QUEUEING THEORY
Worked examples and problems
J. Murdoch
Head of Statistics and Operational Research Unit,
School of Production Studies,
Cranfield Institute of Technology
M
©
J. Murdoch 1978
All rights reserved. No part of this publication may
be reproduced or transmitted, in any form or by any means,
without permission.
First published 1978 by
THE MACMILLAN PRESS LTD
London and Basingstoke
Associated companies in New York Dublin
Melbourne Johannesburg and Madras
ISBN 978-1-349-03313-3
ISBN 978-1-349-03311-9 (eBook)
DOI 10.1007/978-1-349-03311-9
This book is sold subject to the standard conditions of
the Net Book Agreement.
The paperback edition of this book is sold subject to
the condition that it shall not, by way of trade or
otherwise, be lent, resold, hired out, or otherwise
circulated without the publisher's prior consent in any
form of binding other than that in which it is published
and without a similar condition including this condition
being imposed on the subsequent purchaser.
CONTENTS
Preface
v
vi
Glossary of Symbols
Classification of Queueing Systems
1.
2.
3.
BASIC CONCEPTS OF QUEUES
1
1.1
1.2
1.3
1.4
1.S
3
1
3
8
10
2.1
2.2
2.3
2.4
10
Introduction
Resume of Basic Theory and Formulae
Problems
Solutions
M/M/1/ex> SYSTEMS
3.3
3.4
S.
Introduction
The Queueing Situation
Types of Queueing Problem
The Basic Theory
Mathematical Solution of Queueing Problems
BASIC DISTRIBUTIONS IN QUEUEING THEORY
3.1
3.2
4.
viii
10
12
13
18
Introduction
Resume of Basic Theory and Formulae
Problems
Solutions
18
18
19
22
M/M/1/N SYSTEMS
33
4.1
4.2
4.3
4.4
33
Introduction
Resume of Basic Theory and Formulae
Problems
Solutions
M/M/C/ ex> SYSTEMS
5.1
33
38
40
46
Introduction
46
iii
5.2
5.3
5.4
6.
of Basic Theory and Formulae
Problems
Solutions
46
R~sum~
47
50
SYSTEMS WITH ARRIVAL RATE AND/OR SERVICE RATE
62
DEPENDENT ON THE NUMBER IN THE SYSTEM (M /M /-/-)
n
6.1
6.2
6.3
6.4
6.5
7.
n
Introduction
R~sum~ of Basic Theory and Formulae
Special Applications of Theory
Problems
Solutions
SINGLE-CHANNEL SYSTEMS WITH GENERAL SERVICE TIME
DISTRIBUTIONS (M/G/l/~ SYSTEMS)
7.1
7.2
7.3
7.4
Introduction
R~sum~ of Basic Theory and Formulae
Problems
Solutions
62
62
63
65
67
77
77
77
77
78
statistical Tables
Table 1
Poisson Distribution
81
Table 2
Negative Exponential Distribution
86
Table 3
Optimum value of p
for H/H/l/N Systems
87
References
88
PREFACE
The basic concepts and an understanding of mouern queueing theory are requirements not only in the training of
operational research staff, management scientists, etc.,
but also as fundamental concepts in the training of managers or in mangement development programmes.
.
The efficient design and operation of 'service functions' is one of the main problems facing management today and the understanding obtained from a study of queueing theory is ~ssential in the solution of these problems.
Industry and commerce have for too long concentrated
their main resources on designing and operating the 'production units' and little attention has been paid until
recently to the 'service units'. Basic concepts such as
'increased efficiency is achieved when the utilisation
of service units is reduced' are still hard for practical
personnel to understand, brought up as they are on the
concept of 'maximising the utilisation' of their facilities. The ancient Chinese civilisation had a system
based on queueing theory: 'Pay your doctor only when you
are well'. Thus in industry if a system is correctly designed, management should be happy when 'its maintenance
gang is playing cards' since there are no breakdowns to
be repaired!
This book, by concentrating on problems with their
fully worked-out solutions, gives students of queueing
theory not only a chance to test their understanding of
the theory, but also illustrates the wide range of application of the theory.
