Chapter 7 Constraint Management Copyright 2010 Pearson Education

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

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

Theory of Constraints l 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 l It outlines a deliberate process for identifying and overcoming constraints l TOC methods increase the firm’s profits by focusing on materials flow through the entire system Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 3

Theory of Constraints TABLE 7. 1 | HOW THE FIRM’S OPERATIONAL MEASURES RELATE TO ITS | FINANCIAL MEASURES Operational Measures TOC View Relationship 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 4

Theory of Constraints l 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 5

Theory of Constraints l Bottlenecks can both be internal or external to the firm and are typically a process or step with the lowest capacity l 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 l 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 6

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 recordkeeping 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 7

Identifying the Bottleneck Check for credit rating (15 min) Check loan documents and put them order (15 min) Categorize loans (20 min) Complete paperwork for new loan (10 min) Enter loan application into the system (12 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 8

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). Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 9

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

Application 7. 1 Two types of customers enter Barbara’s Boutique shop for customized dress alterations. After T 1, Type A customers proceed to step T 2 and then to any of the three workstations at T 3, followed by steps T 4 and T 7. After step T 1, Type B customers proceed to step T 5 and then steps T 6 and T 7. 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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 11

Application 7. 1 T 3 -a (14) Type A T 1 (12) T 2 (13) T 3 -b (10) T 3 -c (11) Type A or B? Type B T 4 (18) T 5 (15) T 7 (10) T 6 (22) a. For Type A customers, step T 2 can process (60/13) = 4. 62 customers per hour. T 3 has three work stations and a capacity of (60/14) + (60/10) + (60/11) = 15. 74 customer per hour. Step T 4 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, T 4. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 12

Application 7. 1 T 3 -a (14) Type A T 1 (12) T 2 (13) T 3 -b (10) T 3 -c (11) Type A or B? Type B T 4 (18) T 5 (15) T 7 (10) T 6 (22) b. The bottleneck for Type B customers is T 6 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 13

Application 7. 1 T 3 -a (14) Type A T 1 (12) T 2 (13) T 3 -b (10) T 3 -c (11) Type A or B? Type B T 4 (18) T 5 (15) T 7 (10) T 6 (22) c. Type A customers would wait before T 2 and T 4 because the activities immediately preceding them have a higher rate of output. Type B customers would wait before steps T 5 and T 6 for the same reason. This assumes there always new customers entering the shop. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 14

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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 15

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

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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 17

Identifying the Bottleneck 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 18

Identifying the Bottleneck Workstation Load from Product A Load from Product B Load from Product C Load from Product D Total Load (min) V 60 x 30 = 1800 0 1, 800 W 0 0 80 x 5 = 400 100 x 15 = 1, 500 1, 900 X 60 x 10 = 600 80 x 20 = 1, 600 80 x 5 = 400 0 2, 600 Y 60 x 10 = 600 80 x 10 = 800 80 x 5 = 400 100 x 5 = 500 2, 30 Z 0 0 80 x 5 = 400 100 x 10 = 1, 000 1, 400 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 19

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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 20

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

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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 22

Application 7. 2 Work Station Load from Product A Load from Product B Load from Product C Total Load (minutes) W X Y Z Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 23

Application 7. 2 Work Station Load from Product A Load from Product B Load from Product C Total Load (minutes) W (65 x 10)= 650 (70 10)= 700 (80 12)= 960 2310 X (65 9)= 585 (70 12)= 840 (80 10)= 800 2225 Y (65 15)= 975 (70 x 10)= 700 (80 x 5)= 400 2075 Z (65 16)= 1040 (70 13)= 910 (80 10)= 800 2750 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 24

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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 25

Determining the Product Mix Pedro Rodriguez, the newly hired production supervisor, is knowledgeable about theory of constraints and bottleneckbased 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 26

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

Determining the Product Mix Step 1: Calculate the contribution margin per unit of each product as shown here. A B C D Price $75. 00 $72. 00 $45. 00 $38. 00 Raw material and purchased parts – 10. 00 – 5. 00 – 10. 00 = Contribution margin $65. 00 $67. 00 $40. 00 $28. 00 When ordered from highest to lowest, the contribution margin per unit sequence of these products is B, A, C, D. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 28

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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 29

