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Assignment Capacity

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2 Ardavan Asef-Vaziri March, 2015Capacity- Basics A flowchart is a diagram that traces the flow of materials, customers, information, or equipment through the various steps of a process Key Problem1: Flow Chart A B C C D F E B Capacity Metrics: Capacity, Time to Perform the Activity (Unit Load; Tp), Cycle Time Flow Time Metrics: Theoretical Flow Time; Flow Time

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3 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem1: Single-Stage and Two-Stage Process Cycle time = Capacity = Theoretical Flow Time = Ip = Activity A Tp =1 min Activity B Tp =8 min Activity A Tp =10 min Cycle time = Capacity = Theoretical Flow Time = Ip = IpA = IpB = ActA ActB 18 CT CT 48 1 min 1/1 per min, 60 per hour 1 min 1 10 min 1/10 per min, 6 per hour 18 min That is Max Ip indeed

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4 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem1: Two-Stage Process Activity B Tp =10 min Activity A Tp =5 min Cycle time = Capacity = Theoretical Flow Time = Ip = IpA = IpB = ActA ActB 0 15 CT 3525 CT 45 The Resource in charge of Activity A is Specialized and Fast The Resource in charge of Activity B is Specialized and Fast Process Capacity 6 per hour 10 min 6 per hr 15 min

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5 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem1d: Single-Stage Process Lets cross train them and reduce set up time of the operation. They are not fast anymore. Instead of 5+10=15, now it takes 16 to complete a flow unit ActivityAB1 16 Activity AB2 Cycle time = Capacity = Theoretical Flow Time = ActAB1 ActAB2 16 CT CT 40 60(2/16) per hr per hr 8 min Capacity increased from 6 to 7.5. Therefore, pooling and cross-training can increase throughput. We will latter show that flow time will also go down.

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6 Ardavan Asef-Vaziri March, 2015Capacity- Basics MamossaAssaf Inc. fabricates garage doors. Roofs are punched in a roof punching press ( 15 minutes per roof) and then formed in a roof forming press (8 minutes per roof). Bases are punched in a base punching press ( 3 minutes per base) and then formed in a base forming press ( 10 minutes per base), and the formed base is welded in a base welding machine ( 12 minutes per base). The base sub-assembly and the roof then go to final assembly where they are welded together ( 10 minutes per garage) on an assembly welding machine to complete the garage. Assume one operator at each station. Key Problem 2 8 R-Form 10 B-Form 15 R-Punch B- 12 Weld 10 Assembly 3 Roof Base Door

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7 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Flow Time (a) What is the Theoretical Flow Time? (The minimum time required to produce a garage from start to finish.) Roof Path: 15+8 = 23 Base Path: = 25 Max = = 35 Theoretical Flow Time = 35 8 R-Form 10 B-Form 15 R-Punch B- 12 Weld 10 Assembly 3 Roof Base Door Critical Path = Max(33,35) = 35

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8 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Capacity (b) What is the capacity of the system in terms of garages per hour? R-Punch:1/15 per min. or 4 per hr. R-Form:1/8 per min. or 7.5 per hr. B-Punch:1/3 per min. or 20 per hr. B-Form:1/10 per min. or 6 per hr. Welding: 1/12 per min. or 5 per hr. Assembly: 1/10 per min. or 6 per hr. Process Capacity is 4 per hour (c) If you want to increase the process capacity, what is the activity process that you would put some additional resource? R-Punch 8 R-Form 10 B-Form 15 R-Punch B- 12 Weld 10 Assembly 3 Roof Base Door

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9 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Capacity (d) Compute utilization of al the resources at the full process capacity. In other words, assume that the throughput is equal to process capacity. Through put = 4 R-Punch Utilization = 4/4 = 100%. R-Form Utilization = 4/7.5 = 53.33%. B-Punch Utilization = 4/20 = 20%. B-Form Utilization = 4/ 6 = 66.67%. Welding Utilization = 4/5 = 80%. Assembly Utilization = 4/6 = 66.67% No Process can work at 100% capacity. Impossible. In reality, utilization of all the resources will be less than what we have computed. This process can never produce 4 flow units per hour continually.

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10 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Bottleneck Shift Process Capacity is 5 per hour (e) Suppose we double the capacity of the bottleneck by adding the same capital and human resources. What is the new capacity of the system. 8 R-Form 10 B-Form 15 R-Punch B- 12 Weld 10 Assembly 3 Roof Base Door R-Punch R-Punch: 2/15 per min. or 8 per hr. R-Form:1/8 per min. or 7.5 per hr. B-Punch:1/3 per min. or 20 per hr. B-Form:1/10 per min. or 6 per hr. Welding: 1/12 per min. or 5 per hr. Assembly: 1/10 per min. or 6 per hr.