The book covers the steady-state solutions of randomarrival queueing systems. It is designed to meet the
needs not only of management science training programmes,
but also of mangement teaching programmes.
Cranfield, 1976
J. Murdoch
G LOSSAR Y OF SYM SOLS
deterministic distribution
general distribution
negative exponential distribution
negative exponential distribution
with mean dependent on the number
in the system n
number of channels
maximum system size in finite queues
average arrival rate
average inter-arrival time
average service rate
average service time
intensity of traffic for single and
multi-channel queues
traffic offering (multi-channel queues)
variance of ~ervice time
C
N
)..
II)..
~
l/~
p
or
e;
)..
)..
= -~ (or C~
- )
=
)../~
a 2
s
distribution of the time in the system
(steady state)
number in the system
average number in the system
transient state probabilities of n
in the system
steady-state probabilities of n in
the system
number in the queue
average number in the queue
distribution of the waiting time in
the queue in the steady state
average waiting time of all customers
in the queue in the steady state.
d(t)dt
n
-n
q
q
w(t)dt
w
BASIC DISTRIBUTIONS
P(x)
e
-mmx
---x-r-
Poisson
vi
distribution (mean
m)
pet)
1
'fe-tIT dt
Negative exponential distribu-
tion (mean
=
T)
CLASSIFICATION OF QUEUEING SYSTEMS
Queues are classified in the book as follows.
(1)
/
(2)
/
(3)
/
(4)
(1)
Input Distribution
(2)
Service Distribution
e.g. M, D, Mn' G, etc.
e.g. G, D, M, etc.
(3)
Number of Service Channels
e.g. 1, 2,
(4)
Number in the System
Unconstrained "", finite,
maximum size = N
...
C, etc.
EXAMPLES OF USE OF CLASSIFICATION SYSTEM
Thus an M/M/l/~ system is random arrival, negative exponential service time distribution, single-channel, no constraint on queue size.
Again a G/M/C/N system is general arrival distribution,
negative exponential service distribution, C service
channels, maximum number in the system N.
The service mechanism in all problems, is service in
order of arrival, or first-in, first-out (FIFO) system.
BASIC CONCEPTS OF QUEUES
1.1
INTRODUCTION
Queueing situations arise in all aspects of work and life
and are typified by the 'queueing for service'. The
theory of queueing gives a basis for understanding the
various aspects of the problems and enables a quantitative
assessment to be made. Therefore the theory enables these
'service situations' to be more effectively designed and
operated.
Understanding queueing theory and its concepts is thus
basic to all personnel concerned with service situations.
Since a large proportion of both capital and labour is
tied up in service facilities, and these areas have in
the past tended to be neglected for the direct productive
units, there is clearly a large potential area of application of the theory and also large savings to be obtained.
This book, by giving a series of problems with their
worked solutions, aims not only to teach understanding
of the basic theory but also to give readers an insight
into the potential of the theory and its wide field of
application.
1.2
THE QUEUEING SITUATION
A situation in which queueing can occur may be typified
by a shop where customers expect to be served by sales
assistants. If all assistants are busy when a new customer enters, he has to wait and thus forms the beginning
of a queue. In our discussion of queues in general, we
shall call 'customer' the incoming unit, that is, the unit
that enters into a situation in which a queue could form;
such queues need not take the form of 'customers' actually
lining up, all we need, to define a customer as queueing,
is the fact that he has made clear his expectation of
being served, and that the service is not available. By
'service' we shall mean any action necessary to allow
the customer to leave the 'shop', or 'counter' - in general the situation where queueing had been possible.
Thus there are three essential elements of any queueing
situation: (1) input process - the manner in which
customers arrive; (2) queue discipline - the manner
in which customers wait for service after input; and
(3) service mechanism - the manner in which customers
are being served, or the way in which the queue is being
resolved.
Figure 1.1 illustrates the queueing system for a shop
1
Departure on
completion of service
Arrival of customers
Figure 1.1
Customers awaiting service
Diagrammatic Representation of Single-channel
Queue.
with a single server or counter, while figure 1.2 illustrates two different queueing systems for a three-channel
system (three counters in parallel).
la)
Single queue
Counter 3
Counter 2
~
O~·~···.
