + 12(440) 441,380 2018: = = $24.20 18,240 18,240 435,500 + 12(360) 439,820 2019: = = $23.70 18,560 18,560 2015:

2009 : 2010 :

( ) = 441,380 = $24.20

436,100 +12 440 18,240

18,240

( ) = 439,820 = $23.70

435,500 +12 360 18,560

18,560

Improving the quality assurance program has resulted in fewer defective parkas, lower rework costs, and greater productivity. This has lowered the per-unit manufacturing costs without additional capital investment.

2-3.

a. y = (I)(%G) + (I)(1-%G)(%R) = (150)(0.83) + (150)(1-0.83)(.60) = 139.8 file cabinets b.

2-4.

145 = (150) (%G) +(150)(1-%G)(0.60) 145 = 150G + (150-150G)(0.60) 145 = 150g + 90 – 90G 55 = 60 G G = 55 / 60 = 0.916 = 91.6%

Cost with 83% quality = [27(150) + 8(15)] / 139 = $30 per cabinet Note: if don’t round until the end, $29.85 instead of $30 Cost with 90% quality = [27(150) + 8(9)] / 144 = $28.63 per cabinet

2-5.

Manufacturing cost per good product: 2017:

Yield = 32,000 (0.78) + 32,000 (0.22) (0.25) = 26,720 Product Cost = (278,000 + 3520) ÷ 26,720 = $10.54 In this case, total direct manufacturing cost = $278,000 and total direct rework = 32,000(0.22) (0.25) = $3250

2018:

Yield = 34,600 (0.83) + 34,600 (0.17) (0.25) = 30,188.50 Product Cost = (291,000 + 2,941) ÷ 30,188.50 = $9.74

2019:

Yield = 35,500 (0.9) + 35,500 (0.1) (0.25) = 32,837.50 Product Cost = (305,000 + 1,775) ÷ 32,837.50 = $9.34

Percentage change:

2-6.

2017 - 2018: (9.74 – 10.54) / 10.54 = -7.60% 2018 - 2019: (9.34 – 9.74) / 9.74 = - 4.10%

a. Product yield = 300 ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 185 cabinets

)( )( )( )( )( ) roduct yield 300( = 0 87 0 91 0 94 0 93 0 93 0 96 . . . . . . b. For a yield of 300, input would have to be I ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 300

I ( 0.6179 ) = 300 I = 486 cabinets

2-7.

a. Alt. 1: 300 ( 0.93)( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 198 Alt. 2 : 300 ( 0.87 )( 0.96 )( 0.94 )( 0.97 )( 0.93)( 0.96 ) = 204 Greatest yield

Alt. 3 : 300 ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.97 )( 0.98 ) = 197 Alt. 4 : 300 ( 0.87 )( 0.97 )( 0.94 )( 0.93)( 0.93)( 0.96 ) = 198

b. Alternative 2 will result in the highest yield and will be the most effective. 2-8.

320 (1 − 0.12 )(1 − 0.08 )(1 − 0.04 ) = 320 ( 0.88 )( 0.92 )( 0.96 ) = 248 errorless orders

2-9.

a. QPR = b. QPR = c. QPR = d. QPR =

2-10. a. QPR = b. QPR =

( 585 + 16 )(100 )

650 (18 ) + 16 ( 3.75 )

( 720 + 20 )(100 )

800 (18 ) + 20 ( 3.75 )

( 585 + 16 )(100 )

= 5.11 = 5.11

650 (16.50 ) + 16 ( 3.20 )

( 604 + 11)(100 )

650 (18 ) + 11( 3.75 ) 250 (100 )

250 ( 47 ) + 33 (16 ) 320 (100 )

= 5.24

= 2.04

320 ( 42 ) + 19.2 (12 )

2-11. a. Product cost =

= 5.58

= 2.34

( K d ) ( I ) + ( K r )( R ) Y

b. Product cost =

( $6.15)( 680 ) + (1.75 )( 2.72 ) 655.52

4186.76 = $6.39 655.52

( $6.20 )( 680 ) + (1.75)( 0.68)

673.88 4217.19 = = $6.26 673.88

Cost savings = $0.13/ order Annual savings = $0.13/ order  680 orders / day  365 days / year = $32, 266

c. It is likely that some customers who receive defective orders will not return, thus, fewer defective orders will retain more customers and also increase the number of orders. 2-12. a. QPR = b. QPR =

655.52

( 680 )( 6.15) + ( 2.72 )(1.75)

(100 ) = 15.66

673.88 (100 ) = 15.98 ( 680 )( 6.20 ) + ( 0.68)(1.75)

2-13. With defects:

cf p − cv

$350, 000 1, 000 − 600 = 875 units or $875,000 in sales

Without defects (Six Sigma): $350, 000 1, 000 − 540.05 = 760.952 or $760,952 in sales

The slope of the line is steeper with a reduced break-even point; the company can make more money without selling additional units.

2-14. With 8% defect rate: Sales Variable costs Fixed costs Profit

$151,200 61,200 31,000 $59,000

With zero defects: Sales Variable costs Fixed costs Profit

$151,200 56,305 31,000 63,895

Six Sigma results in an 8% reduction in variable costs and a corresponding 8% increase in profit. The return on the Six Sigma investment would be:

100(4, 200 − 695) 25, 000 = 14%

Return =

2-15. A possible version of the cause and effect matrix:

2-16.

Weekly revenue = (18,400 orders) ($47/order) = $864,000 Weekly variable costs = - 365,000 Weekly fixed costs = - 85,000 Total profit = $414,800

Orders returned = (18,400)(.12) = 2,208 Cost of refilling orders = (2,208)(.30)($8) = $5,299 Cost of lost sales = (2,208)(.70)($47)= $72,643 Cost of lost future sales = (2,208)(.70)(.5)(15) = $11,592 Total cost of defects = $5,299 + 72, 643 + 11,592 = $89,534 With quality improvements: Orders returned Cost of refilling orders Cost of lost sales Cost of lost future sales Total cost of defects

= (18,400)(.02) = 368 = (368)(.30)($8) = $883 = (368)(.70)($47) = $12,107 = (368)(.70)(.50)(15) = $1,932 = $883 + 12,107 + 1,932 = $14,922

Savings with quality improvements = $89,534 – 14,922 = $74,612 per week Annual savings = ($74,612)(52) = $3,879,824 Invest in program. Processes likely to be improved include computer ordering system, suppliers, handling, etc. Zero defects are unlikely because some orders will be returned because customers simply change their minds.

CASE 2.1: Designing a Quality Management Program for the Internet at D4Q This can be an instructive, hands-on case project. The students should first search the Internet for different web sites at which retail items can be ordered. They should next develop a list of quality characteristics or dimensions to focus on. These might include the visual appearance of the web site, the friendliness of the language used, the accuracy of instructions, the availability of e-mail or telephone support, etc. An attractive web site should probably include photos of the catalog items, for example book jackets, CD covers and video jackets, instead of just item titles. Instructions for ordering should be detailed and accurate with help icons located at every step. Customer support should be easy to access with e-mail or by telephone. Responses to requests for support should be quick. Service measurement is difficult in this type of operation. If an order center is used, then the company can count the number of customers who enter the center then abort as the result of poor instructions. Follow up surveys of customers who place orders or request hard copy catalogs is a good way to evaluate service. From the server end, the server responses to customer inquiries can be monitored for accuracy, completeness and timeliness. These are just a few of the possible quality initiatives you might suggest to develop a high quality web site ordering system.

CASE 2.2: Quality Management at State University In general, the student should respond to this case by attempting to go through the chapter and to discuss each major topic in terms of a university environment. This will require that they first identify the product and the process, obviously the student and the educational process. However, in a university environment, is the product also the customer? This is an interesting question to begin with. An initial step should be to develop a customer definition of quality—that is, what are the dimensions of quality in an education that parents, students, and legislators expect? This step can be accomplished by surveying alumni, potential employers, parents, students, and legislators. It would also be beneficial to see what the competition (other colleges and universities) does. The various support functions in the university should be identified in terms of a production process. For example, the admissions office is analogous to the purchasing department. A key problem here in a QM approach is that the university has little, if any, control over suppliers (high schools). As a result, admissions must institute inspection and process control procedures to ensure high-quality raw material (i.e., students) is admitted. The product-design function, or curriculum design, is typically decentralized in a university among various colleges and departments. In some cases, the university administration will design a core curriculum for the first two years and college and departments will design the curriculum for the last two. The production process is the movement of students through the curriculum to graduation. Discussion should focus on how to institute process control in order to avoid final product “defects.” This obviously requires a definition as to what a defective item is—a student who enters but fails to graduate, a student who graduates but does not gain employment, or a graduate who indicates disappointment with his or her education five years from now. An area on which to focus is the degree to which quality control tools such as brainstorming, quality circles, histograms, check sheets, and fishbone diagrams can be used to evaluate the process. Customer service would seem to be an integral part of a QM approach in the university. This service would focus on support services such as dining, recreation, housing, advising, counseling, extracurricular activities, entertainment, placement, alumni services, etc.

Depending on the time designated to spend on this case, students might interview various administrators at their own university to determine where QM can be applied in the university and obstacles to a TQM approach.

CASE 2.3: Quality Problems at the Tech Bookstore a. Mr. Watson’s organization of the customer survey categorized the two bookstores and types of customer, i.e., students and non-students. He differentiated between the two stores because they carried different products, and it also was likely that a different population of customers visited the two stores since they were in different locations. Also, the two stores had different managers, staff and employees; in effect they were separate entities. He differentiated between the two customer groups because he knew they likely had different characteristics and different service expectations. Also, it was probable that they shopped for different items. Student would be primarily interested in textbooks, school supplies, computer items and apparel, whereas non-students would have not been as interested in textbooks and school supplies, and they would have had a higher interest in trade books. A customer survey was probably the best way to start in order to see if there was a quality problem and its extent. He might also analyze different processes in the store, such as employee floor service, checkout, etc. He could have probably gotten the help of an OM class on campus to help him analyze various service processes in the store. He could also benchmark other “successful” college bookstores. b. It is possible to develop 9 different Pareto charts—a chart each for students and non-students at the campus store, and a chart each for students and non-students at the off-campus store; a combined chart for students at both stores and a combined chart for non-students at both stores; a chart for each store combining the two customer groups at each one; and a summary chart combining all the data for both stores and both customer categories. Following is a summary of the survey data. c. Using a fitness for use definition, quality should be prompt, knowledgeable and courteous customer service; a pleasant shopping environment with prompt and courteous customer checkout; and quality products at a competitive (or lower) price. d. The most pronounced problem is the discrepancy between the student and non-student groups in their perception of service. This is likely due to the different expectations of the two groups. Non-students are likely to be older and less patient. The student employees’ attitude and demeanor is probably more familiar to other students, i.e., what they are used to, and they are probably more patient with the student employees than the non-student group. Also it is less likely that students will ask questions than the nonstudent group so the problem of not being helpful or knowledgeable does not come up as much with students as with non-students. It also may be that students visit the bookstores more often and know where to find items, and they may also be more familiar with bookstore policies. Many of the nonstudent customers could be visitors. However, this discrepancy does not hide two other potential quality problems—that the off campus store has poorer quality service across all categories than the on-campus store, and there seems to be a significant problem with employee training at both stores. The relatively low percentage of customers who think employees are knowledgeable and helpful, reinforced by the graduate student evaluators and the complaint incidents to the Board, clearly indicates that the student employees are not adequately prepared to do their job. They are probably in need of more training, however it is likely the bookstore has been hesitant to provide additional training because of the high turnover rate, i.e., the return on training investment might be perceived by management to be low when student employees leave after a semester or two. The off campus store may have poorer quality service, in general, because of the

management staff, and the fact that the Executive Director resides at the on campus store. The much lower percentage of students who think that the cost of purchases at the bookstores is reasonable is probably due to the fact that students purchase textbooks, which are expensive. The bookstore probably does not do a very good job of publicizing the fact that it has a very low mark up on textbooks. This is something that could be highlighted on the store web site, which the students access more than non-students. e. Since this is a service, several of the costs of poor quality that relate to manufactured products such as scrap, product rework, returned products, etc., do not apply. Thus, the two primary costs of poor quality are lost sales and customer complaint costs. The costs the bookstore is incurring to conduct the customer survey, hire the graduate student evaluators, and analyze the results are all costs of poor quality. From the limited information provided it is difficult to address the question of lost sales. However, it is noted that that the town and university have been growing while sales have remained steady. Given the ideal location of the on campus store in particular, and the fact that the football team is very successful should mean much higher sales of licensed apparel. It could be that the quality problems are having a very negative impact on sales. f. While the bookstores would benefit from a complete QM program, the most immediate need is for a more extensive employee training program. The bookstore needs to establish a plan for improving its quality that includes employee training as a top priority. The plan needs tangible objectives. For example, every customer question should be answered as promptly as possible. Some form of reward system for employees might be beneficial. They likely would benefit from a set of guidelines for employees for addressing customer questions. Having a resource person available that has intimate knowledge of all aspects of store policies, and knows where all items are in the store, that the student employees could contact at any time would be a good personnel investment. In addition, management needs to establish a process for monitoring employee performance on a routine, daily basis. Benchmarking other college bookstores could provide insight into ways to solve quality problems. A process for measuring customer and employee satisfaction on a regular basis needs to be put into effect, and performance measures could be tied to these surveys. g. Because the bookstores are quasi non-profit, government-type entities, the revenue pressures that a business might feel are not present. Thus, the motivation for improved quality is primarily the store’s reputation. Another factor is the university-invoked policy of hiring students on a part-time basis. This means that to solve the quality problems associated with part time students, bookstore management must look to other solutions besides hiring more-experienced, full-time employees. The fact that students will be serving adults is a situation that cannot be changed. h. The most probable benefit would be an increase in sales and revenue. However, employee satisfaction would also likely increase as a result of a quality management program. Also, the Board of Directors would likely receive fewer complaints.

Campus Store

Were employees courteous and friendly? Were employees knowledgeable and helpful? Was the overall service good? Did you have to wait long for service? Did you have to wait long to checkout? Was the item you wanted available? Was the cost of your purchase(s) reasonable? Have you visited the store web site?

Off Campus Store

Student

Non-student

Student

Non-student

Yes %

No %

Yes %

No %

Yes %

No %

86 79 86 11 12 92 58 50 42

14 21 14 89 88 8 42 50 5

67 52 63 45 46 84 90 12 95

33 48 37 55 54 16 10 88 41

75 77 71 11 16 90 60 58 59

25 23 29 89 84 10 40 42

Yes % 52 41 55 46 45 90 92 5

Total

No %

Yes %

No %

48 59 45 54 55 10 8 95

73 65 71 25 27 89 72 41

27 35 29 75 73 11 28 59

CASE SOLUTION 2.4 - Product Yield at Continental Luggage Company Average Weekly Yield

Stage 1 yield

Stage 2 yield

( )( ) 𝐼( )( ) : 𝑌1 = 𝐼 %𝐺 + 1 %𝐺 %𝑅 ) 500( )( ) 500( − 470 = 0 94 + + 6.0906 0 23 .= . . 476 9 𝑌1 = .

( )( ) 476 9( )( ) 82 + 17 : 𝑌2 = 476 9 =0 457 96 .+ . .36 0 04 0 91 . . . . 475 2 𝑌2 = .

Stage 3 yield

: 𝑌3 =

)( )( ) 475 2( ) ( 451+.44475 . =95 2 9205 67 + 15 . . .. . = 467.6

)( )( ) 467 6( ) ( 453+.57467 . =97 6 4803 89 + 12 . . .. . = 466.1 Stage 5 yield 466 1( ) 466 1( )( ) 456+.78 +.6.71 : 𝑌5 = . =98 02 72 . . . 463 5 = .

Stage 4 yield

: 𝑌4 =

Increasing Good Quality Yield by 1% at Each Stage

Stage 1 yield

Stage 2 yield

Stage 3 yield

Stage 4 yield

Stage 5 yield

: 𝑌1 =

) 500( )( ) 500( 0 95 +480 750 05 0 23 .𝑌1 = . . .

: 𝑌2 =

) 480 75( )( ) 480 75( . 0 97 +479 45 . 0 03 0 91 .𝑌2 = . . .

: 𝑌3 =

) 479 45( )( ) 479 45( . 0 96 +473 12 . 0 04 0 67 .𝑌3 = . . .

: 𝑌4 =

) 473 11( )( ) 473 11( . 0 98 +472 07 . 0 02 0 89 .𝑌4 = . . .

: 𝑌5 =

) 472 07( )( ) 472 07( . 0 99 +470 75 . 0 01 0 72 .𝑌5 = . . .

Difference in yields

Percentage increase =

7 25 436 . 2 .

1 56% = .

470 75 463 5 7 25 units . − . = .

3 Statistical Process Control Answers to Questions 3-1.

In attribute control charts (such as p-charts and c-charts), the measures of quality are discrete values reflecting a simple decision criterion such as good or bad. The quality measures used in variable-control charts (such as x -charts and R-charts) are continuous variables reflecting measurements such as weight, time, or volume.

3-2.

An R-chart reflects the process variability, whereas an x -chart indicates the tendency toward a mean value; thus, the two complement each other. That is, it is assumed that process average and variability must be in control for the process to be in control. When they are used together, the control limits are computed as

x + A2 R. 3-3.

A pattern test is used to determine if sample values from a process display a consistent pattern that is the result of a nonrandom cause, even though control charts may show the process to be in control.

3-4.

Width is determined by the size of the z value used; the smaller the value of z, the narrower the control limits.

3-5.

A c-chart is used when it is not possible to determine a proportion defective (for a p-chart), for example, when counting the number of blemishes on a sheet of material. In a p-chart it must be possible to distinguish between individual defective and nondefective items.

3-6.

Tolerances are product-design specifications required by the customer, whereas control limits are the upper and lower bands of a control chart indicating when a process is out of control.

3-7.

Management usually selects 3-sigma limits because, if the process is in control, they want a high probability that the sample will fall within the control limits. With wider limits management is less likely to erroneously conclude that the process is out of control when points outside the control limits are due to normal, random variations.

3-8.

Process control charts could be used to monitor service time in a restaurant, bank, hospital, store, etc.

3-9.

For example, in a fast food restaurant a control chart could be used to measure service times, defective menu items, out of stock menu items, customer complaints, cleanliness, and order errors, among other things.

3-10. The process capability ration

(C p ) indicates if the process is capable of meeting design specifications. The

process capability index indicates if the process mean is off-center and has shifted toward the upper or lower design specifications.

3-11.

Cp =

tolerance range .14 = = 1.00 process range .14

The tolerance range and process range are equal so the process is capable, but some defects will result.

Solutions to Problems 3-1.

=

453 = 0.151; ( 30 )(100 ) p (1 − p ) = n

( 0.151)( 0.849 ) = 0.0358

UCL = p + z

100

p (1 − p ) n

= 0.151 + 3 LCL = p − z

( 0.151)( 0.849 ) = 0.258 100

p (1 − p )

= 0.151 − 3

( 0.151)( 0.849 ) = 0.044 100

The process does not seem to be out of control, although the decreasing number of defects from sample 8 to sample 17 should probably be investigated to see why the steady improvement occurred; likewise, the steadily increasing number of defects from sample 17 to sample 25 should probably be investigated to see why the quality deteriorated.

3-2.

p = 0.153;  =

p (1 − p ) 0.153 ( 0.847 ) = = 0.036 100 100 UCL = p + z = 0.153 + 2 ( 0.036 ) = 0.225 LCL = p − z = 0.153 − 2 ( 0.036 ) = 0.081

In general, the proportion of defectives increases from sample 6 to sample 20, where it is eventually above the upper control limit. This indicates the process is moving out of control.

3-3.

p = 0.053;  =

p (1 − p ) 0.053 ( 0.947 ) = = 0.016 n 200

UCL = p + 3 = 0.053 + 3 ( 0.016 ) = 0.101 LCL = p − 3 = 0.053 − 3 ( 0.016 ) = 0.005

The process does not seem to be out of control.

3.4

UCL = p + z

p (1 − p ) n

0.03 = 0.02 + z

0.02 ( 0.98 ) n

2 0.02 ( 0.98 )

n = ( 28 ) n = 784

0.01 2

3-5.

742 = 24.73 30

UCL = c + z c = 24.73 + 3 24.73 = 39.65 LCL = c − z c = 24.73 − 3 24.73 = 9.81

With three points outside the control limits, the process appears to be out of control. b.

3-6.

Nonrandom factors that might cause the process to move out of control could include (among other things) problems with the telephone order system, inexperienced operators taking orders, computer system problems, or shipping problems and delays.

86 = 4.3 20

UCL = c + z c = 4.3 + 3 4.3 = 10.52 LCL = c − z c = 4.3 − 3 4.3 = −1.92 or 0.0 ( since the control chart cannot go below zero )

Samples 13 to sample 20 are above the average line and exhibit increasingly nonrandom behavior. Pattern rule 1 is evident, thus the process should be investigated for an out-of-control situation. 3-7. =

= 4.083

UCL = + z = 4.083+ 2 = 8.125 LCL = - z = 4.083- 2 = 0.042 Sample 7 is outside of the control limits; thus, the process should be investigated, and new limits possibly constructed.

3-8.

219 = 7.3 30

Control limits using z = 3.00:

UCL = c + z c = 7.3 + 3 7.3 = 7.3 + 8.11 = 15.41 LCL = c − z c = 7.3 − 3 7.3 = 7.3 − 8.11  0 All the sample observations are within the control limits suggesting that the invoice errors are in control.

3-9.

255 = 12.75 20 UCL = c + z c = 12.75 + 3 12.75 = 23.46

LCL = c − z c = 12.75 − 3 12.75 = 2.04 All the sample observations are within the control limits suggesting that the delivery process is in control. 3-10. Sample 1 2 3 4 5 6 7 8 9 10

Proportion Defective .028 .044 .072 .034 .050 .082 .036 .038 .052 .056

Sample 11 12 13 14 15 16 17 18 19 20

Proportion Defective .076 .048 .030 .024 .020 .032 .018 .042 .036 .024

421 421 = = .0421 ( 20 )( 500 ) 10, 000

UCL = p + z

p (1 − p ) n

= .0421 + 3.00

.0421(1 − .0421) 500

= .0421 + .027 = .069 LCL = p − z

p (1 − p ) n

.0421(1 − .0421) 500 = .0421 − .027

= .0421 − 3 = .015

Samples 3, 6, and 11 are above the upper control limit indicating the process is out of control.

3-11. Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x 1.84 2.08 2.92 1.78 2.70 3.50 2.84 3.26 2.50 4.14 2.12 4.38 2.84 2.70 3.56 2.96 3.34 4.16 3.70 2.72 60.04

R 2.3 2.6 2.7 1.9 3.2 5.0 2.2 4.6 1.3 3.5 3.0 4.0 3.3 1.1 5.6 3.1 6.1 2.4 2.5 2.9 63.3

 R 63.3 = = 3.17 k 20  x 60.20 x= = = 3.01 20 20

R-chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = 2.11( 3.17 ) = 6.69 LCL = D3 R = 0 ( 3.17 ) = 0 There are no R values outside the control limits, which would suggest the process is in control.

x -chart

A2 = 0.58

UCL = x + A2 R = 3.01 + 0.58 ( 3.17 ) = 4.85 LCL = x − A2 R = 3.01 − 0.58 ( 3.17 ) = 1.17 There are no x values outside the control limits, which suggests the process is in control.

3-12. = = .0707 UCL =

= .0707 + 3 = .0950 LCL =

-z

= .0707 - 3 = .0463 Samples 2, 8, 11 and 17 are above the UCL and samples 5, 9, 10 and 14 are below the LCL. The process should be evaluated and corrected, and a new control chart constructed.

3-13.

= .097;

= 0.030

UCL = + 3σ = 0.097 + 3(0.030) = 0.186 LCL = - 3σ = 0.097 – 3(0.030) = 0.008 Sample 24 falls outside of the Upper Control Limit, therefore the process is out of control and an investigation into possible assignable causes for the out-of-control condition is warranted.

3-14. a.

= 7.11

The population process average can be calculated based on the total number of complaints received over the 52-week period, rather than using the sample data to calculate a sample average. UCL = + Z = 7.11 + 3 = 7.11 + 3(2.67) = 15.11 LCL = 0 The process appears to be in control with no discernible patterns. b.

UCL = 7.11 + 2(2.67) = 12.45 LCL = 1.77 The process is still in control.

3-15.

= = .128 UCL =

= .128 + 3.0 = .128 + .142 UCL = .269 LCL = 0 b.

The process appears to be in control.

3-16.

= 338/14 = 24.143

Control limits using z = 3.00: UCL = LCL =

+z -z

= 24.14 + 3 = 24.14 - 3

= 24.14 + 14.73 = 38.88 = 24.14 - 14.73 = 9.41

Day 8 is near the UCL and day 12 exceeds the UCL so the process needs to be examined and corrected, and a new control chart developed. This is especially true since there seems to be an excessive number of errors

3-17. Sample 1 2 3 4 5 6 a.

R 0.67 0.69 0.93 0.52 0.64 0.71

Sample 7 8 9 10 11 12

 R 6.87 = = 0.57 k 12 From Table 3.1 in the text, D3 = 0 and D4 = 2.11 R=

UCL = D4 R = 2.1( 0.57 ) = 1.21

LCL = D3 R = 0 ( 0.57 ) = 0

The process variability is within the control limits.

R 0.45 0.17 1.17 0.99 0.65 0.15

Sample 1 2 3 4 5 6

x 8.89 8.88 8.99 9.19 9.04 8.71

Sample 7 8 9 10 11 12

x 9.05 9.16 9.17 9.06 9.09 9.01

 x 108.04 = = 9.00 12 12 From Table 3.1, A2 = 0.58. x=

UCL = x + A2 R

= 9.00 + 0.58 ( 0.57 ) = 9.33

LCL = x − A2 R

= 9 − 0.58 ( 0.57 ) = 8.67

The process appears to be in control from both the x and R charts, although sample 6 is close to the LCL and perhaps should be investigated. 3-18. Sample 1 2 3 4 5 6 7 8 9 10 11 12

Above/Below B B B A A B A A A A A B

Up/Down — D U U D D U U U D U D

There are no discernible nonrandom patterns.

Zone B B C B C A C B B C C C

3-19. Sample 1 2 3 4 5 6 7 8 9 10

R 8.5 5.8 8.1 6.4 7.1 6.0 9.9 2.5 1.6 9.4

Sample 11 12 13 14 15 16 17 18 19 20

 R 12.7 = = 6.24 k 20 From Table 3.1, D3 = 0.0 and D4 = 2.11 R=

UCL = D4 R = 2.11( 6.24 ) = 13.17

LCL = D3 R = 0.0 ( 6.24 ) = 0

b. The temperature is within the control limits.

c. Sample 1

x 43.9

Sample 11

x 39.2

R 8.1 4.4 5.8 3.9 6.2 5.4 3.6 10.9 7.5 3.6

2 3 4 5 6 7 8 9 10

39.7 37.2 40.4 39.0 41.8 39.4 40.7 41.6 39.0

12 13 14 15 16 17 18 19 20

39.8 42.9 37.8 36.6 37.6 39.9 40.7 38.2 39.0

x = 39.7 From Table 3.1, A2 = 0.58.

UCL = x + A2 R

= 39.7 + 0.58 ( 6.24 ) = 43.32

LCL = x − A2 R

= 39.7 − 0.58 ( 6.24 ) = 36.08

The process appears to be in control, with sample 1 seeming to be an aberration, however, the process should still be checked.

3.20.

Sample 1 2 3 4 5 6 7 8 9 10

Above/ Below A B A A A B A B B A

Up/ Down — D U D U D U D D U

Zone C C C C C C B B A B

Sample 11 12 13 14 15 16 17 18 19 20

Above/ Below A B B B A B B A A B

Up/ Down D D U D U D D U D D

Zone C C C B C C B A C B

Samples 7 through 10 appear to violate a zone pattern test (four out of five points in zone B or beyond) so there may be a nonrandom pattern. 3.21.

Sample 1 2 3 4 5 6 7 8 9 10

Above/ Below A B B A B A B A A B

Up/ Down — D D U D U D U U D

Zone A C A C C B C C B C

Sample 11 12 13 14 15 16 17 18 19 20

The pattern appears to be random.

3-22.

R=2 a.

From Table 3.1, D3 = 0 and D4 = 2.28

UCL = D4 R = 2.28 ( 2 ) = 4.56

LCL = D3 R = 0 ( 2 ) = 0 Sample 1 2 3 4 5 6 7 8 9 10

R 2.1 4.5 7.0 3.0 5.0 4.3 6.5 2.0 3.2 4.0

Above/ Below B B A B B B A A B B

Up/ Down U U U D D U U U D U

Zone C C A B A B C C B C

The process clearly seems to be out of control. There are three of the sample points above the UCL, and all other sample values are above the center line for R , indicating nonrandom variations. c. Sample 1 2 3 4 5 6 .7 8 9 10

x 37.0 35.8 35.3 36.1 36.1 34.3 36.3 35.6 37.7 34.5

x = 35.9 From Table 3.1,

A2 = 0.73. UCL = x + A2 R

= 35.9 + 0.73 ( 2 ) = 37.36

LCL = x − A2 R

= 35.9 − 0.73 ( 2 ) = 34.44

The process may be out of control (sample 6 and 9), although overall the x -chart does not reflect the magnitude of the out of control situation, as does the R-chart in Problem 3-11. The process should be investigated to determine a cause for the out of control samples.

3-23. Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Above/ Below A B A A B B B B A A A A A A A

Up/ Down — D U U D U D D U U D U D D D

Zone C B A A B C C B B A C A A A C

Sample 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Above/ Below B B B B A A B B A A A B B B B

Up/ Down D D U U U U D U U U U D D U D

Zone B A B B C B A B C C A C A B B

The zone pattern test is violated for samples 10 through 14 (2 out of 3 consecutive points in zone A but within limits).

3-24. Sample 1 2 3 4 5 6 7 8 9 10 11 12

Above/ Below B B B A A A B A A A B B

Up/ Down — U D U U U D U D D D U

Sample 16 17 18 19 20 21 22 23 24 25 26 27

Above/ Below B B B B B A A A A A A A

Up/ Down D D U U U U U U D U D U

13 14 15

B B B

U D D

28 29 30

A A A

U D U

There are eight consecutive points on one side of the center line on two occasions—samples 11-20 and samples 21-30—indicating nonrandom patterns may exist. 3-25.

146 = 7.3 20 UCL = c + z c

= 7.3 + ( 3) 7.3 = 7.3 + ( 3)( 2.70 ) = 15.41 LCL = c − z c = 7.3 − ( 3)( 2.70 ) 0

The process appears to be in control however the process may be moving toward an out-of-control situation.

3-26.

x = 3.25, R = 3.86, A2 ( n = 5) = 0.58

UCL = x + A2 R

= 3.25 + ( 0.58 )( 3.86 ) = 5.48

LCL = x − A2 R

= 3.25 − ( 0.58 )( 3.86 ) = 1.01

UCL = D4 R = 2.11( 3.86 ) = 8.14 LCL = D3 R = 0 ( 3.86 ) = 0

The process appears to be in control.

3-27.

718 = 0.36 2000 p (1 − p ) UCL = p + 3 100 = 0.36 + ( 3)( 0.048 ) p=

= 0.504 LCL = p − ( 3)( 0.048 ) = 0.36 − 0.144 = 0.216

The process appears to be out of control.

3-28.

326 = 16.3 20

UCL = c + z c = 16.3 + ( 3)( 4.03) = 28.412 LCL = c − z c = 16.3 + ( 3)( 4.03) = 4.188 The process is in-control.

3-29.

3-30.

x = 7.28, R = 4.25, A2 ( n = 5 ) = 0.58 UCL = x + A2 R

= 7.28 + ( 0.58 )( 4.25 ) = 9.75

LCL = x − A2 R = 7.28 − 2.47 = 4.81 UCL = D4 R = 2.11( 4.25 ) = 8.97 LCL = D3 R = 0 ( 4.25 ) = 0

While the process appears to be in control the mean of 7.28 appears to be significantly lower than the objective of 8 chips per cookie that management has established. Thus, the company should adjust their process to increase the number of chips and construct a new control chart.

3-31.

105 = 0.22 ( 30 )(16 )

UCL = p + 3

p (1 − p )

n = 0.22 + ( 3)( 0.0756 ) = 0.447

LCL = 0.22 − ( 3)( 0.0756 )  0.216 Although the process appears to be in control, the average proportion of “defects’’ among leaving patients, p = 0.22, seems high and the hospital should probably adopt some quality improvement measures.

3-32.

202 = 10.1 20 UCL = c + z c

= 10.1 + ( 3)( 3.18 ) = 19.6 LCL = c − z c = 10.1 − ( 3)( 3.18 ) = 0.57

The process appears to be in control.

3-33.

x = 3.17, R = 3.25, A2 ( n = 5 ) = 0.58

UCL = x + A2 R

= 3.17 + ( 0.58 )( 3.25 ) = 5.04

LCL = x − A2 R = 3.17 − 1.89 = 1.29 UCL = D4 R = 2.11( 3.25 ) = 6.86 LCL = D3 R = 0 ( 3.25 ) = 0

The process appears out of control for sample 10, however, this could be an aberration since there are no other apparent nonrandom patterns or out-of-control points. Thus, this point should probably be “thrown out’’ and a new control chart developed with the remaining eleven samples. 3-34.

x = 3.17, days R = 3.25 3 = A2 R = ( 0.58 )( 3.25 ) = 1.89 Cp =

2.00 2.00 = = 0.53 2 (1.89 ) 3.78

 3.17 − 2.00 4.00 − 3.17  C pk = minimum  ,  1.89  1.89 = minimum (.62, .44 ) = .44 The process is not capable of meeting the company’s design specifications and defects (i.e., late deliveries) will occur.

3-35.

x = 7.28, R = 4.25 3 = ( 0.58 )( 4.25 ) = 2.47 Cp =

4.00 4.00 = = .81 2 ( 2.47 ) 4.94

 7.28 − 6.00 10 − 7.28  C pk = minimum  , 2.47   2.47 = minimum (.52, 1.10 ) = .52 The process is not capable of meeting the design specifications and it appears that cookies will be produced with too few chips. 3-36.

x = 9.00 R = 0.57 3 = A2 R = ( 0.58 )( 0.575 ) = .33 1.00 = 1.52 .66  0.5 0.5  C pk = minimum  ,  .33 .33  = minimum (1.52, 1.52 ) Cp =

= 1.52 X = The process is capable of meeting design specifications. 3-37.

Cp =

420 420 = = 1.27 6 ( 55 ) 330

1, 050 − 915 1,335 − 1, 050  C pk = minimum  ,  3 ( 55 )  3 ( 55 )  = minimum (.82, 1.72 ) = .82 While C p = 1.27 indicates the process is capable, C pk = .82 indicates the process mean has shifted toward the lower design specification, and defective (shorter lived) bulbs will be generated. 3-38.

Cp =

.048 .048 = = 1.00 6 (.008 ) .048

1.281 − 1.251 1.299 − 1.281  C pk = minimum  ,  3 (.008 )   3 (.008 ) = minimum (1.25, 0.75 ) = 0.75

C p = 1.00 means the tolerance range and the process range are virtually the same indicating that some defective parts will occur. C pk = 0.75 indicates that the process mean has shifted toward the upper specification indicating the process will result in some parts that are defective (too large). 3-39. The process mean would need to be shifted back toward the nominal design value of 1,125 hours. To achieve six sigma quality the process range would need to be reduced to one-half of the tolerance range. Since the tolerances are 210 hours, the tolerance range is 420 hours. Thus, the process range would need to be 210 hours, which are 3 control limits of 105 hours. Thus, the process mean would need to be 1,125 hours with an upper control limit of 1,230 hours, and a lower control limit of 1,020 hours. 3-40. Machine 1:

.030 = 1.25 .024  .0995 − .082 .112 − .0995  C pk = minimum  ,  .012 .012  = minimum (1.46, 1.04 ) Cp =

= 1.04 Machine 1 is capable of meeting the design specifications. Machine 2:

.030 = 0.56 .054  .1002 − .0820 .1120 − .1022  C pk = minimum  ,  .027 .027  = minimum (.67, .44 ) Cp =

= .44 Machine 2 is not capable of meeting the design specifications. Machine 3:

.030 = 1.00 .030  .0951 − .0820 .1120 − .0951  C pk = minimum  ,  .015 .015  Cp =

= minimum (.87, 1.13) = .87

Machine 3 is capable of meeting the design specifications, but the process center has shifted too far toward the lower design specification.

3-41.

Month 1 2 3 4 5 6 7 8 9 10 11 12 Avg

7.02 8.16 8.18 9.18 10.32 9.54 6.96 10.38 8.12 10.26 9.66 9.10 8.91

R 5.6 3.7 4.9 5.4 4.1 4.1 3.5 9.0 8.9 6.9 9.2 4.3 5.8

x = 8.91,

R = 5.8, A2 = .58, D3 = 0, D4 = 2.11

X-bar chart

R-Chart

UCL = 12.27 LCL = 5.54

UCL = 12.24 LCL = 0

The process is in control according to both control charts. 3-42.

p = 0.48 UCL = p + 3

p 1− p = 0.48 + 3 n

(

)

0.48(1 − 0.48) = 0.60 150

LCL = p - 3

p 1− p = 0.48 - 3 n

(

)

0.48(1 − 0.48) = 0.36 150

The process is not in control and does not meet the target value of 90%. It appears that improvement occurred in week 7, but the improvement was not consistent and was significantly below the target value. The hospital should reevaluate the process improvements it has implemented using quality tools and, after implementing a new program, a new control chart should be developed.

3-43. Sample1 1 2 3 4 5 6 7 8 9 10

x 145.88 144.86 144.22 145.82 143.10 147.82 143.04 141.44 148.72 142.42

R 6.9 8.6 9.4 5.2 6.7 6.2 5.5 5.9 5.9 8.5

Sample 11 12 13 14 15 16 17 18 19 20

x 144.54 145.48 145.26 148.78 143.58 146.48 145.22 144.80 143.46 145.92

R 6.7 5.5 8.5 4.8 5.1 3.1 2.4 4.6 5.4 7.1

x = 145.06 R = 6.1 A2 = 0.58

x -chart: UCL = 145.06 + 0.58 ( 6.1) = 148.60 LCL = 145.06 − 0.58 ( 6.1)

R -chart: UCL = D4 R = ( 2.11)( 6.1) = 12.87 LCL = D3 R = ( 0 )( 6.1) = 0

= 141.52 Sample 8 is slightly below the LCL and samples 9 and 14 are slightly above it.

Cp =

149 − 142 7.00 = = .99 2 ( 3.54 ) 7.08

145.06 − 142 149 − 145.06  C pk = minimum  ,  3.54 3.54  = minimum (.86, 1.04 ) C pk = .86 The process is not capable of meeting design specifications. Since C p is very close to 1.00, some defective baseballs will be generated, and, C pk = 0.86 indicates they will typically not weigh enough. 3-44. To achieve six sigma quality, the process range would need to be reduced to one-half of the tolerance range. The tolerance range is 7 gms therefore the process range must be 3.5 gms which are 3 control limits of 1.75 gms. Thus, the process mean would need to be 145.5 gms with an upper control limit of 147.25 and a lower control limit of 143.75.

3-45. Sample 1 2 3 4 5

R 7.1 6.4 12.7 6.8 12.4

7.72 6.58 12.90 8.76 11.06

Sample 6 7 8 9 10

x 8.90 11.52 8.02 9.58 9.12

R 6.0 14.6 4.7 8.5 5.6

x = 9.42, R = 8.48 A2 = 0.58

R -chart:

x -chart: x = 9.42 UCL = 14.31 LCL = 4.52

R = 8.48 UCL = 17.94 LCL = 0

The process is in control according to both control charts.

Cp =

12 − 6 6 = = .61 14.31 − 4.52 9.79

 9.42 − 6 12 − 9.42  C pk = minimum  , 4.92   4.92 = minimum ( 0.70, 0.52 ) = 0.52 The process is not capable of meeting the design specifications and the customer waiting times will be greater than the upper specification and lower than the lower specification. Although the lower times might be considered good, it could be that customer service representatives are not devoting enough time to each customer. 3-46. (a) Sample 1 2 3 4 5 6 7 8 9 10

x 21.4 27.0 19.0 24.4 26.6 20.8 24.8 26.4 29.6 25.4 245.4

R 9 23 8 19 33 15 14 23 11 9 164

 R 164 = = 16.4 k 10  x 245.4 x= = = 24.54 10 10

R -chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = ( 2.11)(16.4 ) = 34.604

LCL = D3 R = ( 0 )(16.4 ) =0 There are no R values outside the control limits, which suggests the process is in control.

x -chart

A2 = 0.58

UCL = x + A2 R = 24.54 + (.58 )(16.4 ) = 34.05 LCL = x − A2 R = 24.54 − (.58)(16.4 ) = 15.03 There are no x values outside the control limits, which suggests the process is in control. (b)

upper specification limit − lower specification limit 6 30 − 20 = 19.02 = .52  x − lower specification, upper specification − x  C pk = minumum   3 3  

Cp =

 24.54 − 20 30 − 24.54  = minimum  , 9.51   9.51 = minimum (.47, .57 ) = .47 Since C p = .52, which is less than 1.0, the process range is greater than the tolerance range and the process is not capable of producing within the design specifications all the time. Since Cpk = .47 is less than 1.0, the process has moved closer to the lower design specification and will generate defects.

3-47. (a) Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

x 4.83 5.38 5.38 5.75 5.00 6.40 6.65 4.57 6.18 5.32 4.02 4.57 4.95 5.05 4.47 78.52

R 3.6 3.3 1.3 3.3 2.1 2.9 2.7 3.5 4.6 1.2 4.5 4.2 3.7 2.1 3.6 46.6

 R 46.6 = = 3.107 k 15  x 78.52 x= = = 5.23 15 15 R -chart R=

D3 = 0, D4 = 2.00, for n = 6

UCL = D4 R = ( 2.00 )( 3.107 ) = 6.21 LCL = D3 R = ( 0 )( 3.107 ) = 0 There are no R values outside the control limits, which suggest the process is in control.

x -chart

A2 = .48

UCL = x + A2 R = 5.23 + (.48 )( 3.107 ) = 6.73 LCL = x − A2 R = 5.23 − (.48 )( 3.107 ) = 3.74 There are no x values outside the control limits, which suggest the process is in control. (b)

upper specification limit − lower specification limit 6 6−4 2 = = 2.98 2.98 = .67

Cp =

 x − lower specification, upper specification − x  Cpk = minumum   3 3    5.23 − 4 6 − 5.23  = minimum  , 1.49   1.49 = minimum (.83, .51) = .51 Since C p = .51, which is less than 1.0, the process range is greater than the tolerance range and the process is not capable of producing within the design specifications all the time. Since Cpk = .51 is less than 1.0, the process has moved closer to the lower design and will generate defects. 3-48.

64 = 2.667 24 UCL = c + z c c=

= 2.667 + 3 2.667 = 7.57 LCL = c − z c = 2.667 − 3 2.667 =0 The process is only “out of control” in month 5 when there were zero falls. This month should be investigated to see if the circumstances that resulted in no falls can be repeated. Any number of falls would seem to be poor quality, so even though this “process” is technically “in control,” the process should be improved with a six sigma goal of zero defects. While 2.667 falls per month is not a lot, it is likely too many. 3-49.

(a) Sample 1 2 3 4 5 6 7 8 9 10 11 12

x 88.33 93.33 82.50 91.00 89.83 87.00 89.67 86.00 84.00 93.67 92.83 85.50

 R 230 = = 19.17 k 12  x 1063.67 x= = = 88.64 12 12

R 18 12 28 22 13 15 15 34 31 15 14 13

R-chart

D3 = 0, D4 = 2.00, for n = 6 UCL = D4 R = (2.00)(19.17) = 38.41 LCL = D3 R = (0)(19.17) = 0 There are no R values outside the control limits, which suggests the process is in control. x -chart

A2 = .48 UCL = x + A2 R = 88.64 + (.48)(19.17) = 97.89 LCL = x − A2 R = 88.64 − (.48)(19.17) = 79.38 There are no x values outside the control limits, which suggest the process is in control. (b)

upper specification limit − lower specification limit 6 98 − 92 6 = = (88.64 − 79.38) 18.71 = .32

Cp =

 88.64 − 92 98 − 88.64  C pk = minimum  , 9.27   9.27 = minimum ( −0.36, 1.01) = −0.36 Since C p = 0.32 (1.0), the service department is not currently capable of consistently achieving the desired customer satisfaction score. Since C pk = −0.36 is less than 1.0 the service department will continue to generate less than desired customer scores. The service department would need to make process improvements in order to consistently achieve the desired customer satisfaction scores.

3-50.

(a) Day 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

R 143.00 139.00 112.33 170.67 165.33 135.33 149.33 113.67 143.67 138.00 118.33 176.33 167.67 157.00 165.67 144.67 113.00 167.67 197.33 115.67 148.33 160.33 159.67 107.33 129.67 178.33 155.33 153.67 145.00 132.33 4403.7

40 33 31 88 26 55 32 18 22 57 62 56 92 87 61 31 47 19 34 67 27 81 46 38 61 46 21 78 17 51 1424

 R 1424 = = 47.47 k 30  x 4403.7 x= = = 146.79 30 30

R-chart

D3 = 0, D4 = 2.574, for n = 3 UCL = D4 R = (2.574)(47.47) = 122.18 LCL = D3 R = (0)(47.47) = 0 There are no R values outside the control limits, which suggest the process is in control.

x -chart

A2 = 1.023 UCL = x + A2 R = 146.79 + (1.023)(47.47) = 195.35 LCL = x − A2 R = 146.79 − (1.023)(47.47) = 98.23 There is one x value outside the control limits on day 19, which at least suggests a problem on that day should be investigated, although there is no consistent pattern of being out of control. (b)

upper specification limit − lower specification limit 6 135 − 105 30 = = 97.12 97.12 = .31

Cp =

146.79 − 105 135 − 146.79  C pk = minimum  , 48.56   48.56 = minimum (.86, − 0.24) = −0.24 Since C p = 0.31 and C pk = −0.24, the bed turnaround time is not capable of achieving the hospital’s goal of 120 minutes. Process improvements would be necessary to achieve the desired bed turnaround times.

CASE SOLUTION 3.1: Quality Control at Rainwater Brewery This is basically a discussion question; therefore, the student responses might vary. The owners have stated in the case description that the chances of a batch being spoiled—and, thus, an unhealthy batch—are very unlikely. However, even a very slight risk of a contaminated batch of 1,000 bottles might be too much, given the health consequences of a spoiled batch. Testing only a small sample of even a few bottles would indicate if the batch was, in fact, bad, so a simple testing procedure such as opening and testing 5 to 10 bottles might be prudent. Most of the quality control efforts should focus on process control procedures at the various stages of the brewing process. Obvious candidates are x and R-charts to monitor temperature, specific gravity, and pH during the fermentation and aging stages. Some type of process control testing of the final bottled product probably is warranted also. Quality control methods can also be used at the beginning of the brewing process for checking materials such as bottles and caps and ingredients such as yeast, hops, and grain. Bottles and caps that are not completely clean and sterile can result in a spoiled batch, and poor-quality ingredients can obviously contribute to an “off’’ brew.

CASE SOLUTION 3.2: Quality Control at Grass, Unlimited 249 defects = 4.15 60 samples z = 2.00 UCL = c + Z c = 4.15 + 2 4.15 = 8.22

LCL = c − Z c = 4.15 − 2 4.15 = 0.076 The chart exceeds the control limits for samples 12 and 14, however, the chart appears to have been in control prior to sample 12. Although, the reasons for the out-of-control occurrences must be investigated, the chart based on samples 1 through 11 could be used, or, several observations could be collected once the process is brought back in control and used with samples 1 through 12. Thus, this chart could be implemented for continued use. Other examples of control charts that could be used include p-charts for the number of errors in a sample of orders, or the number of customer complaints for a sample survey of customers. A c-chart could be used for the number of defects found (for cleanliness) during an inspection of facilities.

CASE SOLUTION 3.3: Improving Service Time at Dave’s Burgers x = 1.70, R = 1.40, A2 ( n = 6 ) = 0.48

UCL = x + A2 R

= 1.70 + ( 0.48 )(1.40 ) = 2.37

LCL = x − A2 R

= 1.70 − ( 0.48 )(1.40 ) = 1.03

UCL = D4 R = 2.00 (1.40 ) = 2.80 LCL = D3 R = 0 (1.40 ) = 0

The process is not in control, thus the control chart cannot be used on a continuing basis. The out-of-control situation should be investigated and upon correction, new data should be gathered to establish a revised control chart. If that control chart is valid (i.e., in control), it may be used to monitor quality on a continuing basis. Dave may want to chart % of customers’ orders completed correctly (p-chart) or the number of customer complaints (c-chart). Answers from students, of course, may vary.

Supplement 3 Operational Decision-Making Tools: Acceptance Sampling Answers to Questions S3-1.

Acceptance sampling is an inspection procedure in which samples of inputs, in-process products, or final products are taken and then compared to a quality standard to see if the entire lot the sample came from should be accepted or rejected. Process control is a statistical procedure for testing the production process itself to see if it is functioning properly.

S3-2.

In general, larger sample sizes are needed for attributes because more observations are required to develop a usable quality measure. Alternatively, variable measures require smaller sample sizes because each sample observation provides usable information such as weight or length. Thus, after only a few sample observations it is possible to compute a range or a sample average.

S3-3.

Basically, a trial-and-error approach is required to determine a value for n that (along with c) achieves the desired operating characteristics. Tables and computer programs can make the determination much easier.

S3-4.

The AQL is the fraction of defective items in a lot that is deemed acceptable, and producer’s risk (alpha) is the probability of rejecting a lot that has an acceptable AQL. Alternatively, LTPD is an upper limit of the percentage of defective items that a consumer will accept in a lot, and the consumer’s risk (beta) is the probability of accepting a lot in which the fraction defective items exceeds the LTPD.

S3-5.

In a single sampling plan, only one sample from a lot is taken, and it must be of sufficient size to determine if the whole lot should be rejected. In a double-sampling plan a smaller sample is taken first; if the quality is very good, the lot is accepted, and if it is very bad, it is rejected. However, if the initial sample is inconclusive, a second sample is taken, and the lot is either accepted or rejected based on the combined results of the two samples. A multiple-sampling plan requires the smallest sample size of the sampling plans. An initial small sample is taken; if the results are inconclusive, additional samples are taken until the lot is either accepted or rejected.

S3-6.

Acceptance sampling accepts some level of defects because it is assumed that some defects are inevitable. However, TQM believes that defects can be eliminated altogether. In addition, acceptance sampling is based on inspection of the final product, which according to TQM philosophy, is too late.

S3-7.

Use 100% inspection when the product is a safety threat or a defeat life threatening or when inspection is very cheap compared to the consequences of a defect.

Solutions to Problems These problems are solved using the following Acceptance Sampling Table. To create an OC curve, use OM Tools that accompanies this text. OM Tools creates a sampling plan (that should match the one you calculate by hand) and then uses the Binomial distribution to graph the OC curve associated with that plan. The resulting Type I and Type II errors are also shown. See the Excel Homework solution files for Chapter 3s for further information. Acceptance Sampling Table LTPD/AQL n*AQL c 1.707 22.444 30 1.723 21.594 29 1.739 20.746 28 1.757 19.900 27 1.775 19.058 26 1.795 18.218 25 1.817 17.382 24 1.840 16.548 23 1.865 15.719 22 1.892 14.894 21 1.922 14.072 20 1.954 13.254 19 1.990 12.442 18 2.029 11.633 17 2.073 10.831 16 2.122 10.035 15 2.177 9.246 14 2.240 8.464 13 2.312 7.690 12 2.397 6.924 11 2.497 6.169 10 2.618 5.426 9 2.768 4.695 8 2.957 3.981 7 3.206 3.286 6 3.549 2.613 5 4.057 1.970 4 4.890 1.366 3 6.509 0.818 2 10.946 0.355 1 44.890 0.052 0

S3-1. AQL = 0.02, LTPD = 0.08 LTPD/AQL = 4 The nearest entry in the table is, LTPD / AQL

n*AQL

3.549

2.613

n*.02 = 2.613 n = 2.613/.02 = 130.65 Great Lakes should take a sample of 131 apples from its shipment of 10,000 and reject the sample if more than 5 apples are defective.

S3-2. AQL = 0.04, LTPD = 0.10 LTPD/AQL = 0.10/0.04 = 2.5 The nearest entry in the table is, LTPD / AQL

n*AQL

2.497

6.169

n*.04 = 6.169 n = 6.169/.04 = 154.225 The publishing company should take a sample of 155 bound books from its shipment and reject the sample if more than 10 books are defective.

S3-3. The values in the Acceptance Sampling Table are based on an alpha of .05 and a beta of .10. Thus, if the sampling plan currently used matches the one we create, the alpha and beta standards are met. Current plan: n=150, c= 4 Optimal plan: AQL = 0.06, LTPD = 0.12 LTPD/AQL = 0.12/0.06 = 2 The nearest entry in the table is, LTPD / AQL 1.990

n*AQL 12.442

c = 18 n *.06 = 12.442 n = 12.442 / .06 = 207.37 No, the current plan will not meet the desired standards.

c 18

S3-4.

S3-5.

4 - Product Design Answers to Questions 4-1.

A firm’s products and services define its customers, as well as its competitors. New products and services often involve new markets and require new processes. The design process is the most obvious driver of change in an organization. It capitalizes on a firm’s core competencies and determines what new competencies need to be developed. Organizations can gain a competitive edge through designs that (1) bring ideas to the market quickly, (2) do a better job of satisfying customer needs, and (3) are easier to manufacture, use, and repair than existing products.

4-2.

Quality of design—size; location and amount of lighting; size of blackboards; heat; presentation equipment; Quality of conformance—number of students crowded into room; overhead projector or lights don’t work; too hot or too cold; instructor doesn’t speak loud enough, or students are too loud.

4-3.

Student answers will vary. Bad designs fail to take the user’s point of view, are too confusing to operate, are too complex, and go against the natural tendency of the user.

4-4.

Student answers will vary.

4-5.

Student answers will vary.

4-6.

Student answers will vary.

4-7.

Perceptual maps compare customer perceptions of a company’s products with competitor products. Students will construct a variety of maps in response to this question. Typical dimensions include: (a) cost and reputation, (b) ease of use and features, (c) cost and variety, (d) convenience and cost.

4-8.

Benchmarking refers to finding the best-in-class part, product or process, measuring one’s performance against it, and making recommendations for improvement based on the results. Student answers will vary. Most of the benchmarking studies compare processes, rather than products, across industries.

4-9.

Again, the answers to this question will vary considerably. Some possible characteristics are: size of student body, size of faculty and staff, salary levels, research dollars, publications, number of degrees, types of degrees, SAT or GMAT scores, number of National Merit Scholars, graduation rate, employment rate, and average salary of graduates.

4-10. A market analysis is conducted to assess whether or not sufficient demand for the proposed product exists to merit investment in its further development. If the demand potential exists, an economic analysis is conducted which looks at estimates of production and development costs and compares them to estimated sales volume. Finally, technical and strategic analyses are completed which answer such questions as: Does the new product require new technology? Is the risk or capital investment excessive? Does the company have sufficient labor and management skills to support the technology required? Is sufficient capacity available for production? Does the new product provide a competitive advantage for the company? Does it draw on corporate strengths? Is it compatible with the core business of the firm? 4-11. Performance specifications tell how a product should perform in order to satisfy customer needs. Design specifications detail the measurements and standards to which a product is to be built so that it meets performance specifications. Manufacturing specifications outline how the processes are supposed to operate in order to produce the product to meet design specifications. Sample specs will vary.

4-12. Reliability is the probability that a given part or product will perform its intended function for a specified length of time under normal conditions of use. Maintainability refers to the ease and/or cost with which a product is maintained or repaired. Maintainability and reliability are closely related, and depending on the context, reliability and maintainability may be interdependent. For example, if a product is cheap to manufacture and priced so low that customers throw it away when it fails (such as calculators, telephones, and watches), maintainability may be a moot issue. Similarly, if a product is so reliable that it rarely breaks down then the ease of repair many not be important. On the other hand, it may be less costly to make a product easy to maintain than to increase its reliability. And for some products, both reliability and maintainability are very important (e.g., office machines, computers). 4-13. Simplification attempts to reduce the number of parts, assemblies, or options in a product. Fewer parts and better fitting parts aim to reduce the number of manufacturing steps and provide fewer chances for error in manufacturing. Standardization makes possible the interchangeability of parts among products, resulting in higher volume production and purchasing, lower investment in inventory, easier purchasing and material handling, fewer quality inspections, and less difficulties in production. Modular design consists of combining standardized building blocks or “modules’’ in a variety of ways to create unique finished products. Thus, even though the individual components may be standardized, the finished product is unique. 4-14. Design teams can obtain input from a variety of sources before erroneous decisions are made. There is often a synergistic effect of people working together. Studies have found that the key to a good design is the involvement and interaction of the “create, make, and market’’ functions from the beginning of the design project. That said, working in teams can be difficult. It is usually easier and less conflicting to work solo. Team members must be convinced that their joint effort will produce better results than individual efforts. 4-15. Concurrent design changes the design process from a sequential one, where decisions are made by separate departments, to simultaneous decision making by design teams. Concurrent design attempts to integrate product design and process planning into a common activity. It helps improve the quality of early design decisions and thereby reduces the length and cost of the design process. In a group project, students are responsible for completing an assigned portion of the project. If the project is well designed with clear expectations, this division of labor will work. If not, considerable rework will be required. 4-16. DFM identifies product design characteristics that are inherently easy to manufacture, focuses on the design of component parts that are easy to fabricate and assemble. Techniques for DFM include design for assembly (DFA), failure mode and effect analysis (FMEA), fault tree analysis (FTA), and value analysis (VA). Design for supply chain (DFSC) considers the capabilities of suppliers at each level of the supply chain when designing a product, and looks across products to reduce design variations, increase component commonality and apply universal design concepts. 4-17. Failure Mode and Effects Analysis (FMEA) is a systematic approach to analyzing the causes and effects of product failures. It begins with listing the functions of the product and each of its parts. Failure modes are then defined and ranked in order of their seriousness and likelihood of failure. Each failure is addressed one by one, causes are hypothesized, and design changes are made to reduce the chance of failure. The objective of FMEA is to anticipate failures and design them out of the system. Fault tree analysis (FTA) is similar to FMEA except that it emphasizes the interrelationship between failures and presents the analysis more graphically. Value analysis (also known as value engineering) was developed by General Electric in 1947 to eliminate unnecessary features and functions in product designs. It has re-emerged as an excellent technique for use by multifunctional design teams. An example fault tree analysis for a term paper is given on the next page.

4-18. Students will find references to design for environment (DFE), green products, E3, sustainable materials, sustainable supply chains, and others. The best source of environmental policy for other countries is the European Union’s information site at http://europa.eu 4-19. Student responses will vary. 4-20. The ISO 14000 family of international standards offers a wide-ranging portfolio of standardized sampling, testing and analytical methods to deal with specific environmental challenges. It has developed more than 350 international standards for monitoring the quality of air, water and soil. These standards provide business and government with scientifically valid data on the environmental effects of economic activity and serve in a number of countries as the technical basis for environmental regulations. ISO 14000 also provides a strategic approach to environmental management systems (EMS) that can be implemented in any type of organization in either public or private sector (companies, administrations, public utilities). ISO 14000 was developed in response to discussions of “sustainable development’’ at the United Nations Conference on Environment and Development in Rio de Janeiro in 1992. The International Standards Organization (ISO) put together a committee of scientists to develop the standards which were first published in 1996. A company obtains certification from a 3rd party registrar who follows ISO 14000 document guidelines and performs audits and site visits. Certification allows companies to use Environmental Labeling, reduces insurance, regulatory and operating costs, and enhances market appeal. Although voluntary, ISO 14000 standards are often referenced in technical regulations and other types of legal documents. In addition, companies in certain industries require their suppliers to be ISO 14000 certified. Globally, Japan has the largest number of ISO 14000 registered companies, followed by the UK, Germany, Sweden, and the U.S.

4-21. Quality Function Deployment (QFD) is a structured process that translates the voice of the customer to technical requirements at every stage of design and manufacture. QFD forces management to spend more time defining the new product changes and examining the ramifications of those changes. More time spent in the early stages of design means less time is required later to revise the design and make it work. 4-22. A robust design is one able to withstand variations in environmental and operating conditions and still operate as intended. Example—a cell phone that works after repeatedly being dropped. 4-23. CAD has been especially useful in testing designs (CAE) and as a means of integrating design and manufacture (CAD/CAM). Technology has enabled us to design new products more quickly, consider many more alternatives than would be possible manually, test the design on a computer screen without building a prototype, document the design in different forms, and automatically transfer the design to manufacturing. Design centers or team members can be located in diverse geographic locations and still communicate freely. Experts can collaborate on designs and make changes from a distance via the Internet.

Solutions to Problems 4-1.

Most students will not find these instructions clear. For example, should the center line be folded lengthwise or widthwise? Which way should the airplane be folded in half? How exactly do you fold back the wings? A diagram of each step would definitely help in communicating the design. It’s fun to see how many different prototypes the students come up with from the same design instructions. After the class agrees on a set procedure, revise the instructions, draw a diagram, and test them on an unsuspecting student.

4-2.

Folding the paper lengthwise in alternate directions could be made clearer if you added “like an accordion.’’ Words such as “fan out’’ and “small fold’’ in “nose’’ of plane are imprecise. It is also unclear whether to fold the paper lengthwise or widthwise (either way will do). As with the first design, a diagram would be helpful. This plane was easier to construct but did not look as much like a plane! In terms of flying ability, this plane is supposed to fly further (like a glider). The first plane flies higher and faster.

4-3.

This system has both parallel and series components. Calculate the reliability of the parallel components first:

4-4.

a. x = 0.95 broadcast success 5

x = 0.9898 subsystem reliability

b. x = 0.98 broadcast success 5

x = 0.9960 subsystem reliability

c. x = 0.99 broadcast success 5

x = 0.9980 subsystem reliability

4-5.

Three teams (A, B, and C) are working in parallel and can be considered as a backup to one another. Therefore:

RS = 1 − (1 − R1 )(1 − R2 ) (1 − R3 ) = 1 − ( 0.10 )( 0.20 )( 0.30 ) = 0.9440 4-6.

Vendor 1: .94  .90 + .10 (.86 )   .93 = .8620 Vendor 2: .85  .93 + .07 (.88 )   .95 = .8007 Vendor 3: .92  .95 + .05 (.90 )   .90 = .8239 Choose vendor 1.

b. Vendor 1: .94  .86  .90  .93 = .6766

Vendor 2: .85  .88  .93  .95 = .6609 Vendor 3: .92  .90  .95  .90 = .7079 Yes, choose vendor 3. 4-7.

a. .98  .97  .95  .96  .99 = 0.8583 No, the probability of failure is (1 − .8583) = 0.1417 b. 5 0.96 = 0.9919 Assuming equal components’ reliabilities, each component would need a reliability of .9919

4-8. a. System A reliability = .93 * .93 = .8649 Component cost = $1500 * 2 = $3000 Failure cost = (1-.8649) * $10,000 = $1351 Total cost of system A = $4351 b. System B reliability = .95 * .95 x .95 = .8574 Component cost = $2000 * 3 = $6000 Failure cost = (1-.8574) * $10,000 = $1426 Total cost of system B = $7426 System A is more reliable and less costly than System B.

c. System C reliability = [.95 + (.05) (.90)] * .95 * [.95 + (.05) (.90)] = .9405 Component cost = ($1000 * 2) + ($2000 * 3) = $8000 Failure cost = (1-.9405) * $10,000 = $595 Total cost of system C = $8595

d. System D reliability = [.90 + (.10) (.90)] x [.90 + (.10) (.90)] x [.90 + (.10) (.90)] = .9703 Component cost = $1000 * 6 = $6000 Failure cost = (1-.9703) * $10,000 = $297 Total cost of system D = $6297 System D is more reliable and less costly than System C. This is somewhat surprising because we usually think of a more reliable system as being more costly.

4-9.

System Basic Standard Professional

Component Reliability 0.80 0.90 0.99

No. Components 5 5 5

System Reliability 0.3277 0.5905 0.9510

Purchase Cost $1,000 $2,000 $5,000

Failure Cost $50,000 $50,000 $50,000

Failure Probability 0.6723 0.4095 0.0490

Total Cost $34,616.00 $22,475.50 $7,450.50

No. Components 5 5 5

System Reliability 0.8154 0.9510 0.9995

Purchase Cost $2,000 $4,000 $10,000

Failure Cost $50,000 $50,000 $50,000

Failure Probability 0.1846 0.0490 0.0005

Total Cost $11,231.37 $6,450.50 $10,025.00

Choose the professional system. b.

