This Document Contains Chapters 6 to 8 CHAPTER 6 LEARNING CURVES DISCUSSION QUESTIONS 1. How might the following business specialists use learning curves: accountants, marketers, financial analysts, personnel managers, and computer programmers? Accountants: estimating costs Marketers: setting-selling prices Financial analysts: performing a breakeven analysis for purposes of investment decision making Personnel managers: estimating the number of workers required Computer programmers: estimating times to write programs 2. What relationship is there between learning curves and productivity measurement? As time goes on, the learning effect essentially will increase productivity. Learning will allow a system to take less time to produce a single unit of output, therefore increasing output in a system even though the inputs remain the same. This assumes that demand will be sufficient to justify the higher output rate. 3. What relationship is there between learning curves and capacity analysis? This answer is related to the previous one. The effect of learning will be to increase the best operating level of a system as the time to produce a unit decreases. Even though capital resources may remain constant, the system can produce more due to the effect of learning. 4. Do you thing learning curve analysis has application in a service business like a restaurant? Why or why not? There are limited applications to the employees of that type of business. Cooks will learn to prepare meals more quickly. Servers will learn to take and enter orders more quickly as well as service more customer requests in a given time period. The bus staff will get faster at turning around tables between customers. All of these will tend to increase the system capacity somewhat, but there is still a limit on capacity from the physical number of tables and chairs. Also, there is little you can do to “teach” your customers to finish their meals more quickly. Learning curves can be used to judge the effectiveness of new employees. 5. As shown in the chapter, the effect of learning in a given system eventually flattens out over time. At that point in the life of a system, learning still exists, though its effect continues to diminish. Beyond that point is it impossible to significantly reduce the time to produce a unit? What would it take to do that? If nothing in the system changes except for learning, eventually the production time per unit will become essentially flat. To reduce it significant would require a change in the system. Perhaps the product can be redesigned to make it easier to produce. Or maybe the process is redesigned to include improved or automated equipment that greatly increases the output rate of the system. 6. The learning curve phenomenon has been shown in practice to be widely applicable. Once a company has established a learning rate for a process, they can use it to predict future system performance. Would there be any reason to reevaluate the process’ learning rate once it has been initially established? Like any planning for the future, learning curve analysis faces the unknown quality of what is yet to come. If the rate was established after just a few units of production it would make sense to reevaluate it after a significant number of additional units have been produced. If the rate was estimated based on the assumption of similarity between the new process and similar previous applications, it should be reevaluated after the new process has produced a number of units to confirm if that assumption was valid. The learning rate should be reevaluated if there is a significant change made to the process. 7. As a manager, which learning percentage would you prefer (other things being equal), 110 percent or 60 percent? Explain. Students tend at first glance to erroneously associate higher learning percentages with faster learning. Relative to the 110 percent learning rate, strict interpretation of this would mean that every time output doubles, production time per unit increases by 10 percent. With a 60 percent learning rate, every time output doubles, production time per unit decreases by 40 percent. These statements can be verified by simple arithmetic. 8. Will the Human Resource Management (HRM) policies of a firm have much of an effect on the learning rates the firm may be able to achieve? Much of the learning in a process is done by humans, and not all humans are equal. A firm that is willing to pay more to hire persons with greater technical and intellectual capacity is likely to achieve better learning rates. Whether the benefit from those higher learning rates is worth the additional wages would require further analysis specific to the situation. 