ISYE 6201 Fall 2006 Homework 4 Solution

ISYE 6201 Fall 2006 Homework 4 Solution

ISyE 6201: Manufacturing Systems

Instructor : Spyros Reveliotis

Fall 2006

Solutions for Homework #4

A. Question Set

Chapter 9

Question 1

The conditions under which it is possible for a workstation to operate at 100 percent capacity over the long term and not be unstable is when there is no randomness. In most practical situations, this condition will not be satisfied.

Question 3

Both variability reduction and capacity expansion reduce wait for queue time. The difference is that capacity reduction involves only buying more machines while variability reduction involves understanding the process better – knowledge that has the potential to be transferred to other processes.

Chapter 18

Question 5

A paced assembly line does not have the problems of starving and blocking that an unpaced line would have because it is paced. The pacing mechanism is such that each station can perform its operation before the next unit arrives. This requires that the pacing be slower than the slowest station on the line. Consequently, the bottleneck of the line is the pacing mechanism itself.


For a given cycle time and fixed task processing times, the two objectives are equivalent. The expression for the total idle time is kc - ∑ti. Since ti’s are fixed, minimizing k will also minimize the total idle time for a given value of c. On the other hand, if we consider a more general problem version where the processing times and/or the cycle time are decision variables themselves, then, these two objectives can be different.

B. Problem Set

Chapter 8

Problem 3c

The availability will be A = mf /(mf + mr) = 60/(60+2) = 0.9677. The effective mean will be te = t0/A = 120/0.9677 = 124.

The effective SCV will be

ce2 = c02 + 2(1-A)Amr/t0 = 0.009375 + 2(1-0.9677)(0.9677)(120min)/120 = 0.07181.

So, ce = 0.268.

Problem 4

This is an example of a non-preemptive outage. The average number of jobs between outages will be the number corresponding to 60 hours. Since the average process time for 60 panel jobs is 2 hours, the average number of jobs between outage will be Ns = 60/2 = 30. Other parameters are

ts = 120 min, = 1202 = 14,400 min2 (since repair times are exp. distributed)

Thus, the effective mean, variance, and SCV will be:

te = t0 + ts/ Ns = 120 + 120/30 = 124 min

= 1079 min2

which is not very different from 3(c).

Chapter 9

Problem 10

(a) If we reduce the buffer sizes, the number of jobs that balk will increase, thereby decreasing TH. The maximum WIP level also decreases as does cycle time. Cycle time goes down because it is a convex function of WIP.

(b) Reducing the variability should decrease CT and the average WIP at the different stations. Furthermore, if we assume that the buffer sizes still incurred some blocking in the original setup, there is a possibility that TH will be increased slightly, since the lower WIP levels can lead to a reduction of the experienced balking.

(c) If we unbalance the line without changing rb we must add capacity to the other stations (otherwise a different station would become the bottleneck with a capacity lower than the current value of rb). If we add capacity, we decrease CT (significantly), and there is also a slight chance that we might increase TH (for reasons similar to those explained in part (b)).

(d) The opposite of b.

(e) Decreasing the arrival rate decreases TH (obviously) and also decreases utilization, which thereby decreases CT.

(f) If we decrease the variability enough we might see an increase in TH and a reduction in CT.

Problem 11

We assume there is one machine at this workstation. If demand is 900 units per day, the capacity is 1000 units per day and the utilization is 0.9.

(a) With V=1 and u=0.9 we can compute the CT in terms of te, the raw process time of the single workstation,

(b) Management desires CT to be 1.5. Therefore,

Solving this for u yields u=0.33, or 33% utilization. So (i.e., 1/) would grow from 1.111TH to 3TH and 170% increase.

(c) If V were reduced instead, the V(9) +1 = 1.5 or V = 0.05556. If we were able to reduce ca and ce equally, they would have to be at of their original values, a 76% decrease.

(d) Both a 170% increase in capacity and a 76% decrease in variability were difficult goals. A good strategy might be to request a more reasonable (initial) reduction of CT (e.g. to 3 times the raw process time) and then try to achieve this target through a combined effort of, both, increasing capacity and reducing variability as needed. Such a strategy would be in-line with the spirit of contemporary lean manufacturing for seeking and implementing incremental improvements.

Problem 27 on ALB


Task / Positional Weight
1 / 100
2 / 94
3 / 46
4 / 43
5 / 37
6 / 47
7 / 23
8 / 20
9 / 29
10 / 16
11 / 16
12 / 20
13 / 12
14 / 8
15 / 5

Ranking: 1-2-6-3-4-5-9-7-8-12-10-11-13-14-15

b) 100/30 = 3.33 implying a minimum of 4 stations would be required

Station / 1 / 2 / 3 / 4
Tasks / 1,2,6,3 / 4,5,7,9 / 8,12,10,11 / 13,14,15
Idle Time / 2 / 1 / 2 / 15

c) Start with C=25. A perfect balance would require four stations. Unfortunately this is not possible.

For C=26 we do find the following 4-station balance.

Station / 1 / 2 / 3 / 4
Tasks / 1,2,3,5 / 4,6,7,8 / 9,11,12 / 10,13,14,15
Idle Time / 0 / 0 / 1 / 3

C. Extra Credit

1. Prove Formula 2.55: E[X] = E[L]E[D] = ld


2. Prove Formula 2.56:



Therefore, we get .