A manufacturer plans to assemble electrostatic copiers on an assembly line. The
ID: 382234 • Letter: A
Question
A manufacturer plans to assemble electrostatic copiers on an assembly line. The precedence relationships and activity times in minutes are given below.
Indicate your most efficient arrangement for the assembly line if 5 workstations are to be used.
No. of Stations
Work at Each Station
Cycle Time (CT),
Output in 480 mins,
Efficie-ncy
Idle Time
/cycle,
activity and minutes
mins
pieces
%
mins
5
No. of Stations
Work at Each Station
Cycle Time (CT),
Output in 480 mins,
Efficie-ncy
Idle Time
/cycle,
activity and minutes
mins
pieces
%
mins
5
Explanation / Answer
First, we calculate the Takt time:
Number of Stations = Total task Cycle Time / Takt Time
No. of stations = 5
Total task cycle time = 10+11+5+4+12+3+7+11+3 = 66
Takt time = 66/5 = 13.2
Hence, we will try to keep total time at all stations less than 14 minutes (rounding up to next whole number).
Now, we need to assign 9 tasks to 5 stations
First we need to make a precedence table :
Task Time Precendence
A 10 -
B 11 A
C 5 B
D 4 B
E 12 A
F 3 C,D
G 7 F
H 11 E
I 3 G,H
Now, we make our task assignment table:
Lets start with station 1 and task A.
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1 A A 10 10 4
2
3
4
5
Now since remaining time at station 1 is 4, eligible tasks are D, F, I
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
3
4
5
Now, we assign task B to station 2.
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2 B B 11 11 3
3
4
5
Now since remaining time at station 2 is 3, eligible tasks are F, I
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
B
F,I
B
F
11
3
11
14
3
0
3
4
5
Now, we assign task C to station 3.
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
B
F,I
B
F
11
3
11
14
3
0
3 C C 5 5 9
4
5
Next eligible task is G with task time = 7 and I with task time = 3
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
B
F,I
B
F
11
3
11
14
3
0
3
C
G,I
C
G
5
7
5
12
9
2
4
5
Now, we move on to station 4.
Tasks left to assign are : E, H , I
Hence, we assign E to station 4.
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
B
F,I
B
F
11
3
11
14
3
0
3
C
G,I
C
G
5
7
5
12
9
2
4 E E 12 12 2
5
Now, we move on to station 5.
Tasks left to assign are : H , I
Station Eligible Task Assigned Task Task Time Cumulative Time Remaining Time
1
A
D,F,I
A
D
10
4
10
14
4
0
2
B
F,I
B
F
11
3
11
14
3
0
3
C
G,I
C
G
5
7
5
12
9
2
4 E E 12 12 2
5
H,I
I
H
I
11
3
11
14
3
0
Hence, we have our allocations
Station Tasks Time
1
A
D
14
2
B
F
14
3
C
G
12
4 E 12
5
H
I
14
Critical Path = A-B-C-F-G-I
Cycle time (from critical path)= 10 + 11 + 5 + 3 + 7 + 3 = 39 minutes
Output = 480/39 = 12 units
Efficiency = Sum of all task times / No. of work station * maximum time for a station = 66/5*14 = 94.2%
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