9. One manufacturer has seen a typical learning percentage of 90% in their firm. They have recently found out that a competitor of theirs has a percentage of 85%. What do you think about this? As with comparing productivity figures between firms, we would need to know more about the competitor’s processes. Just because they have a better learning rate does not make them more competitive. That better rate may come from increased expenses in labor, equipment, or materials that are not offset by the increased learning rate. 10. What difference does it make if a customer wants a 10,000 unit order produced and delivered all at one time or in 2,500 unit batches? Aside from the costs of re-setup, undoubtedly some relearning is necessary each time one of the 2,500 unit orders is produced. This would result in additional time and more material and other resource usage. What might be better and cheaper (at least from a learning curve perspective) is to produce the entire 10,000 unit order and simply deliver 2,500 units at a time to the customer. OBJECTIVE QUESTIONS 1. Firm A typically sees a learning percentage of 85% in their processes. Firm B has a learning percentage of 80%. Which firm has the faster learning rate? Firm B 2. Company Z is just starting to make a brand new product they have made before. They have completed two units so far. The first unit took 12 hours to complete and the next unit took 11 hours. Based only on this information, what would be the estimate of the learning percentage in this process? 𝐿𝑒𝑎𝑟𝑛𝑖𝑛𝑔 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 = = 91.67% 3. Omega Technology is starting production of a new supercomputer for use in large research universities. They have just completed the first unit, which took 120 man-hours to produce. Based on their experience, they estimate their learning percentage to be 80%. How many man-hours should they expect the second unit to require to manufacture? 𝑆𝑒𝑐𝑜𝑛𝑑 𝑢𝑛𝑖𝑡′𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 = 120(0.8) = 96 ℎ𝑜𝑢𝑟𝑠 4. a. Units 1 to 2 = 640/970 = 65.98% Units 2 to 4 = 380/640 = 59.38% Units 4 to 8 = 207/380 = 54.47% Average learning rate = 59.94% Round this to 60% to use the tables. b. From Exhibit 6.5 200 units = 12.090 -10 units = 3.813 8.277 Therefore, time for 190 more units = 970(8.277) = 8,029 hours c. For 1,000th unit from Exhibit 6.4: .0062(970) = 6.0 hours 5. LR = 1,800/2,000 = 90% From Exhibit 6.5 6 units = 5.101 -3 units = 2.746 2.355 Therefore, time for 3 more units = 2,000(2.355) = 4,710 hours 6. a. For units 1 to 2 = 10/12 = 83.33% For units 2 to 4 = 6.5/10 = 65 % For units 4 to 8 = 3.6/6.5 = 55.38% For units 8 to 16 = 2.7/3.6 = 75% The average learning rate is = 69.67 Round this to 70%. b. From Exhibit 6.5 120 units = 19.570 -20 units = 7.407 12.163 Therefore, cost for 100 more units = 12 million (12.163) = $145,956,000. LTI should add the appropriate profit margin to this expected cost. c. From Exhibit 6.4: Cost for 120th unit is $12 million(.0851) = $1,021,200 7. For labor the following learning was experienced: Unit 1 to 2 = 1500/2000 = 75%, from units 2 to 4 = 1275/1500 = 85% Based on this, we estimate an average labor learning rate of 80% For cost the following learning was experienced: Unit 1 to 2 = 37050/39000 = 95%, from units 2 to 4 = 31492/37050 = 85% Based on this, we estimate an average cost of parts learning rate at 90% Labor for 12 more units: From Exhibit 6.5 16 units = 8.920 -4 units = 3.142 5.778 Therefore, Labor for 12 more units = 2,000(5.778) = 11,556 hours Cost for 12 more units: From Exhibit 6.5 16 units = 12.040 -4 units = 3.556 8.484 Therefore, cost for 12 more units = 39,000(8.484) = $330,876 8. a. Labor: LR = 3500/5000 = 70% From Exhibit 6.5 12 units = 5.501 -2 units = 1.700 3.801 Therefore, Labor cost for 10 more units = 5,000(3.801)(30) = $570,150 Material: LR = 200000/250000 = 80% From Exhibit 6.5 12 units = 7.227 -2 units = 1.800 5.427 Therefore, Labor cost for 10 more units = 250,000(5.427) = $1,356,750 Total Cost is $570,150 + $1,356,750 = $1,926,900 b. The minimum cost would be as calculated in part a, $1,926,900. This would assume no forgetting. However, the worst case would be total forgetting, which would imply that there was no benefit to having produced units 1 and 2. This cost would be as follows. Complete forgetting: Labor: 4.931(5,000)(30) = $ 739,650 Material: 6.315($250,000) = $1,578,750 Total = $2,318,400 Therefore, the range is from $1,926,900 to $2,318,400. 9. Southwest Honda’s LR = 6.3/9.0 = 70% The breakeven point is where the AVERAGE labor hours on all cars serviced by the mechanic equals 3 hours. According to the following table, that doesn't occur until after car 25. Unless each mechanic is going to service over 25 cars, it's not a good deal. Since the dealer expects to perform the service on 300 vehicles with 6 mechanics (approximately 50 per mechanic), it appears that Honda’s rate is more than fair. Learning Rate: 70% Unit Unit Time Average Time 1 9.0 9.00 2 6.3 7.65 3 5.1 6.80 4 4.4 6.21 5 3.9 5.75 6 3.6 5.39 7 3.3 5.09 8 3.1 4.84 9 2.9 4.63 10 2.8 4.44 11 2.6 4.27 12 2.5 4.13 13 2.4 3.99 14 2.3 3.87 15 2.2 3.76 16 2.2 3.66 17 2.1 3.57 18 2.0 3.49 19 2.0 3.41 20 1.9 3.33 21 1.9 3.26 22 1.8 3.20 23 1.8 3.14 24 1.8 3.08 25 1.7 3.03 26 1.7 2.97 27 1.7 2.92 28 1.6 2.88 29 1.6 2.83 30 1.6 2.79 A more direct solution is to use exhibit 6.5. For 50 units (300 units/6 mechanics), the cumulative value is 12.31. Total hours for each mechanic is therefore 12.31(9) = 110.79 hours. This is an average of 110.79/50 = 2.22 hours per unit. Paying the dealer for 3 hours per unit is definitely fair if the final total number of vehicle repairs per mechanic is as estimated. 