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MEHMET TANLAK Constraint Management For  Operations Management, 9e  by Krajewski/Ritzman/Malhotra  © 2010 Pearson Education PowerPoint Slides by Jeff Heyl
Managing Constraints ,[object Object],[object Object],[object Object],[object Object]
Theory of Constraints ,[object Object],[object Object],[object Object]
Theory of Constraints An increase in U at the bottleneck leads to an increase in net profit, ROI, and cash flows. 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 Utilization (U) A decrease in OE leads to an increase in net profit, ROI, and cash flows. All the money a system spends to turn inventory into throughput Operating  Expense (OE) An increase in T leads to an increase in net profit, ROI, and cash flows. Rate at which a system generates money through sales Throughput (T) A decrease in I leads to an increase in net profit, ROI, and cash flow. All the money invested in a system in purchasing things that it intends to sell Inventory (I) Relationship to Financial Measures TOC View Operational  Measures TABLE 7.1 |  HOW THE FIRM’S OPERATIONAL MEASURES RELATE TO ITS |  FINANCIAL MEASURES
Theory of Constraints 7. Every capital investment must be viewed from the perspective of its global impact on overall throughput (T), inventory (I), and operating expense (OE). 6. Activating a nonbottleneck resource (using it for improved efficiency that does not increase throughput) is not the same as utilizing a bottleneck resource (that does lead to increased throughput). Activation of nonbottleneck resources cannot increase throughput, nor promote better performance on financial measures outlined in Table 7.1. 5. Work, which can be materials, information to be processed, documents, or customers, should be released into the system only as frequently as the bottlenecks need it. Bottleneck flows should be equal to the market demand. Pacing everything to the slowest resource minimizes inventory and operating expenses. 4. Inventory is needed only in front of the bottlenecks in order to prevent them from sitting idle, and in front of assembly and shipping points in order to protect customer schedules. Building inventories elsewhere should be avoided. 3. An hour lost at a bottleneck or a constrained resource is an hour lost for the whole system. In contrast, an hour saved at a nonbottleneck resource is a mirage because it does not make the whole system more productive. 2. Maximizing the output and efficiency of every resource may not maximize the throughput of the entire system. 1. The focus should be on balancing flow, not on balancing capacity. TABLE 7.2  |  SEVEN KEY PRINCIPLES OF THE THEORY OF CONSTRAINTS
Theory of Constraints ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Theory of Constraints ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Identifying the Bottleneck ,[object Object],[object Object]
Identifying the Bottleneck 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. Complete paperwork for new loan  (10 min) 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)
Identifying the Bottleneck ,[object Object],[object Object],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).
Identifying the Bottleneck ,[object Object],[object Object],[object Object],[object Object]
Application 7.1 ,[object Object],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?
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)
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)
Application 7.1 before T2 and T4 because the activities immediately preceding them have a higher rate of output. c. Type A customers would wait  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)
Identifying the Bottleneck ,[object Object],[object Object],[object Object]
Identifying the Bottleneck Figure 7.2 Flowchart for Products A, B, C, and D 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
Identifying the Bottleneck ,[object Object],[object Object],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.
Identifying the Bottleneck Z Y X W V Total Load (min) Load from Product D Load from Product C Load from Product B Load from Product A Workstation
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. 1,800 0 0 0 60    30 = 1800 1,900 100    15 = 1,500 80    5 = 400 0 0 1,400 100    10 = 1,000 80    5 = 400 0 0 2,300 100    5 = 500 80    5 = 400 80    10 = 800 60    10 = 600 2,600 0 80    5 = 400 80    20 = 1,600 60    10 = 600 Z Y X W V Total Load (min) Load from Product D Load from Product C Load from Product B Load from Product A Workstation
Application 7.2 ,[object Object],[object Object]
Application 7.2 Flowchart for Products A, B, and C 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
Application 7.2 ,[object Object],[object Object]
Application 7.2 Z Y X W Total Load (minutes) Load from Product C Load from Product B Load from Product A Work Station
Application 7.2 ,[object Object],2310 (80  12)= 960 (70  10)= 700 (65x10)= 650 2225 (80  10)= 800 (70  12)= 840 (65  9)= 585 2750 (80  10)= 800 (70  13)= 910 (65  16)= 1040 2075 (80x5)= 400 (70x10)= 700 (65  15)= 975 Z Y X W Total Load (minutes) Load from Product C Load from Product B Load from Product A Work Station
Determining the Product Mix ,[object Object],[object Object]
Determining the Product Mix ,[object Object],SOLUTION Decision Rule 1: Traditional Method Select the best product mix according to the highest overall contribution margin of each product.