System Basic Standard Professional

Component w/backup 0.960 0.9900 0.9999

Choose the standard system. 4-10.

a. .90 *.90 *.90 = .729 b. .90 + (.10*.80) = .98; .98 * .98 * .98 = .941

4-11.

a. .75 *.75 * .75 = .422 b. .99 * .50 * .70 = .347 c. (.60 + (.40*.60)) * .80 * (.50 + (.50*.40)) = .470 Choice c. is most reliable

4.12

a. Reliability of Bus Travel = 0.90 * 0.93 = .837 b. Reliability of Air Travel = .90 * .50 = .45

•• c. Reliability of Car Travel = .70 + (.30*.40) = .82 d. Reliability of Car and Bus Travel = (.79 + (.21 *.40)) * .93 = .813 Choose a. Bus Travel

4-13.

a. During the school year: .97 *.97 * .95 *.90 = .80 During the summer months: .90 * .90 * .90 * .90 = .65 b. Summer: [.90 + (1-.90) *.90] * [.90 + (1-.90) *.90] * [.90 + (1-.90) *.90] * [.90 + (1-.90) *.90] = .961 School: [.97 + (1-.97) *.80)] * [.97 + (1-.97) *.80)] * [.95 + (1-.95) *.80)] *[.90 + (1-.90) *.80] = .959 With replacement workers, shift reliability improves by 30% during the summer months and 16% during the rest of the year. c. Yes, Lisa gets a Super Fry award, because her shift is staffed more than 95% of the time during the year.

4-14. Stitching is essentially a backup to the gluing step, therefore:

4-15.

4-16.

SA = MTBF / (MTBF + MTTR) = 100 / (100 + 24) = .8065

4-17. Copier with the highest System Availability should be selected: MTBF (hours) 40 80 240

Provider Able Copy Business Mate Copy Whiz

MTTR (hours) 1 4 8

System Availability 0.9756 0.9524 0.9677

Choose Able Copy

4-18. Provider with the highest System Availability should be selected. MTBF must be calculated first as 1/failure rate. Note: MTBF and MTTR units must match when calculating System Availability. No. of failures MTBF MTTR System Provider per 8 hrs (mins) (mins) Availability Airway 10 48 2 0.9600 Bellular 8 60 4 0.9375 CyCom 3 160 10 0.9412 Choose Airway

4-19. The to Regain service consists of reaching the customer AND fixing the problem, therefore:

Provider JCN Bell Comtron

No. of problems 50 100 250

Mean Time to Reach Customer 3 2 1

MTBF (hours) (8 x 40) / 50 = 6.4 (8 x 40) / 100 = 3.2 (8 x 40) / 250 = 1.3

Mean Time to Fix Customer 2 1 0.5

Choose JCN 4-20. Choose Yourizon. (Note: number of failures observed are over the 8hr work day)

ISP Xceptional

No. Failures 12

Time to regain service, MTTR 2

Failure Rate, MTBF

Uptime as %, SA

0.952

Yourizon

120

0.968

Zelltell

160

0.941

MTTR (hours) 5 3 1.5

System Availability 0.5614 0.5161 0.4604

4-21. Answers will vary. Here is a representative solution.

4-22. Answers will vary. Here is a possible QFD matrix to use as a guide. Provide customer requirements and design characteristics and have students complete the matrix.

4-23. Answers will vary. This is a sample set up for the problem.

4-24. Students will come up with some interesting examples. Here is a sample worksheet, also available as an excel file on the text website.

CASE SOLUTION 4.1 – Not My Fault a. Reliability in sequence = .70 *.90 *.60 *.80 *.75 *.80 *.50 = 0.09 b. Reliability with backups

Student responses will vary. Discuss the problems with too many steps (and the introduction of more errors) versus not checking to make sure corrections are made.

CASE SOLUTION 4.2 – Greening Product Design 1.

2. 3.

Hal expresses the reaction of many who view green design as “the next big thing” in management circles, perhaps a passing fad. However, he does have a point that changing a design so significantly is not a simple process, and that he needs time to explore the consequences of these changes. Perhaps he also needs time to explore the benefits of green design. A good way to justify green design is to quantify the effects on both the environment and the bottom line, shortterm and long-term. Some companies, such as Patagonia, incorporate green design into their mission, i.e., who they are as a company. Green design spurs innovation and becomes a competitive advantage. Other companies, such as WalMart, have found such large cost savings from green design that they also gain an advantage. Customers also pressure companies to act responsibly and their actions can affect sales. These are the most common reasons, other than legal requirements, why companies go green. It is generally easy to purchase green commodity products, like light bulbs. The challenge is choosing a sustainable product when it is costlier, less convenient or less attractive. An interesting Harvard Business School report, entitled “Sweatshop Labor is Wrong Unless the Jeans are Cute” (http://hbswk.hbs.edu/item/firstlook-january-21-2009) says that ethical behavior is influenced by the desirability of the product. Thus, we might buy a green product that costs a little more, but not a lot; or we might opt for a non- green product if it is “cute.”

CASE SOLUTION 4.3 - Lean and Mean

5 - Service Design Answers to Questions 5-1.

Services are acts, deeds, performances or relationships that produce time, place, form or psychological utilities for customers. A smart phone is a product and a service that we use all the time. The product would be of no use if we did not have cellular or Wi-Fi service.

5-2.

(1) & (2) Services are intangible and perishable. This makes design more difficult. Intangibles are harder to specify, and perishable items have to be redesigned frequently. (3) & (4) Services have high customer contact and variable output. Because people are usually involved in services (both on the side of the customer and the service provider), unexpected things can happen. Designs must be flexible but must also include contingency plans. (5) Consumers do not separate the service from the delivery of the service, so everything about the service environment (layout, attitude of employees, etc.) is part of the service design. (6) Services tend to be decentralized and geographically dispersed. Thus, a design must be “workable” in any locale or be adapted to the specific environment in which it must operate. Also, service providers have more authority and may provide the service differently. (7) & (8) Services are “consumed” more often than products and can be easily emulated. This means that improvements to service design are undertaken more frequently than with products, and the results are more immediate. Risks can be taken on a small scale.

5-3.

Bank—types of accounts, credit cards, loans, branches, hours of operation. ATMs, degree of customization, convenience, decor, information on-line, mortgages, automatic deposit and withdrawal, etc. Airline—flights to particular destinations, reservations, check in, baggage handling, accessibility, types of planes, crowding of planes, service personnel, meals, comfort, disruptions, arrives and departs on time, length of flights, number of direct flights, timing of flights, etc. Lawn Service—frequency of service, jobs (cutting lawn, fertilizing, watering, planting, landscaping), equipment, evaluation of lawn needs, demeanor and performance of personnel, ease of payment, etc.

5-4.

Automated banking—more readily available, accessible from home, more services available (stamps, airline tickets, concert tickets), etc. Higher education—break out of semester segments, self-paced learning, access from home or job site, distance learning, expert-a-day (i.e., more than one instructor), etc. Healthcare—more available, access from schools, records more accessible, availability and instruction for selfcare and self-evaluation, etc.

5-5.

A service blueprint maps out a service process and includes several different lanes or ‘lines’ of activities. The line of influence shows activities designed to influence the customer to enter the service facility and the physical evidence to take action throughout the service process. The line of interaction is where the customer interacts with the service provider face-to-face. The line of visibility separates front-office activities from back-office activities, and the line of support is where the service provider interacts with backstage support personnel to complete their tasks. Moving these various lines on the service blueprint allows the designer to experiment with expanding or decreasing tasks or activities in each area.

5-6.

Examples might include a number of different retail stores, banks, movie theaters, service stations, university operations and functions, wireless access, phone calls, etc.

5-7.

The ratio of the arrival rate to the service rate must be less than 1, which means that the service rate must be greater than the arrival rate. If customers arrive faster than they can be served, the system eventually develops an infinite queue.

5-8.

Examples might include a doctor’s office, a faculty advisor, a machine shop, a number of breakdown/repair systems in which the operating units are finite, etc.

5-9.

These elements are queue discipline, calling population, arrival rate, and service rate.

5-10. Waiting lines are an integral part of a multitude of business operations, and their efficient design and management are important to these businesses. 5-11. The mean effective service rate is the number of servers multiplied by the service rate, and it must exceed the arrival rate. 5-12. a. Hair salon: multiple-server; first come, first-served or appointment; calling population can be finite (appointments only) or infinite (off-the-street business). b. Bank: multiple-server; first come, first-served; infinite calling population. c. Laundromat: multiple-server; first come, first-served; infinite calling population. d. Doctor’s office; single- (or multiple-) server; appointment (usually); finite calling population. 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. Mainframe computer: multiple-server; first come, first-served (or priority level); infinite calling population. 5-13. The addition of a new counter created two queues. The multiple-server model is for a single queue with more than one server. 5-14. The traditional cost relationship for waiting line analysis is to provide a level of service that minimizes the total cost of providing service and the cost of customers waiting, which is similar to the traditional quality cost relationship to provide a level of quality that minimizes the total cost of poor quality and the cast of prevention. However, the recent emphasis on quality management suggests that the level of service should be greater than the minimum cost level, approaching a 100 percent service level. 5-15. a. Single-channel, single-phase—a single server; an example is a post office with one postal clerk. b. Multiple-channel, single-phase—two or more servers in parallel; an example is a post office with two or more clerks. c. Single-channel, multiple-phase—a series of single-servers that the customer must pass through; an example is a medical treatment facility (e.g., hospital) wherein a patient must first see a nurse and then see a doctor. d. Multiple-channel, multiple-phase—a series of two or more servers in parallel; an example is a medical facility with two or more series of nurses and doctors. 5-16. Total waiting line cost is the sum of the cost of providing service and the cost of customer waiting,

which generally have an inverse relationship to each other. Thus, as the level of service is increased, the cost of service increases, whereas the cost of waiting decreases. The optimal level of service coincides with the minimum point on the total waiting line cost curve. 5-17. 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. 5-18. If a service facility has a Poisson input with parameter  and an exponential service distribution with parameter  (where    ), then each successive facility in a multiple-phase system will have a Poisson input with . This enables each phase to be analyzed independently; the aggregate operating characteristics at each service facility are summed. 5-19. When arrivals are random, in the short run more customers may arrive than the serving system can accommodate. 5-20. An example is when customers are served according to a prearranged schedule or alphabetically or are picked at random. 5-21. When the arrival rate equals the service rate,  =  . 5-22. Automated equipment (such as car washes) and most robots. 5-23. For traditional rental car services, the customer goes to the rental car site, fills out the paperwork and picks up the vehicle. The rental is then returned to this or another rental site. For Zipcar, customers locate an available vehicle online (or through an app), input their information online, and notify the company where they have left the car after use. Uber comes to the customer, picks them up and delivers them to their desired location. Cars are contacted and paid for through an app. The degree of customer involvement changes with each service concept, as does the method and means of payment. 5-24. Answers will vary by student. This is a good question for class discussion and sharing.

Solutions to Problems 5-1.

Single server model

=

60 = 16.67 customers per hr 3.6

=

60 = 25 customers per hr 2.4

(16.67 ) 2 Lq = = = 1.33 customers waiting  (  −  ) ( 25)( 25 − 16.67 ) 2

Wq =

(16.67 )  = = 0.08 hr or 4.80 minutes waiting in line  (  −  ) ( 25)( 25 − 16.67 )

How long would you be willing to wait at the drive through in a fast food restaurant? A twominute service time is probably too long, as is a wait time of almost five minutes. Providing menu signage before a customer reaches the ordering kiosk, assigning numbers to special meals, adding additional drive-through lanes and increasing the number of servers will all reduce waiting time.

5-2.

Single server model

 = 8/hr  = 12 / hr

5-3.

Wq =

 8 = = 0.167 hr (10 min )  (  −  ) 12 ( 4 )

=

 8 = = 0.667  12

Single server model

 = 5/hr  = 10 / hr

( 5) = 0.5 cars 2 Lq = =  (  −  ) 10 ( 5 ) 2

1 1 = = 0.20 hr (12 min )  − 5

Wq =

 5 = = 0.10 hr ( 6 min )  (  −  ) 10 ( 5)

b. If the arrival rate is increased to 12 per hour, the arrival rate would exceed the service rate; thus, an infinite queue length would result. 5-4.

Single server model The arrival rate must be on an hourly basis.

=

60 = 7.5 per hour 8

 = 10 per hour

( 7.5) = 2.25 parts 2 Lq = =  (  −  ) 10 ( 2.5 ) 2

 = 0.75, I = 0.25 5-5.

Single server model

=

 ;  = 10 per hr ;  = 0.90 

 Therefore, 0.90 = 10 and  = 9 per hour, or 1 part every 6.67 minutes.

5-6.

Single server model

 = 5 per hour

=

60 = 7.5 per hour 8

( 5) = 1.33 2 = a. Lq =  (  −  ) 7.5 ( 2.5 ) 2

b. Wq =

 5 = = 0.26 hr (16 min )  (  −  ) 7.5 ( 2.5) 1 1 = = .4 hr ( 24 min )  −  7.5 − 5

45 , or 0.75, utilization factor.  cannot exceed 0.75. d. 45 minutes per hour is a 60

=

 5 = = 0.67 at present  7.5

Therefore, a new air traffic controller is not needed.

5-7.

Single server model

 = 12 per hour

=

60 = 20 per hour 3

One window:

Wq =

 12 = = 0.075 hr ( 4.5 min )  (  −  ) 20 (8)

Two windows:

 = 20 per hour (does not change) However, the arrival rate for each window is now split.

 = 6 per hour Wq =

 6 = 0.021 hr (1.29 min ) =  (  −  ) 20 (14 )

4.5 − 1.29 = 3.21 min reduction in waiting time 3.21 $2, 000 = $6420 Cost of window = $20, 000 $6420  $20, 000; Therefore, a second drive-in window should not be installed according to cost alone. However, the existing wait time of 4.5 minutes seems excessive, thus in terms of providing quality service, the second window should probably be installed. 5-8.

Single server model

 = 10 per hour

=

60 = 12 per hour 5

a. L =

  −

10 =5 2

(10 ) = 4.17 2 Lq = =  (  −  ) 12 ( 2 ) 2

5-9.

1 1 = = 0.5 hr ( 30 min )  − 2

Wq =

 10 = = 0.42 hr ( 25 min )  (  −  ) 12 ( 2 )

=

 10 = = 0.83 = 83%  12

Single server model

 = 120 per day  = 140 per day

(120 ) = 5.14 trucks 2 = a. Lq =  (  −  ) 140 ( 20 ) 2

1 1 = = 0.05 day  −  20

8 hr / day  60 min / hr = 480 min 0.05  480 = 24 min.

W = 24 min Wq = b.

 120 = = 0.043 day ( 20.6 min )  (  −  ) 140 ( 20 )

24 − 15 = 9 min 9 minutes  $10, 000 = $90, 000 per year loss presently. With a new set of scales, the arrival rate would be split.

 = 60 per day per scale W = 1/ ( − ) =   (−) =  day 8 hr/day * 60 min/hr * 0.0125 day = 6 minutes The entire $90,000 would be saved. Since the scales cost $50,000 per year, they should be installed. 5-10.

Single server model

Arrival rate = 120 trucks per hour Service rate = 140 trucks per hour P (0) = 1-  = 0.143 P (1) = P0 * () = 0.122 P (2) = P0 * () = 0.105 P (3) = P0 * () = 0.090 P (4) = P0 * () = 0.077 0.537 P (< = 4) = 0.537 P (> = 5) = 1 – P (< = 4) = 0.463 This solution assumes that arriving trucks will pass by the station if they see 4 trucks waiting in a line. Since four trucks are waiting, one truck is being served for a total of 5 trucks in the system.

5-11.

Multiple server model

 = 10 per hour  = 12 per hour s=2 1

P0 =

2  1 1  10  1  10   ( 2 )(10 )      +      0 n !  12   2!  12   24 − 12  1 =  1  10 0 1  10 1  1  10 2  20     +   +      0!  12  1!  12   2  12   12  = .412 n

      L=  P0 + 2  ( s − 1)!( s −  )

Lq = L − W=



 = 1.01 − .83 = .177  1.01 = .101 hr ( 6.05 min ) 10

= 1.01

Wq =



.175 = .0175 hr (1.05 min ) 10

The student will probably recommend adding the advisor.

5-12.

Finite calling population

 = 0.25 per hour per patient  = 6 per hour N = 15 patients

P0 =

1 N!     ( N − n )!      n =0 N

1 15

15!  0.25   (15 − n )!  6  n =0

= 0.423

 +  0.25 + 6  Lq = N −   (1 − P0 ) = 15 −   (1 − 0.423)     0.25  = 0.592 patients

L = Lq + (1 − P0 ) = 0.592 + (1 − 0.423) = 1.168 patients Wq =

( N − L) 

0.592 (15 − 1.168)( 0.25)

= 0.171 hr or 10.27 minutes waiting

W = Wq +



= 0.171 +

1 6

= 0.338 hr or 20.27 minutes in system The nurse is correct, the waiting time for a patient who calls does average about 10 minutes. However, the nurse is currently idle a little over 40 percent of the time, thus, the supervisor cannot achieve both objectives, i.e., a reduced waiting time but no more idle time. If the hospital is quality-conscious they would add a second nurse to reduce waiting time regardless of idle time. Perhaps the nurse could be shared with another ward or group of patients. 5-13.

Single server model

Without machine operator:

 = 20/hr  =  = hr

L = 2.0 employees Lq = 1.33 employees

W = 0.10 = 6 min Wq = 0.067 = 4 min With machine operator:

 = 20/hr  =  = hr

L = 1.0 employees Lq = 0.50 employees

W = 0.05 = 3 min Wq = 0.025 = 1.5 min The system operating characteristic required to perform the decision analysis for this problem is L, the mean number in the system. Note that by reducing the mean service time by half a minute, from 2 minutes to 1.5 minutes, which is 25 percent reduction, we have reduced the mean number in the system by 50 percent, from 2 to 1 employee in the system on the average.

Alternative 1. Without operator

Service Cost/Hour $0

2. With operator

Waiting Cost/Hour

( 2)(10.20) = $20.40 (1)(10.20) = $10.20

Total Cost/Hour $20.40 $18.20

The decision analysis, using the cost figures provided, can be summarized as follows, where waiting cost includes time waiting in line and time at the machines: If the number of working hours per day is 8, this analysis can be converted to a daily basis by multiplying all figures by 8, yielding $163.20 per day expected total waiting cost of alternative 1 and $145.60 per day expected total cost of service and waiting for alternative 2. Thus, the firm’s management should assign an operator to the machine, with an expected daily savings of $17.60.

5-14.

Multiple server model

 = 90 per hour  = 60 per hour s=2

P0 =

 1 1  90 n  1  90  2  2 ( 60 )      +      0 n !  60   2!  60   120 − 90 

= .143

      L=  P0 + 2  ( s − 1)!( s −  )

 = 3.43 − 1.50 = 1.93 

Lq = L − W=

3.43 = .038 hr ( 2.29 min ) 90

1.93 = .021 hr (1.29 min ) 90



Wq = 5-15.



Multiple server model

=

60 = 3.75 parties per hour 16

=

60 = 0.75 parties per hour 80

s=6 P0 =

1  6 ( 3.75 ) 1  3.75  1  3.75       +     0 n !  0.75   6!  0.75   ( 6 )(.750 ) − 3.75 

5

n

= 0.005

L = 7.94

 3.75 = 7.94 − = 2.938  .750

Lq = L −

Wq =

 

2.938 = .783 hr or 47 minutes waiting 3.75

7.94 = 2.11 hr or 127 minutes at the restaurant 3.75

A 47 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.” Thus, in this instance lengthy service may be considered as good quality. 5-16.

Multiple server model

 = 300  = 120 s =3 P0 =

1  2 1  300 n  1  300 3  360     +      0 n !  120   3!  120   60 

= 0.045

    3   ( 300 )(120 )( 2.5 )  L=  P0 + = ( 0.045 ) + 2.5 = 6.01 2  2!( 60 ) ( s − 1)!( s −  )2

Lq = L −

Wq =



 = 6.01 − 2.5 = 3.51  =

3.51 = .0117 hr ( 0.70 min ) 300

No, another wrapper is not needed. 5-17.

Multiple server model

 = 6.5 per hour  = 4 per hour s=2 P0 =

1  1 1  6.5 n  1  6.5 2  8     +      0 n !  4   2!  4   1.5  s

= 0.103

 2    ( 6.5 )( 4 )(1.625 )  L=  P0 + = ( 0.103) + 1.625 = 4.78 2  1!(1.5 ) ( s − 1)!( s −  )2

 

Lq = L −

Wq =



 = 4.78 − 1.625 = 3.155  =

3.155 = 0.485 hr ( 29.14 min ) 6.5

For s = 3,

P0 =

n

3  2 1  6.5  1  6.5     +   ( 2.18 )  0 n !  4   3!  4 

= 0.182

    3   ( 6.5 )( 4 )(1.625 )  L=  P0 + = (.182 ) + 1.625 = 1.96 2 2  2! 5.5 s − 1 ! s  −  ( ) ( )( )

Lq = L −

Wq =



 = 1.96 − 1.625 = 0.33  =

0.33 = 0.05 hr ( 3.1 min ) 6.5

Hire a third doctor.

5-18.

Multiple server model

 =9  =6 For s = 2 :

P0 = 0.143 s

1     s  Pw =     P0 s !     s −   =

1 9   2!  6 

 ( 2 )( 6 )    ( 0.143) = 0.643  ( 2 )( 6 ) − 9 

Thus, two salespeople are not enough. For s = 3:

P0 = 0.211 3 1  9   ( 3)( 6 )  Pw =     ( 0.211) = 0.237 3!  6   ( 3)( 6 ) − 9 

Thus, three salespeople are sufficient to meet the company policy that a customer should have to wait no more than 30 percent of the time.

5-19.

Multiple server model

 = 35  = 15 For s = 3;

P0 = 0.064 s

3   35     35 15 ( )( )      15  0.064 + 35 = 4.47 L=  P + = ( ) 0 2  2! ( 3)(15 ) − 35 2 15 ( s − 1)!( s −  )  

Lq = L −

Wq =



 35 = 4.47 − = 2.14  15 =

2.14 = .061 hr = 3.67 min 35

Three servers should be sufficient.

5-20.

Single server model

 = 4 per day  =? If total repair time is 4 days,

1 ;  −

1 days 4.33



Average repair time = 5-21.

1 ; 3 − 12 = 1 ; 3 = 13 ;  = 4.33  −4

8 hour  1.85 hour, or 110.8 minutes. 4.33

Multiple server model Arrival rate:  = 40 units per hour Processing times: 1. Without additional employees:

1

= 1.2 min per unit ; Thus, 1 = 50 units per hour

2. With additional employees:

2

= 0.9 min per unit

Thus,  2 = 66.67 units per hour In-process inventory = number in process and waiting to be processed =

L= number in system



 −

1. Without additional employees:

L1 =

40 40 = = 4 units in system 50 − 40 10

2. With additional employees:

L2 =

40 40 = = 1.5 units in system 66.67 − 40 26.67

Decision analysis: The cost of in-process inventory is 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. 5-22.

Finite Queue Model a.

 =5 =2 s =3 P0 =

1 2 3 1 5 1  5  1  5   1  5  ( 3)( 2 )    +   +   +    0!  2  1!  2  2!  2   3!  2  ( 3)( 2 ) − 5 0

5 ( 5)( 2 )   2

Wq =



( 0.045) +

5 = 3.5 2

3.5 = 0.70 hr, or 42 min 5

6.0 = 1.20 hr, or 72 min 5

b.  = 5

( 3 − 1)!( 3)( 2 ) − 5

Lq = 6 −

= 25 min

 = 2.4 s =3

5 = 6.0 2

= 0.045

P0 = 0.0982 L=

( 5)( 2.4 ) 

5    2.4 

( 3 − 1)! ( 3)( 2.4 ) − 5

Lq = 3.18 −

Wq =

( 0.0982 ) +

5 = 3.18 2.4

5 = 1.1 2.4

1.1 = 0.22 hr, or 13.2 min, waiting time 5

The improvement in average waiting time per truck is 42 − 13.2 = 28.8 minutes. The estimated value of this time savings 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. c. Alternative 1: Add a fourth loading/unloading location at the dock, yielding four locations where each location has a mean service rate of  = 2 per hour. Alternative 2: Add extra employees and equipment at the existing three dock locations to reduce loading/ unloading times from 30 minutes to 23 minutes per truck, yielding  = 2.6 per hour. Decision analysis: The tendency of the student will be to compare the waiting time for both alternatives. However, this is not required, since the alternatives can be evaluated 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 yields the following: Alternative 1: ( no. servers )( mean service rate ) = ( 4 )( 2 ) = 8 trucks per hour Alternative 2: ( no. 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 hour, whereas adding extra resources to the existing dock increases the effective service rate to only 7.8 trucks per hour. 5-23.

Constant service time model

 = 11/hr  = 13.33 / hr

(11) 2 Lq = = = 1.94 drivers 2 (  −  ) ( 2 )(13.33)(13.33 − 11) 2

Wq = 5-24.



1.94 = 0.176 hr, or 10.6 min 11

Constant service time model

 = 50 / hr = .83/min  = 2/min = 0.149 people in line

5-25. Finite Queue Model n

 P ( n  10 ) = 1 −  Pn = 1 −     P0 n n  n

 4  = 1−     (.0762 ) = 1 − .5818 4.33   10 P ( n  10 ) = .4182 5-26.

Constant service time model The appropriate model is Poisson arrivals with constant service times:

 = 149 + 30 = 179 trucks/day  24-hr day 1



= 8 min

Thus,

1  

 =   ( 60 )( 24 ) = 180 / day 8

 179    180   Lq = = 89.003 trucks  179  2 1 −   180  Wq =



89.003 = 0.4972 days 179

0.4972 ( 24 ) = 11.93 hr Yes, the port authority can assure the coal company that their trucks will not have to wait longer than 12 hours each on the average. 5-27.

Multiple server model 7:00 A.M. to 9:00 A.M.:  = 8

 = 2.5

s must equal at least 4 for the mean effective service rate to exceed the arrival rate.

Wq = 0.30 minutes with 4 cashiers, so 4 is sufficient. 9:00 A.M. to Noon:

=4  = 2.5

s must equal at least 2.

Wq = 0.711 minutes with 2 cashiers, so 2 is sufficient. Noon to 2:00 P.M.:

 = 14  = 2.5

s must equal at least 6 cashiers.

Wq = 0.823 minutes with 6 cashiers, so 6 is sufficient. 2:00 P.M. to 5:00 P.M.:

 =8  = 2.5

s must equal at least 4 cashiers. Wq = 0.298 minutes with 4 cashiers, so 4 is sufficient. 5-28.

Multiple server model

=4  = 1.333

s=4 P0 = 0.038

L = 4.53 Lq = 1.53

W = 1.13 hr = 67.97 min Wq = 0.382 hr = 22.96 min b.

s =5 P0 = 0.0466

L = 3.36 Lq = .355

W = .838 hr = 50.33 min Wq = .088 hr = 5.32 min Although the customer waiting time is reduced from 22.96 minutes to 5.32 minutes, 2.3 minutes does not seem excessive for a hair stylist; thus, the impact of adding a fifth stylist may not be significant. 5-29.

Finite calling population

 = .00139/hr  = .08333/ hr Servers = 1 ; N = 8 P0 = .8690 1 − P0 = .1310 −  Lq = N −   (1 − P0 ) = .02   

L = Lq + (1 − P0 ) = .1476 Wq =

( N − L) 

= 1.52 hr

W = Wq +

= 13.52 hr



A patrol car is out of service an average of 13.52 hours when being repaired. Whether or not this is adequate repair service depends on how busy the police department is. 5-30.

Finite calling population

 = 8.57/hr  = 3.0 / hr s=4 s = 12.0 P0  0.046 ( from Table 17.2 ) 1 − P0 = .954 s

      L= P + 2 0  ( s − 1)!( s −  )

 8.57 = 3.98 − = 1.127  3.0

Lq = L − W=



Wq =

5-31.



3.98 = .465 hr = 27.89 min 8.57

1.127 = 0.132 hr = 7.89 min 8.57

Finite calling population, exponential service times.

 1



1 6

= 6 days

 = per day

= 2 days

 = per day

1 2

N =6 First, compute the following values needed in the formula calculations:

0 1 2 3 4 5 6

   

N! ( N − n )!

720 / 720 = 1 720 /120 = 6 720 / 24 = 30 720 / 6 = 120 720 / 2 = 360 720 /1 = 720 720 /1 = 720

P0 =

N!  ( N − n )!  

1 1/3 1/9 1/27 1/81 1/243 1/729

1 N



 ( N − n )!    

n =0

6/3 = 2 30 / 9 = 3.333 120 / 27 = 4.444 360 / 81 = 4.444 720 / 243 = 2.963 720 / 729 = 0.987 Sum = 19.171

1 1 = = 0.05216 Sum 19.171



1 1 + + 6 Lq = N − (1 − P0 ) = 6 − 1 2 (1 − 0.05216 ) = 6 − 4 ( 0.94784 ) = 6 − 3.791 = 2.209  6  2.2 units expected in the queue ( waiting to be repaired )

L = Lq + (1 − P0 ) = 2.2 + 0.94784  3.15 units expected in the system ( being repaired and waiting to be repaired )

Wq =

2.2

( N − L )  ( 6 − 3.15) 1

2.2 2.2 = 2.85 0.475 6

 4.63 days expected time in the queue ( waiting to be repaired )

W = Wq +



= 4.63 + 2

= 6.63 days expected time in the system ( being repaired and waiting to be repaired ) Since the breakdown rate is 16 per day and the number of working days per year is 250, the

= 41.67. expected number of breakdowns per year is 250 6 The expected time at the repair shop per breakdown times the number of breakdowns per year is

6.63  41.67 = 267.27 machine-down days per year. The annual cost of machine downtime is 276.27  50 = $13,813.50. The net difference in the cost of the new service agreement under consideration is

$15, 000 − $3, 000 = $12, 000.

Therefore, the company should select the new service agreement at a net expected cost savings of

$13,813.50 − $12, 000 = $1,813.50.

5-32.

Multiple server model

 = 40/hr  = 12 / hr s=4 s = 48 P0 = 0.021 1 − P0 = .979 s

4   40  ( 40 )(12 )        12  L= P + = 2 0  ( 4 − 1)! ( 4 )(12 ) − 40 2 ( s − 1)!( s −  )  

 

Lq = L − W=

Wq = a.  =

 Lq



(.021) +

40 = 6.62 12

 40 = 6.62 − = 3.289  12 6.62 = .166 hr = 9.93 min 40

3.289 = .082 hr = 4.93 min 40

 40 = = .833 s  48

Idle time 1 − .833 = .167; thus, the postal workers are idle 16.7 percent of the time, which does not seem excessive. b. Wq = 4.93 minutes and Lq = 3.28 customers, neither of which seems excessive although the waiting time borders on too long. c. A customer can expect to walk in and get served without waiting approximately 2 percent (i.e.,

P0 = .021 ) of the time. Overall, the system seems moderately satisfactory from a customer services perspective. However, the post office might want to analyze the system with five stations instead of four because of the somewhat long waiting time. 5-33.

Multiple server model  = 8 manuscripts / week  = 7/10 = .7 manuscripts / week

Lq = 16.35 manuscripts L = 27.78 manuscripts Wq = 2.09 weeks

W = 3.47 weeks  = 0.952 Solution via Excel. 5-34.

Multiple server model a. By testing several different numbers of servers (teams) for the multiple server model using Excel it is determined that at least 4 teams are required to be within the two-week waiting period. The operating characteristics are,

Lq = 1.128 jobs L = 3.98 jobs Wq = 0.28 weeks

W = 1.00 weeks 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 one week period (W = 1.00 ) then that will result in $6,800 in revenue for two weeks with the 4 teams paid $2,000 for two weeks or approximately $4,800 per week.

5-35.

Single server model High-speed copier Regular copier

 = 7 / hr.  = 10 / hr.

 = 7 / hr.  = 20 / hr.

Wq = 14 minutes

Wq = 1.62 minutes

W = 20 minutes

W = 4.62 minutes

Cost of waiting, regular copier = $3.33 per employee = 7  3.33 = $23.31 per hour Cost of waiting, high-speed copier = 7 x $0.77 = $5.39 per hour The lease cost is $8 per hour higher for the high-speed copier, which would still make it more economical than the regular copier, i.e., $13.39 versus $23.31. 5-36.

Single server with finite calling population Current repair service:

 = 0.05 / day  = 1/ day

W = 2.6 days

Downtime Cost = 2.6 ( 24 )( $5) = $312 New repair service:

 = 0.05 / day =2

W = .77 days

Downtime Cost = (.77 )( 24 )(15) = $277.20 Even at $10 per hour more, the new service is better. Solution via Excel. 5-37.

Finite queue model

In this finite queue model, the hotel guests are treated as the servers and the cabs are the customers. The guests have a service time of 8.5 minutes (5 minutes to arrive and 3.5 minutes to load), thus  = 60 / 8.5 = 7.06 cabs per hour are “served.” The cabs arrive at the cab line at the rate of 6 per hour. The queue capacity is 6 cabs. a. Wq = the average time a cab must wait for a fare

b. Probability ( x  6 ) = .0006

n 0 1 2 3 4 5 6 P(n<=6) P(full) =

P(n) 0.22 0.187234 0.159348 0.135615 0.115417 0.098228 0.083598 0.99944 0.00056

Solution via Excel. 5-38.

Single server model One drive-through window:

 = 10 per hour  = 15 per hour

a. No, Wq = 8 minutes b. Yes,  = 24 and and Wq = .53 minutes

c. No, 5-39.

Multiple server model 60/7 = 8.5 Use Excel or OM Tools to try different numbers of servers until W < 12 minutes. For example, # servers

1 2 3 4

 80 40 27 20

 26 26 26 26

infeasible infeasible infeasible .167 hr or 10 minutes

If 4 registers are open,  = 20 and W = 10 minutes which meets the service goal of W = 12 minutes. Thus 4 registers need to be opened. 5-40.

Multiple server model

 = 40 / hr.

 = 15 / hr. s =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 $0.20 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. 5-41.

Finite calling population

=

.7 = .04375 calls per room per day 16

 = 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.

5-42.

Finite Queue Model

=

14 = 1.17 customers / month 12

5 weeks = 1.25 months

=

1 = .80 piece / month 1.25

W = 8.92 months Po = 0.011 1- Po = 0.989 = 98.9% busy Pm = 0.323 1-Pm = .6765

5-43.

Finite calling population

 = 0.0625/ hour  = 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 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 ).

5-44.

Single server model (a)  = 45 passengers per hour  = 52 passengers per hour

Lq =

2  ( −  )

(45)2 = (52)(52 − 45) = 5.56 passengers Wq =

  ( −  )

45 52(52 − 45) = 7.42 minutes =

(b) A multiple server model with 3 security scans can accommodate this increased passenger traffic level.

 = 125 passengers per hour

 = 52 passengers per hour c=3

Po = .066   ( /  )c   L= P + 2 o   (c − 1)!(c  −  )   (125)(52)(125/52)3  125 = .066 +  2 52  (2)!(156 − 125)  = 5.52 passengers Lq = L −

 

= 5.52 − (125/52) = 3.121 passengers Wq =



3.121 = 125 = 1.50 minutes 5-45.

Multiple server model A multiple server model with 4 cranes will accommodate the level of container traffic at the inland port.

 = 9 containers per hour  = 60/25 = 2.4 containers per hour c = 4 cranes Po = .007 L = 16.73 containers Lq = 12.98 containers W = 1.86 hours Wq = 1.44 hours

5-46. Constant service time model  = 36 passengers/(4.1 + 3.5 minutes) = 4.74 passengers/minute

 = 4.4 passengers/minute

Lq =

2 2 (  −  )

(4.4) 2 2(4.74)(4.74 − 4.4) = 6.006 groups of 36 passengers waiting

= (6.006)(36) = 216.21 passengers

5-47. Single server model a. Average number of students in the system i.e. waiting and being served by the cashier L = 25/(30-25) = 25/5 = 5 On average, 5 students would be in the system at any point in time. b. Average number of students in the waiting line Lq = 252 / 30(30-25) = 625/150 = 4.17 There would be an average of 4 students waiting in the checkout line. c. Average time a student spends in the line Wq = 25/30*(30-25) = 25/(30*5) = 25/150 = 0.167 hours or 10.02 minutes Students wait an average of 10 minutes in line. d. Probability that the cashier would be busy, and students would have to wait p = 25/30 = 0.83 There is a 0.83 probability that the cashier will be busy, and the student will have to wait. 5-48. Single server model a. Average number of students in the system i.e. waiting and being served by the cashier L = 30/(40-30) = 30/10 = 3 On average, 3 students would be in the system at any point in time. b. Average number of students in the waiting line Lq = 302 / 40(40-30) = 900/400 = 2.25 There would be an average of 2.25 students waiting in the checkout line. c. Average time a student spends in the line Wq = 30 / [40*(40-30)] = 30 / (40*10) = 30/400 = 0.075 hours = 4.5 minutes

Students wait an average of 4.5 minutes in line. d. Probability that the cashier would busy and student wait p = 30/40 =0.75 There is a 0.75 probability that the cashier will be busy and the student will have to wait.

5-49. Single server model

5-50. Single server model

CASE SOLUTION 5.1 – Streamlining the Refinancing Process

1. Current Process:

Most people assume that bringing in more specialists would speed up the process. But in this case, the loan passes through 7 person’s hands and the customer interacts with several different service providers. A more efficient operation would have only one individual as the point of contact with the customer and reduce the number of people involved with handling the customer file. Notice that each time the file is send to another person, there is a good probability it will wait. Service blueprints will differ based on assumptions and level of detail.

2. Revised Process:

Closing agent handles all tasks except for preliminary loan approval. Electronic approval and transmission of information helps speed process. Loans meeting certain criteria are automatically approved. Attorney performs tasks and verifies closing date. Having a closing date target helps to coordinate activities. Delays are still possible but with fewer handoffs are reduced considerably.

CASE SOLUTION 5.2: Herding the Patient 1. Current process

Analysis: Why does the patient register twice? Can patient pre-register or bring in a hospital card that is scanned? Can undressing/dressing be performed nearer to the exam room? Can patient enter hospital nearer to the department? Can appointment times minimize wait? 2. Revised process

Set up two exam rooms back-to-back so that technician can alternate between the two of them while patients are undressing/dressing. Out-patients can enter diagnostic imaging department directly without having to go through the hospital. Pre-registration by phone, online, or at referring Dr’s office reduces check-in time. As estimates of service time become more accurate, appointment scheduling can be tightened, thereby reducing waiting time. The servicescape should include proper signage describing services and procedures. The rooms should combine a clinical cleanliness with mirrors, comfortable chairs or benches and places to put clothing. A screen or curtain for changing is needed.

CASE SOLUTION 5.3: The College of Business Copy Center A multiple-server queuing model must be evaluated for a center with two copiers and three copiers for the normal academic year and the summer. Normal Academic Year Two Copiers  = 7.5  =5 c=2 W = 0.457 hr = 27.42 min • In an 8-hr day, there are 60 jobs (8  7.5 = 60 )

Three Copiers  = 7.5  =5 c=3 W = 0.232 hr = 13.92 min • In an 8-hr day, there are 60 jobs (8  7.5 = 60 )

• 60 jobs  0.457 hr = 27.42 hr of total secretarial time in the college spent on copying jobs

• 60 jobs  0.232 hr = 13.92 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

• $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  $233.07 = $41, 253.39 per year

•

177 days in the normal academic year  $118.32 = $20.942.64 per year

Summer Months Two Copiers  = 3.75  =5 c=2

W = 0.233 hr = 13.98 min

Three Copiers  = 3.75  =5 c=3

W = 0.204 hr = 12.24 min

• Jobs / day = 30 • 30 jobs  0.233 hr = 6.99 hr

• Jobs / day = 30 • 30 jobs  0.204 hr = 6.12 hr

• $8.50 hr  6.99 hr =

• $8.50 hr  6.12 hr = $52.02

$59.42 / day • 70 days  59.42 = $4.159.05

• 70 days  52.02 = $3, 641.40

The current cost of wages for copying is $41, 253.39 + 4,159.05 = $45, 412.44. The cost with three copiers is $20,942.64 + 3, 641.40 = $24,584.04.

$45, 412.44 − $24,584.00 = $20,828.40

The total annual wage savings by adding a third machine is $20,828 per year. Since a copying machine cost $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 six-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 savings in wages are not really savings to the college but are 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 hard to convince with this argument.

CASE SOLUTION 5.4: Northwoods Backpacker There are four system configurations to be considered, as follows. 1. 5-day, 8-hour per day service 2. 7-day, 8-hour per day service 3. 5-day, 16-hour per day service 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., c  . Remember, the current system has 5 operators (servers), and,  = 60 / 3.6 = 16.67 customers per hour. 5-day, 8-hour service:  = 175,  = 16.67; c   /  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. 7-day, 8-hour service:  = 125,  = 16.67; c   /  or c  125/16.67 = 7.49. Thus, at least 8 operators are required for this system to be feasible. 5-day, 16-hour service:  = 875,  = 16.67; c   /  or c  875/16.67 = 5.24. Thus, at least 6 operators are required for this system to be feasible. 7-day, 16-hour service:  = 625,  = 16.67; c   /  or c  625/16.67 = 3.74. Thus, at least 4 operators are required for this system to be feasible. Therefore, only the first configuration (the current one) is not feasible and is eliminated.

Next the costs of the remaining 3 alternatives are evaluated. 7-day, 8-hour service

 = 125,  = 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 different numbers of operators until the goal of one-half minute waiting time is achieved. 8 operators:

Wq = 5.76 minutes

9 operators:

Wq = 1.20 minutes

10 operators: Wq = 0.42 minutes

( Probability of waiting = 0.18) 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.

5-day, 16-hour service

 = 87.5,  = 16.67 Cost for 16 hour service = $11,500 Cost per extra operator = $4, 700 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 = 0.21) 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, 16-hour service

 = 62.5,  = 16.67 Cost for 16-hour service = $11,500

Cost for 7-day service = $7, 200 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. 5 operators: Wq = 1.32 minutes 6 operators:

Wq = 0.36 minutes

( Probability of waiting = 0.10 ) Since only I 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. 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 80 percent of the time. Thus, other factors may be taken into consideration. For example, both 16-hour service alternatives might be more convenient for customers who work during the day.

5 - Service Design Answers to Questions 5-1.

5-2.

5-3.

5-4.

5-5.

5-6.

Examples might include a number of different retail stores, banks, movie theaters, service stations, university operations and functions, wireless access, phone calls, etc.

5-7.

5-8.

Examples might include a doctor’s office, a faculty advisor, a machine shop, a number of breakdown/repair systems in which the operating units are finite, etc.

5-9.

These elements are queue discipline, calling population, arrival rate, and service rate.

Solutions to Problems 5-1.

Single server model

=

60 = 16.67 customers per hr 3.6

=

60 = 25 customers per hr 2.4

(16.67 ) 2 Lq = = = 1.33 customers waiting  (  −  ) ( 25)( 25 − 16.67 ) 2

Wq =

(16.67 )  = = 0.08 hr or 4.80 minutes waiting in line  (  −  ) ( 25)( 25 − 16.67 )

5-2.

Single server model

 = 8/hr  = 12 / hr

5-3.

Wq =

 8 = = 0.167 hr (10 min )  (  −  ) 12 ( 4 )

=

 8 = = 0.667  12

Single server model

 = 5/hr  = 10 / hr

( 5) = 0.5 cars 2 Lq = =  (  −  ) 10 ( 5 ) 2

1 1 = = 0.20 hr (12 min )  − 5

Wq =

 5 = = 0.10 hr ( 6 min )  (  −  ) 10 ( 5)

b. If the arrival rate is increased to 12 per hour, the arrival rate would exceed the service rate; thus, an infinite queue length would result. 5-4.

Single server model The arrival rate must be on an hourly basis.

=

60 = 7.5 per hour 8

 = 10 per hour

( 7.5) = 2.25 parts 2 Lq = =  (  −  ) 10 ( 2.5 ) 2

 = 0.75, I = 0.25 5-5.

Single server model

=

 ;  = 10 per hr ;  = 0.90 

 Therefore, 0.90 = 10 and  = 9 per hour, or 1 part every 6.67 minutes.

5-6.

Single server model

 = 5 per hour

=

60 = 7.5 per hour 8

( 5) = 1.33 2 = a. Lq =  (  −  ) 7.5 ( 2.5 ) 2

b. Wq =

 5 = = 0.26 hr (16 min )  (  −  ) 7.5 ( 2.5) 1 1 = = .4 hr ( 24 min )  −  7.5 − 5

45 , or 0.75, utilization factor.  cannot exceed 0.75. d. 45 minutes per hour is a 60

=

 5 = = 0.67 at present  7.5

Therefore, a new air traffic controller is not needed.

5-7.

Single server model

 = 12 per hour

=

60 = 20 per hour 3

One window:

Wq =

 12 = = 0.075 hr ( 4.5 min )  (  −  ) 20 (8)

Two windows:

 = 20 per hour (does not change) However, the arrival rate for each window is now split.

 = 6 per hour Wq =

 6 = 0.021 hr (1.29 min ) =  (  −  ) 20 (14 )

Single server model

 = 10 per hour

=

60 = 12 per hour 5

a. L =

  −

10 =5 2

(10 ) = 4.17 2 Lq = =  (  −  ) 12 ( 2 ) 2

5-9.

1 1 = = 0.5 hr ( 30 min )  − 2

Wq =

 10 = = 0.42 hr ( 25 min )  (  −  ) 12 ( 2 )

=

 10 = = 0.83 = 83%  12

Single server model

 = 120 per day  = 140 per day

(120 ) = 5.14 trucks 2 = a. Lq =  (  −  ) 140 ( 20 ) 2

1 1 = = 0.05 day  −  20

8 hr / day  60 min / hr = 480 min 0.05  480 = 24 min.

W = 24 min Wq = b.

 120 = = 0.043 day ( 20.6 min )  (  −  ) 140 ( 20 )

24 − 15 = 9 min 9 minutes  $10, 000 = $90, 000 per year loss presently. With a new set of scales, the arrival rate would be split.

Single server model

5-11.

Multiple server model

 = 10 per hour  = 12 per hour s=2 1

P0 =

      L=  P0 + 2  ( s − 1)!( s −  )

Lq = L − W=



 = 1.01 − .83 = .177  1.01 = .101 hr ( 6.05 min ) 10

= 1.01

Wq =



.175 = .0175 hr (1.05 min ) 10

The student will probably recommend adding the advisor.

5-12.

Finite calling population

 = 0.25 per hour per patient  = 6 per hour N = 15 patients

P0 =

1 N!     ( N − n )!      n =0 N

1 15

15!  0.25   (15 − n )!  6  n =0

= 0.423

 +  0.25 + 6  Lq = N −   (1 − P0 ) = 15 −   (1 − 0.423)     0.25  = 0.592 patients

L = Lq + (1 − P0 ) = 0.592 + (1 − 0.423) = 1.168 patients Wq =

( N − L) 

0.592 (15 − 1.168)( 0.25)

= 0.171 hr or 10.27 minutes waiting

W = Wq +



= 0.171 +

1 6

Single server model

Without machine operator:

 = 20/hr  =  = hr

L = 2.0 employees Lq = 1.33 employees

W = 0.10 = 6 min Wq = 0.067 = 4 min With machine operator:

 = 20/hr  =  = hr

L = 1.0 employees Lq = 0.50 employees

Alternative 1. Without operator

Service Cost/Hour $0

2. With operator

Waiting Cost/Hour

( 2)(10.20) = $20.40 (1)(10.20) = $10.20

Total Cost/Hour $20.40 $18.20

5-14.

Multiple server model

 = 90 per hour  = 60 per hour s=2

P0 =

 1 1  90 n  1  90  2  2 ( 60 )      +      0 n !  60   2!  60   120 − 90 

= .143

      L=  P0 + 2  ( s − 1)!( s −  )

 = 3.43 − 1.50 = 1.93 

Lq = L − W=

3.43 = .038 hr ( 2.29 min ) 90

1.93 = .021 hr (1.29 min ) 90



Wq = 5-15.



Multiple server model

=

60 = 3.75 parties per hour 16

=

60 = 0.75 parties per hour 80

s=6 P0 =

1  6 ( 3.75 ) 1  3.75  1  3.75       +     0 n !  0.75   6!  0.75   ( 6 )(.750 ) − 3.75 

5

n

= 0.005

L = 7.94

 3.75 = 7.94 − = 2.938  .750

Lq = L −

Wq =

 

2.938 = .783 hr or 47 minutes waiting 3.75

7.94 = 2.11 hr or 127 minutes at the restaurant 3.75

Multiple server model

 = 300  = 120 s =3 P0 =

1  2 1  300 n  1  300 3  360     +      0 n !  120   3!  120   60 

= 0.045

    3   ( 300 )(120 )( 2.5 )  L=  P0 + = ( 0.045 ) + 2.5 = 6.01 2  2!( 60 ) ( s − 1)!( s −  )2

Lq = L −

Wq =



 = 6.01 − 2.5 = 3.51  =

3.51 = .0117 hr ( 0.70 min ) 300

No, another wrapper is not needed. 5-17.

Multiple server model

 = 6.5 per hour  = 4 per hour s=2 P0 =

1  1 1  6.5 n  1  6.5 2  8     +      0 n !  4   2!  4   1.5  s

= 0.103

 2    ( 6.5 )( 4 )(1.625 )  L=  P0 + = ( 0.103) + 1.625 = 4.78 2  1!(1.5 ) ( s − 1)!( s −  )2

 

Lq = L −

Wq =



 = 4.78 − 1.625 = 3.155  =

3.155 = 0.485 hr ( 29.14 min ) 6.5

For s = 3,

P0 =

n

3  2 1  6.5  1  6.5     +   ( 2.18 )  0 n !  4   3!  4 

= 0.182

    3   ( 6.5 )( 4 )(1.625 )  L=  P0 + = (.182 ) + 1.625 = 1.96 2 2  2! 5.5 s − 1 ! s  −  ( ) ( )( )

Lq = L −

Wq =



 = 1.96 − 1.625 = 0.33  =

0.33 = 0.05 hr ( 3.1 min ) 6.5

Hire a third doctor.

5-18.

Multiple server model

 =9  =6 For s = 2 :

P0 = 0.143 s

1     s  Pw =     P0 s !     s −   =

1 9   2!  6 

 ( 2 )( 6 )    ( 0.143) = 0.643  ( 2 )( 6 ) − 9 

Thus, two salespeople are not enough. For s = 3:

P0 = 0.211 3 1  9   ( 3)( 6 )  Pw =     ( 0.211) = 0.237 3!  6   ( 3)( 6 ) − 9 

Thus, three salespeople are sufficient to meet the company policy that a customer should have to wait no more than 30 percent of the time.

5-19.

Multiple server model

 = 35  = 15 For s = 3;

P0 = 0.064 s

3   35     35 15 ( )( )      15  0.064 + 35 = 4.47 L=  P + = ( ) 0 2  2! ( 3)(15 ) − 35 2 15 ( s − 1)!( s −  )  

Lq = L −

Wq =



 35 = 4.47 − = 2.14  15 =

2.14 = .061 hr = 3.67 min 35

Three servers should be sufficient.

5-20.

Single server model

 = 4 per day  =? If total repair time is 4 days,

1 ;  −

1 days 4.33



Average repair time = 5-21.

1 ; 3 − 12 = 1 ; 3 = 13 ;  = 4.33  −4

8 hour  1.85 hour, or 110.8 minutes. 4.33

Multiple server model Arrival rate:  = 40 units per hour Processing times: 1. Without additional employees:

1

= 1.2 min per unit ; Thus, 1 = 50 units per hour

2. With additional employees:

2

= 0.9 min per unit

Thus,  2 = 66.67 units per hour In-process inventory = number in process and waiting to be processed =

L= number in system



 −

1. Without additional employees:

L1 =

40 40 = = 4 units in system 50 − 40 10

2. With additional employees:

L2 =

40 40 = = 1.5 units in system 66.67 − 40 26.67

Decision analysis: The cost of in-process inventory is 1. Without additional employees: ( 4 )( $31) = $124.00 / day 2. With additional employees: (1.5) ($31) = $46.50 / day

Finite Queue Model a.

 =5 =2 s =3 P0 =

1 2 3 1 5 1  5  1  5   1  5  ( 3)( 2 )    +   +   +    0!  2  1!  2  2!  2   3!  2  ( 3)( 2 ) − 5 0

5 ( 5)( 2 )   2

Wq =



( 0.045) +

5 = 3.5 2

3.5 = 0.70 hr, or 42 min 5

6.0 = 1.20 hr, or 72 min 5

b.  = 5

( 3 − 1)!( 3)( 2 ) − 5

Lq = 6 −

= 25 min

 = 2.4 s =3

5 = 6.0 2

= 0.045

P0 = 0.0982 L=

( 5)( 2.4 ) 

5    2.4 

( 3 − 1)! ( 3)( 2.4 ) − 5

Lq = 3.18 −

Wq =

( 0.0982 ) +

5 = 3.18 2.4

5 = 1.1 2.4

1.1 = 0.22 hr, or 13.2 min, waiting time 5

Constant service time model

 = 11/hr  = 13.33 / hr

(11) 2 Lq = = = 1.94 drivers 2 (  −  ) ( 2 )(13.33)(13.33 − 11) 2

Wq = 5-24.



1.94 = 0.176 hr, or 10.6 min 11

Constant service time model

 = 50 / hr = .83/min  = 2/min = 0.149 people in line

5-25. Finite Queue Model n

 P ( n  10 ) = 1 −  Pn = 1 −     P0 n n  n

 4  = 1−     (.0762 ) = 1 − .5818 4.33   10 P ( n  10 ) = .4182 5-26.

Constant service time model The appropriate model is Poisson arrivals with constant service times:

 = 149 + 30 = 179 trucks/day  24-hr day 1



= 8 min

Thus,

1  

 =   ( 60 )( 24 ) = 180 / day 8

 179    180   Lq = = 89.003 trucks  179  2 1 −   180  Wq =



89.003 = 0.4972 days 179

0.4972 ( 24 ) = 11.93 hr Yes, the port authority can assure the coal company that their trucks will not have to wait longer than 12 hours each on the average. 5-27.

Multiple server model 7:00 A.M. to 9:00 A.M.:  = 8

 = 2.5

s must equal at least 4 for the mean effective service rate to exceed the arrival rate.

Wq = 0.30 minutes with 4 cashiers, so 4 is sufficient. 9:00 A.M. to Noon:

=4  = 2.5

s must equal at least 2.

Wq = 0.711 minutes with 2 cashiers, so 2 is sufficient. Noon to 2:00 P.M.:

 = 14  = 2.5

s must equal at least 6 cashiers.

Wq = 0.823 minutes with 6 cashiers, so 6 is sufficient. 2:00 P.M. to 5:00 P.M.:

 =8  = 2.5

s must equal at least 4 cashiers. Wq = 0.298 minutes with 4 cashiers, so 4 is sufficient. 5-28.

Multiple server model

=4  = 1.333

s=4 P0 = 0.038

L = 4.53 Lq = 1.53

W = 1.13 hr = 67.97 min Wq = 0.382 hr = 22.96 min b.

s =5 P0 = 0.0466

L = 3.36 Lq = .355

Finite calling population

 = .00139/hr  = .08333/ hr Servers = 1 ; N = 8 P0 = .8690 1 − P0 = .1310 −  Lq = N −   (1 − P0 ) = .02   

L = Lq + (1 − P0 ) = .1476 Wq =

( N − L) 

= 1.52 hr

W = Wq +

= 13.52 hr



A patrol car is out of service an average of 13.52 hours when being repaired. Whether or not this is adequate repair service depends on how busy the police department is. 5-30.

Finite calling population

 = 8.57/hr  = 3.0 / hr s=4 s = 12.0 P0  0.046 ( from Table 17.2 ) 1 − P0 = .954 s

      L= P + 2 0  ( s − 1)!( s −  )

 8.57 = 3.98 − = 1.127  3.0

Lq = L − W=



Wq =

5-31.



3.98 = .465 hr = 27.89 min 8.57

1.127 = 0.132 hr = 7.89 min 8.57

Finite calling population, exponential service times.

 1



1 6

= 6 days

 = per day

= 2 days

 = per day

1 2

N =6 First, compute the following values needed in the formula calculations:

0 1 2 3 4 5 6

   

N! ( N − n )!

720 / 720 = 1 720 /120 = 6 720 / 24 = 30 720 / 6 = 120 720 / 2 = 360 720 /1 = 720 720 /1 = 720

P0 =

N!  ( N − n )!  

1 1/3 1/9 1/27 1/81 1/243 1/729

1 N



 ( N − n )!    

n =0

6/3 = 2 30 / 9 = 3.333 120 / 27 = 4.444 360 / 81 = 4.444 720 / 243 = 2.963 720 / 729 = 0.987 Sum = 19.171

1 1 = = 0.05216 Sum 19.171



1 1 + + 6 Lq = N − (1 − P0 ) = 6 − 1 2 (1 − 0.05216 ) = 6 − 4 ( 0.94784 ) = 6 − 3.791 = 2.209  6  2.2 units expected in the queue ( waiting to be repaired )

L = Lq + (1 − P0 ) = 2.2 + 0.94784  3.15 units expected in the system ( being repaired and waiting to be repaired )

Wq =

2.2

( N − L )  ( 6 − 3.15) 1

2.2 2.2 = 2.85 0.475 6

 4.63 days expected time in the queue ( waiting to be repaired )

W = Wq +



= 4.63 + 2

= 6.63 days expected time in the system ( being repaired and waiting to be repaired ) Since the breakdown rate is 16 per day and the number of working days per year is 250, the

= 41.67. expected number of breakdowns per year is 250 6 The expected time at the repair shop per breakdown times the number of breakdowns per year is

$15, 000 − $3, 000 = $12, 000.

Therefore, the company should select the new service agreement at a net expected cost savings of

$13,813.50 − $12, 000 = $1,813.50.

5-32.

Multiple server model

 = 40/hr  = 12 / hr s=4 s = 48 P0 = 0.021 1 − P0 = .979 s

4   40  ( 40 )(12 )        12  L= P + = 2 0  ( 4 − 1)! ( 4 )(12 ) − 40 2 ( s − 1)!( s −  )  

 

Lq = L − W=

Wq = a.  =

 Lq



(.021) +

40 = 6.62 12

 40 = 6.62 − = 3.289  12 6.62 = .166 hr = 9.93 min 40

3.289 = .082 hr = 4.93 min 40

 40 = = .833 s  48

Multiple server model  = 8 manuscripts / week  = 7/10 = .7 manuscripts / week

Lq = 16.35 manuscripts L = 27.78 manuscripts Wq = 2.09 weeks

W = 3.47 weeks  = 0.952 Solution via Excel. 5-34.

Lq = 1.128 jobs L = 3.98 jobs Wq = 0.28 weeks

5-35.

Single server model High-speed copier Regular copier

 = 7 / hr.  = 10 / hr.

 = 7 / hr.  = 20 / hr.

Wq = 14 minutes

Wq = 1.62 minutes

W = 20 minutes

W = 4.62 minutes

Single server with finite calling population Current repair service:

 = 0.05 / day  = 1/ day

W = 2.6 days

Downtime Cost = 2.6 ( 24 )( $5) = $312 New repair service:

 = 0.05 / day =2

W = .77 days

Downtime Cost = (.77 )( 24 )(15) = $277.20 Even at $10 per hour more, the new service is better. Solution via Excel. 5-37.

Finite queue model

b. Probability ( x  6 ) = .0006

n 0 1 2 3 4 5 6 P(n<=6) P(full) =

P(n) 0.22 0.187234 0.159348 0.135615 0.115417 0.098228 0.083598 0.99944 0.00056

Solution via Excel. 5-38.

Single server model One drive-through window:

 = 10 per hour  = 15 per hour

a. No, Wq = 8 minutes b. Yes,  = 24 and and Wq = .53 minutes

c. No, 5-39.

Multiple server model 60/7 = 8.5 Use Excel or OM Tools to try different numbers of servers until W < 12 minutes. For example, # servers

1 2 3 4

 80 40 27 20

 26 26 26 26

infeasible infeasible infeasible .167 hr or 10 minutes

If 4 registers are open,  = 20 and W = 10 minutes which meets the service goal of W = 12 minutes. Thus 4 registers need to be opened. 5-40.

Multiple server model

 = 40 / hr.

Finite calling population

=

.7 = .04375 calls per room per day 16

 = 2 calls / hour average utilization = 1 − P0 = 1 − .5776 = .4224 Wq = .31 hr. = 19.1 min.

5-42.

Finite Queue Model

=

14 = 1.17 customers / month 12

5 weeks = 1.25 months

=

1 = .80 piece / month 1.25

W = 8.92 months Po = 0.011 1- Po = 0.989 = 98.9% busy Pm = 0.323 1-Pm = .6765

5-43.

Finite calling population

 = 0.0625/ hour  = 0.67 / hour N = 10 (a) Lq = 1.177 athletes waiting

5-44.

Single server model (a)  = 45 passengers per hour  = 52 passengers per hour

Lq =

2  ( −  )

(45)2 = (52)(52 − 45) = 5.56 passengers Wq =

  ( −  )

45 52(52 − 45) = 7.42 minutes =

(b) A multiple server model with 3 security scans can accommodate this increased passenger traffic level.

 = 125 passengers per hour

 = 52 passengers per hour c=3

 

= 5.52 − (125/52) = 3.121 passengers Wq =



3.121 = 125 = 1.50 minutes 5-45.

Multiple server model A multiple server model with 4 cranes will accommodate the level of container traffic at the inland port.

 = 9 containers per hour  = 60/25 = 2.4 containers per hour c = 4 cranes Po = .007 L = 16.73 containers Lq = 12.98 containers W = 1.86 hours Wq = 1.44 hours

5-46. Constant service time model  = 36 passengers/(4.1 + 3.5 minutes) = 4.74 passengers/minute

 = 4.4 passengers/minute

Lq =

2 2 (  −  )

(4.4) 2 2(4.74)(4.74 − 4.4) = 6.006 groups of 36 passengers waiting

= (6.006)(36) = 216.21 passengers

5-49. Single server model

5-50. Single server model

CASE SOLUTION 5.1 – Streamlining the Refinancing Process

1. Current Process:

2. Revised Process:

CASE SOLUTION 5.2: Herding the Patient 1. Current process

Three Copiers  = 7.5  =5 c=3 W = 0.232 hr = 13.92 min • In an 8-hr day, there are 60 jobs (8  7.5 = 60 )

• 60 jobs  0.457 hr = 27.42 hr of total secretarial time in the college spent on copying jobs

• 60 jobs  0.232 hr = 13.92 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

• $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  $233.07 = $41, 253.39 per year

•

177 days in the normal academic year  $118.32 = $20.942.64 per year

Summer Months Two Copiers  = 3.75  =5 c=2

W = 0.233 hr = 13.98 min

Three Copiers  = 3.75  =5 c=3

W = 0.204 hr = 12.24 min

• Jobs / day = 30 • 30 jobs  0.233 hr = 6.99 hr

• Jobs / day = 30 • 30 jobs  0.204 hr = 6.12 hr

• $8.50 hr  6.99 hr =

• $8.50 hr  6.12 hr = $52.02

$59.42 / day • 70 days  59.42 = $4.159.05

• 70 days  52.02 = $3, 641.40

The current cost of wages for copying is $41, 253.39 + 4,159.05 = $45, 412.44. The cost with three copiers is $20,942.64 + 3, 641.40 = $24,584.04.

$45, 412.44 − $24,584.00 = $20,828.40

Next the costs of the remaining 3 alternatives are evaluated. 7-day, 8-hour service

 = 125,  = 16.67 Cost for 7 day service = $3, 600 Cost per extra operator = $3,800

Wq = 5.76 minutes

9 operators:

Wq = 1.20 minutes

10 operators: Wq = 0.42 minutes

( Probability of waiting = 0.18) 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.

5-day, 16-hour service

 = 87.5,  = 16.67 Cost for 16 hour service = $11,500 Cost per extra operator = $4, 700 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 = 0.21) 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, 16-hour service

 = 62.5,  = 16.67 Cost for 16-hour service = $11,500

Cost for 7-day service = $7, 200 Cost per extra operator = $6,300

Wq = 0.36 minutes

7 Capacity and Facilities Design Answers to Questions 7-5.

Student answers will vary. Lighting, size of desks, fixed or adjustable seating, tiered tows, size of blackboard, multimedia equipment, windows, etc. are all factors that affect learning. Classrooms with immobile desks facing forward imply that the class will be taught in a lecture format.

7-6.

The answers to this question will differ based on location and age of the facility. Students should provide a simple sketch of each layout, describe the process of filling a customer’s order, and time how long it takes to be served in each establishment. In general, McDonald’s has multiple servers and multiple lines, Burger King has one line that breaks off into multiple servers, and Taco Bell has one line and one server. Burger King is more efficient but the line seems longer than McDonald’s multiple lines. Burger King’s back office operations are clearly separated, and Taco Bell’s preparation facilities are minimal since the food arrives pre-cooked.

7-7.

