CHAPTER 05 STRATEGIC CAPACITY PLANNING FOR PRODUCTS AND SERVICES

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1 CHAPTER 05 STRATEGIC CAPACITY PLANNING FOR PRODUCTS AND SERVICES Solutions Actual output 7 1. a. Utilizatio n x100% 70.00% Design capacity 10 Actual output 7 Efficiency x100% 87.50% Effective capacity 8 Actual output 4 b. Utilizatio n x100% 66.67% Design capacity 6 Actual output 4 Efficiency x100% 80.00% Effective capacity 5 c. This is not necessarily true. If the design capacity is relatively high, the utilization could be low even though the efficiency is high. 2. Given: Effective capacity =.50 (Design Capacity) Actual output =.80 (Effective capacity) Actual output desired = 8 jobs per week By substitution: Actual output = (.80) x [(.50) (Design capacity)] Actual output = (.40) (Design capacity) Re-arranging terms: Actual output Design Capacity.40 By substitution (Actual output desired = 8 jobs from above): Design capacity jobs 5-1

2 1. Chapter 05 - Strategic Capacity Planning for Products and Services 3. Given: FC = $9,200/month v = $.70/unit R = $.90/unit FC $9,200 a. Q BEP 46,000 units R v $.90 $.70 b. Profit = R x Q (FC + v x Q) 1. P 61,000 = $.90(61,000) [$9,200 + $.70(61,000)] = $3, P 87,000 = $.90(87,000) [$9,200 + $.70(87,000)] = $8,200 Specified profit FC $16,000 9,200 c. Q 126,000 units R v $.90/ unit $.70/ unit Total Revenue $23,000 d. Total Revenue = R x Q, so Q = 25, units R $.90/ unit e. $100,000 TR = Q = 100,000 units TC = Q = 100,000 units TR Cost TC $50,000 $9,200 0 Volume (units) 100, Given: FC R v A: $40,000 $15/unit $10/unit B: $30,000 $15/unit $11/unit a. Q BEP FC R v Q Q BEP, A $40,000 8,000units $15/ unit $10/ unit $30,000 $15/ unit $11/ unit BEP, B 7,500 units 5-2

3 b. Profit = Q(R v) FC [A s Profit] [B s Profit] Q($15 $10) $40,000 = Q($15 $11) $30,000 $5Q - $40,000 = $4Q - $30,000 $5Q - $4Q = - $30,000 + $40,000 Q = 10,000 units c. P A = 12,000($15 $10) $40,000 = $20,000 [A is higher] P B = 12,000($15 $11) $30,000 = $18, Given: Demand = 30,000 = Q FC = $25,000 v = $.37/pen a. Given: R = $1.00/pen Q BEP FC $25,000 39, units R v $1.00 $.37 b. Given: Demand = 30,000 units. Specified profit = $15,000 Specified profit FC Q R v $40,000 Q R $.37 By substitution: $40,000 30,000 R $.37 30,000 x (R - $.37) = $40,000 30,000R - $11,100 = $40,000 30,000R = $40,000 + $11,100 30,000R = $51,100 R = $51,100/30,000 R = $1.71 [rounded up] $15,000 $25,000 R $

4 Monthly cost Chapter 05 - Strategic Capacity Planning for Products and Services 6. a. Cost for Plan A: $20 + $.45(120) + $.20(40) = $82 Cost for Plan B: $20 + $.55(120) + $.15(40) = $92 Cost for Plan C: $20 + $80 + $.40(0) = $100 b. $140 Plan B $120 $100 Plan C Plan A $80 $60 $40 $ Minutes of daytime calls (evening minutes ignored) 5-4

