ISyE 6201: Manufacturing Systems
Instructor : Spyros Reveliotis
Spring 2007
Solutions for Homework #3
ISYE 6201 Spring 2007
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.
Chapter 9
Question 1
The conditions for 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, i.e., we have total
control on the processing times and the inter-arrival times taking place during the system
operation. This is an idealistic situation, not practically realizable.
Consider an asynchronous transfer line loaded at a rate ra, such that the server
utilization is less than one at every station, and with ample buffering capacity at every
station, so that blocking is not an issue. Can you think of any reason(s) that the
output rate of this line might be less than ra ?
The output rate might be less than ra if some of the processed parts are scrapped due to
defects.
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.
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ISYE 6201 Spring 2007
Homework 3 Solution
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
c. 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)(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
CTq 
t e for the case with m machines.
2
m(1  u )
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.
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.
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ISYE 6201 Spring 2007
Machine
TH
te
2
ce
m
u
CTq
CT
Homework 3 Solution
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%.
(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
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 increase TH (slightly by reducing the amount of
bakling and decrease CT.)
(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
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ISYE 6201 Spring 2007
Homework 3 Solution
lower than the current value of rb). If we add capacity, we increase TH (slightly) and
decrease CT (more significantly).
(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.
Two-Station, Single-Machine Production Line Problem:
i.
A
te 
mf
m f  mr
A1 
,
7
 0.824 ,
7  1.5
A2 
5
 0.909
5  0.5
t0
11
11
 13.36 min  0.223hr , t e 2 
 12.1 min  0.202hr
, t e1 
0.824
0.909
A
Station 1 is the effective bottleneck of the line.
ii.
iii.
35 13.36
35 12.1

 0.974 ,
u2 

 0.882
8
60
8 60
The utilization is less than 1 at both stations, so the production line can sustain the
production rate of 35 parts per 8-hour shift.
u1 
u  ra t e ,
For station 1, the SCV of the effective processing time is
m
1.5
ce2  c02  (1  cr2 )(1  A) A r  0.5 2  (1  0.75 2 )(1  0.824)(0.824)
 2.108
t0
11 / 60
 c 2  ce2 u

CT   a

 1t e = 2 hr and 7.815 min = 2.13 hr
1 u 
 2
 0  2.108 ra  0.223

 

 10.223  2.13
2
1  ra  0.223 

 ra  4 parts per hour
The inter-release interval is 15 minutes.
iv.
CTq  CT  t e  1.908hr ,
v.
Mean = 15 minutes =
WIPq  TH  CTq  4  1.908  7.63
1
hrs
4
Variance = variance of inter-departure times
2

1
2
2
= c d21    = u1 ce21  (1  u1 )  c a21
 4
= 0.1045 (hrs)2 = 376.2 (min)2
  14   4  0.223
2
 
2

1
 2.108  0   
 4
2
C. Extra Credit
5
ISYE 6201 Spring 2007
Let
Homework 3 Solution
T0 be the natural processing time
N be the number of failures that occur during the processing of a single part; since
inter-failure times are exponentially distributed with mean equal to mf, N follows a
Poisson distribution with E(N) = var(N)= T0/mf
Yj be the repair time of the j-th failure, j =1, 2, …, N, where E(Yj) = mr, var(Yj)= r
f(·) be the probability density function of T0
n
t
Te be the effective processing time. Te = T0 +  Y j , and E[Te] = 0 .
A
j 1

E[Te2 ] 




 E Te2 | T0  x, N  n  e
x / m f
(x / m f )n
n!
x 0 n 0
f ( x)dx
n

 x / m f (x / m f ) n
2
f ( x)dx
=   E ( x   Y j )   e
n!
j 1
x 0 n 0


 
n
n
 2
 x / m f (x / m f ) n
2
f ( x)dx
=   E  x  2 x  Y j  ( Y j )   e
n!
j 1
j 1
x 0 n 0


n
 
x / m f (x / m f )
2
2
2
f ( x)dx
=   x  2 xnE(Y j )  n(n  1) E (Y j )  nE (Y j )  e
n!
x 0 n 0



 x




2
 2 xE( N ) E (Y j )  E ( N 2  N ) E 2 (Y j )  E ( N ) E (Y j ) f ( x)dx
2
 2 xE( N )mr  [var( N )  E 2 ( N )  E ( N )]mr2  E ( N )( mr2   r2 ) f ( x)dx
2
x 0


 x

x 0
2


2
 x  2  x  2
x
2
 mr  
(mr   r2 ) f ( x)dx
  x  2
mr  
m 
m 
mf

x 0 
 f 
 f 



 m f  mr
 E (T 0 )
 m
f

2
2

  (m r2   r2 ) t 0

mf

(1  A)t 0
1
 ( 02  t 02 )   (mr2   r2 )
Amr
 A
2
since A 

So we have equation (8.5):   E[T ]  E [Te ]   0
 A
2
e
2
e
2
mf
m f  mr
2

(1  A)t 0
  (mr2   r2 )
.

Amr

Taking c 02  t 02 02 and cr2  mr2 r2 , we can derive Equation (8.6):
c 
2
e
 e2
t e2

  0
 A




2
2
 1 
(1  A)t 0

  (mr2   r2 )
Amr
 t0 / A 
2
 1 
A(1  A)mr

  c02  (1  c r2 )
t0
 t0 / A 
6