Facilities can be “laid out” to meet the following objectives: minimize material handling costs, utilize space efficiently, utilize labor efficiently, eliminate bottlenecks, facilitate communication and interaction (between workers, between workers and their supervisors, or between workers and customers), reduce manufacturing cycle time or customer service time, eliminate wasted or redundant movement, facilitate the entry, exit and placement of material, products, or people, incorporate safety and security measures, promote product and service quality, encourage proper maintenance activities, provide a visual control of operations or activities, and provide flexibility to adapt to changing conditions. Efficiency is affected by the time spent moving from place to place, the proximity of resources, and the ease of work and supervision.

7-8.

Easy access – multiple entrances and exits; safety & security – lights at night, one entry point to flights in airport, rooms with keyed entry; flexibility – rooms with movable partitions; eliminate bottlenecks - assembly lines in general; use space efficiently – IKEA shelves from floor to ceiling; facilitate communication – tiered, curved auditorium where the speaker can see the audience.

7-9.

A product layout is a sequential arrangement of machines (usually in a line) used to mass produce standardized products for a stable high-volume market. The equipment is special purpose, the workers have limited skills, work-in-process inventory is low, and material moves along a fixed path (like a conveyor). Product layouts are known for their efficiency. In contrast, a process layout is a functional grouping of machines (usually known as a job shop) used to produce batches of varied products with fluctuating demand and low volume. The equipment is general purpose, the workers have more versatile skills, work-in-process inventory is high, and material moves along a variable path (e.g., with a forklift). Process layouts are known for their flexibility.

7-10. In ship production, the ship stays in one place and workers, materials, and other resources are brought to that location. In personal services like massages or manicures, the customer stays in one location for the duration of the service. 7-11. (a) process, (b) fixed-position (c) product, (d) primarily process 7-12

Fixed-layouts have the lowest fixed costs, but the highest variable costs. Product layouts have the highest fixed costs, but the lowest variable costs. Process layouts are in-between.

7-13. Block diagramming and relationship diagramming are both used to design process layouts. The difference occurs in the type of input data that is allowed. Block diagramming uses quantitative data, while relationship diagramming uses non-quantitative data. 7-14. Service layouts often have different objectives (e.g., maximizing profit per unit of display space), the appearance of the layout is important, and customers or information are tracked, rather than products. Examples will vary. 7-15. Line balancing attempts to equalize the amount of work at each station, eliminate bottlenecks, and minimize the number of stations. Several heuristic approaches are available for line balancing, including (1) ranked positional weight, (2) longest operation time, (3) shortest operation time, (4) most number of following tasks, and (5) least number of following tasks. Elements are assigned to work stations according to the rankings of the particular heuristic used until the cycle time is reached or until all tasks have been assigned. 7-16. Group technology groups parts into families according to similar shapes or processing requirements. Once the families of parts have been determined, the factory can be arranged into “cells” with each cell dedicated to producing a family of parts; thus, the name, cellular layout. Parts are grouped into families and the machines necessary to produce each family are placed into the appropriate cell. Production flow analysis (PFA) is a popular method for determining product families. Large, immovable machines or machines that are used in several cells are located as near as possible to their point of use. Workers in cellular manufacturing may be required to produce many different parts or products, operate several dissimilar machines (so they must be multi-functional) and follow a prescribed worker path through the cell. 7-17. Cellular layouts resemble product layouts “within” the cells and process layouts ‘between” the cells. The objective of this combination of basic layouts is to obtain the efficiency of a product layout with the flexibility of a process layout. Cellular layouts reduce material handling, transit time, setup time and work-in-process inventory, make better use of human resources, and are easier to control and automate than traditional layouts. However, cellular layouts aren’t for everyone. Part families may be inadequate to form meaningful cells, the cells can become poorly balanced, workers need extra training, and increased capital investment may be required. 7-18. A flexible manufacturing system (FMS) consists of programmable equipment connected by an automated material handling system and controlled by a central computer. In cellular layouts, parts are assigned to families based on similar flow paths. Thus, the routing of parts through a manufacturing cell follows the arrangement of machines within the cell (i.e., the product layout concept). The routing of parts in an FMS is highly variable, often referred to as “random” like a job shop or process layout. Processing and material movement are also more automated.

7-19. Mixed-model assembly lines process more than one model of a product on the same assembly line. This causes

line balancing to be based on average processing times across models, adds the decision of model sequencing to the scheduling process, and requires workers to rotate among work stations. 7-20. Student responses will vary. 7-21. See the virtual tours section of the text webpage. 7-22. Real tours are great! Consider using the facility assessment worksheets created by Eugene Goodson and modified here from his article in the May 2002 issue of Harvard Business Review. It works best if a group of students meets directly after the tour to fill it out.

Solutions to Problems 7-1.

7-2.

7-3.

7-4.

7-5. Load Summary Chart

From / To 1 2 3 4 5 6 Composite Movements From <> To Loads 1 2 70 1 3 75 1 4 45 1 5 40 1 6 0 2 3 0 2 4 0 2 5 45 2 6 0 3 4 0 3 5 10 3 6 0 4 5 90 4 6 0 5 6 0

1 50 40 20 20

2 20

Department 3 4 5 35 25 20 35 10 40 50

2 1

4 1

5 1

3 1

There are multiple solutions to this problem where nonadjacent loads equal zero. For example, departments 1 and 5 could be switched in the layout shown above and the layout score would remain the same. Similarly, departments 2, 1, and 3 could be on the bottom rungs of the layout, and 4 and 5 could be on the top.

7-6.

7-7.

7-8.

7-9.

7-10.

7-11.

7-12.

7-13.

Desired cycle time, Cd = 10 Actual cycle time, Ca = 9

Efficiency = 23 / 3 (10 ) = .8519 or 85.19% Daily Output = ( 8  60 ) / 9 = 53.33

7-14.

If volunteers are plentiful, set the cycle time to the maximum task time.

7-15. Task A B C D E F G H I J

Precedence None A A A B C, E D G F, H I

Time (mins) 8 4 7 3 7 11 2 8 5 7 62

7-16. Task A B C D E F

Precedence — A B — C, D E

Time (mins) 30 15 10 5 10 10 80

b. Leadtime = 30 + 15 + 10 + 5 + 10 + 10 = 80 Desired output = 50 cases Working hours = 40 Desired cycle time = (40  60) 50 = 48 minutes c.

Actual cycle time = 45 minutes Efficiency = 80 (2  45) = .8889 or 88.89% There are multiple solutions to this problem. d.

7-17. Task A B C D E F

Precedence — A B A, E — C, D

Efficiency = 16 (4  4) = 1

Time (mins) 1 2 2 4 3 4 16

or 100%

7-18. Task A B C D E F

Precedence — A A B, C D E

Time (mins) 2 1 2 3 3 3 14

b. Demand = 120 pizzas Working hours = 8 c.

d. Demand = 160 pizzas Working hours = 8 Desired cycle time = (8  60) 160 = 3 minutes

Efficiency = 14 (5  4) = .7 or 70% Efficiency = 14/(5 x 3) = 93.33% Efficiency increases, but five people would need to work each night. Alternative solution: A, BC, D, E, F 7-19. Task A B C D E F

Precedence — A A B D C, E

Time (days) 1 4 3 1 4 4 17

Demand = 15 cases

Days working = 75 Desired cycle time = 75 15 = 5 days

(other groupings are possible) Four students should be assigned to each group. The students can complete a case every five days after the first case has been completed. The first case will be completed on day 17. That leaves 58 days to finish the remainder of the cases. At 5 days per case, only 11 more cases can be completed, for a total of 12 cases per semester. No, 15 cases cannot be completed in a semester.

7-20. a.

b. Demand = 125 units Working hours = 40 Cycle time = (40  60) 125 = 19.2 or 19 minutes Min # work stations = 61 19 = 3.2 or 4 workstations c.

Actual cycle time = 17 minutes

Efficiency = 61 (4  17) = .8971 or 89.71% Alternative groupings: ABC, DFG, EHI, JK or ABC, DEF, GHI, JK

7-21.

Efficiency = 7.2 (4  2.4) = .75 or 75% Note: I and K cannot be grouped together without J, thus the theoretical minimum no. of work stations cannot be reached. There are multiple ways to group elements into work stations for this problem.

7-22. Element A B C D E

Precedence — A B A C, D

Time (min) 4 5 2 1 3 15

Quota = 80 claims Working hours = 8 Desired cycle time = (8 x 60) / 80 = 6 minutes There are several ways to balance this line. Output and efficiency vary depending on how the elements are arranged. Option 1:

Actual cycle time = 6 minutes Efficiency = 15 (3  6) = .8333 or 83.3%

Output = (8  60) 6 = 80 claims Option 2:

7-23.

Element A B C D E F G H I J K

Precedence — — — A A, B B D, E F G, H I C, J

Time 4 5 8 4 3 3 5 7 1 7 4 51 Min # work stations = 51 14 = 3.64 or 4 workstations

Actual cycle time = 13 minutes Efficiency = 51 (4  13) = .9808 or 98.08%

8-hr output = (8  60) 13 = 36.92 beds There are alternative groupings that will meet the cycle time, but the solution shown above is the most efficient and productive. Alternative solution:

7-24.

Element A B C D E F G H

Precedence — A B — D, E — F C, E, G

Time (min) 2 4 5 5 3 1 2 4 26

a. Demand = 300 units Working hours = 35

Desired cycle time = (35  60) 300 = 7 minutes b. Min # work stations = 26 7 = 3.714 or 4 workstations

Efficiency = 26 (4  7) = .9286 or 92.86% Alternative solutions: ABF, DG, C, EH AD, BE, FC, GH

c. Desired cycle time = 9 minutes Min # work stations = 26 9 = 2.888 or 3 workstations

7-25.

Efficiency = 46 (4  12) = .9583 or 95.83% Output = (8  60) 12 = 40 units Alternate solutions: ABCD, EF, GJ, HIK; AF, BCE, DGJ, HIK; ADF, EBC, GH, IJK

7-26.

7-27.

e. Yes, it does. If I is increased to .5 minutes, there will need to be 4 workstations, and thus another worker would need to be added to the line.

7-28. Original Matrix:

Hint: Highlight the table, click on Data from the top menu, then Sort in descending order by the processes with the most X’s. Revise the table by rearranging the order of jobs and processes. Matrix Sorted: (answers will vary; this was sorted by 1 then 6 then 15)

Final Matrix: (reorganized and re-grouped)

Cell 1: Processes 1, 2, 3, 7, 9, 12, 14 Jobs A, C, D, G, I, M Cell 2: Processes 5, 10, 15, 16 Jobs H, K, N, O Cell 3: Processes 4, 6, 8, 11, 13 Jobs B, E, F, J, L, P, Q

7-29. Original Matrix:

Hint: Highlight the table, click on Data from the top menu, then Sort in descending order by the processes with the most X's. Revise the table by rearranging the order of jobs and processes. Sorted Matrix: This matrix was sorted by process 6, then process 1.

Final Matrix: The sorted matrix reordered the parts. For the final matrix, reorganize the listing of machines so that they can be grouped into cells that produce a family of parts.

Cell 1: Machines 1, 4, 5, 7 Parts A, D, G Cell 2: Machines 2, 3, 6, 9, 12 Parts H, B, E Cell 3: Machines 8, 10, 11 Parts F, C, I Original Layout:

Revised Layout:

Cell 1

Cell 3

Cell 2

CASE SOLUTION 7.1: Workout Plus 1.

Workout has different types of customers (body builders, athletes, college students, professionals, Moms, senior citizens, rehab patients, youngsters, and so forth) that require different equipment, ambience and services. Needed services include childcare and children’s classes, massage and physical therapy, and nontraditional services such as boxing, Pilates, and yoga. Adding a track, swimming pool or other major multiuse area would be helpful. Instead of a generic process layout, perhaps the faculty could be divided into sections to better serve each set of customers. Shared equipment could be located toward the center of the layout.

Most gyms follow a process layout with similar machines grouped together. Circuit training follows a product layout. Cells are sometimes created with a few pieces of cardio equipment, free weights, and generic strength equipment. Sometimes a large multi-use piece of equipment constitutes a cell.

Student-designed layouts will vary. The layout below responds to customer complaints by: – – – – – –

Removing the cardio area from the main workout areas Putting some cardio machines for warm-ups next to strength machines Always having some cardio machines on the first floor for customers who cannot climb stairs Separating the low impact areas from heavy lifting. Including multi-use equipment, some free weights and basic strength equipment in the low impact area. Locating power lifting, some exercise equipment and hard-core machines away from the entrance and aerobics rooms – If the facility is two-tiered, always locating the least strenuous activities on the top floor so that the less experienced are looking down on the more experienced and not vice versa.

CASE SOLUTION 7.2: Photo Op

a. Yes; the maximum time to complete a task is 60 secs. b. Desired output = 60 per hour Desire cycle time = 1 minute or 60 seconds The first worker would review the application and take the photo. The second worker would verify the information, checking for outstanding debt. The third worker would process the payment, attach the photo and magnetize the passport. Actual cycle time = 60 seconds c. No. workers = 3 Efficiency = 140 (3  60) = .7778 or 77.78%

CASE SOLUTION 7.3 – The Grab ‘n Go Café This case asks you to develop a layout that is both efficient and welcoming to a student population. Customer complaints about lengthy wait times have prompted this analysis. While the open layout described in the case may be more inviting to customers and consistent with the image the company is trying to project, changes can be made to improve the customer’s experience and the efficiency of operations. Better signage, a more apparent customer flow, and better control over customer entrances and exits should be addressed in recommendations. Remember, however, that the objective of a service layout is not all about efficiency. Rather than minimize cost, service layout objectives center on maximizing revenue. So any solution should expose customers to impulse items (such as bakery items), as well as regular purchases (such as fountain drinks). There are several ways to approach this case consistent with chapter material. The layout can be arranged as a process layout using block diagramming techniques or as a cellular layout using production flow analysis. Process Layout. To use the process layout approach, compile the data on movements between food areas. Note the bottled drinks, fresh fruit and salads have been combined into a “refrigerated items” label. Also, be aware that the following movements may have been biased or constrained by the original layout. Nevertheless, this is a good starting point for analysis. 1

1 2 3 4

Entrance Bakery Items Fountain Drinks

Soups Refrig items

Sandwiches (and Wraps)

Coffee or Cookies

6 Sandwiches (and Wraps) 0

7 Coffee or Cookies 1

Bakery Items 14

Fountain Drinks 5

Soups 1

Refrig items 2

3 4

Exit 0

Following is the best schematic diagram that can be developed, resulting in a layout “score” of 11. Alternative layouts are possible.

Schematic Block Diagram 7 8 6

5 3 4

1 2

Nonadjacent load score 1<> 7 1 1<>8 0 1<>6 0 2<>7 0 2<>8 0 2<>6 2 1<>4 1 4<>7 1 4<>5 4 5<>6 1 6<>7 1 Total 11

The final layout would look something like this:

Bottled drinks, fruit, salads & fresh food

Coffee & Cookies

Ck-out

Bakery items

Fountain Drinks

EXIT Ck-out Trays

Soups

Entrance Sandwiches

Trays

Cellular Layout. A cellular layout looks for groups or “families” of customers and identifies their flow paths. The layout is organized so that these different groups can be processed through designated areas of the layout (called cells) efficiently. To begin this analysis, the customer flow data matrix can be sorted in different ways. The first resort was conducted via the data sort command in Excel by bakery items, then fountain drinks, then sandwiches. Following is the re-sorted customer flow matrix.

Customer 20 3 10 6 7 17 22 2 5 1 8 14 19 23 4 9 16 25 11 12 13 18 21 24 15

Bakery Items x x x x x x x x x x x x x x

Fountain Drinks x x x x x x x

Soups x x

Refrig items x

Sandwiches (and Wraps)

Coffee or Cookies

x x x

x x x x x x x

x x

x x x x

x x x

x x x x x x

x x

Total Time (mins) 5.0 3.3 5.2 3.7 4.5 2.0 3.0 4.5 2.5 3.0 8.9 4.5 3.0 2.0 4.5 8.0 10.0 8.0 9.0 10.0 8.0 5.0 1.5 2.2 1.0

Four general customer flows (cells) emerge: Customer Flow 1 – Bakery items and fountain drinks, with an occasional soup or refrigerated item. Current time in the system averages 3.8 minutes. Customer Flow 2 – Bakery items, no fountain drinks, pre-made refrigerated items and some soup, or other items. Current time in the system averages 4.1 minutes. Customer Flow 3 – Made-to-order sandwiches, some soup and fountain drinks. Current time in the system averages 8.3 minutes. Customer Flow 4 – Ready-made items or quick pick-ups. Current time in the system averages 3.5 minutes.

The cellular layout would look something like this:

Sandwich line 1

Bottled drinks, fruit, salads & fresh food

Sandwich line 2

Soups

Fountain Drinks

Bakery items

Coffee & Cookies

Bottled drinks

Bread

Trays

Entrance

Ckout

Entrance

EXIT

Bakery items, coffee and cookies are located next to the entrance/exit. Fountain drinks have been moved to the middle. Bread has been added to the soup area, so customers don’t have to go to the bakery section. Bottled drinks, and other refrigerated items are located in two places in the layout. Sandwiches are divided into two flows. Alternative sorts, by refrigerated items, for example result in different customer flow patterns. Customer flows could also be categorized as made-to-order (e.g. sandwiches, grill), ready-made (e.g., refrigerated items, bakery), and customer served (e.g., fountain drinks, coffee, soup). Obviously, several different layout configurations are possible.

SUPPLEMENT 7 Operational Decision-Making Tools: Facility Location Models Answers to Questions S7-1. For manufacturing and service companies it is often important to be close to materials and have adequate distribution and supply routes and modes of transport. Facilities are usually smaller and less costly for servicerelated businesses and require a sufficient customer population in the vicinity. Although access to customers is a critical location factor for service businesses, for a manufacturing facility important factors include labor climate and wage rates, energy availability and cost, and proximity to other company facilities. S7-2. Service operations tend to locate near malls and interstate interchanges, where there is heavy customer flow, whereas manufacturing facilities locate in Sunbelt regions, often in more rural areas or in smaller towns. S7-3. Factors include lower wage rates, less organized labor, warmer climate, and aggressive marketing tactics by Sunbelt communities. S7-4. Lower wage rates, fewer government regulations, especially related to the environment, and new markets are some of the positive features. Negative features include lack of suppliers, poor modes of distribution and transportation, lack of skilled labor, and lack of markets. S7-5. Customer traffic, access to distribution centers, a labor pool, customer demographics (e.g., income level, youth population), parking or foot traffic, location environment/neighborhood, etc., all are factors. S7-6. Each of these responses will depend on the community in which the students live. S7-7. A labor force that has a strong work ethic, is self-reliant and neighborly. They also have quality health services, low crime rates, a solid infrastructure, quality education and open spaces to expand. S7-8. This answer depends on the facility the student selects. The student should mention a selection of location factors relevant to the type of business it is. S7-9. This answer depends on the 3 sites selected, although proximity to customer markets, traffic, site size, roads, and cost should be mentioned. S7-10. This answer depends on the sites, however, obvious location factors include available space, proximity to student housing concentrations, student traffic, parking, terrain, infrastructure, etc. S7-11. This answer depends on the student selections.

Solutions to Problems S7-1.

Location Factor College proximity Median income Vehicle traffic Mall quality and size Other shopping Total score

Mall 1 12.00 18.75 15.00 9.00 8.00 62.75

Score Mall 2 Mall 3 18.00 27.00 20.00 18.75 22.50 19.75 10.00 8.00 3.00 6.00 73.50 79.50

Mall 4 15.00 17.50 18.50 9.00 7.00 67.00

Mall 3 has the highest location factor score. S7-2.

Location Factors Political stability Economic growth Port facilities Container support Land and construction cost Transportation/distribution Duties and tariffs Trade regulations Airline service Area roads Total Score

Shanghai 12.50 16.20 9.00 5.00 7.20 4.00 4.90 3.50 1.20 1.20 64.70

Weighted Scores Hong Kong Singapore 20.00 22.50 14.40 13.50 14.25 13.50 8.00 9.00 4.00 2.40 6.40 5.60 6.30 6.30 4.75 4.75 1.60 1.40 1.40 1.60 81.10 80.55

Select Hong Kong.

S7-3.

Location Factors Proximity to housing Student traffic Parking availability Plot size, terrain Infrastructure Off-campus accessibility Proximity to dining Visitor traffic Landscape/aesthetics Total Score Select the West A campus site.

South 16.10 16.50 14.40 9.60 5.00 5.40 3.00 2.80 1.00 73.80

Weighted Scores West A West B 20.70 14.95 17.60 13.20 9.60 12.80 8.40 10.80 6.00 4.00 4.20 4.20 4.00 3.50 3.20 2.60 0.80 1.20 74.50 67.25

East 18.40 18.70 11.20 9.00 7.80 4.20 4.00 2.20 1.40 73.90

S7-4.

Location Factors Work ethic Quality of life Labor laws/unionization Infrastructure Education Labor skill and education Cost of living Taxes Incentive package Government regulations Environmental regulations Transportation Space for expansion Urban proximity Total Score

Abbeton 14.40 12.00 10.80 6.00 6.40 5.25 4.20 3.25 4.50 1.20 1.95 2.70 1.80 1.20 75.65

Weighted Scores Bayside Cane Creek 16.20 12.60 13.60 15.20 7.20 7.20 5.00 6.00 7.20 6.80 4.55 4.90 4.80 5.10 3.50 2.75 4.75 3.50 1.50 1.95 1.80 2.10 2.40 2.85 1.90 1.80 1.80 1.40 76.20 74.15

Dunnville 13.50 14.40 8.40 7.00 7.60 5.60 4.50 3.00 4.00 1.65 2.40 2.40 1.80 1.60 77.85

Select Dunnville. S7-5.

S ( C ) = 76

S ( E ) = 75 S ( D ) = 74

S ( B ) = 70 S ( A ) = 69

S7-6. S ( Tysons ) = 85.27 S ( Fairfax ) = 80.94

S ( Alexandria ) = 80.55 S ( Manassas ) = 77.00 S ( Dupont ) = 76.39

S7-7. Score Location Factors Elderly population Income level Land availability Average age Public transportation Crime rate Total score Select Dowling

Ashcroft 75 65 90 80

Brainerd 80 75 70 70

Caffee 65 90 90 80

Dowling 75 85 80 75

95 95 77.5

55 70 75.5

75 85 74.25

95 90 78.75

S7—8.

Location Factors Soccer Interest Entertainment competition Playing facility Population (age 15 to 40) Media market Income level (age 15 to 40) Tax incentives Airline transportation Cultural diversity General sports interest Local government support Community support Total score

Atlanta 70 33 50 100 100 80 20 100 100 75 30 20 65.0

Weighted Scores Birmingham Charlotte 40 75 45 60 65 70 70 90 65 95 70 80 40 75 70 95 80 90 95 85 60 75 35 50 55.8 76.4

Durham 90 95 85 25 40 60 60 65 75 65 90 75 74.3

Select Charlotte

S7-9

Solution depends on the weights the student assigned to the location factors.

S7-10. Solution depends on the weights the student assigned to the location factors.

S7-11. Solution depends on the weights the student assigned to the location factors.

S7-12. x =

(14 )(17, 000 ) + ( 20 )(12, 000 ) + ( 30 )( 9, 000 ) 38, 000

( 30 )(17, 000 ) + (8 )(12, 000 ) + (14 )( 9, 000 ) 38, 000

= 19.68

= 19.26

S7-13. a. 1 x1 = 100

2 x2 = 210

3 x3 = 250

4 x4 = 300

5 x5 = 400

y1 = 300

y2 = 180

y3 = 400

y4 = 150

y5 = 200

w1 = 35

w2 = 24

w3 = 15

w4 = 19

w5 = 38

 xiWi  Wi

(100 )( 35) + ( 210 )( 24 ) + ( 250 )(15 ) + ( 300 )(19 ) + ( 400 )( 38 ) 35 + 24 + 15 + 19 + 38

x = 253.36 y=

 yiWi  Wi

( 300 )( 35) + (180 )( 24 ) + ( 400 )(15 ) + (150 )(19 ) + ( 200 )( 38 )

y = 238.70

131

S7-14. Site A:

Site B:

d1 =

(100 − 350 )2 + ( 300 − 300 )2 = 250.00

d2 =

( 210 − 350 )2 + (180 − 300 )2 = 184.39

d3 =

( 250 − 350 )2 + ( 400 − 300 )2 = 141.4

d4 =

( 300 − 350 )2 + (150 − 300 )2 = 158.11

d5 =

( 400 − 350 )2 + ( 200 − 300 )2 = 111.80

d1 = 70.71 d 2 = 92.20 d3 = 180.28 d 4 = 180.28 d5 = 254.95

Site C:

d1 = 150.00 d 2 = 126.49 d3 = 100.00 d 4 = 158.11 d5 = 180.28

LD ( A ) = 35 ( 250 ) + 24 (184.39 ) + 15 (141.4 ) + 19 (158.11) + 38 (111.8 ) = 22,549.4

LD ( B ) = 35 ( 70.71) + 24 ( 92.9 ) + 15 (180.28 ) + 19 (180.28 ) + 38 ( 254.95 ) = 20, 505.1 LD ( C ) = 35 (150 ) + 24 (126.49 ) + 15 (100 ) + 19 (158.11) + 38 (180.28 ) = 19, 640.5

Site C has the lowest load-distance value and would minimize transportation costs.

Site C is a little more centrally located as the site determined in Problem S7-8; it is closer to site 3, which has the lowest annual truck shipments.

S7-15. x = 935.71 y = 971.27 Note: the size of the bubble indicates weight

Center of Gravity 1800 1600 1400 1200 1000 800 600 400 200 0 -500

500

1000

1500

2000

S7-16. a. 1. Four Corners x1 = 30

y1 = 60

y2 = 40

w1 = 8.5

w2 = 6.1

3. Russellville x3 = 10

2. Whitesburg x2 = 50

4. Whistle Stop x4 = 40

y3 = 70

y4 = 30

w3 = 7.3

w4 = 5.9

 xiWi ( 30 )( 8.5 ) + ( 50 )( 6.1) + (10 )( 7.3) + ( 40 )( 5.9 ) = = 31.3  Wi 8.5 + 6.1 + 7.3 + 5.9

 yiWi ( 60 )( 8.5 ) + ( 40 )( 6.1) + ( 70 )( 7.3) + ( 30 )( 5.9 ) = = 51.9  Wi 27.8

S7-17. x = 1, 665.4 y = 1,562.9

S7-18. x = 21.0 y = 16.2

S7-19. a. x =

(15)(160 ) + ( 42 )( 90 ) + (88 )(105 ) + (125 )( 35 ) 160 + 90 + 105 + 35 + 60 + 75

(135 )( 60 ) + (180 )( 75 ) 160 + 90 + 105 + 35 + 60 + 75

= 78.8 y=

(85)(160 ) + (145)( 90 ) + (145 )(105 ) + (140 )( 35 ) 160 + 90 + 105 + 35 + 60 + 75

(125 )( 60 ) + (18 )( 75 ) 160 + 90 + 105 + 35 + 60 + 75

= 106.0 b.

The closest town to the grid coordinates is Seagrove. However, Ashboro is probably a better location at coordinates (76,116) since it is at the confluence of several highways that lead to Wilmington, Raleigh, Charlotte and Winston-Salem.

c. From a map of North Carolina, Fayetteville is at (145, 72) and Statesville is at (10,125). Fayetteville:

d ( Charlotte ) = 130.7 d ( W − S ) = 126.2

d ( Greensboro ) = 92.6 d ( Durham ) = 70.9

d ( Raleigh ) = 53.9 d ( Wilmington ) = 64.4 d ( Charlotte ) = 40.3

Statesville:

d ( W − S ) = 37.7 d ( Greensboro ) = 80.5 d ( Durham ) = 116.0 d ( Raleigh ) = 125 d ( Wilmington ) = 200.9 LD ( Fayetteville ) = (160 )(130.7 ) + ( 90 )(126.2 ) + (105 )( 92.6 )

+ ( 35 )( 70.9 ) + ( 60 )( 53.9 ) + ( 75 )( 64.4 )

= 52,533.9 LD ( Statesville ) = (160 )( 40.3 ) + ( 90 )( 37.7 ) + (105 )(80.5 )

+ ( 35 )(116 ) + ( 60 )(125 ) + ( 75 )( 200.9 )

= 44,925.4 The best site, according to the load-distance technique is Statesville. S7-20. a. x = 265.23 y = 363.84 b. LD ( Site 1) = 123,859.5 LD ( Site 2 ) = 120,995.3

LD ( Site 3) = 115, 007.4

Site 3 is the best site. S7-21 a.

x = 1559.8 y = 1766.8

b. Site C is the best

LD(Site A) = 298,169 LD(Site B) = 273, 731 LD(Site C) = 247,556

S7-22. a. The first step is to set up a grid map of the eastern U.S. Using a grid map with Miami at the axis (i.e., 0, 0), the east coast shoreline (from Miami to Boston) as the y-axis, and an approximate line along the Gulf (from Miami to New Orleans) as the x-axis, the destination cities have the following coordinates. Boston (0, 1500) New York (0, 1300) Philadelphia (0, 1170) Washington, D.C. (0, 1040) Miami (0, 0) New Orleans (1015, 0) Atlanta (160, 630) Charlotte (200, 720) Nashville (520, 700) Pittsburgh (300, 1200) Chicago (1010, 930) Detroit (820, 940) Next using these coordinates with the truckloads per month in the center-of-gravity formulas results in the following “best site” coordinates,

These (x, y) coordinates on the grid map are in the area of Columbus, OH. b. Belcamp, MD (0,1070)

Columbus, OH (271,893)

d1 = d2 = d3 = d4 = d5 = d6 = d7 = d8 = d9 = d10 = d11 = d12 =

430.0 230.0 100.0 30.0 1070.0 1474.8 468.2 403.1 638.2 327.0 1019.7 830.2

664.7 489.0 387.5 308.3 933.2 1162.3 285.5 187.0 315.0 308.4 739.9 551.0

Although Columbus, OH has a slightly lower LD value it likely would not be worth the cost of building a new distribution center; plus Belcamp is on the coast for overseas shipments from suppliers.

c. The first step is to set up a grid map of the western U.S. Using a grid map with San Diego at the axis (i.e., 0, 0), the west coast shoreline (from San Diego to Seattle) as the y-axis, and an approximate line from San Diego to Houston as the x-axis, the destination cities have the following coordinates. San Diego (0, 0) Los Angeles (0, 150) San Francisco (0, 425) Seattle (0, 1100) Phoenix (300, 0) Houston (1015, 0) Las Vegas (160, 220) Dallas (1200, 220) Denver (980, 600) Kansas City (1500, 600) St. Louis (1700, 600) Minneapolis (1500, 1000)

Next using these coordinates with the truckloads per month in the center-of-gravity formulas results in the following “best site” coordinates,

These (x, y) coordinates on the grid map are in the area of Salt Lake City. (See the ALONG THE SUPPLY CHAIN box, “Adding a New Distribution Center to the Sephora Supply Chain,” in chapter 11.

CASE S7.1: Selecting a European Distribution Center Site for American International Automotive Industries 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 and five potential distribution center sites.

Plant sites (x, y) Vienna (300, 60) Leipzig (180, 225) Budapest (390, 50) Prague (240, 160) Krakow (400, 170) Munich (150, 60) Frankfurt (40, 160)

Load 160 100 180 210 90 120 50

Distribution center sites (x, y) Dresden (225, 225) Lodz (420, 250) Hamburg (90, 340) Gdansk (370, 360) Frankfurt (40, 160)

Using the load-distance technique: LD ( Dresden ) = (160 )(181.2 ) + (100 )( 45 ) + (180 )( 240.5 ) + ( 210 )( 66.7 ) + ( 90 )(183.4 ) + (120 )(181.2 ) + ( 50 )(196.1)

= 138,865.1 LD ( Lodz ) = (160 )( 224.7 ) + (100 )( 241.3) + (180 )( 202.2 ) + ( 210 )( 201.2 ) + ( 90 )( 82.5 ) + (120 )( 330.2 ) + ( 50 )( 390.5 ) = 205,315.2 LD ( Hamburg ) = (160 )( 350 ) + (100 )(146.0 ) + (180 )( 417.3 ) + ( 210 )( 234.3) + ( 90 )( 353.6 ) + (120 )( 286.4 ) + ( 50 )(186.8 ) = 270, 436.5 LD ( Gdansk ) = (160 )( 308.1) + (100 )( 233.1) + (180 )( 310.6 ) + ( 210 )( 238.5 ) + ( 90 )(192.4 ) + (120 )( 372.0 ) + ( 50 )( 385.9 ) = 259,854.1 LD ( Frankfurt ) = (160 )( 278.6 ) + (100 )(154.4 ) + (180 )( 366.9 ) + ( 210 )( 200 ) + ( 90 )( 360.1) + (120 )(148.7 ) + ( 50 )( 0 )

= 218, 296.1

Dresden is clearly the best site using the load-distance technique. Observing a European map it is also centrally located among AIAI’s customers. However, several other factors should be taken into consideration. For example, AIAI currently ships into the port of Hamburg so a warehouse/distribution center there would eliminate transport from the port of entry to a distribution center at Dresden or anywhere else. According to the load-distance technique, Hamburg is the worst site, but its proximity to the port may make it much more attractive. However, the attractiveness of Hamburg as a port could be offset by Dresden’s closer proximity to the plants at Vienna and Budapest that require continuous replenishment. A negative location factor for Dresden is that it’s in the old East Germany and its infrastructure and other facilities may be less than ideal.

Students might be encouraged to research for other location factors, such as transportation between the sites, between country trade regulations, costs, etc.

8 - Human Resources in Operations Management Answers to Questions 8-1.

For companies committed to quality management, their TQM program must be very closely related to their strategic plan. Employee involvement is an integral part of TQM, thus human resources is an important component of strategic planning.

8-2.

Human resources and employee diversity reflect the fact that companies now view employees as more of a valuable “resource” than simply a replaceable part in the production process.

8-3.

Breaking down jobs into their simplest elements and motions render the elements as efficient as possible by eliminating unnecessary motions, and then divide the tasks up among several workers so that each task would require only minimal skill, effectively simplifying job designs to the greatest degree possible.

8-4.

F. W. Taylor’s contributions related to the development of the principles of scientific management and the use of motion analysis and the stopwatch to design jobs and measure job performance. Frank Gilbreth originated motion study as a means to determine the “one best way” to perform a work task, which evolved into widely used principles of motion study.

8-5.

Horizontal job enlargement refers to expanding the scope of a job to include all tasks necessary to complete a product or process, thus providing the worker with a sense of having made something. Vertical job enlargement transfers some supervisory responsibilities for a job from management to the worker, allowing for more job self-determination and control.

8-6.

Tasks are individual activities consisting of one or more elements, which in turn encompass several motions or basic physical movements.

8-7.

The creation of a mass product market and high output volume made the breakdown of work and fragmentation of jobs (and the increase in number of workers) required by scientific management costeffective and economical.

8-8.

Sustainable activities include energy efficient office and workplace facilities and equipment, recycling of office products, energy saving practices, alternative transportation modes (such as public transportation, bicycling and walking), partnering with environmentally friendly suppliers, and telecommuting.

8-9.

This depends on the company the student selects. IBM is a likely candidate.

8-10.

It is difficult for U.S. companies to create sustainable workplaces in countries that have few or ineffective labor laws and regulations, low income, high unemployment, different political systems, and different cultures. However, some companies like Levi Strauss, Target, and P&G insist that their overseas suppliers subscribe to U.S. labor laws regarding worker treatment, and they have established programs and codes for suppliers to improve working conditions, and regularly assess suppliers’ progress in achieving workplace sustainability objectives. Refer students to the Levi Strauss introduction to this chapter and the Target “Along the Supply Chain” boxes in this chapter and in chapter 2.

8-11.

The answer depends on the company that the student selects.

8-12.

Workplace sustainability issues include low wages, poor and unsafe working conditions, substandard worker housing (especially for migrant workers from rural areas), child labor, and excessive working hours. Companies attempt to work with the Chinese government to improve conditions and enforce their own workplace codes and standards on Chinese suppliers. This generally requires on-site monitoring by the U.S. company, and strict enforcement of its expectations.

8-13.

The student should refer to the “Along the Supply Chain” box for Baldrige Award-winning companies in the chapter and the Baldrige Award Web site. Efforts in human resources among award-winning companies include extensive job training, job empowerment, performance recognition, the opportunity for advancement, worker participation and projects, and fair compensation. Measures of employee satisfaction include absenteeism and turnover rates, surveys that measure job satisfaction, benchmark surveys, and training expenditures and opportunities.

8-14.

This answer depends on the award-winning companies the student selects. All of the company summaries describe their human resource efforts.

8-15.

This answer depends on the companies the student selects.

8-16.

Cultural differences create many diversity issues for U.S. companies related to hiring practices, gender and race discrimination, language differences, religion, income levels, etc.

8-17. Social media and the Internet have likely been the major factors in bringing working conditions around the world to the public’s attention. International watchdog organizations plus the U.N. use these various media to expose poor working conditions. 8-18. Industrialization occurred more rapidly in Japan and bypassed the skilled industrial artisan phase that engendered scientific management, plus job flexibility and factory work teams that included management and maintenance were the cultural norm. 8-19. See Figure 8-1. Japanese job design includes such characteristics as horizontal and vertical job enlargement, as opposed to the task specialization and repetition of scientific management. Job design in Japan also encompasses extensive job training, responsibility and control, cross training, job rotation, higher skill levels, and employee involvement, whereas scientific management encompasses minimal job training, low skill levels, tight control and minimal responsibility. 8-20. Jobs were designed so that workers could be easily replaced and trained at low cost; jobs required minimal skills, resulting in a large labor pool; also unskilled, uneducated workers were allowed to gain employment based on their willingness to physically work hard at jobs that were mentally undemanding. 8-21. Advantages include horizontal and vertical job enlargement, individual responsibility for job reliability and quality, job rotation, and communication between workers and workers and management. 8-22. It has shifted more responsibility for quality to the worker and promoted empowerment of the worker to alert management to quality problems and individually act to correct quality problems without fear of reprisal. 8-23. Americans have tended to adopt Japanese management principles such as quality management, JIT systems, and quality circles without adopting their principles of job design, such as job flexibility, cross training, job training and job responsibility. Some of the Japanese job design features and work environments are cultural and cannot be easily adopted. However, some aspects of job design have been successfully adopted by some U.S. companies, including employee involvement programs and quality circles, worker empowerment and job responsibility, more extensive training, and job flexibility. 8-24. The three elements of job design are task analysis, the determination of how to do each task and how all the tasks fit together; worker analysis, the characteristics and responsibilities of the worker who performs the job; and environmental analysis, which involves the physical location of the job and the conditions that must exist to do the job. 8-25. A process flowchart is used to analyze the sequential steps of a job or how a set of jobs fit together to form a production process, whereas a worker-machine chart illustrates the amount of time a worker and a machine are working or idle in a job. 8-26. Depends on student selection of activity. 8-27. For the employee telecommuting allows for a more flexible work schedule, the opportunity to attend to personal matters within a work schedule, reduced travel expenses and time. Disadvantages are a lack of personal contact with co-workers and management, and it is a possible obstacle to advancement because of lack of contact with superiors.

For the manager advantages include a means to motivate employees, and it saves work space and real estate costs. Disadvantages include a feeling of a lack of control because the employee cannot be seen working and some employees may not be self-motivated to work without direct personal supervision. 8-28. Empowerment allows employees to make quality decisions and corrections immediately without delaying to consult with supervisors. It gives employees more control over quality in their job. Empowerment can be abused by the employee and some employees may not be suited to have responsibility and control. 8-29. Depends on article selected. 8-30. The Baldrige Award site includes brief summaries of all award winning companies. Depends on company selected. 8-31. Depends on the workplace the student selects. 8-32. Depends on the job selected by the student. 8-33. Learning curves are useful primarily for measuring work improvement for nonrepetitive, complex jobs that require a relatively long time to complete, as opposed to short, repetitive, routine jobs that show little improvement over time. 8-34. The learning curve reflects the fact that each time the number of units produced doubles, the processing time per unit decreases by a constant percentage. 8-35. Product modifications during the production process can negate the learning curve effect; improvement can result from sources other than worker learning such as modifications of work methods or new equipment; and industry-developed learning rates may not be appropriate for individual companies.

Answers to Problems 8-1.

tn = t1nb t100 = (48)(100)ln(0.88) ln 2

= 20.5 minutes 800 new forms per week will require

(800)(20.5 minutes) = 16,424.1 minutes per week to process 1 employee works (6 hr) (60 mins) (5 days) = 1,800 minutes per week

New employees required = 8-2.

16.424.1 = 9.1 1,800

The time required for the first 10 exams is shown below. Grading Time 12.0 10.8 10.15 9.72 9.39 9.13 8.92 8.74 8.59 8.45

Cumulative Time 12.0 22.8 32.95 42.67 52.07 61.20 70.13 78.88 87.47 95.93

Since the first 10 exams will require approximately 96 minutes, Professor Cook will have 300 − 96 = 204 minutes left. The remaining 25 exams will require 8.46 minutes apiece, or 211.5 minutes to complete. Thus, unless Dr. Cook hurries she will not complete them on time. 8-3.

tn = t1nb

t60 = 80(60)ln(0.92) ln 2 = 48.89 hr t120 = 80(120)ln(0.92) ln 2 = 44.98 hr 8-4.

Using Excel, 9 installations are required to achieve a two-week installation time.

8-5. tn = t1nb

t30 = 1, 600(30)ln(0.87) ln 2 = 807.88 hr t60 = 1, 600(60)ln(0.87) ln 2 = 702.86 hr 8-6. Using Excel, 42 customer calls are required to reduce the average call time to 10 minutes.

8-7.

Using Excel, the learning coefficient is .9024.

8-8.

Using Excel, the total cumulative time required to produce all 120 computers is 1232.9 hours. Thus, it will take the 8 employees, working 20 hours per week, 7.71 weeks to complete the order.

8-9.

tn = t1nb t80 = 126(80)ln(0.85) ln 2 = 45.1 seconds

8-10. Using Excel, the minimum time of 32 minutes is reached at approximately the 90th room, i.e., 90 rooms must be cleaned before complete proficiency is attained. The cumulative time required to clean the first 53 rooms is 3,256.31 minutes. Given that a housekeeper works 390 minutes per day (i.e., 6.5 hours  60 minutes per hour), it will take approximately 8.34 days, or effectively 9 days to become fully proficient. Proof: t90 = 55(90)ln(0.92)/ln2 = 32.01

8-11. 8 books at 17,000 lines per book is 136,000 lines. Thus, after 136,000 lines the scribe is a craftsman.

tn = t1nb t136,000 = 15(136, 000)ln(0.91) ln 2 = 3 minutes per line for an experienced scribe The apprentice scribe could copy 50 lines at 15 minutes per line which means the work day was 50  15 = 750 minutes long or about 12.5 hours. In a 750 minute work day an experienced scribe could copy 750 3 = 250 lines.

8-12. Using Excel, the number of orders required to reach the target value of 3 days is 23.

Case Solution: Maury Mills The following statement about how Anne and Dana reacted when they returned to the business is damning: “They felt that their employees had become complacent and spoiled and weren’t as committed to the business as they had been when they were smaller and were more of a family.” Anne and Dana were accusing their employees of exactly what the two of them had done–they had stepped back from their commitment to running the business and had obviously let others take over. Without them in an on-site leadership role there was nothing to sustain the good employee relations and family atmosphere that had originally existed. It is likely that Maury Mills’s employees were very motivated and committed to the business in the early days because of Anne and Dana’s hands-on, caring managerial style, but as Anne and Dana drifted away from their leadership role, that feeling left. In fact, it appears that most of their original employees may have left the company– “Walking around their facilities they felt they hardly knew anyone that worked for them.”-which should have been a red flag that Anne and Dana did not pick up on. Some of the problems that Maury Mills encountered like the economic recession and increased competition were beyond their direct control. However, they seemed to be very slow to react to these conditions and did not have a long-term contingency plan for handling changing business conditions. It seems likely that Anne and Dana could have turned to their employees and management team for suggestions about how to turn things around in this situation, but instead they went to a consulting firm. While consultants aren’t bad and are often very helpful, they do not have the same close relationship with the company workforce as the owners. Anne and Dana probably biased the consultants against their employees by telling them of their employees’ “malaise.” Also, it is unlikely the consultants would want to openly dispute their client - who they would assume would know more about their employees than they did. Anne and Dana probably should have asked themselves why they thought their employees were not motivated and determined themselves what they expected from their employees to support the business. By rewarding short-term performance with incentives, Maury Mills received short term results. Product quality seemed to be a cornerstone of Maury Mills’s business and the changes that were made obviously created quality problems. While incentive programs and bonuses can be useful, they can also have detrimental effects on quality if they are not well-thought out and part of an overall quality management program. The incentive system in the warehouse and distribution center likely caused employees to be in too much of a hurry to process orders and get them out the door by the end of the day, and this resulted in the incomplete and messed up orders. The late shipments were likely caused by a lack of coordination with the company’s shipper. Problems may have also resulted from increased sales; the sales force offered discounts to increase sales without coordinating with production and distribution to see if those orders could be filled on schedule. The grocer’s complaint about abrupt phone conversations was a result of the directive to reduce call length. The idea of reducing call time so that operators can handle more calls is not a bad one and may be justified, but it must be coupled with an understanding of its impact on quality measures. The emphasis on increased productivity in the food processing area could have led to short-cuts and changes that compromised product quality. Incentive plans based on productivity increases need to be coupled with quality controls and oversight. The company also needed to make sure employees understood the company’s commitment to quality. The quality problems were likely exacerbated by the buyers purchasing inferior ingredients and products in order to reduce purchase costs. Empowering buyers and salespeople with broad discretion without a corresponding understanding of quality goals is an invitation to problems. The conversation of a Maury Mills employee that Anne overheard at Kroger’s and the pending meeting with minority employees are definite signs that employee relations at Maury Mills have eroded. The conversation Anne heard reflects the employee’s lack of pride in the company and the frustration employees were feeling because of the poor quality problems and the perceived state of the company. It seems unlikely that Anne and Dana realized they had a large minority contingent in their business and they were obviously not sensitive to potential diversity issues. Since the cutbacks occurred among a racial grouping it is easy to see how the minority employees might perceive that they were unfairly singled out, especially since other work groups (i.e., in food processing) suffered no cutbacks. The cutbacks did not seem to be racially motivated; however, there was certainly no consideration about how the minority employees might perceive them. At the least, employees should have been informed about the

reason for the cutbacks and why they occurred. There seems to be a lack of communication and diversity management at Maury Mills. Can Maury Mill’s problems be solved, and if yes, what can Anne and Dana do? Maury Mills could probably have weathered its crisis by employing some belt-tightening measures to ride out the recession, although a change in strategy to cope with increased competition seemed to be a good idea. The human resource changes recommended by the consulting firm only made things worse. However, despite these problems, Maury Mills appeared to be a fundamentally sound company before their problems. Although Anne and Dana did not have a background in business they seemed to have solid business instincts and good leadership skills. Their problems did not result because of their lack of business acumen but because they stepped back from their involvement in the company and failed to apply their abilities. They seemed to exhibit a lack of confidence in their own abilities, but events may have shown them that they are the most qualified to turn their business around. In order to turn the company around Anne and Dana need to be more involved and provide more direct leadership. They should develop a plan of action and goals for turning the company around and communicate those plans and goals to their employees both as a whole and in groups. They should return to their commitment to quality and institute process improvement initiatives. While buyers should be allowed to shop for the lowest cost items, they should not compromise quality. It’s likely that Anne and Dana will want to rethink the incentive and bonus plans they introduced and develop some alternative compensation plans that combine performance and productivity with quality. Above all the owners need to return to the employee-friendly environment they originally created.

Supplement 8 Operational Decision-Making Tools: Work Measurement Answers to Questions S8-1. A predetermined motion time system enables a standard time to be developed for a new job before it is implemented, worker cooperation and compliance are not required, the workplace is not disrupted, performance ratings are already included in the motion times, and predetermined times are consistent and not subject to as much uncertainty. However, such a system ignores the job context within which the motion takes place, and it may not reflect the skill level, training, or abilities of workers in a specific company. Also, predetermined motion times are useful only for highly repetitive, simple jobs that can be broken down into basic motions. S8-2. 1. Establish the standard job method; 2. break down the job time into elements. However, elements may not have obvious break points or be very detailed; 3. study the job. (Workers may not be cooperative.); 4. rate the worker’s performance, a subjective process; 5. compute average time; 6. compute normal time; 7. compute standard time. S8-3. The use of time studies for the establishment of wage rates can be counterproductive. The basic premise of an incentive wage system is that pay is the sole motivation for work; however, such wage systems have been shown to inhibit quality improvement efforts. S8-4. Student experiment; students typically rate the dealer’s effort too low. S8-5. By either reducing the confidence level or increasing the allowable error. S8-6. Work sampling is used for jobs that have nonrepetitive tasks. S8-7. 1. Define the job activities; 2. statistically determine the number of observations in the work sample; 3. determine the length of the sampling period; 4. conduct the work sampling study and record the observations; 5. periodically recompute the number of observations. S8-8. Student experiment. A departmental office secretary is a good subject to study.

Solutions to Problems S8-1.

t = 3.62 min RF = 100% AF = 15%

NT = ( t ) ( RF ) = 3.62 (1.00 ) = 3.62 ST = ( NT )(1 + AF ) = 3.62 (1.15 ) = 4.163 min

S8-2. Element

t

1 2 3 4 5 6 7

3.15 8.67 14.25 11.53 6.91 5.72 5.38

0.1575 0.4335 0.7125 0.5765 0.3455 0.2860 0.2690

1.10 1.05 1.10 1.00 0.95 1.05 1.05

0.1734 0.4550 0.7838 0.5765 0.3282 0.3003 0.2825 2.8997 = NT

ST = ( NT )(1 + AF ) = 2.8997 (1.12 ) = 3.2477, or 3.25 min S8-3. Element

t

1 2 3 4 5 6

0.12 0.07 0.86 0.08 0.62 0.16

0.14 0.07 0.89 0.07 0.68 0.14

0.14 0.05 0.82 0.07 0.67 0.14

0.13 0.09 0.86 0.06 0.67 0.15

0.11 0.08 0.87 0.09 0.66 0.14

0.10 0.08 0.83 0.08 0.66 0.16

0.12 0.08 0.85 0.08 0.62 0.17

0.12 0.07 0.84 0.10 0.68 0.15

0.15 0.06 0.82 0.11 0.68 0.15

0.13 0.07 0.84 0.09 0.69 0.16

1.260 0.720 8.480 0.830 6.610 1.540

0.126 0.072 0.848 0.083 0.661 0.154

0.132 0.072 0.933 0.091 0.694 0.154 2.076 = NT

a. Normal cycle time = 2.076

ST = NT (1 + AF ) = 2.076 (1.15 ) = 2.39 min b. Average operator: ST = 2.60 min

Hourly output =

60 = 25.11 caps per hour 2.39

Hourly wage = ( 25.11)( 0.18 ) = $4.52

Subject operator: Nt = 1.944 min Adjusted time = (1.944) (1.15) = 2.24 min

Hourly output =

60 = 26.78 caps per hour 2.24

S8-4. Element

t

0.89

0.089

1.10

0.098

0.86

0.086

0.95

0.082

1.45

0.145

0.90

0.131

0.93

0.093

1.00

0.093

0.67

0.067

0.95

0.064

0.468 = NT

0.480 a. ST = 0.468 (1.10 ) = 0.5148 b. Average worker wage :

60 min = 116.55 0.5148

(116.55)( 0.03) = $3.50 / hr Subject worker: = 0.48(1.10) = 0.528

60 = 113.64 0.528

(113.64 )( 0.03) = $3.41/ hr S8-5. Element

t

10.52

0.5260

1.15

0.6049

18.61

0.9305

1.10

1.0240

26.20

1.3100

1.10

1.4410

16.46

0.8230

1.05

0.8640

3.9339 = NT

3.5940 a. ST = ( 3.94 )(1.15 ) = 4.52 min b. T = 3.59

s = 0.51 2  1.96 ( 0.51)   zs  n=  =  0.05 3.59  ( )   eT  

= ( 5.5688 ) = 31.01 2

The time study should include 31 cycles.

S8-6. Element

t

3.65

0.365

1.05

0.3833

9.00

0.900

0.90

0.8100

4.67

0.467

1.00

0.4670

1.70

0.170

1.05

0.1785

T = 1.902

a. ST = (1.838 )(1.16 ) = 2.13 min 2  1.96 ( 0.1142 )   zs  b. n =   =  0.04 1.902  ( )  eT  

= ( 2.94 ) = 8.65, or 9 cycles 2

Note:

 ( xi − x ) 2 , where n −1

xi = the job cycle times i = 1 to 10

x = 1.902

n = 10 S8-7.

s = 0.25

z = 2.33 T = 2.7805 2  2.33 ( 0.25 )  2  zs  n=  =  0.06 2.7805  = ( 3.94 ) ( )  eT   2

= 12.2, or 13 cycles S8-8.

s = 0.0341 z = 1.96

T = 1.944 e = 0.02 2  1.96 ( 0.0341)  2  zs  n=  =  0.02 1.944  = (1.719 ) ( )  eT    2

= 2.96, or 3 cycles

1.8388 = NT

S8-9. Element

t

2.04

0.204

1.05

0.2142

1.07

0.107

1.00

0.1070

1.72

0.172

1.05

0.1806

2.78

0.278

0.95

0.2641

1.04

0.104

1.00

0.1040

0.81

0.081

1.00

0.0810

2.36

0.236

1.05

0.2478

1.182

a. b. s = 0.0418

T = 1.182

z = 1.96 e=

0.03 min T

= 0.025

 1.96 ( 0.0418 )  2 n= = ( 2.776 )  0.025 (1.182 )    2

= 7.70, or 8 cycles c. Putting inspectors on a piece-rate incentive to inspect more items may result in a less stringent quality inspection and the passage of a greater number of inferior products. S8-10. z = 2.33

e = 0.02

97 = 0.162 idle 600

1 − p = 0.838 busy 2

z  2.33  n =   p (1 − p ) =   ( 0.838 )( 0.162 ) e  0.02 

= 13,572.25 ( 0.838 )( 0.162 ) = 1,842.5, or 1,843 observations

1,843 = 61.4, or 62 observation trips 30

= 1243 additional observations [1,843 required – (30 x 20)]

1243 = 41.4 or 42 additional observations trips 30

S8-11. a. z = 1.96

e = 0.03

p = 0.20 1 − p = 0.80 2

z n =   p (1 − p ) e 2

 1.96  =  ( 0.20 )( 0.80 )  0.03 

= 682.5 or 683 observations 2

z  1.96  b. n =   p (1 − p ) =   ( 0.31)( 0.69 ) e  0.03 

= 913.02, or 914 observations Therefore, 230 additional observations are needed.

z = 1.96

S8-12.

e = 0.03 p = 0.30 1 − p = 0.70 2

z  1.96  n =   p (1 − p ) =   ( 0.30 )( 0.70 ) e  0.03 

= 896.25, or 897 observations b.

p = 0.162 (84 idle / (26 cabs x 20 observations)

1 − p = 0.838 2

 1.96  n=  ( 0.838 )( 0.162 )  0.03 

= 579.4, or 580 observations 2

S8-13. a.

z n =   p (1 − p ) e 2

 z  300 =   ( 0.88 )( 0.12 )  0.03 

2,840.91 =

z2 0.0009

2.556 = z 2

z = 1.59, which corresponds to a confidence level of 88.8%. b. For 95% confidence, 2

 1.96  n=  ( 0.88 )( 0.12 )  0.03 

= 450.7, or 451 observations Therefore, 151 additional observations are needed. S8-14. a. ( 480 min )( 0.78 ) = 374.40 min talking to customers 374.4 min /120 calls = 3.12 min per call (for this operator)

NT = ( 3.12 min )(1.10 ) = 3.432 min per call

ST = ( 3.432 )(1.15 ) = 3.946 min per call 2

z n =   p (1 − p ) e 2

 z  160 =   ( 0.78 )( 0.22 )  0.04 

932.4 =

z2 0.0016

1.49 = z 2

z = 1.22 which corresponds to a confidence level of 77.75%. For a confidence level of 0.95, 2

 1.96  n=  ( 0.78 )( 0.22 )  0.04 

= 412.01, or 413 observations Therefore, 253 additional observations are needed. S8-15. a.

400

( 50 )(17 )

= .47

z  1.65  n =   p (1 − p ) =   ( 0.47 )( 0.53) e  0.05 

 271 observations b. The manager might consider combining the warehouse and packaging areas so that an employee would do both jobs. This would not only allow him to downsize but it would probably speed up the operation.

S8-16. a.

84 = .56 150 2

z  2.58  n =   p (1 − p ) =   ( 0.56 )( 0.44 ) e    0.01 

= 16, 401 observations b. Number of days to conduct the study = 328 (16,401 / 50 observations per day)* *150 observations / 3 days = 50 observations per day c. Either reduce the confidence level or increase the required level of accuracy.

S8-17. a.

34 = 0.57 60

z = 1.96

e = 0.05 2

z  1.96  n =   p (1 − p ) =   ( 0.57 )( 0.43)  377 e    0.05 

Additional observations = 377 − 30 = 347 2

 z  100 =   ( 0.57 )( 0.43)  0.05  2

 z   0.05  = 408.16  

z = 20.2 0.05

z = 1.01 A z value of 1.01 corresponds to a confidence level of approximately 69 percent.

CASE: Measuring Faculty Work Activity at State University First determine the number of observations that would be required: 2

z n =   p (1 − p ) e 2

 1.96  n=  ( 0.2 )( 0.8 )  0.03 

= 682.95, or 683 observations Using a one-digit random number table between 1 and 10 minutes, the average time between observations would be about 5 minutes. Thus, approximately 84 observations could be taken in one 7-hour day (from 9:00 A.M. to 4:00 P.M.). At that rate it would require 8.1 days to complete the work sample. This is about the maximum amount of time you might expect a faculty member to participate in such an experiment. For a 5-student team, this would require that a schedule be set up (using random numbers) so a student could come by on the average of every 5 minutes to check the faculty member’s activity. One student would be required to make 16.8, or 17, observations each day. The work sample could be altered by observing other activities—for example, the time spent on teaching related activities such as making out tests, preparing for class, helping students, etc., or the proportion of time spent of research.

.99769 Project Management Answers to Questions 9-1.

CPM/PERT is popular because it provides a picture of the steps of a project, it is easy for managers and participants to understand, and it is easy to apply.

9-2.

The goal is to show the precedence relationship of activities in a project.

9-3.

A dummy activity is used in an ADA network to show a precedence relationship without the passage of time. It is used to complete a precedence relationship so that two activities will not have the same start and end nodes.

9-4.

The critical path is the longest path in the network. It can be computed by summing the activity times along each path and then seeing which path is the longest. It also is the path with no slack available.

9-5.

Slack is the amount of time an activity can be delayed without affecting the overall project duration. It is computed by subtracting the earliest start time for an activity from the latest start time or the earliest finish time from the latest finish time for an activity.

9-6.

The mean activity time is computed as t = (a + 4m + b) 6, where a is the optimistic activity time, m is the most likely time, and b is the pessimistic time. The variance is computed as v 2 = [(b − a ) 6]2 .

9-7.

Total project variance is computed by summing the variances of the critical path activities.

9-8.

The purpose of project crashing is to shorten the project duration at the least possible cost.

9-9.

See which activity on the critical path has the minimum crash cost, and reduce this activity duration by the maximum amount or until another path becomes critical. If more than one path is critical, both paths must be reduced by the same amount simultaneously. Repeat this process until the crashing objective is reached.

9-10. The preference should depend on the project, including the perceived degrees of variability in project activities, the ability to determine probabilistic time estimates, and the degree to which probabilistic analysis is required. 9-11. The Gantt chart is a graphical technique with a bar or time line displayed for each activity in the project. However, while it will indicate the precedence relationships between activities, these relationships are usually not as easy to discern (i.e., visualize) as with a network. 9-12. Indirect costs include the cost for facilities, equipment, and machinery, interest on investment, utilities, labor, personnel costs, etc. Direct costs are financial penalties for not completing a project on time. In general, project crashing costs and indirect costs have an inverse relationship; crashing costs are highest when the project is shortened, whereas indirect costs increase as the project duration increases. 9-13. A heavy reliance by the project manager on the network can mask errors in the precedence relationships or missing activities can be overlooked. Attention to critical path activities can be excessive to the point of neglecting other crucial project activities. Obtaining both deterministic and probabilistic time estimates can be difficult. The time estimates can be overly optimistic or pessimistic. 9-14. For an activity-on-node network, nodes represent project activities and branches show precedence relationships, whereas on an activity-on-arrow network, branches represent activities and nodes are events specifying the end of one activity and the beginning of another. 9-15. The project team consists of a group of individuals selected because of their special skills, expertise, and experience related to the project. Project planning includes the identification of all project activities and their precedence relationships the determination of activity times, the determination of project duration, comparison of the project time with objectives, and the determination of resource requirements to meet objectives. Project control includes making sure all activities have been identified, making sure the activities

are completed in their proper sequence, identifying resources as they are required, and adjusting the schedule to reflect changes. 9-16. The WBS depends on the “project” the student uses. Figure 9.2 should be used as a guideline. 9-17. Work Breakdown Structure for Dinner Project

9-18. Student answer based on project selected. 9-19. Student answer based on country selected.

Solutions to Problems 9-1.

9-2.

Project completion time = 17 weeks

Activity 1 2 3 4 5 6 7 9-3.

Slack (weeks) 12 0 12 4 0 4 0

Project completion time = 23 weeks

Activity 1 2 3 4 5 6 7 8 9 9-3.

Paths:

1→ 3 → 7 4 + 8 + 2 = 14 1→ 3 → 6 →8 4 + 8 + 5 + 6 = 23* 1→ 4 → 8 4 + 3 + 6 = 13 2→5→8 7 + 9 + 6 = 22

2→9 7 + 5 = 12

Slack (weeks) 0 1 0 10 1 0 9 0 11

9-4.

Paths: 2 → 5 → 7 2→4→6→7 1→ 3 → 6 → 7

10 + 4 + 2 = 16 10 + 5 + 3 + 2 = 20* 7 + 6 + 3 + 2 = 18

Activity 1 2 3 4 5 6 7

EF 7 10 13 15 14 18 20

9-5. Time 7 10 6 5 4 3 2

ES 0 0 7 10 10 15 18

LS 2 0 9 10 14 15 18

Critical path activities have no slack. Critical path = 2-4-6-7 = 20

LF 9 10 15 15 18 18 20

Slack 2 0 2 0 4 0 0

9-6.

Critical path = 2-6-9-11-12 = 38 months

9-7.

9-8.

a, b, c.

Paths: a-b-d-f a-c-d-f a-c-e-f

3 + 3 + 4 + 2 = 12 3 + 5 + 4 + 2 = 14* 3 + 5 + 3 + 2 = 13

9-9.

9-10. Activity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Time 8 12 3 9 3 2 12 7 30 21 20 5 16 17 5 6 3

ES 0 0 0 12 3 3 21 21 21 21 33 51 42 58 58 53 75

EF 8 12 3 21 6 5 33 28 51 42 53 56 58 75 63 59 78

Critical path = 2-4-10-13-14-17 Project completion time = 78 wk.

LS 4 0 9 12 18 19 40 46 23 21 52 53 42 58 70 72 75

LF 12 12 12 21 21 21 52 53 53 42 72 58 58 75 75 78 78

Slack 4 0 9 0 15 16 19 25 2 0 19 2 0 0 12 19 0

9-11.

Critical path = a-b-f-h = 15 weeks 9-12.

Activity a b c d e f g h i j k l m n o p q

Early Start

Early Finish

0 3 3 3 3 8 11 7 13 11 14 13 22 11 11 18 27 Project

Late Start

3 8 7 13 5 11 13 9 14 12 22 18 27 16 14 20 28 28

Late Finish 0 3 7 10 9 8 11 11 13 19 14 20 22 20 24 25 27

Slack

3 8 11 20 11 11 13 13 14 20 22 25 27 25 27 27 28

0 0 4 7 6 0 0 4 0 8 0 7 0 9 13 7 0

9-13. Activity

Slack

2

1 2 3 4 5 6 7 8 9 10 11

6 2 4 3 7 4 3 5 3 4 2

10 7 8 10 9 12 6 9 20 12 9

15 16 11 15 20 15 9 16 35 16 14

10.16 7.66 7.83 9.66 10.50 11.16 6.00 9.50 19.66 11.33 8.66

0 0 0 10.16 10.16 10.16 19.83 7.83 20.66 7.83 25.83

10.16 7.66 7.83 19.83 20.66 21.33 25.83 17.33 40.33 19.16 34.50

0 18.00 14.33 16.00 10.16 20.50 25.66 22.16 20.66 29.00 31.66

10.16 25.66 22.16 25.66 20.66 31.66 31.66 31.66 40.33 40.33 40.33

0 18.00 14.33 5.83 0 10.33 5.83 14.33 0 21.16 5.83

2.25 5.43 1.35 4.00 4.67 3.35 1.00 3.35 28.41 4.00 4.00

Expected project completion time = 40.33 wk

 = 5.95 Critical path = 1-5-9

9-14.

