The Action Toy Company has decided to manufacture a new train set, the productio
ID: 357739 • Letter: T
Question
The Action Toy Company has decided to manufacture a new train set, the production of which is broken into six steps. The demand for the train is 4,600 units per 40-hour workweek: Performance Time (seconds) 24 27 12 15 12 28 Task Predecessors D, E This exercise only contains parts b, c, d, e, and f b) Given the demand, the cycle time for the production of the new train set = 31.3 seconds (round your response to one decimal place) c) The theoretical minimum number of workstations4 (round your response up to the next whole number) d) Using the longest operation time rule, the assignment of tasks to workstations should be: (Hint: Number workstations sequentially in terms of precedence relationships and combine any applicable tasks.)Explanation / Answer
b) Given,
Demand = 4600 units / 40 hr week
Average throughput rate = 40/4600 hrs = 31.3 sec
So, cycle rate = 31.3 sec
c) Minimum number of workstations= sum of total task times / cycle time
= (24+27+12+15+12+28)/31.3 = 3.7 = 4 workstations
Task Table
Task
Task time
Precedence
A
24
-
B
27
A
C
12
A
D
15
A
E
12
B.C
F
28
D,E
The total station time is nothing but the cycle time. So, total station time of each station is 31.3 secs.
In order of longest operation time, tasks: F>B>A>D>C>E
Workstation 1:
First task =A (as F and B are dependent on A)
Time left= 31.3-24 = 7.3 secs
So, workstation 1: A
Workstation 2:
First task =B
Time left= 31.3-27 = 4.3 sec
So, workstation 2: B
Workstation 3:
First task=D
Time left= 31.3 – 15= 16.3 sec
Second task=C
Time left=16.3-12 = 4.3 sec
So, workstation 3: D->C
Workstation 4:
First task= E
Time left=31.3 – 12= 19.3 secs
So, workstation 4: E
No workstation is left for assignment of task F, if following the theoretical minimum rule.
Workstation 5(going against the minimum workstation rule):
First task= F
Time left=31.3 – 28= 3.3 secs
Work Station
Tasks
1
A
2
B
3
D->C
4
E
5
F
Were you able to assign all the tasks to the theoretical minimum no of workstations: No
e) Total idle time = 7.3 + 4.3 + 4.3 + 19.3 +3.3 = 38.5 sec
f) Efficiency with 5 workstations = (sum of all tasks) / (no of workstations * Cycle time)
= (118)/ (5*31.3) = 0.754 or 75.9 %
Efficiency with 6 workstations = (sum of all tasks) / (no of workstations * Cycle time)
= (118)/ (6*31.3) = 0.628 or 62.8 %
Task
Task time
Precedence
A
24
-
B
27
A
C
12
A
D
15
A
E
12
B.C
F
28
D,E
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