Problem 9.15 The Action Toy Company has decided to manufacture a new train set,
ID: 354372 • Letter: P
Question
Problem 9.15 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 15 15 10 27 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 workstts (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
Available time = 40 hours = 40 x 60 x 60 = 144,000 seconds
Demand = 4600 units
Cycle time = available time/ demand = 144,000/4600
Cycle time = 31.30 seconds per unit
Total Work Content = 118 seconds
Number of workstation required = total wok content/cycle time = 118/31.30
N = 3.77
Theoretical number of WS required = 4
Group tasks such that work content of workstation is equal to or less than 31.30 seconds
Work station #
Cycle time remaining
Eligible tasks (according to precedence relationship)
Assigned task
Idle time remaining
1
31.30
A
A
7.3
None
7.3
2
31.30
B, C, D (predecessor A is assigned)
B (27)
(longest duration)
4.3
4.3
3
31.30
C, D (predecessor A is assigned)
C (15)
16.3
D (predecessor A is assigned)
E (predecessors B and C are assigned)
D (15)
1.3
1.3
4
31.30
E (predecessors B and C are assigned)
E (10)
21.30
21.30
5
31.30
F (predecessors D and E are assigned)
F (27)
4.3
4.3
WS
Tasks
1
A
2
B
3
C, D
4
E
5
F
Actual number of workstation required = 5
e. Total Idle time per cycle = 38.5 seconds
f. Efficiency with 5 workstation = (Total work content)/(cycle time x number of workstation)
Efficiency = 118/(31.30 x 5)
Efficiency = 0.7540
Efficiency = 75.40%
Efficiency with 6 workstation = (Total work content)/(cycle time x number of workstation)
Efficiency = 118/(31.30 x 6)
Efficiency = 0.6283
Efficiency = 62.83%
Work station #
Cycle time remaining
Eligible tasks (according to precedence relationship)
Assigned task
Idle time remaining
1
31.30
A
A
7.3
None
7.3
2
31.30
B, C, D (predecessor A is assigned)
B (27)
(longest duration)
4.3
4.3
3
31.30
C, D (predecessor A is assigned)
C (15)
16.3
D (predecessor A is assigned)
E (predecessors B and C are assigned)
D (15)
1.3
1.3
4
31.30
E (predecessors B and C are assigned)
E (10)
21.30
21.30
5
31.30
F (predecessors D and E are assigned)
F (27)
4.3
4.3
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