CHAPTER 9
DISCUSSION QUESTIONS
4.
(a)
(b)
5.
(a)
(b)
The advantages of a product layout are:

The use of special purpose equipment can make the overall process more efficient

Individual workloads can usually be made to be relatively equivalent

It usually results in low variable cost per unit

It tends to keep material handling costs relatively low

It reduces work-in-process inventories

It makes training and supervision easier than with other layout strategies
The disadvantages are:

High volume is required because of the large investment needed to set-up the process

Work stoppage at any point ties up the whole operation

There is a lack of flexibility in handling a variety of products or production rates
The advantages of a process layout are:

It can simultaneously handle a wide variety of products or services, especially in terms
of “batches” or “job lots”

It has considerable flexibility with respect to equipment and labor assignments
The disadvantages of a process layout are:

The use of general purpose rather than special purpose equipment tends to make the
overall process somewhat less efficient

Orders take more time and money to move through the system because of the difficult
scheduling, setting up the process for a wide variety of orders, and considerable
material handling

Labor skill requirements tend to be high because of the use of general purpose equipment

Work-in-process inventories tend to be high
6.
The advantages of work cells are:

Reduction in work-in-process inventory

Reduction in required floor space

Reduced raw material and finished goods inventory

Reduced direct labor cost

Heightened sense of employee participation

Increased utilization of equipment and machinery

Reduced investment in machinery and equipment
The disadvantages are:

Similar to a product layout

High volume is required because of the large investment needed to set-up the process

There is a lack of flexibility in handling a variety of products or production rates

