ISYE 6201 Spring 2006 Homework 6 Solution

ISyE 6201: Manufacturing Systems

Instructor : Spyros Reveliotis

Spring 2006

Solutions for Homework #6

A. Questions

Chapter 7

Question 5

By the definition of the critical WIP, , where rb denotes the bottleneck rate and ti denotes the processing time at the i-th station of the line. Also, by the definition of the bottleneck rate rb,

(i)rb = Nb/tb, where Nb is the number of machines at a bottleneck workstation and tb is the processing time at that station;

(ii)rb  ri = Ni/ti,  rbti  Ni, i=1,…,n, where Niand ti are respectively the number of machines and the processing time at station i.

Hence,

More intuitively, W0 denotes the minimum WIP that can ensure 100% utilization of the system bottlenecks, in a deterministic operational setting. Keeping a WIP which is larger than the total number of the machines in the system, will only increase the waiting time of the jobs, without enhancing the utilization of the system bottlenecks.

Finally, notice that only when rbti = Ni, i=1,…,n, i.e., only when the line is balanced.

Chapter 8

Question 1

The coefficient of variation represents relative variability, i.e., it characterizes the st. dev. of the considered random variable as a percentage of its mean value. For instance, the standard deviation, alone, is not very meaningful when considering process times that can range from fractions of seconds to several hours.

Question 6

For the M/M/1 case, the number of customers is adequate to fully define the state of the system because process and inter-arrival times are memoryless. Thus, knowledge of how long a station has been working on a particular job or how much time has passed since the last arrival, are irrelevant. This is not the case for the G/G/1 system. In a G/G/1 queuing station, the time that a particular job has spent in processing and the time since the last arrival are important pieces of information for analyzing the future evolution of the system, and therefore, they should be part of the state definition.

Chapter 9

Question 3

Both variability reduction and capacity expansion are mechanisms that can be employed in order to reduce the job cycle time, in particular, the experienced waiting time. The difference is that capacity expansion is a rather “brute force” approach, since it relies on buying more machines, while variability reduction involves understanding and managing the process better (with all the potential benefits from such an exercise).

Question 5

While the instance of “worst case” performance materializes under the assumption of deterministic processing times, it does involve variability when considering the inter-arrival times experienced by the various single jobs in the batch. More specifically, at any given station i, the inter-arrival time between the last job of batch j and the first job of batch j+1 is equal to Wti-1, while the inter-arrival times between the remaining jobs in batch j are equal to zero.

Question 6

In this case, the variability will prevent the effective synchronization of the plant operation with the upstream and downstream processes. More specifically, although there will be maximum throughput with a minimum WIP of 4 jobs, even with variable process times, there is still the issue of raw material and finished goods inventories. If customers showed up just when we completed the production of one product, there would be no finished goods inventory. Of course, this is hardly likely and therefore we would either have finished goods inventory or customers would have to wait. In either case, there would be a buffer. We would need similar coordination for raw materials or else we would need to keep a raw material inventory buffer or else periodically starve the system for WIP.

B. Problems

Chapter 7

Problem 1

rb = 0.5 job/hr and T0= 5 hr

Adding another machine at station two makes the parameters

rb = 1 job/hr and T0= 5 hr

so the bottleneck rate increases and the raw process time stays the same. This will decrease cycle time and increase throughput when WIP is greater than the critical WIP.

Speeding up station two makes the parameters

rb = 1 job/hr and T0= 4 hr

and so it increases the bottleneck rate and decreases the raw process time. In this case, throughput will go up and cycle time will decrease regardless of the WIP level.

There is no change in the parameters from (a). Since there is no variability, there will be no change in performance.

This will make the parameters

rb = 0.5 job/hr and T0= 4.5 hr

and it will reduce the cycle time and increase the throughput and, but only when WIP is below the critical WIP. So, in this (deterministic) case, it is true that saving an hour at a non-bottleneck does not help much.

Problem 2

When all jobs are processed before moving, we have the worst case performance with cycle time given by CT = wT0 and TH=1/T0.

In this new setting, the base parameters rb and T0 will be the same with those in Problem 1, for all four cases (a-d). However, the impact of the suggested changes on the system performance can be different, as discussed below.
Since T0 remains the same as in (a), there is no change from (a) regarding cycle time and throughput. This is in contrast with Problem 1.
Speeding up station 2 reduces the raw process time and therefore reduces cycle time and increases throughput for all WIP levels.
If all the jobs are worked on by only one machine at station 1 (as assumed in the worst-case scenario, since the batch moves as a single block), there is absolutely no change in performance.
If station one is speeded up this will reduce the raw process time as in Problem 1 and so will improve performance for all WIP levels.

Problem 3

The practical worst case assumes exponential process times and a balanced line using only a single machine at each station.