5 c. Considering only daytime cost functions: Plan A = $20 + $.45D Plan B = $20 + $.55D Plan C (up to 200 minutes) = $20 + $80 = $100 Plan C (over 200 minutes) = $20 + $80 + $.40 (D - 200) Collecting terms: Plan C (over 200 minutes) = $100 + $.40D - $50 = $50 + $0.40D Note: We can see that Plan A dominates Plan B over all volumes; therefore, we can omit Plan B from our analysis. Comparing Plan A to Plan C over the range minutes: Determine the indifference point between Plan A & Plan C: $20 + $.45D = $100 $.45D = $80 D = $80/$.45 D = minutes Plan A is optimal for 0 to less than daytime minutes. Plan C is optimal for more than minutes up to 200 minutes. Comparing Plan A to Plan C over the range > 200 minutes: Looking at the graph, we can see that Plan C is preferred for more than 200 minutes. Conclusion: Plan A is preferred from 0 to less than daytime minutes. Plan C is preferred for more than daytime minutes. We are indifferent between the two plans at daytime minutes. d. Given: Plan A Cost for Daytime & Evening Minutes: $20 + $.45D + $.20E Plan B Cost for Daytime & Evening Minutes: $20 + $.55D + $.15E Setting these equal and solving for D: $20 + $.45D + $.20E = $20 + $.55D + $.15E $.45D + $.20E = $.55D + $.15E $.45D - $.55D = $.15E - $.20E -$.10D = -$.05E D = $-.05E/-$.10 D =.5E D is half as much as E is. For example, if E = 100 minutes, then D = 50 minutes. This is the same as stating that D = 1/3 [(50/( )] of the total minutes. Conclusion: At 33.33% of total call minutes made in the daytime, she would be indifferent between Plan A & Plan B. 5-5

6 7. Given: Source FC v TC Process A $160,000 $5 160, Q Process B 190, , Q Vendor 7 7Q We can begin by graphing the three total cost functions as shown below. We can see the Vendor costs less until its function intersects with that of Process B. Find the indifference point between the Vendor & Process B: 7Q = 190, Q 7Q 4Q = 190,000 3Q = 190,000 Q = 190,000/3 Q = 63, units Beyond this point, we can see that Process B will be preferred due to its lower cost. The Vendor is preferred between 1 to less than 63, units. Process B is preferred for more than 63, units. Cost ($000) 500 A B Vendor Q (x1000) 5-6

7 8. Given: Source FC v Internal 1 $200,000 $17 Internal 2 240, Vendor A Vendor B Vendor C 20 up to 30,000 units 22 for 1 to 1,000; 18 each if larger amount 21 for 1 to 1,000; 19 each for additional units a. TC for 10,000 units TC for 20,000 units Int. 1: 200, (10,000) = $370, , (20,000) = $540,000 Int. 2: 240, (10,000) = $380, , (20,000) = $520,000 Vend A: 20(10,000) = $200,000 20(20,000) = $400,000 Vend B: 18(10,000) = $180,000 18(20,000) = $360,000 Vend C: 21, (9,000) = $192,000 21, (19,000) = $382,000 At 10,000 units, total cost is lowest when using Vendor B. At 20,000 units, total cost is lowest when using Vendor B. b. Given: Cost functions for each alternative: Internal 1: $200,000 + $17Q Internal 2: $240,000 + $14Q Vendor A: $20Q (Q 30,000) Vendor B: $22Q (Q 1,000) $18Q for all units when Q > 1,000 Vendor C: $21Q (Q 1,000) $21Q + $19(Q - 1,000) when Q > 1,000 First, we analyze the range of 1-1,000 units: Vendor A exhibits lower total cost over this range than do Vendor B and Vendor C; therefore, we can eliminate Vendors B & C from consideration for this range. Next, we could graph the costs functions of the remaining three options for the range of 1 1,000 units: Internal 1: $200,000 + $17Q Internal 2: $240,000 + $14Q Vendor A: $20Q As shown in the Excel chart below, Vendor A provides the lowest total cost over this entire range. If the manager is going to purchase between 1 to 1,000 units, Vendor A is preferred. 5-7

8 300, , ,000 Int. 2 Int. 1 $ 150, ,000 50,000 0 Vend A Units Second, we analyze the range of 1,001 units or more to determine the total costs if we purchase > 1,000 units: Total Cost Functions (when purchasing 1,001 units or more): Internal 1: $200,000 + $17Q Internal 2: $240,000 + $14Q Vendor A: $20Q ( 30,000 units) Vendor B: $18Q Vendor C: $21Q + $19(Q 1,000) Looking at the cost functions above, we can see that Vendor B dominates Vendor A and Vendor C. Therefore, we can eliminate Vendor A and Vendor C from further consideration. We must compare the total costs of Internal 1, Internal 2, and Vendor B when purchasing more than 1,000 units. We can plot these costs functions on a graph as shown in the Excel chart below: 5-8