Determining the Product Mix ,[object Object],= Contribution margin Raw material and purchased parts Price D C B A
Determining the Product Mix ,[object Object],When ordered from highest to lowest, the contribution margin per unit sequence of these products is B, A, C, D. $38.00 $45.00 $72.00 $75.00 – 10.00 – 5.00 – 5.00 – 10.00 $28.00 $40.00 $67.00 $65.00 = Contribution margin Raw material and purchased parts Price D C B A
Determining the Product Mix ,[object Object],Z Y X W V Can Only Make 100 D Can Only Make 40 C Minutes Left After Making 60 A Minutes Left After Making 80 B Minutes at the Start Work Center
Determining the Product Mix ,[object Object],The best product mix according to this traditional approach is then 60 A, 80 B, 40 C, and 100 D. 600 600 600 2,400 2,400 700 2,200 2,400 2,400 2,400 1,200 2,200 2,400 2,400 2,400 300 800 1,000 1,600 2,400 0 0 200 800 2,400 Z Y X W V Can Only Make 100 D Can Only Make 40 C Minutes Left After Making 60 A Minutes Left After Making 80 B Minutes at the Start Work Center
Determining the Product Mix ,[object Object],Profit Overhead Labor Materials Revenue Profits
Determining the Product Mix ,[object Object],Manufacturing the product mix of 60 A, 80 B, 40 C, and 100 D will yield a profit of $1,560 per week. = $15,860 (60    $75) + (80    $72) + (40    $45) + (100    $38) Profit Overhead Labor Materials Revenue Profits = – $2,200 (60    $10) + (80    $5) + (40    $5) + (100    $10) = – $3,600 (5 workers)    (8 hours/day)    (5 days/week)    ($18/hour) = = $1,560 – $8,500
Determining the Product Mix ,[object Object],[object Object],Step 1: Calculate the contribution margin/minute of processing time at bottleneck workstation X:
Determining the Product Mix Contribution margin per minute Time at bottleneck Contribution margin Product D Product C Product B Product A
Determining the Product Mix ,[object Object],$28.00 $40.00 $67.00 $65.00 0 minutes 5 minutes 20 minutes 10  minutes Not defined $8.00 $3.35 $6.50 Contribution margin per minute Time at bottleneck Contribution margin Product D Product C Product B Product A
Determining the Product Mix ,[object Object],Z Y X W V Can Only Make 100 D Can Only Make 40 C Minutes Left After Making 60 A Minutes Left After Making 80 B Minutes at the Start Work Center
Determining the Product Mix ,[object Object],The best product mix according to this bottleneck based approach is then 60 A, 70 B, 80 C, and 100 D. 600 600 2,400 2,400 2,400 500 500 500 900 2,400 1,000 1,000 1,000 1,400 2,400 200 900 1,500 1,900 2,400 0 1,400 2,000 2,400 2,400 Z Y X W V Can Only Make 100 D Can Only Make 40 C Minutes Left After Making 60 A Minutes Left After Making 80 B Minutes at the Start Work Center
Determining the Product Mix ,[object Object],Profit Overhead Labor Materials Revenue Profits
Determining the Product Mix ,[object Object],Manufacturing the product mix of 60 A, 70 B, 80 C, and 100 D will yield a profit of $2,490 per week. = $16,940 (60    $75) + (70    $72) + (80    $45) + (100    $38) = – $2,350 (60    $10) + (70    $5) + (80    $5) + (100    $10) = – $3,600 (5 workers)    (8 hours/day)    (5 days/week)    ($18/hour) Profit Overhead Labor Materials Revenue Profits = = $2,490 – $8,500
Application 7.3 ,[object Object]
Application 7.3 ,[object Object],[object Object],Step 1: Calculate the profit margin per unit of each product as shown below = Contribution Profit Margin Labor Raw Material & Purchased Parts Price C B A  