10. Labor LR = 3,200/4,000 = 80% Material LR = 21,000/30,000 = 70% a. For the 22nd unit: Labor: 4,000(.3697)($18 per hour) = $26,618.40 Material: $30,000(.2038) = $ 6,114.00 Total = $32,732.40 b. From Exhibit 6.5 22 units = 11.230 -2 units = 1.800 9.430 Labor time for 20 more units = 4,000(9.430) = 37,720 hours Average time = 37,720/20 = 1,886 hours c. From Exhibit 6.5 22 units = 7.819 -2 units = 1.700 6.119 Therefore, material cost for 20 more units = $30,000(6.119) = $183,570 Average material cost = $183,570/20 = $9,178.50 Average labor cost = 1,886($18 ) = $33,948.00 Average total cost = $9,178.50 + $33,948.00 = $43,126.50 11. Which type of system is likely to have a faster learning rate – one with primarily highly automated equipment or one that is very labor intensive? Labor intensive 12. Which industry will typically have a faster learning rate: a repetitive electronics manufacturer or a manufacturer of large complex products such as a shipbuilder? Shipbuilder 13. True or False: The only learning for an organization comes from the individual learning of its employees. False 14. A company has just tested the skills of two applicants for the same job. They found that applicant A had a higher learning rate than applicant B. Should they definitely higher applicant A? No CHAPTER 7 MANUFACTURING PROCESSES Discussion Questions 1. What is meant by a process? Describe its important features. A process means a set of tasks that transform input into useful outputs. The important features of process are (a) tasks, (b) flow (of material and information), and (c) storage (of material and information). 2. What is a customer order decoupling point? Why is it important? Essentially it is where inventory is stored awaiting demand from the customer. It is important because it affects the lead time to fulfill the customer’s order and the amount of inventory investment necessary. 3. What’s the relationship between the design of a manufacturing process and the firm’s strategic competitive dimensions (Chapter 2)? There is a natural relationship between the location of the customer order decoupling point, the level of customization the manufacturer provides its customers, and delivery speed of the product to the customer. At one end of the spectrum we have make-to-stock processes which produce in anticipation of demand, allowing inventory to be stored close to the customer for often instantaneous delivery. However, there is virtually no customization available in make-tostock products. On the other end of the spectrum are engineer-to-order processes, where the product is designed from the start to exactly satisfy the customer’s unique needs. Customization is maximized in an engineered-to-order product, but lead time is quite extensive. 4. What does the product-process matrix tell us? How should the kitchen of a Chinese restaurant be structured? The Chinese restaurant case might be debatable since it involves both high volume and high variety. Probably a work cell would be best. 5. It has been noted that during World War II Germany made a critical mistake by having its formidable Tiger tanks produced by locomotive manufacturers, while American car manufacturers produced the less formidable U.S. Sherman tank. Use the product-process matrix to explain that mistake and its likely result. The locomotive manufacturers likely used project technology and processes. This is low volume, high cost production. On the other hand, mass-producing automakers had the technology to make high volume at low per unit cost. 6. How does the production volume affect break-even analysis? A break-even analysis takes into account the production volume and the relevant cost of producing the volume by the available alternative processes. It calculates the relative profit or loss of the alternative processes, thus helping to decide which alternative to choose for a certain volume of production. 7. What is meant by manufacturing process flow? In a manufacturing process, material and information must move throughout the facility between manufacturing points and storage locations. The path that both material and information take as they move through the facility defines the process flow. 8. Why is it that reducing moves, delays, and storages in a manufacturing process is a good thing? Can they be completed eliminated? While unavoidable to some extent, all three of these add time to the process while adding no value to the product. Reduction of these will reduce the time it takes to manufacture a product and thereby improve the process’ flow. Objective Questions 1. What is the first of the three simple steps in the high-level view of manufacturing? Sourcing the parts we need. 2. The customer order decoupling point determines the position of what in the supply chain? Inventory 3. Dell Computers’ primary consumer business takes orders from customers for specific configurations of desktop and laptop computers. Customers must select from a certain model line of computer, and choose from available parts, but within those constraints may customize the computer as they desire. Once the order is received, Dell assembles the computer as ordered and delivers it to the customer. What type of manufacturing process is described here? Assemble-to-order 4. What term is used to mean manufacturing designed to achieve high customer satisfaction with minimum levels of inventory investment? Lean manufacturing 5. Issue Workcenter (Job Shop) Assembly Line Number of Changeovers Many Few Labor content of product High Low Flexibility High Low 6. Workcenter Assembly Line Engineering Emphasis Product variety and improvements Process improvements. General Workforce Skill Skilled workforce. Lower skill levels. Statistical Process Control Less important. More important. Facility Layout Process, functional. Product, line, flow. WIP Inventory Level Depends on the product. Lower. 