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. Minutes at Minutes left after the start making 80 B Minutes left after making 60 A Can only make 40 C Can still make 100 D V 2, 400 2400 – 1800 = 600 600 W 2, 400 2400 – 200 = 2200 – 1500 = 700 X 2, 400 2400 – 1600 = 800 – 600 = 200 – 200 = 0 0 Y 2, 400 2400 – 800 = 1600 – 600 = 1000 – 200 = 800 – 500 = 300 Z 2, 400 2400 – 200 = 2200 – 1000 = 1200 The best product mix according to this traditional approach is then 60 A, 80 B, 40 C, and 100 D. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 30

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

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

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 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: Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 33

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

Determining the Product Mix Product A Product B Product C Product D Contribution margin $65. 00 $67. 00 $40. 00 $28. 00 Time at bottleneck 10 minutes 20 minutes 5 minutes 0 minutes Contribution margin per minute $6. 50 $3. 35 $8. 00 Not defined 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 35

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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 36

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. Minutes at Minutes left after the start making 100 D V 2, 400 W 2, 400 2400 – 1500 = 900 X 2, 400 2400 Y 2, 400 2400 – 500 = 1900 Z 2, 400 2400 1000 = 1400 Minutes left after making 80 C Minutes left after Can only make 70 B making 60 A 600 2400 – 1800 = 600 500 900 – 400 = 500 1400 – 1400 = 0 2400 – 400 = 2000 – 600 = 1400 900 – 700 = 200 1900 – 400 = 1500 – 600 = 900 1000 1400 – 400 = 1000 The best product mix according to this bottleneck based approach is then 60 A, 70 B, 80 C, and 100 D. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 37

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

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

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 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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 40

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 A B C Price Raw Material & Purchased Parts Labor = Contribution Profit Margin Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 41

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 A B C Price $90. 00 $85. 00 $80. 00 Raw Material & Purchased Parts – 13. 00 – 14. 00 – 15. 00 Labor – 10. 00 – 9. 00 – 7. 40 = Contribution Profit Margin $67. 00 $62. 00 $57. 60 When ordering from highest to lowest, the profit margin per unit order of these products is ABC. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 42

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 Center Starting After 65 A After 70 B Can Only Make 45 C W X Y Z Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 43

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 Center Starting After 65 A After 70 B Can Only Make 45 C W 2400 1750 1050 510 X 2400 1815 975 525 Y 2400 1425 725 500 Z 2400 1360 450 0 The best product mix is 65 A, 70 B, and 45 C Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 44

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

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

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 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 A Product B Product C Contribution Margin Time at Bottleneck Contribution Margin per minute Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 47

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 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 A Product B Product C Contribution Margin $67. 00 $62. 00 $57. 60 Time at Bottleneck 16 minutes 13 minutes 10 minutes Contribution Margin per minute 4. 19 4. 77 5. 76 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 48

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 Center Starting After 80 C After 70 B Can Only Make 43 A W X Y Z Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 49

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 Center Starting After 80 C After 70 B Can Only Make 43 A W 2400 1440 740 310 X 2400 1600 760 373 Y 2400 2000 1300 655 Z 2400 1600 690 2 The best product mix is 43 A, 70 B, and 80 C Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 50

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 43 A, 70 B, and 80 C. Profits Revenue Materials Overhead Labor Profit Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 51

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 43 A, 70 B, and 80 C. Profits Revenue $16220 Materials –$2739 Overhead –$8000 Labor –$1920 Profit $3561 Manufacturing the product mix of 43 A, 70 B, and 80 C will yield a profit of $3561. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 52

Drum-Buffer-Rope Systems l 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 l The buffer is the time buffer that plans early flows into the bottleneck and thus protects it from disruption l 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 53

Drum-Buffer-Rope Systems Constraint Buffer CCR (Bottleneck) PROCESS A Capacity 800 units/wk Time Buffer Inventory PROCESS B Capacity 500 units/wk Rope Buffer Drum Nonconstraint Material Release Schedule Shipping Buffer Nonconstraint PROCESS C Capacity 700 units/wk Finished Goods Inventory Shipping Schedule Market Demand 650 units/wk Figure 7. 3 – Drum-Buffer-Rope Systems Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 54

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

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 Time Immediate Description Element (sec) Predecessor(s) A Bolt leg frame to hopper 40 None B Insert impeller shaft 30 A C Attach axle 50 A D Attach agitator 40 B E Attach drive wheel 6 B F Attach free wheel 25 C G Mount lower post 15 C H Attach controls 20 D, E I Mount nameplate 18 F, G Total 244 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 56

