Assignment Capacity

AssignmentCapacity

2Ardavan 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

8R-Form

10B-Form

15R-Punch

B-Punch12

Weld 10Assembly

3

Roof

BaseDoor


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 = 23Base Path: 3+10+12 = 25

Max = 25 + 10 = 35

Theoretical Flow Time = 35

8R-Form

10B-Form

15R-Punch

B-Punch12

Weld 10Assembly

3

Roof

BaseDoor

Critical Path = Max(33,35) = 35

Flow Time Theoretical Flow Time


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

8R-Form

10B-Form

15R-Punch

B-Punch12

Weld 10Assembly

3

Roof

BaseDoor


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. Throughput = 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.


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.

8R-Form

10B-Form

15R-Punch

B-Punch12

Weld 10Assembly

3

Roof

BaseDoor

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.


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 Welding2) Diminishing Marginal Return


Key Problem 2: Pooling and Cross Trainingg) 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

8R-

Form

B-Form

10B-

Form 12

WeldWeld

10Assembl

y

R-Punch

15R-Punch

B-Punch B-Punch

3

R-Punch

15Punch

B-Punch Punch

3

Assembly


Key Problem 2: Pooling and Cross TrainingR-Form

8R-

Form

B-Form

10B-

Form 12

WeldWeld

10Assembl

y

R&B-Punch Punch

Punch15,3

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.



Key Problem 2: Productivity Improvement -Method, Training, Technology, Managementh) 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 ?


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.


R-Form

8R-

Form

B-Form

10B-

Form 10

WeldWeld

10Assembl

y

R&B-Punch Punch

Punch15,3

Assembly


Key Problem 2: Critical ChainR-Form

8R-

Form

B-Form

10B-

Form

R-Punch

15Punch

B-Punch Punch

10

WeldWeld

Assembly

10Assembl

y

3

Assembly

10Assembl

y

23Path

2

23Path

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.


Lessons Learned1. When we relax a Bottleneck Resource, the Bottleneck

shifts to another resource2. 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.


Lessons Learned5. 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.


You May STOP Here


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

CD

F

EB

Capacity Metrics: Capacity, Time to Perform the Activity (Unit Load; Tp), Cycle Time

Flow Time Metrics: Theoretical Flow Time; Flow Time


Key Problem1: Single-Stage and Two-Stage Process Cycle time = Capacity = Theoretical Flow Time

= Ip =

Activity ATp =1 min

Activity BTp =8 min

Activity ATp =10 min

Cycle time = Capacity = Theoretical Flow Time

= Ip =

IpA = IpB =

ActA

ActB

18

CT

0 3828

CT CT

48

1 min1/1 per min, 60 per

hour 1 min1

10 min1/10 per min, 6 per hour 18 min

1.810.8

That is Max Ip indeed




Key Problem1: Two-Stage ProcessActivity BTp =10 min

Activity ATp =5 min

Cycle time = Capacity = Theoretical Flow Time

= Ip =

IpA = IpB =

ActA

ActB

0 15

CT

3525

CT 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 min6 per hr

15 min1.5

0.51


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

16Activity AB2

Cycle time = Capacity =

Theoretical Flow Time =

ActAB1

ActAB2

16CT

0 3224

CT CT

40

60(2/16) per hr

167.5 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.


Flow Time; Parallel Tasks

A B C D2 2 4 1

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

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

M1 H1 H2 M2

H1

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

A

B

C D2

2

4 1

H2 M2M1


Parallel Operations – Resource Pooling & Splitting Activities

A

B

C D2

2

4 1T = Max{5,7} = 7

H

H M2M1

Capacity M1 =(1/2)60= 30M2= (1/1)60 = 60H= (1+1)/(2+4) = 1/3 per minH=(1/3)60 = 20 per hour

A

B

C1 D2

2

3 1

H

H M2M1

C21

H

T = Max{6,6} = 6

Still two machine H But Operation C is cut in twoM1 =(1/2)60= 30M2= (1/1)60 = 60H= (1+1)/(2+1+3) = 1/3 per minH=(1/3)60 = 20 per hour


Capacity CapacityTheoretical Capacity

Flow Time Theoretical Flow Time Very Theoretical Flow Time


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 CB) Station B and CC) Stations C and DD) Stations D and E E) Station C and E

B75

A60 D

100

C 45

E90

AA

BC


Problem 7If 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 BCWe 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%

B75

A60 D

100

C 45

E90

AA

BC


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) AB) BC) CD) DE) J

B60

A50

D 80

C 70

J90

X

Y


Problem 8 Which of the following is 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%

B60

A50

D 80

C 70

J90

X

Y


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 minutes 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


TakeOrder

WashMix

6

Spoon2

loadSet 1

Cool5

Bake9

Unload

Pack2

Pay1

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 = 1+ 2+1 = 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.


d) Compute utilization at full capacity operation (if possible).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


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 5.2f) 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).


Problem 6. Problem 5.4 in the bookA company makes two products, A and B, using a single resource pool. The resource is available for 900 min per day. The contribution margins (P-V) 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 minutesCapacity is 1/14 per minute or 900(1/14) = 64.29 per

dayb) 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) = $26

64.29(26) = $1671.5 per day


Process analysis is the detailed understanding and documentation of how work is performed and how it can be redesigned and improved.

What is a ProcessIdentify

opportunity1

Implement changes

6

Definescope

2

Documentprocess

3

Evaluateperformance

4

Redesignprocess

5

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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Assignment Capacity