7. a. throughput time (time to convert raw material into product) Throughput time decreases as you move from a work center to assembly line environment. b. capital/labor intensity Capital intensity increases as you move to an assembly line (need for more machinery). Labor intensity (assuming you mean # of workers) would probably increase as well (depends on the level of automation). Ratio would increase in an assembly line because capital requirements are so much greater. c. bottlenecks Would probably decrease in an assembly line. 8. a. FC = (P - VC) * Break-even (where FC = fixed cost, P = price, and VC = variable cost) $300,000 = ($23.00 - $8.00) * Break-even Break-even = 20,000 books b. Higher c. Lower 9. FC = (P - VC) * Break-even (where FC = fixed cost, P = price, and VC = variable cost) $150,000 = ($90 - $70) * Break-even Break-even = 7,500 units. 10. a. FC = (P - VC) * Break-even (where FC = fixed cost, P = price, and VC = variable cost). $900 = ($5.50 - $4.50) * Break-even Break-even = 900 units. b. FC + profit = (P - VC) * V (where FC = fixed cost, P = price, and VC = variable cost, and V = Volume) $900 + $500 = ($5.50 - $4.50) * V V = 1400 units. c. $0.25 Profit per unit = (P-VC)*V - FC)/V $.25 = (($5.50 - $4.50)*V - $900)/V .25V = V - 900 .75V = 900 V = 1,200 units. $0.50 Profit per unit = ((P-VC)*V - FC)/V $.50 = (($5.50 -$4.50)*V - $900)/V .50V = V - $900 .50V = 900 V = 1,800 units. $1.50 Profit per unit = ((P-VC)*V - FC)/V $1.50 = (($5.50 -$4.50)*V - $900)/V 1.5V = V - 900 .50V = -900 V = -1,800 units. Not possible. 11. FC = (P - VC) * Break-even (where FC = fixed cost, P = price, and VC = variable cost). $2052 = ($.36 - $.144) * Break-even Break-even = 9,500 miles. 12. a. FC = (Pc - VCc) * Vc + (Pb -VCb) * Vb where FC = fixed cost Pi = price for product i VCi = variable cost for product i Vi = volume for product i where i is c = chair b = bar stool V = Vc = Vb $20,000 = ($50 - $25) * V + ($50 - $20) * V 20,000 = 25V + 30 V 20,000 = 55V V = 364 units of chairs and bar stools. Break-even in dollars = $50(364 + 364) = $36,400. b. FC = (Pc - VCc) * Vc + (Pb -VCb) * Vb where FC = fixed cost Pi = price for product i VCi = variable cost for product i Vi = volume for product i where i is c = chair b = bar stool V = Vc = 0.25Vb $20,000 = ($50 - $25) * V + ($50 - $20) *4V 20,000 = 25V + 120 V 20,000 = 145V V = 138 units of chairs and (4 * 138) = 552 bar stools. Break-even in dollars = $50(138 + 552) = $34,500. 13. BEP (dollars) = $10,000/(1-(($3.50 + $4.50)/$12.50)) = 10,000/.36 = $27,778 BEP (units) = $10,000/(12.50-8) = 10,000/4.50 = 2,222 units 14. Existing process BEP = $14,000/(1-.5) = 28,000 units 30,000-28,000 X $.50 profit = $1,000 New equipment BEP = $20,000/(1-.6) = 50,000 units No profit as volume equals BEP 15. a. Process Flow Diagram Capacity of assembly line 1 = 140 units/hour X 8 hours/day X 5 days/week = 5,600 units/week. Capacity of drill machines = 3 drill machines X 50 parts/hour X 8 hours/day X 5 days/week = 6,000 units/week. Capacity of final assembly line = 160 units/hour X 8 hours/day X 5 days/week = 6,400 units/week. The capacity of the entire process is 5,600 units per week, with assembly line 1 limiting the overall capacity. b. Capacity of assembly line 1 = 140 units/hour X 16 hours/day X 5 days/week = 11,200 units/week. Capacity of drill machines = 4 drilling machines X 50 parts/hour X 8 hours/day X 5 days/week = 8,000 units/week. Capacity of final assembly line = 160 units/hour X 16 hours/day X 5 days/week = 12,800 units/week. The capacity of the entire process is 8,000 units per week, with drilling machines limiting the overall capacity. c. Capacity of assembly line 1 = 140 units/hour X 16 hours/day X 5 days/week = 11,200 units/week. Capacity of drill machines = 5 drilling machines X 50 parts/hour X 8 hours/day X 5 days/week = 10,000 units/week. Capacity of final assembly line = 160 units/hour X 12 hours/day X 5 days/week = 9,600 units/week. The capacity of the entire process is 9,600 units per week, with final assembly machines limiting the overall capacity. d. Cost per unit when output = 8,000 units. Item Calculation Cost Cost of part A $.40 X 8,000 $3,200 Cost of part B $.35 X 8,000 2,800 Cost of part C $.15 X 8,000 1,200 Electricity $.01 X 8,000 80 Assembly 1 labor $.30 X 8,000 2,400 Final assembly labor $.30 X 8,000 2,400 Drilling labor $.15 X 8,000 1,200 Overhead $1,200 per week 1,200 Depreciation $30 per week 30 Total Cost $14,510 Cost per unit = Total cost per week/Number of units produced per week = $14,510/8,000 = $1.81 Cost per unit when output = 9,600 units. Item Calculation Cost Cost of part A $.40 X 9,600 $3,840 Cost of part B $.35 X 9,600 3,360 Cost of part C $.15 X 9,600 1,440 Electricity $.01 X 9,600 96 Assembly 1 labor $.30 X 9,600 2,880 Final assembly labor $.30 X 9,600 2,880 Drilling labor $.15 X 9,600 1,440 Overhead $1,200 per week 1,200 Depreciation $30 per week 30 Total $17,166 Cost per unit = Total cost per week/Number of units produced per week = $17,166/9,600 = $1.79 e. Let X = the number of units that each option will produce. When the company buys the units, the cost is $3.00 per unit (3X). When it manufactures the units, they incur a fixed cost of $120,000 (4 drilling machines at $30,000 apiece) and a per-unit cost of $1.81 (from part d). Therefore, 120,000 + 1.81X is the cost of this option. Set them equal to each other and solve for X to determine the breakeven point. 