Based on the operations and their times, the capacity of this process is 10 per hour and the action that would yield the greatest increase would be 2. increase operation 2 by 14%.
The capacity of the process is the least operation time in the process which is 10/ hr.
Check all the alternatives given. The alternative with the lowest capacity is best.
Increase operation 1 by 12%:
= 11 x (1 + 12%)
= 12.32 per hour
Increase operation 2 by 14%:
= 10 x (1 + 14%)
= 11.4 per hour
Increase operation 3 by 10%:
= 12 x (1 + 10%)
= 13.2 per hour
The best alternative is therefore to increase operation 2 by 14%.
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