9 1,600,000 1,400,000 Int. 1 1,200,000 1,000,000 $ 800, , ,000 Int. 2 Vend B 200, Units We can see in the chart above that Vendor B has the lowest total cost until its total cost function intersects the total cost function of Internal 2. Our next step is to determine the indifference point between Vendor B and Internal 2. Set the two cost functions equal and solve for Q: $18Q = $240,000 + $14Q $18Q - $14Q = $240,000 $4Q = $240,000 Q = $240,000/$40 Q = 60,000 units Therefore, Vendor B has lower total cost in the range of 1,001 units 59,999 units. Internal 2 has lower total cost > 60,000 units. Summary: Purchase Quantity: 1 1,000 units Prefer Vendor A 1,001 59,999 units Prefer Vendor B 60,000 units Indifferent between Vendor B & Internal 2 > 60,000 units Prefer Internal 2 Note: Internal 1 and Vendor C are never best. 5-9

10 9. Given: Actual output will be 225 per day per cell. 240 working days/year. Projected annual demand = 150,000 within 2 years. Annual capacity per cell = 225 units/day x 240 days/year = 54,000 Cells 150,000 : 2.78, roundup to 3cells 54, Given: Our objective is to select one type of machine to purchase. We are given the data below: Purchasing Machine Type Cost/Machine 1 $10,000 2 $14,000 Annual Demand (units) Process Time per Unit on Type 1 (min.) Process Time per Unit on Type 2 (min.) Product , , , a. Number of machines of each type needed if the machines will operate 60 minutes per hour, 8 hours per day, 250 days per year. Using Machine Type 1: Each Machine Type 1 is available 250 x 8 x 60 = 120,000 minutes per year Processing Requirements using Machine Type 1: Product 001: 12,000 x 4 min. = 48,000 min. Product 002: 10,000 x 9 min. = 90,000 min. Product 003: 18,000 x 5 min. = 90,000 min. Total = 228,000 min. Number of Machine Type 1 Needed = processing time needed / processing time capacity per unit = 228,000 / 120,000 = 1.9 = 2 machines (round up) Capacity = 2 x 120,000 minutes = 240,000 minutes Capacity cushion = 240, ,000 = 12,000 minutes Using Machine Type 2: Each Machine Type 2 is available 250 x 8 x 60 = 120,000 minutes per year Processing Requirements using Machine Type 2: Product 001: 12,000 x 6 min. = 72,000 min. Product 002: 10,000 x 9 min. = 90,000 min. Product 003: 18,000 x 3 min. = 54,000 min. Total = 216,000 min. Number of Machine Type 2 Needed = processing time needed / processing time capacity per unit = 216,000 / 120,000 = 1.8 = 2 machines (round up) Capacity = 2 x 120,000 minutes = 240,000 minutes Capacity cushion = 240, ,000 = 24,000 minutes 5-10

11 b. If we faced high uncertainty of annual demand, we would select the type of machine with the higher capacity cushion (Machine Type 2). If we faced low uncertainty of annual demand, we would select the type of machine with the lower capacity cushion (Machine Type 1). c. Given: Operating costs = $6/hour for Type 1 & $5/hour for Type 2 Purchase Cost for Machine Type 1 = 2 machines x $10,000/machine = $20,000 Total Operating Time for Machine Type 1 = 228,000 minutes = 3,800 hours Total Operating Cost = 3,800 hours x $6/hour = $22,800 Total Cost = $20,000 + $22,800 = $42,800 Purchase Cost for Machine Type 2 = 2 machines x $14,000/machine = $28,000 Total Operating Time for Machine Type 2 = 216,000 minutes = 3,600 hours Total Operating Cost = 3,600 hours x $5/hour = $18,000 Total Cost = $28,000 + $18,000 = $46,000 Conclusion: Machine Type 1 would minimize total cost. 11. a. Given: 10 hrs. or 600 min. of operating time per day per machine. 250 days x 600 min. = 150,000 min. per year operating time per machine. Machine purchase costs: A = $40,000; B = $30,000; C = $80,000. Total processing time by machine Product A B C 1 48,000 64,000 32, ,000 48,000 36, ,000 36,000 24, ,000 60,000 30,000 Total 186, , ,000 N N N A B C 186, machines 150, , machines 150, , machine 150,000 Options: Buy two A machines at a total purchase cost of 2 x $40,000 = $80,000. Buy 2 B machines at a total purchase cost of 2 x $30,000 = $60,000. Buy 1 C machine at a total purchase cost of $80,000. Conclusion: We should buy 2 of the B machines at a total cost of $60,