Application 7.3 ,[object Object],[object Object],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. $80.00 $85.00 $90.00 – 15.00 – 14.00 – 13.00 = Contribution Profit Margin Labor Raw Material & Purchased Parts Price C B A   $57.60 $62.00 $67.00 – 7.40 – 9.00 – 10.00
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. Z Y X W Can Only Make 45 C After 70 B After 65 A Starting Work Center
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 510 1050 1750 2400 525 975 1815 2400 0 450 1360 2400 500 725 1425 2400 Z Y X W Can Only Make 45 C After 70 B After 65 A Starting Work Center
Application 7.3 Step 3: Compute profitability for the selected product mix. Profit Labor Overhead Materials Revenue Profits  
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. Profit Labor Overhead Materials Revenue Profits   $15400 – $2500 $2980 – $1920 – $8000
Application 7.3 ,[object Object],Step 1: Calculate the contribution/minute of processing time at bottleneck workstation Z: Contribution Margin per minute Time at Bottleneck  Contribution Margin Product C Product B Product A
Application 7.3 ,[object Object],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.  $57.60 $62.00 $67.00 10 minutes 13 minutes 16 minutes 5.76 4.77 4.19 Contribution Margin per minute Time at Bottleneck  Contribution Margin Product C Product B Product A
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. Z Y X W Can Only Make 43 A After 70 B After 80 C Starting Work Center
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 310 740 1440 2400 373 760 1600 2400 2 690 1600 2400 655 1300 2000 2400 Z Y X W Can Only Make 43 A After 70 B After 80 C Starting Work Center
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. Profit Labor Overhead Materials Revenue Profits  
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. Profit Labor Overhead Materials Revenue Profits   $16220 – $2739 $3561 – $1920 – $8000
Drum-Buffer-Rope Systems ,[object Object],[object Object],[object Object]
Drum-Buffer-Rope Systems Figure 7.3  –  Drum-Buffer-Rope Systems Buffer Drum Market Demand 650 units/wk Shipping Schedule Rope Shipping Buffer Finished Goods Inventory Nonconstraint PROCESS C Capacity 700 units/wk PROCESS B Capacity 800 units/wk CCR (Bottleneck) Constraint Buffer Time Buffer Inventory Nonconstraint PROCESS A Capacity 800 units/wk Material Release Schedule
A Line Process ,[object Object],[object Object],[object Object],[object Object]
Precedence Diagram ,[object Object],[object Object],Total 244 F, G 18 Mount nameplate I D, E 20 Attach controls H C 15 Mount lower post G C 25 Attach free wheel F B 6 Attach drive wheel E B 40 Attach agitator D A 50 Attach axle C A 30 Insert impeller shaft B None 40 Bolt leg frame to hopper A Immediate Predecessor(s) Time (sec) Description Work Element
Precedence Diagram ,[object Object],[object Object],Figure 7.4  – Precedence Diagram for Assembling the Big Broadcaster D 40 I 18 H 20 F 25 G 15 C 50 E 6 B 30 A 40
A Line Process ,[object Object],[object Object],where c  = cycle time in hours r  = desired output rate c  = 1 r
A Line Process ,[object Object],where  t  = total time required to assemble each unit TM =  t c
A Line Process ,[object Object],Idle time =  nc  –   t where n  = number of stations Balance delay (%) = 100  –  Efficiency Efficiency (%) =  (100)  t nc
Calculating Cycle Time, TM, Efficiency ,[object Object],[object Object],a. What should be the line’s cycle time? b. What is the smallest number of workstations that she could hope for in designing the line for this cycle time? c. Suppose that she finds a solution that requires only five stations. What would be the line’s efficiency?