Requires the use of group technology

Requires a high level of training and flexibility on the part of employees

Either considerable staff support or imaginative employees are needed for the initial
development of the work cells
17.
A heuristic is a “rule of thumb” method of problem solving. Line balancing heuristics include:
longest task time, most following tasks, ranked positional weights, shortest task time, and least
number of following tasks.
END-OF-CHAPTER PROBLEMS
Chapter 9: Layout Strategy
1
9.1
Interstation Activity Matrix
Pickup
Paperwork
Advising
and Forms
Station
(A)
(B)
0
450
250
0
0
0
0
0
Pickup
Class
Cards
(C)
550
200
0
0
Verification
of Status
and Payment
(D)
50
0
750
0
Interstation Distance Matrix-Original Layout
Pickup
Pickup
Paperwork
Advising
Class
and Forms
Station
Cards
(A)
(B)
(C)
Paperwork/Forms (A)
0
30
60
Advising (B)
30
0
30
Class Cards (C)
60
30
0
Verification/Payment (D)
90
60
30
Verification
of Status
and Payment
(D)
90
60
30
0
Paperwork/Forms (A)
Advising (B)
Class Cards (C)
Verification/Payment (D)
Load  Distance
A  B:
450  30  13,500
A C:
550  60  33,000
A  D:
50  90  4,500
B  A:
250  30  7,500
B  C:
200  30  6,000
C  D:
750  30  22,500
87,000
Initial layout:
A
2
B
C
D
Instructor’s Solutions Manual t/a Operations Management
Interstation Distance Matrix-Improved Layout
Pickup
Pickup
Paperwork
Advising
Class
and Forms
Station
Cards
(A)
(B)
(C)
Paperwork/Forms (A)
0
30
30
Advising (B)
30
0
60
Class Cards (C)
30
60
0
Verification/Payment (D)
60
90
30
Load  Distance
A  B:
450  30 
A  C : 550  30 
A  D:
50  60 
B  A:
250  30 
B  C:
200  60 
C  D:
750  30 
Verification
of Status
and Payment
(D)
60
90
30
0
13,500
16,500
3,000
7,500
12,000
22,500
75,000
Improved layout:
B
9.2
A
C
D
This matrix includes movements in both directions:
M
W
D
L
G
B
M
–
Chapter 9: Layout Strategy
W
125
–
D
75
0
–
L
0
75
0
–
G
50
0
150
20
–
B
60
0
20
0
0
–
3
Here are two optimal layouts:
Room 1
Room 2
Room 3
B
M
W
D
G
L
Room 4
Room 5
Room 6
Room 1
Room 2
Room 3
W
M
B
L
G
D
Room 4
Room 5
Room 6
or
M  W : 125  20 
M  D:
75  40 
M  G:
50  20 
M  B:
60  20 
W  L:
75  20 
D  G : 150  20 
D  B:
20  20 
L  G:
20  20 
9.3
Movement-Distance Calculations
2,500
3,000
1,000
1,200
1,500
3,000
400
400
13,000 = Minimum distance movement for both (symmetrical layouts)
Current layout:
Entrance
1
Exam I
2
Exam II
3
X-ray
4
Lab, EKG
5
O.R.
6
R.R.
7
Casts
8
Patient movement = 6,700 feet
4
Instructor’s Solutions Manual t/a Operations Management
Improved layout:
Entrance
1
Exam I
2
Lab, EKG
5
O.R.
6
Exam II
3
X-ray
4
R.R.
7
Casts
8
Patient movement = 4,800 feet (shown in Figure 9.17 of text)
Improved layout:
Entrance
1
Exam I
2
Lab, EKG
5
O.R.
6
Exam II
3
X-ray
4
Casts
8
R.R.
7
Patient movement = 4,700 feet
More improved layout (with Exam rooms swapped):
Entrance
1
Exam II
3
Lab, EKG
5
O.R.
6
Exam I
2
X-ray
4
Casts
8
R.R.
7
Patient movement = 4,500 feet
Note that this final solution reflects “common sense,” or the result of one’s intuition. We simply
allocate the spaces in the order in which the tasks are usually performed. We usually start at the
“entrance,” proceed to an examination room; from the examination room, proceed for either x-rays or
lab tests, then (a) leave the facility, (b) go to get a cast put on, or (c) go for surgery.
Chapter 9: Layout Strategy
5
9.4
Layout 1:
Refrig.
(1)
1
0
5
3
3
0
1
2
3
4
5
Counter
(2)
2
8
0
12
0
8
Trip Matrix
3
4
13
0
3
3
0
4
0
0
4
10
Sink
(3)
5
0
8
0
5
0
Storage
(4)
1
2
3
4
5
1
0
4
8
12
16
Stove
(5)
Distance Matrix
2
3
4
4
8
12
0
4
8
4
0
4
8
4
0
12
8
4
5
16
12
8
4
0
  Tij  Dij  600 with rooms fixed (504, if not fixed; 560, if the sink is fixed in one location)
Layout 2:
Sink
(3)
Refrig.
(1)
1
2
3
4
5
1
0
5
3
3
0
Storage
(4)
Counter
(2)
2
8
0
12
0
8
Trip Matrix
3
4
13
0
3
3
0
4
0
0
4
10
5
0
8
0
5
0
Stove
(5)
1
2
3
4
5
1
0
7
8
12
14
Distance Matrix
2
3
4
7
8
12
0
5
6
5
0
4
6
4
0
7
9
6
5
14
7
9
6
0
  Tij  Dij  602 (if rooms are fixed; 566 if not; and 595 if sink is fixed at one location)
9.5
Layout 3:
Refrig.
(1)
Counter
(2)
Sink
(3)
Storage
(4)
Stove
(5)
6
Instructor’s Solutions Manual t/a Operations Management
1
2
3
4
5
1
0
5
3
3
0
2
8
0
12
0
8
Trip Matrix
3
4
13
0
3
3
0
4
0
0
4
10
5
0
8
0
5
0
1
2
3
4
5
1
0
4
8
12
14
Distance Matrix
2
3
4
4
8
12
0
4
8
4
0
3
8
3
0
10
8
4
5
14
10
8
4
0
  Tij  Dij  564 (if rooms are fixed; 474 if unfixed)