12 b. Given: Operating Costs: A = $10/hour/machine; B = $11/hour/machine; C = $12/hour/machine. Total cost for each type of machine: A (2): 186,000 min. / 60 min./hour = 3, hrs. x $10 = $31,000 + $80,000 = $111,000 B (2): 208,000 min. / 60 min./hour = 3, hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 min. / 60 min./hour = 2, hrs. x $12 = $24,400 + $80,000 = $104,400 Conclusion: Buy 2 Bs these have the lowest total cost. 12. Given: R = $45 per customer, v = $20 per customer, each machine can process 100 customers per day, fixed cost for one machine = $2,000 per day total; and fixed cost for two machines = $3,800 per day total. a. FC Range FC Q BEP R v One machine $2,000 1 to / (45 20) = 80 Two machines 3, to / (45 20) = 152 b. Given: Estimated demand = 90 to 120 customers per day Because BEP for one machine is 80 and 80 < 90, and because BEP for two machines is 152 and 152 > 120, we should purchase one machine, because even at the upper limit of demand (120 customers), we have not reached the break-even point associated with two machines. Conclusion: Purchase one machine. 5-12

13 13. Given: R = $5.95/car, v = $3/car, Fixed Cost for one line = $6,000/month, Fixed Cost for two lines = $10,500/month, each line can process 15 cars/hour, & the car wash is open 300 hours/month. Determine the break-even for each option: To do this, we will need to convert fixed costs per month to fixed costs per hour. One line fixed cost per hour = $6,000/300 = $20 Two line fixed cost per hour = $10,500/300 = $35 One Line: Q BEP FC $ cars R v $5.95 $3.00 Two Lines: Q BEP FC $ cars R v $5.95 $3.00 If demand averages between 14 and 18 cars an hour, either option would break even. Therefore, we need to determine the net profit per hour for each possible demand value as shown below. Volume No. of Lines Used Net Profit per Hour 14 1 $21.30 = 14 (5.95 3) = 15 (5.95 3) = 15 (5.95 3) = 15 (5.95 3) = 15 (5.95 3) 20 Volume No. of Lines Used Net Profit per Hour 14 2 $6.30 = 14 (5.95 3) = 15 (5.95 3) = 16 (5.95 3) = 17 (5.95 3) = 18 (5.95 3) 35 Conclusion: Choose one line. Net profit per hour always is higher using one line for the given demand range of 14 to 18 cars per hour. 5-13

14 14. Given : We have a 4-step process with the following effective capacity for each operation: Operation 1 = 12/hr, Operation 2 = 15/hr, Operation 3 = 11/hr, Operation 4 = 14/hr. a. The capacity of the process is determined by the operation with the lowest effective capacity: Operation 3 = 11/hr. b. Given: Option 1: Increase the capacity of Operation 1 by 15%. Option 2: Increase the capacity of Operation 2 by 10%. Option 3: Increase the capacity of Operation 3 by 10%. Option 1 Process Capacity: Operation 1 New Capacity = 12 x 1.15 = 13.8/hr. Operation 1 = 13.8/hr, Operation 2 = 15/hr, Operation 3 = 11/hr, Operation 4 = 14/hr. The capacity of the process is determined by the operation with the lowest effective capacity: Operation 3 = 11/hr. Increase in capacity = 0/hr. Option 2 Process Capacity: Operation 2 New Capacity = 15 x 1.10 = 16.5/hr. Operation 1 = 12/hr, Operation 2 = 16.5/hr, Operation 3 = 11/hr, Operation 4 = 14/hr. The capacity of the process is determined by the operation with the lowest effective capacity: Operation 3 = 11/hr. Increase in capacity = 0/hr. Option 3 Process Capacity: Operation 3 New Capacity = 11 x 1.10 = 12.1/hr. Operation 1 = 12/hr, Operation 2 = 15/hr, Operation 3 = 12.1/hr, Operation 4 = 14/hr. The capacity of the process is determined by the operation with the lowest effective capacity: Operation 1 = 12/hr. Increase in capacity = 1/hr (12/hr 11/hr). Conclusion: Select Option 3. Increasing the capacity of Operation 3 by 10% yields the greatest increase in process capacity (1/hr). 15. Given: Two parallel lines feed their combined output to Operation 7. Upper Line Capacities: Operation 1 = 18/hr, Operation 2 = 15/hr, Operation 3 = 16/hr. Lower Line Capacities: Operation 4 = 17/hr, Operation 5 = 15/hr, Operation 6 = 17/hr. Capacity of Operation 7 = 20/hr. Capacity of Operation 8 = 24/hr. a. Capacity of the system: Capacity of Upper Line = 15/hr. Capacity of Lower Line = 15/hr. Combined Capacity = 30/hr. The two lines feed 30/hr to Operation 7. Operation 7 has capacity of 20/hr and Operation 8 has capacity of 24/hr. Operation 7 has the lowest capacity it can handle only 20/hr. Conclusion: Capacity of the system = 20/hr. 5-14

