Consider the following four-step process. Assume demand arrives at a rate of 6 j

Question

Consider the following four-step process. Assume demand arrives at a rate of 6 jobs per hour, suppose it is possible to add a dedicated worker to one of the four stations.

Number of Workers

At which station should the additional worker be placed? Clearly explain your answer.

With the additional worker placed where you suggest in part a, answer the following questions with “paper-and-pencil” calculations,

i. What is the utilization of each station? What is the bottleneck?
ii. What is the average flow time?
iii. With sufficient demand, what is the maximum throughput rate achievable by the process?
iv. What is the average throughput rate?
v. What is the average time between items?
vi. What is the average work-in-process?

Number of Workers

Explanation / Answer

The effective capacity of a station = (60 * No. of workers) / Processing time

So, for A, the effective capacity = (60*1) / 2.5 = 24 per hour

For B, the effective capacity = (60*1) / 5.0 = 12 per hour

For C, the effective capacity = (60*1) / 3.0 = 20 per hour

For D, the effective capacity = (60*1) / 4.0 = 15 per hour

Since the capacity of B is the least, the additional worker should be added in station B.

After adding one additional worker at B, the capacity of B will be (60*2) / 5.0 = 24 per hour.

(i)

Utilization = Throughput rate / capacity

For A, Utilization = 6 / 24 = 25%
For B, Utilization = 6 / 24 = 25%
For C, Utilization = 6 / 20 = 30%
For D, Utilization = 6 / 15 = 40%

As off now, there is no bottleneck as the demand is less than the capacity. The potential bottleneck is, however, D with minimum capacity of 15 per hour.

(ii)

Average flow time = 2.5 + 5.0/2 + 3.0 + 4.0 = 12 minutes

(iii)

The minimum capacity station is thus D and the capacity of the line is 15 per hour. Had there been a sufficient demand, the throughput rate would have been 15 per hour.

(iv)

Throughput rate = min {Demand, Capacity} = min (15, 6) = 6 per hour.

(v)

Time between items = 60/6 = 10 minutes

(vi)

Average WIP (use Little's Law) = Throughput rate x Flow time = 6 x (12/60) = 1.2 units

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