Please Help Marlyn Hart, operations manager at Hart Electronics, faces assembly-
ID: 392754 • Letter: P
Question
Please Help
Marlyn Hart, operations manager at Hart Electronics, faces assembly-Ine balancing prob ems. She hasbeen told that the firm needs to complete 1,400 elactronic relays par day. Thera are 420 minures of productve time n each working day (which is equalant to 25,200 seconds). Th assemblytine actties are gvan below Must Follow Must Fallow Task Time (seconds Task Task Time (seconds) Task F, G 12 15 C. K Using longest task (operating) tme heuristic, group the assembly-ine activities into appropriate workstations. Complete the folowing table. H, I,J The actuel efficiency of the balance is % enter your response rounded" one decin s' place.Explanation / Answer
a) Given,
Demand = 1400 units / day = 1400 units/25200 sec day
Average throughput rate = 25200/1400 = 18 sec
So, cycle rate = 18 sec
Minimum number of workstations= sum of total task times / cycle time
= (13+4+9+11+6+12+5+6+7+5+4+15)/18 = 97/18 = 5.38 workstations = 6 workstations
b) Task Table
Task
Task time
Precedence
A
13
-
B
4
A
C
9
B
D
11
-
E
6
D
F
12
E
G
5
E
H
6
F,G
I
7
H
J
5
H
K
4
I,J
L
15
C,K
‘
The total station time is nothing but the cycle time. So, total station time of each station is 18 sec
In order of longest task time, tasks: L>A>F>D>C>I>E>H>G>J>B>K
Workstation 1:
First task =A (As L is dependent on C)
Time left= 18-13= 5 sec
Second task =B
Time left= 5-4= 1 sec
So, workstation 1: A->B
Workstation 2:
First task =D
Time left= 18-11 = 7 sec
Second task =E
Time left= 7-6 = 1 sec
So, workstation 2: D->E
Workstation 3:
First task =F
Time left= 18-12 = 6 sec
Second task =G
Time left= 6-5 = 1 sec
So, workstation 3: F->G
Workstation 4:
First task =C
Time left= 18-9 = 9 sec
Second task =H
Time left= 9-6 = 3 sec
So, workstation 4: C->H
Workstation 5:
First task =I
Time left= 18-7 = 11 sec
Second task =J
Time left= 11-5= 6 sec
Third task =K
Time left= 6-4= 2 sec
So, workstation 5: I->J->K
Workstation 6:
First task =L
Time left= 18-15 = 3 sec
So, workstation 6: L
Work Station
Tasks
Workstation Time
Idle Time
1
A->B
17 sec
1 sec
2
D->E
17 sec
1 sec
3
F->G
17 sec
1 sec
4
C->H
15 sec
3 sec
5
I->J->K
16 sec
2 sec
6
L
15 sec
3 sec
Were you able to assign all the tasks to the theoretical minimum no of workstations: Yes
Total idle time = 1+1+1+3+2+3= 11 sec
Efficiency with 6 workstations = (sum of all tasks) / (no of workstations * Cycle time)
= (97)/ (6*18) = 0.898 or 89.8 %
Task
Task time
Precedence
A
13
-
B
4
A
C
9
B
D
11
-
E
6
D
F
12
E
G
5
E
H
6
F,G
I
7
H
J
5
H
K
4
I,J
L
15
C,K