1. 7 – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Constraint Management Chapter 7

2. 7 – 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Managing Constraints Constraints are factors that limit performance Capacity is the maximum rate of output Three types of constraints A bottleneck is any resource whose capacity limits the organization’s ability to meet volume, mix, or fluctuating demand requirements

3. 7 – 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Theory of Constraints TOC is a systematic management approach that focuses on actively managing those constraints that impede a firm’s progress toward its goal of maximizing profits and effectively using its resources It outlines a deliberate process for identifying and overcoming constraints TOC methods increase the firm’s profits by focusing on materials flow through the entire system

4. 7 – 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Theory of Constraints TABLE 7.1| HOW THE FIRM’S OPERATIONAL MEASURES RELATE TO ITS | FINANCIAL MEASURES Operational Measures TOC ViewRelationship to Financial Measures Inventory (I)All the money invested in a system in purchasing things that it intends to sell A decrease in I leads to an increase in net profit, ROI, and cash flow. Throughput (T)Rate at which a system generates money through sales An increase in T leads to an increase in net profit, ROI, and cash flows. Operating Expense (OE) All the money a system spends to turn inventory into throughput A decrease in OE leads to an increase in net profit, ROI, and cash flows. Utilization (U)The degree to which equipment, space, or workforce is currently being used, and is measured as the ratio of average output rate to maximum capacity, expressed as a percentage An increase in U at the bottleneck leads to an increase in net profit, ROI, and cash flows.

5. 7 – 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Theory of Constraints TOC involves the implementation of these five steps 1.Identify the System Bottleneck(s) 2.Exploit the Bottleneck(s) 3.Subordinate All Other Decisions to Step 2 4.Elevate the Bottleneck(s) 5.Do Not Let Inertia Set In

6. 7 – 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Theory of Constraints Bottlenecks can both be internal or external to the firm and are typically a process or step with the lowest capacity Throughput time is the total elapsed time from the start to the finish of a job or a customer being processed at one or more workcenters A bottleneck can be identified in several different ways 1.If it has the highest total time per unit processed 2.If it has the highest average utilization and total workload 3.If a reduction of processing time would reduce the average throughput time for the entire process

7. 7 – 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck EXAMPLE 7.1 Managers at the First Community Bank are attempting to shorten the time it takes customers with approved loan applications to get their paperwork processed. The flowchart for this process, consisting of several different activities, each performed by a different bank employee, is shown in Figure 7.1. Approved loan applications first arrive at activity or step 1, where they are checked for completeness and put in order. At step 2, the loans are categorized into different classes according to the loan amount and whether they are being requested for personal or commercial reasons. While credit checking commences at step 3, loan application data are entered in parallel into the information system for record- keeping purposes at step 4. Finally, all paperwork for setting up the new loan is finished at step 5. The time taken in minutes is given in parentheses.

8. 7 – 8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Complete paperwork for new loan (10 min) Identifying the Bottleneck Check for credit rating (15 min) Enter loan application into the system (12 min) Categorize loans (20 min) Check loan documents and put them order (15 min) Figure 7.1 – Processing Credit Loan Applications at First Community Bank Which single step is the bottleneck? The management is also interested in knowing the maximum number of approved loans this system can process in a 5-hour work day.

9. 7 – 9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck SOLUTION We define the bottleneck as step 2, where a single-minute reduction in its time reduces the average throughput time of the entire loan approval process. The throughput time to complete an approved loan application is 15 + 20 + max(15, 12) + 10 = 60 minutes. Although we assume no waiting time in front of any step, in practice such a smooth process flow is not always the case. So the actual time taken for completing an approved loan will be longer than 60 minutes due to nonuniform arrival of applications, variations in actual processing times, and the related factors. The capacity for loan completions is derived by translating the “minutes per customer” at the bottleneck step to “customer per hour.” At First Community Bank, it is 3 customers per hour because the bottleneck step 2 can process only 1 customer every 20 minutes (60/3).

10. 7 – 10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck Services may not have simple line flows and demand may vary considerably Bottlenecks can be identified by using average utilization Variability creates floating bottlenecks Variability increases complexity

11. 7 – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.1 Two types of customers enter Barbara’s Boutique shop for customized dress alterations. After T1, Type A customers proceed to step T2 and then to any of the three workstations at T3, followed by steps T4 and T7. After step T1,Type B customers proceed to step T5 and then steps T6 and T7. The numbers in the parentheses are the minutes it takes that activity to process a customer. a.What is the capacity per hour of Type A customers? b.If 30 percent of the customers are Type A customers and 70 percent are Type B customers, what is the average capacity? c.When would Type A customers experience waiting lines, assuming there are no Type B customers in the shop? Where would Type B customers have to wait, assuming no Type A customers?