-......
U".
. . 69-.. . :. . ~;. .".
Arrival of
customers
. ...
I
f'\fV"\O
........
,
,0
V
.
: : :. .'~ .......
~
Departure on
-~~~
completion of service
Customers queueing and going
to first free counter
Ib)
Independent queues
Arrival of
customers
Ifree to join
any queue)
Figure 1.2
Departure on
completion of service
" ~.....
:>
Diagrammatic Representation of Three-channel
Queue.
2
Queueing situations
rently quite different
Figure 1.3 shows a few
of which have been the
1.3
are very widespread and many appaexamples can be found in practice.
of the more important ones, many
subject of published investigations.
TYPES OF QUEUEING PROBLEM
Although all queueing situations are basically similar,
there are an almost infinite number of different situations that can arise in practice.
As previously stated, the three basic elements of a
queueing problem are
(1)
(2)
(3)
input process
queueing discipline
service mechanism.
These elements have within themselves a large number of
possible variations, which give rise to a large number
of different queueing situations. Figure 1.4 gives a
list of possible variations, although this itself is not
exhaustive.
1.4
THE BASIC THEORY
Queueing problems arise, as has been seen, in any activity
where demands for service arise from a multiplicity of
sources acting more or less independently of each other.
Where customer arrivals, or demands, for service can be
scheduled exactly, then it is relatively easy to provide
appropriate service facilities, and this is really a
trivial problem compared with the ones that occur more
usually in practice and to which queueing theory give
the basis for solution.
In developing the theory, it is convenient to imagine
customers arriving at a counter and queueing for service,
if the service mechanism is busy (see figur~l.l and
1.2). This method can also cover situations where a
physical queue does not in practice exist, for example,
machines awaiting service from an overhead crane, callers
waiting on different lines for a connection by the telephone operator, etc.
1.4.1
Measures of Effectiveness
It has been found possible to set up mathematical models
to describe queueing situations specified by different
forms of the three basic elements. These models can
then be manipulated to show what the service system under
investigation should be capable of achieving and how any
two or more systems compare. In order to make a decision
3
..,.
Jobs requiring
movements
Planes arriving to
land
Arrival of batches
of goods from
supplier
Factory handling
system
Airport
Stocking of goods
Customers arriving
Traffic
Bus stops
Taxi ranks
Patients arrival
for treatment
Customers or
clients arriving
for service
Shop
Booking office
Post Office
Bank
Hairdressing salon
Doctor's or hospital
outpatients waiting
room
Input to Queue
Stocks of goods in
store
Planes circling overhead waiting for free
runway
Jobs waiting at
various points
for movement
Patients waiting
their turn
Customers queueing
waiting for bus or
taxi
Waiting for counter
to be free
Queue
Usage or purchases
of goods from store
Planes landing on
runway at airport
Actual movement of
job by transport
Treatment by doctor
Arrival of bus,
taxi, etc.
