(A) Answer: Figure: 2
Ratuonal: Total cycle time = 10 minutes
For Figure 1:
One of the stations has tasks G-D-E-F.
Task time of G-D-E-F = 2.8 + 3.6 + 3 + 2 = 11.4
The task time of a station is greater than cycle time which is not possible, therefore figure 1 is incorrect.
For figure 3:
One of the stations has tasks C-D-E-F.
Task time of C-D-E-F = 3 + 3.6 + 3 + 2 = 11.6
The task time of a station is greater than cycle time which is not possible, therefore figure 3 is incorrect.
Therefore, correct answer is figure 2 because all its work stations have task time of less than 10 minutes.
(B) Answer: 96.33%
Rationale:
Total task time = 5+1.8+3+3.6+3+2+2.8+3.7+2+2
Total task time (TT) = 28.9 minutes
Total cycle time (CT) = 10 minutes
Number of workstations (N) = 3
Efficiency = (100 × TT)/(N × CT)
Efficiency = (100 × 28.9)/(3 × 10)
Efficiency = 96.33%
(C) Answer: YES
Rationale: The efficiency can be increased to 100% if the idle time of each workstation is reduced to zero by increasing task time of tasks.
(D) Answer: 3
Rationale:
Theoretical number of workstations = Total task time/ Cycle time
Theoretical number of workstations = 28.9/10
Theoretical number of workstations = 2.89
Theoretical number of workstations = 3 (rounded off)
Therefore, theoretical number of workstations = 3
J Packing 2.0 a-b) One of the possible assignment of tasks to workstations is shown correctly...
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