12. 7 – 12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.1 a.For Type A customers, step T2 can process (60/13) = 4.62 customers per hour. T3 has three work stations and a capacity of (60/14) + (60/10) + (60/11) = 15.74 customer per hour. Step T4 can process (60/18) = 3.33 customers per hour. The bottleneck for type A customers is the step with the highest processing time per customer, T4. T1 (12) T7 (10) T4 (18) T3-a (14) T3-c (11) T3-b (10) Type A or B? Type A Type B T2 (13) T6 (22) T5 (15)

13. 7 – 13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.1 b.The bottleneck for Type B customers is T6 since it has the longest processing time per customer. The capacity for Type B customers is (60/22) = 2.73 customers per hour. Thus the average capacity is 0.3(3.33) + 0.7(2.73) = 2.9 customers per hour T1 (12) T7 (10) T4 (18) T3-a (14) T3-c (11) T3-b (10) Type A or B? Type A Type B T2 (13) T6 (22) T5 (15)

14. 7 – 14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. before T2 and T4 because the activities immediately preceding them have a higher rate of output. c.Type A customers would wait Application 7.1 before steps T5 and T6 for the same reason. This assumes there are always new customers entering the shop. Type B customers would wait T1 (12) T7 (10) T4 (18) T3-a (14) T3-c (11) T3-b (10) Type A or B? Type A Type B T2 (13) T6 (22) T5 (15)

15. 7 – 15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck EXAMPLE 7.2 Diablo Electronics manufactures four unique products (A, B, C, and D) that are fabricated and assembled in five different workstations (V, W, X, Y, and Z) using a small batch process. Each workstation is staffed by a worker who is dedicated to work a single shift per day at an assigned workstation. Batch setup times have been reduced to such an extent that they can be considered negligible. Figure 7.2 is a flowchart of the manufacturing process. Diablo can make and sell up to the limit of its demand per week, and no penalties are incurred for not being able to meet all the demand. Which of the five workstations (V, W, X, Y, or Z) has the highest utilization, and thus serves as the bottleneck for Diablo Electronics?

16. 7 – 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck Product A $5 Raw materials Purchased parts Product:A Price:$75/unit Demand:60 units/wk Step 1 at workstation V (30 min) Finish with step 3 at workstation X (10 min) Step 2 at workstation Y (10 min) $5 Product C Raw materials Purchased parts Product:C Price:$45/unit Demand:80 units/wk Finish with step 4 at workstation Y (5 min) Step 2 at workstation Z (5 min) Step 3 at workstation X (5 min) Step 1 at workstation W (5 min) $2 $3 Product B Raw materials Purchased parts Product:B Price:$72/unit Demand:80 units/wk Finish with step 2 at workstation X (20 min) Step 1 at workstation Y (10 min) $3 $2 Product D Raw materials Purchased parts Product:D Price:$38/unit Demand:100 units/wk $4 Step 2 at workstation Z (10 min) Finish with step 3 at workstation Y (5 min) Step 1 at workstation W (15 min) $6 Figure 7.2Flowchart for Products A, B, C, and D

17. 7 – 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck SOLUTION Because the denominator in the utilization ratio is the same for every workstation, with one worker per machine at each step in the process, we can simply identify the bottleneck by computing aggregate workloads at each workstation. The firm wants to satisfy as much of the product demand in a week as it can. Each week consists of 2,400 minutes of available production time. Multiplying the processing time at each station for a given product with the number of units demanded per week yields the workload represented by that product. These loads are summed across all products going through a workstation to arrive at the total load for the workstation, which is then compared with the others and the existing capacity of 2,400 minutes.

