9
C H A P T E R
Layout Strategies
DISCUSSION QUESTIONS
1. The seven layout strategies are:
◼ Fixed-position or project layout
LO 9.5: Explain how to achieve a good process-oriented facility
2. Fixed-position layouts are complicated by: limited space
3. The advantages of a process layout are:
◼ It can simultaneously handle a wide variety of products or
equipment tends to make the overall process somewhat less
efficient.
4. Most organizations have some procedure for documenting
movement of their product. For instance, a hospital has doctor’s
orders indicating the tests and procedures that a patient is to un-
5. The advantages of a product layout are:
◼ It usually results in low variable cost per unit.
The disadvantages are:
6. Preconditions for high-volume, low-variety products are:
CHAPTER 9 LA Y O U T ST R A T E G I E S 135
Copyright ©2017 Pearson Education, Inc.
7. Three types of work cells are the small work cell within a
plant, focused work center, and focused factory.
LO 9.6: Define work cell and the requirements of a work cell
8. The advantages of work cells are:
◼ Reduction in work–in-process inventory
◼ Reduction in required floor space
The disadvantages are:
of employees
9. The requirements for a focused work center or focused factory
11. Some of the layout variables you might want to consider as
particularly important in an office where computer programs are to
be written are:
13. Retail store layout variables that a manager can manipulate
are:
◼ Overall arrangement or pattern through the store
14. Each student will sketch the layout of a local supermarket.
They should observe the long aisles, power items at aisle caps, and
spread of staples at corners of store (fruit/meat/dairy/bakery).
LO 9.2: Define the objectives of retail layout
AACSB: Application of knowledge
15. “Random stocking” is placing a unit (product) anywhere in a
16. Random stocking works best with sophisticated information
systems that rapidly identify items by bar codes (or other scannable
17. Cross-docking means that units are never put in storage in an
the shipping dock for transmittal to that final destination.
LO 9.8: Explain how to balance production flow in a repetitive or
product-oriented facility
major companies.
◼ The retailer is interested in products that will sell and pro-
vide margin—a strong inducement for any good product—
136 CHAPTER 9 LA Y O U T ST R A T E G I E S
◼ Slotting fees are the inevitable result of the free market at
work, one more item in the give and take between buyer
and seller.
◼ Slotting fees may make up half of the profit of a grocer.
◼ Since the fees are revenue for the retailers, the fees allow
lower prices.
◼ U.S. District Judge Gene Carter held in 2001, in Maine,
that slotting fees in a news company distribution case
were not illegal.
The authors see no problem with the ethics of slotting fees but
would encourage firms not to hide the nature of the payments.
ACTIVE MODEL EXERCISE
ACTIVE MODEL 9.1: Process Layout
W
0
0
0
2. Assembly (A) and Machine Shop (M) have the highest degree
G
0
B
3. Use the Swap button one swap at a time. If the swap helps,
move to the next pair. If not, click Swap to put the departments
4. Look at the two data tables and use the yellow shaded column
to put processes in rooms. What room assignments lead to the
9.1
=
+ + +
(a) Plan A movements (20 × 6) + (12 ×18) + (8 × 2)
(6 × 4) (10 × 2) (4 × 18)
= 120 + 216 + 16 + 24 + 20 + 72
9.2
= 468 (in 100s) = 46,800
Cost = 46,800 × $0.50 = $23,400
(b) Plan B movements = (20 × 6) + (8 ×18) + (12 × 2)
+ + + (10 × 4) (6 × 2) (4 × 18)
= 412 (in 100s) = 41,200
Cost = 41,200 × $0.50 = $20,600
(c) Plan C movements = (20 × 6) + (10 ×18) + (6 × 2)
+ (8 × 4) + (12 × 2) + (4 ×18)
= 120 +180 +12 + 32 + 24 + 72
= 440 (in 100s)
= 44,000
Cost = 44,000 × $0.50 = $22,000
(d) Plan B is the lowest cost, at $20,600.
9.3 (23 × 10) + (32 × 5) + (20 × 8) = 230 + 160 + 160 = 550
Cost = 550 × $2 = $1,100
9.4 This matrix includes movements in both directions:
Movement–Distance Calculations
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 =
400
M
W
D
L
G
B
M
—
125
75
0
50
60
D
—
0
150
20
L
—
20
0
Movements = (4 × 8) + (9 × 7) + (7 × 4) + (6 × 3) + (8 × 2) + (10 × 6)
CHAPTER 9 LA Y O U T ST R A T E G I E S 137
9.5
Parts Moved Between Work Centers A, B, C, and D
A
B
C
D
A
0
450
550
50
B
350
0
200
C
0
0
0
9.6 (a) Layout 1:
CHAPTER 9 LA Y O U T ST R A T E G I E S 139
9.11 (a) Takt time = Minutes available per day/Units demanded
per day
9.12
D
15
A
E
10
F
30
(e) Total idle time/cycle = 4 + 1 + 2 + 3 + 2 + 1 + 2 = 15 min.
per cycle.
Efficiency = 76.6%
**Longest operation time
9.14
5 12 60×
9.15 (a)
Performance Time
Task
(in seconds)
Predecessors
A
20
None
B
30
A
C
15
A
3,600 20seconds per PDA
(60)(60 sec)
(a) Cycle time =
180 PDAs
cycle time 9.6
number of stations
task times
(a) Theoretical minimum number of stations = cycle time
140 CHAPTER 9 LA Y O U T ST R A T E G I E S



60 min 60 sec
40 hr
(b) cycle time =
4,800 units hr min
= 30 sec per unit
Time Left
A
B
0
C
D
0
E
B
C, E
2
C
3
7
E, D
D
4
3
E
E
3
0
3
F
2
8
H
Time
Time Left
Ready
Station
Task
(minutes)
(minutes)
Tasks
1
A
5
5
B, G
G
3
2
B
CHAPTER 9 LA Y O U T ST R A T E G I E S 141
Note: Four stations with a 7-minute cycle time is possible;
Efficiency would become:
25 89.3%
47
=
(a)
(b)
or Efficiency = 7.6% with a cycle time of 3.33.
=
(c) Theoretical minimum number of
work stations Cycle time
i
t
9.19 (a) Resolving Problem 9.18 with a production time of
300 minutes per day:
13
==
13 minutes
Theoretical efficiency 0.9758
4 stations 3.33 minutes
300 minutes
Cycle time 60 units
5 minutes unit
=
=
=
==
=
6.67 minutes unit
13
Minimum number of stations Cycle time 6.675
1.95 or 2 workstations
i
t
CHAPTER 9 LA Y O U T ST R A T E G I E S 143
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.
9.22 Note: This problem does not have a computer icon next to
it, but it is possible for students to manipulate the input so that
POM for Windows or Excel OM can solve the problem.
(a) One possible layout is:
9.23 (a)
(b)
seconds = cycle tim e From (a)
N um ber of stations is at least 5
60
==
= = → = =
60(60)seconds
Cycletim e 60seconds (per iScan)
60 iScans
274
so 4.5667 5 Theoretical m inim um
60 num ber of w orkstations
nn
(seconds)
Cycletim e Cycletim e (seconds)
==
i
t
n274
274 0 7611
6 60
==
(d) O perating efficiency =
(N o. w orkstations)(Cycle tim e)
.
()
i
t