9-15.

Activity a b c d e f g h i j

Activity Predecessor — — a, b c — d e d f g

Time Estimates (days) 3 5 4 10 2 3 2 2 1 1

0 0 5 9 0 19 2 19 22 4

3 5 9 19 2 22 4 21 23 5

2 0 5 9 18 19 20 21 22 22

5 5 9 19 20 22 22 23 23 23

2 0 0 0 18 0 18 2 0 18

0 6 0 10 6 10 15 15 25

6 10 6 15 9 14 25 19 33

0 6 4 10 12 11 15 21 25

6 10 10 15 15 15 25 25 33

0 0 4 0 6 1 0 6 0

Critical path = b - c - d - f - i Project duration time = 23

9-16.

Activity a b c d e f g h i

Activity Predecessor — a — b, c a b, c d, e, f d g, h

Time Estimates (days) 6 4 6 5 3 4 10 4 8

Critical path = b - c - d - f - i Project duration time = 33

9-17. a, b, c, d.

Activity

Slack

2

1 2 3 4 5 6 7 8 9 10 11 12

4 6 2 1 3 3 2 9 5 7 5 3

8 10 10 4 6 6 8 15 12 20 6 8

12 15 14 13 9 18 12 22 21 25 12 20

8.00 10.16 9.33 5.00 6.00 7.50 7.66 15.16 12.33 18.66 6.83 9.16

0 0 0 8.00 8.00 10.16 9.33 17.66 17.66 13.00 30.00 36.83

8.00 10.16 9.33 13.00 14.00 17.66 17.00 32.83 30.00 31.66 36.83 46.00

3.66 0 8.33 22.33 11.66 10.16 22.33 21.66 17.66 27.33 30.00 36.83

11.66 10.16 17.66 27.33 17.66 17.66 30.00 36.83 30.00 46.00 36.83 46.00

3.66 0 8.33 14.33 3.66 0 13.00 4.00 0 14.33 0 0

1.77 2.25 4.00 4.00 1.00 6.25 2.76 4.67 7.08 9.00 1.35 8.01

Critical path = 2-6-9-11-12 Expected project completion time = 46 mo ;  = 5 mo

9-18

a, b, c, d.

Activity

Slack

2

1 2 3 4 5 6 7 8 9 10 11 12

1 1 3 3 2 2 1 1 1 2 1 1

2 3 5 6 4 3 1.5 3 1 4 2 1

6 5 10 14 9 7 2 5 5 9 3 1

2.50 3.00 5.50 6.83 4.50 3.50 1.50 3.00 1.66 4.50 2.00 1.00

0 2.50 2.50 2.50 8.00 8.00 8.00 9.50 12.50 12.50 12.50 17.00

2.50 5.50 8.00 9.33 12.50 11.50 9.50 12.50 14.16 17.00 14.50 18.00

0 7.50 2.50 2.66 10.50 9.00 8.00 9.50 15.33 12.50 15.00 17.00

2.50 10.50 8.00 9.50 15.00 12.50 9.50 12.50 17.00 17.00 17.00 18.00

0 5.00 0 0.16 2.50 1.00 0 0 2.83 0 2.50 0

0.694 0.436 1.35 3.35 1.35 0.689 0.026 0.436 0.436 1.35 0.109 0

Critical path = 1-3-7-8-10-12

Expected project completion time = 18 months ;  = 1.97

9-19.

x−



50 − 46 4 = = 0.80 5 5

From normal table, p = 0.2881 0.5000 − 0.2881 = 0.2119 probability that the project will exceed 50 mo.

9-20.

Activity

Slack

2

a b c d e f g h I j k

1 2 1 4 3 10 5 2 1 2 2

2 5 3 10 7 15 9 3 4 5 2

3 8 5 25 12 25 14 7 6 10 2

2.00 5.00 3.00 11.50 7.16 15.83 9.16 3.50 3.83 5.33 2

0 0 0 2.00 2.00 5.00 3.00 13.50 20.83 20.83 26.16

2.00 5.00 3.00 13.50 9.16 20.83 12.16 17.00 24.66 26.16 28.16

7.33 0 8.66 9.33 13.66 5.00 11.66 22.66 22.33 20.83 26.16

9.33 5.00 11.66 20.83 20.83 20.83 20.83 26.16 26.16 26.16 28.16

7.33 0 8.66 7.33 11.66 0 8.66 9.16 1.50 0 0

0.11 1.00 0.436 12.25 2.25 6.25 2.25 0.689 0.689 1.77 0

Critical path = b-f-j-k

Expected project completion time = 28.17 weeks;  = 3.00

x−



35 − 28.16 6.84 = = 2.28 3 3.00

From normal table, p = 0.4887 0.5000 − 0.4887 = 0.0113 probability that the company will be fined.

9-21.

x−



15 − 18 = −1.52 1.97

From normal table, p = 0.4357 0.5000 − 0.4357 = 0.0643 probability that preparations would be in time.

9-22. Activity

Slack

2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1 4 20 4 2 8 10 5 1 6 5 5 4 5 5 1

3 6 35 7 3 12 16 9 2 8 8 10 7 7 9 3

5 10 50 12 5 25 21 15 2 14 12 15 10 12 20 7

3.00 6.33 35.00 7.33 3.16 13.50 15.83 9.33 1.83 8.66 8.16 10.00 7.00 7.50 10.16 3.33

0 0 0 3.00 10.33 13.50 13.50 13.50 35.00 10.33 35.00 36.83 36.83 19.00 43.83 54.00

3.00 6.33 35.00 10.33 13.50 27.00 29.33 22.83 36.83 19.00 43.16 46.83 43.83 26.50 54.00 57.33

7.50 14.66 0 10.50 17.83 23.33 21.00 27.50 35.00 41.16 49.16 44.00 36.83 49.83 43.83 54.00

10.50 21.00 35.00 17.83 21.00 36.83 36.83 36.83 36.83 49.83 57.33 54.00 43.83 57.33 54.00 57.33

7.50 14.66 0 7.50 7.50 9.83 7.50 14.00 0 30.83 14.16 7.16 0 30.83 0 0

0.436 1.00 25.00 1.77 0.25 8.01 3.35 2.76 0.029 1.77 1.36 2.77 1.00 1.36 6.25 1.00

Expected completion time = 57.33 days

 = 5.77 Z=

x−



67 − 57.33 = 1.68 5.77

P( x  67) = 0.9535

9-23. Activity a b c d e f g h I j k l m n o

ES 0 5.33 5.33 5.33 10.33 10.33 9.17 9.17 11.67 11.67 21.17 19.17 17.83 26 34.5

EF 5.33 10.33 9.17 11.67 17.83 19.83 21.17 18.33 19.17 26 34 30 25.17 34.5 42.33

LS 0 15 13.67 5.33 27.5 20 17.5 22.33 24 11.67 29.5 31.5 35 26 34.5

LF 5.33 20 17.5 11.67 35 29.5 29.5 31.5 31.5 26 42.33 42.33 42.33 34.5 42.33

Slack 0 9.67 8.33 0 17.17 9.67 8.33 13.17 12.33 0 8.33 12.33 17.17 0 0

Variance 1 1 0.69 1 2.25 3.36 7.11 1.36 3.36 5.44 3.36 2.25 1.78 4.69 0.69

Critical path = a-d-j-n-o Expected project completion time = 42.3 weeks

 = 3.58 Since probability is 0.90, Z = 1.29. 1.29 =

x − 42.3 3.58

x − 42.3 = 4.61

x = 46.92

To be 90 percent certain of delivering the part on time, RusTech should probably specify at least 50.3 or 51 weeks in the contract

9-24.

a b c d e f g h I j k l m n o p q

a b c d e f

Activity Time 15 8.83 24.16 19.5 8.16 13.66 20.16 25 14.66 23 8.66 7.16 5 4.33 7 5.5 20.833

Early Start 0 0 15 39.16 39.16 47.33 61 58.66 58.66 58.66 81.66 83.66 73.33 90.83 90.83 90.33 97.83

1.66 1.16 2.5 2.16 1.16 2.33

Activity std dev g 1.5 h 3.33 I 2 j 2.33 k 1.33 l 1.5

Critical path = a-c-d-h-l-o-q

Early Finish 15 8.83 39.16 58.66 47.33 61 81.16 83.66 73.33 81.66 90.33 90.83 78.33 94.66 97.83 95.83 118.66

m n o p 1

0.66 1 1 0.83 2.5

Late Start 0.0 64.5 15 39.16 51.5 59.66 73.33 58.66 78.16 61.83 84.83 83.66 92.83 93.5 90.83 113.16 97.83

Late Finish 15 73.33 39.16 58.66 59.66 73.33 93.5 83.66 92.83 84.83 93.5 90.83 97.83 97.83 97.83 118.66 118.66

Slack 0 64.5 0 0 12.33 12.33 12.33 0 19.5 3.16 3.16 0 19.5 3.16 0 22.83 0

 = 118.67  = 5.85 Z=

x−



120 − 118.67 = 0.227 5.85

P( x  120) = 0.59 9-25. a.

Present (normal) critical path = 1-4 Normal critical path time = 30 wk Crash critical path (all crash time) = 1-4 Maximum possible project crash time = 20 wk

Normal cost = 3950 Crashed project cost = 4700

Activity a b c d e

Normal Time 20 24 14 10 11

Crash Time 8 20 7 6 5

Normal Cost 1,000 1,200 700 500 550 3950

a b c d e

Crash Cost/pd 40 50 70 80 30

Crash By 10 4 0 0 5

Crashing Cost 400 200 0 0 150 750

Crash Cost 1,480 1,400 1,190 820 730 5620

9-26.

Normal critical path = a-d-h Normal critical path time = 36 wk Project completion time = 36 Normal cost = 14400 Minimum project completion time = 22 Crash cost = 23,250

a b c d e f g h

Normal Time 16 14 8 5 4 6 10 15

a b c d e f g h

Crashing Cost 2,400 800 200 700 750 0 1,000 3,000

Crash Time 8 9 6 4 2 4 7 10

Normal Cost 2,000 1,000 500 600 1,500 800 3,000 5,000

Crash Cost 4,400 1,800 700 1,300 3,000 1,600 4,500 8,000

Crash Cost/pd 300 160 100 700 750 400 500 600

Crash By 8 5 2 1 1 0 2 5

9-27. Project completion time = 23 Normal cost = 1500 Minimum project completion time = 15 Crash cost = 2650 Activity 1 2 3 4 5 6 7

Normal Time 8 10 5 3 6 3 4

Crash Time 5 7 3 1 4 3 3

Normal Cost 100 250 400 200 150 100 300

Activity 1 2 3 4 5 6 7

Crashing Cost/pd 100 50 200 100 75 0 200

Crash By 3 2 2 0 2 0 1

Crashing Cost 300 100 400 0 150 0 200

Crash Cost 400 400 800 400 300 100 500

9-28. Project completion time = 33 Normal cost = 28800 Minimum project completion time = 26 Crash cost = 33,900

a b c d e f g h i j k

Normal Time 9 11 7 10 1 5 6 3 1 2 8

Crash Time 7 9 5 8 1 3 5 3 1 2 6

Critical path = a-d-g-k Crashing cost = $5,100 Total network cost = $33,900

Normal Cost 4,800 9,100 3,000 3,600 0 1,500 1,800 0 0 0 5,000

Crash Cost 6,300 15,500 4,000 5,000 0 2,000 2,000 0 0 0 7,000

Normal Cost 750 3,200 500 700 0 250 200 0 0 0 1,000

Crash By 2 0 0 2 0 0 1 0 0 0 2

Crashing Cost 1,500 0 0 1,400 0 0 200 0 0 0 2,000

9-29.

Project a b c d e f g h I j

Activity Time 23 3 3.167 4.167 2.833 5 1.833 5.833 3.833 4.167 2.167

Slack

0 0 0 3 5.833 10.833 10.833 12.667 16.667 20.833

3 3.167 4.167 5.833 10.833 12.667 16.667 16.5 20.833 23

0 7.667 6.667 3 5.833 15.167 10.833 17 16.667 20.833

3 10.833 10.833 5.833 10.833 17 16.667 20.833 20.833 23

0 7.667 6.667 0 0 4.333 0 4.333 0 0

Probability the project will be completed in 21 days? Z=

x−



21 − 23 = −1.18 1.70

P( x  21) = .119

Standard Deviation 1.70 0.667 0.5 0.833 0.5 1 0.167 0.833 0.5 0.5 0.5

9-30.

Activity a b c d e f g h i j k

Earliest Start 0.00 7.00 18.50 23.50 18.50 27.33 33.67 42.83 49.50 63.83 86.67

Earliest Finish 7.00 18.50 23.50 30.83 27.33 33.67 42.83 49.50 63.83 86.67 104.83

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

Latest Start 0.00 7.00 21.33 26.33 18.50 27.33 33.67 42.83 49.50 63.83 86.67

Latest Finish 7.00 18.50 26.33 33.67 27.33 33.67 42.83 49.50 63.83 86.67 104.83

Activity Slack 0.00 0.00 2.83 2.83 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Critical Path Variance 1.00 2.25

3.36 1.00 2.25 1.00 2.78 6.25 4.69

9-31.

Activity a b c d e f g h I j k l m n o p q r s t u

Time 7.17 18.00 10.17 22.17 30.00 8.67 7.00 21.33 20.17 13.33 7.83 12.33 10.17 7.33 18.33 7.17 8.17 7.00 27.83 17.50 8.17

ES 0 0 0 0 7.17 37.17 45.83 18 18 52.83 39.33 38.17 37.17 45.83 66.17 84.5 66.17 66.17 45.83 47.33 64.83

EF 7.17 18 10.17 22.17 37.17 45.83 52.83 39.33 38.17 66.17 47.17 50.5 47.33 53.17 84.5 91.67 74.33 73.17 73.67 64.83 73

The critical path is: a-e-f-g-j-o-p

LS 0 0 27 15 7.17 37.17 45.83 37 33.67 52.83 58.33 53.83 55.83 58.83 66.17 84.5 83.5 84.67 63.83 66 83.5

LF 7.17 33.67 37.17 37.17 37.17 45.83 52.83 58.33 53.83 66.17 66.17 66.17 66 66.17 84.5 91.67 91.67 91.67 91.67 83.5 91.67

Slack 0 15.67 27 15 0 0 0 19 15.67 0 19 15.67 18.67 13 0 0 17.33 18.5 18 18.67 18.67

Variance 0.69 7.11 1.36 12.25 0 1.78 1 7.11 3.36 4 0.69 1.78 1.36 1.78 2.78 0.69 1.36 1 6.25 4.69 1.36

Project duration = 91.667

 = 3.3082 From January 20 to April 29 is 101 days.

P( x  101) =

x− Z 101 − 91.667 3.3082

= 2.82

P( x  101) = .9976 Activity “n,” send out acceptance letters, has ES = 45.83 (March 6) and LF = 66.17 (March 20), so it appears the club would meet the deadline of March 30 to send out acceptance letters. 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.

9-32.

Activity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Activity time 11.67 36.67 17.67 9.17 9.5 12 16.17 26.67 8 10.67 9.67 11.17 14.67 14.67 17 4.83

Early Start 0 0 11.67 11.67 36.67 29.33 20.83 36.67 46.17 20.83 54.17 31.5 63.33 63.83 63.83 80.83

Early Finish 11.67 36.67 29.33 20.83 46.17 41.33 37 63.33 54.17 31.5 63.83 42.67 78 78.5 80.83 85.67

Late Start 4.83 0 16.5 20.83 36.67 34.17 30 39.5 46.17 42 54.17 52.67 66.17 66.17 63.83 80.83

Late Finish 16.5 36.67 34.17 30 46.17 46.17 46.17 66.17 54.17 52.67 63.83 63.83 80.83 80.83 80.83 85.67

Slack 4.83 0 4.83 9.17 0 4.83 9.17 2.83 0 21.17 0 21.17 2.83 2.33 0 0

Standard Deviation 3 6.67 4.67 3.17 1.5 1.67 3.5 4 1 2 1.33 3.5 1.33 1.67 2.33 .5

Critical path = 2-5-9-11-15-16 Project completion time = 85.66 months Project standard deviation = 7.43 P(x<=96) = .947

9-33.

Project completion = 32.83 σ = 1.86

Activity

Mean

Slack

Variance

a b c d e f g h i j k l m

1.83 2.00 5.83 2.00 1.00 2.17 4.00 3.83 3.00 3.83 6.33 5.83 3.17

0 1.83 3.83 3.83 5.83 3.83 9.67 1.83 13.67 13.67 17.50 23.83 29.67 Project

Critical Path = a-b-c-g-j-k-l-m Z = (24 – 32.833) / 1.86 = -4.74 P (x<= 6 months or 24 weeks) = 0

Z = (52 – 32.833 / 1.86 = 10.31 P(x<= a year or 52 weeks) = 1.00

1.83 3.83 9.67 5.83 6.83 6.00 13.67 5.67 16.67 17.50 23.83 29.67 32.83 32.833

0.00 1.83 3.83 10.67 12.67 11.50 9.67 10.67 14.50 13.67 17.50 23.83 29.67

1.83 3.83 9.67 12.67 13.67 13.67 13.67 14.50 17.50 17.50 23.83 29.67 32.83

0.00 0.00 0.00 7 6.83 7.67 0.00 9 1 0 0.00 0 0 Project Std.dev

0.027778 0.111111 0.694444

0.44

0.25 1.00 0.69 0.25 3.47 1.86

9-34.

a b c d e f g h i j k l m n

Mean 2.00 3.00 2.00 14.33 7.00 7.00 24.33 6.33 12.33 12.67 2.00 3.83 2.17 1.83

ES 0.00 2.00 5.00 5.00 5.00 5.00 12.00 7.00 36.33 48.67 61.33 63.33 67.17 69.33 Project

EF 2.00 5.00 7.00 19.33 12.00 12.00 36.33 13.33 48.67 61.33 63.33 67.17 69.33 71.17 71.17

LS 0.00 2.00 28.00 22.00 5.00 29.33 12.00 30.00 36.33 48.67 61.33 63.33 67.17 69.33

Critical path = a – b – e – g – i- j – k – l – m – n Z = (52-71.17) / 3.51= - 5.46 P(x<=52) = 0

Z = (76-71.17) / 3.51= 1.376 P(x<=76) = .500 + .4147 = .9147

LF 2.00 5.00 30.00 36.33 12.00 36.33 36.33 36.33 48.67 61.33 63.33 67.17 69.33 71.17

Slack 0.00 0.00 23.00 17.00 0.00 24.33 0.00 23.00 0.00 0.00 0.00 0.00 0.00 0.00 Project Std.dev

Variance 0.11 0.11

1.00 5.44 1.00 4.00 0.11 0.25 0.25 0.03 12.31 3.51

9-35. Activity a b c d e f g h I j k l m n

Time 8.17 5.83 21.50 31.00 7.00 17.33 5.33 2.17 35.83 6.33 15.17 6.83 6.17 4.00

ES 0 0 0 0 8.17 15.17 21.50 31.00 33.17 69.00 15.17 75.33 82.17 88.33

EF 8.17 5.83 21.50 31.00 15.17 32.50 26.83 33.17 69.00 75.33 30.33 82.17 88.33 92.33

LS 0.67 3.00 6.33 0.00 8.83 15.83 27.83 31.00 33.17 69.00 60.17 75.33 82.17 88.33

The “suggested” network is as follows:

Critical path = d-h-i-j-l-m-n Expected project completion time = 92.33

 = 5.93 P( x  90) =

90 − 92.33 = .39 5.93

P( x  90) = .5000 − .1517 = .348

LF 8.83 8.83 27.83 31.00 15.83 33.17 33.17 33.17 69.00 75.33 75.33 82.17 88.33 92.33

Slack 0.67 3.00 6.33 0 0.67 0.67 6.33 0 0 0 45.00 0 0 0

Variance 1.36 0.69 6.25 13.44 1.00 4.00 1.00 0.25 17.36 1.00 3.36 0.69 1.36 1.00

9-36. Activity a b c d e f g h I j k l m n o p q r s t u v w

Time 9.33 4.83 5.00 14.50 9.50 2.17 5.00 13.67 6.17 22.17 9.67 25.67 12.50 15.83 45.00 6.83 5.00 12.33 4.17 17.17 2.00 0.50 0.50

ES 0 9.33 9.33 14.33 28.83 14.33 16.5 9.33 23 23 9.33 23.00 48.67 61.17 77.00 48.67 55.50 60.50 122.00 55.50 55.50 45.17 126.17

EF 9.33 14.17 14.33 28.83 38.33 16.50 21.50 23.00 29.17 45.17 19.00 48.67 61.17 77.00 122.00 55.50 60.50 72.83 126.17 72.67 57.50 45.67 126.67

LS 0 120.83 96.67 101.67 116.17 118.5 120.67 9.33 119.5 103.5 116 23 48.67 61.17 77 102.17 109.33 114.33 122 109 124.17 125.67 126.17

LF 9.33 125.67 101.67 116.17 125.67 120.67 125.67 23.00 125.67 125.67 125.67 48.67 61.17 77.00 122.00 109.00 114.33 126.67 126.17 126.17 126.17 126.17 126.67

Slack 0 111.50 87.33 87.33 87.33 104.17 104.17 0 96.50 80.50 106.67 0 0 0 0 53.50 53.83 53.83 0 53.50 68.67 80.50 0

Variance 5.44 1.36 0.44 3.36 1.36 0.25 0.44 5.44 1.36 6.25 4.00 18.78 4.69 6.25 25.00 1.36 1.00 1.00 0.69 3.36 0.11 0.00 0.00

Critical path = a-h-l-m-n-o-s-w Expected project duration ( ) = 126.67 days

 = 8.14 days P ( x  150 days): Z =

x−

 150 − 126.67 8.14

Z = 2.87

P( x  150) = .5000 + .4979 = .9979

The “suggested” network is as follows (although the student’s version may vary).

CASE: Bloodless Coup Concert

Activity 1 2 3 4 5 6 7 8 9 10 11 12

a 2 4 3 1 1 2 1 2 2 2 1 1

m 4 5 5 3 2 4 3 3 6 3 2 5

b 7 8 10 8 4 7 5 4 12 6 3 12

t 4.16 5.33 5.50 3.50 2.16 4.16 3.00 3.00 6.33 3.33 2.00 5.5

ES 0 0 4.16 4.16 4.16 7.66 7.66 9.66 6.33 11.83 10.66 6.33

EF 4.16 5.33 9.66 7.66 6.33 11.83 10.66 12.66 12.66 15.16 12.66 11.83

Critical path = 1-4-6-10 Expected project completion time = 15.17 days

 = 1.79 Z=

x−



18 − 15.17 = 1.58 1.79

P( x  18) = 0.9429

LS 0 3.50 6.66 4.16 6.66 7.66 10.16 12.16 8.83 11.83 13.16 9.66

LF 4.16 8.83 12.16 7.66 8.83 11.83 13.16 15.16 15.16 15.16 15.16 15.16

Slack 0 3.50 2.50 0 2.50 0 2.50 2.50 2.50 0 2.50 3.33

sd( ) 0.83 0.66 1.16 1.16 0.50 0.83 0.66 0.33 1.66 0.66 0.33 1.83

CASE SOLUTION: Moore Housing Contractors

a b c d e f g h i j k l m n o p q r s t u v w x

Activity time 4.16 3.16 3.83 2.16 2 3.83 3.16 4.16 2.83 2.16 5.16 6.5 8.33 3.33 2.33 3.5 4.16 6.33 5.83 4.33 3.33 6.33 5 2.83

Early Start 0 4.16 7.33 7.33 7.33 11.16 9.33 9.33 15 15 12.5 17.83 17.66 24.33 27.66 30 26 33.5 30.16 30.16 30.16 34.5 40.83 40.83

Early Finish 4.16 7.33 11.16 9.5 9.33 15 12.5 13.5 17.83 17.16 17.66 24.33 26 27.66 30 33.5 30.16 39.83 36 34.5 33.5 40.83 45.83 43.66

Late Start 0.0 4.16 7.83 33.83 7.33 11.66 9.33 13.5 21 15.5 12.5 23.83 17.66 30.33 33.66 36 26 39.5 35 30.16 31.17 34.5 40.83 43

Late Finish 4.16 7.33 11.66 36 9.33 15.5 12.5 17.66 23.83 17.66 17.66 30.33 26 33.66 36 39.5 30.16 45.83 40.83 34.5 34.5 40.83 45.83 45.83

Slack 0.0 0.0 0.5 26.5 0.0 0.5 0.0 4.16 6 0.5 0.0 6 0.0 6 6 6 0 6 4.83 0 1 0 0 2.16

Activity std dev 0.5 0.5 0.5 0.5 0.33 0.5 0.5 0.83 0.5 0.5 0.83 0.83 1 0.66 0.66 0.83 0.5 1 1.5 1 0.66 1 1 0.5

Project completion time = 45.83 Project standard deviation = 2.40 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.

10 Supply Chain Management Strategy and Design Answers to Questions 10-1. McDonald’s: The company is supplied by food distributors and restaurant product suppliers (for plates, napkins, etc.). Production is located in retail sites that are usually small and are near large easily accessible customer markets. Their inventory levels are typically small because food cannot be stored in large quantities. Their primary mode of transportation is trucking. Toyota: Toyota suppliers include raw materials and auto parts. They receive some items on-demand for JIT production and some is stored in warehouses. Production is in large plants with heavy-machinery in close proximity to good transportation sources. Toyota plants are located all over the world as are their distribution systems. Transportation is by all modes of transportation. 10-2. The strategic goals are low cost and customer service. Purchasing from suppliers must be on time or the entire supply chain is delayed, creating late deliveries to customers. Erratic and poor quality supply can also increase costs. If facilities are not located properly it can delay product or service flow through the supply chain, and increase costs for longer deliveries. Production inefficiency and poor quality can cause delays in product or service flow and it also creates the need for more inventory which increases cost. Inefficient transportation can also result in higher inventories to offset delays and raise costs, and, causes delayed delivery to customers. 10-3. Answer depends on the businesses selected. 10-4.

Some service providers actually incorporate manufacturing aspects, for example, a restaurant “assembles” menu items from raw materials and parts (food items), and provides the customer with a final product. A hospital takes an “unfinished” product (a patient) and then it flows through a set of processes to produce an end result. As a result, there aren’t as many “differences” in SCM between manufacturing operations and some types of service providers. A “pure” service provider like a hotel, while not “assembling” a product will still include supply chain processes like procurement, quality management, scheduling, etc., and the same approaches to efficient management of these processes will apply.

10-5. Answer depends on the example the student selects. One example is a grocery chain that uses POS terminals to send information sales to distribution center who, in turn, sends mixed loads to the store by truck weekly. 10-6. The answer depends on the company the student selects. 10-7. Answer depends on the e-marketplace selected. 10-8. Answer depends on the ERP provider selected. 10-9. The answer depends on the article the student selects. 10-10. The answer depends on the article the student selects. 10-11. Toyota is one automobile company that has begun a limited BTO program. The student should go tan automobile company website to see what they are doing with BTO options. There are traditionally many more options with automobiles than with a product like a computer, which means more suppliers to work with. Customers must be reconditioned to make selections from a smaller list of options—like colors, seat type and fabric, sound system, trim, tires, etc. Also the component parts for a computer can generally be stored in one location however, component parts for autos tend to be more geographically dispersed. 10-12. RFID allows items to be scanned and identified without direct sight lines which enables items to be checked much more rapidly. Since bills-of-lading must not be much more detailed than previously (listing virtually all items in a shipment) this can be accomplished much more quickly and thoroughly with RFID technology and the information can be transported via satellite. This allows for much quicker security approvals, thus

allowing ships and planes to load and unload more quickly. 10-13. Figure 10.6 shows some of the advantages that RFID technology would provide for a retailer like Wal-Mart. The primary obstacle to the use of RFID is (currently) cost. 10-14. It provides a set of performance evaluators and an apparatus for comparing supply chains, which enables generic supply chain evaluation. These performance evaluators apply to supply chains of different sizes and cut across supply chains from different industries, which could allow them to be used to establish standards that can be used as a certification tool. 10-15. Universities have traditional suppliers for MRO goods like supplies, telephone and computer services, furniture, maintenance equipment and products, food and beverages, etc. They also have non-traditional suppliers like high schools that provide students. The producers are the teachers and other academic and support services that educate the student and provide for daily student life and well-being. Distributors include placement services and companies and organizations students are hired by. Inventory exists for MRO goods and services. Students are also a form of inventory as are classrooms, dorm rooms, parking spaces, etc. Students as inventory incur a carrying cost. The longer a student remains in school the greater the cost to the institution or state. Not only is there a direct cost for maintaining a student, but a student who stays too long also takes up a spot for other incoming students, thus creating shortages of facilities like dorm rooms and classrooms, which in-turn require expenditures for capital expansion. 10-16.

The bullwhip effect occurs when demand variability becomes magnified as demand information is transmitted back upstream in the supply chain causing suppliers to compensate for demand uncertainty with larger than necessary buffer inventories.

10-17.

Global supply chain risks include natural and/or man-made disasters that temporarily disrupt the flow of supply, such as earthquakes and floods, and, labor strikes and political instability.

10-18.

Companies manage supply chain risks by building resiliency into the supply chain by developing processes that identify conditions that could cause supply disruptions and developing contingency plans for them, including a pool of alternative suppliers, logistics providers and energy sources. Another approach is risk pooling achieved by aggregating risks to reduce their overall impact, such as combining inventory locations from numerous risky locations to a few. Also, risk can be managed by reducing product variability allowing companies to meet demand with fewer suppliers.

10-19.

Student answers will vary, however, examples include: a. Overseas suppliers – defective products b. Container ships – delays and unreliable delivery c. Transportation – late deliveries and product damage d. Packaging – defective packaging, labeling errors e. Distribution – order errors, incorrect deliveries f. Warehousing – theft and product damage

10-20. Answer depends on student responses. 10-21. Answer depends on company student identifies. 10-22. Answer depends on student responses. 10-23. Sustainability to a large extent is about conserving resources, which coincides with the quality management focus of eliminating waste. In the human resources area, quality management focuses on employee satisfaction, which also coincides with sustainability goals. To a certain extent, sustainability also supports the quality goal of customer satisfaction since customers these days seem to desire sustainable products. 10-24.

As a result of cultural, economic and political differences companies often encounter workplace sustainability issues related to worker wages, treatment and living conditions. Also, companies tend to

reap the negative media impact of environmental pollution, greenhouse gas emissions, and waste created by their overseas suppliers. 10-25.

In China sustainability issues are prominent in the workplace as a result of a huge labor pool and a lack of government control that allows suppliers to mistreat workers. In Mexico workplace issues are less pervasive, however, environmental pollution laws are more relaxed than in the U.S.

10-26.

By introducing programs to reduce waste and conserve energy, which also saves costs. Telecommuting is a specific example of how a company can reduce facility and transportation costs by having employees work from home.

10-27.

The answer depends on the companies the student selects; however, examples might include Apple, WalMart, Levi Strauss, P&G and Target.

10-28. Service companies can achieve sustainability by reducing waste and conserving resources, like reducing paper usage, conserving on heat and electricity in workplace facilities, and telecommuting. However, some service companies, like UPS and Wal-Mart, can also reduce greenhouse gas emissions like carbon dioxide by using more energy-efficient vehicles. Manufacturing firms can reduce their carbon footprint, as well as eliminate waste and conserve energy and other sustainable resources. 10-29.

Answer depends on student responses.

10-30.

Answer depends on student responses.

10-31. In a digital supply chain, products move and are distributed electronically so there are no physical products, modes of transport, inventory or facilities; i.e., the things that make up a physical supply chain. However, there are still suppliers of digital content, the movement of products, distributors, stores and end-use customers. The supply chain tends to be embedded in the Internet. 10-32.

Common characteristics include excellent financials, adoption of industry best practices, product innovation and simplification, effective supply chain planning and execution, lower cycle times, high order fulfillment rates, a focus on total supply chain costs and inventory reduction, a commitment to quality, effective risk management, a focus on core competencies, collaboration with suppliers, well-defined strategies, advanced IT capabilities, collaboration with suppliers, and supply chain visibility, among others.

10-33.

The Gartner score is a composite based on five measures including the opinion in the form of a ranking from an in-house Gartner panel, the rankings of a peer panel, the three-year return on assets (ROA), inventory turns, and the three-year weighted revenue growth.

10-34.

In 2012 IBM and Microsoft dropped out of the top 25 primarily due to the way Gartner’s computes its scores. The ranking includes companies that operate predominantly physical supply chains, and one of the criteria looks at the proportion of revenue related to physical product versus services or software, and these two companies fell below the cutoff (of 50%) for this revenue proportion.

10-35.

This answer depends on which companies the student selects.

10-36.

This answer depends on which company applications the student selects.

10-37.

Some advantages of blockchains include increased security for information sharing among supply chain members; faster financial transactions; more consistent and secure record keeping; transaction visibility among supply chain members; more accurate real time product tracking; improved document flow and customs brokerage at ports;

Solutions to Problems 10-1.

Inventory turns =

470 = 14.2 17.5 + 9.3 + 6.4

Days of supply =

33.2 = 25.94 days 1.28

10-2. Average aggregate value of inventory = $12,042,000 Inventory turns = Days of supply

= 5.3 = = = 69

10-3. Average aggregate value of inventory:

Raw material = 82, 500 WIP = 61, 300 Finished goods = 220, 000 363, 800

Inventory turns =

3, 700, 000 = 10.17 363,800

Days of supply =

363,800 = 35.88 ( 3, 700, 000 ) / ( 365)

10-4. Aggregate value of inventory: Raw materials = 817,500 WIP = 67,800 Finished goods = 64,000 949,300 17,500, 000 Inventory turns = = 18.4 949,300

Weeks of supply =

949,300 = 2.7 weeks (17,500, 000 ) / ( 50 )

10-5. Average aggregate value of inventory: Raw materials = 281,090

WIP = 3,435,000 Finished goods = 3,087,000 6,803,090 Inventory turns =

18,500, 000 = 2.72 6,803, 090

Days of supply =

6,803, 090 = 134.22 (18,500, 000 ) / ( 365)

10-6. (a)

Inventory turns Days of supply

1 8.0 45.7

2 8.3 44.1

Year 3 11.1 32.8

4 12.0 30.5

(b) The company’s supply chain performance has marginally improved. However, the number of turns still seems excessive for a computer firm. (c) For a BTO company the average work-in-process and finished goods inventory seems excessive; the company should work to reduce these levels, possibly by working with its suppliers.

10-7.

Inventory turns Weeks of supply

Supplier 1 22.6 2.3

Supplier 2 9.25 5.6

Supplier 1 has the better supply chain performance. Other factors that might influence Delph could be quality and delivery speed.

10-8.

219 = 7.3 30

Control limits using Z = 3.00: UCL = c + Z c = 7.3 + 3 7.3 = 7.3 + 8.11 = 15.41

LCL = c − Z c = 7.3 − 3 7.3 = 7.3 − 8.11  0 All the sample observations are within the control limits suggesting that the invoice errors are in control.

10-9.

255 = 12.75 20

UCL = c + Z c = 12.75 + 3 12.75 = 23.46

LCL = c − Z c = 12.75 − 3 12.75 = 2.04 All the sample observations are within the control limits suggesting that the delivery process is in control. 10-10. Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Proportion Defective .028 .044 .072 .034 .050 .082 .036 .038 .052 .056 .076 .048 .030 .024 .020 .032 .018 .042 .036 .024

421

( 20 )( 500 )

UCL = p + Z

421 = .0421 10, 000

p (1 − p ) n

= .0421 + 3.00

.0421(1 − .0421) 500

= .0421 + .027 = .069 LCL = p − Z

p (1 − p )

= .0421 − 3

.0421(1 − .0421) 500

= .0421 − .027 = .015

Samples 3 and 11 are above the upper control limit indicating the process may be out of control. 10-11. x 2.00 2.08 2.92 1.78 2.70 3.50 2.84 3.26 2.50 4.14 2.12 4.38 2.84 2.70 3.56 2.96 3.34 4.16 3.70 2.72 60.20

Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

 R = 63.3 = 3.17

 x = 60.20 = 3.01

R 2.3 2.6 2.7 1.9 3.2 5.0 2.2 4.6 1.3 3.5 3.0 4.0 3.3 1.1 5.6 3.1 6.1 2.4 2.5 2.9 63.3

R-chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = 2.11( 3.17 ) = 6.69 LCL = 0 R = 0 ( 3.17 ) = 0

There are no R values outside the control limits, which would suggest the process is in control.

x -chart

A2 = 0.58 UCL = x + A2 R = 3.01 + 0.58 ( 3.17 ) = 4.85 LCL = x − A2 R = 3.01 − 0.58 ( 3.17 ) = 1.17

There are no x values outside the control limits, which suggests the process is in control.

10-12. x = 3.17, R = 3.25, A2 ( n = 5 ) = 0.58

UCL = x + A2 R = 3.17 + ( 0.58)( 3.25 ) = 5.04 LCL = x − A2 R = 3.17 − 1.89 = 1.29 UCL = D4 R = 2.11( 3.25 ) = 6.86

LCL = D3 R = 0 ( 3.25 ) = 0

The process appears out of control for sample 10, however, this could be an aberration since there are no other apparent nonrandom patterns or out-of-control points. Thus, this point should probably be “thrown out” and a new control chart developed with the remaining eleven samples.

CASE PROBLEM SOLUTION: Somerset Furniture Company’s Global Supply Chain This case is based on a real furniture company the authors are familiar with. The company essentially reinvented itself from a furniture manufacturer to a supply chain management and distribution company. In doing so it outsourced its primary area of expertise, i.e., manufacturing, thereby losing direct control of the thing it knew most about, and adopted the role of supply chain manager, something it had little prior expertise with. The company suffered a number of inadequacies that left it ill prepared to assume its new role. The case as presented addresses many of the problems faced by companies that are confronted by a global supply chain, and provides students with a general platform for discussing supply chain problems and their remedy. Following is the variable timeline for product lead time: 1. Develop purchase order: 12 to 25 days 2. Process purchase order overseas: 10 to 20 days 3. Manufacturing process: 60 days 4. Transport from plants to ports: 1 to 14 days 5. Arrange for containers and paperwork: 5 to 10 days 6. Wait for containers: 1 to 7 days 7. Load containers: 3 to 6 days 8. Security checks: 7 to 21 days 9. Overseas shipment: 28 days 10. Clear customs and loading trucks: 7 to 14 days 11. Transport to warehouse: 1 to 3 days 12. Receive and warehouse: 1 to 30 days Total lead time: 119 to 238 days Following is a list of discussion points for this case: • The company identified a set of characteristics of their global supply chain that seem to be endemic to many companies, as follows: Lower product cost Greater information requirements Longer delivery times to customers Quality problems Decrease in forecast accuracy Increase in inventory Replacement parts and repair service challenges Complex multi-step, inter-modal logistics Clash of different business cultures Hiring challenges Longer product to cash cycle Collaboration with multiple business partners • Primary among its deficiencies was its lack of expertise in the information technology area and an insufficient computing system. Much of the variability in the company’s supply chain was attributable to information technology problems. After the company made new hires to acquire IT expertise, and upgraded their computing technology, they were able to successfully address many of their problems in logistics and order processing and fulfillment, including the reduction of system variability. • The company adopted better, more innovative forecasting methods, which improved their scheduling. • They contracted with an international trade logistics (ITL) company to improve their overseas logistics, which combined with their improved information technology capabilities, allowed them to directly monitor and

control overseas product movement. This reduced scheduling variability. • In the quality area they worked with their Chinese suppliers to improve quality management including training local plant managers in the use of quality control techniques. More quality auditors were hired to make more frequent visits to the plants, and quality was monitored more closely at various key points in the manufacturing process. In the case of the humidity problem the production of some lines was moved to other plants in China. • Suppliers were selected based on the following criteria: quality, capacity, delivery and service. • Many furniture products were redesigned with the objective of creating full container loads. It was discovered that removing even a few inches from the dimensions of a piece of furniture would allow additional pieces to be loaded into a container. • The company decided to keep its own spare parts inventory in the U.S., ordering spare parts from their Chinese suppliers when the product line is initially produced. This required a more accurate forecast of spare parts demand. However, improved product quality eventually reduced the need for spare parts. • To improve service the company determined required inventory levels based on customer demand with an objective of on-time delivery while at the same time seeking ways to reduce inventory levels. The reduction in system variability simultaneously worked to reduce inventory requirements and improve customer service. • The company implemented a warehouse management system (WMS) to improve its local distribution service to customers. • The company is also exploring moving some production to other countries to limit its vulnerability to anticipated restrictions on furniture imports from China. Also, the SARS epidemic has caused the company to reconsider sourcing its products from only one country.

11 Global Supply Chain Procurement and Distribution Answers to Questions 11-1. McDonalds: Distribution to McDonalds franchises takes place from its regional distribution centers, as well as directly from suppliers. 11-2. A few suppliers or carriers are easier to coordinate. If a company gives one or two suppliers all of their business they can be more demanding on quality and deliveries. Single-sourcing provides the supplier or carriers with economy of scale that enables them to reduce costs. 11-3. The strategic goals are low cost and customer service. Procured items must be delivered on time or the entire supply chain is delayed, creating late deliveries to customers. Erratic and poor quality supply can also increase costs. If facilities are not located properly it can delay product or service flow through the supply chain, and increase costs for longer deliveries. Inefficient transportation can also result in higher inventories to offset delays and raise costs, and, causes delayed delivery to customers. 11-4. Answer depends on the businesses selected. 11-5. The answer depends on example/business the student selects. One example is a grocery chain that ships foodstuffs from several warehouse/distribution centers to various stores. 11-6. Answer depends on the company the student selects. 11-7. Answer depends on the company the student selects. 11-8. Answer depends on the e-marketplace selected. 11-9. Answer depends on the transportation exchange selected. 11-10. Answer depends on the ERP provider selected. 11-11. Answer depends on the international logistics provider selected. 11-12. The answer depends on the article the student selects. 11-13. The answer depends on the article the student selects. 11-14. Companies like L.L. Bean and Sears use sophisticated IT systems and a network of distribution centers to supply customers and stores. One of their biggest problems is “reverse logistics,” i.e. what to do with items returned from the customer. 11-15. RFID allows items to be scanned and identified without direct sight lines which enables items to be checked much more rapidly. Since bills-of-lading must not be much more detailed than previously (listing virtually all items in a shipment) this can be accomplished much more quickly and thoroughly with RFID technology and the information can be transported via satellite. This allows for much quicker security approvals, thus allowing ships and planes to load and unload more quickly. The primary obstacle to the use of RFID is (currently) cost. 11-16. In vendor-managed inventory a supplier monitors and maintains inventory levels for its customer and has the authority to replenish when necessary. Postponement moves some of the final manufacturing steps like assembly into the warehouse or distribution center. Both concepts shift some area of supply chain responsibility away from the manufacturer or retailer to another party—a supplier or a distributor. 11-17. Cross-docking allows a company to move goods coming into a distribution center or warehouse directly to a shipping dock for outgoing orders without storing them first in a warehouse. This increases the speed with which goods flow through the supply chain and it reduces inventory and handling costs. 11-18. Universities have traditional suppliers for MRO goods like supplies, telephone and computer services, furniture, maintenance equipment and products, food and beverages, etc. They also have non-traditional suppliers like high schools that provide students. The producers are the teachers and other academic and

support services that educate the student and provide for daily student life and well-being. Distributors include placement services and companies and organizations students are hired by. Inventory exists for MRO goods and services. Students are also a form of inventory as are classrooms, dorm rooms, parking spaces, etc. Students as inventory incur a carrying cost. The longer a student remains in school the greater the cost to the institution or state. Not only is there a direct cost for maintaining a student, but a student who stays too long also takes up a spot for other incoming students, thus creating shortages of facilities like dorm rooms and classrooms, which in-turn require expenditures for capital expansion. 11-19. Answer depends on student responses. 11-20.

Using more energy efficient transport vehicles, improving transportation routes, reducing the use of packaging materials, and using less energy to heat, cool, and light warehouse and distribution facilities.

11-21. Depends on the article and company the student identifies. 11-22. Quality problems, transportation costs, communication problems, and intellectual property theft. 11-23. Answer depends on student responses. 11-24.

Spend analysis is the formal process of collecting, cleansing, classifying and analyzing procurement data in order to reduce procurement costs and improve the efficiency of the procurement process. It attempts to assess the who, what, when, where and how of a company’s expenditure process and determine how much is being spent, with which suppliers and is the promised value being realized. It is typically conducted by an automated system (i.e., software) or by a third party provider.

11-25.

China has low labor costs and a large labor force and it is a future emerging market, however, it also has an unreliable and often inadequate infrastructure and transportation and distribution system, a restrictive government environment, and problems with quality. Mexico has higher wage rates than China but it is close to the U.S. thus resulting in lower transportation and distribution costs, and because of its nearness supply flow is more reliable. Since many Americans speak Spanish and many Mexicans speak English, language is less of a barrier to effective supply chain management, and cultural differences are less pronounced. Brazil has recently emerged as a favored supply source because of its closeness to the U.S., its low energy costs and favorable climate, and its relatively low labor costs, however it has limited industrial capacity and an inferior infrastructure, and a labor force limited in technical skills.

11-26.

Here is what the 2012 Gartner Top 25 report said about the three companies: “Maintaining its record in the No. 1 slot is Apple, a master at delivering total solutions to its customers through tightly integrated design of hardware components, firmware, a proprietary operating system and an ecosystem of applications that run on top of that platform. Stellar financials, which further improved this year, supported by the highest voting scores point to its combination of operational and innovation excellence, a zealous focus on starting with the consumer experience and working back through the design of its supply network, and mastery in orchestrating its end-to-end value network.” “Jumping three spots to No. 2 this year, Amazon is a great example of an "orchestrator" that goes beyond simply borrowing and adapting others' best practices and consistently defies conventional wisdom. With a three-year weighted revenue growth of 38%, the online retailing juggernaut has led in offering services where the profit potential was uncertain, such as cloud computing services, intraday delivery for major markets and use of pickup lockers at selected 7-Eleven stores, all requiring robust demand management and supply delivery capabilities. Moving into the media tablet business with the Kindle Fire, Amazon has shifted its model so that nine of its top 10 offerings are now digital content, augmenting its vast physical supply chain.” “At No. 5 is P&G, a perennial winner in our Supply Chain Top 25, a supply chain thought leader and a standard bearer of brand management. Its ability to optimize decisions across the supply network allows it to orchestrate demand and connect the supply chain to the shelf. Using its world-class, open innovation

platform, combined with an impressive new product organizational capability that is tightly integrated with supply chain, it taps a deep well of understanding its consumers' "moments of truth" to continue to deliver new products, such as the highly popular Tide Pods.” The report further noted four key among the leaders: a return to growth, continued focus on supply chain resiliency, simplification and "multilocal" operations. The three companies differ in that Apple manufactures many of its products, Amazon purchases finished goods, which it resells (i.e., it’s a marketplace), and P&G also manufactures its products that it sells to retailers. All three have an extensive network of suppliers in China and thus experience similar supply chain problems. 11-27.

Workplace sustainability issues are a constant problem for many of Wal-Mart’s Chinese suppliers. Excessive working hours, poor workplace environmental conditions, and substandard living conditions are still pervasive, and expose Wal-Mart to potential media scrutiny. In addition, environmental pollution problems are also a source of concern.

11-28.

This answer depends on the companies the student selects for each mode of transportation.

11-29.

This answer depends on the 3PL company the student selects. A list of the “Top 100 3PL Providers” can be found at ww.inboundlogistics.com (http://www.inboundlogistics.com/cms/top-100-3pls/).

11-30.

Among the numerous problems that can be encountered in a global supply chain, prominent is the loss of direct managerial control, which can detrimentally affect product quality. Another major problem is unreliable transportation and distribution, and excessive lead times for product delivery. Supply chain risk, in terms of major disruptions in supply due to natural or man-made disasters can be the source of numerous problems.

11-31.

The WTO Web site at www.wto.org has links on its home page that answers the questions, “Who we are,” and “What we do.” Quoting the site’s response to “who we are”: “At its heart are the WTO agreements, negotiated and signed by the bulk of the world’s trading nations. These documents provide the legal ground rules for international commerce. They are essentially contracts, binding governments to keep their trade policies within agreed limits. Although negotiated and signed by governments, the goal is to help producers of goods and services, exporters, and importers conduct their business, while allowing governments to meet social and environmental objectives. The system’s overriding purpose is to help trade flow as freely as possible — so long as there are no undesirable side effects — because this is important for economic development and well-being. That partly means removing obstacles. It also means ensuring that individuals, companies and governments know what the trade rules are around the world, and giving them the confidence that there will be no sudden changes of policy. In other words, the rules have to be ‘transparent’ and predictable.” Quoting “what we do”: While the WTO is driven by its member states, it could not function without its Secretariat to coordinate the activities. The Secretariat employs over 600 staff, and its experts — lawyers, economists, statisticians and communications experts — assist WTO members on a daily basis to ensure, among other things, that negotiations progress smoothly, and that the rules of international trade are correctly applied and enforced.” Further, the WTO engages in trade negotiations, implementation and monitoring, dispute settlement, building trade capacity, and outreach.

11-32.

The ten benefits listed on the WTO Webs site are:

1. 2. 3. 4.

The system helps promote peace Disputes are handled constructively Rules make life easier for all Freer trade cuts the costs of living

5. 6. 7. 8. 9. 10.

It provides more choice of products and qualities Trade raises incomes Trade stimulates economic growth The basic principles make life more efficient Governments are shielded from lobbying The system encourages good government

Each of these is explained in greater detail on the WTO Website. 11-33.

A tariff is either (1) a tax on imports or exports (trade tariff) in and out of a country, or (2) a list or schedule of prices for such things as rail service, bus routes, and electrical usage. A duty is a kind of tax, often associated with customs, a payment due to the revenue of a state, levied by force of law. It is a tax on certain items purchased abroad. A duty differs from a tax in that it is levied on specific commodities, financial transactions, estates, etc., and not on individuals. Landed cost is the total cost of a product once it has arrived at the buyer’s door. The list of components that are needed to determine landed costs include the original cost of the item, all brokerage and logistics fees, complete shipping costs, customs duties, tariffs, taxes, insurance, currency conversion, crating costs, and handling fees. Not all of these components are present in every shipment, but all that are must be considered part of the landed cost. A value added tax (VAT) is a form of consumption tax. From the perspective of the buyer, it is a tax on the purchase price. From that of the seller, it is a tax only on the value added to a product, material, or service, from an accounting point of view, by this stage of its manufacture or distribution. The manufacturer remits to the government the difference between these two amounts, and retains the rest for themselves to offset the taxes they had previously paid on the inputs. A VAT is like a sales tax in that ultimately only the end consumer is taxed. It differs from the sales tax in that, with the latter, the tax is collected and remitted to the government only once, at the point of purchase by the end consumer. With the VAT, collections, remittances to the government, and credits for taxes already paid occur each time a business in the supply chain purchases products.

11-34.

On the WTO Web site home page there is a link to “”Documents and resources,” and from that link the student can access the “Statistics database,” and from that they can access “Trade Profiles” of individual countries. The students’ answer depends on which country they select.

11-35.

From the EU Web site at http://europa.eu: “The EU is a unique economic and political partnership between 27 European countries that together cover much of the continent. It was created in the aftermath of the Second World War. The first steps were to foster economic cooperation: the idea being that countries who trade with one another become economically interdependent and so more likely to avoid conflict. Since then, the EU has developed into a huge single market with the euro as its common currency. What began as a purely economic union has evolved into an organization spanning all policy areas, from development aid to environment. It has delivered half a century of peace, stability, and prosperity, helped raise living standards, and launched a single European currency. Thanks to the abolition of border controls between EU countries, people can travel freely throughout most of the continent. And it's also become much easier to live and work abroad in Europe. The EU is based on the rule of law. This means that everything that it does is founded on treaties, voluntarily and democratically agreed by all member countries. These binding agreements set out the EU's goals in its many areas of activity…. The single market is the EU's main economic engine, enabling most goods, services, money and people to move freely. Another key objective is to develop this huge resource to ensure that Europeans can draw the maximum benefit.” A more extensive description of the EU’s history, membership, operation, processes and organization can be found at the website.

11-36.

This answer depends on the international trade specialist the student selects.

11-37.

Intermodal shipping involves the transportation of freight in an intermodal container or vehicle, using various modes of transportation - rail, ship, and truck - without any handling of the freight itself when changing modes. The method reduces cargo handling, and improves security, reduces damage and loss, and allows freight to be transported faster. Reduced costs over road trucking is the key benefit for intercontinental use, which may be offset by reduced timings for road transport over shorter distances. Containers, also known as intermodal containers or ISO containers because the dimensions have been defined by ISO, are the main type of equipment used in intermodal transport, particularly when one of the modes of transportation is by ship. Containers are 8-foot (2.4 m) wide by 8-foot (2.4 m) high. And the most common lengths are 20 feet (6.1 m). Container variations exist, including open-topped versions covered by a fabric curtain are used to transport larger loads. A container called a tanktainer, with a tank inside a standard container frame, carries liquids. Refrigerated containers are used for perishables. Container ships are used to transport containers by sea, some of which can hold thousands of containers. Their capacity is often measured in TEU (twenty-foot equivalent unit), or FEU (forty-foot equivalent unit). For example, a vessel that can hold 1,000 40-foot containers or 2,000 20-foot containers can be said to have a capacity of 2,000 TEU. Ships are typically stacked up to seven units high. It costs around $5000 to ship a container from China to the US subject to the price of oil. Once container ships reach their destination ports of entry, the containers are transported to distribution points or final destinations via rail, truck or barge. The largest shipping liner company in terms of TEU capacity is A.P. Moeller-Maersk Group. Containers are often shipped by rail in container well cars. These cars resemble flatcars but the have a container-sized depression, or well, that provides sufficient clearance to allow two containers to be loaded in the car in a "double stack" arrangement. It is also common in North America to transport semitrailers on railway flatcars or spine cars, an arrangement called "piggyback". Trucking is frequently used to connect the ocean and rail segments of a global intermodal freight movement. This specialized trucking that runs between ocean ports, rail terminals, and inland shipping docks, is often called drayage, and is typically provided by dedicated drayage companies or by the railroads.

11-38.

Gartner’s 2012 “Top 25” Report describes the following (paraphrased) “trends’ in global supply chain management: Supply Chain Risk Management and Resilience: Surviving and Thriving Many companies are looking to improve the resiliency of their supply chains to mitigate this risk. The past year (2012) brought global-scale supply chain disruptions that impacted multiple industries, from chemicals to semiconductors and electronics to automotive. Increased demand uncertainty and more complex global supply networks dependent on high-risk geographic zones placed additional pressures on the ability of supply chains to deliver predictable results. These disruptions called into question whether supply chains have become too lean, requiring a fundamental change in approach. In turbulent times, and in the face of growing complexity and risk, leading companies need sustainable, resilient supply chains that support profitability and drive industry leadership. This requires managers to re-evaluate the layout of their supply network designs to make them more resilient to future catastrophes. It may also include designing products that allow more flexibility in supply and manufacturing, increasing long-term alternative sources of raw materials and logistics capabilities, and expanding outsourced manufacturing capacity. Finally, with recent crises fresh in management mind share, now is a prime opportunity to push for more robust and funded risk management, including a "sense and respond" capability to recover quickly and profitably from disruptions. Many of the Supply Chain Top 25 companies were impacted by natural disasters, such as the Japanese earthquake and tsunami, and the massive flooding in Thailand. Leading companies such as Intel, P&G and Unilever improved multitier supply chain visibility and advanced network management capabilities to be agile in the face of disruptions. Overall, leaders focused on building resiliencies, and structural changes to streamline the flow of supply, and eliminate less profitable product and portfolio complexity. Simplification Supply chain leaders are adopting complexity optimization strategies to infrequently used product features, service offerings, suppliers and distribution network capacity that does not add sufficient value to

customers. Supply chain segmentation has emerged as a critical enabler of supply chain simplification, delivering only the level of service required by each customer type and nothing more. Alternatively, delivering on open-ended service requirements from a one-size-fits-all supply chain drives increased complexity and inefficiency. The concept of segmenting end-to-end supply chains to address customer-driven needs, such as cost efficiency, personalization and speed to market, has been around for several years. A Shift Toward Multilocal Operations Manufacturers and retailers have long sought ways to balance the trade-off in their supply network designs between global economies of scale and the demand for local responsiveness. Leading companies are reassessing their sourcing and manufacturing networks, and rebalancing their supply network strategies in favor of multilocal design, supply and support. Specifically, they are shifting from a centralized model, where these functions support global markets, to a regionalized approach, where capabilities are placed locally, but architected globally. Several factors are driving this trend. Tax and other government incentives, coupled with meaningful concessions from organized labor, are enticing manufacturers to set up or expand operations in mature markets. Annual wage increases between 9% and 35% in China, combined with rising logistics expenses, are leading to higher core supply chain costs in a traditionally low-cost country. An ever-increasing demand to be responsive to local markets is further fueling the trend, and a growing sophistication with techniques, such as cost-to- serve analysis, is enabling it. Even within emerging markets, manufacturers are shifting capacity based on regional wage and logistics expense differentials. 11-39.

There are a number of Internet sites that provides lists of the top global ports, so answers could vary. In general, Shanghai, Singapore, Hong Kong, Rotterdam, Kobe, Dover, Shenzhen, Ningbo-Zhoushan, Tianjin, Guangzhou, Qingdao, and Busan are included on most lists.

CASE PROBLEM 11.1 - Somerset Furniture Revisited Following is the variable timeline for product lead time: 1. Develop purchase order: 12 to 25 days 2. Process purchase order overseas: 10 to 20 days 3. Manufacturing process: 60 days 4. Transport from plants to ports: 1 to 14 days* 5. Arrange for containers and paperwork: 5 to 10 days* 6. Wait for containers: 1 to 7 days* 7. Load containers: 3 to 6 days* 8. Security checks: 7 to 21 days* 9. Overseas shipment: 28 days 10. Clear customs and loading trucks: 7 to 14 days* 11. Transport to warehouse: 1 to 3 days* 12. Receive and warehouse: 1 to 30 days Total lead time: 119 to 238 days This case is based on an actual company. The company contracted with an international trade logistics (ITL) company to improve their overseas logistics, which combined with their improved information technology capabilities, allowed them to directly monitor and control overseas product movement. This reduced scheduling variability. Transportation within China is primarily by truck, although rail is sometimes used. Poor infrastructure means shipments can be delayed or damaged. Transportation to the U.S. is by ship. Delays are primarily at the sending and receiving ports. Transportation within the U.S. is predominantly by truck, although rail is sometimes used. Shipments are sent directly to dealers, stores, and customers. Few shipments are sent to Somerset directly.

Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful.

Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. *

SUPPLEMENT 11 Operational Decision-Making Tools: Transportation and Transshipment Models Answers to Problems S11-1.a. Bethlehem − Detroit = 130 Bethlehem − St. Louis = 50 Birmingham − Detroit = 30

Birmingham − Norfolk = 220 Gary − St. Louis = 70 Gary − Chicago = 170 Z = $84,100 b. No effect. The Gary Mill has 70 tons left over as surplus; reducing the capacity at Gary to 260 tons still leaves 20 tons. S11-2.

Tampa − New York = 160 Tampa − Chicago = 130 Miami − Philadelphia = 140 Fresno − Philadelphia = 70 Fresno − Boston = 180 Z = $6, 220

S11-3.

There is no effect

S11-4.

Phoenix − Louisville = 10 Seattle − Louisville = 5 St. Louis − Charlotte = 15 Detroit − Greensboro = 10 Detroit − Charlotte = 5 Z = $215,000

S11-5 a. Atlanta − New Orleans = 45 Atlanta − Buffalo = 25 St. Louis − Buffalo = 25 St. Louis − Dallas = 90 Charlotte − New Orleans = 35 Charlotte − Pittsburgh = 25

Z = $17,750

b. Atlanta − Buffalo = 25 Atlanta − Pittsburgh = 25 St. Louis − Buffalo = 25 St. Louis − Dallas = 90 Z = $10,500 S11-6. WV, KY, VA to Norfolk = 180 tons IL, IN to Mobile = 130 tons MO, IA, NE to New Orleans = 90 tons Z = $4,460 S11-7. a. Tampa − GA = 2100 Tampa − SC = 1050 St. Louis − TN = 950 St. Louis − NC = 1700 St. Louis − KY = 2350 Milwaukee − TN = 850 Milwaukee − VA = 1400 Z = $3,635 b. No change; the total minimum cost remains $3,635 although are multiple optimal solutions (routes). S11-8.

Richmond − A & M = 280 Richmond − Central = 140 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 140

D.C. − Central = 200 Z = 21, 720 S11-9.

Alternative 1: Richmond − A & M = 280 Richmond − Central = 320 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 320 D.C. − Central = 20 Z = $28,920

Alternative 2: Richmond − A & M = 280 Richmond − Central = 60 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 340 Charlotte − State = 20 Charlotte − Central = 28

Z = $25,600 Select alternative 2; add a warehouse at Charlotte. S11-10. 1 − B = 60 2 − A = 55 2 − B = 10 2 − C = 40 3 − B = 10 Z = 1660 units

S11-11. 1 − 1 = 27 1− 2 = 5 1 − 4 = 10 2 − 2 = 21 3 − 3 = 18 4 − 4 = 23 5−4 = 4 5 − 5 = 31 6−3 = 4 6−4 = 7 6 − 6 = 16 Z = 467 miles

S11-12. Charlotte − Atlanta = 30 Memphis − St. Louis = 30 Louisville − NY = 30

Z = $159,000 S11-13. A − 3 = 1 C −1 = 1 D −1 = 1 E −3 =1 F−2 =1 G −2 =1 H −1 = 1 Z = 1, 070 Multiple optimal solutions exist.

S11-14. 1 − Bloomington = 1 2 − Columbus = 1 3 − Bloomington = 1 4 − Columbus = 1 5 − Madison = 1 6 − Madison = 1 7 − Columbus = 1 8 − Madison = 1

9 − Bloomington = 1 Z = 28.5 hours

S11-15. North − Addison = 270 South − Beeks = 120 South − Canfield = 190 East − Addison = 130 East − Canfield = 190 West − Daley = 220 Central − Beeks = 280

Z = 20,550 minutes (multiple optimal solutions) S11-16. North − Addison = 270 South − Beeks = 200 South − Canfield = 110 East − Addison = 80 East − Canfield = 240 West − Daley = 220 Central − Beeks = 150 Central − Daley = 130 Optimal. Total travel time = Z = 21,050 minutes. The overall travel time increased by 500 minutes, which divided by all 1,400 students is only an increase of 0.357 minutes per student. This does not seem to be a significantly large increase. S11-17.

Optimal, Total Profit = $1,528 with multiple optimal solutions at (B, 2), (E, 4) and (B, Dummy)

S11-18. If Atlantic 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 S11-17. 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 S11-17. 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 is shown as follows.

Total profit = $1,545 with multiple optimal solutions S11-19. 1 − E = 8 2 − C = 12 3− E = 2 4−D =8 5−A = 9 6−B = 6 Z = 1, 263 hours S11-20. GA1 − NCSW = 2 GA1 − NCW = 2 GA1 − VASW = 10 GA2 − NCSW = 3 GA2 − VAC = 9 SC1 − NCSW = 2 SC1 − NCP = 6 SC1 − VAT = 7 FL1 − NCE = 9 FL1 − NCW = 6 Z = $806,000 S11-21. Sacremento − St. Paul = 15 Sacremento − Topeka = 1 Bakersfield − Denver = 10 San Antonio − Topeka = 12 Montgomery − Denver = 10 Jacksonville − Akron = 15 Jacksonville − Topeka = 7 Ocala − Louisville = 15 Z = $276, 200 It is cheaper for TransAm Foods to continue to operate it own trucking firm.

