b) Cycle Time= 60*420/500 = 50.4 sec
c) Minimum number of workstations= sum of total task times / cycle time
= 195/50.4 = 3.872 = 4 workstations
Task Table
Task |
Task time |
Followers |
Followers # |
Precedence |
A |
45 |
B,C,F,G,J,K |
6 |
- |
B |
11 |
C,F,G,J,K |
5 |
A |
C |
9 |
F,G,J,K |
4 |
B |
D |
50 |
E,H,I,J,K |
5 |
- |
E |
15 |
H,I,J,K |
4 |
D |
F |
12 |
J,K |
2 |
C |
G |
12 |
J,K |
2 |
C |
H |
12 |
J,K |
2 |
E |
I |
12 |
J,K |
2 |
E |
J |
8 |
K |
1 |
F,G,H,I |
K |
9 |
- |
0 |
J |
The total station time is nothing but the cycle time. So, total station time of each station is 50.4 secs.
In order of longest task time and then on the largest no of following tasks, tasks:
D>A>E>H>I>B>C>F>G>J>K
Workstation 1:
First task =D
Time left= 50.4-50 = 0.4 secs
So, workstation 1: D
Workstation 2:
First task =A
Time left= 50.4-45 = 5.4 sec
So, workstation 2: A
Workstation 3:
First task=E
Time left= 50.4-15= 35.4 sec
Second task=H
Time left=35.4-12 = 23.4 sec
Third task=I
Time left=23.4-12 = 11.4 sec
Fourth task=B
Time left=11.4-11 = 0.4 sec
So, workstation 3: E->H->I->B
Workstation 4:
First task= C
Time left=50.4 – 9= 41.4 secs
Second task=F
Time left=41.4-12 = 29.4 sec
Third task=G
Time left=29.4-12 = 17.4 sec
Fourth task=J
Time left=17.4-8 = 9.4 sec
FIFTH task=K
Time left=9.4-9 = 0.4 sec
So, workstation 4: C->F->G->J->K
Work Station |
Tasks |
Workstation time |
Idle Time |
1 |
D |
50 |
0.4 |
2 |
A |
45 |
5.4 |
3 |
E->H->I->B |
50 |
0.4 |
4 |
C->F->G->J->K |
50 |
0.4 |
e) Total idle time = 0.4+5.4+0.4+0.4 = 6.6 sec
f) Efficiency with 4 workstations = (sum of all tasks) / (no of workstations * Cycle time)
= (195)/ (4*50.4) = 0.967 or 96.7 %
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