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11 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Diminishing Marginal Return (f) We doubled the capacity of the bottleneck but the capacity of the system increased by only 25%. This situation is an example of what managerial experiment? 1) Bottleneck shifts from R-Punch to Welding 2) Diminishing Marginal Return

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12 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Pooling and Cross Training g) Now suppose we return back to the original situation where we have a single machine and a single operator at each operation. However, also suppose that we pool R-Punch and B- Punch machines and we cross-train their operations and form a new resource pool named Punch where both R-Punch and B- Punch operations can be done in this resource pool. What is the new capacity of the system? R-Form 8 B-Form 10 B-Form 12 Weld 10 Assembly R-Punch 15 R-Punch B-Punch 3 R-Punch 15 Punch B-Punch Punch 3 Assembly

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13 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Pooling and Cross Training R-Form 8 B-Form 10 B-Form 12 Weld 10 Assembly R&B-Punch Punch 15,3 Assembly Punch: 2/18 per min. or 6.67 per hr. R-Form:1/8 per min. or 7.5 per hr. B-Form:1/10 per min. or 6 per hr. Welding: 1/12 per min. or 5 per hr. Assembly: 1/10 per min. or 6 per hr. Process Capacity is 5 per hour

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14 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Productivity Improvement -Method, Training, Technology, Management h) This situation is an example of what managerial experiment? 1) Cross training and pooling can increase the capacity 2) Usually cost of cross training and pooling is lower than the cost of adding the second resource unit. i) Now suppose by investing in improved jigs and fixtures ( technology ), and also by implementing a better method of doing the job, and also training, we can reduce the welding time from 12 minutes to 10 minutes. What is the new capacity of the system ?

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15 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: More Than One Bottleneck j) Why it is impossible to work at 100% of capacity? There are 3 bottlenecks. This is a risky situation. Any of the bottlenecks could cause the throughput of the system to fall below 6 per hour. The more bottlenecks in the system, the higher the probability of not meeting the capacity. Suppose punch fail to provide input to B-Form for 1 hour, or B-Form fails to provide Weld, or Weld fails to provide Assembly- That hour of capacity perishes. Process Capacity is 6 per hour R-Form 8 B-Form 10 B-Form 10 Weld 10 Assembly R&B-Punch Punch 15,3 Assembly

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16 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: Critical Chain R-Form 8 B-Form 10 B-Form R-Punch 15 Punch B-Punch Punch 10 Weld Assembly 10 Assembly 3 10 Assembly 23 Path 2 23 Path 1 Not only the system has two bottlenecks, but one bottleneck feeds the second. Furthermore. Both paths to the last bottleneck are critical. They can both increase the flow time.

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17 Ardavan Asef-Vaziri March, 2015Capacity- Basics STOP The Rest Not needed This Semester

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18 Ardavan Asef-Vaziri March, 2015Capacity- Basics Lessons Learned 1. When we relax a Bottleneck Resource, the Bottleneck shifts to another resource 2. By doubling the bottleneck resource, the capacity usually does not double. This could be interpreted as diminishing marginal return situation. 3. One other way to increase capacity is cross training (for human resources ) and pooling (for Capital Resources). 4. Usually cost of cross training and pooling is lower than the cost of adding the second resource unit. 5. One other way to increase capacity in to reduce unit load. This is done by better management, (a) better methods, (b) training, (c) replacing human resources by capital resources (more advanced technology), and (d) better management. 6. Processes cannot work at 100% capacity. Capacity is perishable- it is lost if input is not ready. The more the bottleneck recourses the lower the utilization. 7, Convergence points are important in managing the flow time. The more convergence points the high the probability of the flow time exceed the average flow time.

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19 Ardavan Asef-Vaziri March, 2015Capacity- Basics Kristen and her roommate are in the business of baking custom cookies. As soon as she receives an order by phone, Kristen washes the bowl and mixes dough according to the customer's order - activities that take a total of 6 minutes. She then spoons the dough onto a tray that holds one dozen cookies ( 2 minutes ). Her roommate then takes 1 minute to set the oven and place the tray in it. Cookies are baked in the oven for 9 min utes and allowed to cool outside for 5 minutes. The roommate then boxes the cookies (2 minutes) and collects payment from the customer (1 minute). Determine the unit load on the three resources in the process – Kristen, her roommate and the oven. Assuming that all three resources are available 8 hours a day 100% of the time. Problem 3: Problem 5.2 book

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20 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take Order wash Mix 6 Spoon 2 load Set 1 Cool 5 Bake 9 Un load Pack 2 Pay 1 Problem 3: Problem 5.2: Flow unit = 1 order of 1 dozen. a) Compute the unit load of each resource Kristen = 6+ 2 = 8 min/unit. Roommate = = 4 min/unit. Oven = 1+9 = 10 min/unit. b) Compute the capacity of each resources. Kristen = 1/8 = per min = 7.5 orders per hour. RoomMate = 1/4 per min = 15 orders/hour. Oven = 1/10 =per min = 6 orders/hour min. c) Compute the process capacity. Capacity = min {7.5, 15, 6} = 6 orders of 1 dozen/hr. The oven is the theoretical bottleneck.