15 b. Given: The capacity of one operation could be increased. Process capacity is limited by Operation 7; therefore, increase the capacity of Operation 7 by 4 units/hour from 20 units/hour to 24 units/hour at which time Operation 8 also becomes a bottleneck. 13. Given: Three parallel lines feed their combined output to Operation 10. Upper Line Capacities: Operation 1 = 22/hr, Operation 2 = 17/hr, Operation 3 = 18/hr. Middle Line Capacities: Operation 4 = 20/hr, Operation 5 = 18/hr, Operation 6 = 18/hr. Lower Line Capacities: Operation 7 = 22/hr, Operation 8 = 17/hr, Operation 9 = 15/hr. Capacity of Operation 10 = 51/hr. Capacity of Operation 11 = 54/hr. Capacity of Upper Line = 17/hr. Capacity of Middle Line = 18/hr. Capacity of Lower Line = 15/hr. Combined Capacity = 50/hr. The three lines feed 50/hr to Operation 10. Operation 10 has capacity of 51/hr and Operation 11 has capacity of 54/hr. The lowest capacity is the combined capacity of the three lines = 50/hr. Conclusion: Capacity of the system = 50/hr. 17. Given: Two parallel lines feed their combined output to Operation 7. Upper Line Capacities: Operation 1 = 15/hr, Operation 2 = 10/hr, Operation 3 = 20/hr. Lower Line Capacities: Operation 4 = 5/hr, Operation 5 = 8/hr, Operation 6 = 12/hr. Capacity of Operation 7 = 34/hr. Capacity of Operation 8 = 30/hr. a. Capacity of Upper Line = 10/hr. Capacity of Lower Line = 5/hr. Combined Capacity = 15/hr. The two lines feed 15/hr to Operation 7. Operation 7 has capacity of 34/hr and Operation 8 has capacity of 30/hr. The combined capacity of the two lines is lowest at 15/hr. Conclusion: Capacity of the system = 15/hr. b. Increasing Capacity of One Operation: Process capacity is limited by the combined capacity of the two lines (15/hr). The Upper Line Capacity = 10/hr and is determined by Operation 2. If the capacity of Operation 2 were increased by 5/hr to 15/hr (the same as Operation 1), then the Upper Line Capacity would increase by 5/hr, the Combined Capacity would increase by 5/hr, and process capacity would increase by 5/hr to 20/hr. The Lower Line Capacity = 5/hr and is determined by Operation 4. If the capacity of Operation 4 were increased by 3/hr to 8/hr (the same as Operation 5), then the Lower Line Capacity would increase by 3/hr, the Combined Capacity would increase by 3/hr, and process capacity would increase by 3/hr to 18/hr. Conclusion: Increase the capacity of Operation 2 by 5/hr. The resulting process capacity would be 20/hr. 5-15

16 18. Given: New equipment initial cost = $12,000. Annual savings = $1, Given: New equipment initial cost = $18,000. Annual savings = $2, Given: Initial cost of remodeling = $25,000. Annual savings = $3,000 in Year 1, $4,000 in Year 2, & $5,000 thereafter. Savings in 2 years = $3,000 + $4,000 = $7,000. That leaves us with $18,000 to recoup at $5,000 per year. Conclusion: Total time to recoup initial cost = = 5.6 years. Case: Outsourcing of Hospital Services 1. The advantages of having the outsourced work performed within the hospital are that the hospital s workers felt a connection with the hospital, still felt a sense of ownership in their jobs, and did not need to be trained to perform the food service work. Perhaps in a larger hospital with a larger staff, this connection between workers and the hospital might not be an issue. In addition, there might be large cost savings involved. 2. There could be a cost savings in having an outside firm manage the service, or the motivation for outsourcing could be avoidance of the burden of managing housekeeping. 3. The reason for asking another hospital to join it would be economies of scale and lower costs for laundry service. 5-16