3X = 120,000 + 1.81X X = 100,840 units Therefore, it is better to buy the units when you produce less than 100,840, and better to produce them when demand is greater than 100,840 units. 16. 60/.75 = 80 units/hour 17. Determine capacity analysis, look at moving from job shop to continuous flow production, price low and make up profits in volume, analyze production deadlines in order to determine labor needs (i.e. can the shop get by with a part time employee to meet the extra capacity). Case: Circuit Board Fabricators, Inc. - Teaching note A good way to use this case is to discuss product/process design concepts in the first part of the class and then work on the case. This is a batch process and it is good to spend some time discussing how a batch process works. All of the tasks are given in exhibit 7.8. Describe in some detail how the first set of activities work (i.e. the machine is loaded, then the board is automatically cleaned and coated, finally an operator must unload a board from the machine). This is done for each board in the batch. The batch is then moved to the expose area. Draw a diagram on the board as you go through the process (if strapped for time, just put the following diagram up. One issue that you need to discuss is the case when there are multiple machines. Will all the machines be setup for a job or will only one machine be setup. This can dramatically impact the capacity of the system. The calculations below assume that only one machine is setup for each order. Things do not look very good even with this assumption. Go through the calculations for the first task in the process by first calculating the “operation time” for the task. This is the setup and run time for a typical order. The case states that they normally start with 75 boards in an order, so the operation time is the sum of the setup + 75 times the run time. Next ask the question, “Given that we know how long it takes to process each order, how many orders can we process per day?” This is calculated by taking the total time available in a day (number of machines x 7.5 hours x 60 minutes/hour) and dividing by the operation time. Stress that here we are assuming that only one machine is being used for each job. This is a “best case” analysis. Now you can discuss the impact of the losses in the system. After “inspect” the size of the order drops to 63.75 (75 x .85) and after “final test” only 60.56 (63.75 x .95) boards are left in each batch. Your spreadsheet should look like that shown below: Circuit Board Fabricators - Process Data Required Output per shift 1000 Average Job Size (boards) 60 75 Production hours per day 7.5 Working Days per Week 5 Average Number Setup Run Size of Operation Capacity of Number of (minutes (minutes Job time per per day Process/Machine Machines Employees per job) per part) (boards) order (jobs) Load 1 1 5 0.33 75.00 29.75 15.13 Clean 1 0.5 75.00 37.5 12.00 Coat 1 0.5 75.00 37.5 12.00 Unload 1 1 0.33 75.00 24.75 18.18 Expose 5 5 15 1.72 75.00 144 15.63 Load 1 1 5 0.33 75.00 29.75 15.13 Develop 1 0.33 75.00 24.75 18.18 Inspect 2 2 0.5 75.00 37.5 24.00 Bake 1 0.33 63.75 21.0375 21.39 Unload 1 1 0.33 63.75 21.0375 21.39 Drilling 6 3 15 1.5 63.75 110.625 24.41 Copper Plate 1 2 5 0.2 63.75 17.75 25.35 Final Test 6 6 15 2.69 63.75 186.4875 14.48 Output 60.56 12.00 Total capacity (boards) 726.75 Next, get into a discussion of where the bottleneck is in this process. The “clean” and the “coat” tasks are the bottlenecks. You can talk about blocking and starving at this time. Sum this discussion up by calculating the capacity of the system by taking the number of orders that can be processed by the bottleneck (i.e. 12) and multiplying by the typical order size of 60. It looks like the capacity is only about 720 boards per day. Ask the question, “How many orders do we need to process to get 1,000 boards per day out of the system?” With an average order size of 60 this is equal to 1000/60 or 16.6667 orders per day. Finish up with a discussion of how to get the capacity up to 17 orders per day. First, you would need to work on the bottleneck process (clean and coat). You could add another machine, but this would result in much idle time on that machine. Another idea would be to run overtime on that machine. Experimentation shows that you need about 3 hours of overtime there. Except for the final test, it would probably be good to do some reengineering on the other below 17 order/day tasks. You will probably need to buy some additional machines for final test to get sufficient capacity. Let the students be