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21 Ardavan Asef-Vaziri March, 2015Capacity- Basics d) Compute utilization at full capacity Kristen = 6/7.5 = 80% RM = 6/15 = 40% Oven = 6/6 = 100%. e) What is the impact of buying another Oven? Doubles the oven resources pool capacity to 12 orders per hour. Oven = 2/10 per min = 12 orders/hour min Capacity = min { 7.5, 15, 12} = 7.5 orders of 1 dozen/hr. The bottleneck shifts to Kristen. Doubling the capacity of oven does not double the process capacity. The process capacity is only increased to 7.5 orders per hour. That is 25% improvement. This is an example of (1) shift in the bottleneck, (2) diminishing marginal return. Problem 3: Problem 5.2

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22 Ardavan Asef-Vaziri March, 2015Capacity- Basics The unit load of Worker Resource Pool is 8+4 = 12 min. per unit. The capacity of Workers Resource Pool is increased to 2/12 per min. = 10 orders of 1 dozen/hr. Capacity = min { 10, 6 } = 6 orders of 1 dozen/hr. With one oven, cross training does not affect the theoretical process capacity. The Oven remains the bottleneck. The capacity is 6 dozen per hour. g) Now suppose we have two ovens. With two ovens, capacity = min { 10, 2* 6 } = 10 per hr. The bottleneck shifted to the Workers Resource Pool. 6 + Two Ovens 7.5, 6 + Cross Train 6, 6 + Two Ovens + Cross Train 10. Problem 3: Problem 5.2 f) Lets go back to one oven case. What is the impact of cross training of Kristen and RM? Cross training pools Kristen and RM into a single resource pool (Workers).

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23 Ardavan Asef-Vaziri March, 2015Capacity- Basics Resource-Activity match High Utilization (Low Safety Capacity) Division of Labor (Job-Simplification) High Standardization and Modularization Effective Facility Layout Clear Material Flow Pattern Flow-Shop Value Analysis- Value Added and Non-Value Added Training Method Improvement Technology Key Problems: Think About Internal Measures Throughout the Semester : Cost

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24 Ardavan Asef-Vaziri March, 2015Capacity- Basics Cross-trained Workers Short Set-up Time Delayed Differentiation (postponement) Small Batch Size Job-Shop U-Shaped Layout Internal Uniformity vs. External Variability Key Problems: Think About Internal Measures Throughout The Semester : Flexibility

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25 Ardavan Asef-Vaziri March, 2015Capacity- Basics Small Batch Size –uniform operations – short flow time - low setup time Small number of suppliers Long term relationship with suppliers. Suppliers located in short distances Inventory Turnover Reliability in flow time No Starvation or Blockage Centralization Commonality Pooling Variance Reduction Not high utilization Key Problems: Think About Internal Measures Throughout The Semester : Flow Time

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26 Ardavan Asef-Vaziri March, 2015Capacity- Basics Conformance of Design and Manufacturing Quality at Source Reliability (quality over time) Service Level No Defect and Re-work Training Method Improvement Management Key Problems: Think About Internal Measures Throughout The Semester : Quality

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27 Ardavan Asef-Vaziri March, 2015Capacity- Basics Process analysis is the detailed understanding and documentation of how work is performed and how it can be redesigned and improved. What is a Process Capacity Metrics: Capacity, Time to perform the process (Unit Load; Tp) Quality Metrics: Defective rate, Customer satisfaction rate Efficiency Metrics: Cost, Productivity, Utilization Flexibility Metrics: Setup time, Cross Training

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28 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 6. Problem 5.4 in the book A company makes two products, A and B, using a single resource pool. The resource is available for 900 min per day. The profit margins for A and B are $20 and $35 per unit respectively. The total unit loads are 10 and 20 minutes. a) The company wishes to produce a mix of 60% As and 40% Bs. What is the effective capacity (units per day)? An aggregate product will need 0.6(10) + 0.4(20) = 14 minutes Capacity is 1/14 per minute or 900(1/14) = per day b) What is the financial throughput per day? Financial throughput is the rate at which a firm is generating money. An aggregate product will generate 0.6(20) + 0.4(35) = $ (28) = $ per day

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29 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 7 The following graph shows a production process for two products AA and BC. Station D and E are flexible and can handle either product. No matter the type of the product, station D can finish 100 units per day and station E can finish 90 units per day. Station A works only for Product A and have a capacity of 60 units per day. Station B and C are only for Product BC and have capacity of 75 and 45 units per day, respectively. The demands for each product is 50 units per day. Which station(s) is the bottleneck? A)Stations A and C B)Station B and C C)Stations C and D D)Stations D and E E)Station C and E B 75 A 60 D 100 C 45 E 90 AA BC

