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Process Capacity. 2 Ardavan Asef-Vaziri June-2013Capacity- Basics A Process; Three Sequential Activities You Oven Friend 15 minutes 205 3 sequential activities;

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Presentation on theme: "Process Capacity. 2 Ardavan Asef-Vaziri June-2013Capacity- Basics A Process; Three Sequential Activities You Oven Friend 15 minutes 205 3 sequential activities;"— Presentation transcript:

1 Process Capacity

2 2 Ardavan Asef-Vaziri June-2013Capacity- Basics A Process; Three Sequential Activities You Oven Friend 15 minutes sequential activities; A (preparation), B (bake), and C (package and label). 3 resources; you, oven, and your friend. To produce each batch of muffin, you prepare the material, then put the batch in the oven (there is only a single oven and can bake one batch at a time), then your friend take the batch out and does packaging and labeling. The processing time at each operation is given above. This system works for four hours. 4×60 = 240 (Estimating processing times is the subject of motion and time studies.) Resource Activity Time Operation A Operation BOperation C

3 3 Ardavan Asef-Vaziri June-2013Capacity- Basics Resources, and Resource Units Capital Resources – Fixed Assets such as land, buildings, facilities, machinery, oven, etc. Human Resources – People such as engineers, operators, assemblers, chefs, customer-service representatives, you, your friend, etc. Each activity may require one or more resources and each resource may be allocated to one or more activities. A resource, a baker, may be used by several activities such as mixing, kneading and forming dough. An activity like loading an oven, may require multiple resources such as a baker and an oven. Resource Unit – An individual resource (chef, mixer, oven), or a combination of different individual resources (an operating room).

4 4 Ardavan Asef-Vaziri June-2013Capacity- Basics Capacity of a Process with Sequential Activities Capacity (per hour) = 60/15 = 4= 60/20 = 3 = 60/5 = 12 Process Capacity = Min {4,3,12} = Capacity of the bottleneck = 3 Each hour we produce 3 units. Starting from the second unit, every 60 mins a total of 3 units may enter, pass, and leave the process. 60/3 = 20 interarrival time and interdeparture time (cycle Time). You Oven Friend Operation A Operation BOperation C 15 minutes 205

5 5 Ardavan Asef-Vaziri June-2013Capacity- Basics Per Minute and Per Hour Reach the Same Results We computed capacity /hr, we could have computed the capacity per min Capacity (per min) = 1/15= 1/20 = 1/5 Process Capacity = Min{1/15. 1/20. 1/5} Capacity of the bottleneck = 1/20 per min Each min we produce 1/20 units. In 20 mins we can send out or take in one product. 20 mins interarrival time and interdeparture time (Cycle Time). You Oven Friend Operation A Operation BOperation C 15 minutes 205

6 6 Ardavan Asef-Vaziri June-2013Capacity- Basics Cycle Time Flow Time = 40 a) How long does it take to produce a batch of muffin? In a formal term, what is the flow time in this process? b) How often a batch of muffin enter (exit) this process? In a formal term, what is the Cycle time of this system? You prepare a batch and pass it to the oven at min 15. You then start the next batch and complete it at min 30. Oven is still baking the first batch. It will be done at min = 35. You need to wait for 5 minutes to put the 2nd batch in the oven. You Oven Friend Operation A Operation BOperation C 15 minutes 205

7 7 Ardavan Asef-Vaziri June-2013Capacity- Basics Cycle Time Oven is the bottleneck. Batches exit the oven every 20 mins. You also can put batches into the oven every 20 mins. At min 35 your friend can take the first batch out of the oven, and after 5 mins at min 40 he is done. First batch exits at min 40. At min 35 you put the second batch in the oven. Your friend takes it out of oven at min = 55 and send it out of the process at min 60. That is 60-40=20 mins after the first batch. Therefore the time between exit of two consecutive batches is? Cycle time is 20 min. The oven is the bottleneck. Cycle time = Max{15, 20, 5} = 20 You Oven Friend Operation A Operation BOperation C 15 minutes 205