Similar with Problem 2, the base parameters for the system do not change, but the observed performance sometimes does.

Throughput will increase and cycle times will decrease for all WIP levels.

Speeding up station 2 reduces the raw process time and therefore reduces cycle time and increases throughput for all WIP levels.

Adding a second machine at a non-bottleneck will improve throughput and reduce cycle times for all WIP levels above WIP = 1.

Speeding up station one will improve performance for all WIP levels.

Problem 6

rb = 2000 per day = 125 per hour, and T0= 0.5 hr, W0 = 62.5 cases, TH = 1700 cases/day = 106.25 per hour, CT = 3.5 hours.

Average WIP level = TH*CT = 106.25 cases per hour * 3.5 hours = 372 cases
At w = 372, we have THPWC(372) = cases/hour = 107.27 and CTPWC =hours, so we are roughly operating at the PWC level.

Throughput would increase (or at least not decrease), because bottleneck would be blocked/starved less. Unbalancing the PWC line causes it to perform better.

Throughput would increase (or at least not decrease), because bottleneck would be blocked/starved less. Replacing single machine stations with parallel machine stations in the PWC causes it to perform better. This is an example of performance improvement through flexibility enhancement.

Moving cases in batches would further inflate cycle time by adding “wait for batch” time, and increasing the variability experienced in the flow of the line.

Chapter 8

Problem 1

The mean is 5 and the variance is 0. The coefficient of variation is also zero. These process times could be from a highly automated machine dedicated to one product type.

The mean is 5, the standard deviation is 0.115 and the CV is 0.023. These process times might be from a machine that has some slight variability in process times.

The mean of these is 11.7 and the standard deviation is 14.22 so the CV is 1.22. The times appear to be from a highly regular machine that is subject to random outages.

The mean is 2. If this pattern repeats itself over the long run, the standard deviation will be 4 (otherwise, for these 10 observations it is 4.2). The CV will be 2.0. The pattern suggests a machine that processes a batch of 5 items before moving any of the parts.

Problem 3

The natural CV is 1.5/2 = 0.75.

The mean would be (60)(2) = 120 min. The variance would be (60)(1.5)2 = 135. The CV will be CV = √135/120 = 0.0968

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 5

In the tables below TH is the throughput that is given, te, is the mean effective process time, ce2 is the effective SCV of the process times, u is the utilization given by (TH)(te), CTq is the expected time in queue given by

for the single machine case and

for the case with m machines.

Finally, CT is the sum of CTq and te

In the first case, even though B has greater capacity, A has shorter cycle time since its SCV is much smaller.

Doubling the arrival rate (TH) and the number of tools makes station A have a longer average cycle time than B.

Note the large increase in cycle time with the modest increase in throughput as compare to (a).

We now consider Machine A only.

i) First we increase TH by 1% from 0.5. The increase in cycle time is less than one percent.

ii) Next we increase TH by 1% from 0.95. The increase in cycle time is almost 23%.

Problem 9

The same computations as in problem 5 are displayed for each option below. It is clear that Option a is the best since it not only has the lowest cost but it results in less than 2 days average cycle time.

Chapter 9

Problem 10

TH decreases. CT decreases.
TH increases. CT decreases.
TH increases. CT decreases.
TH decreases. CT increases.
TH decreases. CT decreases.
TH increases. CT decreases. Reducing the buffer size will reduce the WIP that keeps the machine utilizations and the TH high. However, if the process times variability is reduced at the same time, such protection may not be necessary and the TH will still increase.

Chapter 10

Problem 1

It can be seen that the CONWIP line achieves a higher throughput for this same WIP level. This effect results from the fact that the push system may release work into the line when the queue is very long, causing congestion that inflates the cycle time and (by Little’s law) reduces the throughput.

Problem 5

(ii)

Individual stations are affected by having bottleneck at station 1 or 3, but overall line performance is not.

(ii)

It is more effective (i.e., it improves TH more) to reduce variability at bottleneck.

(ii)

Speeding up non-bottleneck here is better than reducing variability at nonbottlenecks. Note that while this will often be the case, it will not always occur. Depending on the specifics of the problem, including the costs, reducing variability can be more attractive than increasing capacity.

C. Extra Credit – Chapter 9 Problem 6

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 125 parts divided by the heat treat time of 6 hours which is 20.833 per hour or 333.333 per 16-hour day.

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

The utilization would be .

The SCV of arrivals would be and

the SCV of process time is (3h/6h)2 = 0.25.

Using the VUT equation we get

Add these together with the process time and you get 6.2h + 0.714h + 6h = 12.91h.

(d) WTBT61 = , , . Hence

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

(e) Searching over the range reveals a minimum at batch size of 91 with a cycle time of 12.02 hours.