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 D interpretation. Work element F, H 40 B for example, can be done 20 anywhere on the line after E 30 element C is completed. 6 A However, element I must F 40 await completion of C 25 elements F and G. 50 I Figure 7. 4 – Precedence Diagram for Assembling the Big Broadcaster Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. G 18 15 7 – 57

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

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

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

A Line Process l The theoretical minimum number of stations is t TM = c = 244 seconds/60 seconds = 4. 067 It must be rounded up to 5 stations Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 61

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

Calculating Cycle Time, TM, Efficiency Idle time = nc – t = 5(60) – 244 = 56 seconds t Efficiency = (100) = nc 244 = 81. 3% 5(60) Balance Delay - amount by which efficiency falls short of 100%. (100 − 81. 3) = 18. 7% Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 63

Finding a Solution l 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 l The work content for each station is equal (or nearly so, but less than) the cycle time for the line l Trial-and-error can be used but commercial software packages are also available l The decision rules used by POM for Windows are described in Table 7. 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 64

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 40 30 20 E 6 A 40 H C F 25 Cumm Station Candidate Choice Time 50 Idle Time I G 18 15 7 – 65

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 6 A 40 E C F 25 Cumm Station Candidate Choice Time S 1 50 A A 40 Idle Time 20 I G 18 15 7 – 66

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 6 A 40 E C F 25 Cumm Station Candidate Choice Time S 1 50 A A 40 Idle Time 20 I G 18 15 7 – 67

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 40 20 30 S 1 E 6 A 40 H C F 25 Cumm Station Candidate Choice Time S 1 50 A A 40 Idle Time 20 I G 18 15 7 – 68

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 40 20 30 S 1 E 6 A 40 H C 50 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C C I 50 10 G 18 15 7 – 69

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 40 20 30 S 1 E 6 A 40 H C 50 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C C I 50 10 G 18 15 7 – 70

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C C I 50 10 G 18 15 7 – 71

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G C I 30 30 G B 18 15 7 – 72

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G C I 30 30 G B 18 15 7 – 73

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G C I 30 30 G B 18 15 7 – 74

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G D, E, F, G C I 30 55 30 5 G B 18 F 15 7 – 75

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 1 A S 2 40 C 50 E 6 F 25 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G D, E, F, G C I 30 55 30 5 G B 18 F 15 7 – 76

Line Balancing Green Grace, Inc. c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B H 40 20 30 S 3 S 1 A S 2 40 C 50 F 25 E 6 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G D, E, F, G C I 30 55 30 5 G B 18 F 15 7 – 77

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% Green Grace, Inc. D B H 40 20 30 S 3 S 1 A S 2 40 C 50 F 25 E 6 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G D, E, G C I 30 55 40 30 5 20 G S 4 15 B 18 F D 7 – 78

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% Green Grace, Inc. D B H 40 20 30 S 3 S 1 A S 2 40 C 50 F 25 E 6 Cumm Station Candidate Choice Time Idle Time S 1 A A 40 20 S 2 B, C 50 10 S 3 B, F, G D, E, G C I 30 55 40 55 30 5 20 5 G S 4 15 B 18 F D G 7 – 79

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 30 S 3 S 1 A S 2 40 C 50 Green Grace, Inc. F 25 H 40 20 E Cumm 6 Choice Time Station Candidate Idle Time S 1 A A 40 20 S 2 B, C C 50 10 S 3 B, F, G D, E, G E, I B I F D 18 G I 30 55 40 55 18 30 5 20 5 42 S 4 G S 5 15 7 – 80

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 30 S 3 S 1 A S 2 40 C 50 Green Grace, Inc. F 25 H 40 20 E Cumm 6 Choice Time Station Candidate Idle Time S 1 A A 40 20 S 2 B, C C 50 10 S 3 B, F, G D, E, G E, I E B I F D 18 G I E 30 55 40 55 18 24 30 5 20 5 42 36 S 4 G S 5 15 7 – 81

Line Balancing c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% D B 30 S 3 S 1 A S 2 40 C 50 Green Grace, Inc. F 25 H 40 20 E Cumm 6 Choice Time Station Candidate Idle Time S 1 A A 40 20 S 2 B, C C 50 10 S 3 B, F, G D, E, G E, I E H B I F D 18 G I E H 30 55 40 55 18 24 44 30 5 20 5 42 36 16 7 – 82 S 4 G S 5 15