17 Enrichment Module: Solving Capacity Planning Problems Capacity-planning problems can be classified in a number of different ways. One such classification for intermediate and short-range problems is given below: 1. Output capacity determination 2. Input capacity determination 3. Capacity-demand match (input or output) The categories listed above can involve either manufacturing or service problems. The solutions to the following realistic examples will provide an easy and an intuitive way to comprehend and solve capacityplanning problems. Problem 1 Manufacturing Example (Output capacity determination and capacity-demand match) A battery manufacturing plant normally operates two eight-hour shifts per day and 6 days per week. The manufacturer can produce 375 units per hour. Over the next four weeks, the aggregate demand for the batteries is given in the following table. Week Demand 30,000 32,000 38,000 40,000 a. Calculate the weekly capacity of the plant. b. If the firm attempts to produce the demanded quantity, at what percentage of the capacity would it be operating each week? c. Determine the Level production schedule and the resulting average inventory for the 4-week period. Assume that no shortages are allowed and the current inventory is zero and desired ending inventory in week 4 also is zero. d. Determine the Chase production schedule and the resulting average inventory for the 4-week period. Assume that no shortages are allowed and the current and desired ending inventory in week 4 is zero. e. Based on your answers to part c and d, discuss the trade-off between Level and the Chase production plans. Note: Part a of this problem can be classified as output capacity determination while parts b through e deal with capacity-demand match. 5-17

18 Problem 2 Service Example (Output capacity determination) A small grocery store has a total of four regular checkout lines and one express checkout line. Recently on Sundays the store has been experiencing either excessive idle time for cashiers or excessively long customer waiting lines. The results of a recent time study performed by a management consulting company showed that the average service time for express and regular checkout lines are 3 and 10 minutes respectively. As the next step in analyzing the problem, the manager of the grocery store wants to determine the estimated capacity of the store on Sundays in terms of total number of customers. Currently the store is open from 6 a.m. to midnight on Sundays. The express checkout line is always open while there is only one regular line open from 6 a.m. to 9 a.m. and also only one regular line open from 9 p.m. to midnight. There are two regular checkout lines open from 9 a.m. to noon and also from 6 p.m. to 9 p.m. All four regular lines are open between noon and 6 p.m. a. Determine the current capacity of the store in total number of customers for Sundays. b. Assume that the store manager decides to reduce the number of regular lines from 2 to 1 between 7 p.m. and 9 p.m. and closes the express line between 6 a.m. and 8 a.m. and 10 p.m. and midnight. What is the revised capacity for Sundays? Problem 3 Manufacturing Example (Input capacity determination number of resources needed) A video equipment manufacturer produces videotapes and DVDs. The manufacturing facility operates two eight-hour shifts per day for 6 days a week. The unit manufacturing time is 6 minutes for each videotape and 8 minutes for each DVD. a. Given that machine operators work at 80% efficiency, determine the number of workers needed to produce 5000 videotapes and 2500 DVDs per week. b. Given that machines have 95% efficiency, determine the number of machines needed to produce 5000 videotapes and 2500 DVDs per week. c. Assume that the number of workers is sufficient, what is the maximum number of videotapes and the maximum number of DVDs that can be manufactured with 15 machines. d. Assume that the number of machines is sufficient, what is the maximum number of videotapes and the maximum number of DVDs that can be manufactured with 20 workers. 5-18

