# Chapter 10: Solutions Manual

```Chapter
10
Layout
11. Assembly-line balancing. Longest work element rule to produce 40 units per hour.
1 1 hour
3600 sec
sec

 90
a. c  
r 40 units 40 units
unit
 t  415  4.611 or 5
b. TM 
c
90
c. Sl = {A, C, E}, S2 = {B}, S3 = {G, D}, S4 = {H, F, I}, S5 = {J, K}
FI
HK
Station
S1
S2
S3
S4
S5
Candidate(s)
A
C
E
B
D, F, G
D, F, I
F, H, I
F, I
I
J
K
Work Element
Time (sec)
40
30
20
80
60
25
45
15
10
75
15
Choice
A
C
E
B
G
D
H
F
I
J
K
Cumulative
Time (sec)
40
70
90
80
60
85
45
60
70
75
90
 t 
415
100%  
 92.2%
nc
590
Balance delay %   100%  Efficiency
d. Efficiency (%) 
 100%  92.2%
 7.8%
12. Johnson Cogs
D
40
B
E
H
30
6
20
A
G
40
15
J
C
30
50
F
I
25
18
211
Idle Time
( c  90 sec)
50
20
0
10
30
5
45
30
20
15
0
212
PART FOUR

Capacity, Location, and Layout
a. Before calculating the theoretical minimum number of stations, we find the cycle time as:
t
3600 sec hr
274
c
 60 sec unit . Then we find TM   
 4.56 or 5
60 sec unit
c
60
b. Task assignments using longest work-element time rule:
F
IJ
G
HK
Station
S1
S2
S3
S4
S5
S6
Candidates
A
B, C
B, F, G
E, F, G
D, E, G
E, G
E, I
E
H
J
Assignment
A
C
B
F
D
G
I
E
H
J
Cumulative
Time
40
50
30
55
40
55
18
24
44
30
Idle Time
( c  60 )
20
10
30
5
20
5
42
36
16
30
Six workstations are required.
c. Largest number of followers. The number of followers for each work element are:
Work
Element
A
B
C
D
E
Number
of Followers
9
4
4
2
2
Work
Element
F
G
H
I
J
Number
of Followers
2
2
1
1
0
Task assignments using largest number of followers rule:
Station
S1
S2
S3
S4
S5
S6
Candidates
A
B, C (tie)
B, F, G
E, F, G (tie)
D, E, G (tie)
E, G
E, I
H, I (tie)
I
J
Assignment
A
C
B
F
D
G
E
H
I
J
Cumulative
Time
40
50
30
55
40
55
6
26
44
30
Idle Time
( c  60 )
20
10
30
5
20
5
54
34
16
30
d. Efficiency with 5 workstations:
 t 100%   274 100%   9133%
.
Efficiency 
c
560
FI
HK
13. Baxter Bicycle Company
a. We must first determine the cycle time, expressed in seconds because the work element
times are also in seconds.
Layout
a

CHAPTER TEN
f
r  576 units day 8 hours shift @3 shifts day
 24 units per hour
c  1 24 3600  150 sec unit
a f
The sum of the work element times (t) is 515 seconds.
TM 
F t I  515  343
Hc K 150 . or 4
C
J
30
B
D
F
H
50
25
55
50
A
E
G
75
45
70
90
I
25
b. Solution using longest work-element time rule: S3 = {H,E,C,I}
Station
S1
S2
S3
S4
Candidates
A
B
D
C, E, F, G
C, E, F
C, E, H
C, E
C, I
I
J
Choice
A
B
D
G
F
H
E
C
I
J
Cumulative
Time
75
125
150
70
125
50
95
125
150
90
The third workstation would be assigned elements H, E, C, and I.
Idle Time
( c  150 )
75
25
0
80
25
100
55
25
0
60
213
```