18. 7 – 18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Workstation Load from Product A Load from Product B Load from Product C Load from Product D Total Load (min) V W X Y Z Identifying the Bottleneck

19. 7 – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Identifying the Bottleneck These calculations show that workstation X is the bottleneck, because the aggregate work load at X exceeds the available capacity of 2,400 minutes per week. 60 x 30 = 1800 0 60 x 10 = 600 0 0 0 80 x 20 = 1,600 80 x 10 = 800 0 0 80 x 5 = 400 0 100 x 15 = 1,500 0 100 x 5 = 500 100 x 10 = 1,000 1,800 1,900 2,600 2,30 1,400 Workstation Load from Product A Load from Product B Load from Product C Load from Product D Total Load (min) V W X Y Z

20. 7 – 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.2 O’Neill Enterprises manufactures three unique products (A, B, C) that are fabricated and assembled in four different workstations (W, X, Y, Z) using a small batch process. Each of the products visits every one of the four workstations, though not necessarily in the same order. Batch setup times are negligible. A flowchart of the manufacturing process is shown below. O’Neill can make and sell up to the limit of its demand per week, and there are no penalties for not being able to meet all the demand. Each workstation is staffed by a worker dedicated to work on that workstation alone, and is paid $12 per hour. Variable overhead costs are $8000/week. The plant operates one 8-hour shift per day, or 40 hours/week. Which of the four workstations W, X, Y, or Z has the highest total workload, and thus serves as the bottleneck for O’Neill Enterprises?

21. 7 – 21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.2 Product B Raw materials Purchased part Product:B Price:$85/unit Demand:70 units/wk Finish with step 4 at workstation Z (13 min) Step 2 at workstation W (10 min) Step 3 at workstation Y (10 min) Step 1 at workstation X (12 min) $9 $5 Product A Raw materials Purchased part Product:A Price:$90/unit Demand:65 units/wk Finish with step 4 at workstation Z (16 min) Step 2 at workstation Y (15 min) Step 3 at workstation X (9 min) Step 1 at workstation W (10 min) $7 $6 Product C Raw materials Purchased part Product:C Price:$80/unit Demand:80 units/wk Finish with step 4 at workstation Z (10 min) Step 2 at workstation X (10 min) Step 3 at workstation W (12 min) Step 1 at workstation Y (5 min) $10 $5 Flowchart for Products A, B, and C

22. 7 – 22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.2 SOLUTION Identify the bottleneck by computing total workload at each workstation. The firm wants to satisfy as much of the product demand in a week as it can. Each week consists of 2400 minutes of available production time. Multiplying the processing time at each station for a given product with the number of units demanded per week yields the capacity load. These loads are summed across all products going through that workstation and then compared with the existing capacity of 2400 minutes.

23. 7 – 23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Work Station Load from Product A Load from Product B Load from Product C Total Load (minutes) W X Y Z Application 7.2

24. 7 – 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.2 These calculations show that workstation Z is the bottleneck, because the aggregate work load at Z exceeds the available capacity of 2400 minutes per week. (65x10)= 650 (70  10)= 700(80  12)= 960 2310 (65  9)= 585(70  12)= 840(80  10)= 800 2225 (65  15)= 975 (70x10)= 700(80x5)= 4002075 (65  16)= 1040(70  13)= 910(80  10)= 800 2750 Work Station Load from Product A Load from Product B Load from Product C Total Load (minutes) W X Y Z

25. 7 – 25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix EXAMPLE 7.3 The senior management at Diablo Electronics (see Exercise 7.2) wants to improve profitability by accepting the right set of orders, and so collected some additional financial data. Variable overhead costs are $8,500 per week. Each worker is paid $18 per hour and is paid for an entire week, regardless of how much the worker is used. Consequently, labor costs are fixed expenses. The plant operates one 8-hour shift per day, or 40 hours each week. Currently, decisions are made using the traditional method, which is to accept as much of the highest contribution margin product as possible (up to the limit of its demand), followed by the next highest contribution margin product, and so on until no more capacity is available.

26. 7 – 26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Pedro Rodriguez, the newly hired production supervisor, is knowledgeable about the theory of constraints and bottleneck- based scheduling. He believes that profitability can indeed be improved if bottleneck resources were exploited to determine the product mix. What is the change in profits if, instead of the traditional method used by Diablo Electronics, the bottleneck method advocated by Pedro is used to select the product mix? SOLUTION Decision Rule 1: Traditional Method Select the best product mix according to the highest overall contribution margin of each product.