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30 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 7 If the system can work at the process capacity, which of the following is NOT true? A)The utilization of machine A is at least 75% B)The utilization of machine B at least about 53% C)The utilization of machine B is at most 60% D)The utilization of machine D is 90% E)All of the above. E We can produce at most 90 AA and BC. C We can produce at most 45 BC We may produce all combinations from 50AA and 40 BC to 45AA and 45 BC A)We produce at least 45 AA: 45/60 = 75% B)We produce at least 40 BC: 40/75 = 53.33% C)45/75 = 60% D)90/100 = 90% B 75 A 60 D 100 C 45 E 90 AA BC

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31 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 8 A company has five machines and two products. Product X will be processed on Machine A, then J, then B. Product Y will be processed on Machine C, then J, then D. The demands for both products are 50 units per week. The capacities (units/week) of the machines are marked in the graph on the right. Which machine is the bottleneck? A)A B)B C)C D)D E)J B 60 A 50 D 80 C 70 J 90 X Y

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32 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 8 Which of the following is NOT true? A)The utilization of machine A is at least 80% B)The utilization of machine B at least about 66% C)The utilization of machine D is at least 50% D)The utilization of machine C is at most about 72% E)All of the above. We can produce at most 90 X and Y. We may produce all combinations from 50 X and 40 Y to 40 X and 50Y A)We produce at least 40 X: 40/50 = 80% B)We produce at least 40 X: 40/60 = 66.67% C)40/80 = 50% D)50/70 = 71.43% B 60 A 50 D 80 C 70 J 90 X Y

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33 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 9. Multiple Choice Which of the following 2 statements is true? I. A process can have more than 1 bottleneck resource. II. Having flexible equipment can increase utilization. A)Only I B)Only II C)Both I and II D)Neither I nor II E)Cannot be determined Which of the following statement is false? A)Throughput rate is always smaller than or equal to the capacity B)Customers may wait even if the utilization rate of the service process is smaller than 100% C)Bottleneck resource(s) always has 100% utilization rate D)Increasing WIP may increase utilization rate E)None of the above

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34 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 9. Multiple Choice To improve the utilization rate, we can I. Cross-train the workers II. Adopt flexibility equipment III. Shift from MTS systems to MTO system Choose the most appropriate. A)I B)II C)III D)I and II E)I, II, and III

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35 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 2. Problem 5.3 in the book Three hairstylists and a receptionist. On average it takes 10 minutes to shampoo, 15 minutes to style the hair and 5 minutes to bill a customer. The customer first checks in with the receptionist. This takes only 3 minutes. One of the three stylists then takes charge and performs all the three activities—shampooing, styling and billing—consecutively. ReceptionistHair Stylist 3 minutes =30 Check-in Sh-Sty-Bill a) How many customers can be serviced per hour in this salon? The capacity of each stylists is 1/30 per min. Capacity of the stylist pool is 3(1/30) = 3/30 = 1/10 per min. Capacity of the stylist pool is 60(1/10) = 6 customers per hour. Capacity of the receptionist is 1/3 per min or (60) = 20 per hour.

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36 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 2. Problem 5.3 in the book The capacity of the process is min (6, 20) = 6 customers per hour. The bottleneck is the stylist pool. ReceptionistHair Stylist 3+5 = =25 ChinBill Sh-Sty b) What would be the impact on the theoretical capacity if The capacity each stylists is 1/25 = per min. Stylist pool has 3 stylists. Capacity of pool is 3/25 per min or 60(3/25) = 7.2 per hr. Capacity of the receptionist is (1/8)(60) = 7.5 customers per hour. The capacity is min (7.2, 7.5) = 7.2 customers per hour. Cross training increased capacity form 6 to 7.2. The bottleneck is still the stylist pool. the billing operation is transferred to the receptionist?

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37 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 3 Eastern Coffee follows the flow chart below to serve its customers. It takes a worker two minutes to take order and receive payment, two minutes to prepare coffee, and three minutes to clean equipment. Eastern Coffee has two workers: worker A takes order and prepares coffee, while worker B handles the cleaning. Order processing takes 2 min., preparing the coffee takes 2 min., Cleaning takes 3 min. a) How many customers can Eastern Coffee serve per hour? Take Order & Receive Payment Prepare Coffee Clean Equipment

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38 Ardavan Asef-Vaziri March, 2015Capacity- Basics Problem 3 TpA = 2+2=4, TpB = 3 Capacity of worker A: 1/4 customers / hr or 60(1/4) = 15 / hr. Capacity of worker B: 1/3 / min, or 60(1/3) = 20 / hr. Capacity of the bottleneck is 15 customers/hour Western Coffee follows the same flow chart above, and each activity takes the same amount of time as Eastern. Western Coffee also has two workers: worker C only takes order and payment, while worker D handles the coffee preparation and cleaning. b) How many customers can Eastern Coffee serve per hour? TpC = 2, TpD = 2+3=5 Capacity of worker C: 60(1/2) = 30 customers/hour Capacity of worker D: 60(1/5) = 12 customers/hour Capacity of the bottleneck is 12 customers/hour