19 Problem 4 Manufacturing Example with Multiple Products and Multiple Machines (Input capacity determination number of resources needed) Among many other products, a firm manufactures three different electronic components (A, B, C) on any of three different machines (1, 2, 3). The quarterly forecasted demand for the three components is given in Table 1. Table 1 Quarterly Forecasted Demand by Product Type Season Component Winter Spring Summer Fall A 8,000 20,000 12,000 6,400 B 4,000 12,000 8,000 5,600 C 9,600 19,200 14,400 7,200 Table 2 displays the unit production time for each product on each machine. Table 2 Unit Production Time (in hours) Component Machine A B C Interpreting Table 2, we can state that each unit of product A takes 15 minutes (.25 x 60 min.) to process on machine 1, while it takes 12 minutes (.20 x 60 minutes) to process one unit of product B on machine 3. a. Determine the maximum number of machine hours demanded for each quarter machine combination. b. The production manager has determined that the amount of productive time available for each machine per quarter is 600 hours. Determine the maximum number of each machine type needed to be dedicated to produce all components in each quarter. c. Does there appear to be seasonal variation in demand? Explain. 5-19

20 Solution to Problem 1 Manufacturing Example (Output capacity determination and capacity-demand match) a. The number of units/week = (375 units/hr.) (8 hrs/shift) (2 shifts/day) (6 days/week) The number of units/week = 36,000 batteries b. Week Forecasted demand 30,000 32,000 38,000 40,000 Capacity 36,000 36,000 36,000 36,000 % of capacity utilized 83.33% 88.89% % % c. In determining the Level production plan, if the demand is less than or equal to the production capacity, we simply determine the average demand for the four-week period and use the average demand as our production quantity. However, if the average demand is above capacity, then we can try either to expand capacity, to delay the order, or to reduce the quantity. Because in this instance the average demand is less than capacity in each week, we can use the average demand as our production quantity. Average demand 30,000 32,000 38,000 40,000 35,000 4 The Level production plan and the resulting ending inventory for each week are given in the following table. Week Forecasted demand 30,000 32,000 38,000 40,000 Capacity 36,000 36,000 36,000 36,000 Production 35,000 35,000 35,000 35,000 Ending Inventory 0 5,000 8,000 5,000 0 Average inventory = 18,000 / 4 = 4,500 units. d. In determining the Chase production plan, we attempt to match production with demand unless there is insufficient capacity. The amount of shortage from the latest period with insufficient capacity is scheduled for production in the latest period with excess supply. In our problem, week 4 has a potential shortage of 4,000 units and week 2 is the latest period with excess capacity of 4,000 units. Therefore, week 4 s shortage is scheduled for production in week 2. Likewise, week 3 has a potential shortage of 2,000 units, which is scheduled for production in week

21 The Chase production plan and the resulting ending inventory for each week are given in the following table. Week Forecasted demand 30,000 32,000 38,000 40,000 Capacity 36,000 36,000 36,000 36,000 Production 32,000 36,000 36,000 36,000 Ending Inventory 0 2,000 6,000 4,000 0 Average inventory = 12,000/4 = 3,000 units. e. The Chase production plan results in fewer units in inventory, while the Level production plan results in more uniform production, thus less hiring and layoff costs. Solution to Problem 2 Service Example (Output capacity determination) a. Hourly capacity of the express line = (60 minutes) / (3 minutes per cust.) = 20 customers Hourly capacity of the regular line = (60 minutes) / (10 minutes per cust.) = 6 customers Capacity of the express line for Sundays = (20 customers) x (18 hours) = 360 customers Capacity of the regular line: From 6 a.m. to 9 a.m. = (6 customers/hr) (3 hours) (1 line) = 18 customers From 9 a.m. to noon = (6 customers/hr) (3 hours) (2 lines) = 36 customers From noon to 6.p.m. = (6 customers/hr) (6 hours) (4 lines) = 144 customers From 6 p.m. to 9 p.m. = (6 customers/hr) (3 hours) (2 lines) = 36 customers From 9 p.m. to midnight = (6 customers/hr) (3 hours) (1 line) = 18 customers Sunday total regular line capacity = = 252 customers Overall Sunday capacity = Total regular line capacity + Express line capacity Therefore overall capacity for Sundays = = 612 customers b. Reduction in express line capacity = (4 hours) (20 customers / hour) = 80 customers Reduction in regular line capacity = (2 hours) (6 customers per hour) = 12 customers Revised Sunday capacity = 612 ( ) = 520 customers 5-21