27. 7 – 27 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 1:Calculate the contribution margin per unit of each product as shown here. ABCD Price Raw material and purchased parts = Contribution margin

28. 7 – 28 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 1:Calculate the contribution margin per unit of each product as shown here. When ordered from highest to lowest, the contribution margin per unit sequence of these products is B, A, C, D. $75.00$72.00$45.00$38.00 –10.00–5.00 –10.00 $65.00$67.00$40.00$28.00 ABCD Price Raw material and purchased parts = Contribution margin

29. 7 – 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 2:Allocate resources V, W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation X) is encountered. Subtract minutes away from 2,400 minutes available for each week at each stage. V W X Y Z

30. 7 – 30 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 2:Allocate resources V, W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation X) is encountered. Subtract minutes away from 2,400 minutes available for each week at each stage. The best product mix according to this traditional approach is then 60 A, 80 B, 40 C, and 100 D. Minutes at the start 2,400 Minutes left after making 80B 2,400 2400 – 1600 = 800 2400 – 800 = 1600 2,400 Minutes left after making 60A 2400 – 1800 = 600 2400 800 – 600 = 200 1600 – 600 = 1000 2400 Can only make 40C 600 2400 – 200 = 2200 200 – 200 = 0 1000 – 200 = 800 2400 – 200 = 2200 Can still make 100 D 600 2200 – 1500 = 700 0 800 – 500 = 300 2200 – 1000 = 1200 V W X Y Z

31. 7 – 31 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 3:Compute profitability for the selected product mix. Profits Revenue Materials Labor Overhead Profit

32. 7 – 32 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 3:Compute profitability for the selected product mix. Manufacturing the product mix of 60 A, 80 B, 40 C, and 100 D will yield a profit of $1,560 per week. (60  $75) + (80  $72) + (40  $45) + (100  $38) =$15,860 Profits Revenue Materials Labor Overhead Profit (60  $10) + (80  $5) + (40  $5) + (100  $10) =–$2,200 (5 workers)  (8 hours/day)  (5 days/week)  ($18/hour) =–$3,600 =–$8,500 =$1,560

33. 7 – 33 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Decision Rule 2: Bottleneck Method Select the best product mix according to the dollar contribution margin per minute of processing time at the bottleneck workstation X. This method would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck. Step 1:Calculate the contribution margin/minute of processing time at bottleneck workstation X:

34. 7 – 34 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Product AProduct BProduct CProduct D Contribution margin Time at bottleneck Contribution margin per minute

35. 7 – 35 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix When ordered from highest to lowest contribution margin/ minute at the bottleneck, the manufacturing sequence of these products is D, C, A, B, which is reverse of the earlier order. Product D is scheduled first because it does not consume any resources at the bottleneck. $65.00$67.00$40.00$28.00 10 minutes20 minutes5 minutes0 minutes $6.50$3.35$8.00Not defined Product AProduct BProduct CProduct D Contribution margin Time at bottleneck Contribution margin per minute

36. 7 – 36 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 2:Allocate resources V, W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation X) is encountered. Subtract minutes away from 2,400 minutes available for each week at each stage. V W X Y Z

37. 7 – 37 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 2:Allocate resources V, W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation X) is encountered. Subtract minutes away from 2,400 minutes available for each week at each stage. The best product mix according to this bottleneck based approach is then 60 A, 70 B, 80 C, and 100 D. Minutes at the start 2,400 Minutes left after making 100D 2,400 2400 – 1500 = 900 2400 2400 – 500 = 1900 2400 1000 = 1400 Minutes left after making 80C 2400 900 – 400 = 500 2400 – 400 = 2000 1900 – 400 = 1500 1400 – 400 = 1000 Minutes left after making 60A 2400 – 1800 = 600 500 2000 – 600 = 1400 1500 – 600 = 900 1000 Can only make 70 B 600 500 1400 – 1400 = 0 900 – 700 = 200 1000 V W X Y Z

38. 7 – 38 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 3:Compute profitability for the selected product mix. Profits Revenue Materials Labor Overhead Profit

39. 7 – 39 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining the Product Mix Step 3:Compute profitability for the selected product mix. Manufacturing the product mix of 60 A, 70 B, 80 C, and 100 D will yield a profit of $2,490 per week. (60  $75) + (70  $72) + (80  $45) + (100  $38) =$16,940 (60  $10) + (70  $5) + (80  $5) + (100  $10) =–$2,350 (5 workers)  (8 hours/day)  (5 days/week)  ($18/hour) =–$3,600 Profits Revenue Materials Labor Overhead Profit =–$8,500 =$2,490

40. 7 – 40 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 The senior management at O’Neill Enterprises wants to improve the profitability of the firm by accepting the right set of orders. Currently, decisions are made to accept as much of the highest contribution margin product as possible (up to the limit of its demand), followed by the next highest contribution margin product, and so on until no more capacity is available. Since the firm cannot satisfy all the demand, the product mix must be chosen carefully. Jane Hathaway, the newly hired production supervisor, is knowledgeable about the theory of constraints and bottleneck based scheduling. She believes that profitability can indeed be approved if bottleneck resources were exploited to determine the product mix. What is the change in profits if instead of the traditional method that O’Neill has used thus far; a bottleneck based approach advocated by Jane is used instead for selecting the product mix?

