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

Process Capacity

2Ardavan Asef-Vaziri June-2013Capacity- Basics

A Process; Three Sequential Activities

You Oven Friend

15 minutes 20 53 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.)

ResourceActivityTime

Operation A Operation B Operation C


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).


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 = 3Each 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 FriendOperation A Operation B Operation C

15 minutes 20 5


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 minEach 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).


15 minutes 20 5


Cycle Time

Flow Time 15 + 20 + 5 = 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

15+20 = 35. You need to wait for 5 minutes to put the 2nd batch in

the oven.


15 minutes 20 5


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 35+20 = 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


15 minutes 20 5


Flow Time


15 minutes 20 5

0 7555

Operation A

95

Operation B

Operation C

35

CT CT CT


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 240 - 40 = 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.


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) = 12By now we should know the followingsFlow 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 friend’s utilization is 5/20 = 0.25 or 25%.


Utilization

We can compute Utilization in an alternative way.

= 60/15 = 4 = 60/20 = 3 = 60/5 = 12

Activity Time

CapacityProcess Capacity = min{4,3,12} = 3Utilization = 3/4 = 0.75 = 3/3 = 1 = 3/12 = 0.25


15 minutes 20 5


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.

OvenOperation B

20

Resource Pool – A collection of interchangeable resource units that can perform an identical set of activities. Operation B


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 minYou are the bottleneck.

Capacity of Resource Pool batch/min

= (1/15)(1) =1/15 = (1/20)(2) = 1/10 = (1/5)(1)=1/5

YouOven

FriendOperation A

Operation BOperation C

15 minutes20

5Operation B



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

YouOven

FriendOperation A


15 minutes20

5Operation 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.


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.

YouOven

FriendOperation A


15 minutes20 5

Operation B

U = 4/4=1 4/6 = 0.67 4/12=0.33


Two Ovens Plus Cross Functional Workers

Resource Human OvenTime 5+15 20Capacity 2*60/20 =6 2*60/20 =6

HumanOven

Human

Operation AOperation B

Operation C

15 mins20

5Operation 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.


Resources, Resource Pools and Resource PoolingResource 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.


Christine, Roommate, Mixer, and Oven

Christine, Roommate, Mixer, and One OvenResource Christine Roommate Mixer OvenTime 8 3 6 10Capacity/hr 60/8 =7.5 60/3=20 60/6=10 60/10

= 6Christine, Roommate, Mixer, and Two OvensResource Christine Roommate Mixer OvenTime 8 3 6 10Capacity 60/8 = 7.5 60/3=20 60/6=10

120/10 =12One Oven and Cross Functional WorkersResource Human Mixer OvenTime 8+3 6 10Capacity 120/11 =10.9 60/6 = 10

120/10 =12


Christine, Roommate, Mixer, and Oven

Christine, Roommate, Mixer, and One OvenResource Christine Roommate Mixer OvenTime 8 3 6 10Capacity 1/8 1/3 1/6

1/10Christine, Roommate, Mixer, and Two OvensResource Christine Roommate Mixer OvenTime 8 3 6 10Capacity 1/8 1/3 1/6 2(1/10)

=1/5Cross Functional Workers and Two Ovens Resource Human Mixer OvenTime 8+3 6 10Capacity 2(1/11) =2/11 1/6 2(1/10)

= 1/5


Network of ActivitiesThe 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 CB) Station B and CC) Stations C and DD) Stations D and E E) Station C and E

B: 80

A: 60 D:

100

C: 45

E: 90



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 BCWe 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 = 0.5625

B: 80

A: 60 D:

100

C: 45

E: 90


Flow Time; Parallel Tasks

A B C D

2 2 4 1

Flow Time = 2+2+4+1 = 9

A

B

C D

2

2

4 1

Path 1 = 2+2+1 = 5Path 2 = 2+4+1 = 7Flow Time = 7

M1 H1 H2 M2

H1

H2 M2M1

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


Parallel Operations – Resource Pooling & Splitting Activities

A

B

C D

2

2

4 1

T = Max{5,7} = 7

H

H M2M1

Capacity M1 =30, H=10*2=20, M2=60

A

B

C1 D

2

2

3 1

H

H M2M1

C2

1

H

T = Max{6,6} = 6 Capacity M1 =30, H=10*2=20, M2=60

Documents

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