Assistant, teller,
etc., serving at
counter
Service Mechanism
Some Typical Situations for the Application of Queueing Theory
Situation
Figure 1.3
<.n
Input to Queue
Customers picking
up phone
Possible contracts
for pricing for
tendering
Ships arriving at
port for loading
or unloading
Operators and/or
machine breakage,
etc., requiring
skilled setter (or
maintenance operator)
Plant breaks down
Situation
Telephone switchboard
Estimating
(tendering)
Harbour design
Semi-skilled operators in machine
shop
Maintenance
department
Ships being loaded
or unloaded
Pricing by estimator
and sending off
tender
Switching at s~itch­
board or exchange
Service Mechanism
Plant awaiting
repair
Plant repair
Operators and machines Adjustment (or repair)
by skilled setter, or
waiting for skilled
maintenance)
setter (or maintenance)
Ships awaiting
berth
Contracts awaiting
pricing
Customers awaiting
telephone switchboard operator·' s
response
Queue
0\
Infinite
Finite
Singly
In batches of constant
number
In batches of variable
number
(c)
(a)
(b)
Constant
Completely random
(Poisson input)
Other distribution of
intervals
3. Intervals between arrivals
(c)
(a)
(b)
2. Number arriving at one time
(a)
(b)
1. Number of potential customers
Can vary as follows in
Input Process
Single queue
Several queues
(e)
(b)
(c)
(d)
(a)
Service in order of
arrival
At random
Priority
Service in reverse order
of queueing (unfair queue)
Service time-dependent
2. Queue discipline
(a)
(b)
Can vary as follows in
1. Number of queues
Queueing Discipline
One
Several
Number variable
One at a time
In batches of constant
number
In batches of variable
number
(a)
(b)
Permanently
Intermittently
3. Service available
(c)
(a)
(b)
2. Number served
(a)
(b)
(c)
Can vary as follows in
1. Number of service points or
servers
Service Mechanism
"
Constant
Varying with time
Influenced by a state of
queue
(a)
(b)
Figure 1.4
No outside influence
Input is the output of a
previous queueing
situation
5. Outside influence
(a)
(b)
(c)
4. Average rate of arrival
Input Process
Service Mechanism
Constant
Exponentially distributed
(times of beginning and end
of service distributed
independently and at
random)
Other distributions
Dependertt on time customer
has spent in queue
(a)
(b)
Constant
Varying with time or state
5. Average rate of service
(c)
(d)
(a)
(b)
4. Duration of service
Queueing Systems: Types of variation
in System.
Queueing Discipline
on which system is the 'best', certain 'measures of effectiveness' are required.
be
Useful measures of effectiveness have been found to
(1) The probability of having n customers waiting at a
time t, given the initial state of the system. Knowing this probability distribution, the size of the
queue that wili be exceeded for only 5 per cent, say,
or 1 per cent of the time, can be determined. This
could be useful, for example, in determining what
size of waiting room needs to be provided for customers so that only rarely will there be an overflow of
customers and possible loss of business if there is
an alternative service point they can go to. Alternatively, with a given size of waiting space, the
service facility required could be determined such
that the waiting space will be adequate most of the
time.
(2) The distribution of waiting time of customers. From
this distribution can be found the average waiting
time of customers and the proportion of customers
who have to wait longer than a certain time t, say,
If the probability of waiting longer than t is high,
then customers may be discouraged from joining the
queue, which would result in a loss of potential
business in something like a petrol station or a
supermarket. The cost of providing more or faster
service facilities may be more than compensated for
by the extra business produced by a reduction of
customers' waiting time. Again, in the case of an
internal stores in a factory, provision for an extra
storekeeper, say, may pay handsome dividends in
reducing the lost production time of skilled men
who have to queue for a long time for service.
These measures of effectiveness (depending which, if any,
is appropriate to the problem) can be used to decide, for
instance (usually on a cost basis) whether to speed up
the existing service rate of each channel or whether to
provide extra channels working at the same rate as the
present ones or whether even a reduction in service facility can be contemplated.
1.S
1.5.1
MATHEMATICAL SOLUTION OF QUEUEING PROBLEMS
Transient and Steady-state Solutions
The solution of queueing problems is considered in two
parts, namely the transient or time-dependent solution,
and the steady-state solution. Briefly, provided that
the service channel is capable of serving at a faster
8
average rate than that at which customers arrive, then
the steady-state is reached when the queue behaves independently of the initial state of the system, and the
probability of having a given number, n, say, in the
queue remains constant with time. This situation exists,
more or less, in a machine shop where the queue of demands
for the overhead crane unsatisfied at the end of the day
is carried over to the next day (assuming that the crane
only operates during normal working hours and does not
work off the backlog of jobs during the night, say). The
state of the queue soon becomes independent of the starting conditions when the present pattern of production
was begun.
On the other hand, in somewhere like a bank, the system
starts every day with no people at all either being served
or queueing, and the chance of finding, say, 6 people
queueing depends on how soon after opening time the observation is made. Assuming that there are always the same
number of clerks on duty and customers arrive at a constant average rate throughout the day, the chance of
finding 6 people queueing immediately after the bank
opens is likely to be very small indeed. As the day
proceeds, the queue gradually achieves a steady state
and eventually the queue fluctuates about a fixed average size, the probability of finding 6 people queueing
now being higher than it was at the very start of the
day's business.
