Chapter 1



ISyE 6201: Manufacturing Systems

Instructor : Spyros Reveliotis

Fall 2006

Solutions for Homework #3

A. Questions

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.

B. Problems

Chapter 8

Problem 1

a. 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.

b. 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.

c. 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.

d. 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

a. The natural CV is 1.5/2 = 0.75.

b. 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

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

[pic]for the single machine case and

[pic] for the case with m machines.

Finally, CT is the sum of CTq and te

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

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

[pic]

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

[pic]

d. 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%.

[pic]

C. Four-station assembly line case study:

To identify good candidates for consideration, we first look at the performances and prices of each option. We can rule out some options from further consideration since they are dominated by some other candidate(s). For instance, among the options for station 1, option 3 is obviously slower than option 1 and option 2 and it operates with high variance and with a much higher cost. A similar situation arises at station 4, where option 2 can be eliminated.

Next, we attempt to find a good configuration that meets the demand and performance requirements while minimizing the deployment cost.

The easiest way to solve this problem is to build a spreadsheet. The tricky parts are converting rate in piece/hour to batch process times and computing the SCV (squared coefficient of variation) for the batch processing.

Let: b = batch size

te = effective processing time for one piece

ts = time to perform a setup (to be taken equal to 0 in our case)

tb = effective processing time for a batch

Ce2 = SCV for the processing time for a piece

Cb2 = SCV for the processing time for a batch

ra = arrival rate (equals to throughput, TH, in our case)

We compute the mean effective processing time for a batch as:

tb = b * te + ts = b * te = 50 * (1/ speed)

e.g., for option 1 of station 1, tb = 50 / 42 = 1.19 hr/batch

Similarly, the fact that the batch processing time is the sum of the processing times of the b pieces in the batch, which are identically distributed, implies that σb2 = b * σe2. Since C2 = σ2 / t2, we have that Cb2 = σb2 / tb2 = (b * σe2) / (b * te)2 = Ce2/ b

e.g., for option 1 of station 1, Ce2 = (2.0)2 / 50 = 0.08

We get the arrival rate in batches per hour as (assume that there are 8 working hrs/day):

ra = (1000 prt/day) / (50 prt/batch)(8 hr/day) = 2.5 batches / hour

We can now start our calculation as follows:

i. Min. machines required = ra * tb

ii. Actual machine required, m = roundup (ra * tb)

iii. Utilization, u = ra * tb / m

iv. Average time in queue,

[pic]

v. Total waiting time at the station,

[pic]

vi. Since the SCV for inter-departure time, Cd2 , of previous station is the SCV for the arrivals, Ca2, to the next station. The linking equation for multiple machine station is

[pic]

We must also determine the SCV Ca2 for the inter-arrival times experienced at the first station. Since we are told that batches are released to the line at a constant rate, there is no variability in the inter-arrival times, and therefore, Ca2 is equal to zero (This is typical in repetitive, high throughput manufacturing).

We will start with a configuration, that selects the option with the lowest cost at each workstation, and utilizes the minimum number of machines that provides a machine utilization lower than one for the considered throughput.

The result is that we have 3, 3, 6, and 3 machines at station 1, 2, 3, and 4 respectively. From this configuration, we see that resulting total cycle time is higher than the cycle time constraint (less than 8 hrs) with a total cycle time of 21.40 hrs.

The largest portion of cycle time is at station 2, 13.34 hrs, so we will work on this first. By adding one more machine of the same technology option, the total cycle time goes down to 8.25 hrs and the cost goes up to $1,010, with a resulting benefit of 0.26 hr/$-added (You can check by editing the number of machine at station 2 in the spreadsheet in the previous page to be 4 machines). We now try to replace the machines at station 1 by a machine of option 2, which has a smaller variability. This selection is entered in our spreadsheet by setting the corresponding part processing CV to be 1 and the corresponding cost per machine equal to $85. With four machines of option 1 at station 2, the cycle time goes down to 6.75 hrs, i.e., our cycle time requirement is met. At the same time the total cost increases to $1,115.

Alternatively, we consider the possibility of adding one more machine of option 1 at station 1. The results for this configuration are given below:

[pic]

We see that the total cycle time is equal to 6.20 hrs and the resulting total cost is $1,060. Both of these numbers are better than the previously considered configuration of using three machines of option 2 for the station 1. Hence, we decide to stick with this configuration.

We note that the procedure described above is based on trial and error and it will not necessarily lead to an optimal solution, unless we perform an exhaustive search over all possible configurations. This is a typical use of queueing theory as a “design” tool; i.e., queueing theory essentially allows the systematic evaluation of the performance resulting from each considered alternative.

In the more specific context of our example, it is unclear without any further analysis and/or computation whether the generated solution is optimal; all we can say is that it is a configuration that meets the posed production requirements at a fairly low cost.[pic][pic][pic]

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