creative here, but show how each suggestion impacts the capacity of the system and make sure they realize the impact on machine utilization and overall capacity of the process. Finish off by reviewing what they have learned: (1) they have learned how to analyze process engineering data typical of what would be available about a process, (2) they have learned how to analyze a process where setups are significant, (3) the can see how process losses are considered, and (4) how the bottleneck limits capacity of the entire system. CHAPTER 8 FACILITY LAYOUT Discussion Questions 1. What kind of layout is used in a physical fitness center? Process layout—similar equipment or functions are grouped together, such as rowing machines in one area, and weight machines in another. The exercise enthusiasts move through the fitness center, following an established sequence of operations. 2. What is the objective of assembly-line balancing? How would you deal with the situation where one worker, although trying hard, is 20 percent slower than the other 10 people on a line? The objective is to create an efficient balance between the tasks and workstations to minimize idle time. If the employee is deemed valuable, training may enhance his/her speed. It is also possible to place him/her in the “choice” job, i.e., that workstation which has most idle time to adjust for the slowness. Also, faster workers may assist the slowpoke if the balance and physical features of the line permit. 3. How do you determine the idle-time percentage from a given assembly-line balance? Idle-time percentage is given as “balance delay” in the chapter. It is simply one minus efficiency, where efficiency is equal to the sum of the task times divided by the number of workstations times the cycle time. 4. What is the essential requirement for mixed-model lines to be practical? The need to develop a cycle mix that minimizes inventory build-up while keeping cycle time constant. Sufficient capacity and short changeovers are also necessary. 5. Why might it be difficult to develop a manufacturing cell? a. Distinct parts families must exist. This requires developing and maintaining a computerized parts classification and coding system. This can be a major expense. b. Several of each type of machine must be available. This could be an expensive proposition, given the cost of purchasing and maintaining duplicate sets of machinery. c. Taking a machine out of a cluster should not rob a cluster of all of its capacity. d. There may be parts that cannot be associated with a family and specialized machinery that cannot be placed in a cell because of its general use. e. Training personnel to perform multiple types of tasks may be initially difficult. Union regulations and interpersonal problems within a group working in a cell must be resolved before the cell is implemented. 6. In what respects is facility layout a marketing problem in services? Give an example of a service system layout designed to maximize the amount of time the customer is in the system. The facility layout must be designed to meet customer expectations. Unlike fast food outlets, many finer restaurants will try to maximize the time that a customer is in system via requiring a waiting period before seating customers (even those with reservations). Many customers will patronize the cocktail lounge (thus enhancing profits). 7. Consider a department store. Which departments probably should not be located near each other? Would any departments benefit from close proximity? Ask the students to list the various types of departments in a department store. For example, typical departments include: women’s clothing, men’s clothing, children’s clothing (both male and female), shoes, health and beauty aids, tools, camping supplies, home supplies, music, toys, etc. There are obvious non-complimentary departments such as tools and health and beauty aids. An example of complimentary departments would be men’s clothing and shoes. 8. How would a flowchart help in planning the servicescape layout? What sorts of features would act as focal points or otherwise draw customers along certain paths through the service? In a supermarket, what departments should be located first along the customers’ path? Which should be located last? The flowchart aids in monitoring the flow of customers through the service area. This would provide a means for providing a layout that minimizes the distance required for customers to reach the product. Focal points could include K-mart style blue lights and “end cap” displays. In a supermarket, the first things in the customer’s path should be shopping carts and convenience items such as a delicatessen. It should be remembered that certain popular items such as milk and eggs are kept at the rear of a store to cause the customer to walk through the store and increase impulse purchases. Objective Questions 1. Cyprus Citrus Cooperative has many alternatives. Below is the existing layout with the steps in the process numbered. Students will identify many possible alternatives. Assuming that no jobs are combined, the following is a possibly: This layout allows an efficient “U” shaped flow through the department. 