In this case, the utilization of effective capacity for each operation represent
ID: 464655 • Letter: I
Question
In this case, the utilization of effective capacity for each operation represents the percent of maximum capacity each operation is actually expected to maintain for next month. For example, for next month, Operation B's effective capacity is expected to be 77 percent of its maximum capacity, 455 units (= 100 + 80 + 150 + 125). Thus, Operation B is expected to product 350.35 units per hour (= 455 units * 77 percent) next month. Operation's D effective capacity can be calculated in the same way: 308.125 units per hour (= 425 units * 72.5 percent).
The following table shows the company's overall production process information for next month.
Maximun capacity
(units/ hour)
D1
D2
190
235
72.5% for D
For the company's overall production process next month:
1.Calculate maximum capacity. For this question, you need to take a look at the maximum capacity of each operation so that you can find the bottleneck in this process.
2.Calculate effective capacity. For this question, you need to take a look at the effective capacity of each operation so that you can find the bottleneck in this process.
Operation MachineMaximun capacity
(units/ hour)
Utilization of effective capacity for each operaction A A1 400 80% for A B B1 100 77% for C B2 80 B3 150 B4 125 C C1 350 95% for C DD1
D2
190
235
72.5% for D
Explanation / Answer
1. Maximum Capacity Calculation:
Operation
Machine
Maximum Capacity for current month (units/hour)
Maximum Capacity for next month (units/hour)
A
A1
400
0.8 x 400 = 320
B
B1
100
0.77 x (100 + 80 + 150 + 125) = 0.77 x 455 = 350.35
B2
80
B3
150
B4
125
C
C1
350
0.95 x 350 = 332.50
D
D1
190
0.725 x (190 + 235) = 0.725 x 425 = 308.125
D2
235
By looking at Maximum Capacity for current month for each operation, we can see that operation C has lowest Maximum Capacity, i.e., 350 units/hour.
So operation C is the bottleneck in the overall production process.
2. Effective Capacity Calculation:
Operation
Machine
Maximum Capacity for next month (units/hour)
Effective Capacity for next month (units/hour)
A
A1
320 (80% of 400)
0.8 x 320 = 256
B
B1
77 (77% of 100)
0.77 x (77 + 61.6 + 115.5 + 96.25) =
0.77 x 350.35 = 269.77
B2
61.6 (77% of 80)
B3
115.5 (77% of 150)
B4
96.25 (77% of 125)
C
C1
332.5 (95% of 350)
0.95 x 332.5 = 315.875
D
D1
137.75 (72.5% of 190)
0.725 x (137.75 + 170.375) =
0.725 x 308.125 = 223.39
D2
170.375 (72.5% of 235)
By looking at Effective Capacity for each operation for next month, we can see that operation D has lowest Effective Capacity, i.e., 223.39 units/hour.
So in this case operation D is the bottleneck in the overall production process.
Operation
Machine
Maximum Capacity for current month (units/hour)
Maximum Capacity for next month (units/hour)
A
A1
400
0.8 x 400 = 320
B
B1
100
0.77 x (100 + 80 + 150 + 125) = 0.77 x 455 = 350.35
B2
80
B3
150
B4
125
C
C1
350
0.95 x 350 = 332.50
D
D1
190
0.725 x (190 + 235) = 0.725 x 425 = 308.125
D2
235
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