THE NEXT THREE QUESTIONS REFER TO THE FOLLOWING:
A five stage process is depicted below. Each task must be performed in the order given.
Task time in seconds
Employees assigned
V 
15
1
W  X 
30
20
1
1
Y 
10
1
Z
45
2
1. What is the minimal possible throughput time of this process?
A.
B.
C.
D.
E.
30 seconds
45 seconds
97.5 seconds
120 seconds
165 seconds
2. If the process operates at maximum capacity (i.e., the bottleneck is always busy), what is the capacity
utilization at task Z?
A. 1
B. More than 0.8 but less than 1
C. More than 0.6 but less than 0.8
D. More than 0.4 but less than 0.6
E. 0.4 or less
3. Which of the following alternatives is best in terms of efficiency?
A.
B.
C.
D.
E.
Add a third worker to task Z
Combine tasks V and W into one work station and eliminate one worker
Combine tasks X and Y into one work station and eliminate one worker
Indifferent between answer B and C
Indifferent between answers A, B, and C
1. D
Throughput time = time per job from start to finish = 15 + 30 + 20 + 10 + 45 + any wait times. So
minimum throughput time = 120 seconds. (Note that even though there are 2 workers at Z, each
individual job still takes 45 seconds.)
2. C
Because there are two workers at Z, the average cycle time for this task is 45 / 2 = 22.5 sec and it is
NOT the bottleneck. The bottleneck is task W with a cycle time of 30 sec. In each cycle of 30
seconds, the workers at Z will be busy on average for 22.5 / 30 = 75% of the time. (Alternatively, we
could calculate the capacity per hour at W as 60 min/hr * 60 sec/min / 30 sec/unit = 120 units and the
capacity per hour at Z as 60 min/hr * 60 sec/min / 22.5 sec/unit = 160 units. Then utilization at Z =
capacity required / capacity available = 120 / 160 = 75%.)
3. C
Adding a 3rd worker to Z will increase costs but NOT capacity since Z is not the
bottleneck. Combining tasks V and W will eliminate a worker but DECREASE capacity since it will
increase the bottleneck cycle time to 45 seconds. Combining tasks X and Y will eliminate a worker
WITHOUT reducing capacity since the combined task time of 30 seconds is not any longer than the
current bottleneck of 30 seconds at task W. Consequently this last option is the best of the three.