In most practical situations only the steady-state
queue need be considered, but occasionally only the
transient solution is applicable since the queue never
reaches a steady state. This latter is generally true
if the arrival rate is greater than the service rate
(A > ~) and applies in some cases when the system is
not in operation long enough before reverting to the
starting state, usually with no customers at all in the
system.
This book deals only with the steady-state solution
results, and the use of the theory is demonstrated with
a range of problems for each system. The queueing
systems covered relate only to random-arrival systems.
2
BASIC DISTRIBUTIONS IN
QUEUEING THEORY
2.~
INTPODUCTION
The Poisson distribution and the negative exponential
distribution are the two basic distriblltions in queueing
theory. Their theory and general fields of application
can be studied in general statistical theory texts and
this book will deal primarily with their application to
queueing theory.
2.2
2.2.1
RESUME OF BASIC THEORY AND
FOR~IULAE
Poisson Distribution
General La 1<-
If the chance of an event occurring at any instant of
time is constant in a continuum of time and if the average number of successes in time t is m, then the probability of x successes in time t is
pex)
x -m
m e
-x;
with mean of distribution = m
and variance of distribution = m
The Poisson distribution is tabulated in table 1, in the
statistical tables at the end of this book.
2.2.2
Negative Exponential Distribution
General Law
If the chance of an event occurring at any instant of
time is constant in a continuum of time, then if the
average time interval between successes is T, then the
probability of an interval t between successes is
P (t)
Thus both these distributions describe the same random
situation. Also, the probability of a time interval
exceeding t is
10
e -tIT
This function is tabulated in table 2 of the statistical
tables at the end of this book.
2.2.3
Special Property of the Negative Exponential
Distribution
The negative exponential distribution of service times
is important in queueing theory because of the following
special property, namely that the time a service has been
in progress does not affect the probability of its completion. This proof is given below.
Let service have been in progress for time t. Then
it must be 'at least' time t long, the probability of
this being
<lO
P(>t)
=)(
f e-t/Tdt
= e -tIT
If the service now ends during interval dt, it must
therefore have had time between t and (t + dt), this
probability being
P(t)dt
Now P (t)dt
dt e-t/T
T
(probability of call lasting till time t)
conditional probability that it does
not last beyond (t + dt)
x
Therefore given that the call has lasted till t
probability that it finishes
in interval (t) to (t + dt)
P(t)dt
P(>t)
(dt/T)e-t/T
e -tIT
1:.T
dt
or this probability is independent of length of time
(t) the call has been in progress.
11
2.3
PROBLEMS
1. Customers arrive at a store for service at an average
rate of 10 per hour. Given that customers arrive randomly, what is the probability of
(a) more than 15 customers arriving in one hou"
(b) exactly 10 customers arriving in one hou"
(c) more than 6 cu~tomers arriving in half an hou"
2. Draw the distribution of number of arrivals per hour
given that average number of arrivals per hour = 3, and
that customers arrive randomly.
3. Customers arrive randomly at a service point with an
average rate of 4 per hour. Draw the distribution of
the interval between successive arrivals.
4. The time taken to repair a machine is distributed as
the negative exponential distribution with a mean of 10
hours. What is the probability that
(a) a machine takes longer than 6.9 hours to repair?
(b) a machine takes longer than 10 hours to repair?
What is the repair time that is exceeded by chance once
in 100 repairs?
5. The failure rate for a television receiver is 0.02
failures per hour. Calculate the average time between
failures. What is the probability of it failing within 4
ho vrs?
6. The time interval, in minutes, between the arrival
of successive customers at a cash desk of a self-service
store was measured over 56 customers and the results are
given below.
Time Interval between Arrivals (min)
0.05
0.21
1.14
0.57
1.16
0.15
0.43
3.12
1. 68
2.71
0.16
0.65
0.58
0.42
2.16
0.62
0.78
2.12
0.31
4.60
0.57
0.25
2.68
3.70
1.10
2.81
0.91
1.72
0.04
0.05
0.08
1.48
0.32
3.30
0.18
0.52
1.19
1.18
4.20
2.08
1. 61
0.15
0.04
2.32
0.11
3.90
0.09
1. 76
0.10
0.54
1.16
0.08
0.05
0.01
0.63
1. 21
Fit a negative exponential distribution to the data.
