Question

A heat-treatment operation takes 6 hours to process a batch of parts with a standard deviation of 3 hours. The maximum that the oven can hold is 125 parts. Currently there is demand for 160 parts per day (16-hour day). These arrive to the heat-treatment operation one at a time according to a Poisson stream (i.e., with ca=1).

(a)What is the maximum capacity (parts per day) of the heat-treatment operation? (b)If we were to use the maximum batch size, what would be the average cycle time through the operation? (c)What is the minimum batch size that will meet demand? (d) If we were to use the minimum feasible batch size, what would be the average cycle time through the operation?

Answer 1

The demand is 160 parts per day. With a 16-hour day, this becomes = 160/16 = 10 parts per hours.

(a) The maximum capacity is k = 125 parts divided by the heat treat time of 6 hours which is 20.83 parts per hour or 333.33 parts per 16-hour day.

(b) If we assume we always do a full batch we must consider the “wait -to-batch- time” (WTBT). We can model this as a two process with a “batcher” before the oven. Once the batcher accumulates a batch, a queue of batches forms in front of the oven.

The wait-to-batch time (WTBT) would be   

The utilization would be

The SCV of arrivals would be

the SCV of process time is

Using the VUT equation we get

Adding these together with the process time, we get 6.2h + 0.714h + 6h = 12.91h

(c) The minimum batch size to meet demand is given by

Therefore,

(d)   

Hence,

Average cycle time = 3 + 47.95 + 6 = 56.95 hours

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