Calculating Cycle Time, TM, Efficiency ,[object Object],[object Object],c  = 1/ r  =  b. Now calculate the theoretical minimum for the number of stations by dividing the total time,   t , by the cycle time,  c  = 60 seconds. Assuming perfect balance, we have 1/60 (hr/unit) = 1 minute/unit = 60 seconds/unit TM =  t c 244 seconds 60 seconds =  = 4.067 or 5 stations
Calculating Cycle Time, TM, Efficiency ,[object Object],Efficiency =  (100)  =  t nc 244 5(60) = 81.3%
Finding a Solution ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Finding a Solution Picking the candidate with the fewest followers is the opposite of the most followers rule. Fewest 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. Most followers 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. Shortest 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. Longest work element Logic Decision Rule 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. 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 TABLE 7.3 |  HEURISTIC DECISION RULES IN ASSIGNING THE NEXT WORK ELEMENT TO A |  WORKSTATION BEING CREATED
Finding a Solution ,[object Object],Figure 7.5  –  Big Broadcaster Precedence Diagram Solution D 40 I 18 H 20 F 25 C 50 E 6 B 30 A 40 G 15
Application 7.3 ,[object Object],435 Total  = F, J 12 K I 60 J G, H 35 I ― 72 H ― 14 G B, E 72 F C, D 38 E ― 24 D ― 36 C A 60 B ― 12 A Immediate Predecessor Time (sec) Work Element
Application 7.3 ,[object Object],A C G H D K 435 Total  = F, J 12 K I 60 J G, H 35 I ― 72 H ― 14 G B, E 72 F C, D 38 E ― 24 D ― 36 C A 60 B ― 12 A Immediate Predecessor Time (sec) Work Element F J B E I
Application 7.3 ,[object Object],c  =  = 1 r 1 30 (3600) = 120 sec/unit TM =  t c =  = 3.6  or 4 stations 435 120
Application 7.3 ,[object Object],Idle time =  nc  –   t Balance delay (%) = 100  –  Efficiency = 4(120) – 435 = 45 seconds = 100 – 90.6 = 9.4% Efficiency (%) =  (100)  t nc =  (100) = 90.6% 435 480
Application 7.3 ,[object Object],5 4 3 2 1 Idle Time ( c  = 120) Cumulative Time Work Elements Assigned Station
Application 7.3 ,[object Object],0 120 H, C, A 22 98 B, D, G A fifth station is not needed 13 107 I, J, K 10 110 E, F 5 4 3 2 1 Idle Time ( c  = 120) Cumulative Time Work Elements Assigned Station
Managerial Considerations ,[object Object],[object Object],[object Object],[object Object]
Solved Problem 1 ,[object Object],A8 (10) A7 (12) A6 (20) A5 (5) Deluxe A4 (15) A3 (12) Standard Standard or Deluxe A2 (6) A1 (5)
Solved Problem 1 ,[object Object],[object Object],[object Object],[object Object]
Solved Problem 1 ,[object Object],[object Object],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.