S11-22. 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. 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. S11-23. L.A. − Singapore = 150

L.A. − Taipei = 300 Savannah − Hong Kong = 400 Savannah − Taipei = 200 Galveston − Singapore = 350 Shortages − Hong Kong = 200

Z = $723,580 z=$723500 Penalty costs = 200  $800 = $160,000 S11-24. x1C (New Orleans – Nashville) = 72 x1D (New Orleans – Louisville) = 13 x2D (Jackson – Louisville) = 76 x2I (Jackson – Wilmington) = 34 x3A (Savannah – Atlanta) = 34 x3B (Savannah – Jacksonville) = 39 x4B (Mobile – Jacksonville) = 45 x5A (Birmingham – Atlanta) = 29 x6F (Houston – Roanoke) = 96 x6G (Houston – Norfolk) = 8 x6H (Houston – Memphis) = 8 x7I (Charleston – Wilmington) = 88 x8B (Shreveport – Jacksonville) = 18 x8D (Shreveport – Louisville) = 27 x8E (Shreveport – Richmond) = 41 x8H (Shreveport – Memphis) = 57

Z = $ 15618

S11-25. a. x1G (Truck 1 – Customer G) = 1 x2B (Truck 2 – Customer B) = 1 x3H (Truck 3 – Customer H) = 1 x4L (Truck 4 – Customer L) = 1 x5K (Truck 5 – Customer K) = 1 x6F (Truck 6 – Customer F) = 1 x7D (Truck 7 – Customer D) = 1 x8A (Truck 8 – Customer A) = 1 Z = $4,260 b. x1G (Truck 1 – Customer G) = 1 x2F (Truck 2 – Customer F) = 1 x3D (Truck 3 – Customer D) = 1 x4L (Truck 4 – Customer L) = 1 x5C (Truck 5 – Customer C) = 1 x6E (Truck 6 – Customer E) = 1 x7J (Truck 7 – Customer J) = 1 x8A (Truck 8 – Customer A) = 1 Z = $4,840 Average capacity = 87.3%

S11-26. The variables for the amounts shipped from each prospective farm to each plant are non-integer, defined as follows: xij = potatoes (tons, 1,000s) shipped from farm i to plant j, where i = 1, 2, 3, 4, 5, 6 and j = A, B, C The objective function (Z) combines shipping costs and the annual fixed costs, where Z = $1,000s: Minimize Z = 18x1A + 15x1B +12x1C +13x2A + 10x2B + 17x2C + 16x3A + 14x3B + 18x3C + 19x4A + 15x4B + 16x4C +17x5A + 19x5B + 12x5C + 14x6A + 16x6B + 12x6C + 405y1 + 390y2 + 450y3 + 368y4 + 520y5 + 465y6 The constraints for production capacity require that potatoes be shipped (i.e., xij variables have positive values) only from a farm if it is selected; for example, for farm 1. x1A + x1B + x1C ≤ 11.2y1 In this constraint, if y1 = 1, then 11.2 thousand tons of potatoes are available at farm 1 for shipment to one or more of the plants. Similar constraints are developed for each of the five other farms. Constraints are also necessary for the production capacity to be filled at each plant. The complete model is formulated as; Minimize Z = 18x1A + 15x1B +12x1C +13x2A + 10x2B + 17x2C + 16x3A + 14x3B + 18x3C + 19x4A + 15x4B + 16x4C +17x5A + 19x5B + 12x5C + 14x6A + 16x6B + 12x6C + 405y1 + 390y2 + 450y3 + 368y4 + 520y5 + 465y6 subject to x1A + x1B + x1C – 11.2y1 ≤ 0 x2A + x2B + x2C – 10.5y2 ≤ 0 x3A + x3B + x3C -12.8y3 ≤ 0 x4A + x4B + x4C – 9.3y4 ≤ 0 x5A + x5B + x5C – 10.8y5 ≤ 0 x6A + x6B + x6C – 9.6y6 ≤ 0 x1A + x2A + x3A + x4A + x5A+x6A = 12 x1B + x2B + x3B + x4B + x5B + x6B = 10 x1C + x2C + x3C + x4C + x5C + x6C = 14 xij ≥ 0 yi = 0 or 1

Solution: x1C = 11,200; x2A = 2,400; x2B = 8,100; x4B = 1,900; x4C = 2,800; x6A = 9,600; y1 = 1 (farm 1), y2 = 1 (farm 2), y4 = 1 (farm 4), y6 = 1 (farm 6); Z = $2,082,300

S11-27. Louisiana – Charleston = 12,000 bales Mississippi – Savannah = 19,000 bales Mississippi – New Orleans = 14,000 bales Texas – Houston = 26,000 bales Z = $1,197,000 S11-28. Houston – Singapore = 26,000 bales Savannah – Singapore= 8,000 bales Savannah – Karachi = 10,000 bales Savannah – Mumbai = 1,000 bales New Orleans – Mumbai = 14,000 bales Charleston – Karachi = 12,000 bales Z = $1,748,000 S11-29. Singapore – China = 5,425,000 yards Singapore– Japan = 1,675,000 yards Singapore– Turkey = 3,950,000 yards Karachi – India = 6,350,000 yards Karachi – Turkey = 800,000 yards Mumbai – Japan = 1,775,000 yards Mumbai – Italy = 3,100,000 yards Z = $882,950 S11-30. China – New York = 3,149,000 China – New Orleans -= 468,000 India – New York = 3,978,000 India – Marseilles = 255,000 Japan – Liverpool = 368,000 Japan – New Orleans = 1,932,000 Turkey – Marseilles = 3,167,000 Italy – Liverpool = 2,067,000 Z = $6,850,280 S11-31.

x17 (Brazil – Mobile) = .3 x19 (Brazil – Savannah) = 5.6 x28 (Columbia – New Orleans) = 4.3 x38 (Indonesia – New Orleans) = 1.2 x3,10 (Indonesia – Jacksonville) = 2.6 x47 (Kenya – Mobile) = 2.8 x4,10 (Kenya – Jacksonville) = 3.9 x59 (Cote d’Ivoire – Savannah) = 2.5 x68 (Guatemala – New Orleans) = 4.8 x7,11 (Mobile – New York) = 3.1 x8,11 (New Orleans – New York) = 10.3 x9,11 (Savannah – New York) = 8.1 x10,11 (Jacksonville – New York) = 6.5 Z = $2,206,000

S11-32. x14 ( Hamburg − Norfolk ) = 42

x59 ( NY − Chicago ) = 50 x26 ( Marseilles − Savannah ) = 63

x35 ( Liverpool − NY ) = 37 x48 ( Norfolk − St. Louis ) = 42 x15 ( Hamburg − NY ) = 13

x67 ( Savannah − Dallas ) = 60 x68 ( Savannah − St. Louis ) = 3

Z = $77,362

HND = 38

HNS = 17 MSD = 22

S11-33. x16 ( Mexico − Houston ) = 18

x24 ( Puerto Rico − Miami ) = 11

x34 ( Haiti − Miami ) = 23 x47 ( Miami − NY ) = 20

x48 ( Miami − St. Louis ) = 12 x49 ( Miami − LA ) = 2 x69 ( Houston − LA ) = 18 Z = $479 or $479,000

S11-34. (a)

x14 = 72 x25 = 105 x34 = 83

x46 = 75 x47 = 80 x56 = 15 x58 = 90

Z = $10,043,000

(b)

Adding a capacity constraint at plants in Indiana and Georgia

x14 = 72 x25 = 105 x34 = 48 x35 = 35

x46 = 40 x47 = 80 x56 = 50 x58 = 90

Z = $10,043,000 S11-35.

x1C = 70

xBA = 10

x2B = 80

xCB = 30

x3A = 50

Z = 1, 490 or $14,900

S11-36. x37 ( Italy − Texas ) = 2.1 x15 ( Germany − Mexico ) = 5.2

x26 ( Belgium − Panama ) = 6.3 x59 ( Mexico − Ohio ) = 5.2

x68 ( Panama − Virginia ) = 3.7 x69 ( Panama − Ohio ) = 2.6 Z = $27.12 million

S11-37. 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 = 1.09 x16 + 1.26 x17 + .89 x25 + 1.32 x26 + 1.17 x27 + .78 x35 + 1.22 x36 + 1.36 x37 + 1.19 x45 +1.25 x46

+ 1.42 x47 + 4.56 x58 + 3.98x59 + 4.94 x510 + 3.43x68 + 5.74 x69 + 4.65x610 + 5.01x611 + 5.39x78 + 6.35x79 + 5.70 x710 + 4.87x711

Subject to

x16 + x17  1, 600 x25 + x26 + x27  1,100 x35 + x36 + x37  1, 400 x45 + x46 + x47  1900 x25 + x35 + x45  1,800 x16 + x26 + x36 + x 46  2, 200 x17 + x27 + x37 + x 47  1, 600 x58 + x68 + x78 = 1, 200 x59 + x69 + x79 = 900 x510 + x610 + x710 = 1,100 x611 + x711 = 1,500

x25 + x35 + x45 = x58 + x59 + x510 x16 + x26 + x36 + x 46 = x68 + x69 + x610 + x 611 x17 + x27 + x37 + x 47 = x78 + x79 + x710 + x 711 xij  0 Solution:

x16 (Idaho − Kansas) = 1, 600 x27 (Nebraska − Arkansas) = 1,100 x35 (South Dakota − Indiana) = 1, 400 x46 (Michigan − Kansas) = 600 x59 (Indiana − Ohio) = 900 x510 (Indiana − South Carolina) = 500 x68 (Kansas − Maryland) = 1, 200 x610 Kansas − South Carolina) = 600 x611 (Kansas − Alabama) = 400 x711 = 1,100

Z = $25,192

S11-38. 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 + 1,800 x15 + 2,345x16 + 2,700 x17 + 1,825x24 + 1,750x25 + 1,945x26

+ 2,320 x27 + 2,060 x34 + 2,175x35 + 2,050 x36 + 2, 475x37 + 825x48 + 545x49 + 320 x410 + 750 x58 + 675x59 + 450 x510 + 325x68 + 605x69 + 690 x610 + 270 x78 + 510 x79 + 1,050 x710 + 450 x811 + 830 x812 + 565 x813 + 420 x814 + 960 x815 + 880 x911 + 520 x912 + 450 x913 + 380x914 + 660x915 + 1,350x1011 + 390 x1012 + 1, 200 x1013 + 450 x1014 + 310 x1015 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 x410 + x510 + x610 + x710  140 x811 + x911 + x1011 = 85 x812 + x912 + x1012 = 60 x813 + x913 + x1013 = 105 x814 + x914 + x1014 = 50 x815 + x915 + x1015 = 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 (Antwerp – Boston) = 85 x15 (Antwerp – Savannah) = 40 x25 (Barcelona – Savannah) = 70 x26 (Barcelona – Mobile) = 15 x27 (Barcelona – Houston) = 125 x36 (Cherbourg – Mobile) = 85 x410 (Boston – NC) = 85 x59 (Savannah – Texas) = 55 x510 (Savannah – NC) = 55 x68 (Mobile – Ohio) = 100 x78 (Houston – Ohio) = 70 x79 (Houston – Texas) = 55 x811 (Ohio – Phoenix) = 85 x813 (Ohio – KC) = 35 x814 (Ohio – Louisville) = 50 x912 (Texas – Columbus) = 40 x913 (Texas – KC) = 70 x1012 (NC – Phoenix) = 20 x1015 (NC – Memphis) = 120 Z = $1,179,400

S11-39. (a)

The model is the same as problem #S11-32 except the available shipments from the European ports are 5 each:

x14 + x15 + x16 + x17  5 x24 + x25 + x26 + x27  .5 x34 + x35 + x36 + x37  5 and the demand constraints for the U.S. distributor 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 Phoenix: Barcelona – Boston – Texas (31) Columbus: Barcelona – Boston – North Carolina (28) Kansas City: Barcelona – Boston – Texas (29) Louisville: Barcelona – Boston – Texas (27) Memphis: Barcelona – Boston – Texas (29)

(b)

x14 = 2 x24 = 3 x49 = 2 x410 = 3 x911 = 1 x913 = 1 x1012 = 1 x1014 = 1 x1015 = 1 Z = 154 days

S11-40. Minimize Z = ∑ cij xij + ∑ cjk xjk + ∑vi yj 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) xij = containers shipped from Asian port “i” to distribution center “j” 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) xjk = containers shipped from distribution center “j” to U.S. port “k” vj = cost of distribution center “j” yj = 0 if distribution center “j” is selected, 1 if not selected subject to:

Solution: yH (Antwerp) = 1 yI (Bremen) = 1 yJ (Valencia) = 1 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

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

xEJ (Kaoshiung – Valencia) = 105

Z = $47,986,050 Shipping cost = $6,215,050

S11-41.

yF (Rotterdam) = 1 yH (Antwerp) = 1 yK (Lisbon) = 1 xAH (Hong Kong – Antwerp) = 235 xBK (Shanghai – Lisbon) = 170 xCK (Busan – Lisbon) = 165

xFM (Rotterdam – New York) = 395 xFO(Rotterdam - Miami) = 70 xHN (Antwerp - Savannah) = 305

xDF (Mumbai – Rotterdam) = 60

xHO (Antwerp - Miami) = 120

xDH (Mumbai – Antwerp) = 190

xKM (Lisbon – New York) = 45

xDK (Mumbai – Lisbon) = 75 xEF (Kaoshiung – Rotterdam) = 405 Z = total shipping cost = $5,843,715 Distribution center cost = $44,302,000 Total cost: $50,145,715 Shipping cost is reduced S11-42. Arkansas – Charleston = 12,000 bales Mississippi – Savannah = 19,000 bales Mississippi – New Orleans = 14,000 bales Texas – Houston = 26,000 bales Houston – Shanghai = 26,000 bales Savannah – Shanghai = 8,000 bales Savannah – Karachi = 10,000 bales Savannah – Saigon = 1,000 bales New Orleans – Saigon = 14,000 bales Charleston – Karachi = 12,000 bales Z = $2,945,000

xKP (Lisbon – New Orleans) = 365

CASE S11.1: 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

Kingsport → 3.

Macon = 19 bbls

Danville

→ B.

Los Canos = 80 bbls

Macon

→ C.

Duras = 78 bbls

Selma

→ 3.

Macon = 17 bbls

Selma

→ 5.

Columbus = 36 bbls

Columbus → A. Whitewater = 65 bbls

Allentown → 2.

Danville = 38 bbls

CASE S11.2: Global Supply Chain Management at Centrex International European Port to U.S. Port Lisbon − Jacksonville = 2062.5 lb. goat Lisbon − Savannah = 2000 lb. goat Lisbon − Jacksonville = 1350 lb. lamb

Marseilles − Savannah = 2500 lb. goat Marseilles − Savannah = 3000 lb. lamb

Caracas − New Orleans = 4200 lb. goat Caracas − Jacksonville = 1237.5 lb. goat Caracas − New Orleans = 2375 lb. lamb Caracas − Jacksonville = 475 lb. lamb

U.S. Port to Distribution Center New Orleans − TN = 4200 lb. goat New Orleans–TN = 2375 lb. lamb Jacksonville–OH = 3000 lb. goat

Jacksonville–NY = 300 lb. goat Jacksonville − OH = 1825 lb. lamb

Savannah − NY = 4500 lb. goat Savannah − OH = 125 lb. lamb Savannah − NY = 2875 lb. lamb

Tanning Factory to Plant Mende − Limoges = 4000 lb. goat Mende − Limoges = 4400 lb. lamb Foggia − Naples = 3000 lb. lamb Saragossa − Madrid = 6500 lb. goat Saragossa − Madrid = 1300 lb. lamb Feira − Sao Paulo = 5100 lb. goat El Tigre − Caracas = 3600 lb. goat El Tigre − Sao Paulo = 2500 lb. lamb El Tigre − Caracas = 3200 lb. lamb

Plant to Port Madrid − Lisbon = 4062.5 lb. goat Madrid − Lisbon = 650 lb. lamb

Naples – Marseilles 1500 lb. lamb

Limoges − Marseilles = 2500 lb. goat Limoges − Lisbon = 700 lb. lamb Limoges − Marseilles = 1500 lb. lamb Sao Paulo − Caracas = 3187.5 lb. goat Sao Paulo − Caracas = 1250 lb. lamb

Caracas – Caracas = 2250 lb. goat Caracas – Caracas = 1600 lb. lamb

Total cost = $606,965.63

12 Forecasting Answers to Questions 12-1.

Production planning, such as scheduling, inventory, process, facility layout and design, work force, and material purchasing, financial planning, including the development of budgets and capital expenditures; and various marketing functions are dependent on forecasting demand.

12-2.

Qualitative forecasting methods are subjective estimates based on judgment, opinion, past experience, and so on, whereas quantitative methods are mathematical, based on formulas.

12-3.

Short-range forecasts typically encompass the immediate future, in other words, several months, and are concerned with daily operations; medium-range forecasts encompass anywhere from several months up to several years and are used for annual budgets and production plans or the development of a project or program; long-range forecasts usually are for periods longer than one or two years and are used for strategic planning, such as new product development or new programs.

12-4.

All the elements of the supply chain including purchasing, inventory, production, scheduling, facility location, transportation and distribution are affected by forecasting. An inaccurate forecast can result in excessive costly inventories or frequent stockouts and late deliveries.

12-5.

Continuous replenishment requires that suppliers replenish a company’s inventory levels as products are 900.44 requires very accurate forecasts by suppliers to always be able to meet customer demand on very short notice. Without accurate inventories, suppliers must maintain high inventory levels themselves.

12-6.

Quality customer service means having products or services available when customers demand them, and, being able to deliver products and services on time. Without accurate forecasting of customer demand them is difficult to keep the appropriate amount of inventory on hand to meet demand in a timely manner without excessive costs.

12-7.

Qualitative methods are most often used for long-range strategic planning. Often called, “the jury of executive opinion” it uses judgment, expertise and opinion of knowledgeable people in a company. Other methods include consumer research, the Delphi method and consulting firms.

12-8.

The Delphi method uses the informed opinions, expertise and judgments of knowledgeable individuals and experts. A questionnaire is used to develop a consensus forecast of future trends and events. It’s especially useful for predicting technological advances.

12-9.

A trend is a gradual, long-term, up or down movement of demand; a cycle is an undulating up-and-down movement that repeats itself over a lengthy time span; a seasonal pattern is an oscillating movement in demand that occurs periodically and is repetitive.

12-10.

Exponential smoothing is a moving average that weights the most recent past data more strongly than more distant past data.

12-11.

Other ways used to obtain initial forecasts include taking an average of demand in preceding periods or making a subjective estimate. If forecasting has been a continual process, then preceding forecasts might be used.

12-12.

The higher the smoothing constant, the more sensitive the forecast will be to changes in recent demand.

12-13.

Adjusted exponential smoothing is the simple exponential smoothing forecast with a trend adjustment factor added to it.

12-14.

It is a judgmental choice, but in general, a high smoothing constant  reflects trend changes more than a lower 

12-15.

In a linear trend line equation, the independent variable, x, is always time.

12-16.

This question requires an opinion of the student, but in general, the appropriate model is determined

primarily by the extent of any trend pattern. 12-17.

A linear trend model will not adjust to a change in trend as the adjusted exponential smoothing model will, thus limiting the trend line method to a shorter time frame.

12-18.

By summing the differences between the actual forecast and demand; a large positive value indicates the forecast is probably consistently low, whereas a large negative value implies the forecast is consistently high.

12-19.

The movement of a tracking signal is compared to control limits, as long as the tracking signal is within the control limits, the forecast is in control and not biased.

12-20.

This question requires a subjective estimate on the part of the student. A particular method might be viewed as being superior because it is easier to use (compared to the other methods), it is easier to interpret, it makes more sense, it can be used alone rather than in comparison, or it seems to fit the data better.

12-21.

Linear regression relates demand to one other independent variable, whereas multiple regression reflects the relationship between a dependent variable and two or more independent variables.

12-22.

y is the dependent variable, x is the independent variable, a is the intercept, and b is the slope of the line.

12-23.

The Delphi method might be an appropriate method to use to forecast technological advances in video equipment, whereas market/consumer research could be used to forecast consumer demand. Various individuals in-house might also be able to assist in developing a forecast.

Solutions to Problems 12-1. a., b.

Month January February March April May June July August September October November December January c.

Sales 9.00 7.00 10.00 8.00 7.00 12.00 10.00 11.00 12.00 10.00 14.00 16.00 —

3-Month Moving Average — — — 8.67 8.33 8.33 9.00 9.67 11.00 11.00 11.00 12.00 13.33

5-Month Moving Average — — — — — 8.20 8.80 9.40 9.60 10.40 11.00 11.40 12.60

3-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.

12-2. a., b.

Month 1 2 3 4 5 6 7 8 9 c.

Sales 5 10 6 8 14 10 9 12 —

3-Month Moving Average — — — 7.00 8.00 9.33 10.67 11.00 10.33

Weighted 3-Month Moving Average — — — 17.20 7.58 11.06 11.08 9.93 10.77

3-month MAD = 2.07; weighted 3-month MAD = 2.49. The 3-month moving average forecast appears to be slightly more accurate.

12-3. a., b., and c.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13 d.

Demand 105.00 150.00 93.00 121.00 140.00 170.00 105.00 150.00 150.00 170.00 110.00 130.00 —

3-Quarter Moving Average Forecast — — — 116.00 121.33 118.00 143.67 138.33 141.67 135.00 156.67 143.33 136.67

Error — — — 5.00 18.67 52.00 –38.67 11.67 8.33 35.00 –46.67 –13.33 —

5-Quarter Moving Average — — — — — 121.80 134.80 125.80 137.20 143.00 149.00 137.00 142.00

Error — — — — — 48.20 –29.80 24.20 12.80 27.00 –39.00 –7.00 —

Weighted 3-Quarter Moving Average — — — 113.95 116.69 125.74 151.77 132.40 138.55 142.35 160.00 136.69 130.20

Error — — — 7.15 23.31 44.26 –46.77 17.60 11.45 27.65 –50.00 –6.60 —

Cumulative errors are: 3-quarter moving average, E = 32.0 5-quarter moving average, E = 36.4 Weighted 3-quarter moving average, E = 28.05 The weighted 3-quarter forecast appears to be the most accurate. All the forecasts exhibit a low bias. 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.

12-4. a., b.

Semester 1 2 3 4 5 6 7 8 9 c.

Enrollment 270 310 250 290 370 410 400 450 —

3-Semester Moving Average — — — 276.67 283.33 303.33 356.67 393.33 420.00

Exponentially Smoothed Forecast — 270.00 278.00 272.40 275.92 294.74 317.79 334.23 357.38

3-semester MAD = 61.33; exponentially smoothed MAD = 70.42; 3-semester moving average appears to be slightly more accurate.

12-5. a., b.

Month

Demand

Exponentially Smoothed Forecast (  = 0.30 )

October November December January February March April May June July August

800 725 630 500 645 690 730 810 1200 980 —

800.00 800.00 777.50 733.25 663.27 657.79 667.45 686.21 723.35 866.34 900.44

Adjusted Exponentially Smoothed Forecast (  = 0.30,  = 0.20 ) — 800.00 773.00 720.70 639.23 637.46 653.18 678.55 724.64 895.98 930.96

Exponentially smoothed MAPD = 1, 282.86 / 7,710 = 0.166 = 16.6%; MAPD = 1282.86/6910 = 18.6%

Adjusted forecast MAPD = 1, 264.59 / 7,710 = 0.1640 = 16.4%. MAPD = 1264.59/6910 = 18.3% Both forecasts appear to be approximately equally accurate. 12-6.

Month

Price

Exponentially Smoothed Forecast (  = 0.40 )

1 2 3 4 5 6 7 8 9 10 11

62.70 63.90 68.00 66.40 67.20 65.80 68.20 69.30 67.20 70.10 —

62.70 62.70 63.18 65.10 65.62 66.25 66.07 66.92 67.87 67.60 68.60

Cumulative Error MAD

Adjusted Exponentially Smoothed Forecast (  = 0.40,  = 0.30 )

Linear Trend Line y = 63.54 + 0.607x

— 62.70 63.32 65.78 66.25 66.88 66.46 67.45 68.53 67.98 69.17

64.15 64.75 65.36 65.97 66.57 67.18 67.79 68.39 69.01 69.61 70.22

Exponentially Smoothed

Adjusted Exponentially Smoothed

Linear Trend

14.75 1.89

10.73 1.72

— 1.09

The linear trend line forecast appears to be the most accurate.

12-7.

Year

Occupancy Rate

Exponentially Smoothed Forecast (  = 0.20 )

1 2 3 4 5 6 7 8 9 10

.75 .70 .72 .77 .83 .81 .86 .91 .87 —

.75 .75 .74 .74 .74 .76 .77 .79 .81 .82

E MAD

Exponentially Smoothed Forecast .046 .064

Adjusted Forecast .044 .061

Adjusted Exponentially Smoothed Forecast (  = 0.20,  = 0.20 )

Linear Trend Line y = .683 + .024x

— .75 .74 .73 .74 .76 .77 .80 .82 .83

.71 .73 .76 .78 .80 .83 .85 .88 .90 .92

Linear Trend Trend Forecast — 0.026

The linear trend line forecast appears to be the most accurate. 12-8. a.

3-month moving average forecast for month 21 = 74.67 MAD = 3.12

3-month weighted moving average forecast for month 21 = 75.875 MAD = 2.98

Exponentially smoothed forecast ( = 0.40 ) for month 21 = 74.60 MAD = 2.87 MAD = 2.99

12-9.

The lowest MAD values are with both the weighted 3-month moving average forecast and the exponentially smoothed forecast.

Group data into 3-month periods to forecast periods 19, 20 and 21. Possible models include the following: Exponential smoothing forecasts ( = 0.3) . F19, 20, 21 = 51.67, MAD = 18.93 Linear trend line forecasts

F19, 20, 21 = 69.27, MAD = 1.10

12-10. a.

Ice Cream Sales 350 510 750 420 370 480 860 500 450 550 820 570 —

Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13

Exponentially Smoothed Forecast (  = 0.50 ) 350.00 350.00 430.00 590.00 505.00 437.50 458.75 659.37 579.69 514.84 532.42 676.21 623.11

E = 26.30

E = 289.336 The forecast seems to be biased low. b.

Seasonal factors: Quarter 1: 1,170 / 6,630 = 0.18 Quarter 2: 1,540 / 6,630 = 0.23 Quarter 3: 2, 430 / 6,630 = 0.37 Quarter 4: 1, 490 / 6,630 = 0.22

Forecast for 2005: y = 1,850 + 180 x = 2,570 Seasonally adjusted forecasts: Quarter 1: 2,570 ( 0.18 ) = 453.53 Quarter 2: 2,570 ( 0.23) = 596.95 Quarter 3: 2,570 ( 0.37 ) = 941.95 Quarter 4: 2,570 ( 0.22 ) = 577.57 The seasonal factor seems to provide a more accurate forecast. 12-11. Seasonal factors: Quarter 1: 395/1,594 = 0.25 Quarter 2: 490 /1,594 = 0.31 Quarter 3: 308/1,594 = 0.19 Quarter 4: 401/1,594 = 0.25

Adjusted Exponentially Smoothed Forecast (  = 0.50,  = 0.50 ) — 350.00 470.00 690.00 512.50 407.50 454.37 757.50 588.91 487.03 527.30 745.55 631.22

Error — 160.00 280.00 –270.00 –142.50 72.50 405.62 –257.50 –138.91 62.97 292.69 –175.55 —

Forecast year 4: y = 440.33 + 45.5x = 622.33 Seasonally adjusted forecasts: Quarter 1: 622.33 ( 0.25 ) = 155.6 Quarter 2: 622.33 ( 0.31) = 192.9 Quarter 3: 622.33 ( 0.19 ) = 118.2 Quarter 4: 622.33 ( 0.25 ) = 155.6 12-12. Day 1 2 3 4 5 6 7 8

Daily Demand 212 182 215 201 158 176 212 188  D = 1,544

S1 (10 A.M. − 3 P.M.) = S2 ( 3 P.M. − 7 P.M.) =

D

567 = 0.37 1,544



S3 ( 7 P.M. − 11 P.M.) = S4 (11 P.M. − 2 A.M.) =

389 = 0.25 1,544

D

320 = 0.21 1,544

268 = 0.17 1,544

D

Linear trend forecast for day 9: y = 202.54 − 2.12 x = 183.46 Day 9 forecast for 10 A.M. − 3 P.M. : 183.46 ( 0.25 ) = 45.87

3 P.M. − 7 P.M. : 183.46 ( 0.37 ) = 67.88 7 P.M. − 11 P.M. : 183.46 ( 0.21) = 38.52 11 P.M. − 2 A.M. : 183.46 ( 0.17 ) = 31.18

12-13.

Year

Sales

1 2 3 4 5 6 7 8 9

4,260.00 4,510.00 4,050.00 3,720.00 3,900.00 3,470.00 2,890.00 3,100.00 —

MAD E

Adjusted Exponentially Smoothed Forecast 431.71 –2.522

Exponentially Smoothed Forecast (  = 0.30 ) — 4,260.00 4,335.00 4,249.50 4,090.65 4,033.45 3,864.42 3,572.09 3,430.37

Adjusted Exponentially Smoothed Forecast (  = 0.30,  = 0.20 ) — 4,260.00 4,350.00 4,244.40 4,054.80 3,993.34 3,798.52 3,460.91 3,313.19

Linear Trend Line 166.25

The linear trend line forecast appears to be the most accurate. 12-14. a.

Seasonally adjusted forecast

January − March :

D1 = 106.8

April − June :

D2 = 135.6

July − September :

D3 = 109.0

October − December : D4 = 233.6

106.8 = 0.18 585

D

135.6 = 0.23 585

D

109.0 = 0.19 585

233.6 = 0.40 585

s1 =



s2 = s3 = s4 =

D

Linear trend line forecast: y = 96.33 + 6.89 x = 137.67 January–March forecast: SF1 = S1 F6 = 0.18 (137.67 ) = 24.78 April–June forecast: SF2 = S2 F6 = 0.23 (137.67 ) = 31.66

Linear Trend Line y = 4,690 − 211.67x 4,478.33 4,266.67 4,055.00 3,843.33 3,631.67 3,420.00 3,208.33 2,996.67 2,785.00

July–September forecast: SF3 = S3 F6 = 0.19 (137.67 ) = 26.16 October–December forecast: SF4 = S4 F6 = 0.40 (137.67 ) = 55.07 b.

Linear trend line forecast for January–March 2005:

y = 16.29 + 1.69 x = 26.43 Linear trend line forecast for April–June 2005:

y = 21.75 + 1.79 x = 32.49 Linear trend line forecast for July–September 2005:

y = 18.35 + 1.15x = 25.25 Linear trend line forecast for October–December 2005:

y = 39.94 + 2.26 x = 53.50 c.

Year/Quarter

Orders

Seasonally Adjusted Forecast

Jan–Mar Apr–June Jul–Sep Oct–Dec Jan–Mar Apr–June Jul–Sep Oct–Dec Jan–Mar Apr–June Jul–Sep Oct–Dec Jan–Mar Apr–June July–Sep Oct–Dec Jan–Mar Apr–June July–Sep Oct–Dec

18.6 23.5 20.4 41.9 18.1 24.7 19.5 46.3 22.4 28.8 21.0 45.5 23.2 27.6 24.4 47.1 24.5 31.0 23.7 52.8

18.58 23.74 19.61 41.29 19.82 25.32 20.92 44.04 21.06 26.91 22.23 46.80 22.30 28.49 23.54 49.56 23.54 30.08 24.85 52.31

Seasonally adjusted forecast MAD =

Linear Trend Line Forecast

Dt − Ft

0.02 0.24 0.79 0.61 1.72 0.62 1.42 2.26 1.34 1.89 1.23 1.30 0.90 0.89 0.86 2.50 0.96 0.92 1.15 0.49  | Dt − Ft | = 22.07

17.98 23.54 19.50 42.20 19.67 25.33 20.65 44.46 21.36 27.12 21.80 46.72 23.05 28.91 22.95 48.98 24.74 30.70 24.10 51.24

Dt − Ft

0.62 0.04 0.90 0.30 1.57 0.63 1.15 1.84 1.04 1.68 0.80 1.22 0.15 1.31 1.45 1.88 0.24 0.30 0.40 1.56  | Dt − Ft | = 19.08

|D − F | = 22.07 = 1.10

Linear trend forecast for seasons MAD =

|D − F | = 19.08 = 0.954 1

Although both forecasts seem to be relatively accurate, the linear trend line forecast for each season is slightly more accurate according to MAD.

S ( Fall ) =

173.3 = .215 806.1

S ( Winter ) =

155.9 = .193 806.1

S ( Spring ) =

203.5 = .252 806.1

S ( Summer ) =

273.4 = .339 806.1

12-15.

y = 195.55 + 2.39 ( 5 ) = 207.5 Forecasts for year 5: Fall: ( 207.5)(.215) = 44.61

( 207.5)(.193) = 40.13 ( 207.5)(.252 ) = 52.38 ( 207.5)(.339 ) = 70.34

Winter: Spring: Summer:

Yes, there does appear to be a seasonal pattern. 12-16. a.

Year Time 7:00 AM 8:00 9:00 10:00 11:00 Noon 1:00 PM 2:00 3:00 6:00 7:00 8:00 9:00 Total

1 56 31 15 34 45 63 35 24 27 31 25 14 10 410

2 64 41 22 35 52 71 30 28 19 47 35 20 8 472

3 66 37 24 38 55 57 41 32 24 36 41 18 16 485

4 60 44 30 31 49 65 42 30 23 45 43 17 14 493

Linear Trend Line:

410.40 21.03 b= Linear trend line forecast for year 7 = 557.6 a=

5 72 52 19 28 57 75 33 35 25 40 39 23 15 513

6 65 46 26 33 50 70 45 33 27 46 45 27 18 531

Total 383 251 136 199 308 401 226 182 145 245 228 119 81 2904

Year 7 Forecasts:

SF1 (7:00)= SF2 (8:00)= SF3 (9:00)= SF4 (10:00)= SF5 (11:00)= SF6 (noon)= SF7 (1:00)= SF8 (2:00)= SF9 (3:00)= SF10 (6:00)= SF11 (7:00)= SF12 (8:00)= SF13 (9:00) =

73.54 48.19 26.11 38.21 59.14 77.00 43.39 34.95 27.84 47.04 43.78 22.85 15.55

Year 1 2 3 4 5 6 9

Pool Attendance 410 472 485 493 513 531

Exponentially Smoothed Forecast 410.00 410.00 428.60 445.52 459.76 475.73 492.31

Error

Trend

62.00 56.40 47.48 53.24 55.27

0.0000 3.7200 6.3600 7.9368 9.5436 10.951

12-17. Week 1 2 3 4 5 6 7 8 9 10 Total

Patients per Period Weekend Weekdays 105 73 119 85 122 89 128 83 117 96 136 78 141 91 126 100 143 83 140 101 1277 879

Total 178 204 211 211 213 214 232 226 226 241 2156

Adjusted Smoothed Forecast 410.00 432.32 451.88 467.70 485.28 503.27 MAD =

Error 62.00 52.68 41.12 45.30 45.72 49.364

Linear Trend Line: a = 186.93 b = 5.21 Linear trend line forecast for Week 11 = 244.27 SF1 (weekend) = 144.68 SF2 (weekdays) = 99.59 12-18. Linear trend line: y = 347.33 + 3.856x 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 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. 12-19. Linear Trend Line: F(25) = 1109.23 MAD = 119.83 Exponential Smoothing: F(25) = 998.76 MAD = 164.02 3-Month Weighted Moving Average: F(25) = 1057.89 MAD = 109.18

12-20.

 D = 2,156

S1 =

1, 277 = .592 2,156

S2 =

879 = .408 2,156

Linear trend line forecast for week 11: y = 186.9 + 5.21x = 244.27 Weekend forecast: ( 244.27 )(.592 ) = 144.68 Weekday forecast: ( 244.27 )(.408 ) = 99.59

Month

Actual Demand

Forecast Demand

1 2 3 4 5 6 7 8 9 10

160 150 175 200 190 220 205 210 200 220

170 165 157 166 183 186 203 204 207 203

Cumulative error Bias MAD MAPD

86.00 8.60 15.00 0.08

Error

Dt − Ft

–10 –15 18 34 7 34 2 6 –7 17 86

10 15 18 34 7 34 2 6 7 17 150

Running MAD

Cumulative Error

Tracking Signal

10.00 12.50 14.33 19.25 16.80 19.67 17.14 15.75 14.78 15.00

–10 –25 –7 27 34 68 70 76 69 86

–1.00 –2.00 –0.49 1.40 2.02 3.46 4.08 4.83 4.67 5.73

There is really no way to determine if this is an accurate forecast method unless it is compared with some other method.

See (b) solution for tracking signal values.

The forecast appears to be biased low (i.e., actual demand exceeds the forecast). 12-21.  = 239.38; There does not seem to be any high or low bias in the forecast.

12-22..

Year 1 2 3 4 5 6 7 8

Error — 250 –300 –524.4 –154.8 –523.34 –908.52 –360.91

Cumulative Error — 250 –50 –574.4 –729.4 –1,252.54 –2,161.06 –2,521.97

Running MAD — 250.00 275.00 358.13 307.30 350.51 443.51 431.71

Tracking Signal 1.00 –0.18 –1.60 –2.37 –3.57 –4.87 –5.84

The control chart suggests the forecast is not performing accurately and is consistently biased high (i.e., the actual demand is consistently lower than the forecast).

12-23. a. Month March April May June July August September October November

Demand

Forecast

Error

Dt − Ft

120 110 150 130 160 165 140 155

— 120.0 116.0 129.6 129.7 141.8 151.1 146.7 150.0

— –10.00 34.00 0.40 30.30 23.20 –11.10 8.30 75.10

— 10.00 34.00 0.40 30.30 23.20 11.10 8.30 117.30

Bias MAD MAPD Cumulative error

10.73 16.76 0.1038 75.10

Month

Demand

3-Month Moving Average

March April May June July

120 110 150 130 160

— — — 126.67 130.00

Error

Dt − Ft

— — — 3.33 30.00

Running MAD

Cumulative Error

Tracking Signal

— 10.00 22.00 14.80 18.67 19.58 18.17 16.76

— –10.0 24.0 24.4 54.7 77.9 66.8 75.1

— –1.00 1.09 1.65 2.93 3.98 3.67 4.48

Month

Demand

August September October November

165 140 155

3-Month Moving Average 146.67 151.67 155.00 153.33 8.00 12.67 0.08 276.64 39.99

Bias MAD MAPD MSE Cumulative error

Error

Dt − Ft

18.33 –11.67 0.00 39.99

18.33 11.67 0.00 63.33

The 3-month moving average seems to provide a better forecast. c.

The tracking signal moves beyond the 3 MAD control limit for July and continues increasing indicating the forecast is consistently biased low.

12-24. The 3-month moving average forecast appears to be more accurate. Month

Demand

Forecast

Dt − Ft

Month

Demand

Forecast

Dt − Ft

January February March April May June July

9 7 10 8 7 12 10

9.00 9.00 8.60 8.88 8.70 8.36 9.09

— 2.00 1.40 0.88 1.70 3.64 0.91

August September October November December January

11 12 10 14 16 —

9.27 9.62 10.09 10.07 10.86 11.88

1.73 2.38 0.09 3.92 5.14 —

Mad

Moving Average Forecast (Prob. 12-1a) 1.89

Exponentially Smoothed (Prob. 12-20) 2.16

12.25. MAD = 1.79 Cumulative error = 12.36

According to these measures, the forecast appears to be fairly accurate.

Year 1 2 3 4 5 6 7 8

Demand 16.8 14.1 15.3 12.7 11.9 12.3 11.5 10.4

Forecast 16.8 16.8 15.7 15.5 14.4 13.4 12.9 12.4

Error 0 −2.7 −0.4 −2.8 −2.5 −1.1 −1.4 −1.6

Running MAD — 2.70 1.55 1.97 2.10 1.90 1.82 1.79

Cumulative Error — −2.7 −3.1 −5.9 −8.4 −9.5 −10.9 −12.5

Tracking Signal — −1.00 −2.00 −3.00 −4.00 −5.00 −6.00 −7.00

The tracking signal goes outside of the control limits beginning in year 4, indicating a forecast that is biased high.

Linear trend model: Year 1 2 3 4 5 6 7 8 9

Demand 16.80 14.10 15.30 12.70 11.90 12.30 11.50 10.80 —

Forecast 15.87 15.10 14.33 13.56 12.79 12.02 11.24 10.47 9.70

MAD = 0.68

The linear trend line forecast appears to be more accurate for MAD. 12-26. The forecast for 2016 is used as actual demand for the forecast for 2017, and so on for 2017 and 2018.

2016: Linear trend line: y = -214.36 + 203.27x y (2016) = 1,615 2017: Linear trend line: y = -214.33 + 203.27x y (2017) = 1,818

2018: Linear trend line: y = -214.26 + 203.25x y (2018) = 2,022

Smartphone sales in 2018 are forecast to be around 1.9 billion by 2018 so the linear trend line forecast seems to be relatively accurate, although it may be slightly high due to predicted slower market growth in the future by most forecasters. 12 – 27. y = 3643.35 – 15.67x y (115) = 1841.78 r = - 0.655 A moderately strong relationship

12-28. y = 49.95 + 0.428x, where y = occupancy rate and x = wins.

Forecast if the Blue Sox win 85 games: 49.95 + .428 ( 85 ) = 86.3 occupancy rate. yes; r = .8626 12-29. a.

y = .51 + .403x, where y = sales and x = permits. Forecast if 25 permits are filed: .51 + .403 ( 25 ) = 10.57

b. 12-30. a.

The correlation coefficient is .914 indicating a strong causal relationship.

y = −50.21 + 2.056 x, where y = gallons of ice cream and x = temperature. Forecast for temperature of 80 : − 50.21 + 2.056 (80 ) = 114.29 gal.

The correlation coefficient is 0.833, indicating a strong causal relationship.

c. Coefficient of determination = ( 0.833) = 0.694, indicating that 69.4% of the variation of ice cream sales 2

can be attributed to the temperature. 12-31. a.

y = 7039.24 − 0.337 x where y = applications and x = tuition. If tuition is $10,000 forecast is 7039.24 −.337 (10, 000 ) = 3, 670.12 applications. If tuition is $7,000, forecast is 7039.24 − .337 ( 7000 ) = 4, 680.86 applications.

b. c.

The correlation coefficient is −.808, indicating a fairly strong linear relationship between tuition costs and number of applicants. Number of class sections, number of dormitory rooms, number of persons per class, plus numerous budgeting decisions.

12-32. y = 15.864 − 0.575x r = −0.785

r 2 = 0.616 There seems to be a relatively strong relationship between production time and defects. Forecast for “normal” production time of 23 minutes:

y = 15.864 − 0.575 ( 23)

y = 2.64% defects 12-33. y = 1.317 + 0.151x r = .744, which indicates a fairly strong relationship between hits and orders

r 2 = 0.553, which means 55.3% of the variation in orders can be attributed to the number of web site hits.

At 60,000 hits/month, y = 1.317 + .151( 60 ) =10.4 or 10,400 orders

12-34.

Year 1 2 3 4 5 6 7 8 9 10 11

Application 6,010 5,560 6,100 5,330 4,980 5,870 5,120 4,750 4,615 4,100 —

Linear Trend Line Forecast 6,069.72 5,886.12 5,702.51 5,518.91 5,335.30 5,151.69 4,968.09 4,784.78 4,600.88 4,417.27 4,233.66

y = 6253.33 − 183.606 x Correlation coefficient = −0.850 a.

The linear regression forecast (from Problem 12-32) has a MAD value of 310 whereas the MAD value for the linear trend line forecast in this problem is 256, indicating that the linear trend line forecast is somewhat better.

The correlation coefficient is −0.850, indicating a strong relationship between applications and time.

12-35. The slope, b = 2.98, indicates the rate of change, that is, the number of gallons sold for each degree increase in temperature. 12-36. a.

y = 380.93 + 16.03x y (11) = 380.93 + 16.03 (11)

= 557.26 printers b.

y = −22.07 + .41x y = −22.07 + .41(1500 ) = 594.10

MAD for the linear trend line forecast in a. equals 85.69 while MAD for the linear regression forecast in b. equals 45.20. In addition, the correlation coefficient for the linear trend is r = 0.426 whereas the correlation coefficient for the linear regression is r = .848. This evidence seems to indicate the forecast model in b is best.

d. Year 1 2 3 4 5 6 7 8 9 10 11

Demand 381 579 312 501 296 415 535 592 607 473

Forecast — 381.00 440.40 401.88 431.62 390.93 398.85 439.21 485.04 521.63 507.04

The exponential smoothing forecast ( MAD = 104.54 ) appears to be less accurate than the linear regression forecast ( MAD = 45.20 ) developed in 12-41a. 12-37. a.

Seasonally adjusted forecast. Quarter 1: D1 = 306 Quarter 2: D2 = 334 Quarter 3: D3 = 404 Quarter 4: D4 = 348 S1 =

306 = 0.220 1392

S2 =

354 = 0.240 1392

S3 =

404 = 0.290 1392

S4 =

348 = 0.250 1392

Linear trend line forecast for year 6: ; y = 271.2 + 2.4 x ; y ( 6 ) = 271.2 + 2.4 ( 6 ) = 285.6

SF1 = ( 0.220 )( 285.6 ) = 62.78 SF2 = ( 0.240 )( 285.6 ) = 68.53 SF3 = ( 0.290 )( 285.6 ) = 82.82

SF4 = ( 0.250 )( 285.6 ) = 71.40 b.

Quarter 1: y = 52.69 + 0.3973x y ( 20 ) = 56.69 + 0.3973 ( 20 )

= 64.64 =60.64

Quarter 2: y = 91.62 − 0.71x

y ( 36 ) = 91.62 − ( 0.71)( 36 ) = 66.06

Quarter 3: y = 73.57 + 0.29 x

y ( 25 ) = 73.57 + 0.29 ( 25) = 80.82

Quarter 4: y = 37.47 + 1.06 x y ( 30 ) = 37.47 + 1.06 ( 30 ) = 69.27

This is an intuitive assessment, which managers must do on occasion. In general, the linear regression forecast provides a more conservative estimate. The adjusted exponentially smoothed forecast ( = 0.4,  = 0.4 ) has a first quarter forecast for year 6 of 75.68 % seat occupancy. It has a E (bias) value of 1.08 and a MAD value of 8.6, which seem low. Thus, this may be the best overall forecast model compared to the one developed in 12-413a.

12-38.

The following table shows several different forecast models developed using Excel and selected measures of forecast accuracy. Year 25 Forecast 5.89

MAD 1.58

E (bias) .127

8.22

1.86

0.000

6.64

1.59

−0.330

Exponential smoothing ( = 0.5 )

6.13

1.29

.031

Exponential smoothing ( = 0.3,  = 0.4 )

6.24

1.33

−0.020

Exponential smoothing ( = 0.4,  = 0.5)

5.94

1.22

0.003

Forecast Method Moving average ( n = 3) Linear trend line Exponential smoothing ( = 0.3)

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 (  = 0.4 and  = 0.5 ).

12-39.

y = 13.8399 + .00150 x y (12000 ) = 31.79 r = 0.95

12-40.

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 57.54 whereas the MAD value if a seasonally adjusted linear trend line forecast (see Excel solution) has a MAD value of 699.65. 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.

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

3,415

6,627

0.113

Total

29,357

29,308

Units Month #

Month

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

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

3,415

0.113

3,314

100.71 1376

MAD = 1376 / 24 = 57.33 12-41.

Si = monthly service calls / 21,850 Forecast (i) = (Si)(7655.33) where,

y = 673.33+231x

y (4) = 7655.33 Monthly Calls

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

12-42.

The linear regression model is:

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, 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). 12-43.

The answer depends on the student’s approach to developing a solution. However, one possible 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. This results in the following linear trend line equation: y = 40,200 + 251.65x and to forecast for planning purposes in the 5 th year, use 48 months (i.e., after 4 years) in the linear equation: y(48) = 40,200+251.65x = 52,279

12-44. (a) Forecast of applicants:

y = 13,803.07 + 470.55 x y (11) = 18,979.12 applicants Forecast of % acceptances:

y = 37.72 + .247 x 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 − 1.68 x 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. 12-45. (a) y = 381.32 + 68.40 x

y (11) = 1, 270.48 (b) r = .973 There appears to be a very strong linear relationship 12-46. (a) y = 219.27 + 12.28x

y (16) = 415.67

(b) y = −5349.77 + .147 x y (39,300) = 438.31 r = .966 y (40, 000) = 541.41

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. 12-47. (a) y = 116.12 − 1.28x

y (70) = 116.12 − 1.28(70) = 26 r 2 = .537 (b) y = 116.9 − 1.24 x1 − 0.14 x2

y (70, 40) = 116.9 − 1.24(70) − 0.14(40) = 25 r 2 = .538 Very little difference between the two forecasts. Annual budget appears to replicate endowment. 12-48. y = 1.704 + 0.269 x, where y = sales and x = promotional expenditures. Correlation coefficient = 0.546 The correlation coefficient indicates a weak linear relationship between sales and promotion, thus a linear regression model should not be used. 12-49. We tested 3 forecasting methods, as follows.

Month 1 2 3 4 5 6 7 8 9 10 11 12 13 15

Demand 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 11.70

Linear Trend Line Forecast 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 10.74

3-Month Moving Average — — — 7.93 8.30 8.83 9.30 9.67 9.87 10.10 10.17 9.67 9.57 10.23

Adjusted Exponentially Smoothing Forecast (  = 0.30,  = 0.50 ) 18.20 8.20 7.99 8.02 8.40 8.61 8.88 9.34 9.44 9.67 9.95 9.42 9.57 10.00

16 17 14 18 19 20 21 22 23 24 25

9.80 10.80 10.50 11.30 12.60 11.50 10.80 11.70 12.50 12.80 —

10.92 11.09 10.56 11.27 11.45 11.63 11.81 11.99 12.17 12.34 12.52

Forecasting Alternatives

10.83 10.67 9.47 10.77 10.63 11.57 11.80 11.63 11.33 11.67 12.33

10.51 10.30 9.79 10.45 10.70 11.27 11.34 11.18 11.33 11.68 12.29

MAD

Linear trend line

0.546

—

3-month moving average

0.825

9.2

Adjusted exponential smoothing

0.817

9.47

All three methods we chose to evaluate appear to be relatively accurate. The student might select another method that will be more accurate. 12-50. a.

y = 745.91 − 2.226 x1 + 0.163x2

r 2 = 0.992

y = 7,186.91

12-51. a.

y = 608.795 + 0.215x1 − 0.985x2

r 2 = 0.766

y = 608.795 + 0.215 (1,500 ) − 0.985 ( 300 ) = 635.79

12-52. a.

y = 219.167 − 0.027 x1 + 233.871x2

r 2 = 0.956

y = 219.67 − 0.027 (10, 000 ) + 233.871( 4 ) = $882.82

12-53. Selected forecast models 5-day moving average forecasts for day 21: 11 − 12 = 82.11, MAD = 12.40 12 − 1 = 128.4, MAD = 28.32 1 − 2 = 93.0, MAD = 15.82

Exponentially smoothed ( = 0.3) forecasts for day 21: 11 − 12 = 82.11, MAD = 12.40

12 − 1 = 129.7, MAD = 26.36 1 − 2 = 99.61, MAD = 14.23

Linear trend line forecasts for day 21: 11 − 12 = 81.86, MAD = 11.25 12 − 1 = 132.42, MAD = 22.14 1 − 2 = 103.5, MAD = 12.44

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. 12-54. a.

y = 43.09 + .0007 x1 + 1.397 x2

where y = SOL scores

x1 = average teacher salary

x2 = average tenure b.

r 2 = 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.

y = 43.09 + .0007 ( 30, 000 ) + 1.397 ( 9 ) = 76.66 No, the SOL score would only increase to 76.66.

CASE 12.1: Forecasting at State University 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. 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—that is, a definition of quality according to students, parents, and legislators. In-house 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.

CASE 12.2: The University Bookstore Student Computer Purchase Program The following table shows several different forecast models developed using POM for Windows and selected measures of forecast accuracy. Year 25 Forecast 1,004.66

MAD 96.96

E (bias) 66.00

1,020.07

73.24

0.00

941.53

126.88

108.59

Exponential smoothing ( = 0.5 )

1,003.70

104.95

74.72

Exponential smoothing ( = 0.3,  = 0.4 )

983.22

109.58

62.19

Exponential smoothing ( = 0.4,  = 0.5)

1,031.09

105.13

61.31

Forecast Method Moving average ( n = 3) Linear trend line Exponential smoothing ( = 0.3)

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 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.

CASE 12.3: Cascades Swim Club Attendance

Week Day

Total

139

198

341

287

303

242

194

197

275

246

224

258

235

3,139

273

310

291

247

223

177

207

273

241

177

239

130

218

3,006

172

347

380

356

315

245

215

213

190

161

274

195

271

3,334

275

393

367

322

258

390

304

303

243

308

205

238

259

3,865

337

421

359

419

193

284

331

262

277

256

361

224

232

3,956

402

595

463

516

378

417

407

447

241

391

411

368

317

5,353

487

497

578

478

461

474

399

384

400

419

541

369

5,886

Total

2,085

2,761

2,779

2,625

2,131

2,229

2,057

2,094

1,851

1,939

2,133

1,954

1,901

28,539

The seasonal factors for each weekday are as follows: S1 ( Monday ) =

3,139 = .110 28,539

S 2 ( Tuesday ) =

3, 006 = .105 28,539

S3 ( Wednesday ) =

S 4 ( Thursday ) = S5 ( Friday ) =

3,334 = .117 28,539

3,865 = .135 28,539

3,956 = .139 28,539

S6 ( Saturday ) =

S7 ( Sunday ) =

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 − 57.5604 x 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,

y = 2,598.2308 − 57.5604 (1) = 2,540.67 Week 1 Forecasts Monday = (.110 )( 2,540.67 ) = 279.5

Tuesday = (.105 )( 2,540.67 ) = 266.8 Wednesday = (.117 )( 2,540.67 ) = 297.3 Thursday = (.135 )( 2,540.67 ) = 343.0

Friday = (.139 )( 2,540.67 ) = 353.2 Saturday = (.188 )( 2,540.67 ) = 477.6 Sunday = (.206 )( 2,540.67 ) = 523.4

The remaining 12 weeks of daily forecasts would be developed similarly. 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 like forecast to compute forecasted weekly attendance.

CASE 12.4 – FORECASTING ARIPORT 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

6-8pm

34,600/677,200 = .051

8-10pm

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 point: y = 11,413.3 + 5580 x y (4) = 33,733.3 (B) Linear trend line forecast for year 4 developed using all 30 sample data points: y = 14,893 + 503.62 x y (31) = 30,505.2 Seasonally Adjusted Forecast (A)

Seasonally Adjusted Forecast (B)

4-6 am

4,926.50

4455.06

6-8 am

5529.24

5000.12

8-10 am

5783.28

5229.85

10- Noon

3247.80

2937.01

Noon – 2 pm

4019.91

3635.22

2-4 pm

4249.05

3842.43

4-6 pm

3726.01

3369.45

6-8pm

1723.53

1558.60

8-10pm

528.02

477.59

13 Inventory Management Answers to Questions 13-1. In general, independent demand items are final or finished products that are not dependent upon internal production activity; that is, the demand is usually external and beyond the direct control of the organization. Alternatively, dependent demand is usually a component part or material used to produce a final product. An example of independent demand for a pizza restaurant would be a final product such as a pizza, whereas dependent demand would be any of the ingredients (cheese, tomato sauce, dough, etc.) and perhaps complementary items such as drinks. 13-2. In a fixed-order-quantity system, an order is placed for the same constant amount whenever the inventory on hand decreases to a certain level, whereas in a fixed-time-period system, an order is placed for a variable amount after an established passage of time. 13-3. The customer service level is the ability to meet internal or external demand at a specified level of efficiency. High quality service is often perceived as always being able to meet demand, which normally requires high inventory levels and can be costly, or efficient management of the inventory system such that demand is met most of the time. 13-4. An ABC system is a method for classifying inventory according to its dollar value. In general, about 5 to 15 percent of all inventory items will account for 70 to 80 percent of the total dollar value of inventory. Each level of inventory requires different levels of inventory control. That is, the higher the value of inventory, the higher the control. Such a system generally requires less record-keeping and focuses managerial attention on the most important inventory items. 13-5. The two basic inventory decisions are how much to order and when to order items for inventory. In a continuous order system, whenever inventory decreases to a specific level (referred to as the reorder point), a new order is placed for a fixed amount. Alternatively, in a periodic inventory system, inventory on hand is counted at specific time intervals and an order is placed for an amount that will bring inventory back to a desired level. 13-6. The categories are ordering cost, carrying cost, and shortage costs. As the order size increases, ordering costs and shortage costs decrease while carrying costs increase. 13-7. The optimal order quantity occurs when the ordering cost equals carrying cost; thus, the order quantity can be determined by equating these two cost functions and solving for the optimal value. 13-8. Demand is known with certainty, shortages are not allowed, lead time for order receipts is constant, and orders are received all at once. These assumptions are limiting to the extent that they eliminate all uncertainty and potential variation in the model. 13-9. In a continuous inventory system, the reorder point is the inventory level at which a new order is placed, and lead time is the time required to receive an order after it has been placed. 13-10. In a noninstantaneous receipt model, the order quantity is received gradually over time and the inventory level is depleted at the same time it is being replenished, whereas in the basic EOQ model orders are received all at once. 13-11. The total purchase price of all items demanded must be included in the model, since the differences in prices for different order sizes must be reflected in the model. 13-12. Price has no real impact on the optimal order size; it is a constant value that would not alter the basic shape of the EOQ total cost curve. 13-13. The noninstantaneous receipt EOQ model would approach the basic EOQ model with instantaneous receipt. 13-14. The service level is the probability that the amount of inventory on hand during the lead time is sufficient to meet expected demand. The safety stock is the amount of inventory to keep on hand necessary to achieve

this probability.

13-15. The answer to this question is included in the “Along the Supply Chain” box for Procter & Gamble in this chapter. The inventory management models in this chapter are single-stage in that they optimize inventory of a single item at a single location. Multi-echelon inventory optimization employs software designed to determine the minimum inventory levels for multiple materials, parts, subassemblies and finished goods across an entire supply chain.

Solutions to Problems 13-1.

D = 1500 Co = $625 Cc = $130

2 ( 625)(1500 ) 2Co D = = 120.1 Cc 130

TC =

Co D CcQ + Q 2

( 625)(1500 ) + (130 )(120.1) = $15, 612.49 120.1

D 1500 = = 12.49 orders Q 120.1

364 = 29.14 days 12.49

13-2. Case a b c d 13-3.

D = 16,500 Co = $70 Cc = $27

Q 120.1 120.1 132.8 108.6

TC $14,051.25 $17,173.74 $15,534.24 $15,534.24

2 ( 70 )(16,500 ) 2Co D = = 292.5 Cc 27

TC =

Co D CcQ ( 70 )(16,500 ) ( 27 )( 292.5) + = + Q 2 292.5 2

= $7,897.47

13-4.

D 16,500 = = 56.41 orders Q 292.5

320 = 5.67 days 56.41

D = 45, 000 Co = $1,500 Cc = $0.70

2 (1500 )( 45, 000 ) 2Co D = Cc 0.70

= 13,887.3 yd

TC =

Co D CcQ + Q 2

( 0.70)(13,887.3) + (1500 )( 45, 000 ) 2

13,887.3

= $9, 721.11

Number of orders =

D 45, 000 = = 3.24 per year Q 13,887.3

Time between orders = 13-5.

D = 1, 415, 000 Co = $2, 200

365 = 112.6 days 3.24

Cc = $0.08 a.

2Co D Cc

2 ( 2, 200 )(1, 415, 000 )

0.08

TC =

= 278,971.3 yd

Co D CcQ + Q 2

( 2, 200 )(1, 415, 000 ) + ( 0.08)( 278,971.3) 278,971.3

= $22,317.71

13-6.

D 1, 415, 000 = = 5.07 per year Q 278,971.3

365 = 72.0 days 5.07

D = 6, 000 d = 23.08 / day p = 116 / day Co = $700 Cc = $9

2 ( 700 )( 6, 000 ) 2Co D =  d  23.08  5 1 − Cc 1 −   116  p  

= 1, 079.41 b.

TC =

Co D CcQ  d  + 1 −  Q 2  p

( 700 )( 6, 000 ) + ( 9 )(1, 079.4 ) 1 − 23.08  1079.4

= $7, 782.84

D 6, 000 = = 5.96 runs = 5.56 rund Q 1079.41

260 = 46.67 working days 5.96 Q 1079.41 = = 9.31 working days p 116

13-7.

D = (18)( 52 ) = 936 Co = ( $300 )( 0.25) = $75 Cc = $250

2 ( 250 )( 936 ) 2Co D = Cc 75

= 79 bicycles

TC =

Co D CcQ + Q 2

( 250 )( 936 ) + ( 75)( 79 ) 79

= $5,924.53 13-8.

D = 7, 000 Co = $3, 600

Cc = $50

L = 10 Q=

2Co D Cc

 

 116 

2 ( 3, 600 )( 7, 000 )

= TC =

= 1004 gallons

Co D CcQ + Q 2

 7, 000   1004  = 3, 600   + 50    1004   2 

= $50,199.6  7, 000  R = dL =   (10 ) = 225.81 gallons  310  13-9.

D = 5, 000 Co = $80 Cc = $0.50 L=4 a.

2 (80 )( 5, 000 ) 2Co D = = 1, 264.9 boxes Cc 0.50

TC =

Co D CcQ  5, 000   1264.9  + = 80   + 0.50   Q 2  1264.9   2 

= $632.46 c.

13-10.

 5, 000  R = dL =   ( 4 ) = 54.79 boxes  365 

d = 205 lb / day p = 350 lb / day

D = 74,825 Co = $175 Cc = $12

TC =

2 (175 )( 74,825 ) 2Co D = = 2295.18  d  205  12 1 − Cc 1 −   p  350  

Co D CcQ  d + 1 −  Q 2  p

(175)( 74,825) + (12 )( 2295.18) 1 − 205  2295.18

 

 350 

= $11, 410.32 13-11.

d = 1,800 p = 3, 000

D = 657, 000 Co = $7,500

Cc = $60

TC =

2 ( 7,500 )( 657, 000 ) 2Co D = = 20, 263.88  d  1,800  50 1 − Cc 1 −   p  3, 000  

Co D CcQ  d + 1 −  Q 2  p

= ( 7,500 )

657, 000 20, 263.88  1,800  + ( 60 ) 1 − 3, 000  20, 263.88 2  

= $486,333.22 13-12. Operates 360 days/year 12 converters 5 tons coal/day/converter

D = ( 5 tons )(12 converters )( 360 days ) = 21, 600 tons / year

Co = $80 Cc = 20%of average $ inventory level

Cc = ( 0.20 )( $12 ) = $2.40 2 (80 )( 21, 600 ) 2Co D = = 1, 444, 000 Cc 2.4

= 1, 200 tons b.

TC =

Cc D Co D ( 2.4 )(1, 200 ) (80 )( 21, 600 ) + = + 2 Q 2 1, 200

= 1, 440 + 1, 440 = $2,880 R=

13-13.

LD 5 ( 21, 600 ) = = 300 tons 360 360

D = 10, 000 logs / year T = 250 days / year p = 60 / day

R = 60 ( 250 ) = 15, 000 / year Co = $1, 600

Cc = $15

2 (1, 600 )(10, 000 ) 2Co D = D   10, 000  Cc 1 −  15 ) 1 − (  R   15, 000 

= 6, 400, 000 = 2,529.8 logs b.

TC =

Co D Cc Q  D + 1 −  Q 2  R

(1, 600 )(10, 000 ) +  (15)( 2,529.8) 1 − 10, 000   2,592.8

 

 

 15, 000  

= 6,324.5 + 6,324.5 = $12, 648 c.

D 10, 000 = = 3.95 = 4 orders per year Q 2,529.8

Tb =

T 250 = = 63.3 days between orders N 3.95

Q = 2,529.8, R = 60 The number of operating days to receive the entire order is

Q 2,529.8 = = 42.2 days r 60 13-14. a.

D = 12, 400 Cc = $3.75 Co = $2, 600

TC =

2 ( 2, 600 )(12, 400 ) 2Co D = = 4146.6 Cc 3.75 Co D CcQ + Q 2

( 2, 600 )(12, 400 ) + ( 3.75)( 4,912.03) 4,912.03

= $15,549.92 b.

Co = $1,900

Cc = $4.50

TC =

2 (1,900 )(12, 400 ) 2Co D = = 3235.9 Cc 4.50 Co D CcQ + Q 2

(1,900 )(12, 400 ) + ( 4.50 )(3,833.19 ) 3,833.19

= 14,561.59 Select the new location. 13-15. a.

d = 220, 000 Cc = $0.12 / lb Co = 620

p = 305, 000

2 ( 620 )( 220, 000 ) 2Co D = 220  d ( 0.12 ) 1 −  Cc 1 −  p  305  

= 90,317.52

 d maximum level = Q 1 −  = ( 90,317.52 )( 0.2787 ) p  = 25,170.46

TC =

Co D CcQ  d + 1 −  Q 2  p

( 620 )( 220, 000 ) + ( 0.12 )( 90,317.52 ) 0.2787 ( ) 90,317.52

= $3020.45 b.

P = 360, 000 Q=

TC =

2 ( 620 )( 220, 000 ) = 76, 457.27  220  ( 0.12 ) 1 −   305 

( 620 )( 220, 000 ) + ( 0.12 )( 76, 457.27 ) 0.25 ( ) 76, 457.27

= $3,568.01 No, the total inventory cost increases.

D = 1400

13-16.

Co = $7, 600

Cc = ?

Q = 120 2Co D Cc

2 ( 7, 600 )(1400 )

120 =

(120 )2 =

Cc 2 ( 7, 600 )(1400 ) Cc

Cc = $1, 477.78 13-17.