41. 7 – 41 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 SOLUTION Decision rule 1: Traditional method - Select the best product mix according to the highest overall profit margin of each product. Step 1:Calculate the profit margin per unit of each product as shown below ABC Price Raw Material & Purchased Parts Labor = Contribution Profit Margin

42. 7 – 42 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 SOLUTION Decision rule 1: Traditional method - Select the best product mix according to the highest overall profit margin of each product. Step 1:Calculate the profit margin per unit of each product as shown below When ordering from highest to lowest, the profit margin per unit order of these products is ABC. $90.00$85.00$80.00 –13.00–14.00–15.00 ABC Price Raw Material & Purchased Parts Labor = Contribution Profit Margin –10.00–9.00–7.40 $67.00$62.00$57.60

43. 7 – 43 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 2:Allocate resources W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation Z) is encountered. Subtract minutes away from 2400 minutes available for each week at each stage. Work CenterStartingAfter 65 AAfter 70 BCan Only Make 45 C W X Y Z

44. 7 – 44 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 2:Allocate resources W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation Z) is encountered. Subtract minutes away from 2400 minutes available for each week at each stage. The best product mix is 65 A, 70 B, and 45 C 240017501050510 24001815975525 24001425725500 240013604500 Work CenterStartingAfter 65 AAfter 70 BCan Only Make 45 C W X Y Z

45. 7 – 45 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 3:Compute profitability for the selected product mix. Profits Revenue Materials Overhead Labor Profit

46. 7 – 46 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Profits Revenue Materials Overhead Labor Profit Application 7.3 Step 3:Compute profitability for the selected product mix. Manufacturing the product mix of 65 A, 70 B, and 45 C will yield a profit of $2980. $15400 –$2500 –$8000 –$1920 $2980

47. 7 – 47 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Decision rule 2: Bottleneck-based approach - Select the best product mix according to the dollar contribution per minute of processing time at the bottleneck workstation Z. This rule would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck. Step 1:Calculate the contribution/minute of processing time at bottleneck workstation Z: Product AProduct BProduct C Contribution Margin Time at Bottleneck Contribution Margin per minute

48. 7 – 48 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Decision rule 2: Bottleneck-based approach - Select the best product mix according to the dollar contribution per minute of processing time at the bottleneck workstation Z. This rule would take advantage of the principles outlined in the theory of constraints and get the most dollar benefit from the bottleneck. Step 1:Calculate the contribution/minute of processing time at bottleneck workstation Z: When ordering from highest to lowest contribution margin/minute at the bottleneck, the manufacturing sequence of these products is CBA, which is reverse of the traditional method order. $67.00$62.00$57.60 16 minutes13 minutes10 minutes 4.194.775.76 Product AProduct BProduct C Contribution Margin Time at Bottleneck Contribution Margin per minute

49. 7 – 49 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 2:Allocate resources W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation Z) is encountered. Subtract minutes away from 2400 minutes available for each week at each stage. Work CenterStartingAfter 80 CAfter 70 BCan Only Make 43 A W X Y Z

50. 7 – 50 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 2:Allocate resources W, X, Y, and Z to the products in the order decided in step 1. Satisfy each demand until the bottleneck resource (workstation Z) is encountered. Subtract minutes away from 2400 minutes available for each week at each stage. The best product mix is 43A, 70B, and 80C 24001440740310 24001600760373 240020001300655 240016006902 Work CenterStartingAfter 80 CAfter 70 BCan Only Make 43 A W X Y Z

51. 7 – 51 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Step 3:Compute profitability for the selected product mix. The new profitability figures are shown below based on the new production quantities of 43A, 70B, and 80C. Profits Revenue Materials Overhead Labor Profit

52. 7 – 52 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Profits Revenue Materials Overhead Labor Profit Application 7.3 Step 3:Compute profitability for the selected product mix. The new profitability figures are shown below based on the new production quantities of 43A, 70B, and 80C. Manufacturing the product mix of 43 A, 70 B, and 80 C will yield a profit of $3561. $16220 –$2739 –$8000 –$1920 $3561