Solved Problem 1 ,[object Object]
Solved Problem 2 ,[object Object],Total 720 C, I 115 J H 130 I G 145 H A 120 G B 15 F B 20 E B 25 D D, E, F 30 C A 80 B None 40 A Immediate Predecessor(s) Time (sec) Work Element
Solved Problem 2 ,[object Object],[object Object],[object Object],[object Object],SOLUTION a. Substituting in the cycle-time formula, we get c  =  = 1 r 8 hours 192 units (3,600 sec/hr) = 150 sec/unit
Solved Problem 2 ,[object Object],which may not be achievable. TM =  t c =  = 4.8  or 5 stations 720 sec/unit 150 sec/unit-station
Solved Problem 2 ,[object Object],Figure 7.6  –  Precedence Diagram J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40 C, I J H I G H A G B F B E B D D, E, F C A B None A Immediate Predecessor(s) Work Element
Solved Problem 2 S5 S4 S3 S2 S1 Idle Time (c= 150 sec) Cumulative Time (sec) Work-Element Time (sec) Choice Candidate(s) Station J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40
Solved Problem 2 J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40 110 40 40 A A 30 120 80 B B 5 145 25 D D, E, F 5 145 115 J J 120 30 30 C C 5 145 15 F F 20 130 130 I F, I 5 145 145 H F, H 10 140 20 E E,  F 30 120 120 G E, F, G S5 S4 S3 S2 S1 Idle Time (c= 150 sec) Cumulative Time (sec) Work-Element Time (sec) Choice Candidate(s) Station
Solved Problem 2 ,[object Object],Thus, the balance delay is only 4 percent (100–96). =  96% Efficiency (%) =  (100)  t nc = 720 sec/unit 5(150 sec/unit)
 

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mehmet tanlak - 2

  • 1. MEHMET TANLAK Constraint Management For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education PowerPoint Slides by Jeff Heyl
  • 2.
  • 3.
  • 4. Theory of Constraints An increase in U at the bottleneck leads to an increase in net profit, ROI, and cash flows. 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 Utilization (U) A decrease in OE leads to an increase in net profit, ROI, and cash flows. All the money a system spends to turn inventory into throughput Operating Expense (OE) An increase in T leads to an increase in net profit, ROI, and cash flows. Rate at which a system generates money through sales Throughput (T) A decrease in I leads to an increase in net profit, ROI, and cash flow. All the money invested in a system in purchasing things that it intends to sell Inventory (I) Relationship to Financial Measures TOC View Operational Measures TABLE 7.1 | HOW THE FIRM’S OPERATIONAL MEASURES RELATE TO ITS | FINANCIAL MEASURES
  • 5. Theory of Constraints 7. Every capital investment must be viewed from the perspective of its global impact on overall throughput (T), inventory (I), and operating expense (OE). 6. Activating a nonbottleneck resource (using it for improved efficiency that does not increase throughput) is not the same as utilizing a bottleneck resource (that does lead to increased throughput). Activation of nonbottleneck resources cannot increase throughput, nor promote better performance on financial measures outlined in Table 7.1. 5. Work, which can be materials, information to be processed, documents, or customers, should be released into the system only as frequently as the bottlenecks need it. Bottleneck flows should be equal to the market demand. Pacing everything to the slowest resource minimizes inventory and operating expenses. 4. Inventory is needed only in front of the bottlenecks in order to prevent them from sitting idle, and in front of assembly and shipping points in order to protect customer schedules. Building inventories elsewhere should be avoided. 3. An hour lost at a bottleneck or a constrained resource is an hour lost for the whole system. In contrast, an hour saved at a nonbottleneck resource is a mirage because it does not make the whole system more productive. 2. Maximizing the output and efficiency of every resource may not maximize the throughput of the entire system. 1. The focus should be on balancing flow, not on balancing capacity. TABLE 7.2 | SEVEN KEY PRINCIPLES OF THE THEORY OF CONSTRAINTS
  • 6.
  • 7.
  • 8.
  • 9. Identifying the Bottleneck 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. Complete paperwork for new loan (10 min) 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)
  • 10.
  • 11.
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  • 13. 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)
  • 14. 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)
  • 15. Application 7.1 before T2 and T4 because the activities immediately preceding them have a higher rate of output. c. Type A customers would wait 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)
  • 16.
  • 17. Identifying the Bottleneck Figure 7.2 Flowchart for Products A, B, C, and D 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
  • 18.