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39 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: The Impact Converging Activities 10 Activity 20 Path 2 20 Path 1 10 Activity 15 Path 2 20 Path 1 Which process has a longer flow time?. In a deterministic world they both have a flow time of 30 mins. The situation differs in real world where nothing is deterministic. Suppose instead of 20, 15, and 10, we have 16-24, 13 to 17, and 6 to 12. That means all the paths and activities have a coefficient of variations of %

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40 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem 2: The Impact Converging Activities 10 Activity 20 Path 2 20 Path 1 10 Activity 15 Path 2 20 Path 1 Which process has a longer flow time?. In a deterministic world they both have a flow time of 30 mins. The situation differs in real world where nothing is deterministic. Suppose instead of 20, 15, and 10, we have 16-24, 13 to 17, and 6 to 12. That means all the paths and activities have a coefficient of variations of %

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41 Ardavan Asef-Vaziri March, 2015Capacity- Basics Path 1 Time = 16+8rand() rand() is a random number between 0 and 1. If it is 0, Path 1 Time is 16, if it is 1, Path1 Time is 24. For all possible rands 16 ≤ Path 1 Time ≤ 24 On the same token Path 2 Time = 12+6rand() For all possible rands 12 ≤ Path 2 Time ≤ 18 The last activity time = 8+4rand() For all possible rands 8 ≤ Activity Time ≤ 12 The following is an instantiation of this situation Key Problem Flow Time: The Impact Converging Activities

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42 Ardavan Asef-Vaziri March, 2015Capacity- Basics Now suppose Path 2 is exactly the same as path 1. That is Path 1 Time = 16+8rand() Path 2 Time = 16+8rand() The last activity time = 8+4rand() The following is an instantiation of this situation Key Problem Flow Time 5: Random Number Generation Now let’s see what is the results for 10,000 instances of each if these two processes. The average flow time in the first process was mins. While in the second process was In 64% of the instancess flow time of the second process was larger than that of the first process.

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43 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem Flow time: 10,000 Instances

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44 Ardavan Asef-Vaziri March, 2015Capacity- Basics Key Problem Flow Time: Converging Activities + Common Resources Activity R2 Activity B2 Activity B1 Activity R1 Activity G All activities are [8,12] minutes. One Recourse Red, One Resource Blue, One Resource Green.

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45 Ardavan Asef-Vaziri March, 2015Capacity- Basics A local grocery store has 2 cashier stations, and 1 experienced cashier and 1 novice cashier. During a typical working day (8 hours), 120 customers will show up. The novice cashier will serve 48 customers and the experienced cashier will serve 72 customers. On average it takes 6 minutes for the novice cashier to serve one customer and 3 minutes for the experienced cashier to serve one customer. Problem 10 STOP

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46 Ardavan Asef-Vaziri March, 2015Capacity- Basics a) During the rush hours, approximate 25 customers will show up in an hour. Does the store have enough capacity for the rush hour? a)Yes, the capacity of the store is 25 customers per hour. b)Yes, the capacity of the store is 30 customers per hour. c)No, the capacity of the store is 10 customers per hour. d)No, the capacity of the store is 20 customers per hour. e)None of the above. Problem 10 On average it takes 6 minutes for the novice cashier to serve a customer 60/6 = 10 customers/hr On average it takes 3 minutes for the experienced cashier to serve a customer 60/3 = 20 customers/hr Capacity = = 30 STOP

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47 Ardavan Asef-Vaziri March, 2015Capacity- Basics b) During the day, both cashier stations on average have 2 customers waiting. On average, how long does a customer stay in the novice cashier’s waiting line? a)8 minutes b)11 minutes c)16 minutes d)20 minutes e)None of the above Problem 10 The novice cashier will serve 48/8 = 6 customers/hr R = 6 /hr T iN R = I iN 6T iN = 2 T iN = 1/3 hr or 20 min STOP

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48 Ardavan Asef-Vaziri March, 2015Capacity- Basics c) On average, how long does a customer stay in the experienced cashier’s line? a)7 minutes b)9.4 minutes c)13.3 minutes d)15 minutes e)None of the above Problem 10 On the same token the novice cashier will serve 72/8 = 9 customers/hr R = 9 /hr TR = I 9T iE = 2 T iE =2/9 hr or min STOP

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49 Ardavan Asef-Vaziri March, 2015Capacity- Basics d) On average, how long does a customer spend in the store? Problem 10 The novice and experienced cashier serve 6, and 9 customers per hour, respectively. Customers served by Novice: 6 min service time, 20 min waiting time. T = 26 min for 6/15 customers. Customers served by Experienced: 3 min service time, min waiting time. T = for 9/15 customers. (6/15)(26) + (9/15)(16.33) = 20.2 STOP