53. 7 – 53 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Drum-Buffer-Rope Systems The bottleneck schedule is the drum because it sets the beat or the production rate for the entire plant and is linked to market demand The buffer is the time buffer that plans early flows into the bottleneck and thus protects it from disruption The rope represents the tying of material release to the drum beat, which is the rate at which the bottleneck controls the throughput of the entire plant

54. 7 – 54 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Drum-Buffer-Rope Systems BufferDrum Market Demand 650 units/wk Shipping Schedule Rope Shipping Buffer Finished Goods Inventory Nonconstraint PROCESS C Capacity 700 units/wk PROCESS B Capacity 500 units/wk CCR (Bottleneck) Constraint Buffer Time Buffer Inventory Nonconstraint PROCESS A Capacity 800 units/wk Material Release Schedule Figure 7.3 – Drum-Buffer-Rope Systems

55. 7 – 55 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process Line Balancing  Assignment of work to stations in a line so as to achieve the desired output rate with the smallest number of workstations  Achieving the goal is similar to the theory of constraints but it differs in how it addresses bottlenecks Precedence diagram – AON network

56. 7 – 56 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Precedence Diagram EXAMPLE 7.4 Green Grass, Inc., a manufacturer of lawn and garden equipment, is designing an assembly line to produce a new fertilizer spreader, the Big Broadcaster. Using the following information on the production process, construct a precedence diagram for the Big Broadcaster. Work Element Description Time (sec) Immediate Predecessor(s) ABolt leg frame to hopper40None BInsert impeller shaft30A CAttach axle50A DAttach agitator40B EAttach drive wheel6B FAttach free wheel25C GMount lower post15C HAttach controls20D, E IMount nameplate18F, G Total 244

57. 7 – 57 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Precedence Diagram SOLUTION Figure 7.4 shows the complete diagram. We begin with work element A, which has no immediate predecessors. Next, we add elements B and C, for which element A is the only immediate predecessor. After entering time standards and arrows showing precedence, we add elements D and E, and so on. The diagram simplifies interpretation. Work element F, for example, can be done anywhere on the line after element C is completed. However, element I must await completion of elements F and G. D 40 I 18 H 20 F 25 G 15 C 50 E 6 B 30 A 40 Figure 7.4 –Precedence Diagram for Assembling the Big Broadcaster

58. 7 – 58 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process The desired output rate is matched to the staffing or production plan Cycle time is the maximum time allowed for work at each station is c = 1r1r where c =cycle time in hours r =desired output rate

59. 7 – 59 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process Green Grass, Inc. Desired output rate, r = 2400/week Plant operates 40 hours/week r = 2400/40 = 60 units/hour  Cycle time, c = = 1/60 = 1 minute/unit = 60 seconds/unit 1r1r

60. 7 – 60 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process The theoretical minimum number of stations is TM = tctc where  t =total time required to assemble each unit

61. 7 – 61 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process The theoretical minimum number of stations is TM = tctc = 244 seconds/60 seconds = 4.067 It must be rounded up to 5 stations

62. 7 – 62 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A Line Process Idle time, efficiency, and balance delay Idle time = nc –  t where n =number of stations Efficiency (%) = (100)  t nc Balance delay (%) = 100 – Efficiency

63. 7 – 63 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating Cycle Time, TM, Efficiency Efficiency = (100) =  t nc 244 5(60) = 81.3% Idle time = nc –  t = 5(60) – 244 = 56 seconds Balance Delay - amount by which efficiency falls short of 100%. (100 − 81.3) = 18.7%

64. 7 – 64 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Finding a Solution The goal is to cluster the work elements into workstations so that 1.The number of workstations required is minimized 2.The precedence and cycle-time requirements are not violated The work content for each station is equal (or nearly so, but less than) the cycle time for the line Trial-and-error can be used but commercial software packages are also available The decision rules used by POM for Windows are described in Table 7.3

65. 7 – 65 Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime

66. 7 – 66 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 Green Grace, Inc.

67. 7 – 67 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 Green Grace, Inc.

68. 7 – 68 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 Green Grace, Inc.

69. 7 – 69 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 Green Grace, Inc.

70. 7 – 70 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 Green Grace, Inc.

71. 7 – 71 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 Green Grace, Inc.

72. 7 – 72 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 S3B,F,GB3030 Green Grace, Inc.

73. 7 – 73 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 S3B,F,GB3030

74. 7 – 74 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 S3B,F,GB3030 Green Grace, Inc.