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. Green Grass, Inc. Line Balancing Solution D B 30 S 3 S 1 A S 2 40 C 40 © 2007 Pearson Education 20 E S 4 F 25 50 c = 60 seconds/unit TM = 5 stations Efficiency = 81. 3% H 6 S 5 I G 15 18 7 – 83

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 Rule Logic Longest work element Picking 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 element This 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 followers When 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 followers Picking the candidate with the fewest followers is the opposite of the most followers rule. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 84

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 Immediate Work Element Time (sec) Predecessor predecessors for the A 12 ― work elements are B 60 A as follows: C 36 ― D 24 ― E 38 C, D F 72 B, E G 14 ― H 72 ― I 35 G, H J 60 I K 12 F, J Total = Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 435 7 – 85

Application 7. 3 Draw a precedence diagram, complete I, F, J, and K Work Element Time (sec) Immediate Predecessor A 12 ― B 60 A C 36 ― D 24 ― E 38 C, D F 72 B, E G 14 ― H 72 ― I 35 G, H J 60 I K 12 F, J Total = 435 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A C B F D K E J G I H 7 – 86

Application 7. 3 If the desired output rate is 30 units per hour, what are the cycle time and theoretical minimum? 1 c = = r 1 (3600) = 120 sec/unit 30 t 435 TM = = = 3. 6 or 4 stations c 120 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 87

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 = 4(120) – 435 = 45 seconds t 435 = (100) = 90. 6% Efficiency (%) = (100) nc 480 Balance delay (%) = 100 – Efficiency = 100 – 90. 6 = 9. 4% Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 88

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

Application 7. 3 Using trial and error, one possible solution is shown below. Station Work Elements Assigned Cumulative Time 1 H, C, A 120 0 2 B, D, G 98 22 3 E, F 110 10 4 I, J, K 107 13 5 Idle Time (c = 120) A fifth station is not needed Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 90

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

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 A 1 and A 2. The Standard wash then goes through steps A 3 and A 4 while the Deluxe is processed through steps A 5, A 6, and A 7. Both offerings finish at the drying station (A 8). The numbers in parentheses indicate the minutes it takes for that activity to process a customer. A 3 (12) Standard A 1 (5) A 2 (6) A 4 (15) A 8 (10) Standard or Deluxe Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. A 5 (5) A 6 (20) A 7 (12) 7 – 92

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 A 1, A 2, or A 8. 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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 93

Solved Problem 1 SOLUTION a. Step A 4 is the bottleneck for the Standard car wash process, and Step A 6 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 A 4 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 94

Solved Problem 1 d. Standard wash customers would wait before steps A 1, A 2, A 3, and A 4 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 A 1, A 2, and A 6 for the same reasons. A 1 is included for both types of washes because the arrival rate of customers could always exceed the capacity of A 1. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 95

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 Element Time (sec) Immediate Predecessor(s) A 40 None B 80 A C 30 D, E, F D 25 B E 20 B F 15 B G 120 A H 145 G I 130 H J 115 C, I Total 720 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 96

Solved Problem 2 a. What is the desired cycle time (in seconds)? b. What is 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 1 8 hours (3, 600 sec/hr) = 150 sec/unit c = = r 192 units Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 97

Solved Problem 2 b. The sum of the work-element times is 720 seconds, so t 720 sec/unit TM = = = 4. 8 or 5 stations c 150 sec/unit-station which may not be achievable. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 98

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. D 25 B E C 80 20 30 A 40 G F J 15 115 Work Element Immediate Predecessor(s) A None B A C D, E, F D B E B F B G A H G I H J C, I 120 H I 145 130 Figure 7. 6 – Precedence Diagram Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 99

Solved Problem 2 D 25 B E C 80 20 30 40 Candidate(s) Choice Work-Element Time (sec) 15 G 120 Station 115 F A Cumulative Time (sec) H J I 130 145 Idle Time (c= 150 sec) S 1 S 2 S 3 S 4 S 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7 – 100

Solved Problem 2 D 25 B E C 80 20 30 40 15 G H 120 Station Candidate(s) Choice Work-Element Time (sec) 115 F A I 130 145 Cumulative Time (sec) Idle Time (c= 150 sec) A A 40 40 110 B B 80 120 30 D, E, F D 25 145 5 E, F, G G 120 30 E, F E 20 140 10 S 3 F, H H 145 5 S 4 F, I I 130 20 F F 15 145 5 C C 30 30 120 J J 115 145 5 S 1 S 2 S 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. J 7 – 101

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