Assignment
Capacity
Key Problem 2
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.
Roof
Base
R-Punch
15
R-Form
B-Punch
3
B-Form
Capacity- Basics
8
10
Assembly
Weld
12
Ardavan Asef-Vaziri
Door
10
March, 2015
2
Key Problem 2: Flow Time
Roof
Base
R-Punch
15
R-Form
B-Punch
3
B-Form
Weld
10
12
8
Assembly
Door
10
(a) What is the Theoretical Flow Time? (The minimum time
required to produce a garage from start to finish.)
Roof Path: 15+8 = 23
Max = 25 + 10 = 35
Base Path: 3+10+12 = 25
Critical Path = Max(33,35) = 35
Theoretical Flow Time = 35
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
3
Key Problem 2: Capacity
(b) What is the capacity of the system in terms of garages per hour?
Roof
Base
R-Punch
15
R-Form
B-Punch
3
B-Form
8
Assembly
10
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.
Weld
12
Door
10
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?
Assembly: 1/10 per min. or 6 per hr. R-Punch
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
4
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
In reality, utilization of
R-Punch Utilization = 4/4 = 100%.
all the resources will be
R-Form Utilization = 4/7.5 = 53.33%.
less than what we have
B-Punch Utilization = 4/20 = 20%.
computed.
B-Form Utilization = 4/ 6 = 66.67%.
Welding Utilization = 4/5 = 80%.
This process can never
produce 4 flow units per
Assembly Utilization = 4/6 = 66.67%
hour continually.
No Process can work at 100% capacity. Impossible.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
5
Key Problem 2: Bottleneck Shift
(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.
Roof
Base
R-Punch
R-Punch
15
B-Punch
3
R-Form
8
B-Form
10
Assembly
Weld
12
Door
10
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. Process Capacity is 5 per hour
Assembly: 1/10 per min. or 6 per hr.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
6
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
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
7
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 BPunch machines and we cross-train their operations and form a
new resource pool named Punch where both R-Punch and BPunch operations can be done in this resource pool. What is the
new capacity of the system?
R-Form
R-Punch
R-Form
R-Punch
Punch
Assembly
8
15
Assembly
B-Punch
B-Punch
B-Punch
Punch
3
Capacity- Basics
B-Form
B-Form
10
Weld
Weld
10
12
Ardavan Asef-Vaziri
March, 2015
8
Key Problem 2: Pooling and Cross Training
R&B-Punch
Punch
Punch
15,3
R-Form
R-Form
Assembly
8
B-Form
B-Form
Assembly
Weld
Weld
10
12
10
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. Process Capacity is 5 per hour
Assembly: 1/10 per min. or 6 per hr.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
9
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 ?
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
10
Key Problem 2: More Than One Bottleneck
R&B-Punch
Punch
Punch
15,3
R-Form
R-Form
Process Capacity is 6 per hour
Assembly
8
B-Form
B-Form
Assembly
Weld
Weld
10
10
10
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
11
Key Problem 2: Critical Chain
R-Punch
Punch
R-Form
R-Form
15
8
B-Form
B-Punch
B-Form
Punch
3
10
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.
Capacity- Basics
Assembly
Assembly
Weld
Weld
10
10
Path 1
23
Assembly
Assembly
10
Path 2
23
Ardavan Asef-Vaziri
March, 2015
12
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
13
Lessons Learned
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 perishableit 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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
14
Problem 3: Problem 5.2 book
You May
STOP Here
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
15
Key Problem1: Flow Chart
A flowchart is a diagram that traces the flow of materials,
customers, information, or equipment through the various steps
of a process
E
C
B
A
D
B
C
F
 Capacity Metrics: Capacity, Time to Perform the Activity (Unit
Load; Tp), Cycle Time
 Flow Time Metrics: Theoretical Flow Time; Flow Time
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
16
Key Problem1: Single-Stage and Two-Stage Process
 Cycle time =
1 min
 Capacity = 1/1 per min, 60 per hour
 Theoretical Flow Time = 1 min
 Ip = 1 That is Max Ip indeed
Activity A
Tp =1 min
Activity B
Activity A
 Cycle time = 10 min
Tp =10 min
 Capacity = 1/10 per min, 6 per hour
 Theoretical Flow Time = 18 min
ActA
 Ip = 1.8 That is Max Ip indeed
 IpA = 1
ActB

IpB = 0.8
That is Max Ip indeed
CT
0
Capacity- Basics
Tp =8 min
Ardavan Asef-Vaziri
18
28
March, 2015
CT
CT
38
48
17
Key Problem1: Two-Stage Process
Activity B
Tp =10 min
Activity A
Tp =5 min
 Cycle time = 10 min
 Capacity = 6 per hr
 Theoretical Flow Time = 15 min
 Ip = 1.5


IpA = 0.5
IpB = 1
ActA
ActB
CT
0
15
25
CT
CT
35
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
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
18
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
 Cycle time = 8 min
 Capacity = 60(2/16) per hr
7.5 per hr
 Theoretical Flow Time = 16
Activity AB2
16
Capacity increased from 6 to ActAB1
7.5. Therefore, pooling and
ActAB2
cross-training can increase
throughput. We will latter
show that flow time will also
0
go down.
Capacity- Basics
Ardavan Asef-Vaziri
CT
16
24
March, 2015
CT
32
CT
40
19
Problem 3: Problem 5.2 book
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
20
Problem 3: Problem 5.2: Flow unit = 1 order of 1 dozen.
Take
Order
Wash
Mix 6
Spoon
2
load
Set 1
Bake
9
Un
load
Cool
5
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
Pack
2
Pay
1
21
Problem 3: Problem 5.2
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
22
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).
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.
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
23
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 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 minutes
Capacity is 1/14 per minute or 900(1/14) = 64.29 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) = $26
64.29(26) = $1671.5 per day
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
24
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
Capacity- Basics
A
60
AA
BC
Ardavan Asef-Vaziri
B
75
D
100
E
90
C
45
March, 2015
25
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
AA
D
A) We produce at least 45 AA: 45/60 = 75% 60
100
B) We produce at least 40 BC: 40/75 = 53.33%
C) 45/75 = 60%
B
C
BC
75
45
D) 90/100 = 90%
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
E
90
26
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?
B
A
X
A) A
60
50
B) B
J
C) C
90
D) D
C
D
E) J
Y
70
80
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
27
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
X
A)
B)
C)
D)
We produce at least 40 X: 40/50 = 80%
We produce at least 40 X: 40/60 = 66.67%
40/80 = 50%
C
Y
50/70 = 71.43%
70
Capacity- Basics
Ardavan Asef-Vaziri
B
60
A
50
J
90
March, 2015
D
80
28
What is a Process
Process analysis is the
detailed understanding and
documentation of how
work is performed and
how it can be redesigned
and improved.
Identify
opportunity
1
Define
scope
2
Document
process
3
Implement
changes
6
Redesign
process
5
Evaluate
performance
4
 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
Capacity- Basics
Ardavan Asef-Vaziri
March, 2015
29