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50 Ardavan Asef-Vaziri March, 2015Capacity- Basics To shorten the waiting time, the manager does a detailed study and finds that half of the customers purchase 5 items or less, and half of the customers purchase more than 5 items. The manager decides to let the novice cashier only serve the customers that purchase 5 items or less. After the change, it turns out that both cashier stations have 1.75 customers waiting on average. Assume that the novice cashier serves all of the customers purchasing 5 items or less and the experienced cashier serves all of the customers purchasing more than 5 items. Problem 10 STOP

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51 Ardavan Asef-Vaziri March, 2015Capacity- Basics e) What is the average waiting time in the novice cashier’s line? a)13 minutes b)14 minutes c)17 minutes d)19 minutes e)None of the above Problem 10 Each of the two cashiers will serve 120/2 = 60 customers per day or 60/8 =7.5 customers/hr T iN R = I iN 7.5T iN = 1.75 T iN = 1.75/7.5 = hr or 14 min STOP

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52 Ardavan Asef-Vaziri March, 2015Capacity- Basics f) What is the average waiting time in the experienced cashier’s line? a)9 minutes b)11 minutes c)12 minutes d)14 minutes e)None of the above Problem 10 R and I are the same for Naïve and Experience Therefore, T is the same; 14 min STOP

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53 Ardavan Asef-Vaziri March, 2015Capacity- Basics Angels Inc. fabricates garage doors. Roofs are punched in a roof punching press (10 minutes per roof) and then formed in a roof forming press (5 minutes per roof). Bases are punched in a base punching press (10 minutes per base) and then formed in a base forming press (15 minutes per base), and the formed base is welded in a base welding machine (5 minutes per base). The base sub-assembly and the roof then go to final assembly where they are welded together (10 minutes per garage) on an assembly welding machine to complete the garage. Assume one operator at each station. (a) Draw a flowchart of the process. (b) What is the minimum time required to produce a garage (from starting an order to finishing it)? (c) What is the capacity of the factory in terms of garages per hour? (d) If you want to increase the capacity, what is the stage that you would put some additional resource? Problem 4 STOP

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54 Ardavan Asef-Vaziri March, 2015Capacity- Basics Flow Chart R-PunchR-Form R-Punch R-Form B-PunchB-Form B-Punch B-Form Weld Assembly STOP

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55 Ardavan Asef-Vaziri March, 2015Capacity- Basics (a) What is the Theoretical Flow Time? (The minimum time (the required to produce a garage from start to finish.) Roof: =25 Base: = 40 Problem 4 Max(25, 40) = 40 min (b) What is the capacity of the factory in terms of garages per hour? R-Punch:1/10 per min. or 6 per hr R-Form:1/5 per min. or 12 per hr B-Punch:1/10 per min. or 6 per hr B-Form:1/15 per min. or 4 per hr Welding: 1/5 per min. or 12 per hr Assembly: 1/10 per min. or 6 per hr 4 Process Capacity is 4 per hour (c) If you want to increase the capacity, what is the sub-process that you would put some additional resource? B-Form STOP

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56 Ardavan Asef-Vaziri March, 2015Capacity- Basics (d) Suppose the machines and operations of B-Punch and R-Bunch can do both Base-punching and Roof-punching. Also suppose there will be no change in the activity times. Further, make the same assumption regarding Forming operations. Problem 4 R-PunchR-Form 10 5 R-Punch R-Form B-PunchB-Form B-Punch B-Form Weld 5 Assembly 10 Assembly Punch Form 10 5 R-Punch R-Form Punch Form B-Punch B-Form Weld 5 Assembly 10 Assembly STOP

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57 Ardavan Asef-Vaziri March, 2015Capacity- Basics Punching (R+B) = 10+10=20 Forming (R+B) = = 20 Punching Capacity = 2/20 = 1/10 per min = 6 per hour Forming Capacity = 2/20 = 1/10 per min = 6 per hour Welding: 1/5 per min = 12 per hour Assembly: 1/10 per min = 6 per hour Process Capacity is 6 per hour (c) If you want to increase the capacity, what is the stage that you would put some additional resource? All the three departments except Welding. Problem 4 STOP

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58 Ardavan Asef-Vaziri March, 2015Capacity- Basics Dr. Asef has enlisted the help of one of his students, Hansen, to help him bake cookies to raise money for his newly created “Race to the Top” program at CSUN. The money will be used to equip the graduate launch with better capital resources and encourage graduate students to spend more time in the launch and help each others to learn quantitative concepts. Orders are taken over the phone in the Grad Lounge. Prof. cleans the bowl and prepares cookie dough for each specific order. It takes a total of 7 minutes for these activities. Then the Prof. spends 3 minutes to artistically spoon the dough in the shape of a Matador. Hansen (in need of extra credit) takes 2 minutes to set the oven and insert the tray. Cookies bake for 13 minutes and cool for 5 minutes. Hansen spends 3 minutes packing cookies (careful not to break Matador shaped hats on each cookie), and then collects payment from the customer – a 1 minute activity. a) Draw a flowchart and define the types of resources used. Problem 9 STOP