2. There will be multiple good answers to this problem. Following are examples. a. b. c. Excess 20' 60' d. Unlikely. The nurses’ station is not centrally located, and while patients may not go to the lab (department 5) very often, the nurses do. Thus, they will do a great deal of walking. It would be interesting to have the nursing staff develop closeness ratings and contrast them with that of administrators and M.D.s. 3. a. There are 8! = 40,320 assignments possible if no constraints are applied. b. There are only two layouts that satisfy all of the constraints. Layout 1 A B G H 1 2 3 4 Courtyard E F C D 5 6 7 8 Layout 2 A C G H 1 2 3 4 Courtyard E F B D 5 6 7 8 The following table contains the total materials handling costs for the two alternative layouts. Layout 1 Layout 2 Layout 1 Layout 2 A-B 2 10 25 20 50 A-C 0 25 10 0 0 A-D 0 34 34 0 0 A-E 5 15 15 75 75 A-F 0 18 18 0 0 A-G 0 20 20 0 0 A-H 0 30 30 0 0 B-C 0 18 18 0 0 B-D 0 25 10 0 0 B-E 0 18 20 0 0 B-F 3 15 10 45 30 B-G 0 10 15 0 0 B-H 2 20 18 40 36 C-D 0 10 25 0 0 C-E 0 20 18 0 0 C-F 0 10 15 0 0 C-G 0 15 10 0 0 C-H 3 18 20 54 60 D-E 4 30 30 120 120 D-F 0 20 20 0 0 D-G 0 18 18 0 0 D-H 0 15 15 0 0 E-F 1 10 10 10 10 E-G 0 25 25 0 0 E-H 0 34 34 0 0 F-G 1 18 18 18 18 F-H 0 25 25 0 0 G-H 4 10 10 40 40 Total 422 439 Based upon total material handling cost, layout 2 is optimal. 4. Mixed model balance. Required output/day: 24B + 24D Process times: 12 minutes per B, and 8 minutes per D. Model sequence B B D D D B B B B D D D Operation time 12 12 8 8 8 12 12 12 12 8 8 8 Mini-cycle time 24 24 24 24 24 5. a. b. C = production time per day/required output per day = (8 hour/day)(3600 seconds/hour)/240 units per day = 120 seconds per unit c. Work station Task Task time Idle time I A D 60 50 10 II B C 80 20 20 III E F 90 30 0 IV G H 30 60 30 T 420 d. Efficiency = = = .875 or 87.5% NaC 4(120) 6. a. b. C = production time per day/required output per day = (450 minutes/day)/360units per day = 1.25 minutes per unit or 75 seconds per unit c. Work station Task Task time Idle time I A C E 30 30 15 0 II F 65 10 III B G 35 40 0 IV D H 35 25 15 T 275 NaC 4(75) d. Efficiency = = = .917 or 91.7% 7. a. C = production time per day/required output per day = (7.5 hour per day) (3600 seconds per hour)/1000units per day = 27 seconds per unit b. Work station Task Task time Idle time 15 A I 6 6 C II B 24 3 E 18 III 2 F 7 I 14 IV 1 D 12 G 11 V H 9 J 7 0 K 15 L 10 2 VI 148 c. Efficiency = = = .914 or 91.4% NaC 6(27) d. Reduce cycle time to 25 ((7.5*3600)/1100 = 24.54 seconds), which requires rebalancing the line or work overtime: (100 units) 27 seconds per unit = 2700 seconds or 45 minutes of overtime. 8. 100' 100' 100' A B C D 50'50' 50' From/to Distances - rectilinear Flow Cost = distance X flow X $2 A to B 100’ 10 $2,000 A to C 200’ 25 10,000 A to D 250’ 55 27,500 B to C 100’ 10 2,000 B to D 150’ 5 1,500 C to D 50’ 15 1,500 Total $44,500 9. a. b. C = production time per day/required output per day = (7.5 hour per day) (3600 seconds per hour)/300 units per day = 90 seconds per unit T 410 C 90 c. Ni = = = 4.56 → 5 work stations d. Work station Task Task time Idle time I a d 70 10 10 II g j 60 25 5 III c b 45 40 5 IV e h 30 50 10 V f i k l 20 15 20 25 10 T 410 NaC 5(90) e. Efficiency = = = .911 or 91.1% f. Reduce cycle time, say, to 81 seconds per unit. This produces (7.5 hours)(3600 seconds per hour)/81 seconds per unit = 333.3 units. Another option is to work 45 minutes overtime (7.5 x 10% = .75 hour or 45 minutes). There are many other options possible that are combinations of these two options. 10. a. C = production time per day/required output per day = (7 hour per day) (3600 seconds per hour)/750 units per day = 33.6 seconds per unit T 118 C 33.6 b. Ni = = = 3.51 → 4 work stations c. d. Work station Task Task time Idle time I A B 20 7 6.6 II D F 22 10 1.6 III C 20 13.6 IV E G 15 16 2.6 V H 8 25.6 T 118 NaC 5(33.6) e. Efficiency = = = .702 or 70.2% f. Reduce cycle time to 32. New production level is (7 hours/day)(3600 seconds per hour)/32 seconds per unit) = 787.5 units per day. Therefore, they are 800 – 787.5 = 12.5 units short. Work (12.5 units )(32 seconds per unit) = 400 seconds or 6.67 minutes overtime. g. 1000 – 787.5 = 212.5 units short, work (212.5 units)(32 seconds per unit) = 6800 seconds or 113.3 minutes or 1.89 hours of overtime. May want to consider rebalancing the line. 11. Student answers will vary. Following is one possible solution a. 1 2 3 4 8 7 6 5 b. Adjacency Flow Distance Flow * Distance 1-2 35 10 350 1-5 1 40 40 2-3 30 10 300 2-7 4 10 40 3-4 45 10 450 3-5 5 20 100 4-5 30 10 300 4-7 3 30 90 5-6 30 10 300 6-7 20 10 200 7-8 15 10 150 Total: 2320 12. a. Cycle time = Production time per day/required output per day 7.5 (60)(60)/900 units = 30 seconds b. Efficiency = sum of task time/(actual number of workstations x cycle time) 120 sec/ (4(30)) = 1 This is 100 percent efficient c. To make 900 units per shift, station 3 will need to be duplicated. Efficiency formula = 135 / (5(30)) = .9 This assumes that each of the other operators still take 30 seconds to complete. The system becomes less efficient due to the idle time at the parallel workstations. 