22 Solution to Problem 3 Manufacturing Example (Input capacity determination number of resources needed) In general, we can express the equation for number of resources using the following notation: N R k pidi i1 ( T )( E) where: N R = Number of resources (machines or workers) required k = number of products produced T = Total time available per resource per scheduled time period i p i = Unit production time for product i D i = Demand for product i for the scheduled time period E = Efficiency of the resource measured as a percentage Therefore, if we know the number of workers and want to determine the maximum demand that can be satisfied for a given product, we can manipulate the formula given above and obtain the following equation: ( N D i R )( T)( E) p i Given the above information, we can now solve Problem 3. T (2 shifts )(6 days)(8 hrs./ shift )(60 min./ hr.) 5,760 min./ week k a. pidi i1 (6 min.)(5,000) (8 min.)(2,500) NW workers ( T)( E) (5,760)(.80) k pidi i1 (6 min. )(5,000) (8 min. )(2,500) b. NM machines ( T)( E) (5,760)(.95) c. d. D D D D videotape DVD videotape DVD ( NM )( T)( E) p videotape ( NM )( T)( E) p DVD ( NM )( T )( E) p videotape ( NM )( T)( E) p DVD (15)(5,760)(.95) 6 (15)(5,760)(.95) 8 (20)(5,760)(.80) 6 (20)(5,760)(.80) 8 13,680videotapes 10,260 DVDs 15,360videotapes 11,520 DVDs 5-22

23 Solution to Problem 4 Manufacturing Example with Multiple Products and Multiple Machines (Input capacity determination number of resources needed) a. First, we need to convert the demand to machine hours for each machine in each season. The demand in the winter is 100, 50, and 120 for components A, B and C respectively and it takes.25 hours,.5 hours, and.4 hours to process components A, B and C respectively on machine 1. Therefore with this information, we can compute the maximum machine hours demanded for machine 1 (M1) in the winter quarter. Max. hrs. for M1 in Winter = (.25)(8000)+(.5)(4,000)+(.4)(9600) = 7,840 hrs. Similarly the quarterly machine hours demanded can be calculated for the rest of the machineseason combinations: Max. hrs. for M1 in Spring = (.25)(20,000)+(.5)(12,000)+(.4)(19,200) =18,680 hrs. Max. hrs. for M1 in Summer = (.25)(12,000)+(.5)(8,000)+(.4)(14,400) = 12,760 hrs. Max. hrs. for M1 in Fall = (.25)(6,400)+(.5)(5,600)+(.4)(7,200) = 7,280 hrs. Max. hrs. for M2 in Winter = (.10)(8,000)+(.30)(4,000)+(.15)(9,600) = 3,440 hrs. Max. hrs. for M2 in Spring = (.10)(20,000)+(.30)(12,000)+(.15)(19,200) = 8,480 hrs. Max. hrs. for M2 in Summer = (.10)(12,000)+(.30)(8,000)+(.15)(14,400) = 5,760 hrs. Max. hrs. for M2 in Fall = (.10)(6,400)+(.30)(5,600)+(.15)(7,200) = 3,400 hrs. Max. hrs. for M3 in Winter = (.45)(8,000)+(.2)(4,000)+(.35)(9,600) = 7,760 hrs. Max. hrs. for M3 in Spring = (.45)(20,000)+(.2)(12,000)+(.35)(19,200) = 18,120 hrs. Max. hrs. for M3 in Summer = (.45)(12,000)+(.2)(8,000)+(.35)(14,400) = 12,040 hrs. Max. hrs. for M3 in Fall = (.45)(6,400)+(.2)(5,600)+(.35)(7,200) = 6,520 hrs. 5-23

24 b. Because (T) (E) = 600 productive hours per quarter, 7,840 hrs. demanded N M 1(Winter) ~14 machine 1s 600 hrs. Therefore we can conclude that at most we need to allocate 14 machine 1s to produce components A, B and C in the winter quarter. Similarly the maximum number of machine 3s needed in the spring quarter to make all three components can be determined as follows: 18,120 hrs. demanded NM 3(Spring) ~ 31machine 3s 600 hrs. The following table summarizes the maximum number of each machine type needed by quarter. Quarterly Maximum number of machine types needed Season Machine Winter Spring Summer Fall 1 14* *All values in the table are rounded up. c. Yes, there appears to be a significant seasonal variation in demand. It appears that the highest demand is experienced in the spring followed by summer. Therefore, most likely the components are used in summer products and because of lead times, the demand peaks in the spring. 5-24