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59 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Flowchart, Resources, and Resource units Capital resources – CSUN’s Oven. It is a fixed asset just like the assets seen on a balance sheet such as property, land, and equipment. Human Resources – Prof. & Hansen. Any labor input used to create a product or deliver a service such as, engineers, chefs, customer service, and etc. Each resource is allocated to one or more activities, and each activity may require one or more resources. For example loading the oven requires Hansen and the Oven. Resource Units – Prof., Hansen, and the Oven are individual resource units. Next: Determine Theoretical Flow Time and Cycle Time 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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60 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Flow time, cycle time, bottlenecks. Theoretical Flow Time = = 34 minutes. The flow time is the amount of time required to produce a batch of cookies from start to finish. Cycle Time = Theoretical Bottleneck The Oven is the bottleneck = = 15 minutes. A batch of cookies can exit every 15 minutes. Therefore, Cycle Time = 15 minutes. Next: Determine unit load for each resource Determine capacity for each resource Determine the process capacity per hour 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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61 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Unit load, capacity, process capacity. Unit load for each resource: Hansen = = 6 min/unit. Prof. = = 10 min/unit. Oven = = 15 min/unit. Capacity for each resource: Hansen = 60/6 = 10 orders/hour. Kristen = 60/10 = 6 orders/hour. Oven = 60/15 = 4 orders/hour min. Process capacity per hour: The oven is the theoretical bottleneck. Therefore, process capacity = 4 orders of 1 dozen/hr. Next: How many batches can be produced in an 8 hour day ? 4 x 8 = 32? 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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62 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Production capacity per day: Starting from 0 How many batches can be produced in an 8 hour day? Starting from zero, if you answered 32 batches, it would be incorrect! Theoretical Flow Time = = 34 minutes. From start to finish it takes 34 minutes for Prof. and Hansen to produce their first batch. 8 hours x 60 minutes = 480 – 34 = 446 minutes left over Cycle Time = Bottleneck = = 15 minutes. 446 minutes left/15 minutes = = 29 batches. *cannot have work in progress when baking so round down to 29. Total batches produced per day First batch + 29 batches = 30 per day Next: Determine the utilization at full capacity for each resource : 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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63 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Utilization Unit load for each resource: Capacity for each resource: Hansen = = 6 min/unit. Hansen = 60/6 = 10 orders/hour. Prof. = = 10 min/unit. Prof. = 60/10 = 6 orders/hour. Oven = = 15 min/unit. Oven = 60/15 = 4 orders/hour min. The utilization at full capacity for each resource: Hansen = 4/10 = 40%; Prof = 4/6 = 67%; Oven = 4/4 = 100%. Alternative method for computing Utilization = Unit load/bottleneck Bottleneck = 15 minutes. Hansen = 6/15 = 40%, Prof = 10/15 = 67%, Oven = 15/15 = 100% Next: What would be the impact if we bought another oven? 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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64 Ardavan Asef-Vaziri March, 2015Capacity- Basics Resource pool, diminishing marginal returns What is the impact of buying another Oven? New oven resource pool capacity = 2 x 4 = 8 orders per hour. Bottleneck has shifted to Prof. so new Process capacity = 6 orders/hour but doubling the capacity of the oven did not double production since the bottleneck shifted to the Prof. This is an example of diminishing marginal return. Resource Pooling – increases process flow rate, capacity, and flow time, Next: What will be the impact if we cross train Prof. and Hansen Oven Bake Operation 4 orders/hour Bake Operation STOP

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65 Ardavan Asef-Vaziri March, 2015Capacity- Basics Take order wash mix spoonload & set cool bakeunload pack get pay Cross training, resource pool, process capacity. Cross training Prof. and Hansen will create a single resource pool. New unit load for resource pool = = 16 minutes Capacity of resource pool = (60)(2)/16 = 7.5/hour Process Capacity will not increase with 1 oven since the bottleneck still remains at 4 batches per hour. With 2 ovens at 8 batches per hour process capacity will increase to 7.5/hour. Next: Managerial considerations - In general, what will happen to capacity & flow time if we increase the set up batch? (if more than one batch of dough is prepared at a time). 7 min3 min2 min 13 min 5 min 3 min 1 min STOP

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66 Ardavan Asef-Vaziri March, 2015Capacity- Basics Determining the correct set up batch is a key managerial consideration. What is the ideal setup batch size? Setup Batch: the number of units produced consecutively after a set up batch. 2 things to consider: Increased Batch Setup = lower unit load and higher capacity, but inventory will increase and with it flow time. Load batch: the number of units processed simultaneously. In our cookie scenario load batch will be constrained by oven capacity. Load batch is often constrained by technological difficulties. Next: Managerial value analysis: identify & eliminate non-value adding activities. What might they be? Setup Batch Size – Managerial Decisions STOP