8 8 Ardavan Asef-Vaziri June-2013Capacity- Basics Flow Time You Oven Friend Operation A Operation BOperation C 15 minutes Operation A 95 Operation B Operation C 35 CT

9 9 Ardavan Asef-Vaziri June-2013Capacity- Basics Cycle Time; Starting from 0 vs. Continual a) How many batches can you produce per day? Case 1, Starting form 0. We have 4×60 minutes. It takes you 40 minutes to produce the first batch. In the remaining = 200 mins, since cycle time is 20 mins, therefore, we produce 1 batch per 20 mins, that is 10 batches in 200 mins. We produce 1+10 = 11 per 4 hours. We could have also said that in the first 40 minutes we produce 1 batch and in the next 200 minutes we produce 1/20 batch per min. That is 1+200(1/20) = 11 per 4 hours. Case 2, Continual. Suppose we are not producing muffins but something else such that at the start of each day there are WIP of the previous day in the system. For example suppose it is a painting oven for a small part and you can make the part ready and put it into the oven at the start of the next day.

10 10 Ardavan Asef-Vaziri June-2013Capacity- Basics Utilization What is the capacity (or maximum Throughput)? The flow time is 40 mins. The cycle time is 20 mins. Therefore, capacity is 1/20 per min. In 4 hours it is 240(1/20) = 12 By now we should know the followings Flow time. Cycle time. Capacity. What is Utilization of the oven? Oven is always working. Every 20 minute 1 batch comes and 1 batch leaves. Utilization of the oven is 1 or 100%. In each 20 minutes you only work 15 mins. Your utilization is 15/20 = 0.75 or 75%. In each 20 minutes your friend only work 5 mins. Your friends utilization is 5/20 = 0.25 or 25%.

11 11 Ardavan Asef-Vaziri June-2013Capacity- Basics Utilization We can compute Utilization in an alternative way. = 60/15 = 4= 60/20 = 3 = 60/5 = 12 Activity Time Capacity Process Capacity = min{4,3,12} = 3 Utilization = 3/4 = 0.75= 3/3 = 1 = 3/12 = 0.25 You Oven Friend Operation A Operation BOperation C 15 minutes 205

12 12 Ardavan Asef-Vaziri June-2013Capacity- Basics Parallel Activities, Resource Pool Processing time = Tp = 20 minutes Resource Pool contains 2 Resource units. c =2 Capacity of a resource unit = 1/20 per min Capacity of a resource pool = 2(1/20) = 1/10 per min After how many minutes a product exist this system Every one minute 0.1 product. After how many minutes 1 product? 1/0.1 = 10 Cycle time is 10 minutes. Oven Operation B 20 Resource Pool – A collection of interchangeable resource units that can perform an identical set of activities. Operation B

13 13 Ardavan Asef-Vaziri June-2013Capacity- Basics Network of Activities Now suppose there are two ovens. Compute flow time. Capacity or maximum accessible throughput, cycle time, and utilization of each resource unit and resource pool. Capacity of Resource Unit batch/min = 1/15= 1/20 = 1/5 Process Capacity = Capacity of the bottleneck = 1/15 per min You are the bottleneck. Capacity of Resource Pool batch/min = (1/15)(1) =1/15= (1/20)(2) = 1/10 = (1/5)(1)=1/5 You Oven Friend Operation A Operation B Operation C 15 minutes 20 5 Operation B

14 14 Ardavan Asef-Vaziri June-2013Capacity- Basics Network of Activities Process Capacity = 1/15 per min Capacity of Resource Pool batch/min (1/15)(1) =1/15(1/20)(2) = 1/10 (1/5)(1)=1/5 U = (1/15)/(1/15)=1 = (1/15)/(1/10) = 0.67 = (1/15)(1/5)=0.33 You Oven Friend Operation A Operation B Operation C 15 minutes 20 5 Operation B Each min the system produces 1/15 units. In 15 mins we can send out or take in one product. Interarrival time and interdeparture time (cycle Time) = 15 min.