D = 200 / day Co = $25 Cc = $0.20 / min. = $120 / day

2 ( 25 )( 200 ) 120

= 9.1 orders or 9

The truck should carry approximately 9 orders each time it makes deliveries.

200 = 24 deliveries per day 9 10 = 0.416 hour = every 25 minutes a delivery truck goes out to deliver orders 24 TC =

( 25)( 200 ) + (120 )(8) 8

= $1095.45

Q=6

200 = 33.33 or 33 deliveries per day. 6 10 = .30 hr. = every 18 minutes a delivery truck is sent out. 33 TC =

( 25)( 200 ) + (120 )( 6 ) 6

= $1193.73 13-18.

Co = $1, 700 Cc = $1.25

D = 21, 000 / yr.

P = 30, 000 / yr.

TC =

2 (1, 700 )( 21, 000 )  21, 000  1.25 1 −   30, 000 

= 13, 798.55

(1, 700 )( 21, 000 ) + 1.25  13, 798.55   1 − 21, 000  13, 798.55

 

    30, 000 

= 5,174.46

Number of production runs =

D 21, 000 = = 1.52 Q 11, 062.62

 d Maximum inventory level = Q 1 −  p   21, 000  = 13, 798.55 1 −   30, 000  = 4,139.57 b.

Maximum inventory level = 2,500

 21, 000  2,500 = Q 1 −   30, 000 

2,500 = Q (.3) Q = 8,333.33

TC =

(1, 700 )( 21, 000 ) + 1.25  8,333.33  1 − .70 )  ( 

8,333.33

= $5,846.50 13-19.

D = 280, 000 C = $7, 000

Cc = $0.80 / ft.3 Q=

2 ( 7, 000 )( 280, 000 ) 0.80

= 70,000 ft.3 TC =

( 7, 000 )( 280, 000 ) + 0.80 ( 70, 000 ) 52,915

= $56, 000

Number of orders =

280, 000 56, 000

13-20. D = 715 Co = 6,000 Cc = 1,200 Q= = 84.56 TC =



= $101,469.20 No. of orders = 8.46 Time between orders = 43.2 13-21. D = (13.33 motorcycles/month)(12) = 160 motorcycles/year Co = $3200 Cc = $375 Q = 52.3 motorcycles TC = $19,595.92 Orders = 3.06 Time between orders = 119.2 days R = 13.33

13-22. D = (11,000 yds/month)(12) = 132,000 yds/year Co = $425 Cc = $0.63/yd p = 1200 yds/day d = 132,000/360 = 366.67 yds/day Q = 16,014.31 yards TC = $7,006.23 R = 2,566.67 yard 13-23. D = 29,330 units Co = $6,500 Cc = $115.75/unit p = 200 units/day d = 80.36 units per day Q = 2,346.63 TC = $162,484.30 Maximum inventory level = 2,346.03 R = 2,036.81

13-24. Co = $19.5Q D = 48,000 tons Cc = $112/ton

19.5Q (48,000) = 112Q Q

Q = 16,714

TC = $1,872,000 R = (131.5/day) (3 days) = 394.5 tons

13 - 25. Q = 603,008.3 bales TC = $9,497,381 Shipments = 2.06 Time between shipments = 176.8 days.

13-26.

D = 40, 000 Co = $800 Cc = $1.90 Without discount:

Q= TC =

2 ( 800 )( 40, 000 ) 1.90

= 5,803.81

Co D Q + Cc + PD Q 2

(800 )( 40, 000 ) + (1.90 )(5,803.81) + 3.40 40, 000 ( )( ) 5,803.81

= 147, 027.24 with discount of Q = 20, 000

TC = $140, 600 Take discount for Q = 20, 000

13-27.

D = 10, 000 Co = $300 Cc = $1.25 Order Size 0–4,999 5,000 +

P $8.00 $6.50

Without discount:

TC =

2 ( 300 )(10, 000 ) 2Co D = = 2,190.9 Cc 1.25 Co D CcQ + + PD Q 2

( 300 )(10, 000 ) + (1.25)( 2,190.9 ) + 10, 000 8 ( ) 2,190.9

= $82, 738.61 With discount:

Q = 5, 000 TC =

Co D CcQ + + PD Q 2

( 300 )(10, 000 ) + (1.25)(5, 000 ) + 10, 000 6.50 ( ) 5, 000

= $68, 725 Select discount; Q = 5, 000.

13-28. Without discount:

Q = 200

TC = $55, 440 With discount:

Q = 300

P = 52 TC =

Co D CcQ + + PD Q 2

(160 )( 900 ) + ( 7.20 )( 300 ) + 52 900 ( )( ) 300

= 48,360 Take the discount, Q = 300.

13-29.

D = 1, 700 Co = $120

Cc = ( 0.25)( $38) = $9.50 Q=

2 (120 )(1, 700 ) 2Co D = = 207 Cc ( 9.5)

TC =

(120 )(1, 700 ) + ( 9.50 )( 207 ) + 38 1, 700 ( )( ) 207

= $66,568.76 Q = 300 :

TC =

(120 )(1, 700 ) + ( 9.31) 300 + 37.24 1, 700 ( )( ) 300

= $65,384.80

Q = 500 : TC =

(120 )(1, 700 ) + ( 9.12 ) 500 + 36.48 1, 700 ( )( ) 500

= $64, 704 Q = 800 :

TC =

(120 )(1, 700 ) + ( 9.025 ) 800 + 36.10 1, 700 ( )( ) 800

= $65, 235 Select Q = 500; TC = $64, 704.

13-30.

2 (120 )(1, 700 )

TC =

(8)

= 226

(120 )(1, 700 ) + (8)( 226 ) + 38 1, 700 ( )( ) 226

= $66, 406.65 Q = 300 :

TC =

(120 )(1, 700 ) + (8)( 300 ) + 37.24 1, 700 ( )( ) 300

= $65,188 Q = 500 :

TC =

(120 )(1, 700 ) + (8)( 500 ) + 36.48 1, 700 ( )( ) 500

= $64, 424

Q = 800 : TC =

(120 )(1, 700 ) + (8)(800 ) + 36.10 1, 700 ( )( ) 800

= $64,825 Select Q = 500; TC = $64, 424.

13-31.

D = 6,500 Co = $28 Cc = $3

2 ( 28)( 6,500 ) 2Co D = = 348.32 = 348 Cc 3

TC =

( 28)( 6,500 ) + ( 3)( 348) + 16 6,500 ( )( ) 348

= $105, 045 Q = 1, 000 :

( 28)( 6,500 ) + ( 3)(1, 000 ) + 14 6,500 ( )( )

TC =

1, 000

= $92, 682

Q = 3, 000 :

( 28)( 6,500 ) + ( 3)( 3, 000 ) + 13 6,500 ( )( )

TC =

3, 000

= $89, 060.67 Q = 6, 000 :

TC =

( 28)( 6,500 ) + ( 3)( 6, 000 ) + 12 6,500 ( )( ) 6, 000

= $87, 030.33 Select Q = 6, 000; TC = $87, 030.33.

13-32.

( 2 )( 28)( 6,500 ) = 337.26 = 337 boxes

TC =

( 28)( 6,500 ) + ( 3.20 )( 337 ) + 16 6,500 ( )( )

3.20

337

= $105, 079.20 Q = 1, 000 :

TC =

( 28)( 6,500 ) + ( 2.80 )(1, 000 ) + 14 6,500 ( )( ) 1, 000

= $92,582

Q = 3, 000 :

TC =

( 28)( 6,500 ) + ( 2.60 )(3, 000 ) + 13 6,500 ( )( ) 3, 000

= $88, 460.67 Q = 6, 000 :

TC =

( 28)( 6,500 ) + ( 240 )( 6, 000 ) + 12 6,500 ( )( ) 6, 000

= $85, 230.33

Select Q = 6, 000; TC = $85, 230.33. 13-33.

D = 2,300, 000 /100 = 23, 000 boxes Co = $320 Cc = $1.90

2 ( 320 )( 23, 000 ) 2Co D = = 2, 783.4  2, 784 boxes Cc 1.90

TC =

( 320 )( 23, 000 ) + (1.9 )( 2, 784 ) + 47 23, 000 ( )( ) 2, 784

= $1, 086, 228.50

Q = 7, 000 :

TC =

( 320 )( 23, 000 ) + (1.9 )( 7, 000 ) + 43 23, 000 ( )( ) 7, 000

= $996, 701.43 Q = 12, 000 :

TC =

( 320 )( 23, 000 ) + (1.9 )(12, 000 ) + 41 23, 000 ( )( ) 12, 000

= $955, 013.33 Q = 20, 000 :

TC =

( 320 )( 23, 000 ) + (1.9 )( 20, 000 ) + 38 23, 000 ( )( ) 20, 000

= $893,368 Select Q = 20, 000; TC = $893,368.

13-34.

2 ( 320 )( 23, 000 )

TC =

2.35

= 2,502.76  2,503 boxes

( 320 )( 23, 000 ) + ( 2.35)( 2,503) + 47 23, 000 ( )( ) 2,503

= $1, 086,881.50 Q = 7, 000 :

TC =

( 320 )( 23, 000 ) + ( 2.15)( 7, 000 ) + 43 23, 000 ( )( ) 7, 000

= $997,576.43 Q = 12, 000 :

TC =

( 320 )( 23, 000 ) + ( 2.05)(12, 000 ) + 41 23, 000 ( )( ) 12, 000

= $955,913.33

Q = 20, 000 :

TC =

( 320 )( 23, 000 ) + (1.90 )( 20, 000 ) + 38 23, 000 ( )( ) 20, 000

= $893,368

Select Q = 20, 000 boxes; TC = $893,368.

13-35.

d = 4, 000

L=7

 d = 600 R = dL + Z d L R = 4, 000 ( 7 ) + 1.64 ( 600 ) 7 = 30, 603.42 Safety stock = 2, 603.42 yd If safety stock = 2, 000,

Z ( 600 )

( 7 ) = 2,000, Z = 1.26,

which corresponds to a 90% service level. 13-36.

d = 9, 000

 d = 1,900 L =8

Z = 2.05 R = dL + Z d L = ( 9, 000 )(8 ) + ( 2.05 )(1,900 ) 8 = 83, 016.7 lbs 13-37.

d = 20

d = 4 L = 2 L = 12

Z = 1.28 R = dL + Z d L = ( 20 )( 2 ) + 1.28 ( 4 ) 2 = 40 + 7.30 (20)(12) + 1.28(4)sqrt(12)

= 257.87 gal

Safety stock =17.87 gal. For service level of 95%

Z = 1.65 R = 262.86 gal Safety stock = 22.86 gal. 13-38. R = dL + Z d L = 30 (8) + 0.68 (10 )

( 8 ) = 259.2 gal

For a 95% service level,

Safety stock Z d L = 1.65 (10 )

( 8 ) = 46.67

The reorder point increases to 286.7 gal. 13-39.

d = 3.5

 d = 1.2 L = 25

Z = 1.29 R = dL + Z d L = ( 3.5)( 25) + 1.29 (1.2 ) 25 = 87.5 + 7.74

R = 95.24 Safety stock = 7.74 13-40. R = dL + Z d L = 3.5 (8) + 1.29 (1.2 )

( 8 ) = 32.38

Decision would be based on inventory holding cost, desire for low inventory, importance of reliable delivery, cost of the monitors from each source, etc.

13-41.

d = 200 tb = 30 L=4

 d = 80 I = 60 Q = d ( tb + L ) + Z d tb + L − I = 200 ( 30 + 4 ) + 2.33 (80 ) 30 + 4 − 60 = 7,826.89oz 13-42.

d =8 tb = 10 L=3  d = 2.5 I =0 Q = d ( tb + L ) + Z d tb + L − I

= 8 (10 + 3) + 2.33 ( 2.5 ) 10 + 3 − 0 = 122 pizzas 122 − 5 = 117 pizzas

13-43.

d = 18 tb = 30 L=2 d = 4 I = 25 Q = d ( tb + L ) + Z d tb + L − I

= 18 ( 30 + 2 ) + 1.65 ( 4 ) 30 + 2 − 25

= 588.3 bottles 13-44.

Item 25 23 20 22 24 16 5 10 12 2 4 1 27 9 29 26 28 13 30 18 6 7 21 17 19 8 3 11 15 14

Usage 870 30 19 12 24 60 18 67 682 510 300 36 750 344 46 244 45 95 165 270 500 710 910 120 45 80 50 510 820 10 8,342

Unit Cost 105 2,710 3,200 4,750 1,800 610 1,900 440 35 30 45 350 15 28 160 30 110 50 25 15 8 4 3 20 50 26 23 2 1 3

Annual Usage $91,350 81,300 60,800 57,000 43,200 36,600 34,200 29,480 23,870 15,300 13,500 12,600 11,250 9,632 7,360 7,320 4,950 4,750 4,125 4,050 4,000 2,840 2,730 2,400 2,250 2,080 1,150 1,020 820 30 571,957

% Annual Value 15.97% 14.21 10.63 9.97 7.55 6.40 5.98 5.15 4.17 2.68 2.36 2.20 1.97 1.68 1.29 1.28 0.87 0.83 0.72 0.71 0.70 0.50 0.48 0.42 0.39 0.36 0.20 0.18 0.14 0.01 100.00%

% Annual Usage 10.43% 0.36 0.23 0.14 0.29 0.72 0.22 0.80 8.18 6.11 3.60 0.43 8.99 4.12 0.55 2.92 0.54 1.14 1.98 3.24 5.99 8.51 10.91 1.44 0.54 0.96 0.60 6.11 9.83 0.12 100.00%

Class A A A A A A A B B B B B B B B B B C C C C C C C C C C C C C

13-45. Class A B C

13-46.

Items 4, 5, 6, 13, 20, 22, 23, 30 7, 10, 14, 15, 19, 21 1, 2, 3, 8, 9, 11, 12, 16, 17, 18, 24, 25, 26, 27, 28, 29, 31, 32

Quantity 52

$ Value 59,835

% Value 51

30,640

8.2

811

27,493

86.3

940

$117,968

100%

d = 2.6 packages/day  d = 0.8 packages/day

% Quantity 5.5

L = 2 days  L = 0.5 days Z = 2.33

R = d L + Z  d2 L +  L2 d 2 = ( 2.6 )( 2 ) + 2.33 (.8) 2 + (.5) ( 2.6 ) 2

= 9.22 packages of paper 13-47.

d = 6/hr.  d = 2.5/hr.

L = 0.5 hr.  L = .133 hr. Z =? a.

R = d L + Z  d2 L +  L2 d 2 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 ) = 6.977 pizzas

CASE SOLUTION 13.1: The A to Z Office Supply Company

D = $17, 000 / day = $5,185, 000/year ( 305 day year ) Cc = $0.09 / dollar / year = $0.09 Co = $1, 200 / loan + 0.0225Q

L = 15 days Optimal loan amount per loan:

2Co D = Cc

( 2 )(1, 200 )( 5,185, 000 ) 0.09

= $371,842.26 loan amount per loan Memo: The 0.0225Q 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., 0.0225  D.

TC = Co

D Q + Cc + 0.0225D Q 2

 5,185, 000   371,842  = (1, 200 )  + ( 0.09 )    2    371,842 

+ ( 0.0225)( 5,185, 000 ) = $150,128.30 total cost of borrowing

D 5,185, 000 = = 13.944 loans / year Q 371,842

 14 loans/year for about $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.

TC = Co

D Q + Cc + 0.02 D Q 2

 5,185, 000   500000  = (1, 200 )  + ( 0.09 )     2   500, 000 

+ ( 0.02 )( 5,185, 000 ) = 12, 444 + 22,500 + 103, 700 = $138, 644

Since this option yields a lower TC of $11, 484 (150,128 − 138, 644 ) ; it should be accepted.

CASE SOLUTION 13.2: The Texas Stadium Store The objective of this case problem is to determine the reorder point with variable demand. The first step is to complete the average demand 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. d = 42.57 hats per week

 = 10.41 hats per week L = 20 days = 2.86 weeks The first question is, if R = 140, what level of service does this correspond to. Thus, we are seeking Z as follows

140 = dL + Z d L 140 = ( 42.57 )( 2.86 ) + Z (10.41) 2.86

140 = 121.75 + Z (17.6 ) Z = 1.04 This Z value corresponds to a normal probability value of 0.8508, thus, the service level is approximately 85.1 percent. The desired service level is 99 percent ( Z = 2.33) . The reorder point and safety stock for this service level is determined as follows.

R = dL + Z d L R = ( 42.57 )( 2.86 ) + 2.33 (10.41) 2.86 140 = 121.75 + 41.02

R = 162.76 or 163 hats 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 hats are shipped from Jamaica while it would probably not be very expensive to store hats (given their small size and weight).

CASE SOLUTION 13.3: Pharr Food Company This problem requires the development of a forecast for product demand in year 4 (see chapter 11). 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

2696

651

487

660

263

370

537

2968

685

539

672

302

411

572

3181

Total

1943

1514

1811

821

1123

1633

8845

S1 =

0.220

S2 =

0.171

S3 =

0.205

S4 =

0.093

S5 =

0.127

S6 =

0.185

Linear trend line forecast: y = 2463.3 + 242.5 x

y ( 4 ) = 3433.33 cases

SF1 =

754.21

SF 2 =

587.68

SF 3 =

702.97

SF 4 =

318.68

SF 5 =

435.91

SF 6 =

633.88

Total

3433.33

2 ( 4700 )( 3433.33) 116

= 527.5

Comparing monthly forecasts (with seasonal pattern) with order size, Q, using order frequency of 2 months:

No. of orders =

D 3433.3 = = 6.5 orders Q 527.5

52 weeks = 8 weeks ( 2 months ) per order 6.5 orders Monthly Forecast 792

528

Balance

January

251

541

February

251

289

March

251

566

April

294

272

May

294

506

June

234

272

July

234

565

August

234

331

September

318

541

October

436

105

November

317

316

December

317

−1

Total

3433

Note that the “.5” order was added to the first month. The order size ( Q = 528 ) seems to be adequate to offset seasonal patterns.

Ingredient orders: Chocolate

D = 108,140 lbs. (3433.3 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.) Co = $5, 700 Cc = $0.45 / lb.

Q = 52,340

TC (1) = $353,380, Q = 52,340 TC ( 2 ) = $337,159, Q = 52,340 TC ( 3) = 326, 048, Q = 100, 000 TC ( 4 ) = $319, 023, Q = 150, 000*Optimal Nuts

D = 77, 242 lb. Co = $6,300 Cc = $0.63 / lb.

Q = 39,304 lbs.

TC (1) = $526,834, Q = 39,304 TC ( 2 ) = $507,524, Q = 39,304 TC ( 3) = $488,591, Q = 70, 000*Optimal Filling

D = 61, 794 lb. Co = $4,500 Cc = $0.55

Q = 31, 799

TC (1) = $110,180, Q = 31, 799 TC ( 2 ) = $101,373, Q = 40, 000*Optimal TC ( 3) = $102, 718, Q = 80, 000 Pharr 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.

Supplement 13 Operational Decision-Making Tools: Simulation Answers to Questions S13-1. The Monte Carlo technique is a technique for selecting numbers randomly from a probability distribution for use in a trial run of a simulation. Random numbers are organized into ranges to reflect the probabilities of the different possible occurrences of a random variable. When the random numbers are selected in the simulation, they represent the value of a random variable. S13-2. They are achieved by repeating the simulation many times. S13-3. In general, simulation provides descriptive, probabilistic information about a system. It can be used to find an optimal set of operating characteristics by searching through all possible sets of operating characteristics until the best set is found.

Solutions to Problems S13-1. Time Between Calls(hours) 1 2 3 4 5 6

Cumulative Probability 0.05 0.15 0.45 0.75 0.95 1.00

RN 65 71 20 15 48 89

Time Between Calls 4 4 3 2 4 5

Random Numbers 01–05 06–15 16–45 46–75 76–95 96–99, 00

a. Cumulative Clock 4 8 11 13 17 22

RN 18 83 08 90 05 89 18 08 26 47 94 06 72 40 62

Time Between Calls 3 5 2 5 1 5 3 2 3 4 5 2 4 3 4

Cumulative Clock 25 30 32 37 38 43 46 48 51 55 60 62 66 69 73

=

73 = 3.48 hours between calls 21

EV = 1( 0.05 ) + 2 ( 0.10 ) + 3 ( 0.30 ) = 3.65 + 4 ( 0.30 ) + 5 ( 0.20 ) + 6 ( 0.05 )

The results are different because there were not enough simulations to enable the simulated average to approach the analytical result. c.

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.

S13-2. Machine Breakdowns per Week 0 1 2 3 4 5

Cumulative Probability 0.10 0.20 0.40 0.65 0.95 1.00

Random Numbers 01–10 11–20 21–40 41–65 66–95 96–99, 00

a. Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

=

RN 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Breakdowns 1 2 5 2 0 3 3 5 3 2 4 4 4 4 3 4 4 3 3 0 59

59 = 2.95 breakdowns per week 20

First three columns are from part a. Select as many RN2’s as there are breakdowns from a different random number stream.

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

RN1 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Breakdowns

RN 2

Repair Time

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, 32, 44 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+2+2 =8 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 = 111 hours

=

111 = 5.55 hr / wk 20

It could bias the results. Selecting a high random number, such as 98, 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 hours). Thus, a relationship will result wherein a 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. c.

Total weekly breakdown cost = 111 hours x $50 = $5,550

=

5,550 = $277.50 per week 20

d. Breakdowns per Week 0 1 2 3 4 5

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Cumulative Probability 0.20 0.50 0.70 0.85 0.95 1.00

RN1 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Random Numbers 01–20 21–50 51–70 71–85 86–95 96–99, 00

Breakdowns 0 1 5 1 0 2 1 5 2 1 3 3 3 3 1 3 4 1 2 0

Total repair time = 77 hours

RN 2 — 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, 32, 44 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+2+2 = 6 1+ 3 + 2 +1 = 7 2 1+ 3 = 4 0

Total breakdown cost = 77  $50

= $3,850

=

3,850 20

Average weekly breakdown cost = $192.50 Reduction in average weekly repair cost = 277.50 − 192.50 = $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.

S13-3

Demand/Week 0 1 2 3 4 5 6

Random Numbers 01–04 05–12 15–40 41–80 81–96 97–98 99, 00

Lead Time 1 2 3

Random Numbers 01–60 61–90 91–99, 00

=

Week 0 1 2 3 4 5 6

RN1 — 39 72 37 87 98 10

RN 2 — 73 75 02 — — 47

5, 020 = $251 20

Demand

Order Placed

Order Received

Balance

Carrying Cost

Order Cost

Stockout Cost

Total

— 2 3 2 4 5 1

— 5 5 5 — — 5

— — — 5 5+5 — —

5 3 0 3 9 4 3

— 120 0 120 360 160 120

— 100 100 100 — — 100

— — — — — — —

— 220 100 220 360 160 220

7 8 9 10 11 12 13 14 15

93 21 97 41 80 67 59 63 87

— 95 69 91 — — — 78 47

4 2 5 3 3 3 3 3 4

— 5 5 5 — — — 5 5

5 — — — 5+5 — 5 — —

4 2 0 0 7 4 6 3 0

160 80 0 0 280 160 240 120 0

— 100 100 100 — — — 100 100

— — 400 400 — — — — 400

160 180 500 500 280 160 240 220 500

Week

RN1 56 22 19 78 03

RN 2 — — 16 — —

Demand

Order Placed

Order Received

Balance

Carrying Cost

Order Cost

Stockout Cost

Total

3 2 2 3 0

— — 5 — —

5+5 — — 5 —

7 5 3 5 5

280 200 120 200 200

— — — — —

16 17 18 19 20

S13-4. White Sox Play Designation No advance Groundout Double play Long fly Very long fly Walk Infield single Outfield single Long single Double Long double Triple Home run

Cumulative Probability 0.03 0.42 0.48 0.57 0.65 0.71 0.73 0.83 0.86 0.90 0.95 0.97 1.00

Random Numbers (r) 01–03 04–42 43–48 49–57 58–65 66–71 72–73 74–83 84–86 87–90 91–95 96–97 98–99, 00

280 200 120 200 200 5,020

Yankees Play Designation No advance Groundout Double play Long fly Very long fly Walk Infield single Outfield single Long single Double Long double Triple Home run

Cumulative Probability 0.04 0.42 0.46 0.56 0.62 0.69 0.73 0.83 0.87 0.92 0.95 0.96 1.00

Random Numbers (r) 01–04 05–42 43–46 47–56 57–62 63–69 70–73 74–83 84–87 88–92 93–95 96 97–99, 00

Inning 1:

White Sox

Yankees

RN 39 73 72 75 37 02

Play Groundout I. single I. single O. single Groundout Out

87 98 10 47 93 21

Single Home run Groundout Fly Double Groundout

Outs 1

RBI

1 1 1

____ 1 2

1 1 1

____ 2

Inning 2:

White Sox

Yankees

RN 95 97 69 41 91 80 67 59 63

Play Double Triple Walk Groundout L. Double O. Single Walk V. long fly V. long fly

Outs

RBI

78 87 47 56 22

O. single L. single L. fly L. fly Groundout

RN 19 16 78 03

Play Groundout Groundout O. single Out

Outs 1 1

RBI

____ 0

04 61 23

Out V. long fly Groundout

1 1 1

1 1

1 1 1

1 1

____ 4

1 1 1

1 ____ 1

Inning 3:

White Sox

Yankees

____ 0

Inning 4:

White Sox

Yankees

RN 15 58 93 78 61

Play Groundout V. long fly L. double O. single V. long fly

Outs 1 1

42 77 65 71 18 12

Groundout O. single Walk I. single Groundout Groundout

1 1

RBI

1 ____ 1

Inning 5:

White Sox

Yankees

RN 17 48 89 18

Play Groundout Out Double Groundout

83 08 90 05 89 18

O. single Groundout Double Groundout Double Groundout

Outs 1 1

RBI

____ 0

1 1 1 1 ____ 2

Inning 6:

White Sox

Yankees

RN 08 26 47

Play Groundout Groundout Out

94 06 72 62 47

L. double Groundout I. single V. long fly L. fly

Outs 1 1 1

RBI

____ 0

1 1 1 1

____ 1

Inning 7:

White Sox

Yankees

RN 68 60 88 17 36

Play Walk V. long fly Double Groundout Groundout

77 43 28

O. single Double play Groundout

Outs

RBI

1 1 1

2 1

1 ____ 1

____ 0

Inning 8:

White Sox

Yankees

RN 31 06 68 39

Play Groundout Groundout Walk Groundout

71 22 76 81 88 94 76 23 47

I. single Groundout O. single O. single Double L. double O. single Groundout Out

RN 25 79 08 15

Play Groundout O. single Groundout Groundout

Outs 1 1

RBI

____ 0

1 1 1 2 1 1 1

____ 5

Inning 9:

White Sox

White Sox Yankees

1 2

4 1

Outs 1 1 1

1 1

Line Score: 0 0 2 1

Cumulative Probability 50 .80 1.00

Random Numbers 01–50 51–80 81–99, 00

0 0

S13-5. Months to Receive an Order 1 2 3

Demand per Month 1 2 3 4

Cumulative Probability 0.10 0.40 0.80 1.00

Random Numbers 01–10 11–40 41–80 81–99, 00

RBI

0 0

1 0

0 5

0 X

7 12

Reorder Number

RN1 21 41 14 59 28 68 13 09 20 73 77 29 72 89 81 20 85 59 72 88 11

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Reorder Number 22 23 24 25 26 27 28 29 30

RN1 89 87 59 66 53 45 56 22 49

Months Lead Time

RN 2

1 1 1 2 1 2 1 1 1 2 2 1 2 3 3 1 3 2 2 3 1

70 38 11 70, 52 88 25, 17 18 00 70 04, 11 76, 29 01 11, 42 70, 17, 48 87, 20, 16 56 35, 94, 91 65, 20 97, 46 10, 98, 32 57

Months Lead Time 3 3 2 2 2 1 2 1 1

RN 2 92, 55, 35 57, 99, 20 52, 45 01, 73 58, 24 72 84, 35 41 32

Total demand = 143 Average demand during lead time = 143/30 = 4.76 Thus, should reorder at 5-car level.

Demand per Month 3 2 2 3, 3 = 6 4 2, 2 = 4 2 4 3 1, 2 = 3 3, 2 = 5 1 2, 3 = 5 3, 2, 3 = 8 4, 2, 2 = 8 3 2, 4, 4 = 10 3, 2 = 5 4, 3 = 7 2, 4, 2 = 8 3

Demand per Month 4, 3, 2 = 9 3, 4, 2 = 9 3, 3 = 6 1, 3 = 4 3, 2 = 5 3 4, 2 = 6 3 2 143

S13-6.

Customers/ Day 0 1 2 3

Random Numbers 01–20 21–40 41–90 91–99, 00

Day

Available Cars

Duration 1 2 3 4 5

Customers

RN1 62

3 4

1 2

96 86

3 2

6 7 8 9 10

1 0 1 2 3

29 79 22 08 62

1 2 1 0 2

Random Numbers 01–10 11–40 41–80 81–90 91–99, 00

RN 2 19 66 27 43 20 92 22 91 46 49 — 66 — 09

Duration/ Car

Car Day Available

2 3 2 3 2 5 2 5 3 3 — 3 — 1 81

3 4 4 5 5 9 6 10 8 9 — 11 — 11 4

Customers Not Served

Probability of not having a car available = customers not served/ total customers = 4/17 = 0.235. 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.

S13-7. Time Between Arrivals 5 10 15 20 25 30

Cumulative Probability 0.06 0.16 0.39 0.68 0.86 1.00

Cumulative

RN Ranges 01–06 07–16 17–39 40–68 69–86 87–99,00

Service Doctor (D) Nurse (N) Both (B)

Time 10 15 20 25 30

Patient 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Probability 0.50 0.70 1.00

Doctor Cumulative Probability 0.22 0.53 0.78 0.90 1.00

RN1 — 24 48 27 79 81 64 34 65 22 26 15 52 12 40 27 76 51 38 40

Ranges 01–50 51–70 71–99,00

RN Ranges 01–22 23–53 54–78 79–90 91–99,00

Arrival Clock — 30 50 65 90 115 135 150 170 185 200 210 230 240 260 275 300 320 335 355

Time 5 10 15 20

Service

RN 2 31 01 00 74 77 92 06 53 68 30 43 85 46 26 33 13 55 57 83 71

P ( wait ) =

Nurse Cumulative Probability 0.08 0.32 0.83 1.00

D D B B B B D N N D D B D D D D N N B B

15 = 0.75 20

RN 3 98 56 58 76 48 48 94 88 79 53 52 87 47 56 31 06 13 31 79 94

RN Ranges 01–08 09–32 33–83 84–99,00

Time 15 20 25 30 35 40

Both Cumulative Probability 0.07 0.23 0.44 0.72 0.89 1.00

Time of Service (mins.) Doctor Nurse Wait? 30 20 30 35 30 30 30

30 35 30 30 20 15

15 15 35 15 20 15 10 10 10 35 40

Average waiting time =

35 40

15 25 25 35 20 20 15 15 20 35 60 60 60

10 25

RN Ranges 01–07 08–23 24–44 45–72 73–89 90–99,00

Departure clock D N 30 50 80 115 140 170 200

190 205 215 230 265 300 320 335 345

380 420

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.

S13-8.

Direction

Probability

Cumulative Probability

RN Ranges

80 115 140 170

265

310 330 380 420

( x = +1) ( x = −1) West ( y = +1) North ( y = −1) South East

0.25

01–25

0.25

0.50

26–50

0.25

0.75

51–75

0.25

1.00

76–99, 00

Monte Carlo Simulation (using 16th row of random numbers from Table S12.3). 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. End of Block 1 2

RN 58 47

Trial 1 (x, y) (0, 1) ( −1, 1)

Trial 2 RN (x,y) 68 (0,1) 13 (1,1)

RN 20 85

Trial 3 (x,y) (1,0) (1, − 1)

Trial 4 RN (x, y) 53 (0,1) 45 − ( 1, 1)

RN 24 83

Trial 5 (x, y) (1,0) 1, ( − 1)

(0,1)

(2,1)

(1,0)

( −1, 2 )

( 2, − 1)

(0,2)

(3,1)

(1,1)

(0,2)

( 2, − 2 )

( −1, 2 )

(3,2)

(1,0)

( −1, 2 )

( 3, − 2 )

(0,2)

(3,1)

(2,0)

(0,2)

( 3, − 3)

( −1, 2 )

(2,1)

( 2, − 1)

(0,1)

( 3, − 4 )

(0,2)

(2,2)

( 2, − 2 )

(0,2)

( 2, − 4 )

(0,3)

(2,1)

( 2, − 1)

( −1, 2 )

( 2, − 3)

( −1, 3)

(2,0)

( −2, 2 )

( 3, − 3)

Within 2 blocks?

yes

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. S13-9. Stock Price Movement (+) Increase Same (0) (−) Decrease

Change in Stock Price 1/8 1/4

Probability 0.45

Cumulative Probability 0.45

RN Ranges 01–45

0.30 0.25

0.75 1.00

46–75 76–99, 00

Probability Increase 0.40 0.17

Cumulative Probability 0.40 0.57

RN Ranges 01–40 41–57

Probability Decrease 0.12 0.15

Cumulative Probability 0.12 0.27

RN Ranges 01–12 13–27

3/8 1/2 5/8 3/4 7/8 1

0.12 0.10 0.08 0.07 0.04 0.02

0.69 0.79 0.87 0.94 0.98 1.00

58–69 70–79 80–87 88–94 95–98 99, 00

0.18 0.21 0.14 0.10 0.05 0.05

0.45 0.66 0.80 0.90 0.95 1.00

28–45 46–66 67–80 81–90 91–95 96–99, 00

Monte Carlo Simulation (using the third column of random numbers from Table S12.3)

Day

Stock Price movement

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 + +

Price +, − Change ( ) −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

Stock Price 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.

S13-10. Sales Volume 300 400 500 600 700 800

RN 1 Range 1–12 13–30 31–50 51–73 74–91 91–99, 00

Price $22 23 24 25 26 27

RN2 Range 1–7 8–23 24–47 48–72 73–90 91–99, 00

Variable Cost $8 9 10 11 12

RN3 Range 1–17 18–49 50–78 79–92 93–99, 00

c f = $9, 000

Month

RN1

Sales Volume (V)

RN 2

Price (p)

RN 3

Variable Cost ( C v )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

58 69 41 28 09 77 89 85 88 87 53 22 82 49 05 78 53 79 37 65

600 600 500 400 300 700 700 700 700 700 600 400 700 500 300 700 600 700 500 600

47 35 14 68 20 29 81 59 11 59 45 49 55 24 81 92 04 61 45 37

$24 24 23 25 23 24 26 25 23 25 24 25 25 24 26 27 22 25 24 24

23 21 59 13 73 72 20 72 89 66 56 08 27 83 07 36 75 44 18 30

9 9 10 8 10 10 9 10 11 10 10 8 9 11 8 9 10 9 9 9

Z = VP − 9,000 − VCv 0 0 −2, 200 −2,500 −5,100 +800 +2,900 +1,500 −600 +1,500 −600 −2, 200 +2, 200 −2,500 −3, 600 +3, 600 −1,800 +2, 200 −1,500 0

Probability of at least breaking even =

10 = 0.50 20

S13-11.

Attractiveness 1 2 3 4 5

Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

RN1 Range 1–27 28–62 63–76 77–85 86–99, 00

RN1 95 93 09 95 56 79 61 14 85 09 60 86 69 96 78 61 04 16 82 03

Intelligence 1 2 3 4 5

Attractiveness 5 5 1 5 2 4 2 1 4 1 2 5 3 5 4 2 1 1 4 1

RN 2 Range 1–10 11–26 27–71 72–88 89–99, 00

RN 2 30 28 54 36 26 14 81 24 49 53 98 74 08 06 22 18 23 20 88 65

Personality 1 2 3 4 5

Intelligence 3 3 3 3 2 2 4 2 3 3 5 4 1 1 2 2 2 2 4 3

RN 3 Range 1–15 16–45 46–78 79–85 86–99, 00

RN 3 59 72 66 98 60 50 84 75 08 45 90 55 10 62 99 45 63 99 85 33

Personality

Average Rating

3 3 3 5 3 3 4 3 1 2 5 3 1 3 5 2 3 5 4 2

3.67 3.67 2.33 4.33 2.33 3.00 3.33 2.00 2.67 2.00 4.00 4.00 1.67 3.00 3.67 2.00 2.00 2.67 4.00 2.00

Average overall rating of Salem dates = 2.92 S13-12. There are several ways to access 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 stimulated result. E ( Attractiveness ) = 2.50 E ( Intelligence ) = 3.05

E ( Personality ) = 2.77 Average rating =

2.50 + 3.05 + 2.77 = 2.77 3

This is relatively close to the simulated result of 2.92 which tends to verify that result. Confidence limits can also be developed for the average rating. However, this is best done with Excel.

14 - Sales and Operations Planning Answers to Questions 14-1.

The purpose of sales and operations planning (S&OP) is to determine the resource capacity a firm will need to meet its demand over an intermediate time horizon. A S&OP process is a structured collaborative decision making process involving the top management and all functional areas of the firm to develop a company “game plane” under which all functional areas operate. The game plan is created by updating and reconciling a sales plan and a production plan monthly to iron out any differences in supply, demand and new product plan.

14-2.

Capacity can be adjusted by using overtime and undertime, hiring or firing workers, subcontracting out or in, using part-time or temporary workers, adding an extra shift, working more or less days in a week, and making current resources more flexible. Demand can be managed by shifting it to other periods through incentives to buy or use the product or service at off-peak times, days, and seasons. Demand can be smoothed by finding another product or service with opposite demand patterns. Demand distortion can be eliminated through collaborative planning.

14-3.

The output of aggregate production planning is a production plan that specifies how many workers are needed in each period, the amount and type of production (i.e., regular, overtime, subcontracting, etc.), and the units to be produced, stored, and back ordered per month or per quarter. The “units” of production capacity are expressed in general, or aggregate terms.

14-4.

Linear programming provides an optimum solution using linear cost functions. The linear decision rule provides an optimum solution using quadratic cost functions. The search decision rule provides a satisfactory solution using any type of cost function. The management coefficients model does not use any cost functions; rather, it bases its decision on past experience. There is no guarantee or measure of optimality.

14-5.

Seasonal clothing has a highly variable demand pattern. Since styles change, inventory is not always a good alternative. Most stores offer deep discounts to extend the buying season. Fruits and vegetables have highly variable supply patterns.

14-6. a.

b. c.

14-7.

Classroom sizes could be expanded, schools could be open longer hours, more days of the week, more days during the year, or more seasons of the year, students could be bused to less crowded schools, learning could take place off-site, students could be allowed to advance at their own pace, teaching could be compressed, and students could attend school in shifts (i.e., morning, afternoon, evening). More prisoners per cell, early release, temporary housing, alternative waking/sleeping hours, reduced sentences, and alternatives to incarceration are all possibilities. More flights on a certain route, more routes, larger aircraft, and more utilization of aircraft are all possible.

Part-time workers are very flexible, and they cost less than full-time workers because of fewer benefits, lower wage rate, etc. However, turnover is high, availability may be a problem, training may be costly or lengthy, and the quality of work may not be as high or as consistent as with a full-time work force. Subcontracting can get work out faster, but care must be taken to ensure acceptable quality levels. Also, trade secrets may be a risk with subcontracting, and over an extended period of time, the company may lose its ability to perform the subcontracted work. Inventory is expensive to maintain. It must be moved, sorted, loaded, unloaded, and tracked. Money tied up

in inventory that’s just sitting around could be invested for a higher return. Inventory can “go bad” through obsolescence, pilferage, or expired shelf life. It’s hard to predict the consumer’s demand preferences ahead of time, and companies may find themselves stuck with excess inventory. 14-8.

The hierarchical planning process for production moves from aggregate production planning for product lines, to master production scheduling for individual products, to material requirements planning for components, and shop floor scheduling for manufacturing operations. The hierarchical planning process for capacity is similar, moving from resource requirements planning for plants, to rough-cut capacity plans for critical work centers, to capacity requirements planning for all work centers, and input/output control for individual machines.

14-9.

Collaborative planning involves selecting the products to be jointly managed, creating a single forecast of customer demand, and synchronizing production across the supply chain. Each partner has access to an Internet-enabled planning book in which forecasts, customer orders and production plans are visible. Events trigger responses by partners. Available-to-promise, for example, executes a series of rules when assessing product availability and alerts the planner when customer orders exceed or fall short of forecasts.

14-10.

Student responses will vary. Some vendors have special modules for these topics, others incorporate them into a collaborative planning module.

14-11.

Services typically experience more dramatic demand swings over shorter periods of time. Building and depleting inventory may not be an option. Demand is difficult to predict, and capacity is not easy to measure. Resources, however, are more flexible, since most services are labor intensive and most workers are cross trained.

14-12.

Revenue management involves determining the percentage of seats or rooms to be allocated to different fare classes in order to maximize profit. It also determines the optimal number of overbookings to accept. Restaurants, doctors’ offices, cruise ships, and most professional services would benefit from some type of revenue management. Consumers usually find the practice confusing, frustrating, and unfair.

Solutions to Problems 14.1 Beg. Wkrs

Production

Units/wkr

10,000

Hiring

$700

Beg. Inv.

Firing

$700

Inventory

a) Chase Demand

COST =

$9,024,500

Quarter

Demand

Production

Inventory

# Workers

# Hired

# Fired

150,000

250,000

350,000

250,000

1,000,000

b) Level Production

COST =

$9,403,500

Quarter

Demand

Production

Inventory

# Workers

# Hired

# Fired

150,000

250,000

100,000

250,000

100,000

350,000

250,000

1,000,000

200,000

Choose chase demand

14.2

Input:

Month Apr May Jun July Aug Sept Total b. Month Apr May Jun July Aug Sept Total

Pure strategy – Chase Demand (Note: Level production is not feasible because of April’s demand) Beg. Wkrs Units/wkr Beg. Inv.

25 1000 0

Regular Overtime Subk

$9 $10 $12

Hiring Firing Inventory

$1,500 $1,500 $2

Demand 60,000 22,000 15,000 46,000 80,000 15,000 238,000

Reg 60,000 22,000 15,000 46,000 80,000 15,000 238,000

OT 0 0 0 0 0 0 0

Subk 0 0 0 0 0 0 0

Inv 0 0 0 0 0 0 0

#Wkrs 60 22 15 46 80 15

$2,457,000

#Hired 35 0 0 31 34 0 100

#Fired 0 38 7 0 0 65 110

Mixed Strategy – Keep current workforce and supplement with OT and Subk Demand 60,000 22,000 15,000 46,000 80,000 15,000 238,000 Cost: $2,367,000

Month Apr May Jun July Aug Sept Total

Cost:

Demand 60,000 22,000 15,000 46,000 80,000 15,000 238,000

Cost: $2,346,999

Reg 25,000 25,000 25,000 25,000 25,000 15,000 140,000

OT 25,000 0 0 8,000 25,000 0 58,000

Subk 10,000 0 0 0 30,000 0 40,000

Inv 0 3,000 13,000 0 0 0 16,000

#Wkrs 25 25 25 25 25 15

#Hired 0 0 0 0 0 0 0

#Fired 0 0 0 0 0 10 10

This is very near to the optimal solution. Reg 32,600 32,600 32,600 32,600 32,600 15,000 178,000

OT 27,400 0 0 0 32,600 0 60,000

Subk 0 0 0 0 0 0 0

Inv 0 10,600 28,200 14,800 0 0 53,600

#Wkrs 33 33 33 33 33 15

This is the optimum solution to the problem.

#Hired 8 0 0 0 0 0 8

#Fired 0 0 0 0 0 18 18

14.3 a.

Input:

Level production

Beg. Wkrs Units/wkr Beg. Inv.

40 1250 0

Regular Inventory

$10 $1

Hiring Firing

$500 $500

QTR 1 2 3 4

Demand 70,000 100,000 50,000 150,000

Reg 92,500 92,500 92,500 92,500

Inv 22,500 15,000 57,500 0

#Wkrs 74 74 74 74

#Hired 34 0 0 0

#Fired 0 0 0 0

Total

370,000

95,000

Input:

Beg. Wkrs

Regular

$10

Hiring

$500

Chase Demand

Units/wkr

1250

Inventory

Firing

$500

Cost:

Beg. Inv.

QTR

Demand

Reg

Inv

#Wkrs

#Hired

#Fired

70,000

100,000

0 40

50,000

150,000

120

Total

370,000

120

Input:

Beg. Wkrs

Regular

$10

Hiring

$500

Units/wkr

1250

Inventory

Firing

$500

Beg. Inv.

Cost:

$3,812,000

$3,780,000

Mixed Cost:

$3,770,000

Best solution

QTR

Demand

Reg

Inv

#Wkrs

#Hired

#Fired

70,000

100,000

50,000

100,000

50,000

150,000

100,000

Total

370,000

50,000

Input:

Beg. Wkrs

Regular

$10

Hiring

$500

Mixed

Units/wkr

1250

Inventory

Firing

$500

Cost:

Beg. Inv.

QTR

Demand

Reg

Inv

#Wkrs

#Hired

#Fired

70,000

90,000

20,000

100,000

90,000

10,000

50,000

90,000

50,000

150,000

100,000

Total

370,000

80,000

$3,800,000

14.4 a.

Input:

Beg. Wkrs Units/wkr Beg. Inv.

Month

Demand

Reg

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 24,000

8 250 0

Regular Overtime Subk

$15 $25 $30

Hiring Firing Holding Backordering

Subk

Inventory

Backorder

1,000 2,500 4,000 4,000 3,000 1,000 0 0 1,000 2,500 4,000 3,000 26,000

0 0 0 0 0 0 0 0 0 0 0 0 0

2,000 1,000

3,000

$100 $200 $0.50 $10

#Wkr s 8 8 8 8 8 8 8 8 8 8 8 8 96

#Hired

#Fired

0 0 0 0 0 0 0 0 0 0 0 0 0

Cost: $448,000 Note: Demand cannot be met with level production alone. Overtime is needed in July and August. Ending inventory is also greater than zero.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

Reg 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 24,000

Subk

Inventory 1,000 2,500 4,000 4,000 3,000 1,000 0 0 0 0 1,000 0 16,500

Backorder 0 0 0 0 0 0 2,000 3,000 2,000 500 0 0 7,500

#Wkrs 8 8 8 8 8 8 8 8 8 8 8 8 96

#Hired 0 0 0 0 0 0 0 0 0 0 0 0 0

#Fired 0 0 0 0 0 0 0 0 0 0 0 0 0

Inventory 0 0 0 0 0 0 0 0 0 0 0 0 0

Backord 0 0 0 0 0 0 0 0 0 0 0 0 0

#Wkrs 4 2 2 8 12 16 20 12 4 2 2 12 96

#Hired 0 0 0 6 4 4 4 0 0 0 0 10 28

#Fired 4 2 0 0 0 0 0 8 8 2 0 0 24

Inventory 0 0 0 1,000 1,000 0 0 0 2,000 2,000 2,000 0 8,000

Backord 0 0 0 0 0 0 0 0 0 0 0 0 0

#Wkrs 4 2 2 12 12 12 12 12 12 2 2 4 88

#Hired 0 0 0 10 0 0 0 0 0 0 0 2 12

#Fired 4 2 0 0 0 0 0 0 0 10 0 0 16

Cost: $443,250 c.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

Reg 1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

Subk

Cost: $367,600 d.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

Reg 1000 500 500 3000 3000 3000 3000 3000 3000 500 500 1000 22,000 Cost:

Subk

2,000

$388,400

**Chase demand is the best of the strategies tested.

14-5. a.

Level Production Input:

Month Mar Apr May June July Aug Sept Oct Nov Dec Jan Feb Total

Demand 2,000 1,000 1,000 1,000 1,000 1,500 2,500 3,000 9,000 7,000 4,000 3,000 36,000

Beg. Wkrs Units/wkr Beg. Inv.

10 200 0

Regular Overtime Subk

$30 $40 $50

Hiring Firing Holding

Reg 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 36,000

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 1,000 3,000 5,000 7,000 9,000 10,500 11,000 11,000 5,000 1,000 0 0 63,500

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 0 0 0 0 0 0 0 0 0 0 0 0

$5,000 $8,000 $2

#Wkrs 15 15 15 15 15 15 15 15 15 15 15 15

Cost:

$1,232,000

#Hired 5 0 0 0 0 0 0 0 0 0 0 0 5

#Fired 0 0 0 0 0 0 0 0 0 0 0 0

#Hired 0 0 0 0 0 3 5 3 30 0 0 0 40

#Fired 0 5 0 0 0 0 0 0 0 10 15 5 35

b. Chase Demand Month Mar Apr May June July Aug Sept Oct Nov Dec Jan Feb Total

Demand 2,000 1,000 1,000 1,000 1,000 1,500 2,500 3,000 9,000 7,000 4,000 3,000 36,000

Reg 2,000 1,000 1,000 1,000 1,000 1,500 2,500 3,000 9,000 7,000 4,000 3,000 36,000 Cost: $1,560,000

#Wkrs 10 5 5 5 5 8 13 15 45 35 20 15

Current Workforce, supplement with OT and Subk as needed Month Mar Apr May June July Aug Sept Oct Nov Dec Jan Feb Total

Demand 2,000 1,000 1,000 1,000 1,000 1,500 2,500 3,000 9,000 7,000 4,000 3,000 36,000

Reg 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 24,000

OT 0 0 0 0 0 1,000 2,000 2,000 2,000 2,000 2,000 1,000 12,000

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 1,000 2,000 3,000 4,000 5,500 7,000 8,000 3,000 0 0 0 33,500

#Wkrs 10 10 10 10 10 10 10 10 10 10 10 10

#Hired 0 0 0 0 0 0 0 0 0 0 0 0 0

#Fired 0 0 0 0 0 0 0 0 0 0 0 0 0

Cost: $1,267,000 For this problem, it is cheaper to use overtime in previous periods and hold it in inventory rather than subcontract. The difference in subcontracting and overtime is $10 a unit. Holding cost is $2 per period. Thus, overtime can be used up to 5 periods back before the cost exceeds subcontracting. 14.6 a.

Input:

Beg. Wkrs Units/wkr Beg. Inv.

Month Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Total

Demand 1500 1000 600 600 600 800 1000 1000 4000 6500 6000 4000 27,600

25 100 0

Reg 2,300 2,300 2,300 2,300 2,300 2,300 2,300 2,300 2,300 2,300 2,300 2,300 27,600

Cost : $1,638,800 Level Production

Regular Overtime Subk

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

$40 $60 $70

Inv 800 2,100 3,800 5,500 7,200 8,700 10,000 11,300 9,600 5,400 1,700 0 66,100

Hiring Firing Inventory

#Wkrs 23 23 23 23 23 23 23 23 23 23 23 23

#Hired 0 0 0 0 0 0 0 0 0 0 0 0 0

$2,000 $3,000 $8

#Fired 2 0 0 0 0 0 0 0 0 0 0 0 2

Month Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Total

Demand 1500 1000 600 600 600 800 1000 1000 4000 6500 6000 4000 27,600

Reg 1500 1000 600 600 600 800 1000 1000 4000 6500 6000 4000 27,600

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

Cost = $1, 354,000

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 0 0 0 0 0 0 0 0 0 0 0 0

#Wkrs 15 10 6 6 6 8 10 10 40 65 60 40

#Hired 0 0 0 0 0 2 2 0 30 25 0 0 59

#Fired 10 5 4 0 0 0 0 0 0 0 5 20 44

Inv 500 1,500 2,900 4,300 5,700 6,900 7,900 8,900 6,900 2,400 400 0 48,300

#Wkrs 20 20 20 20 20 20 20 20 20 20 20 20

#Hired 0 0 0 0 0 0 0 0 0 0 0 0 0

#Fired 5 0 0 0 0 0 0 0 0 0 0 0 5

Chase Demand

Savings = $284,800 c.

Month Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Total

Demand 1500 1000 600 600 600 800 1000 1000 4000 6500 6000 4000 27,600

Reg 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 24,000

OT 0 0 0 0 0 0 0 0 0 0 2,000 1,600 3,600

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Cost = $1, 577, 400 Steady workforce Savings = $61,400

14.7 a. Current Workforce with Overtime and Subcontracting as needed Input:

Beg. Wkrs Units/wkr Beg. Inv.

10 200 0

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 6400 7000 1500 500 600 1400 1600 2000 1400 1500 5200 6900 36,000

Reg 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 24,000

Regular Overtime Subk

$8 $12 $16

OT 2,000 2,000

Subk 2,400 3,000

2,600 6,600

5,400

Hiring Firing Inventory

$500 $500 $2

Inv 0 0 500 2,000 3,400 4,000 4,400 4,400 5,000 5,500 2,300 0 31,500

#Wkrs 10 10 10 10 10 10 10 10 10 10 10 10 120

#Hired 0 0 0 0 0 0 0 0 0 0 0 0 0

#Fired 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 0 0 0 0 0 0 0 0 0 0 0 0

#Wkrs 32 35 8 3 3 7 8 10 7 8 26 35 180

#Hired 22 3 0 0 1 4 1 2 0 1 19 9 60

#Fired 0 0 28 5 0 0 0 0 3 0 0 0 36

Cost: $421,800 b. Chase Demand Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 6400 7000 1500 500 600 1400 1600 2000 1400 1500 5200 6900 36,000

Reg 6400 7000 1500 500 600 1400 1600 2000 1400 1500 5200 6900 36,000

Cost: $335,750

Subk

c. Input:

Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Total

Beg. Wkrs Units/wkr Beg. Inv.

10 200 0

Regular Overtime Subk

Demand 6400 7000 1500 500 600 1400 1600 2000 1400 1500 5200 6900 36,000

Reg 6,600 6,600 1,400 600 600 1,400 1,600 1,600 1,600 1,600 5,800 6,000 35,400

OT 0 200 100 0 0 0 0 300 0 0 0 0 600

$8 $12 $16

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Hiring Firing Inventory

$500 $500 $2

Inv 200 0 0 100 100 100 100 0 200 300 900 0 2,000

#Wkrs 33 33 7 3 3 7 8 8 8 8 29 30

Cost:

#Hired 23 0 0 0 0 4 1 0 0 0 21 1 50

$334,400

#Fired 0 0 26 4 0 0 0 0 0 0 0 0 30

This solution assumes no part-time workers are allowed; i.e., an integer constraint has been added.

Input:

Beg. Wkrs Units/wkr Beg. Inv.

Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Total

Demand 6400 7000 1500 500 600 1400 1600 2000 1400 1500 5200 6900 36,000

10 200 0

Reg 2,000 2,000 1,400 600 600 1,200 1,600 1,600 1,600 1,600 1,600 1,600 17,400

Regular Overtime Subk

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

$8 $12 $9

Hiring Firing Inventory

Subk 4,400 5,000 100 0 0 100 0 400 0 0 3,300 5,300 18,600

Inv 0 0 0 100 100 0 0 0 200 300 0 0 700

$500 $500 $2

#Wkrs 10 10 7 3 3 6 8 8 8 8 8 8

Cost:

#Hired 0 0 0 0 0 3 2 0 0 0 0 0 5

$314,000

#Fired 0 0 3 4 0 0 0 0 0 0 0 0 7

Savings $20,400 Other factors to consider include the quality of the product, the leadtime to obtain the product, damage in shipment, extra inventory to cover demand during leadtime, intellectual property rights, shipping costs, and stability. Note: this solution assumes part-time workers are not allowed. If part-time workers are allowed, the cost would be $313,375

14-8. a.

Input:

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Beg. Wkrs Units/wkr Beg. Inv.

Demand 500 500 1000 1200 2000 400 400 1000 1000 1500 7000 500 17,000

10 50 0

Reg 500 500 1,000 1,000 1,000 400 400 1,000 1,000 1,500 3,500 500 12,300 Cost:

OT 0 0 0 200 1,000 0 0 0 0 0 3,500 0 4,700

Regular Overtime Subk

$16 $24 $33

Hiring Firing Inventory

$500 $500 $20

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 0 0 0 0 0 0 0 0 0 0 0 0

#Wkrs 10 10 20 20 20 8 8 20 20 30 70 10 246

#Hired 0 0 10 0 0 0 0 12 0 10 40 0 72

#Fired 0 0 0 0 0 12 0 0 0 0 0 60 72

Subk 0 0 0 0 800 0 0 0 0 0 5,500 0 6,300

Inv 0 0 0 0 0 150 300 0 0 0 0 0 450

#Wkrs 10 10 12 12 12 11 11 14 15 15 15 10 147

#Hired 0 0 2 0 0 0 0 3 1 0 0 0 6

#Fired 0 0 0 0 0 1 0 0 0 0 0 5 6

$381,600

b. Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand 500 500 1000 1200 2000 400 400 1000 1000 1500 7000 500 17,000

Reg 500 500 600 600 600 550 550 700 750 750 750 500 7,350

OT 0 0 400 600 600 0 0 0 250 750 750 0 3,350

Cost: $432,900 No, more work is subcontracted with the higher penalty for firing.

14-9. Input:

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Beg. Wkrs Units/wkr Beg. Inv.

Demand 1000 1200 1200 3000 3000 3000 2200 2200 4000 4000 2200 3000 30,000

8 100 0

Reg 1,000 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 1,500 17,500

Regular Overtime Subk

$36 $54 $70

Subk 0 0 0 0 750 750 400 1,000 1,000 1,000 0 700 5,600

Inv 0 300 750 0 0 0 450 1,500 750 0 50 0 3,800

OT 0 0 150 750 750 750 750 750 750 750 750 750 6,900

Hiring Firing Inventory

#Wkrs 10 15 15 15 15 15 15 15 15 15 15 15 175

$0 $0 $10

#Hired 2 5 0 0 0 0 0 0 0 0 0 0 7

#Fired 0 0 0 0 0 0 0 0 0 0 0 0 0

Cost: $1,432,600 With a limit on the workforce size, neither chase demand nor level production is viable. This solution uses the maximum workforce size of 15 after the first month and supplements with overtime and subcontracting as needed. Remember that overtime is limited to half of regular production and subcontracting is limited to 1000 units. Also, using overtime in the previous period is cheaper than subcontracting in the current period.

14-10. Input:

Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Total

Beg. Wkr Units/wkr Beg. Inv.

8 250 0

Demand 1000 500 500 2000 3000 4000 5000 3000 1000 500 500 3000 24,000

Reg 1,000 500 500 2,000 3,000 4,500 4,500 3,000 1,000 1,000 1,000 2,000 24,000

Regular Overtime Subk

$15 $25 $30

OT 0 0 0 0 0 0 0 0 0 0 0 0 0

Inventory 0 0 0 0 0 500 0 0 0 500 1,000 0 2,000

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Hiring Firing Holding Backordering

Backordrs 0 0 0 0 0 0 0 0 0 0 0 0 0

$100 $200 $0.50 $10

#Wkrs 4 2 2 8 12 18 18 12 4 4 4 8 20

Cost:

#Hired 0 0 0 6 4 6 0 0 0 0 0 4 20

$367,000

#Fired 4 2 0 0 0 0 0 6 8 0 0 0 20

14-11. Input:

Beg. Wkrs Units/wkr Beg. Inv.

Month Mar Apr May June July Aug Sept Oct Nov Dec Jan Feb Total

Demand 2,000 1,000 1,000 1,000 1,000 1,500 2,500 3,000 9,000 7,000 4,000 3,000 36,000

Input:

Beg. Wkrs Units/wkr Beg. Inv.

10 200 0

Reg 2,000 2,000 2,000 2,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 32,000

Regular Overtime Subk

OT 0 0 0 0 0 0 0 0 0 3,000 1,000 0 4,000

$30 $40 $50

Hiring Firing Holding

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 1,000 2,000 3,000 5,000 6,500 7,000 7,000 1,000 0 0 0 32,500

$5,000 $8,000 $2

#Wkrs 10 10 10 10 15 15 15 15 15 15 15 15

Cost:

$1,210,000

#Hired 0 0 0 0 5 0 0 0 0 0 0 0 5

#Fired 0 0 0 0 0 0 0 0 0 0 0 0 0

14-12.

Month Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Total

Demand 1500 1000 600 600 600 800 1000 1000 4000 6500 6000 4000 27,600

25 100 0

Reg 1,500 1,000 650 650 650 650 1,000 3,500 4,000 4,000 4,000 4,000 25,600

Regular Overtime Subk

OT 0 0 0 0 0 0 0 0 0 0 2,000 0 2,000

$40 $60 $70

Hiring Firing Inventory

Subk 0 0 0 0 0 0 0 0 0 0 0 0 0

Inv 0 0 50 100 150 0 0 2,500 2,500 0 0 0 5,300

$2,000 $3,000 $8

#Wkrs 15 10 7 7 7 7 10 35 40 40 40 40

Cost:

$1,308,900

#Hired 0 0 0 0 0 0 4 25 5 0 0 0 34

#Fired 10 5 4 0 0 0 0 0 0 0 0 0 19

14-13.

Period of Production 1

Unused Capacity

600

1,000

Overtime

200

500

300

500

Regular

1,000

Overtime

500

800

200

1,000

Overtime

100

500

400

Subk

500

Regular

1,000

Overtime

500

8,000

2,200

Total Cost

$45,900

Regular

Subk Units Produced

600

2,100

800

1,800

5,300

Demand

600

2,100

800

1,800

5,300

Demand 600 2,100 800 1,800

Production Plan Regular OT 1,000 200 1,000 500 1,000 100 1,000 500 4,000 1,300

Subk 0 0 0 0 0

Inv 600 0 300 0 900

Total Cost

$45,900

Unmet Demand

Qtr 1 2 3 4

Capacity

400

Subk 3

Units Produced

Regular

Subk 2

Period of Use 2 3

NOTE: This optimum solution was obtained through Solver. If the problem is worked by hand beginning in the NW corner and allocating the most units possible to the lowest cost cells, the solution would be $46,200.

14-14.

There are multiple optimal solutions to this problem. Use Solver to find the alternate optimum solution.

14-15.

There are multiple optimal solutions to this problem. Use Solver to find the alternate optimum solution.

14-16.

14-17.

14-18.

14-19.

14-20.

14-21.

14-22.

14-23. On hand =

ATP in period 1 = ( 60 + 250 ) − (85 + 125 + 95 ) = 5 ATP in period 4 = 250 − ( 85 + 45 + 15 ) =

105

14-24. On hand =

ATP in period 1 = (10 + 100 ) − ( 56 + 17 ) = 37

ATP in period 3 = 100 − ( 75 + 50 ) = −25 0

ATP in period 5 = 100 − (16 + 14 ) = 70 14-25. On hand =

ATP in period 1 = ( 30 + 100 ) − 75 = 55

ATP in period 2 = ( 50 − 50 ) = 0 ATP in period 3 = 100 − 116 = −16 ATP in period 4 = 50 − 73 = −23 ATP in period 5 = 100 − 45 = 55 ATP in period 6 = 50 − 23 = 27

0 0

14-26.

On hand = 100

Period

Forecast

100

125

175

MPS

200

ATP

ATP in period 1 = (100 + 200) - (50 + 125 + 75) = 50 15 ATP in period 4 = 200 - (175 + 45 + 15 ) = -35 0

14-27. On hand =

ATP in period 1 = (10 + 100 ) − 25 = 85 ATP in period 2 = 100 − 50 = 50

ATP in period 3 = 100 − 137 = −37 0 ATP in period 4 = 100 − 72 = 28 ATP in period 5 = 100 − 23 = 77 ATP in period 6 = 100 − 5 = 95

An order for 250 B’s could not be filled until period 6.

14-28.

P(N  X ) =

Cu = 50/ ( 50 + 100 ) = 0.333 Cu + Co

14-29. a.

Co = 450

Cu = 300 − ( 6000 /120 ) = 250 P ( N  X ) = 250 / ( 250 + 450 ) = 0.36 Choose the next lowest probability.

.36

Overbook by 2 seats. b.

Co = 450

Cu = 200 − ( 6000 /120 ) = 150

P ( N  X ) = 150 / (150 + 450 ) = 0.25 Overbook by 1 seat. 14-30.

Cu = 100 − 30 = 70 Co = 80

P ( N  X ) = 70 / ( 70 + 80 ) = 0.467

Blue Roof should change its overbooking policy to one room.

14-31.

Cu = 169 − 89 = 80 Co = 169

P ( N  X ) = 80 / (80 + 169 ) = 0.321

Reserve 55 seats for full fare passengers. 14-32.

Cu = 350 − 120 = 230 Co = 350 P (N < X) = 230 / (230 + 350) = .3965 = .40

Reserve 30 seats for full-fare passengers, and 30 seats for discounted fare passengers. See Table 14.4 for premium ticket cost descriptions. 14-33.

Cu = 25 − 10 = 15

Co = 9

P ( N  X ) = 15 / (15 + 9 ) = 0.63

Cut 20 trees

14-34.

Cu = 5 − 2 = 3 Co = 2 − 1 = 1

P ( N  X ) = 3/ ( 3 + 1) = 0.75

14-35.

a. b. c.