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67 Ardavan Asef-Vaziri March, 2015Capacity- Basics Management has determined that waste factor is much higher when cookies are made in the shape of a Matador, especially during peak hours – Hansen stresses out and breaks too many Matador cookie hats and legs. Management has also determined that customers buy CSUN cookies because of taste and sales are not affected when they are shaped in a circle. Furthermore, Prof. will be able to increase his capacity when spooning dough. The Prof. will need to find another activity to fill his artistic passions. Eliminating non-value-adding activities: it increases resource utilization, and eliminates waste. ↓ Tp or ↑ Net availability. Reducing Resource Capacity Waste ↓ Tp STOP

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68 Ardavan Asef-Vaziri March, 2015Capacity- Basics Prof. and Hansen have developed new skills baking cookies for the “No Student Left Behind” program and now they have the entrepreneurial itch and are currently trying to decide if they should buy the a very popular cookie and ice cream hangout in Westwood – Diddy Riese. The wait line is staggering, averaging between 15 and 70 people any time during mid- afternoon or at night (real case scenario, real waiting lines!). Investment Opportunity Diddy Riese STOP

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69 Ardavan Asef-Vaziri March, 2015Capacity- Basics The owner of Diddy Riese is retiring and has offered to sell his business to Prof. and Hansen at an exorbitantly high price. New businesses often fail in this neighborhood where rents are exceptionally high. Prof. and Hansen have determined they will make zero economic profit if they buy the store and are looking for a bigger return on their investment. The owner claims his store has the capacity to produce more than 3 times the number of cookies they currently produce, but they just can’t process orders fast enough! Diddy Riese may actually become much more profitable if process capacity can be increased. Mentor and student begin sharpening their pencils and look over their list of internal measures to see if they can increase process capacity by reducing unit load for processing customer orders. The two of them identify 6 key internal measures that can be re-worked to increase overall capacity for the Diddy Riese system. Internal Measures: Resource-Activity matchHigh Utilization (Low Safety Capacity) Division of Labor (Job-Simplification)High Standardization and Modularization Effective Facility LayoutClear Material Flow Pattern Flow-ShopTraining Method ImprovementTechnology Job-shopU-shaped layout Cross-trained workersShort setup time Small batch sizeInternal uniformity vs external variability Internal Measures STOP

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70 Ardavan Asef-Vaziri March, 2015Capacity- Basics Floor space is limited and Hansen noticed the entrance doors open inward and recommends they be re-configured to open outward and against the front of the building to make room for a more effective facility layout, and clear material flow pattern by using a U-shaped layout for the counters. The U-shape will increase flexibility and give Prof. and Hansen the ability to train their employees to use a flow shop design part of the time and a job shop design at other times. Internal Measures STOP

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71 Ardavan Asef-Vaziri March, 2015Capacity- Basics The existing system for taking orders makes use of job shop technique. 75% of sales are ice cream cookie sandwiches the rest are Hawaiin shaved ice & ice cream sundaes. One employee takes an order selects type of cookies, and walks around the cashier station to the other side of the store where the ice cream station is located and customer must follow the employee to select desired ice cream. The employee scoops the ice cream and makes the sandwich. There are 5 workers performing these same multiple activities. A cashier or an employee rings up the sale and collects payment. Only one cash register! Current Inefficient System STOP

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72 Ardavan Asef-Vaziri March, 2015Capacity- Basics The new layout will make use of flow shop part of the time and job shop part of the time using a U-shaped layout. 75% of orders are for cookie sandwiches. One employee will take the cookie order, place the cookies in a cup and pass it down the counter to the next employee at the ice cream station. Once the sandwich is made, the worker slides it down the counter to the cashier. Both sides of the counter will employ the same techniques. We have 2 cash registers. Future Efficient System STOP

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73 Ardavan Asef-Vaziri March, 2015Capacity- Basics When the cookie worker is shaving ice, the worker from the ice cream station will work the cookie station, take the order, and step over to the ice cream station. While these activities are transpiring, the cashier will take the next order at the cookie station. When the sandwich is prepared the worker will ring up the sale and go back to the cookie station or the ice cream station depending on whether or not the shaved ice order has been completed. Since 75% of the orders are for cookie sandwiches a portion of activities can be performed using a more efficient flow shop technique and at other times job shop. Future Efficient System STOP

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74 Ardavan Asef-Vaziri March, 2015Capacity- Basics If any station bottleneck’s, the upstream or downstream worker can immediately double the capacity at the bottleneck to quickly free the constraint and resume smooth flow. If Prof. and Hansen can decrease customer waiting time, sales should increase. Prof. calculated lost sales due to long lines at ~48.5% revealing the hidden value of Diddy Riese. Lost sales do not appear on a cash flow statement or balance sheet when appraising the value of a firm. Conclusion STOP

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