9.6
(a)
Layout 4:
Counter
(2)
Sink
(3)
Storage
(4)
Refrig.
(1)
1
2
3
4
5
Stove
(5)
1
0
5
3
3
0
2
8
0
12
0
8
Trip Matrix
3
4
13
0
3
3
0
4
0
0
4
10
5
0
8
0
5
0
1
2
3
4
5
1
0
5
8
11
13
Distance Matrix
2
3
4
5
8
11
0
4
8
4
0
4
8
4
0
11
8
5
5
13
11
8
5
0
  Tij  Dij  609
Chapter 9: Layout Strategy
7
(b)
Layout 5
ref-1
coun-2
sink-3
stor-4
stov-5
1
0
5
3
3
0
2
8
0
12
0
8
Trip Matrix
3
4
13
0
3
3
0
4
0
0
4
10
5
0
8
0
5
0
ref-1
coun-2
sink-3
stor-4
stov-5
1
0
4
12
12
4
Distance Matrix
2
3
4
4
12
12
0
3
4
3
0
4
4
4
0
4
4
3
5
4
4
4
3
0
  Tij  Dij  478
Solution: Refrig in Room 3 (where sink was); Counter in Room 4 (where storage was); Sink in
Room 2 (where counter was); Storage in Room 1 (where Refrig was); Stove in Room 5 (no
change)
9.7
Task
A
B
C
D
E
F
(b)
(c)
Performance Time
(in minutes)
1
1
4
1
2
4
13
Task Must Follow
This Task
—
—
A, B
B, C
D
E
400 minutes
 5 minutes unit
80 units
 ti  13  2.6  3 stations
Minimum number of stations 
CT
5
Cycle time 
(a) &
(d)
Station 2
(1)
B
(1)
A
Station 4
(1)
D
(4)
C
(2)
E
(4)
F
Station 3
Station 1
(e)
Efficiency 
13 minutes
 0.65  65%
4 stations  5 minutes
Multiple layouts are possible.
8
Instructor’s Solutions Manual t/a Operations Management
9.8
Performance Time
(in minutes)
4
7
6
5
6
7
8
6
49
Task
A
B
C
D
E
F
G
H
Task Must Follow
This Task
—
—
A, B
C
D
E
E
F, G
480 minutes
 9.6 minutes
50 units
 ti  49  51
Minimum number of stations 
.  6 stations
cycle time 9.6
Cycle time 
Station 2
(7)
B
Station 1
(4)
A
Station 4
(5)
D
Station 5
(6)
E
(6)
C
Station 3
Station 7
(7)
F
Station 8
(6)
H
(8)
G
Station 6
Efficiency 
49 minutes
 0.638
8 stations  9.6 minutes*
or
Efficiency = 63.8%
*
9.9
(a)
Longest operation time
A
B
C
D
E
F
G
(b)
H
There are multiple alternatives. Here is one that does not violate any procedures.
Station
Chapter 9: Layout Strategy
Tasks
9
1
2
3
4
(c)
A, D
B, G
C, F
H, E
Each alternative has an overall efficiency of 92.5%.
9.10
Task
A
B
C
D
E
F
G
H
I
J
Performance Time
(in minutes)
1
3
2
4
1
3
2
5
1
3
25
Task Must Follow
This Task
—
A
B
B
C, D
A
F
G
E, H
I
400 minutes
 6.67 minutes unit
60 units
 ti  25  3.75 or 4 workstations
Minimum number of stations 
cycle time 6.67
Cycle time 
Note: Four stations with a 7-minute cycle time is possible. However, efficiency becomes
25
 89.3%
47
and production drops to 57 from the required 60 units. One five-station solution (there are multiple
answers) is:
Station 2
(1)
A
(2)
C
(3)
B
(1)
E
(4)
D
Station 1
(3)
F
Station 5
(2)
G
(1)
I
(3)
J
Station 3
(5)
H
Station 4
Theoretical:
10
Instructor’s Solutions Manual t/a Operations Management
Efficiency 
25 minutes
 0.749  75%
5 stations  6.67 minutes
Efficiency 
25 minutes
 0.837  84%
5 stations  6.00 minutes
Actual:
9.11
Performance Time
(in minutes)
1
1
2
1
3
1
1
2
1
13
Task
A
B
C
D
E
F
G
H
I
(1)
B
Task Must Follow
This Task
—
A
A
C
C
C
D, E, F
B
G, H
(2)
H
Station 3
(1)
A
(1)
I
(1)
D
(2)
C
Station 2
(3)
E
Station 1
(1)
F
Efficiency 
(1)
G
Station 5
Station 4
13 minutes
 0.78
5 stations  3.33 minutes
or Efficiency = 78%. Multiple layouts with this efficiency exist.
9.12 (a)
Resolving Problem 9.11 with a production time of 300 minutes per day:
300 minutes
 5 minutes unit
60 units
 ti  13  2.6 or 3 workstations
Minimum number of stations 
cycle time 5
Cycle time 
Chapter 9: Layout Strategy
11
(1)
B
(2)
H
(1)
A
(1)
I
(1)
D
(2)
C
(3)
E
Station 1
Station 2
(1)
G
Station 3
(1)
F
Efficiency 
13 minutes
 0.867
3 stations  5 minutes
or Efficiency = 86.7%. Multiple layouts with this efficiency exist.
(b)
Resolving Problem 9.11 with a production time of 400 minutes per day:
400 minutes
 6.67 minutes unit
60 units
 ti  13  1.95 or 2 workstations
Minimum number of stations 
cycle time 6.675
Cycle time 
(1)
B
(2)
H
(1)
A
Station 3
(1)
I
(1)
D
Station 1
(2)
C
(3)
E
Station 2
(1)
G
(1)
F
Efficiency 
13 minutes
 0.649
3 stations  6.67 minutes
or Efficiency = 64.9%. Multiple layouts with this efficiency exist.
operating time
CT
min. cycle time = 3 minutes (Activity E time)
400
 133.33 or 133
output 
3
9.13 Output 
12
Instructor’s Solutions Manual t/a Operations Management
9.14
Performance Time
(in minutes)
0.1
0.1
0.1
0.2
0.1
0.2
0.1
0.1
0.2
0.1
0.2
0.2
0.1
1.8
Task
A
B
C
D
E
F
G
H
I
J
K
L
M
Cycle time 
Minimum number of stations 
Task Must Follow
This Task
—
A
B
B
B
C, D, E
A
G
H
I
F
J, K
L
5 hours  5 days  60 minutes hours 1,500