  • 19. Identifying the Bottleneck Z Y X W V Total Load (min) Load from Product D Load from Product C Load from Product B Load from Product A Workstation
  • 20. 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. 1,800 0 0 0 60  30 = 1800 1,900 100  15 = 1,500 80  5 = 400 0 0 1,400 100  10 = 1,000 80  5 = 400 0 0 2,300 100  5 = 500 80  5 = 400 80  10 = 800 60  10 = 600 2,600 0 80  5 = 400 80  20 = 1,600 60  10 = 600 Z Y X W V Total Load (min) Load from Product D Load from Product C Load from Product B Load from Product A Workstation
  • 21.
  • 22. Application 7.2 Flowchart for Products A, B, and C 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
  • 23.
  • 24. Application 7.2 Z Y X W Total Load (minutes) Load from Product C Load from Product B Load from Product A Work Station
  • 25.
  • 26.
  • 27.
  • 28.
  • 29.
  • 30.
  • 31.
  • 32.
  • 33.
  • 34.
  • 35. Determining the Product Mix Contribution margin per minute Time at bottleneck Contribution margin Product D Product C Product B Product A
  • 36.
  • 37.
  • 38.
  • 39.
  • 40.
  • 41.
  • 42.
  • 43.
  • 44. 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. Z Y X W Can Only Make 45 C After 70 B After 65 A Starting Work Center
  • 45. 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 510 1050 1750 2400 525 975 1815 2400 0 450 1360 2400 500 725 1425 2400 Z Y X W Can Only Make 45 C After 70 B After 65 A Starting Work Center
  • 46. Application 7.3 Step 3: Compute profitability for the selected product mix. Profit Labor Overhead Materials Revenue Profits  
  • 47. 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. Profit Labor Overhead Materials Revenue Profits   $15400 – $2500 $2980 – $1920 – $8000
  • 48.
  • 49.
  • 50. 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. Z Y X W Can Only Make 43 A After 70 B After 80 C Starting Work Center
  • 51. 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 310 740 1440 2400 373 760 1600 2400 2 690 1600 2400 655 1300 2000 2400 Z Y X W Can Only Make 43 A After 70 B After 80 C Starting Work Center
  • 52. 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. Profit Labor Overhead Materials Revenue Profits  
  • 53. 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. Profit Labor Overhead Materials Revenue Profits   $16220 – $2739 $3561 – $1920 – $8000
  • 54.
  • 55. Drum-Buffer-Rope Systems Figure 7.3 – Drum-Buffer-Rope Systems Buffer Drum Market Demand 650 units/wk Shipping Schedule Rope Shipping Buffer Finished Goods Inventory Nonconstraint PROCESS C Capacity 700 units/wk PROCESS B Capacity 800 units/wk CCR (Bottleneck) Constraint Buffer Time Buffer Inventory Nonconstraint PROCESS A Capacity 800 units/wk Material Release Schedule
  • 56.
  • 57.
  • 58.
  • 59.
  • 60.
  • 61.
  • 62.
  • 63.
  • 64.
  • 65.
  • 66. Finding a Solution Picking the candidate with the fewest followers is the opposite of the most followers rule. Fewest 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. Most followers 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. Shortest 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. Longest work element Logic Decision Rule 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. 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 TABLE 7.3 | HEURISTIC DECISION RULES IN ASSIGNING THE NEXT WORK ELEMENT TO A | WORKSTATION BEING CREATED
  • 67.
  • 68.
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  • 70.
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  • 80.
  • 81.
  • 82.
  • 83. Solved Problem 2 S5 S4 S3 S2 S1 Idle Time (c= 150 sec) Cumulative Time (sec) Work-Element Time (sec) Choice Candidate(s) Station J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40
  • 84. Solved Problem 2 J 115 C 30 D 25 E 20 F 15 I 130 H 145 B 80 G 120 A 40 110 40 40 A A 30 120 80 B B 5 145 25 D D, E, F 5 145 115 J J 120 30 30 C C 5 145 15 F F 20 130 130 I F, I 5 145 145 H F, H 10 140 20 E E, F 30 120 120 G E, F, G S5 S4 S3 S2 S1 Idle Time (c= 150 sec) Cumulative Time (sec) Work-Element Time (sec) Choice Candidate(s) Station
  • 85.
  • 86.