75. 7 – 75 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 Green Grace, Inc.

76. 7 – 76 S1 S2 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% 40 6 20 50 15 18 E 30 25 40 H I D B F C A G CummIdle StationCandidateChoiceTimeTime S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 Green Grace, Inc.

77. 7 – 77 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

78. 7 – 78 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 S4D,E,GD4020 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

79. 7 – 79 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 S4D,E,GD4020 E,GG555 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

80. 7 – 80 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 S4D,E,GD4020 E,GG555 S5E,II1842 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

81. 7 – 81 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 S4D,E,GD4020 E,GG555 S5E,II1842 EE2436 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

82. 7 – 82 Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% S1 S2 S3 40 6 20 50 15 18 E 30 25 40 H I D B F C A G S1AA4020 S2B,CC5010 S3B,F,GB3030 D,E,F,GF555 S4D,E,GD4020 E,GG555 S5E,II1842 EE2436 HH4416 CummIdle StationCandidateChoiceTimeTime Green Grace, Inc.

83. 7 – 83 © 2007 Pearson Education S1 S2 S3 S5 S4 6 E 20 H 18 I 40 D 30 B 25 F 50 C 40 A 15 G Green Grass, Inc. Line Balancing Solution c = 60 seconds/unit TM = 5 stations Efficiency = 81.3% The goal is to cluster the work elements into 5 workstations so that the number of work-stations is minimized, and the cycle time of 60 seconds is not violated. Here we use the trial-and-error method to find a solution, although commercial software packages are also available.

84. 7 – 84 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Finding a Solution TABLE 7.3| HEURISTIC DECISION RULES IN ASSIGNING THE NEXT WORK ELEMENT TO A | WORKSTATION BEING CREATED Create one station at a time. For the station now being created, identify the unassigned work elements that qualify for assignment: They are candidates if 1. All of their predecessors have been assigned to this station or stations already created. 2. Adding them to the workstation being created will not create a workload that exceeds the cycle time. Decision RuleLogic Longest work elementPicking the candidate with the longest time to complete is an effort to fit in the most difficult elements first, leaving the ones with short times to “fill out” the station. Shortest work elementThis rule is the opposite of the longest work element rule because it gives preference in workstation assignments to those work elements that are quicker. It can be tried because no single rule guarantees the best solution. It might provide another solution for the planner to consider. Most followersWhen picking the next work element to assign to a station being created, choose the element that has the most followers (due to precedence requirements). In Figure 7.4, item C has three followers (F, G, and I) whereas item D has only one follower (H). This rule seeks to maintain flexibility so that good choices remain for creating the last few workstations at the end of the line. Fewest followersPicking the candidate with the fewest followers is the opposite of the most followers rule.

85. 7 – 85 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 A plant manager needs a design for an assembly line to assembly a new product that is being introduced. The time requirements and immediate predecessors for the work elements are as follows: Work ElementTime (sec) Immediate Predecessor A12― B60A C36― D24― E38C, D F72B, E G14― H72― I35G, H J60I K 12F, J Total =435

86. 7 – 86 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. K Application 7.3 Draw a precedence diagram, complete I, F, J, and K Work Element Time (sec) Immediate Predecessor A12― B60A C36― D24― E38C, D F72B, E G14― H72― I35G, H J60I K 12F, J Total =435 F J B E I A C G H D

87. 7 – 87 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 If the desired output rate is 30 units per hour, what are the cycle time and theoretical minimum? c = = 1r1r 1 30 (3600) = 120 sec/unit TM = tctc = = 3.6 or 4 stations 435 120

88. 7 – 88 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Suppose that we are fortunate enough to find a solution with just four stations. What is the idle time per unit, efficiency, and the balance delay for this solution? Idle time = nc –  t Efficiency (%) = (100)  t nc Balance delay (%) = 100 – Efficiency = 4(120) – 435 = 45 seconds = 100 – 90.6 = 9.4% = (100) = 90.6% 435 480

89. 7 – 89 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Station Work Elements AssignedCumulative Time Idle Time ( c = 120) 1 2 3 4 5 Application 7.3 Using trial and error, one possible solution is shown below.