Room rate = $125 Cost per room = $50 Replacement room = $100 Co = $100 Cu = (125-50) = $750 P(N<X) = Cu / (Cu + Co) = 75 / (75 + 100) = 0.43

No-shows 0 1 2 3

Frequency 10 5 6 9

Probability 0.33 0.17 0.20 0.30

P(N<X) 0.00 0.33 0.50 0.70

Overbook by 1 room

14-36. Discounted room rate Lost revenue Replacement room

$95 $600 $450

Co = Cu =

$450 (600 - 95) =

a. b. c.

$505

P(N<X) = Cu / (Cu + Co) = 505 / (505 + 450) = 0.53 No-shows 0 1 2 3

Overbook by 2 rooms.

Frequency 1 2 3 4 10

Probability 0.10 0.20 0.30 0.40

P(N<X) 0.00 0.10 0.30 0.60

CASE SOLUTION 14.1 – SEATS FOR SALE a)

Cost of regular seat Cost of premium seat No. of seats in stadium

$40 $200 75,000

Cu = $200 − $40 = $160 Co = $40 P ( N  X) =

160 0.80 160 + 40

Revenue management produces $130,000 more revenue than the expected value alone.

CASE SOLUTION 14.2: Erin’s Energy Plan

Total cost = $706,948

Supplement 14 Operational Decision-Making Tools: Linear Programming Answers to Questions S14-1.

A linear programming model includes only linear relationships in the objective function and constraints.

S14-2.

First, define the decision variables; then, develop the objective function; finally, develop the constraints.

S14-3.

a. b. c. b. c.

Plot the model constraints as equations and then determine the feasible solution space. Plot the objective function and move this line out from the origin until the optimal solution point is reached. Solve simultaneous equations at the solution point to find the optimal solution values. Or: Solve simultaneous equations at each corner point. Substitute the values of the decision variables determined at each comer point to find the solution that results in the maximum or minimum value of Z.

S14-4.

The objective function line will fall directly on the constraint line, which indicates that all the points along the constraint line are optimal.

S14-5.

It is limited to linear programming models with only two variables, it is cumbersome, and the graph itself will not always be accurately drawn. It provides a picture of how the solution of a linear programming model is derived and thus helps one understand the mathematical solution better.

S14-6.

The feasible solution area is the area that satisfies all the constraints simultaneously.

S14-7.

It is the last corner point the objective function touches as it leaves the feasible solution space.

S14-8.

It represents an unused resource or activity and thus contributes nothing to the achievement of the objective function.

Solutions to Problems S14-1.

x1 = no. of yards of denim x2 = no. of yards of corduroy Maximize Z = $2.25x1 + 3.10 x2 subject to

5.0 x1 + 7.5 x2  6,500 lb cotton 3.0 x1 + 3.2 x2  3,000 lb cotton x2  510

x1 , x2  0

x1 = 0

  x2 = 510 

x1 = 456   Z = $2, 607 optimal x2 = 510 

x1 = 1000   Z = $2, 250 x2 = 0 

Z = $1,581

x1 = 456; x2 = 510; Z = $2,607

c. S14-2.

Points

a. x1 = no. of clocks

x2 = no. of radios x3 = no. of toasters Maximize Z = 7 x1 + 10 x2 + 7 x3 subject to

8x1 + 10 x2 + 5 x3  2,000 2 x1 + 3x2 + 2 x3  660 x1  200 x2  300 x3  150 x1 , x2 , x3  0 b.

x1 = 37.5; x2 = 95; x3 = 150; Z = $2, 262.50

S14-3.

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

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

x1a + x1b + x1c = 40 x2a + x2b + x2c = 40 x3a + x3b + x3c = 40 x1a + x2a + x3a  40 x1b + x2b + x3b  60 x1c + x2c + x3c  50 xij  0 Solution: x1b = 40; x2a = 40; x3c = 40; s4 = 10; s5 = 20; s6 = 10; Z = $212,000 S14-4.

x1 = no. of sofas x2 = no. of tables x3 = no. of chairs Maximize Z = 320 x1 + 275x2 + 190 x3 subject to

7 x1 + 5x2 + 4 x3  2, 250 12 x1 + 7 x3  1, 000 9 x1 + 7 x2 + 5 x3  240 2 x1 + x2 + x3  650 x1 , x2 , x3  0

S14-5.

x2 = 34; s1 = 2078.57; s2 = 1000; s4 = 466; Z = $9, 428.57

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)

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

subject to

0.6 xcs − 0.4 xbs − 0.4 xos − 0.4 xms  0 −0.3xbs + 0.7 xos − 0.3xcs − 0.3xms  0 0.4 xbd − 0.6 xod − 0.6 xcd − 0.6 xmd  0 −0.1xbd − 0.1xod − 0.1xcd + 0.9 xmd  0 −0.6 xbr − 0.6 xor − 0.6 xcr + 0.4 xmr  0 0.7 xbr − 0.3xor − 0.3xcr − 0.3xmr  0 xbs + xbd + xbr  110 xos + xod + xor  80 xcs + xcd + xcr  70 xms + xmd + xmr  150 xi  0 b.

xos = 52.5 Special : xos + xcs + xms = 175 lbs. xcs = 70 Dark : xbd + xmd = 87 lbs. xms = 52.5 Regular : xbr + xor + xmr = 148 lbs. xbd = 52.2 xod = 26.1 xmd = 8.7 xbr = 57.8 xor = 1.4 xmr = 88.8

Z = $1, 251 S14-6.

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 = 22 x1 + 18x2 + 35x3 + 29 x4 + 30 x5 + 28 x6 + 25 x7 + 36 x8 + 18 x9 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 Solution: x2 = 1; x4 = 1; x9 = 1; Z = 65 S14-7. a. Minimize Z = 11x1 + 16x2 s.t. x1 + x2 = 500 .07x1 + .02x2 ≤ 25 ≥ .20 ≥ .20 x1 ,x2 ≥ 0

x1 = 300 x2 = 500. Z = 6,500

b. Minimize Z = 11x1 + 16x2 s.t. x1 + x2 = 500 .05x1 + .02x2 ≤ 25 ≥ .20 ≥ .20 x1 ,x2 ≥ 0

x1 = 400 x2 = 100. Z = 6,000

c. Minimize Z = .07x1 + .02x2 s.t. 11x1 + 16x2 ≤ 7,000 x1 + x2 = 500 ≥ .20 ≥ .20 x1, x2 ≥ 0

x1 = 200 x2 = 300 Z = 20

S14-8.

x1 = no. of hours molding x2 = no. of hours smoothing x3 = no. of hours painting Maximize Z = 175 ( 7 x1 ) = 1225 x1 subject to

8x1 + 5x2 + 6.5x3  3, 000 x1 + x2 + x3  120 90 ( 7 x1 )  10, 000

 7 x1 − 12 x2 = 0  12 x − 10 x = 0  3  2   7 x1 − 10 x3 = 0 

7 x1 = 12 x2 = 10 x3 x1 , x2 , x3 ,  0 (The number of units going through each process must be equal. More bathtubs cannot be smoothed than are first molded.) Solution: x1 = 15.87; x2 = 9.56; x3 = 11.11;

Z = $19, 444.45 S14-9.

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 S (super). Regular: Premium: Super:

x1R + x2 R + x3R + x4 R x1P + x2 P + x3P + x4 P x1S + x2 S + x3S + x4 S

Maximize Z = 3x1S + 5x2 S + 6 x4 S + 9 x1P + 11x2 P + 6 x3P + 12 x4 P + 1x1R + 3x2 R + 4 x4 R − 2 x3R subject to

x1S + x2 S + x3S + x4 S  3,000 x1P + x2 P + x3P + x4 P  3,000 x1R + x2 R + x3R + x4 R  4,000 x1S + x1P + x1R  5,000 x2S + x2 P + x2 R  2, 400 x3S + x3P + x3R  4, 000

x4 S + x4 P + x4 R  1,500 0.6 x1S − 0.4 x2S − 0.4 x3S − 0.4 x4S  0 −0.2 x1S + 0.8x2 S − 0.2 x3S − 0.2 x4 S  0 −0.3x1S − 0.3x2S + 0.7 x3S − 0.3x4S  0 −0.4 x1P − 0.4 x2 P + 0.6 x3P − 0.4 x4 P  0 −0.5x1R + 0.5 x2 R − 0.5 x3R − 0.5 x4 R  0 0.9 x1R − 0.1x2 R − 0.1x3R − 0.1x4 R  0 x ji  0

Solution:

x1S = 1200, x1R = 3800, x2 P = 2200 x2 R = 200, x3S = 900, x3P = 3100, x4S = 900, x4 P = 600, z = $63, 400 S14-10.

First, all possible patterns that contain the desired lengths must be determined. Pattern Length

7 ft

9 ft

10 ft

Total used (ft)

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

3x1 + 2 x2 + 2 x3 + x4 = 700 x3 + 2 x4 + x5 = 1, 200 x2 + x5 + 2 x6 = 500 xi  0 b.

Length 7 ft 9 ft 10 ft

1 3 0 0

2 2 0 1

3 2 1 0

Pattern 4 1 2 0

5 0 1 1

6 0 0 2

Trim loss (ft)

xi = no. of standard length boards to cut using pattern i coefficients of objective function = trim loss using pattern i Minimize Z = 4 x1 + x2 + 2 x3 + 0 x4 + 6 x5 + 5x6 subject to

3x1 + 2 x2 + 2 x3 + x4  700 x3 + 2 x4 + x5  1, 200 x2 + x5 + 2 x6  500 xi  0 Solution:

(a) x2 = 50 boards

(b) x2 = 500 boards

x4 = 600 boards x6 = 225 boards

x4 = 600 boards Z = 500 boards

Z = 875 boards

S14-11.

Minimize Z = $190 ( r1 + r2 + r3 + r4 + r5 + r6 ) +$260 ( o1 + o2 + o3 + o4 + o5 + o6 ) +5 ( i1 + i2 + i3 + i4 + i5 ) subject to r j  160 ( j = 1, 2, 3, 4, 5, 6 ) o j  50 ( j = 1, 2, 3, 4, 5, 6 )

r1 + o1 − i1  105 r2 + o2 + i1 − i2  170 r3 + o3 + i2 − i3  230 r4 + o4 + i3 − i4  180 r5 + o5 + i4 − i5  150 r6 + o6 + i5  250 rj , o j , i j  0

Solution:

r1 = 160 computers produced in regular time in week 1 r2 = 160 computers produced in regular time in week 2 r3 = 160 computers produced in regular time in week 3 r4 = 160 computers produced in regular time in week 4 r5 = 160 computers produced in regular time in week 5 r6 = 160 computers produced in regular time in week 6 o3 = 25 computers produced with overtime in week 3 o4 = 20 computers produced with overtime in week 4

o5 = 30 computers produced with overtime in week 5 o6 = 50 computers produced with overtime in week 6 i1 = 55 computers carried over in inventory in week 1 i2 = 45 computers carried over in inventory in week 2 i3 = 40 computers carried over in inventory in week 3

z = $215,600 S14-12.

xi = employee i assigned to department j; i = 1, 2, 3, 4 and j = a (lamps), b (sporting goods), c (linen) Maximize Z = 130 x1a + 190 x1b + 90 x1c + 275x2a + 300 x2b + 100 x2c + 180x3a + 225x3b + 140x3c

+200 x4a + 120 x4b + 160 x4c subject to

x1a + xab + 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 Solution: x1b = 1; x2a = 1; x3b = 1; x4c = 1; Z = $850 Multiple optimal solutions exist. S14-13.

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 = 2 AM-4 AM, etc.

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 x1 + x2 + x3 + x4  50 x2 + x3 + x4 + x5  60 x3 + x4 + x5 + x6  80 x4 + x5 + x6 + x7  90 x5 + x6 + x7 + x8  70 x6 + x7 + x8 + x9  70 x7 + x8 + x9 + x10  60

x8 + x9 + x10 + x11  50 x9 + x10 + x11 + x12  40 xij  0 Solution:

x1 = 40 x4 = 20 x5 = 40 x6 = 20 x7 = 10 x9 = 40 x10 = 10 Z = 180

S14-14.

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 + xt 2 + xt 3 ) + 2.80 ( xm1 + xm 2 + xm3 ) + 3.25 ( xb1 + xb 2 + xb3 ) subject to

0.50 xt1 − 0.50 xm1 − 0.50 xb1  0 −0.20 xt1 + 0.80 xm1 − 0.20 xb1  0 −0.30 xt 2 − 0.30 xm2 + 0.70 xb 2  0 −0.30 xt 2 + 0.70 xm2 − 0.30 xb 2  0 0.80 xt 2 − 0.20 xm2 − 0.20 xb 2  0 0.50 xt 3 − 0.50 xm3 − 0.50 xb3  0 0.30 xt 3 − 0.70 xm3 − 0.70 xb3  0 −0.10 xt 3 − 0.10 xm3 + 0.90 xb3  0 xt1 + xm1 + xb1  1,500 xt 2 + xm 2 + xb 2  900 xt 3 + xm3 + xb3  2, 400 xij  0 Solution:

xt1 = 750 xt 2 = 180 xt 3 = 1, 680 xm1 = 750 xm2 = 450 xm3 = 480 xb1 = 0

xb 2 = 270

xb3 = 240

Z = $10,798.50 S14-15.

A: x1 = 0

* C: x1 = 4

x2 = 4

x2 = 3

Z = $1,600

Z = $2, 400 D: x1 = 6

B: x1 = 2

x2 = 0 Z = $1,800

x2 = 4 Z = $2, 200 b.

Optimal solution:

x1 = 4 necklaces x2 = 3 bracelets

The maximum demand is not achieved by one bracelet. c.

The solution point on the graph that 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 + 2 x2 = 18. Transforming this constraint into the form y = a + bx enables us to compute the slope: 2 x2 = 18 − 3x1

x2 = 9 −

3 x1 2

From this equation, the slope is -3/2. Thus, the slope of the objective function must be at least -3/2.

The slope of the objective function is -3/4.

400 x2 = Z − 300 x1 x2 =

Z 3 − x1 400 4

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

400 x2 = Z − 600 x1 x2 =

Z 3 − x1 400 2

However, this creates a situation where both points C and D are optimal (i.e., multiple optional solutions), as are all points on the line segment between C and D. S14-16.

A: x1 = 0

x2 = 6 Z = $960 * B: x1 = 1

x2 = 3 Z = $680 b.

x1 = 1; x2 = 3; s3 = 16; Z = $680

C: x1 = 3

x2 = 1 Z = $760 D: x1 = 6

x2 = 0

Z = $1, 200

S14-17.

x1 = 0 x2 = 8 Z = $560

x1 = 5.3 x2 = 4.7 Z = $488

* B:

x1 = 3.3 x2 = 6.7 Z = $568

x1 = 8 x2 = 0 Z = $240

The slope of the original objective function is computed as follows:

Z = 30 x1 + 70 x2 70 x2 = Z − 30 x1 x2 =

Z 3 − x1 70 7

Slope = −

3 7

The slope of the new objective function is computed as follows:

Z = 30 x1 + 70 x2 70 x2 = Z − 30 x1 x2 =

Z 3 − x1 70 7

Slope = −

3 7

A steeper slope changes the optimal point.

S14-18.

(a) Maximize Z = $7,600x1 + 22,500x 2 Subject to

x1 + x 2  3,500

x 2 / ( x1 + x 2 )  .40

.12x1 + .24x 2  600 x1 , x 2  0 (b)

S14-19.

(a) x1 = $ invested in stocks

x 2 = $ invested in bonds Maximize Z = $0.18x1 + 0.06x 2 (average annual return) Subject to

x1 + x 2  $720, 000 ( available funds )

x1 / ( X −X )  .65 ( % of stocks ) 1

.22x1 + .05x 2  100, 000 ( total possible loss )

x1 , x 2  0

(b)

S14-20. (a)

x1 = exams assigned to James x2 = exams assigned to Ann Minimize Z = .12x1 + .05x2 Subject to x1 + x2 = 130 x1  840/8.4 (or 100) x2  720/15 (or 48) x1, x2  0

(b)

(d)

If the constraint for Sarah’s time became x2  52 with an additional hour then the solution point at A would move to x1 = 78, x2 = 52 and Z = 9.36. If the constraint for Brad’s time became x1  107.14 with an additional hour then the solution point (A) would not change. All of Brad’s time is not 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 12 to 9.36, whereas increasing Brad by one hour would have no effect on the solution. This indicates that Ann’s additional time is of more value to Professor Wang than James’s.

S14-21. (a)

x1 = # cups of Morning Blend x2 = # cups of Study Break Maximize Z = $1.95x1 + 1.70x2 subject to 16x1 + 16x2  3,200 oz. (or 25 gal. x 128 oz.) (.25)(.0625)x1 + (.55)(.0625)x2  5 lbs. Brazilian (.30)(.0625)x1 + (.15)(.0625)x2  5 lbs. Tanzanian

(.45)(.0625)x1 + (.30)(.0625)x2  5 lbs. Guatemalan x1/x2 = 3/2 x1,x2  0 (b)

Solution: x1 = 120 cups x2 = 80 cups Z = $370

(b) x2

(d)

None of the constraints for Brazilian, Tanzanian and Guatemalan coffee are binding and there is already extra, or slack, pounds of these coffees available. Thus, getting more Brazilian would not affect the optimal solution.

(e)

Increasing the brewing capacity to 30 gallons would affect the solution; it would change the optimal solution to x1 = 123.1 cups, x2 = 82.1 cups and sales (Z) would increase from $370 to $379.53.

(f)

If the café increased the demand ratio of Morning Blend to Study Break from 3 to 2 to 2 to 1 it would only increase daily sales to $373.33, so the café should not spend the extra $25 per day on advertising.

S14-22. 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 x1 = 2,651.5 x2 = 1,325.8 Z = 1,683.71 b. Minimize Z = .09x1 + .18x2 Subject to .46x1 + .35x2 ≤ 2,000 x1 ≥ 1,000 x2 ≥ 1,000 .91x1 - .82x2 = 3,500 x1,x2 ≥ 0 x1 = 2,945.05 x2 = 1,000 Z = 445.05 477-445 = 32 fewer defective items

S14-23. a. Minimize Z = $3700x1 + 5100x2 Subject to x1 + x2 = 45 (32x1 + 14x2) / (x1 + x2) ≤ 21 .10x1 + .04x2 ≤ 6 ≥ .25 ≥ .25 x1, x2 ≥ 0 x1 = 17.5 x2 = 27.5 Z = $205,000 b. The solution would not change if shipping costs are reduced. The solution would not change if damaged orders were reduced. Yes, the solution would change to China (x1) = 22.5, Brazil (x2) = 22.5, Z = $198,000.

S14-24.

Minimize Z = 0.18x1 + 0.22 x2 + 0.10 x3 + 0.12 x4 + 0.10x5 +0.09 x6 + 0.40 x7 + 0.16 x8 + 0.50 x9 +0.07x10

subject to

90 x1 + 110 x2 + 100 x3 + 90 x4 + 75x5 + 35x6 + 65x7 + 100x8 + 120x9 + 65x10  420

2 x2 + 2 x3 + 2 x4 + 5x5 + 3x6 + 4 x8 + x10  20 270 x5 + 8x6 + 12 x8  30 6 x1 + 4 x2 + 2 x3 + 3x4 + x5 + x7 + x10  5 20 x1 + 48x2 + 12 x3 + 8x4 + 30 x5 + 52 x7 + 250 x8 + 3x9 + 26 x10  400 3x1 + 4 x2 + 5x3 + 6 x4 + 7 x5 + 2 x6 + x7 + 9 x8 + x9 + 3x10  20 5x1 + 2 x2 + 3x3 + 4 x4 + x7 + 3x10  12 xi  0 x3 = 1.025 cups of oatmeal x8 = 1.241 cups of milk x10 = 2.975 slices of wheat toast Z = $0.509 cost per meal S14-25.

xij = number of units of products i ( i = 1, 2, 3) produced on machine j ( j = 1, 2, 3, 4 )

Maximize Z = $7.8 x11 + 7.8 x12 + 8.2 x13 + 7.9 x14 + 6.7 x21 + 8.9 x22 + 9.2 x23 + 6.3 x24 +8.4 x31 + 8.1x32 + 9.0 x33 + 5.8x34 subject to 35x11 + 40 x21 + 38x31  9,000 41x12 + 36 x22 + 37 x32  14, 400 34 x13 + 32 x23 + 33x33  12,000 39 x14 + 43x24 + 40 x34  15,000 x11 + x12 + x13 + x14 = 450 x21 + x22 + x23 + x24 = 600 x31 + x32 + x33 + x34 = 320 xij  0 b.

x11 = 65.385 x14 = 384.615 x22 = 400.00 x23 = 170.00 x31 = 150.3 x33 = 169.7

Z = $11,738.28 S14-26.

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

x11 + x12 + x13 + x14  220 x21 + x22 + x23 + x24  190 x31 + x32 + x33 + x34  280 x11 + x21 + x31 = 110 x12 + x22 + x32 = 160 x13 + x23 + x33 = 90 x14 + x24 + x34 = 180

Ash : 0.03x11 − 0.01x21 − 0.02 x31  0 0.04 x12 − 0.0 x22 − 0.01x32  0 0.04 x13 − 0.0 x23 − 0.01x33  0 0.03x14 − 0.01x24 − 0.02 x34  0 Sulfur : 0.01x11 − 0.01x21 − 0.02 x31  0 0.01x12 − 0.01x22 − 0.02 x32  0 0.01x13 − 0.03x23 − 0.04 x33  0 0.04 x14 − 0.02 x24 − 0.03x34  0 xij  0

Solution:

x11 = 38 x13 = 18 x14 = 72 x21 = 30 x22 = 160 x31 = 42 x33 = 72 x34 = 108

Z = $41,594

S14-27.

( )

xij = space ft 2 rented in month i for j months, where i = 1, 2,

, 6 and j = 1, 2,

Minimize Z = 1.70 x11 + 1.40 x12 + 1.20 x13 + 1.10 x14 +1.05 x15 + 1.00 x16 + 1.70 x21 + 1.40 x22 +1.20 x23 + 1.10 x24 + 1.05 x25 + 1.70 x31 +1.40 x32 + 1.20 x33 + 1.10 x34 + 1.70 x41 +1.40 x42 + 1.20 x43 + 1.70 x51 + 1.40 x52 +1.70x61

subject to

x11 + x12 + x13 + x14 + x15 + x16 = 35,000 x12 + x13 + x14 + x15 + x16 + x21 + x22 + x23 + x24 + x25 = 47,000 x13 + x14 + x15 + x16 + x22 + x23 + x24 + x25 + x31 + x32 + x33 + x34 = 52,000 x14 + x15 + x16 + x23 + x24 + x25 + x32 + x33 + x34 x41 + x42 + x43 = 27,000 x15 + x16 + x24 + x25 + x33 + x34 + x42 + x43 + x51 + x52 = 19,000 x16 + x25 + x34 + x43 + x52 + x61 = 15,000

Solution:

x13 = 25, 000 x14 = 8,000 x16 = 2,000 x24 = 4, 000 x25 = 8,000 x34 = 5,000

Z = $59,100 b.

x16 = 52, 000

Z = $52,000 It is cheaper to rent all the space for the entire six month period in April and have excess or “surplus” space.

S14-28.

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

y

Minimize Z = 0.11x1 + 0.05

i =1

subject to:

July : August : September : October : November : December : End :

x1 + y1 + 20,000 − c1 = 60,000 c1 + y2 + 30,000 − c2 = 60,000 + y1 c2 + y3 + 40,000 − c3 = 80,000 + y2 c3 + y4 + 50,000 − c4 = 30,000 + y3 c4 + y5 + 80,000 − c5 = 30,000 + y4 c5 + y6 + 100,000 − c6 = 20,000 + y5 x1 + y6  c6 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 b.

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

S14-29.

xij = production in month i to meet demand in month j, where i = 1, 2,

, 7 and = 4, 5, 6 and 7

y j = overtime production in month j where j = 4, 5, 6, 7.

Minimize Z = 150 x14 + 100 x24 + 50 x34 + 200 x15 + 150 x25 + 100 x35 + 250 x16 + 200 x26 +150 x36 + 300 x17 + 250 x27 + 200 x37 + 400 y4 + 400 y5 + 400 y6 + 400 y7 subject to

x14 + x24 + x34 + x44 + y4 = 60 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  40 x44  40 x55  60 x66  90 x77  50 y4  20 y5  20 y6  20 y7  20 Solution:

x14 = 20 x44 = 40 x35 = 20 x55 = 60 x66 = 90 x17 = 10 x27 = 30 x77 = 50 b.

x34 = 20 x44 = 40 x25 = 5 x35 = 20 x55 = 60 x26 = 10 x66 = 90 x17 = 40

y5 = 5 y6 = 10 y7 = 20 Z = $31,500

x27 = 25 x77 = 50 y7 = 5

Z = $26,000 S14-30.

x = full-time operators y j = part-time operators hired in week i, where

i = 1, 2,

, 8. 8

y

Minimize Z = $4,880 x + $450

i =1

subject to

360 x + 270 y1 = 19,500 360 x + 270 y2 = 21,000 360 x + 270 y3 = 25,600 360 x + 270 y4 = 27, 200 360 x + 270 y5 = 33, 400 360 x + 270 y6 = 29,800 360 x + 270 y7 = 27,000 360 x + 270 y8 = 31,000 1.1x + 2.7 yi  200, i = 1, 2,

x, yi  0 b. Solution

x = 53.6 full-time operators y1 = .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

S14-31.

Maximize Z = $0.97 x1 + 0.83x2 + 0.69 x3 subject to

x1 + x2 + x3  324 cartons x3  x1 + x2

x3 3 x1 x2  120 x1 = 54, x2 = 108, x3 = 162, Z = $253.80 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.

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.

The discount would change the objective function to, Maximize Z = 0.86 x1 + 0.83x2 + 0.69 x3 and the constraint for relative demand would change to, x3  1.5 x1

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

S14-32.

x1 = road racing bikes x2 = cross country bikes x3 = mountain bikes Maximize Z = 600 x1 + 400 x2 + 300 x3 subject to

1, 200 x1 + 1,700 x2 + 900 x3  $12,000 x1 + x2 + x3  20 8x1 + 12 x2 + 16 x3  120 x3  2 ( x1 + x2 )

x1 , x2 , x3  0 Solution:

x1 = 3 x3 = 6

Z = $3,600 a.

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

The additional net sales would be $900. Since the cost of the labor is $300, the additional profit would be $600.

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 .

S14-33.

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

0.45x1 + 0.41x2 + 0.50 x3  960 x1 + x2 + x3  2, 000 x1  200 x2  200 x3  200 x1  x2 + x3 x1 , x2 , x3  0

Solution:

x1 = 1,000 x2 = 800 x3 = 200 Z = $760

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 in additional profit.

x1 = 1,000 x2 = 1, 000 Z = $770

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. c.

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.

S14-34.

xi = In-state freshmen in college i; i = 1, 2, 6 yi = Outofstate freshmen in college i; i = 1, 2, 6 Maximize Z = 8, 600 (  xi ) + 19, 200 (  yi ) subject to

x1 + y1  470 x2 + y2  1300 x3 + y3  240 x4 + y4  820 x5 + y5  1060 x6 + y6  610  SATIi ( xi ) +  SATOi ( yi )  xi +  yi

 1150

 yi  .47  xi +  yi xi  .30, i = 1, 2, xi + yi

 xi +  yi  4,500 xi , yi  0 Solution:

x1 ( in-state Architecture ) = 470 x2 ( in-state A&S) = 557 x3 ( in-state Agriculture ) = 103

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

y1 ( out-of-state Architecture ) = 0 y2 ( out-of-state A&S) = 743 y3 ( out-of-state Agriculture ) = 137

y4 ( out-of-state Business ) = 281 y5 ( out-of-state Engineering ) = 606 y6 ( out-of-state Human Resources ) = 349

Z = $61,119,000

S14–35. (a). Xij = consultant i hours assigned to project j;

, F and j = 1,

i = A,

Minimize Z =

i=A

j =1

  ( ranking )  X

Subject to 8

 X  available hours, i = A, , F ij

j =1 F

 X = projected hours, j = 1, ,8 ij

i=A F

 ( hourly rate )  X  Budget, j = 1, , 8 ij

i=A

Solution:

x A3 = 400 x A4 = 50 x B4 = 250

x D2 = 131.7 x D7 = 15.93 x E1 = 208.33

x B5 = 350 x C4 = 175 x C7 = 274.1 x C8 = 50.93

x E8 = 149.07 x F1 = 291.67 x F2 = 108.3 x F6 = 460

Z = 12,853.33

(b) X A4 = 257.2

XA5 = 192.8 XB4 = 95 XC1 = 333.3 XC8 = 166.7 X D3 = 130 XD4 = 122.8 XD5 = 47.2

Z = $576, 250

XE1 = 166.7 XE2 = 240 X E3 = 270 XE8 = 33.3 XF5 = 110 X F6 = 460 X F7 = 290

Minimize Z = 1x1D + 3x1E + 5x1F + 6 x1G + 9 x1H + 10 x1I + 12 x1J + 1x2 E + 3x2 F + 4 x2G

S14-36.

+7 x2 H + 8x2 I + 10 x2 J + 2 x3F + 3x3G + 6 x3H + 7 x3I + 9 x3 J + 1x4 I + 3x4 J + 1x5 J + 6 x A4 + 8x A5 +10 xA6 + 11xA7 + 5xB 4 + 7 xB5 + 9 xB6 + 10 xB7 + 4 xC 4 + 6 xC 5 + 8xC 6 + 9 xC 7 + 2 xD 4 +4 xD5 + 6 xD6 + 7 xD7 + 2 xE 5 + 4 xE 6 + 5xE 7 + 2 xF 6 + 3xF 7 + 1xG 6 + 2 xG 7 Subject to

1  x1D + x1E + x1F + x1G + x1H + x1I + x1J  2 1  x2 E + x2 F + x2G + x2 H + x2 I + x2 J  2 1  x3F + x3G + x3H + x3I + x3 J  2 x4 I + x4 J  2 x5 J = 1 1  xA4 + xA5 + xA6 + xA7  2 1  xB 4 + xB5 + xB6 + xB7  2 1  xC 4 + xC 5 + xC 6 + xC 7  2 1  xD 4 + xD5 + xD6 + xD7  2 xE 5 + xE 6 + xE 7  2 xF 6 + xF 7  2 xG 6 + xG 7  2 1  x1D + xD 4 + xD5 + xD6 + xD7  2 1  x1E + x2 E + xE 5 + xE 6 + xE 7  2 1  x1F + x2 F + x3F + xF 6 + xF 7  2 1  x1G + x2G + x3G + xG 6 + xG 7  2 1  xA4 + xB 4 + xC 4 + xD 4 + x4 I + x4 J  2 1  xA5 + xB5 + xC 5 + xD5 + xE 5 + x5 J  2 1  xA6 + xB6 + xC 6 + xD6 + xE 6 + xF 6 + xG 6  2 1  xA7 + xB7 + xC 7 + xC 7 + xD7 + xE 7 + xF 7 + xG 7  2 x1H + x2 H + x3H  1 x1I + x2 I + x3I + x4 I  1 x1J + x2 J + x3 J + x4 J + x5 J  1 xA4 + xB 4 + xC 4 + xD 4  1 xA5 + xB5 + xC 5 + xD5 + xE 5  1 xA6 + xB6 + xC 6 + xD6 + xE 6 + xF 6 + xG5  1 xA7 + xB7 + xC 7 + xD7 + xE 7 + xF 7 + xG 7  1 x1E + x2 E  1 x1F + x2 F + x3F  1 x1G + x2G + x3G  1 x1H + x2 H + x3H  1 x1 l + x2 l + x3 l + x4 l  1 x1J + x2 J + x3 J + x4 J + x5 J  1 xi  0

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) C–6 (8 hr) G–7 (2 hr) Z = 35 hours ground time.

6 crews originate in Pittsburgh and 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 flies 4 back to Orlando while a crew ferries on flight 4 and flies I back to Pittsburgh. There are multiple optimal solutions.

S14-37.

xi = flow of product from node i to node j Minimize 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 x47 = x57 .25x28 + .25x38 + .25x48 + .25x58 = x89 + x810 + x811 x28 = x38 x38 = x48 x48 = x58 x69 + x79 + x89 = x912 x610 + x710 + x810 = x1012

x611 + x711 + x811 = x1112 x912 + x1012 + x1112 = 37,000 x12  60, 000 x13  50,000 x14  55,000 x15  60, 000 x26 + x27 + x28  50,000 x36 + x37 + x38  40,000 x46 + x47 + x48  46,000 x56 + x57 + x58  50,000 x69 + x610 + x611  12,000 x79 + x710 + x711  14,000 x89 + x810 + x811  19,000 x912  14, 000 x1012  16, 000 x1112  12, 000 x ij  0

Solution:

x12 = x13 = x14 = x15 = 37,000 x26 = x36 = x46 = x56 = 12,000 x27 = x37 = x47 = x57 = 6,000 x28 = x38 = x48 = x58 = 19,000 x69 = 5,000 x79 = 6,000 x89 = 3,000 x810 = 16, 000 x611 = 7,000 x912 = 14, 000 x1012 = 16, 000 x1112 = 7, 000

Z = $40,680

S14-38. a. Maximize Z = $8.65x1 + 10.95x2 s.t. x1 + x2 ≤ 250 x1 + x2 ≥ 120 x1 – 2x2 ≤ 0 -x1 + 1.2x2 ≤ 0 x1 ≤ 150 x2 ≤ 110 x1, x2 ≥ 0

x1 = 140 x2 = 110 Z = $2,415.50

beef; a higher dual value of $2.30/lb., compared to no dual value for pork; there is pork left over.

c. profit increases to $2,423.86, however, they should not do this because profit will increase to this amount only if it orders 3.64 additional lbs., the upper limit of the sensitivity range for beef supply. d. No; the upper limit ratio of pork BBQ to beef is a non-binding constraint.

S14-39. Maximize Z = 18x1A + 20x1B +21x1C +17x1D + 19x2A + 15x2B + 22x2C + 18x2D + 20x3A + 20x3B + 17x3C + 19x3D +24x4A + 21x4B + 16x4C + 23x4D + 22x5A + 19x5B + 21x5C + 21x5D 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 + 17x1D+ 19x2A + 15x2B + 22x2C + 18x2D + 20x3A + 20x3B + 17x3C + 19x3D + 24x4A + 21x4B + 16x4C + 23x4D + 22x5A + 19x5B + 21x5C + 21x5D) ≤ .04 x1A + x1B + x1C + x1D ≤ 1 x2A + x2B + x2C + x2D ≤ 1 x3A + x3B + x3C + x3D ≤ 1 x4A + x4B + x4C + x4D ≤ 1 x5A + x5B + x5C + x5D ≤ 1 x1A + x2A + x3A + x4A + x5A = 1 x1B + x2B + x3B + x4B + x5B = 1 x1C + x2C + x3C + x4C + x5C = 1 x1D + x2D + x3D + x4D + x5D = 1 xij = 0 or 1 x1C = 1, x3D = 1, x4B = 1, x5A = 1, Z = 83 parts S14-40. The variables in “Solver” need to be set to binary. In the constraints at least one of the variables must equal 1 in order for a city to be assured a hub within 300 miles Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 subject to Atlanta: x1 + x3 + x8 ≥ 1 Boston: x2 + x9 ≥ 1 Charlotte: x1 + x3 + x11 ≥ 1 Cincinnati: x4 + x5 + x6 + x8 + x10 ≥ 1 Detroit: x4 + x5 + x6 + x7 + x10 ≥ 1 Indianapolis: x4 + x5 + x6 + x7 + x8 + x12 ≥ 1 Milwaukee: x5 + x6 + x7 ≥ 1 Nashville: x1 + x4 + x6 + x8 + x 12 ≥ 1 New York: x2 + x9 + x11 ≥ 1 Pittsburg: x4 + x5 + x10 + x11 ≥ 1 Richmond: x3 + x9 + x10 + x11 ≥ 1 St. Louis: x6 + x8 + x12 ≥ 1 xi = 0 or 1 x2 (Boston), x3 (Charlotte), x5 (Detroit), x12 (St. Louis) = 1; Z = 4 There are multiple optimal solutions

S14-41. Minimize Z = x1 + x2 + x3 + x4 + x5 +x6 + x7 + x8 + x9 +x10 subject to: Atlanta: x1 + x2 + x7  1 Charlotte: x1 + x2 + x9  1 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 Nashville: x1 + x3 + x5 + x6 +x7 + x10  1 Pittsburgh: x3 + x4 + x8 +x9  1 Richmond: x2 + x8 +x9  1 St. Louis: x5 + x6 +x7 + x10  1 276x1 + 253x2 + 394x3 + 408x4 + 282x5 + 365x6 + 268x7 + 323x8 + 385x9 + 298x10  900 xi = 0 or 1 a.

x1 (Atlanta) = 1 x7 (Nashville) = 1 x8 (Pittsburgh) = 1 Z=3

x4 (Cleveland) = 1 x7 (Nashville) = 1 x9 (Richmond) = 1 Z=3

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

S14-42.

a. Minimize Z = 211x1 + 173x2 + 410x3 + 152x4 + 263x5 + 414x6 + 302x7 s.t. 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 x1 = 4, x2 = 3, x4 = 2, x5 = 5, Z = 2,982 miles

b. x1 = 4, x2 = 3, x4 = 2, x6 = 3, x7 = 2, Z = $51,110

S14-43. The solution to this problem requires that the variables be designated as integer (binary) in “Solver.” a. Maximize Z = 3y1 + 2y2 + 1y3 + 1y4 + 2y5

Where yi = 1, if project i is selected = 0, if project i is not selected

s.t. $2,600,000x1 + 950,060x2 + 38,000x3 + 365,000x4 + 175,000x5 ≤ 30,060,000 17,500x1 + 8,600x2 + 25x3 + 1,700x4 + 900x5 ≥ 250,000 220,000x1 + 125,000x2 + 26,000x3 + 75,000x4 + 45,000x5 ≥ 4,000,000 400,000x1 + 150,000x2 + 34,000x3 + 1,200x3 + 55,000x5 ≥ 5,000,000 x1 ≤ 75 x2 ≤ 75 x3 ≤ 165 x4 ≤ 75 x5 ≤ 75

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 x6 = 7 Z=9 b. Change objective function to: Maximize Z = 220,000x1 + 125,000x2 + 26,000x3 + 75,000x4 + 45,000x5 Add constraint: 3y1 + 2y2 + 1y3 + 1y4 + 2y5 ≥ 7 Delete media/public relations constraint.

Solution: y2 = 1, y3 = 1 x2 = 29, x3 = 64 Z = $5,289,000 S14-44. 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. Minimize Z = .42x14 + .56x15 + .39x24 + .44x25 + .52x34 + .61x35 + .24x46 + .12x47 + .21x48 + .16x56 + .18x57 + .19x58 subject to:

x14 + x15  27, 000 x24 + x25  16, 000 x34 + x35  35, 000 x14 + x24 + x34  45,000 bushels

x15 + x25 + x35  34,000 bushels ( x14 + x24 + x34 )/2 = x46 + x47 + x48 gals. ( x15 + x25 + x35 )/2 = x56 + x57 + x58 gals. x46 + x56 = 10,000 gals. x47 + x57 = 14, 000 gals. x48 + x58 = 16,000 gals. xij  0 Solution:

x15 = 27, 000 x14=27000 x24 = 10, 000 x25=16000 x25 = 6, 000 x34=18000 x34 = 35, 000 x35=17000 x46 = 9,500 x47=14000 x47 = 13,000 x48=8500 x56 = 500 x56=10000 x58 = 16,000 x58=6500

Z = $46,820 44410 S14-45. 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. Minimize Z = $24 ( xA1 + xA2 + xA3 + xA4 ) + 27 ( xB1 + xB2 + xB3 + xB4 )

+ 26 ( xC1 + xC2 + xC3 + xC4 ) + 33 ( xD1 + xD2 + xD3 + xD4 ) + 28 ( xE1 + xE2 + xE3 + xE4 ) + 36 ( xF1 + xF2 + xF3 + xF4 ) + 12.10 xA1 + 14.35xA2 + 11.25xA3 + 15.05xA4 + 10.95 xB1 + 13.75 xB2 + 11.65 xB3 + 14.55 xB4 + 15.15xC1 + 16.75 xC2 + 12.70 xC3 + 12.10 xC4 + 14.35xD1 + 11.80 xD2 + 16.15xD3 + 11.45xD4 + 12.75xE1 + 9.85xE2 + 10.25 xE3 + 9.75 xE4 + 16.55xF1 + 14.75xF2 + 13.60 xF3 + 14.70 xF4 subject to: xA1 + xA2 + xA3 + xA4 = 180,000

xB1 + xB2 + xB3 + xB4 = 315,000 xC1 + xC2 + xC3 + xC4 = 340,000 xD1 + xD2 + xD3 + xD4 = 115,000 xE1 + xE2 + xE3 + xE4 = 105, 000 xF1 + xF2 + xF3 + xF4 = 185,000 27.2 xA1 + 28.1xB1 + 26.6 xC1 + 22.4 xD1 + 19.7 xE1 + 24.6xF1  4,700,000 27.2 xA2 + 28.1xB2 + 26.6 xC2 + 22.4 xD2 + 19.7 xE2 + 24.6 xF2  6,300,000 27.2 xA3 + 28.1xB3 + 26.6 xC3 + 22.4 xD3 + 19.7 xE3 + 24.6 xF3  5,600,000

27.2 xA4 + 28.1xB4 + 26.6 xC4 + 22.4 xD4 + 19.7 xE4 + 24.6 xF4  7, 400,000 xij  0

Solution:

xA2 = 34,558.82 xA3 = 145, 441.18 xB1 = 167, 259.79 xB2 = 147, 740.21 xC3 = 61,804.51 xC4 = 278,195.49 xE2 = 61,345.18

Z = $34,133,052.63 S14-46. xij = team i, where i = 1, 2,  16, assigned to field j, where j = 1, 2,  12. Minimize Z =  xij (priority ij) where priority ij = 1, 2, 3 or 4 (for a field not selected) subject to 12

 xij = 1, where i = 1, 2,  16 j =1

 xij  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 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.

S14-47. Maximize Z =

 (profit) ∙ x i

subject to

 (load-lbs i) ∙ x  80,000 lbs for j = A, B, C ij

 (load-ft i) ∙ x  5,500 ft for j = A, B, C 3

 (time i) ∙ x  90 hrs. for all j = A, B, C ij

 x  1, for i = 1, 2, …, 12 ij

xij = 0 or 1 for customer shipment i (i = 1, 2, …, 12) assigned to truck j (j = A, B, C) Solution: x1B = 1 x2A = 1 x3A = 1 x4B = 1 x7C = 1 x11C = 1 Z = $78,000

S14-48. a. Minimize Z =

 (mileage ) x ij

where xij = truck “i” assigned to customer “j” subject to:

 x = 1 for all j = 1, ...,L ij

 x  1 for all i = 1, ..., 8 ij

xij  0 Solution: x1G = 1 x2B = 1 x3H = 1 x4L = 1 x5K = 1 x6F = 1 x7D = 1 x8A = 1 Total mileage = 4,260

b. Add the constraint:

 ((capacity ) x ) / 8  85 ij

Solution: x1G = 1 x2B = 1 x3D = 1 x4L = 1 x5K = 1 x6E = 1 x7A = 1 x8C = 1 Total mileage = 4,420 Mileage not significantly different; only 160 additional miles.

S14-49. Minimize Z =

x

subject to 3rd Street: x1 + x5 ≥ 1 10th Street: x2 + x3 + x10 ≥ 1 South Street: x3 + x2 + x8 ≥ 1 Mulberry Avenue: x4 + x12 ≥ 1 Rose Street: x1 + x5 + x11 ≥ 1 Wisham Avenue: x6 + x9 + x10 ≥ 1 Richmond Road: x7 + x10 ≥ 1 Hill Street: x3 + x8 ≥ 1 23rd Avenue: x6 + x9 ≥ 1 Broad Street: x6 + x7 + x10 ≥ 1 Church Street: x5 + x11+ x12 ≥ 1 Beamer Boulevard: x4 + x11 + x12 ≥ 1 xi = 0 or 1 Solution: x3 (South Street) = 1 x5 (Rose Street) = 1 x9 (23rd Avenue) = 1 x10 (Broad Street) = 1 x12 (Beamer Boulevard) = 1 Z = 5 stores Total cost = $899,000

S14-50. xi = 0 if facility “i” is selected, and “0” if it is not. (a) Maximize Z =

 (annual usage i ) • x

subject to

 (acreage i ) • x  55 i

 (cost i ) • x  $550,000 i

 (priority i) • x − ( x • 1.75)  0 i

xi = 0 or 1 Solution:

football fields, playground, walking/running trails and softball fields. Z = 123,500

(b) Minimize Z =

 x (priority i) i

subject to

 (acreage i ) • x  55 i

 (cost i ) • x  $550,000 i

 (expected usage i) • x  120,000 i

xi = 0 or 1 Solution: soccer fields, playground, walking/running trails Z = 4.0 or 1.33 average priority (c) Maximize Z =

 (acreage i ) • x

subject to:

 (acreage i ) • x  55 i

 (cost i ) • x  $550,000 i

 (priority i) • x − ( x • 1.75)  0 i

xi = 0 or 1 Solution: rugby fields, soccer fields, walking/running trails Z = 52 acres Expected annual usage = 83,700

S14-51. (a) Minimize Z =

 x p 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. subject to: 3

 x = 1, for i = 1, 2, … 15 ij

j 15

x 8 i1

i 15

 x  10 i2

i 15

x  7 i3

10 ∙

 xi 1 + 20 ∙  xi 2 + 35  xi 3 ≤ 22 xij = integer and ≥ 0

Solution: Victim 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Hospital County General County General All Souls County General Memorial County General All Souls Memorial Memorial County General All Souls All Souls All Souls County General All Souls

Average Priority = 1.07 (b)

Minimize Z = 10 ∙

 xi 1 + 20 ∙  xi 2 + 35  xi 3

Subject to: 3

 x = 1, for i = 1, 2, …, 15 ij

j 15

x 8 i1

i 15

 x  10 i2

i 15

x  7 i3

/ 15 ≥ 1.50

Solution:

Victim 1 2 3 4 5 6

Hospital County General County General County General County General County General County General

7 8 9 10 11 12 13 14 15

All Souls Memorial Memorial Memorial County General Memorial Memorial County General All Souls

Z = avg. transport time = 16.67 mins. Average priority = 1.47 S14-52. This is an example of the classic “traveling salesman” problem. Minimize Z = 10x12+ 15x13 +20x14 +40x15 + 10x21 + 12x23 + 16x24 + 24x25 + 15x31 + 12x32 + 10x34 + 20x35 +20x41 + 16x42 + 10x43 + 16x45 + 40x51 + 24x52 + 20x53 + 16x54 subject to

x21 + x31 + x41 + x51 = 1 x12 + x32 + x42 + x52 = 1 x13 + x23 + x43 + x53 = 1 x14 + x24 + x34 + x54 = 1 x15 + x25 + x35 + x45 = 1 x12 + x13 + x14 + x15 = 1 x21 + x23 + x24 + x25 = 1 x31 + x32 + x34 + x35 = 1 x41 + x42 + x43 + x45 = 1 x51 + x52 + x53 + x54 = 1 u2 − u3 + 5 x23  4 u2 − u4 + 5 x24  4 u2 − u5 + 5 x25  4 u3 − u2 + 5 x32  4 u3 − u4 + 5 x34  4 u3 − u5 + 5 x35  4 u4 − u2 + 5 x42  4 u4 − u3 + 5 x43  4 u4 − u5 + 5 x45  4 u5 − u2 + 5 x52  4 u5 − u3 + 5 x53  4 u5 − u4 + 5 x54  4 xij = 0 or 1 Solution: 1 - 3 - 4 - 5 - 2 - 1 Z = 75 miles

CASE S14.1: Mosaic Tile Company a.

Maximize Z = $190 x1 + 240 x2 subject to

0.30 x1 + 0.25x2  60 hr.-molding 0.27 x1 + 0.58 x2  105 hr.-baking 0.16 x1 + 0.20 x2  40 hr.-glazing 32.8 x1 + 20 x2  6, 000 lb.-clay x1 , x2  0 b.

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

0.30 x1 + 0.25 x2 + s1 = 60 0.27 x1 + 0.58x2 + s2 = 105 0.16 x1 + 0.20 x2 + s3 = 40 32.8x1 + 20 x2 + s4 = 6,000 x1 , x2 , s1 , s2 , s3 , s4  0 c.

x1 = 0 x2 = 181.103

Z = $44, 234.80

Z = $43, 447.20 *B :

x1 = 56.70 x2 = 154.64 Z = $47,886.60

x1 = 100 x2 = 120

x1 = 136.36 x2 = 76.36

x1 = 182.93 x2 = 0 Z = $34.756.70

Z = $47,800 d.

x1 = 56.70, x2 = 154.64 0.30 ( 56.7 ) + 0.58 (154.64 ) + s1 = 60

s1 = 4.33 hr. of molding time 0.27 ( 56.7 ) + 0.58 (154.64 ) + s2 = 105

s2 = 0 hr. of baking time 0.16 ( 56.7 ) + 0.20 (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 e.

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,

−190 / c2 = 0.27 / 0.58 c2 = $408.14 f.

Z = 47886.598 x1 = 56.701

x2 = 154.639 s1 = 4.330 s2 = 0.000 s3 = 0.000 s4 = 1047.423 g.

Since there is 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.

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

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

A loss of 3 kiln hours will reduce profit by ( 3) (1,170.103) = $3,510.31.

CASE S14.2: Summer Sports Camp at State University Model Formulation:

wi = new sheets purchased in week i ( i = 1, 2,

, 8)

xi = sheets cleaned at laundry at end of week i yi = sheets cleaned by Mary 's friends at end of week i Minimize Z = 10 ( w1 + w2 + w3 + w4 + w5 + w6 + w7 + w8 ) + 4 ( x1 + x2 + x3 + x4 + x5 + x6 )

+ 2 ( y1 + y2 + y3 + y4 + y5 ) subject to

w1 = 115 x1 + y1 = 115 w2 = 210 x2 + y2 = 210 w3 + 0.8 x1 = 250 x3 + y3 = 250 w4 + 0.8 x2 + 0.8 y1 = 230 x4 + y4 = 230 w5 + 0.8 x3 + 0.8 y2 = 260 x5 + y5 = 260 w6 + 0.8 x4 + 0.8 y3 = 300 x6 = 300 w7 + 0.8x5 + 0.8 y4 = 250 w8 + 0.8x6 + 0.8 y5  190 wi , xi , yi  0

Model Solution:

w1 = 115 w2 = 210 w3 = 250 w4 = 138 w5 = 50 w6 = 0 w7 = 0 w8 = 0

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

Z = $11,940

CASE S14.3: Spring Garden Tools Model Formulation:

Let i = 1 ( trowel ) , 2 ( hoe ) , 3 ( rake ) , 4 ( shovel )

Ri = regular production of product i in stage 1 Si = subcontracted production of product i in stage 1 xi = overtime production of product i in stage 1 Ai = regular production of product i in stage 2 yi = overtime production of product i in stage 2 Minimize Z = $6 R1 + 10 R2 + 8R3 + 10 R4 + 7.2S1 + 12S2 + 9.6S3 + 12S4 + 6.2 x1 + 10.7 x2 + 8.5 x3 + 10.7 x4 + 3 A1 +5 A2 + 4 A3 + 5 A4 + 3.1y1 + 5.4 y2 + 4.3 y3 + 5.4 y4 subject to 0.04 R1 + 0.17 R2 + 0.06 R3 + 0.12 R4  500 hrs. 0.04 x1 + 0.17 x2 + 0.06 x3 + 0.12 x4  100 hrs. 0.05R1 + 0.14R2 + 0.14R4  400 hrs. 0.05x1 + 0.14 x2 + 0.14 x4  100 hrs. 1.2 R1 + 1.6 R2 + 2.1R3 + 2.4 R4 + 1.2 x1 + 1.6 x2 + 2.1x3 + 2.4 x4  10, 000 ft 2

0.06 A1 + 0.13 A2 + 0.05 A3 + 0.10 A4  600 hrs. 0.06 y1 + 0.13 y2 + 0.05 y3 + 0.10 y4  100 hrs. 0.05 A1 + 0.21A2 + 0.02 A3 + 0.10 A4  550 hrs. 0.05 y1 + 0.21y2 + 0.02 y3 + 0.10 y4  100 hrs. 0.03 A1 + 0.15 A2 + 0.04 A3 + 0.15 A4  500 hrs. 0.03 y1 + 0.15 y2 + 0.04 y3 + 0.15 y4  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 = 0 1691.95 R2 = 1, 222.96 1319.54 R3 = 1, 600 R4 = 1,634.29 933.33

S1 = 0 S2 = 105.71 0 S3 = 0 S4 = 165.71 866.67

A1 = 1,800 A2 = 1, 400 A3 = 1, 600 A4 = 1,146.67 Z = $84,614.20 85472.6

y1 = 0 y2 = 0 y3 = 0 y4 = 653.33

108.05

x2 = 71.43 80.46 x3 = 0 x4 = 0

CASE S14.4: 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), i (Indiana).

yij = tons of grape juice processed into product i at

plant j where i = j ( juice ) , k ( concentrate ) , l ( jelly )

Minimize Z = $850 xnv + 720 xnm + 910 xnt + 750xni + 970 x pv + 790 x pm + 1, 050 x pt + 880 x pi

+ 900 xov + 830 xom + 780 xot + 820 xoi + 2,100 y jv + 2,350 y jm + 2, 200 y jt + 1,900 yi + 4,100 ykv + 4,300 ykm + 3,950 ykt + 3,900 yki + 2,600 yiv + 2,300 ylm + 2,500 ylt + 2,800 yli subject to

xnv + xnm + xnt + xni  1, 400 x pv + x pm + x pt + x pi  1,100 xov + xom + xot + xoi  1,700 xnv + x pv + xov  1, 200 xnm + x pm + xom  1,100

xnt + x pt + xot  1, 400

xni + x pi + xoi  900

y jv + y jm + y jt + y ji = 1, 200

ykv + ykm + ykt + yki = 900 ylv + ylm + ylt + yli = 700 y jv + 2 ykv + 1.5 ylv = xnv + x pv + xov ym + 2 ykm + 1.5 ylm = xnm + x pm + xom yt + 2 ykt + 1.5 ylt = xnt + x pt + xot yi + 2 yki + 1.5 yli = xni + x pi + xoi

xi , yi  0

Solution:

xni = 1400 x pm = 1100

yi = 1200 ykv = 75

xov = 150 xot = 1, 400 Z = $10,606,000

ykm = 25 ykt = 700 yki = 100 ylm = 700

CASE S14.5: Julia’s Food Booth a.

x1 = pizza slices, x2 = hot dogs, x3 = barbecue sandwiches The model is for the first home game, Maximize Z = $0.75x1 + 1.05x2 + 1.35x3 subject to:

$0.75 x1 + 0.45 x2 + 0.90 x3  $1,500 24 x1 + 16 x2 + 25 x3  55, 296 in 2 of oven space

x1  x2 + x3 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 in 2 , by 8, or approximately 24 in 2 per slice. The total space available is the dimension of a shelf, 36 in.  48 in. = 1, 728 in 2 , multiplied by 16 shelves, 27, 648 in 2 , which is multiplied by 2, the times before kickoff

and halftime the oven will be filled = 55, 296 in 2 .

x1 = 1, 250 pizza slices x2 = 1, 250 hot dogs x3 = 0 barbecue sandwiches

Solution:

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. 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 $238.52 or a total profit of $2,488.32.

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.

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 just 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.

CASE S14-6: The Sea Village Amusement Park xi = experienced employees in week i yi = new employees in week i Zi = pool of employees available in week i Minimize Z =

y

subject to:

10 yi + 30 xi  weekly hours needed, where i = 1, , 16 30 xi  weekly hours needed, where i = 17, 18, 19, 20 x1 = 800 xi = .85xi-1 + yi , where i = 2, 3, , 16 x17 = .25 x16 xi = .90 xi-1 , where i = 18, 19, , 20 yi  zi , where i = 2, 3, , 16 z1 = 1,600 z2 = 1, 600 − y1 + 210

z3 = z2 − y2 + 210 z4 = z3 − y3 + 210 z5 = z4 − y4 + 210 z6 = z5 − y5 + 210 z7 = z6 − y6 + 210 z8 = z7 − y7 + 210 z9 = z8 − y8 + 120 z10 = z9 − y9 + 120 z11 = z10 − y10 + 120 z12 = z11 − y11 + 120 z13 = z12 − y12 + 120 z14 = z13 − y13 + 120 z15 = z14 − y14 + 120 z16 = z15 − y15 + 120 Solution: y1 = 0

y2 = 195.3 y3 = 280.1 y4 = 387.8 y5 = 46.1 y6 = 440.8 y7 = 92.3 y8 = 416.5 y9 = 219.5 y10 = 218.2 y11 = 221.0 y12 = 214.9 y13 = 327.9 y14 = 0.0 y15 = 843.4 y16 = 0

Z = 3,904

x1 = 800.0 x2 = 680 x3 = 773.3 x4 = 937.4 x5 = 1184.6 x6 = 1053.1 x7 = 1335.9 x8 = 1227.8 x9 = 1460.2 x10 = 1460.6 x11 = 1459.7 x12 = 1461.7 x13 = 1457.4 x14 = 1566.7 x15 = 1331.7 x16 = 1975.3 x17 = 493.8 x18 = 444.4 x19 = 400.0 x20 = 360.0

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

15 Resource Planning Answers to Questions 15-1.

MRP is useful for dependent and discrete demand items, complex products, job shop production, and assemble-to-order environments. MRP is not needed for very simple products with few components and no variability in output.

15-2.

The demand for cars is independent. The demand for tires is dependent. If a manufacturer makes 100 cars and four tires are needed for each car, he or she would need 400 tires.

15-3.

The objective of an MRP system is to ensure the availability of material while maintaining the lowest possible level of inventory. The major inputs to MRP are: the master schedule, the product structure record, and the item master file. The output from MRP can take the form of work orders, purchase orders, and various reports, such as planned order reports or action reports.

15-4.

The master production schedule is created within the confines of the aggregate production plan. It specifies which end items are to be produced, how many are needed, and when they are needed. The MPS drives the MRP process by providing final product needs, from which the MRP system determines component needs.

15-5.

Phantom bills are used for transient subassemblies. K-bills group small parts together under one part number. Modular bills are used to simplify forecasting and planning.

15-6.

The item master file contains four main sections: description, inventory policy, physical inventory, and usage/sales. It includes lot sizes, lead times, inventory on hand and inventory on order.

15-7.

Cycle counting selects some items of inventory to count each day. Discrepancies are reconciled immediately instead of at the end of the year. The system is geared toward keeping tight control on some items and looser control on others.

15-8.

Netting is subtracting out what’s on hand from the demand requirements. Explosion is the determination of lower-level requirements by multiplying the quantity per assembly by the parent requirements. Time phasing is the subtracting of lead time from a due date to get a start date (i.e., order release date). The MRP process consists of exploding down a bill of material, netting out inventory, and offsetting for lead times to determine order quantities and release dates.

15-9.

The inputs to CRP are the planned order releases from the MRP schedule, a routing file, and an open orders file. Overloads can be leveled by producing the items in earlier time periods, producing them in later time periods, using overtime or subcontracting, or hiring more workers. Underloads can be leveled by pulling work forward, becoming a subcontractor, cutting worker hours, or laying off workers.

15-10. MRP systems assume material is the most constraining resource, leadtimes are fixed, and every movement and transaction must be reported. These assumptions can be relaxed with ERP systems. Companies constrained by a particular process may reconcile capacity issues first and use bills of labor or bills of resources instead of bills of material. ERP processors today are able

to handle variable lead-times, incorporate JIT concepts with MRP, and utilize more sophisticated scheduling systems with less data requirements. 15-11. (a) requirements for labor, equipment, surgical tools and supplies can be exploded from a bill of resources established for each type of surgery; (b) classes can be scheduled based on a bill of courses for students in a particular major; (c) food purchases can be exploded from a bill of menu offerings; (d) hotel renovations can be scheduled from a bill of materials by style of room. 15-12. MRP systems focus on an individual plant’s operation; ERP systems manage the resources of an entire enterprise. For example, an MRP system is concerned with customer demand, production schedules, inventory levels, and available capacity at work centers within a plant. An ERP system is concerned with customer demand and available capacity at its plants worldwide, and production schedules and inventory levels along its supply chain as well as throughout the company. Support for multiple languages and currencies, foreign taxes and accounting rules, integrated logistics, and electronic commerce are also common. ERP systems typically contain production-oriented MRP modules as well as modules for business planning, customer service, financial management, and accounting. 15-13. Student responses will vary. 15-14. Customer relationship management software plans and executes business processes that involve customer interaction, such as marketing, sales, fulfillment and service. CRM uses ERP to quote and schedule delivery dates, process customer orders, report order status, and establish after-sales service. 15-15. Supply chain management software handles all interactions with suppliers along the supply chain including supplier certification, e-procurement, order fulfillment and network design. SCM uses ERP’s data base and works with ERP in demand planning and collaborative production planning. The distinction between ERP and SCM software is becoming increasingly blurred. 15-16. Student responses will vary. 15-17. Collaborative product commerce is concerned with new product design and development, as well as product life cycle management. CRM-CPC-SCM collaboration on design can reduce time to market for new products and service. CRM-ERP-SCM collaboration gets the product to the customer faster. 15-18. ERP systems from different vendors can converse via EAI software, EDI, and XML. Standards of communication are currently being developed. 15-19. Student responses will vary. 15-20. ERP organizes and manages a company’s business processes by sharing information across functional areas. IT serves as the backbone for an organization’s information needs, as well as its e-business initiatives. In addition to the four major modules of finance and accounting, sales and distribution, production and material management, and human resources, there are hundreds of support modules. Implementation is hindered by overambitious time schedule, lack of process knowledge, and erroneous decisions on the appropriate degree of standardization.

15-21. Student responses will vary. 15-22. Student responses will vary. 15-23. Student responses will vary.

Solutions to Problems 2 1 3 = 6 b. (11 3) + ( 2  2 ) = 7 c. LLC = 3

15-1. a.

15-2. Level 0 –1 – –2 ––2 –1 ––2 ––2

Item Z X U V Y U W

Quantity 1 2 1 2 3 1 4

(

) (

)

No. of U’s per Z = 1 2 + 1 3 = 5 No. of W’s per Z = 4  3 = 12 15-3.

15-4.

15-5. a.

[ (1 x 2) + (2 x 3)] x 100 = 800

c. Orders for C3 should be released at the beginning of day 1.

C3 C2

C1 A C2 S1 C3

15-6.

4  3  28 = 3072 different scooters

4 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 23 options in a modular bill of material. Options allow a product to better match customer needs, but options are also more difficult and costly to forecast, produce, and inventory than standard products.

15-7.

Item Deck Grip tape Wheel assembly* Nuts & bolts Riser Skateboard

Ea $54.90 $4.95 $71.60 $1.99 $3.95

Qty 1 1 2 4 2

Profit Margin Selling Price

Total $ 54.90 $ 4.95 $143.20 $ 7.96 $ 7.90 $218.91 10% $240.80

*Cost of Wheel Assembly Item Wheels Bearings Truck

15-8. a.

