Question: SCV = 1 . The rst machine can always pull in material, and the second machine can always push material to nished goods. Between the

SCV =1. The rst machine can always pull in material, and the second machine can always
push material to nished goods. Between the two machines is a buffer that can hold only 10
jobs (see Sections 8.7.1 and 8.7.2).
a. Model the system using an M/ M/1/b queue. (Note that b =12 considering the two
machines.)
i. What is the throughput?
ii. What is the partial WIP (i.e., WIP waiting at the rst machine or at the second
machine, but not in process at the rst machine)?
iii. What is the total cycle time for the line (not including time in raw material)?(Hint:
Use Little's law with the partial WIP and the throughput and then add the process
time at the rst machine.)
iv. What is the total WIP in the line? (Hint: Use Little's law with the total cycle time
and the throughput.)
b. Reduce the buffer to one (so that b =3) and recompute the above measures. What
happens to throughput, cycle time, and WIP? Comment on this as a strategy.
c. Set the buffer to one and make the process time at the second machine equal to 10
minutes. Recompute the above measures. What happens to throughput, cycle time, and
WIP? Comment on this as a strategy.
d. Keep the buffer at one, make the process times for both stations equal to 20 minutes (as
in the original case), but set the process CVs to 0.25(SCV =0.0625).
i. What is the throughput?
ii. Compute an upper bound on the WIP in the system.
iii. Compute an approximate upper bound on the total cycle time.
iv. Comment on reducing variability as a strategy.

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related General Management Questions!