90. 7 – 90 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 7.3 Using trial and error, one possible solution is shown below. H, C, A1200 B, D, G9822 E, F11010 I, J, K10713 A fifth station is not needed Station Work Elements AssignedCumulative Time Idle Time ( c = 120) 1 2 3 4 5

91. 7 – 91 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Managerial Considerations Pacing is the movement of product from one station to the next Behavioral factors such as absenteeism, turnover, and grievances can increase after installing production lines The number of models produced complicates scheduling and necessitates good communication Cycle times are dependent on the desired output rate

92. 7 – 92 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 Bill’s Car Wash offers two types of washes: Standard and Deluxe. The process flow for both types of customers is shown in the following chart. Both wash types are first processed through steps A1 and A2. The Standard wash then goes through steps A3 and A4 while the Deluxe is processed through steps A5, A6, and A7. Both offerings finish at the drying station (A8). The numbers in parentheses indicate the minutes it takes for that activity to process a customer. A8 (10) A7 (12) A6 (20) A5 (5) Deluxe A4 (15) A3 (12) Standard Standard or Deluxe A2 (6) A1 (5)

93. 7 – 93 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 a.Which step is the bottleneck for the Standard car wash process? For the Deluxe car wash process? b.What is the capacity (measured as customers served per hour) of Bill’s Car Wash to process Standard and Deluxe customers? Assume that no customers are waiting at step A1, A2, or A8. c.If 60 percent of the customers are Standard and 40 percent are Deluxe, what is the average capacity of the car wash in customers per hour? d.Where would you expect Standard wash customers to experience waiting lines, assuming that new customers are always entering the shop and that no Deluxe customers are in the shop? Where would the Deluxe customers have to wait, assuming no Standard customers?

94. 7 – 94 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 SOLUTION a.Step A4 is the bottleneck for the Standard car wash process, and Step A6 is the bottleneck for the Deluxe car wash process, because these steps take the longest time in the flow. b.The capacity for Standard washes is 4 customers per hour because the bottleneck step A4 can process 1 customer every 15 minutes (60/15). The capacity for Deluxe car washes is 3 customers per hour (60/20). These capacities are derived by translating the “minutes per customer” of each bottleneck activity to “customers per hour.” c.The average capacity of the car wash is (0.60  4) + (0.40  3) = 3.6 customers per hour.

95. 7 – 95 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 d.Standard wash customers would wait before steps A1, A2, A3, and A4 because the activities that immediately precede them have a higher rate of output (i.e., smaller processing times). Deluxe wash customers would experience a wait in front of steps A1, A2, and A6 for the same reasons. A1 is included for both types of washes because the arrival rate of customers could always exceed the capacity of A1.

96. 7 – 96 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 A company is setting up an assembly line to produce 192 units per 8-hour shift. The following table identifies the work elements, times, and immediate predecessors: Work ElementTime (sec)Immediate Predecessor(s) A40None B80A C30D, E, F D25B E20B F15B G120A H145G I130H J115C, I Total 720

97. 7 – 97 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 a.What is the desired cycle time (in seconds)? b.What is the theoretical minimum number of stations? c.Use trial and error to work out a solution, and show your solution on a precedence diagram. d.What are the efficiency and balance delay of the solution found? SOLUTION a.Substituting in the cycle-time formula, we get c = = 1r1r 8 hours 192 units (3,600 sec/hr) = 150 sec/unit

98. 7 – 98 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 b.The sum of the work-element times is 720 seconds, so TM = tctc = = 4.8 or 5 stations 720 sec/unit 150 sec/unit-station which may not be achievable.

99. 7 – 99 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 c.The precedence diagram is shown in Figure 7.6. Each row in the following table shows work elements assigned to each of the five workstations in the proposed solution. J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40 Figure 7.6 – Precedence Diagram Work Element Immediate Predecessor(s) ANone BA CD, E, F DB EB FB GA HG IH JC, I

100. 7 – 100 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. StationCandidate(s)Choice Work-Element Time (sec) Cumulative Time (sec) Idle Time (c= 150 sec) S1 S2 S3 S4 S5 Solved Problem 2 J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40

101. 7 – 101 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40 AA 110 BB8012030 D, E, FD251455 E, F, GG120 30 E, FE2014010 F, HH145 5 F, II130 20 FF151455 CC30 120 JJ1151455 StationCandidate(s)Choice Work-Element Time (sec) Cumulative Time (sec) Idle Time (c= 150 sec) S1 S2 S3 S4 S5

102. 7 – 102 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 d.Calculating the efficiency, we get Thus, the balance delay is only 4 percent (100–96). Efficiency (%) = (100)  t nc = 720 sec/unit 5(150 sec/unit) = 96%