13. a. 5hours x 5days x 60min/ hr C = = 0.5 min/unit 3000units/ day 1.9 min Nt = = 3.8 round to 4 0.5 min The solution is: Station Solution 1 1, 2, 7,4 2 8, 9, 10 3 3, 5, 6 4 11, 12, 13 Efficiency = = 95% 14. Cycle time = 4 minutes 13 1 or 4 stations Minimum number of stations = 4 = 3 3 There are two solutions: Station Solution 1 Solution 2 1 A, B, C A, B, C 2 F, D, G F, G, H 3 E D, E 4 H, I I Efficiency = = 81.25% 15. There are multiple solutions. Following is one example. a. b. (60 minutes/hr)/(15 units/hr) = 4 minutes per unit. C = 4. c. Work station Task Task time Idle time I A C 1 3 0 II E 3 1 III B F 2 2 0 IV D G 1 3 0 Here are some other solutions. Station Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 1 A-B-C A-B-D-F A-B-F-G A-B-F-G A-B-F-G 2 F-G-H C-G C-H C-D D-H 3 D-E E D-E E C 4 I H-I I H-I E-I T 15 Efficiency = = = .9375 or 93.75% NaC 4(4) 16. In manufacturing layout design, the key concern is the resulting efficiency of the operation. In retail service operations, what is the primary concern or objective? Maximizing net profit per square foot of store space. 17. What is the term used to refer to the physical surroundings in which service operations occur and how these surroundings affect customers and employees? Servicescape 18. What are the three terms used to describe the parts of a service operation that have social significance? Sign, symbols, and artifacts ANALYTICS EXERCISE: Designing a Manufacturing Process 1. What is the daily capacity of the assembly line designed by the engineers? The line operates for 7.5 hours per day. Workstation 9 is the bottleneck in the initial line balance, limiting the cycle time to 2 minutes, so output is limited to 30 units/ hours * 7.5 hours = 225 units per day. 2. When it is running at maximum capacity, what is the efficiency of the line relative to its use of labor? Assume that the “supporter” is not included in efficiency calculations. For this efficiency calculation, only consider the tasks that are performed at workstations using labor, not the 310 seconds for the software load. Therefore the sum of the task times for this calculation is 583 seconds. T 583 Efficiency = = =.8097 or 80.97% NaC 6(120) 3. How should the line be redesigned to operate at the initial 250 units per day target, assuming that no overtime will be used? What is the efficiency of your new design? Without using overtime, the cycle time will have to be reduced. The maximum cycle time that will meet this production rate is: Production Time per Day (seconds) 7.5*60*60 C= = =108 seconds/unit Required Output per Day 250 All current stations are under that cycle time except for station 6 (position 9). Because of the precedence relationships for tasks 16 and 17, they must be split across two stations to meet the new cycle time. A simple way to meet this cycle time is to just put task 17 into position 10 and add an additional worker. The efficiency of this solution is: T 583 Efficiency = = =.7712 or 77.12% NaC 7(108) 4. What about running the line at 300 units per day? If overtime were used with the engineer’s initial design, how much time would need to be run each day? With the original design, output is 30 units per hour. To reach output of 300 units would require 2.5 hours of overtime per day. 5. Can the assembly line produce 300 units per day without using overtime? The cycle time to meet this production rate without overtime is: Production Time per Day (seconds) 7.5*60*60 C= = = 90 seconds/unit Required Output per Day 300 This may be possible with a redesigned line, but we might not have enough Line Positions to accommodate the new design. The following is a design that was constructed using the “longest task” priority run. This is actually a pretty good design and it would be possible to make 300 units per day. The labor efficiency of this line is 583/(8(90)) = .809 or 81%. [Since stations 7-9 are not staffed, they are not included in the efficiency calculation.] Station Task Task time Labor Idle time Station idle time 1 1 75 15 15 2 2 3 61 24 5 5 3 4 8 36 44 10 10 4 6 7 39 32 15 15 5 9 5 10 29 22 26 13 15 11 12 6 13 14 15 52 7 5 11 15 15 0 7 15 90 0 0 8 15 90 0 0 9 15 90 0 0 15 10 16 25 60 30 5 11 17 60 30 30 6. What other issues might Toshihiro consider when bringing the new assembly line up to speed? The total costs of the various options should be considered. For example, is it less expensive to work overtime to meet increased demand or to add another workstation requiring another full-time worker? Also, the quality of demand forecasts should be assessed. Redesigning the line to achieve higher output will be expensive, and should not be done unless there is strong confidence in demand forecasts. Finally, the cost of redesigning the line once it is operational and resultant downtime should be considered. A thorough analysis of demand and line design options should be performed to minimize the risk of having to redesign the line once operational in the near future. Perhaps the best option would be the balance from part 3, allowing maximum production of about 273 per day with no overtime, and the ability to reach 300 units per day with less than one hour overtime per day. Solution Manual for Operations and Supply Chain Management F. Robert Jacobs, Richard B. Chase 9780078024023, 9780077824921, 9781260238907, 9780077228934, 9781259666100
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