= 0.5 minutes bottle
3,000 bottles
3,000
.
 ti  18
 3.6 or 4 workstations
cycle time 0.5
(0.1)
E
Station 3
Station 4
(0.1)
A
(0.1)
B
(0.2)
D
(0.2)
F
(0.1)
G
(0.2)
L
(0.1)
M
(0.1)
J
(0.1)
C
Station 1
(0.2)
K
(0.2)
I
(0.1)
H
Efficiency 
Station 2
18
. minutes
 0.90
0.5 minutes  4 stations
or
Efficiency = 90%
9.15 Resolving Problem 9.14 with double the original demand:
Cycle time 
Minimum number of stations 
Chapter 9: Layout Strategy
5 hours  5 days  60 minutes
 0.25 minutes unit
6,000 units
 ti
cycle time

18
.
 7.2 or 8 workstations
0.25
13
Station 6
(0.1)
E
Station 9
(0.1)
A
(0.1)
B
(0.2)
D
Station 1
(0.2)
F
Station 2
(0.2)
K
Station 7
(0.1)
C
(0.2)
L
Station 8
(0.1)
M
Station 10
(0.1)
J
Station 5
(0.2)
I
(0.1)
G
(0.1)
H
Station 4
Station 3
Efficiency 
18
. minutes
 0.72
0.25 minutes  10 stations
or Efficiency = 72%. Multiple layouts exist with the same efficiency.
9.16
Performance Time
(in minutes)
5
3
4
3
6
1
4
2
28
Task
A
B
C
D
E
F
G
H
Task Must Follow
This Task
—
A
B
B
C
C
D, E, F
G
Minimum number of stations 
 ti
cycle time

28
 4 workstations
7
The work activities may be grouped, however, into no fewer than five workstations without violating
precedence requirements.
(4)
C
Station 4
Station 1
(5)
A
(6)
E
(3)
B
(1)
Station 3 F
(4)
G
(2)
H
Station 5
(3)
D
Station 2
14
Instructor’s Solutions Manual t/a Operations Management
Efficiency 
28
28

 0.80
5  7 35
or
Efficiency = 80%
Several other balances are also possible. One of them is to place A alone, tasks B and C together, D
and F together, E by itself, and G and H together.
9.17
Task
A
B
C
D
E
F
G
H
I
J
K
L
Performance Time
(in minutes)
13
4
10
10
6
12
5
6
7
5
4
15
97
Task Must Follow
This Task
—
A
B
—
D
E
E
F, G
H
H
I, J
C, K
Cycle time 
Minimum number of stations 
25,200 seconds
 18 seconds unit
1,400 units
 ti  97  5.4 or 6 workstations
cycle time 18
Station 1
(13)
A
(4)
B
Station 6
(15)
L
(10)
C
(12)
F
Station 4
(7)
I
Station 2
(10)
D
(6)
E
Station 3
(5)
G
Chapter 9: Layout Strategy
(6)
H
Station 5
(4)
K
(5)
J
15
Total idle time  11 seconds
Efficiency 
97 seconds
 0.898
6 stations  18 seconds
or
Efficiency  89.8%
9.18 There are three alternatives:
Station
1
2
3
4
5
Alternative 1 Tasks Alternative 2 Tasks Alternative 3 Tasks
A, B, F
A, B
A, F, G
C, D
C, D
H, B
E
F, G
C, D
G, H
E
E
I
H, I
I
Each alternative has an efficiency of 86.67%.
CT 
16
OT
480
implies 3 
, so output  160
Demand
output
Instructor’s Solutions Manual t/a Operations Management