15 15 Ardavan Asef-Vaziri June-2013Capacity- Basics Per Minute and Per Hour Reach the Same Results Process Capacity = 4 per hour Capacity of Resource Pool batch/hr (60/15)(1) =4(60/20)(2) = 6 (60/5)(1)=12 Each hr the system produces 4 units. In 15 mins we can send out or take in one product. Interarrival time and interdeparture time (cycle Time) = 15 min. You Oven Friend Operation A Operation B Operation C 15 minutes 20 5 Operation B U = 4/4=1 4/6 = /12=0.33

16 16 Ardavan Asef-Vaziri June-2013Capacity- Basics Two Ovens Plus Cross Functional Workers ResourceHuman Oven Time Capacity2*60/20 =62*60/20 =6 Human Oven Human Operation A Operation B Operation C 15 mins 20 5 Operation B Cross functional workers and resource pooling are great operational strategies. However, in this specific example we need to be careful. We did not increase throughput. Furthermore, U of all resources is now 100%. Very risky, a small variation can reduce the capacity.

17 17 Ardavan Asef-Vaziri June-2013Capacity- Basics Resources, Resource Pools and Resource Pooling Resource Pooling – Combining separate resource pools into a single more flexible pool that is able to perform several activities. Transforming specialized resources into general purpose resources. Cross-trained workers. General purpose machines. It is a powerful operational concept that can significantly affect not only process flow rate and capacity but also flow time.

18 18 Ardavan Asef-Vaziri June-2013Capacity- Basics Christine, Roommate, Mixer, and Oven Christine, Roommate, Mixer, and One Oven ResourceChristine RoommateMixer Oven Time Capacity/hr60/8 =7.560/3=2060/6=10 60/10 = 6 Christine, Roommate, Mixer, and Two Ovens ResourceChristine RoommateMixer Oven Time Capacity 60/8 = /3=2060/6=10 120/10 =12 One Oven and Cross Functional Workers ResourceHuman MixerOven Time Capacity 120/11 = /6 = /10 =12

19 19 Ardavan Asef-Vaziri June-2013Capacity- Basics Christine, Roommate, Mixer, and Oven Christine, Roommate, Mixer, and One Oven ResourceChristine RoommateMixer Oven Time Capacity1/8 1/31/6 1/10 Christine, Roommate, Mixer, and Two Ovens ResourceChristine RoommateMixer Oven Time Capacity 1/8 1/31/6 2(1/10) =1/5 Cross Functional Workers and Two Ovens ResourceHuman MixerOven Time Capacity 2(1/11) =2/111/6 2(1/10) = 1/5

20 20 Ardavan Asef-Vaziri June-2013Capacity- Basics Network of Activities 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 AA and have a capacity of 60 units per day. Station B and C are only for Product BC and have capacity of 80 and 45 units per day, respectively. The demands for each product is 50 units per day. Which station 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: 80 A: 60 D: 100 C: 45 E: 90

21 21 Ardavan Asef-Vaziri June-2013Capacity- Basics Network of Activities Which of the following is NOT correct? A)The utilization rate of station A is at least 75% B)The utilization rate of station B at least 50% C)The utilization rate of station B is at most 56.25% D)The utilization rate of station D is 90% E)None 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 Station A- At least 45/60 = 75% Station D : 90/100 = 90% Station B - At least 40/80 = 50% Station B – At most 45/80 = B: 80 A: 60 D: 100 C: 45 E: 90

22 22 Ardavan Asef-Vaziri June-2013Capacity- Basics Flow Time; Parallel Tasks AB C D Flow Time = = 9 A B C D Path 1 = = 5 Path 2 = = 7 Flow Time = 7 M1H1H2M2 H1 H2 M2 M1 Capacity /hr M1 =30, H1=30, H2=15, M2=60 Capacity M1 =30, H1=30, H2=15, M2=60 Capacity /min M1 =1/2, 1/2, 1/4, 1/1 Capacity = 15 per hour

23 23 Ardavan Asef-Vaziri June-2013Capacity- Basics Parallel Operations – Resource Pooling & Splitting Activities A B C D T = Max{5,7} = 7 H H M2 M1 Capacity M1 =30, H=10*2=20, M2=60 A B C1 D H H M2 M1 C2 1 H T = Max{6,6} = 6 Capacity M1 =30, H=10*2=20, M2=60


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