Ea $8.95 $4.95 $33.90

Qty 2 4 1

2  2  3 = 12 different bikes

10, 000 

2 = 6, 700 26-inch bikes, 10, 000  0.40 3 = 4, 000 10-speed bikes

Total $17.90 $19.80 $33.90 $71.60

15-9. Item B C D E F G H I

Qty for 50 X’s 50 x 2 50 x 1 50 x 1 (50 x 4) + (50 x 2 x 2 x 1) (50 x 1 x 2) + (50 x 2 x 2 x 2 x 1 ) + (50 x 2 x 4) 50 x 2 x 2 50 x 2 x 1 50 x 1 x 2 x 1

100 50 50 400 900 200 100 100

15-10. Item:

Lot Size: L4L 1

LLC: 0 LT:

Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

8 100

0 90 90

185

Item:

LLC: 0

Lot Size:

L4L 1

LT : 2

Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

Period

Period 1

200 5

Item: C LLC: 0 Lot Size: Min 150 LT: 4 Gross Requirements Scheduled Receipts Projected on Hand 140 Net Requirements Planned Order Receipts Planned Order Releases

Item: D LLC: 1 Lot Size: Mult 250 LT: 2 Gross Requirements Scheduled Receipts Projected on Hand 200 Net Requirements Planned Order Receipts Planned Order Releases

0 195 195

195

140

200

250 450

0 450

250

Period 4 5 270 0 140 20 130 150

150

4 585 0 115 135 250

Period 5 6 180 0 185 185 65 250

Planned Order Report 1 A B C D

Period 3

5 90

195 150 250

250

15-11. Level 0 —1 —1 —1 ——2 ———3 ———3 ————4 ————4 ———3 ————4 ————4 ———3 ————4 ————4 ——2 — — —3 ————4 ————4 ———3 ————4 ————4 ———3 ————4 ————4 ———3 ——2 ——2 ———3 ———3

Item Sandbox Seat Nails Box assembly Left corner assembly Nuts and bolts 5-foot side Unprepared wood Water sealer 7-foot side Unprepared wood Water sealer 1-foot wood strip Unprepared strip Water sealer Right Corner Assembly 5-foot side Unprepared wood Water sealer 7-foot side Unprepared wood Water sealer 1-foot wood strip Unprepared strip Water sealer Nuts and bolts Nuts and bolts 1-foot wood strip Unprepared wood Water sealer

UM Ea Ea Ea Ea Ea Ea Ea Ea Oz Ea Ea Oz Ea Ea Oz Ea Ea Ea Oz Ea Ea Oz Ea Ea Oz Ea Ea Ea Ea Oz

Qty 1 4 16 1 1 4 1 1 2 1 1 2 1 1 2 1 1 1 2 1 1 2 1 1 2 4 8 2 1 2

15-12. Item: A LLC: 1 Lot Size: Mult 50 LT : 2 Gross Requirements Scheduled Receipts Projected on Hand 50 Net Requirements Planned Order Receipts Planned Order Releases

1 10

2 15

3 25

100

Period 4 5 75 60

6 85

7 45

8 60

25 75 100 100

30 70 100 50

35 15 50

25 25 50

15 35 50 50

15-13. Item: X LLC: 0 Lot Size: Min 50 LT: 2 Gross Requirements Scheduled Receipts Projected on Hand 30 Net Requirements Planned Order Receipts Planned Order Releases

1 25 5

2 30 50 25

3 56

Period 4 5 25 100

19 31 50 56

44 6 50 50

0 56 56 50

6 40

7 30

8 20

10 40 50

30 20 50

Release orders in periods 1 through 5 for quantities of 50, 50, 56, 50, and 50 respectively.

15-14.

Item: D Lot Size: Min 100

LLC: 1 LT: 2

Gross Requirements

150

180

270

120

Scheduled Receipts Projected on Hand

150 120

160

270

120

Planned Order Receipts

100

160

270

120

100

160

270

120

Periods 1, 3, 4, 5 0 160 Item: D

LLC: 1

Lot Size: LFL

LT: 2

150

180

270

120

Net Requirements

180

270

120

Planned Order Receipts

180

270

120

180

270

120

Gross Requirements Scheduled Receipts Projected on Hand

Planned Order Releases d.

120

Net Requirements Planned Order Releases a. b. c.

150 120

120

Orders would be placed in periods 1, 2, 3, 4, and 5.

e. Item: D

LLC: 1

Lot Size: Mult 100

LT: 2

Gross Requirements

150

180

270

120

Scheduled Receipts

150

Projected on Hand

120

Net Requirements

160

230

Planned Order Receipts

100

200

300

100

Planned Order Releases

100

200

300

100

Orders would be placed in the same time periods but their amounts would change in periods 3, 4 and 5 to 200, 300, and 100, respectively.

15-15.

Item: X Lot Size: Mult 25 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: 0 LT: 1

Item: Y Lot Size: Mult 100

LLC: 1 LT:

Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

Period 1

1 2

150 160

4 54

5 75

6 50

7 80

8 100

106

1 74 75 100

1 99 100

6 19 25 75

150

200

75 25 100 200

25 75 100

25 175 200

Period 1

100

100 Period

Planned Order Report:

1 Product X Part Y

100

25 200

100

15-16.

b. Level 0 -

1 1 1 -

2 2 2 -

2 2 2 2 2

3 3

3 3 3

3 3 3 3 3

Continued….

Item Bungee Chair Base Assembly Molded base w legs & casters Molded base w legs Casters Rod Plate assembly Large plate Seat support rods Seat adjustment bar 2" screws Chair Assembly RH arm LH arm Seat assembly Molded seat sides Seat webbing Back webbing Small plates 1.5" screws Fasteners Top back rail

Quantity 1 1 1 1 4 1 1 1 2 1 4 1 1 1 2 2 9 10 38 76 4 1

c. One alternative is to eliminate all level 3 items by assembling them before delivery to the customer. Another alternative is to pre-assemble the ‘seat assembly’ as shown below. Level 0 -

1 1 1 -

2 2 2 -

2 2 2 2 2

3 3

3 3 3

Quantity 1 1 1 1 4 1 1 1 2 1 4 1 1 1 2 4 1

15-17.

5 days

no;

Inventory the back panel, drawer guide, side panel and legs

15-18.

Beg Inventory = Co = $100 Cc = $1.00 d = 29

25 EOQ = POQ =

Exact 76.16 2.63

Rounded 77 3

1 20

2 40

Period 3 30

4 10

5 45

0 30 30 10

0 10 10 45

0 45 45

0 35 35 30

1 20

2 40

Period 3 30

4 10

5 45

42 35 77

34 43 77

a. Item: File Cabinet Lot Size: L4L 1 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: 1 LT: 1

Cost of LFL = (4 * $100) + (5 * $1) = $405 b. Item: File Cabinet Lot Size: EOQ 77 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: LT:

1 1

Cost of EOQ = (2 * $100) + (5+42+12+2+34) *$1 = $295 c. Item: X Lot Size: POQ 3 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: 1 LT: 1

1 20

2 40

40 35 75

Period 3 30

4 10

5 45

0 45 45

Cost of POQ = (2 * $100) + (5+40+10) *$1 = $255

45 Choose POQ

15-19.

Beg Inventory Co = Cc =

150 $400.00 $2.00

155.63

1 100

2 90

3 85

0 40 40 85

Item: A LLC: 0 Lot Size: L4L 1 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 150 Net Requirements Planned Order Receipts Planned Order Releases

Exact 249.50

Rounded 250

1.60

4 70

Period 5 150

6 200

7 300

8 250

0 85 85 70

0 70 70 150

0 150 150 200

0 200 200 300

0 300 300 250

0 250 250 0

EOQ = POQ =

Cost of L4L = $2,900

Item: A LLC: 0 Lot Size: EOQ 250 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 150 Net Requirements Planned Order Receipts Planned Order Releases

100

Period 5 70 150

210 40 250

125

250

155 95 250 250

200

300

250

205 45 250 250

155 95 250 250

155 95 250

Cost of EOQ = $4, 220

Item: A LLC: 0 Lot Size: POQ 5 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 150 Net Requirements Planned Order Receipts Planned Order Releases

Cost of POQ = $2, 770

100

85 40 125

125

Choose POQ

220

Period 5 70 150 150 70 220

500

200

300

250

300 200 500

0 250 250

250

15-20. a.

Beg Inventory =

Co = $200 Cc = $2.00 d = 50.00

Item: A Lot Size: L4L 1 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

EOQ = POQ =

LLC: LT:

0 0

Exact 100.00 2.00 Period

1 50

2 50

3 50

4 50

0 50 50 50

Cost of L4L = $200  4 = $800 Item: A Lot Size: EOQ 100 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: LT:

0 0

1 50

50 50 100 100

Period 2 3 50 50

4 50

50 50 100 100

Cost of EOQ = ( $200  2 ) + (100  $2 ) = $600

Item: A Lot Size: POQ 2 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: LT:

0 0

Period 1 50

2 50

3 50

4 50

50 50 100 100

Cost of POQ = ( $200  2 ) + (100  $2 ) = $600 Choose EOQ or POQ

Beg Inventory =

0 $200 $2.00

Co = Cc = d = 30.00

Item: A Lot Size: L4L 1 Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

LLC: LT:

EOQ = POQ =

0 0

Exact 77.46

Rounded 78

2.58

1 50

2 10

3 50

Period 4 10

0 50 50 50

0 10 10 10

0 50 50 50

0 10 10 10

Cost of L4L = $800.00

Item: A LLC: 0 Lot Size: EOQ 78 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

Period 1 50

2 10

3 50

4 10

28 50 78 78

46 32 78 78

Cost of EOQ = $656.00 Item: A LLC: 0 Lot Size: POQ 3 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

Cost of POQ = $620.00

Choose POQ

Period 1 50

2 10

3 50

4 10

60 50 110 110

0 10 10 10

Beg Inventory =

Co = $200 Cc = $2.00 d = 30.00

Item: A LLC: 0 Lot Size: L4L 1 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

EOQ = POQ =

Exact 77.46

Rounded 78

2.58

1 50

Period 2 50

3 10

4 10

0 50 50 50

0 10 10 10

Cost of L4L = $800.00 Item: A LLC: 0 Lot Size: EOQ 78 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

Period 1 50

2 50

3 10

4 10

28 50 78 78

56 22 78 78

1 50

2 50

3 10

4 10

60 50 110 110

0 10 10 10

Cost of EOQ = $732.00 Item: A LLC: 0 Lot Size: POQ 3 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

Cost of POQ = $540.00

Choose POQ

Period

Beg Inventory =

0 $200 $2.00

Co = Cc = d = 30.00

Item: A LLC: Lot Size: L4L 1 LT: Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

0 0

EOQ = POQ =

Exact 77.46

Rounded 78

2.58

Period 1 50

2 10

3 10

4 50

0 50 50 50

0 10 10 10

0 50 50 50

1 50

2 10

3 10

4 50

28 50 78 78

36 42 78 78

1 50

2 10

3 10

4 50

20 50 70 70

0 50 50 50

Cost of L4L = $800.00 Item: A LLC: 0 Lot Size: EOQ 78 LT: 0 Gross Requirements Scheduled Receipts Projected on Hand 0 Net Requirements Planned Order Receipts Planned Order Releases

Period

Cost of EOQ = $580.00 Item: A LLC: Lot Size: POQ 3 LT: Gross Requirements Scheduled Receipts Projected on Hand Net Requirements Planned Order Receipts Planned Order Releases

0 0

Cost of POQ = $460.00

Choose POQ

Period

15-21.

Co = $100.00 Cc = $0.50 d = 45.50

Item: X LLC: 1 Lot Size: L4L 1 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 70 Net Requirements Planned Order Receipts Planned Order Releases

EOQ = POQ =

Exact 134.91

Rounded 135

2.96

1 50

2 30

3 25

4 35

Period 5 6 40 50

7 35

8 45

9 70

10 75

0 10 10 25

0 25 25 35

0 35 35 40

0 40 40 50

0 35 35 45

0 45 45 70

0 70 70 75

0 75 75

0 50 50 35

Cost of L4L = $910.00

Item: X LLC: 1 Lot Size: EOQ 135 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 70 Net Requirements Planned Order Receipts Planned Order Releases

1 50

2 30

3 25

4 35

Period 5 6 40 50

7 35

8 45

9 70

10 75

125 10 135

100

95 40 135

135

110 25 135

135

Cost of EOQ = $632.50

Item: X LLC: 1 Lot Size: POQ 3 LT: 1 Gross Requirements Scheduled Receipts Projected on Hand 70 Net Requirements Planned Order Receipts Planned Order Releases

Cost of POQ = $527.50

Choose POQ

1 50

2 30

3 25

4 35

60 10 70

125

Period 5 6 40 50 85 40 125

7 35

8 45

9 70

10 75

145 45 190

190

15-22. Requirements Matrix Item Product A Component B Component C

1 100 100 200

2 150 150 300

3 100 100 200

4 200 200 400

5 100 100 200

6 100 100 200

1 200 400 40

2 300 600 60

3 200 400 40

4 400 800 80

5 200 400 40

6 200 400 40

Load Summary Work Center Machining Heat Treat Assembly

Load Profile Charts

Pulling work ahead results in inventory costs and may not be possible when operations have precedence requirements. Pushing work back means due dates will not be met. It may be preferable to use overtime in peak periods rather than completely level the load. 15-23.

Period 1

2 3 4

Item A B D C B C

Setup 15 30 45 50 30 50

Run Time 3 15 4 10 15 10

Units 90 100 60 600 250 160

Load (mins) 285 1530 285 6050 3780 1650 Average

Load (hrs)

35.00 100.83 63.00 27.50 56.58

The initial load profile is not doable because more than 40 hours of OT are needed in period 2.

Completely Leveled Load Period

Load

Reg

Postponed

Inventory

16.58

21.58

100.83

16.58

22.67

16.58

27.5

16.58

29.08

Total

226.33

160

66.32

22.67

50.66

$1,600

$994.80

$453.40

$151.98

Total $ =

$3,200

Cost

Completely Leveled Load Profile Overtime capacity

Regular capacity

By considering the cost of regular, overtime, inventory and postponement, a more economical load profile can be determined as shown below. Economically Leveled Load Period

Load

Reg

Postponed

Inventory

15.83

20.83

100.83

27.5

Total

147.5

78.83

20.83

Cost

$1,475.00

$1,182.45

$0.00

$62.49

Total Cost =

$2,719.94

Economically Leveled Load Profile Overtime capacity

Regular capacity

15-24. a.

Effective capacity = (# workers) (# hours) (utilization) (efficiency) = () () () () =  hours or 570 minutes Output = 570/10 = 57 pizzas

Breakeven point = $100/($5 – $2) = 33 pizzas Yes, the promotion is worth it.

15-25. Tasks Government American Lit Spanish

Setup Time 15 30 10

Processing Time 40 120 30

# Tasks 2 1 3

Load 95* 150 100 345 minutes

*This assumes that all of tasks of one type are performed together, so there is only one setup. Total time = (12 − 5) (60) = 420 minutes Total nonworking time = 60 + 30 + 30 + (5  7) = 155 minutes Utilization = (420 − 155)/420 = .63095 or 63% Efficiency = 80% a.

Effective capacity = 420  63%  80% = 212 minutes

Load percent = 345/212 = 162,74%

If Amy were to become more motivated (i.e., efficiency of 90%) and cut out all nonworking activities except dinner, she could reduce the load percent to 98% and finish her work on time. As another alternative, Amy could finish her work if she stayed awake another hour and 20 minutes.

15-26. Model B2610 B2003 B2001 Model B2610 B2003 B2001 Efficiency Utilization Hours/wk

Depts Welding Assembly

Welding Assembly

Week 1 50 15 20

2 100 30 40

Welding 0.20 0.15 0.07 0.95 0.90 40.00

1 13.65 13.25

3 195 65 80

4 150 45 60

Assembly 0.18 0.15 0.10 0.90 0.92 40.00

Load per Week 2 3 27.30 54.35 26.50 52.85

4 40.95 39.75

Load Percent per Week 1 2 3 40% 80% 159% 40% 80% 160%

Capacity 34.20 33.12

4 120% 120%

To level the load, overloads in weeks 3 & 4 could be moved back to weeks 1 & 2 respectively.

15-27.

Class 1 2 3 4

Sleep Eat Drive Check-in Pack Total

Time to create key (mins) 10 15 5 20

5 1 1 1 1 9

Time to grade each paper (mins) 2 5 1 10

hours hour hour hour hour hours non-grading

# Papers 35 50 60 25

Total Grading Time (mins) 10 + (2*35) = 80 15 + (5*50) = 265 5 + (1*60) = 65 20 + (10*25) = 270 Total = 680 mins, or Load = 11.33 hrs

Utilization = 0.63 Efficiency = 0.80 Capacity = 12 hours Load % = 11.33/12 = 0.94 Yes, Bryan can finish his job and make it to the airport in time.

15-28. Product A B C Total

MPS 60 220 140

Wk Ctr 100 (1.70* 60) = 102 (1.30*220) = 286 (1.80*140) = 252 640

Wk Ctr 200 (0.50* 60) = 30 (0.90*220) = 198 (0.66*140) = 92.4 320.4

Monthly capacity per worker = (40 hours * 4 weeks ) = 160 hours

Wk Ctr 100 Wk Ctr 200

Load 640 320.4

Capacity 160 160

# Workers 640 / 160 = 4.00 320.4 / 160 = 2.00

CASE SOLUTION 15.1 – Just Sofas 1. The case says that the ERP system was needed to handle customer orders, factory schedules and supply chain coordination. These are specific needs and require that the software actually be able to make good decisions with accurate data. A smaller system, more focused in the furniture industry is probably what the company needs, instead of a large generic system. 2. Just Sofas began as many companies do, with the finance and accounting module, when a major sales promise had been issued that affected operations’ ability to respond. It does not appear that JS reviewed its process before automation, or selected processes to implement that had a high impact on the customer or that would benefit from integration. There are many different ERP vendors available, some who specialize in manufacturing or particular industries, and some for small-to-medium sized businesses, like Just Sofa. Choosing the same one as major companies could result in an overly complicated system, or one does not match the needs of JS. If JS is having trouble with scheduling, look for a vendor that specializes in scheduling, etc. 3. Ruffner developed a “strategy” of quick response without an understanding of operations’ role in executing the strategy. Nor did Ruffner understand the manufacturing capabilities of the firm and its suppliers, or the relationships between load, capacity and leadtime. Wishing to complete customer orders quickly does not make it so. Operations was treated as an after-thought – we’ll promise a four-day turnaround and operations will deliver. 4. Overpromising, installing a new IT system at the same time a major initiative is underway, leaving manufacturing until last in the install, not having the basic prep work done in terms of data and database structure, not pre-testing the scheduling system, a poor selection process, and possible mismatch between system and needs all contributed to the disappointing results. JS needs to scale back, look at its processes and establish reliable data before proceeding with the system install.

CASE SOLUTION 15.2

Snowfall

3" 6" 1' > 2'

b. Miles Lanes* Passes Total miles

Efficiency Utilization

Hours

# Plows

Mph

Capacity (miles)

5 5 5 5

35 35 35 35

1654 1378 919 551

0.900 0.750 0.500 0.300

0.875 0.875 0.875 0.875

12 12 12 12

Major Highways

Primary roads

Secondary roads

10 12 3 360

25 8 2 400

50 4 1 200

960

*No. of lanes is doubled to account for N and S movement.

Snowfall

Capacity (miles)

Load

Load %

3" 6" 1'

1654 960 58% 1378 960 70% 919 960 104% 551 960 174% > 2' GDOT can handle a 3" and 6" snowfall with its current system. Only a slight adjustment is needed for a 1' snowfall. Consider clearing 10 lanes out of 12 on the highways. A larger adjustment is needed for a snowfall >2'. Consider clearing less lanes out of each category. There are obviously many different solutions that students could recommend. An example solution appears below.

1' snowfall Miles Lanes Passes Total miles Total miles all roads

Major Highways

Primary roads

Secondary roads

10 8 3 240

25 8 2 400

50 4 1 200

840

> 2' snowfall Miles Lanes Passes Total miles

Major Highways

Primary roads

Secondary roads

10 8 3 240

25 4 2 200

50 2 1 100

Capacity (miles)

Load

Load %

1654 1378 919 551

960 960 840 540

58% 70% 91% 98%

540

Total miles all roads

Revised Load

Snowfall

3" 6" 1' > 2'

For a 1' snowfall, we recommend reducing the number of lanes cleared on highways to 8 out of the twelve, for a load percent of 91%. For a snowfall > 2', we recommend clearing only 8 out of the 12 lanes on highways, 4 out of the 8 lanes on primary roads, and 2 out of the 4 lanes on secondary roads for a load percent of 98%.

d. If snowplows were the only adjustment, 4 extra plows would be needed to handle the >2' load.

Snowfall

3" 6" 1' > 2'

Efficiency Utilization

0.900 0.750 0.500 0.300

0.875 0.875 0.875 0.875

Hours

# Plows

Mph

Capacity (miles)

Load percent

12 12 12 12

5 5 6 9

35 35 35 35

1654 1378 1103 992

58% 70% 87% 97%

Other alternatives for improving service may include improving service for the main roads at the expense of the smaller roads. Students will generate multiple solutions and the discussion of tradeoffs should be interesting.

CASE SOLUTION 15.3: Hosuki 1.

With current inventory levels, the 10 cars can be delivered in 6.5 days. The time-phased assembly chart is shown below.

Hosuki has used up its inventory in the previous order. To fill an order for 5 more cars requires 7.5 days.

At first glance, to produce a custom-made car in 5 days, Hosuki would have to inventory transmissions engine blocks, brakes, body parts (quarter panels, hoods, fenders, roofs, doors), windshields, rims, and tires — not a very practical idea. Examining the activities under our control, if we reduce the batch size to 5 cars, the assembly time could be cut in half and the body work reduced to 3.5 days. In addition, we could negotiate with supplier C to pre-assemble the tires and deliver within 4.75 (rather than 5) days. As shown in the revised assembly chart, we would still need to inventory engine blocks and transmissions; or arrange with our suppliers to keep a one to two day supply of these items.

16 - Lean Systems Answers to Questions 16-1. The purpose of lean production is to eliminate waste and continually improve operations. 16-2. Reducing inventory was the first step in eliminating waste in early lean systems. With no inventory to guard against poor quality, late deliveries, and machine breakdowns, companies implementing lean production are forced to improve their quality systems, supplier relationships, and preventive maintenance systems. In addition, the pull system itself encourages continuous improvement. Inventory is held only to satisfy demand during lead time (i.e., until a replacement order arrives). To reduce inventory further, one must reduce lead time or the components of lead time, such as processing time, transit time, setup time, and inspection time. Thus, lean production touches every area of operations. 16-3. Flexible resources are important because there is no excess inventory to guard against variations in demand or supply. Multifunctional workers, general-purpose machines, and a flexible control system enable lean factories to adapt to changes in demand and supply rather than compensate for them with “just-in-case” inventory. 16-4. Cellular layouts, because of their manageable size, work flow, and flexibility, facilitate the “pull” production element of lean production. They also make problems more visible, improve quality, reduce transit times, and encourage worker ideas and input. Work centers in lean systems must be closely linked together. The cellular concept facilitates and forms that linkage. 16-5. In a push system, a schedule is prepared in advance for a series of workstations and each workstation “pushes” the work they have completed to the next station. With the pull system, workers go back to previous stations and take only those parts or materials they need and can process immediately. 16-6. In the two-bin system, two bins are maintained for each item. The first (and usually larger bin) contains the order quantity minus the reorder point; the second bin contains the reorder point quantity. At the bottom of the first bin is an order card that describes the item and specifies the supplier and the quantity that is to be ordered. When the first bin is empty, the card is removed and sent to the purchasing department to order a new supply of the item. While the order is being filled, the quantity in the second bin is used. If everything goes as planned, when the second bin is empty, the new order will arrive and both bins will be filled again. The kanban system eliminates the first bin, places the order card, or kanban, at the top of the second bin, and continually orders enough inventory to fill the second bin. As the system progresses, a full bin of material arrives just as the current bin is being emptied. 16-7. In a reorder point system, a certain quantity, Q, is ordered whenever the stock on hand falls below a reorder point. The reorder point is determined so that demand can be met while an order for new material is being processed. Thus, the reorder point corresponds to demand during lead time. A kanban is a visual reorder point. When the kanban arrives with an empty container, a new order needs to be placed. The kanban system and reorder point system are different in that the quantity ordered and the reorder point are equivalent in the kanban system. (In terms of the two-bin system, that means the first bin is eliminated.) With kanbans, an order for new material is always

outstanding. 16-8. a. A withdrawal kanban is a request for more input from the preceding workstation. It authorizes the movement of material from one workstation to the next and it starts off attached to an empty container. If the feeding workstation that receives a withdrawal kanban has a full container of material available, it exchanges the kanbans and containers. That is, it takes the production kanban that is attached to the full container and places it on the empty container. Similarly, the withdrawal kanban that originally accompanied the empty container is attached to the full container and sent immediately to the next process. A production kanban is a work order. It signals a workstation to begin producing enough of the item requested to fill the empty container to which the production kanban is attached. b. A kanban square is a marked area that will hold a certain amount of output items (usually one or two). If the kanban square following a worker’s process is empty, the worker knows it is time to begin production again. c. A signal kanban closely resembles the reorder point system. A triangular marker, or signal, is placed at a certain level of inventory. When the marker is reached (a visual reorder point), it is removed from the stack of goods and placed on an order post, thereby generating a replenishment order for the item. d. A material kanban is a square-shaped kanban often used in conjunction with a signal kanban in cases where it is necessary to order the material for a process in advance of the initiation of the process. e. Supplier kanbans are used outside the factory to order material from vendors. The supplier delivers the order directly to its point of use in the factory and then picks up an empty container (with supplier kanban attached) to fill and return later. 16-9. By producing small amounts at a time, processes can be physically moved closer together and transport between stations can be simplified. Quality problems are easier to detect, and workers show less tendency to let poor quality pass (as they might in a system that is producing huge amounts of an item anyway). Lower inventory levels make processes more dependent on each other, revealing errors and bottlenecks that may otherwise go undetected. 16-10. Large lot sizes would slow down a pull system. The system would not be able to adjust to changing demand patterns. Production in large lots would create lumpy demand throughout the manufacturing system. Inventory would actually increase, but the product would not get out the door faster. 16-11. Small lot sizes in a push system would leave no cushion for alterations in the schedule. Forecasts would have to be extremely accurate. Outages would be frequent. In short, the system would always be operating in a crisis environment, always trying to catch up. 16-12. SMED stands for single minute exchange of dies. Its objective is to reduce setup time to under 10 minutes (hence the name single digit). The ideal achievement would be push-button setups or eliminating the need for setups altogether. The principles of SMED are: 1. separate internal setup from external setup; find out which setup activities can be performed in advance, and which must be performed at the machine while it is idle, 2. convert internal setup to external setup; perform as many preparatory tasks as you can while the machine is otherwise occupied, 3. streamline all aspects of setup; use time and motion studies to improve the setup process, 4. perform setup activities in parallel or eliminate them altogether; explore the need for setups; have teams of workers perform setups; practice setup procedures. 16-13. Uniform production is essential for a pull system. It allows component production to be balanced

and work to flow smoothly from one workstation to the next. Uniform production is achieved through accurate forecasts of final demand, frozen master schedules, and mixed-model sequencing of the final assembly line. 16-14. In mixed-model sequencing, daily production is arranged in the same ratio as monthly demand, and jobs are distributed as evenly as possible across the day’s schedule. Thus, at least some quantity of every item is produced daily; and the company will always have some quantity of an item available to respond to variations in demand. The mix of assembly also steadies component production, reduces inventory levels. and supports the pull system of production. 16-15. In lean systems, there is no extra inventory to buffer against defective units, so maintaining high quality levels is important. In one sense, then, quality is a pre-requisite of lean production. On the other hand, the small-lot production characteristic of lean production encourages quality because workers can observe quality problems easier; when problems are detected, they can be traced to their source and remedied without reworking too many units. So in another sense, lean production facilitates quality. Lean six sigma combines lean's principles for eliminating waste with six sigma's reduction of variability 16-16. Breakdown maintenance can be very expensive. It includes the cost of fixing the machine, the cost of downtime while the machine is being fixed, and any lost sales or other opportunities due to downtime. Preventive maintenance is less expensive when performed on a regular basis. The frequency of preventive maintenance affects cost and the probability of a breakdown occurring. A schedule of maintenance activities should consider the condition of the machine, the demand for its services, its maintenance and breakdown record, and the cost of breakdowns versus preventive maintenance. 16-17. Total productive maintenance takes a broader view of preventive maintenance. It is concerned with maximizing the productive potential of every machine over its lifespan. 16-18. Equipment operators maintain their own machines with daily care, periodic inspections, and preventive repair. They compile and interpret maintenance and operating data on their machines, identify signs of deterioration prior to failure, and scrupulously clean equipment, tools, and workspaces. 16-19. Consider a heart attack patient. Breakdown maintenance is the treatment she receives in the hospital that gets her back on her feet. Preventive maintenance is the medication she takes to control the heart condition and a change in her diet and exercise routine. Total productive maintenance is predicting before the heart attack that changes in lifestyle are needed and starting the diet, exercise, etc., before a breakdown occurs. 16-20. Suppliers must deliver fewer orders more frequently. Quality expectations are higher. Partnerships with producers mean longer-term contracts and competition based on more than price. 16-21. To facilitate meeting lean production requirements, suppliers can locate nearer to their customers, use small, sideloaded trucks and ship mixed loads, establish small warehouses near to customer sites or consolidate warehouses with other suppliers, use standardized containers and scheduled delivery times, and become a certified supplier so that quality does not have to be documented with each delivery. 16-22. Visual control enhances quality by making problems visible, making workers aware of their

environment and making it easier to “do things right the first time.” Kanbans, andons, process control charts, and machines or stockpoints painted different colors are examples of visual control in manufacturing. The red area on the speedometer of your car that indicates legal speed is an example of visual control in day-to-day activities. 16-23. Poka-yoke goes further than visual control by preventing defects or mistakes from occurring. For example, a car that sounds an alarm when you withdraw the key from the ignition without placing the gear in park is audio control. A car that will not let you take the key out of the ignition until the gear is in park is a poke-yoke. 16-24. Worker involvement is essential to Kaizen. Who knows an operation better than the people who do the job? Outsiders can make major suggestions for improvement, but day-to-day incremental continuous improvements rely on the worker. The Kaizen Institute web site has many examples. 16-25. Typical benefits from implementing lean production include: lower costs, reduced space requirements, shorter lead time, increased productivity, greater flexibility, better relations with suppliers, simplified scheduling and control activities, increased capacity, better use of human resources, continuous improvement of operations, more product variety, increased customer satisfaction, higher profits, and increased market share. 16-26. U.S. firms typically do not receive deliveries as often from suppliers, and the quantity received is greater (because the lead time is longer). Supplier-producer partnerships are growing stronger in the United States, but they are not as formalized as in other countries. Also, workers are motivated in different ways and may require different work practices. Supplier relations, cellular layouts and total quality are elements of lean production that have experienced widespread adoption and are considered part of a well-run company. The pull production system and severely limited inventory levels are not appropriate for every business, and should be applied cautiously. 16-27. Lean production is most successful in stable, repetitive environments with flexible resources. 16-28. Some lean concepts are more prevalent in services than in manufacturing. Services compete on speed, provide variety, are very flexible, crosstrain their workers, use cellular layouts, have standard routines, and replenish inventory in direct response to customer demand. 16-29. Lean production is widely used in many industries. For information on lean production see https://www.lean.org/

16-30. Lean eliminates waste. This is consistent with environmental concerns about wasting resources such as water, air, and petroleum. Recycling and using less packaging is also consistent with lean.

16-31. a. Lean will not work if the distance between production and supply is lengthy or uncertain. Global supply chains where shipments take weeks or months to travel to their destination require larger orders and extra inventory to guard against delays. Lengthy supply chains prevent quality problems from being addressed in a timely fashion, make it difficult to operate in a collaborative fashion, and preclude demand-driven production. b. During periods of uncertainty or disruption, production may be shut down due to lack of supplies because the segments of the supply chain are so closely linked. This can have a ripple effect throughout an industries. Having alternative sources of supply in different parts of the world can help alleviate this problem. c. Decisions based solely on short-term costs of supply ignore the cost of poor quality that may be incurred at a later date, the cost of poor communication with suppliers, the cost of monitoring a supply chain for human rights violations, and the cost of lost customers and brand damage due to these occurrences.

Solutions to Problems 16-1.

Model SS50 SS100 SS200

Monthly Requirements 7,200 3,600 3,600

Daily Requirements 7200/30 = 240 3600/30 = 120 3600/30 = 120

Demand Ratio 2 1 1

The following sequence spreads out the production of each model evenly. It should be repeated 120 times a day to meet demand. SS50—SS100—SS50—SS200

16-2. Tile Quarry Italian mosaic Bathroom

Monthly Demand 30,000 15,000 45,000

Bathroom

Quarry

Daily Demand 1,000 500 1,500

Mosaic

This sequence would be repeated 500 times a day.

16-3.

d = 100 per hr

Demand Ratio 2 1 3

Bathroom

Quarry

L = 20 min = 0.33 hr S = 0.10 (100  0.33) = 3.3 C = 10 dL + S 100 ( 0.33) + 3.3 N= = = 3.63 C 10

16-4. a.

200 ( 0.33) + 6.66 10

Round up to 4 kanbans.

= 7.26, or 8 kanbans

d and S change in the formula. The number of kanbans doubles. b.

L and S change in the formula

100 ( 0.50 ) + 5 10

55 = 5.5, or 6 kanbans 10

The number of kanbans increases.

c. C changes in the formula. N=

100 ( 0.33) + 3.33 5

= 7.27, or 8 kanbans

The number of kanbans doubles.

d. S changes in the formula.

S = 0.20 (100  0.33) = 6.6 N=

100 ( 0.33) + 6.6 10

= 3.96, or 4 kanbans

The actual number of kanbans remains the same. 16-5.

L = 20 min = 0.33 hr N =5 S = 0.2 ( d  0.33) 5=

0.33d + 0.20 ( 0.33d )

50 = 0.396d 126.26 = d

The demand for widgets is approximately 126 per hour.

d = 600 L = 30 min = 0.5 hours S = 0.10 (600 x 0.5) = 30 C = 100

16-6.

N = (600 x .05) + 30 100

= 3.3 kanbans

3.3 kanbans are needed for the letter-sorting process. This means 3 containers plus 1/3 of another container would be used.

a. N = [(600 x .5) + 0] / 100 = 3 kanbans b. N = [(600 x .25) + 15] / 100 = 1.65 kanbans c. N = [(600 x .5) + 30] / 300 = 1.1 kanbans

16-7.

Eliminating the safety factor does not change the number of kanbans required. Decreasing the time between deliveries or increasing the bin capacity reduces the number of kanbans by 1. There is generally a direct relationship between the number of kanbans and inventory levels; that is, as the number of kanbans circulating between two work centers increases, workin-process increases. Eliminating safety stock or decreasing lead time cuts inventory levels and reduces the number of kanbans. The exception is when the container size is changed. One kanban’s worth of inventory with the smaller container is not equivalent to one kanban’s worth of inventory in a larger container. So, if the office administrator wanted to reduce the number of kanbans to reduce inventory, she should choose option (b). 16-8. d = 250

L = 30 min = 0.5 hr C = 25 S = 0.10 ( 250  0.5) = 12.5 N=

( 250  0.5) + 12.5 = 137.50 = 5.5 or 6 25

5.5 containers are needed, with a kanban attached to each container. The factory could have 5 full containers, plus 1 container with a cardboard insert so that only half of it can be filled. Or, the factory could use 6 kanbans and containers.

16-9. a.

L = 45 min, or 0.75 hr

S = 0.10 ( 250  0.75) = 18.75 N=

( 250  0.75) + 18.75 = 206.25 = 8.25 kanbans 25

The number of kanbans goes up slightly; inventory goes up by at least 50 units.

d = 125

S = 0.10 (125  0.5) = 6.25 N=

(125  0.5) + 6.25 = 68.75 = 2.75 25

The number of kanbans decreases by half and inventory is halved also. c.

C = 10 N=

( 250  0.5) + 12.5 = 137.5 = 13.75 kanbans 10

The number of kanbans increases significantly. There is no increase in inventory. The change in container size may also affect the lead time of the previous process, so this is a tentative calculation.

16-10. d A = 200 boards per hr

LA = 15 min = 0.25 hr

SA = 0

C A = 20

dL + S ( 200  0.25 ) + 0 = C 20

= 2.5 kanbans or containers d B = 200 boards per hr LB = 25 min = 0.42 hr

SB = 0

C B = 10

dL + S ( 200  0.42 ) + 0 = C 10

= 8.4 kanbans or containers With rounding, we have 3 kanbans circulating between process A and the assembly cell, and 9 kanbans circulating between process B and the assembly cell.

16-11.

Capacity Days/Mo

Cars Vans SUVs

200 20 Monthly 2000 1000 500

Daily 100 50 25 175

Ratio 4 2 1

a. C-V-C-S-C-V-C b. Repeat 25 times a day c. By spreading out model production and not stockpiling inventories, manufacturers can change production more easily to meet changing demands. Component demand will be smoothed out by a mixed sequence. 16-12.

Days/Mo

Short Haul Long Haul Vocational

Monthly Daily 750 37.5 500 25 250 12.5

S-L-S-V-S-L

Repeat 12.5 times daily

Ratio 3 2 1

16-13. From annual to monthly to daily demand and beyond, divide demand for the products into the smallest ratio. Then, create a sequence that is as mixed as possible and repeat that sequence throughout the day until demand is met.

Product Tillers (T) Blowers (B) Mowers (M)

Annual Demand 4500 3000 9000

Monthly Demand 375 250 750

Daily Demand 15 10 30

a. Mixed model sequence: M-T-M-B-M-T-M-B-M-T-M b. Repeat this sequence 5 times a day.

CASE PROBLEM 16.1: The Blitz is On

Lowest ratio 3 2 6

This is an open-ended case asking a team of students to conduct a Kaizen blitz at their university, organization, home or place of business. You will find that students, like managers, tend to jump to conclusions about a problem situation. Using a set procedure slows the group down and forces them to make their decisions based on data. An Excel file with sample Kaizen worksheets is included in the Exhibit files that can be downloaded from the text website. The file includes the following worksheets: Kaizen Blitz Charter, Muda Walk Checklist, 5S Workplace Scan, Gripe Interview Sheet, Process Flow Chart, SIOPC Chart, Swimlane Chart, VSM - Push, VSM - Pull, Game Plan, Kaizen Acheivements, Action Plan and Video Clips.

CASE PROBLEM 16.2: Where’s My Cart? 1. Wasted time cutting items that are not immediately used. Supervisor’s time wasted time labeling items and moving them to the waiting area. Inventory piled up waiting to be worked on. Wasted space holding WIP inventory. Time wasted finding the appropriate carts. Time wasted waiting for the carts to be found. 2. The pager suggestion reduces the time it takes to find the appropriate cart.

3. The proposed solution is a quick fix typical of push production systems with too much inventory. Scheduling is difficult because of the overloaded facility. Work at each department is performed without regard to the capabilities of the next department. The root cause of the cart problem has not been identified. 4. The shop needs the discipline of pull production. Work at the saw should be paced with the ability of the next process to complete work. Visual systems need to be in place to alert the saw when more material can be cut. Tape on the floor with a limited number of places or “squares” for carts would limit WIP inventory. The saw would cut more material only when an empty square is available. Creating assembly “cells” for certain types of customers would also help streamline the flow. A large scheduling board (or Gantt chart) would keep workers and management up-to-date on work needing to be done.

CASE PROBLEM 16.3: Leaning the Warehouse 1. The DC manager views lean as a group of techniques; the professor views lean as a philosophy that requires discipline to maintain. One is a short-term view that may lead to some immediate results; the other is a long-term view that may lead to a transformation in how work is performed and ultimately to strategic advantages for the firm. The more extensive training would involve a leadership team and workers buying into the lean philosophy of continuous improvement, worker participation and waste minimization. This type of philosophy will only work if all levels of the firm participate and reinforce its elements.

2. Quick results could be obtained through 5S, cellular layouts, standard procedures and flexible resources. Longer term results from quality at the source and pull systems would also be helpful. 3. In the short term, 5S and visual control techniques would be useful. However, if more substantial changes in layout, quality improvement methods and pull are not made, the system will most likely revert to its former state. Building a lean system around already established standards before improving the work environment can worsen quality as artificially established time constraints tend to reward speed over quality.

4. Short term results can be tempting, but the real power of lean occurs with concerted longterm effort. Warehouse work does lend itself to lean. Look on the Internet for training programs and examples of lean warehousing.

17 Scheduling Answers to Questions 17-1. Projects are scheduled with network planning techniques such as PERT or CPM, which are powerful enough to handle the numerous interrelated decisions that must be made. Many of the scheduling decisions for mass production are made when the assembly line is set up and balanced. Day-to-day scheduling decisions include increasing or decreasing the output rate and model sequencing. Process scheduling involves determining an appropriate mix of ingredients (a linear programming problem) and the optimum length of a production run (an inventory problem). 17-2. Scheduling a job shop is difficult because of the variety of jobs that are processed, each with distinctive routing and processing requirements. Also, there are usually many jobs in the shop at the same time competing for limited resources. 17-3. A production control department checks the availability of resources, assigns jobs to machines, prioritizes the work for each machine, and monitors the progress of each job as it moves through the system. 17-4. A rescue squad wants to minimize response time. A print shop wants to meet customer due dates. A grocery store might want to minimize the time a customer spends in the system. A manufacturer would be concerned about job lateness as well as efficiency measures such as minimizing inventory or overtime. 17-5. The success of a scheduling system is usually measured in terms of mean flow time (the average amount of time a job is in the system), mean tardiness, maximum tardiness, and percent of jobs completed tardy. 17-6. Loading involves assigning jobs to machines or workers to tasks. Usually there are several machines (or workers) that can perform a task, but not at the same level of efficiency. Load leveling tries to smooth out the initial loading assignments to balance the work-load in a facility. Assignments can be made with the assignment method of linear programming. 17-7. Dispatch lists specify the order in which a list of jobs is to be completed. They are usually computergenerated daily according to a priority rule chosen by management. 17-8. a. SPT is most useful when a shop is highly congested. b. Johnson’s rule gives an optimum sequence for minimizing makespan when jobs are processed through two serial processes. c. DDATE should be used when only small values of tardiness can be tolerated. d. FCFS is fine when operating at low-capacity levels. 17-9. DDATE and SLACK are the most common ways to schedule work. However, some students may prefer to complete short assignments first (SPT); others may tackle large projects first (LPT), work on classes they enjoy or those in which their grade is suffering (customer priority), or simply work on assignments as they are assigned (FCFS).

17-10. Critical ratio uses job due date, remaining processing time, and today’s date to construct a ratio of work remaining to time remaining. Priorities can change as work is completed. SLACK uses the same information but arranges it in a different format, time remaining minus work remaining. The sequence of jobs may differ for the two rules. 17-11. Work packages contain all the instructions for manufacture, such as bills of material, routing sheets, and operations sheets. Hot lists specify what jobs need to be finished immediately. Exception reports point out problems and highlight deficiencies. 17-12. Gantt charts plan, or “map out,” work activities. They can also be used to monitor a job’s progress against the plan. Gantt charts are simple to construct and understand. They are a good means of visual control. 17-13. Input/output control (I/O) monitors the input to and output from each work center. Examining only the output from a work center in comparison to the planned output can lead to some erroneous conclusions about the source of a problem. Reduced output at one point in the production process may have been caused by problems at the current work center, or by problems at previous work centers that feed the current work center. The input rate to a work center can readily be controlled only for the initial operations of a job, called the gateway work centers. Input to later operations, performed at downstream work centers, is difficult to control because it is a function of how well the rest of the shop is operating. The backlog, or queue, in front of a process is a measure of congestion and the effectiveness of a scheduling system. 17-14. Infinite scheduling assumes infinite capacity in the initial loading process. Leveling and sequencing decisions are made after overloads or underloads have been identified. Finite scheduling assumes a fixed maximum capacity and will not load the resource beyond its capacity. Loading and sequencing decisions are made at the same time, so that the first jobs loaded onto a work center are the highest priority. 17-15. The theory of constraints considers priority and capacity simultaneously. In concentrates on bottleneck scheduling and treats process batches and transfer batches differently. Bottleneck resources control the flow of work through a system. They should be scheduled first with no slack. The nonbottleneck resources should be scheduled to support the bottleneck activities. Transfer batches refer to the amount that must accumulate before items are transferred to the next workstation. Ideally, this should be one item. Process batches refer to the amount of a product that is produced at one setting of a machine. They are determined by factors associated with the individual process and should have no influence on how finished items are transferred. 17-16. The drum is the bottleneck, beating to set the pace of production for the rest of the system. The buffer is inventory placed in front of the bottleneck to insure it is always kept busy. The rope is the communication signal that tells the processes upstream from the bottleneck when they should begin production. 17-17. Both TOC and lean production signal upstream production and transfer in small lots. TOC sets the pace of production according to the bottleneck operation. Lean production sets the pace according to customer demand. Lean production is a more comprehensive system that includes other factors such as continuous improvement and cellular layouts. TOC is more applicable to batch production; lean production is more applicable to repetitive systems.

17-18. Employee scheduling must consider the needs of the workplace and the availability and suitability of the worker. Absenteeism, performance, personal schedules, and the right mix of workers all affect the scheduling decision. 17-19. Math programming, heuristics, and expert systems have been applied successfully to employee scheduling. The heuristic presented in this chapter is Baker-Magazine. Advanced planning and scheduling systems use mathematical programming, genetic algorithms, neural networks, constraint based programming, simulation, network analysis and expert systems. 17-20. Students will find a variety of scheduling software and techniques.

Solutions to Problems 17-1.

17-2.

None of the nurses are assigned to the same patients as in problem 17-1. This points out the usefulness of multicriteria decisionmaking. It appears that patients like the nurses spending time with them. Oftentimes the producer and the consumer measure quality and efficiency differently. Other criteria: Seniority, nurse preference, skill set, rotation, previous assignment, location

17-3.

17-4.

17-5. Initial matrix: 18 17 15 19 18 16

17 15 15 16 15 16

14 12 13 18 12 16

19 14 17 18 17 18

19 20 20 18 19 20

18 17 18 20 17 17

Minimization matrix (convert max prob to min prob by assigning zero to max value in table and assigning other elements the value of the difference between their value and the max value) 2 3 5 1 2 4

3 5 5 4 5 4

6 8 7 2 8 4

1 6 3 2 3 2

1 0 0 2 1 0

2 3 2 0 3 3

2 5 5 4 4 4

5 8 7 2 7 4

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

0 3 3 2 2 2

3 6 5 0 5 2

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

0 3 3 2 2 2

3 6 5 0 5 2

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

Row reduction: 1 3 5 1 1 4 Column reduction: 0 2 4 0 0 3

Cover all zeroes: 0 2 4 0 0 3

Modify matrix: 2 2 4 2 0 3

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

Cover all zeroes: 2 2 4 2 0 3

Make assignments: 2 2 4 2 0 3

There are multiple optimal solutions. Albertson and Finch could change assignments without affecting total performance. Original Matrix with assignment:

Employee Albertson Bunch Carson Denali Ebersole Finch

Sales

Finance

Logistics

Marketing

Production

Customer Service

18 17 15 19 18 16

17 15 15 16 15 16

14 12 13 18 12 16

19 14 17 18 17 18

19 20 20 18 19 20

18 17 18 20 17 17

Total performance = 18 + 17 + 18 + 18 + 20 + 18 = 109 Avg performance = 109/6 = 18.17

17-6.

Hires Project 1 Project 2 Project 3 Project 4 Alex 8 10 17 9 Bhavna 3 8 5 6 Chen 10 12 11 9 Denise 6 13 9 7

Row reduction: Hires Project 1 Project 2 Project 3 Project 4 Alex 0 2 9 1 Bhavna 0 5 2 3 Chen 1 3 2 0 Denise 0 7 3 1

Column reduction: Hires Project 1 Project 2 Project 3 Project 4 Alex 0 0 7 1 Bhavna 0 3 0 3 Chen 1 1 0 0 Denise 0 5 1 1

Make assignments:

Hires Project 1 Project 2 Project 3 Project 4 Alex 0 0 7 1 Bhavna 0 3 0 3 Chen 1 1 0 0 Denise 0 5 1 1 Calculate time:

Hires Project 1 Project 2 Project 3 Project 4 Alex 8 10 17 9 Bhavna 3 8 5 6 Chen 10 12 11 9 Denise 6 13 9 7 Sum of resources = 10 + 5 + 9 + 6 = 30 weeks Assign Alex to Project 2, Bhavna to Project 3, Chen to Project 4, and Denise to Project 1.

The projects should require a total of 30 weeks worth of resources.

17-7.

17-8. a. FCFS Sequence FCFS Job A B C D E F Average

Start time 0 2 3 7 10 14

Processing time 2 1 4 3 4 5

Completion Time 2 3 7 10 14 19 9.17

Duedate 3 2 12 4 8 10

Tardiness 0 1 0 6 6 9 3.67

b. SPT Sequence

SPT Job B A D C E F Average c.

Start time 0 1 3 6 10 14

Processing time 1 2 3 4 4 5

Completion Time 1 3 6 10 14 19

Duedate 2 3 4 12 8 10 8.83

Tardiness 0 0 2 0 6 9 2.83

DDATE

DDATE Job B A D E F C Average d.

Job A B D E F C

SLACK Job A B D E F C Average

Start time 0 1 3 6 10 15

Processing time 1 2 3 4 5 4

Completion Time 1 3 6 10 15 19 9.00

Duedate 2 3 4 8 10 12

Tardiness 0 0 2 2 5 7 2.67

Completion Time 2 3 6 10 15 19 9.17

Duedate 3 2 4 8 10 12

Tardiness 0 1 2 2 5 7 2.83

SPT 8.83 2.83 3 9

DDATE 9.00 2.67 4 7

SLACK 9.17 2.83 5 7

SLACK Processing time 2 1 3 4 5 4

Start time 0 2 3 6 10 15

Duedate 3 2 4 8 10 12

Processing time 2 1 3 4 5 4

Slack 1 1 1 4 5 8

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

FCFS 9.17 3.67 4 9

17-9. FCFS Start Processing Job time time 1 2 5 2 7 10 3 17 4 4 21 21 5 42 14 Average Mean Flowtime*

Completion Time 7 17 21 42 56 28.6 26.6

Duedate 20 33 25 45 32

Tardiness 0 0 0 0 24 4.8

*Mean flowtime is different from avg. completion time for this problem since the start time was 2. SPT Start Processing Job time time 3 2 4 1 6 5 2 11 10 5 21 14 4 35 21 Average Mean Flowtime

DDATE Start Job time 1 2 3 7 5 11 2 25 4 35 Average Mean Flowtime

Completion Time 6 11 21 35 56 25.8 23.8

Processing time 5 4 14 10 21

Completion Time 7 11 25 35 56 26.8 24.8

SLACK Job 1 5 3 2 4

Processing time 5 14 4 10 21

Duedate 20 32 25 33 45

Duedate 25 20 33 32 45

Slack 15 18 21 23 24

Tardiness 0 0 0 3 11 2.8

Duedate 20 25 32 33 45

Tardiness 0 0 0 2 11 2.6

SLACK Start Job time 1 2 5 7 3 21 2 25 4 35 Average Mean Flowtime

Processing time 5 14 4 10 21

Completion Time 7 21 25 35 56 28.8 26.8

Duedate 20 32 25 33 45

Tardiness 0 0 0 2 11 2.6

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

FCFS 26.6 4.8 1 24

SPT 23.8 2.8 2 11

DDATE 24.8 2.6 2 11

SLACK 26.8 2.6 2 11

Recommendation FCFS completes the most jobs on time; however, the assignment that is late is overdue by 24 days. DDATE and SLACK yield the lowest mean tardiness of assignments. SPT has the lowest mean flowtime. Katie might also want to consider the weight each assignment has in her class, her average grade in each class, her goal in terms of grades, and the penalties for late assignments.

17-10.

Today is day 4.

Job D A B E F G C

Processing time 4 3 10 5 8 7 2

FCFS Start Job time A 4 B 7 C 17 D 19 E 23 F 28 G 36 Average Mean Flowtime

Duedate 8 10 12 15 18 20 25

Processing time 3 10 2 4 5 8 7

Slack 0 3 -2 6 6 9 19

Completion Time 7 17 19 23 28 36 43 24.71 20.71

Duedate 10 12 25 8 15 18 20

Tardiness 0 5 0 15 13 18 23 10.57

SPT Start Job time C 4 A 6 D 9 E 13 G 18 F 25 B 33 Average Mean Flowtime

Processing time 2 3 4 5 7 8 10

Completion Time 6 9 13 18 25 33 43 21.00 17.00

DDATE Start Job time D 4 A 8 B 11 E 21 F 26 G 34 C 41 Average Mean Flowtime

Processing time 4 3 10 5 8 7 2

Completion Time 8 11 21 26 34 41 43 26.29 22.29

SLACK Start Job time B 4 D 14 A 18 E 21 F 26 G 34 C 41 Average Mean Flowtime

Processing time 10 4 3 5 8 7 2

Completion Time 14 18 21 26 34 41 43 28.14 24.14

Summary of Results Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

FCFS 20.71 10.57 5 23

Duedate 25 10 8 15 20 18 12

Tardiness 0 0 5 3 5 15 31 8.43

Duedate 8 10 12 15 18 20 23

Tardiness 0 1 9 11 16 21 18 10.86

Duedate 12 8 10 15 18 20 25

Tardiness 2 10 11 11 16 21 18 12.71

SPT 17.00 8.43 5 31

DDATE 22.29 10.86 6 21

SLACK 24.14 12.71 7 21

SPT yields the lowest mean flowtime, mean tardiness, and number of jobs tardy.

17-11. Today’s date is day 5. Sequencing rule: DDATE Processing Job time A 5 B 8 C 6 D 3 E 10 F 14 G 7 H 3 Start Job time A 5 B 10 C 18 D 24 E 27 F 37 G 51 H 58 Average Mean flowtime

Duedate 10 15 15 20 25 40 45 50 Processing time 5 8 6 3 10 14 7 3

DDATE Performance Measures Mean Flowtime 30.75 Mean Tardiness 8.25 Max Tardiness 13 No. of Jobs Tardy 7 Sequencing rule: SPT Processing Job time D 3 H 3 A 5 C 6 G 7 B 8 E 10 F 14

Duedate 20 50 10 15 45 15 25 40

Completion Time 10 18 24 27 37 51 58 61 35.75 30.75

Duedate 10 15 15 20 25 40 45 50

Tardiness 0 3 9 7 12 11 13 11 8.25

Start Job time D 5 H 8 A 11 C 16 G 22 B 29 E 37 F 47 Average Mean flowtime

Processing time 3 3 5 6 7 8 10 14

SPT Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

23.88 9.75 22 5

Sequencing rule: SLACK Processing Job time A 5 B 8 C 6 E 10 D 3 F 14 G 7 H 3

Duedate 10 15 15 25 20 40 45 50

Start Job time A 5 B 10 C 18 E 24 D 34 F 37 G 51 H 58 Average Mean flowtime

Processing time 5 8 6 10 3 14 7 3

Completion Time 8 11 16 22 29 37 47 61

Duedate 20 50 10 15 45 15 25 40 28.875 23.875

Tardiness 0 0 6 7 0 22 22 21 9.75

Slack 5 7 9 15 17 26 38 47 Completion Time 10 18 24 34 37 51 58 61

Duedate 10 15 15 25 20 40 45 50 36.63 31.63

SLACK Performance Measures Mean Flowtime 31.63 Mean Tardiness 9.13 Max Tardiness 17 No. of Jobs Tardy 7 Recommendation: SPT is the best rule for mean flowtime and number of jobs tardy. DDATE is best for mean tardiness and maximum tardiness.

Tardiness 0 3 9 9 17 11 13 11 9.125

Alice should use whichever rule meets her objectives. For example, more customers will be happy if she uses SPT; however, jobs B, E, and F will be very late. With DDATE no job is over 2 weeks late. 17-12. a.

SPT Job B D A C

Job B D A C Average

Processing time 10 15 20 30 Start time 0 10 25 45

Processing time 10 15 20 30

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness b.

Duedate 15 30 20 50 Completion Time 10 25 45 75 38.75

Duedate 15 30 20 50

Tardiness 0 0 25 25 12.5

38.75 12.5 2 25

SLACK Job A B D C

Job A B D C Average

Processing time 20 10 15 30 Start time 0 20 30 45

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

Duedate 20 15 30 50

Processing time 20 10 15 30

Slack 0 5 15 20 Completion Time 20 30 45 75 42.5

Duedate 20 15 30 50

Tardiness 0 15 15 25 13.75

42.5 13.75 3 25

Recommendation SPT has the same maximum tardiness as SLACK, but outperforms SLACK on every other measure. Choose SPT.

17-13.

SPT Job B A D C

Job B A D C Average

Processing time 3 5 6 9 Start time 0 3 8 14

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

Duedate 5 8 7 18 Processing time 3 5 6 9

Completion Time 3 8 14 23 12

Duedate 5 8 7 18

Tardiness 0 0 7 5 3

12 3 7 2

DDATE Job B D A C

Job B D A C Average

Processing time 3 6 5 9 Start time 0 3 9 14

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

Duedate 5 7 8 18 Processing time 3 6 5 9

Completion Time 3 9 14 23 12.25

12.25 3.25 6 3

SLACK Job D B A C

Processing time 6 3 5 9

Duedate 7 5 8 18

Slack 1 2 3 9

Duedate 5 7 8 18

Tardiness 0 2 6 5 3.25

Job D B A C Average

Start time 0 6 9 14

Processing time 6 3 5 9

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

Completion Time 6 9 14 23 13

Duedate 7 5 8 18

Tardiness 0 4 6 5 3.75

13 3.75 6 3

Recommendation DDATE outperforms SLACK. Choose DDATE if minimizing maximum tardiness is most important. Choose SPT otherwise. 17-14. a.

FCFS

Job A B C D

Processing time 2 5 1 8

Duedate 7 8 15 9

Slack 5 3 14 1

Start time 0 2 7 8

Processing time 2 5 1 8

Completion Time 2 7 8 16 8.25

CALCULATIONS Job A B C D Average

OUTPUT Performance Measures Mean Flowtime 8.25 Mean Tardiness 1.75 Max Tardiness 7 No. of Jobs Tardy 1

Duedate 7 8 15 9

Tardiness 0 0 0 7 1.75

SLACK

Job D B A C

Processing time 8 5 2 1

Duedate 9 8 7 15

Slack 1 3 5 14

Start time 0 8 13 15

Processing time 8 5 2 1

Completion Time 8 13 15 16 13

Processing time 1 2 5 8

Duedate 15 7 8 9

Slack 14 5 3 1

Start time 0 1 3 8

Processing time 1 2 5 8

Completion Time 1 3 8 16 7

CALCULATIONS Job D B A C Average

OUTPUT Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy c.

Duedate 9 8 7 15

Tardiness 0 5 8 1 3.5

13 3.5 8 3

SPT

Job C A B D CALCULATIONS Job C A B D Average

OUTPUT Performance Measures Mean Flowtime 7 Mean Tardiness 1.75 Max Tardiness 7 No. of Jobs Tardy 1

Duedate 15 7 8 9

Tardiness 0 0 0 7 1.75

Recommendation Surprisingly, SLACK performs the worst of the sequencing rules. SPT outperforms FCFS on mean flowtime, but is otherwise the same as FCFS. Choose SPT. d.

Both rules have the same mean tardiness, maximum tardiness and number of jobs tardy. With SPT, however, job C is completed a week earlier, while A and B are completed only one day later. The only job tardy is job D. With both rules it is tardy by a week.

17-15. The Johnson’s rule sequence is: 8, 4, 6, 5, 9, 2, 1, 3, 10, 7

17-16.

17-17.

17-18.

17-19.

Process 1 5 7 3 4 1 3

Process 2 4 3 2 1 2 4

E 1 3

Completion Times F A 4 9 8 13

Job A B C D E F Sequence

Process 1 Process 2

Makespan

17-20. a. Johnson’s Rule sequence: E, B, D, C, A

B 16 19

C 19 21

D 23 24

b. LPT sequence: D, C, B, A, E

17-21. Chapter 1 2 3 4 5 6

Typing 30 90 60 45 75 20

Proofing 20 25 15 30 60 30

17-22. Job A B C D E

Cutting 4 6 1 2 3

Sewing 2 3 3 4 1

C 1 4

D 3 8

Makespan

Sequence

Process 1 Process 2

Completion Times B 9 12

A 13 15

E 16 17

The order can be shipped in 17 days.

17-23.

Work Center 7 Period Planned input Actual input Deviation Planned output Actual output Deviation Backlog

1 50 50 0 65 60

2 55 50

3 60 55

4 65 60

5 65 65 0 65 60

−5

65 60

−5

65 60

−5

65 60

−5

Total 295 280 -15 325 300

−25

It appears that no matter how many units are input, work center 7 can process only 60 units per week. The variation in input has caused the backlog to decrease, but now it is starting to rise again. Unless the input is reduced, the backlog will continue to increase.

17-24. Work Center 6 Period Planned input Actual input Deviation Planned output Actual output Deviation Backlog

1 50 40

2 55 50

3 60 55

4 65 60

−10

−5

50 50 0 0

55 50

60 55

65 60

−5

5 65 65 0 65 65 0 0

Total 295 270

−25 295 280

−15

It appears from the planned I/O report that the scheduler wants to maintain a 10 unit backlog. The backlog actually disappears in the first period. Work center 6 is producing everything it can, given the input. The problem is at the feeding work center until week 5. From this I/O alone, the scheduler should input 10 extra units next week to maintain a 10-unit backlog. If this I/O is considered in conjunction with the one in problem 17-19, however, then this action would aggravate work center 7’s problem. This illustrates the domino effect of scheduling linked systems.

17-25. a. Nurses needed Johnson Swann Coligny Betts Truong

M 3 O O X X X

T 3 X X O O X

W 4 X X X X O

TH 5 X X X X X

F 4 O X X X X

SA 3 X O O X X

SU 3 X X X O O

W 4 X X X X O

TH 5 X X X X X

F 4 X X O X X

SA 3 X X O O X

SU 3 O X X O X

b. Revised for consecutive days off:

Nurses needed Johnson Swann Coligny Betts Truong

M 3 O O X X X

T 3 X O X X O

There are several arrangements that produce two consecutive days off. This is one of them. Note that both M-T and SU-M are consecutive. With everyone working TH, days off would need to include TW, F-SA and SA-SN.

17-26. Volunteers 1. Haynes 2. Tagliero 3. White 4. Cooke 5. Black 6. Romero

M 4 O O X X X X

T 3 X X O O O X

W 2 O O O X X O

TH 3 X X X O O O

F 6 X X X X X X

SA 4 O O X X X X

SU 2 X X O O O O

Wait Staff 1. A. Russell 2. S. Hiller 3. J. Jones 4. T. Turner 5. E. Trice 6. P. Dubois

M 2 O O O O X X

T 3 O X X X O O

W 4 X O O X X X

TH 4 X X X O O X

F 5 X X X X X O

SA 5 O X X X X X

SU 4 X O O X X X

T 3 O X X X O

W 3 X X O O X

17-27.

17-28. Daily Requirements Anna Brent Caleb David Edhas

M 1 X O O O O

TH 5 X X X X X

F 4 X O X X X

S 5 X X X X X

SN 4 O X X X X

17-29. a. Machine 2 is the bottleneck. b. D should be scheduled first because it has the shortest path to the bottleneck that can be sustained. c. C should be scheduled last because it ends at the bottleneck. d. Completion time = 5201 hours. See Gantt chart below.

CASE SOLUTION 17.1 a. Distance to School (in minutes)

Student Amy Brent Calvin Deidre Eliena Franklin

1 45 10 30 20 30 15

Assignment based on distance

Student Amy Brent Calvin Deidre Eliena Franklin Total

1 0 1 0 0 0 0 1

2 30 30 45 30 45 20

School (P.S. or M.S.) 3 4 5 60 30 45 45 20 15 30 60 20 45 15 45 15 45 15 20 30 20 Total Distance

2 0 0 0 0 0 1 1

6 20 10 30 45 30 60

100

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 1

6 1 0 0 0 0 0 1

Total 1 1 1 1 1 1

b. Preference/Need Matchup Student Amy Brent Calvin Deidre Eliena Franklin

School (P.S. or M.S.) 1 2 3 4 5 5 2 1 2 4 2 1 4 1 3 1 2 4 1 1 2 2 1 3 2 2 5 2 3 1 1 3 1 3 2 Scored on a scale of 0 to 5. One point is given for every preference that matches a need.

6 1 3 2 3 2 1

Assignment based on preferences

Student Amy Brent Calvin Deidre Eliena Franklin Total c.

1 1 0 0 0 0 0 1

2 0 0 0 0 1 0 1

Score

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1

6 0 0 0 1 0 0 1

Total 1 1 1 1 1 1

There are several ways distance can be combined with preferences. One simple way is to assign a 1 to the closest schools and add that to the preference/need matrix. Preference /Need Matrix including distance

Student Amy Brent Calvin Deidre Eliena Franklin

1 5 3 1 2 2 2

Assignment

Student Amy Brent Calvin Deidre Eliena Franklin Total

2 2 1 2 2 5 3

School (P.S. or M.S.) 3 4 5 1 2 4 4 1 3 4 1 2 1 4 2 3 3 2 1 3 2

Score

1 1 0 0 0 0 0 1

2 0 0 0 0 1 0 1

6 2 4 2 3 2 1

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1

6 0 1 0 0 0 0 1

Total 1 1 1 1 1 1

d. To prioritize preferences or needs, assign different weights to each factor. The "score" for each school for a student would be a sum of the weighted factors. Alternatively, more points could be assigned to the presence of certain factors.

CASE SOLUTION 17.2 1. Find the bottleneck Machine 1:

A1 C1 D1 D3

4 Machine 2: 5 2 4 15 Machine 2 is the bottleneck.

A3 B2 C2 D4

5 8 4 7 24

Machine 3:

A2 B1 C3 D2

6 4 9 1 20

2. Find the fastest time to the bottleneck A3 10 min B2 4 min C2 5 min D4 7 min Bottleneck sequence: B2, C2, D4, A3 3. Sequence the other machines to support the bottleneck. Machine 1: D1, C1, D3, A1 Machine 3: B1, D2, C3, A2

Items are transferred one at a time, but processed in batches of 50. Machine 2 starts 4 minutes after machine 3. 50 X’s can be assembled in 1204 minutes. Determining the sequence involves trying different arrangements, always keeping the bottleneck busy.