ISyE 6201: Manufacturing Systems
Instructor : Spyros Reveliotis
Fall 2006
Solutions for Homework #3
ISYE 6201 Fall 2006
Homework 3 Solution
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
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ISYE 6201 Fall 2006
Homework 3 Solution
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)(t e),
CTq is the expected time in queue given by
(1  ce2 ) u
CTq 
t e for the single machine case and
2 (1  u)
(1  ce2 ) u 2 ( m 1) 1
t e for the case with m machines.
2
m(1  u )
Finally, CT is the sum of CTq and te
CTq 
a. In the first case, even though B has greater capacity, A has shorter cycle time
since its SCV is much smaller.
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
0.92
1
0.92
0.85
0.25
1
0.92
7.1875
8.1875
4
1
0.782
7.6227
8.4727
b. Doubling the arrival rate (TH) and the number of tools makes station A have a
longer average cycle time than B.
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
1.84
1
1.84
0.85
0.25
2
0.92
3.4616
4.4616
4
2
0.782
3.4125
4.2625
c. Note the large increase in cycle time with the modest increase in throughput as
compare to (a).
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
0.95
1
0.95
0.85
0.25
1
0.95
11.8750
12.8750
4
1
0.8075
8.9140
9.7640
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%.
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ISYE 6201 Fall 2006
Homework 3 Solution
(i)
Machine
TH
te
2
ce
m
u
CTq
CT
% Increase
A
0.5
1
0.25
1
0.5
0.6250
1.6250
(ii)
A
0.505
1
0.25
1
0.505
0.6376
1.6376
0.7770
A
0.95
1
0.25
1
0.95
11.8750
12.8750
A
0.9595
1
0.25
1
0.9595
14.8071
15.8071
22.7736
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
te
ts
tb
Ce2
Cb2
ra
= batch size
= effective processing time for one piece
= time to perform a setup (to be taken equal to 0 in our case)
= effective processing time for a batch
= SCV for the processing time for a piece
= SCV for the processing time for a batch
= 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 b2 = b * e2.
Since C2 = t2, we have that Cb2 = b2 / tb2 = (b * e2) / (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
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ISYE 6201 Fall 2006
Homework 3 Solution
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,
(C 2  Cb2 ) u ( 2( m 1) 1)
CTq  a
tb
2
m(1  u )
v.
Total waiting time at the station,
CT  CTq  t b
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
u2
Cd2  1  (1  u 2 )(Ca2  1) 
(Cb2  1)
m
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.
MEASURE:
Part Processing Speed (prt / hr)
Part Processing CV
Part Processing Time (hr)
STATION:
1 / te
Ce
te
2
Part Processing SCV
Eff Batch Processing Time (failures+setups)
Ce
tb = b * te
Eff Batch Processing SCV (failures+setups)
Arrival Rate (parts/hr)
Min. # of Machines Required
Actual # of Machines Required
Utilization
Arrival CV
cb = C e / b
ra
ratb
m
u
2
ca
Departure SCV
Time in Queue (hr)
Station Cycle Time (hr)
Cost Per Machine
Cost Per Station
2
2
2
cd
CTq
CT = CTq+tb
1
42
2
0.0238
2
42
2
0.0238
3
25
1
0.0400
4
50
0.75
0.0200
4.0000
1.1905
4.0000
1.1905
1.0000
2.0000
0.5625
1.0000
0.0800
2.5
2.976
3
0.992
0.000
0.0800
2.5
2.976
3
0.992
0.461
0.0200
2.5
5.000
6
0.833
0.469
0.0113
2.5
2.500
3
0.833
0.560
0.461
1.971
3.162
$50
$150
0.469
13.340
14.530
$50
$150
0.560
0.296
2.296
$100
$600
0.469
0.409
1.409
$20
$60
Performance Requirement
Total cycle Time =
21.40
does NOT Meet the cycle time requirement
Total Cost =
$960
INPUT VALUE
OUT PUTVALUE
Note:
Assume C a 2 at the first station equals to 0, which is a typical value for repetitive, high-throughput manufacturing
Assume 8 working hour per day
Batching assumes that transfer lots equal process lots.
Multi-machine stations are assumed to share batches, rather than have individual machines dedicated to jobs
in order to reduce number of setups.
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ISYE 6201 Fall 2006
Homework 3 Solution
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:
MEASURE:
Part Process Speed (prt / hr)
Part Process CV
Part Process Time (hr)
STATION:
1 / te
Ce
te
2
Part Process SCV
Eff Batch Process Time (failures+setups)
Ce
tb = b * te
Eff Batch Process SCV (failures+setups)
Arrival Rate (parts/hr)
Min. Machine Required
Actual Machine Required
Utilization
Arrival CV
cb = C e / b
ra
ratb
m
u
2
ca
Departure SCV
Queue Time (hr)
Cycle Time (hr)
Cost Per Machine
Cost Per Station
2
2
2
cd
CTq
CT = CTq+tb
1
42
2
0.02381
2
42
2
0.02381
3
25
1
0.04000
4
50
0.75
0.02000
4.00000
1.19048
4.00000
1.19048
1.00000
2.00000
0.56250
1.00000
0.08000
2.5
2.97619
4
0.74405
0
0.08000
2.5
2.97619
4
0.74405
0.29895
0.02000
2.5
5.00000
6
0.83333
0.43240
0.01125
2.5
2.50000
3
0.83333
0.54873
0.29895
0.02454
1.21502
$50
$200
0.43240
0.11626
1.30673
$50
$200
0.54873
0.27443
2.27443
$100
$600
0.46568
0.40123
1.40123
$20
$60
Performance Requirement
Total cycle Time =
6.20
Meets the Demand
Total Cost =
$1,060
INPUT VALUE
OUT PUTVALUE
Note:
Assume C a 2 at the first station equals to 0, which is a typical value for repetitive, high-throughput manufacturing
Assume 8 working hour per day
Batching assumes that transfer lots equal process lots.
Multi-machine stations are assumed to share batches, rather than have individual machines dedicated to jobs
in order to reduce number of setups.
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.
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ISYE 6201 Fall 2006
Homework 3 Solution
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.
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