12
2.4
SOLUTIONS
l(a) Here average arrival rate per hour m = 10, therefore
Probability of more than
IS customers in one hour
GO
= r
x=16
From table 1
P(>lS)
= 0.0487
(b) Again from table 1
P(lO) = 0.5421 - 0.4170
(c) From table 1, now m
0.1251
=5
P(>6) = 0.2378
2. From table 1
Prob. of no customers arriving P (0)
0.0492
P(>7)
0.0119
P
P
P
P
P
P
P
(1)
0.1493
(2)
0'.2241
(3) = 0.2240
(4)
0.1681
(5)
0.1008
(6)
0.0504
(7)
0.0216
Figure 2.1 shows the histogram.
0.3
~
0.2
:s
.21
e
a..
P(2)
0.1
P(3)
P(4)
PIl)
P(S)
0
P(O)
o
P(6)
2
3
4
S
I
6
Number of Arrivals
Figure 2.1
3. Here the average interval between arrivals
hours. Distribution is
13
0.25
pet)
=
1 e- t / 0 . 25 dt
0.25
All probability of exceeding t
=
roo_l_e-t/0.25dt
0.25
Jt
= e- t / 0 . 25
This exponential function is tabulated in table 2.
Interval between
Arrivals h
(t)
Probability of
Exceeding t
P(>t)
0.00
0.05
0.10
0.15
0.20
0.30
0.40
0.50
0.75
1.00
1.0000
0.8187
0.6703
0.5488
0.4493
0.3012
0.2019
0.1353
0.0498
0.0183
Therefore the probability distribution is as follows.
Interval between
Arrivals (h)
0.00
0.05
0.10
0.15
0.20
0.30
0.40
0.50
0.75
over
Probability
-
0.1813
0.1484
0.1215
0.0995
0.1481
0.0993
0.0666
0.0855
0.0315
0.0183
0.05
- 0.10
- 0.15
- 0.20
- 0.30
- 0.40
- O. SO
-
0.75
1.00
1.00
1.0000
Total
This probability distribution is drawn in figure 2.2.
14
.20
t-.15
,-~
~ .10
-
~0
ci:
-
.05
111---_---,I
oLJ-l-L1-~L--L
.4
.2
.3
o .05 .1
I
__1-______
~======~
.6
.7
.8
.9
.10
.5
Interval between arrivals (mins.)
Figure 2.2
4.
Ca)
Here T = 10 hours.
From table 2
Probability of repair
exceeding 6.9 hours
e- 6 • 9 /l O
e
-0.69
0.5016
(b)
Probability that repair
exceeds 10 hours
e- lO / lO
0.3679
e
-1
ec) From table 2, the value of m in the function e- m
that gives a value of 0.01 to the function is m = 4.61,
therefore
t
4.61
.,. =
and
t =
46.1 h
or a repair time of 46.1 h or more is to be expected
once in 100 times on the average.
5. If the failure rate is 0.02 failures per hour, then
the average time between failures
=
1
o:oz
SO h
Probability of
failing inside 4 h
1 - Probability of failing
in over 4 h
IS
1 - e- 4 / 50
1 - e -0.08
from table 2
1 - 0.9231
0.0769
The data are summarised into a distribution in
the following table, together with the negative exponential distribution
6.
Fitting a negative exponential distribution
Average interval between arrivals
T
19 x 0.25
72.5
--s6
+
11 x 0.75
+ •••
56
+
1 x 4.25
+
1 x 4.75
= 1 • 29
From Table 2
PCt > 0.50 min)
e-O. 5/1. 29
e
-0.39
therefore
Probability of interval
between 0 and 0.50 min = 1 - 0.6771
0.6771
= 0.3229
Again from Table 2
PCt > 1.00 min)
e-l.00/l.29
e
-0.78
0.4584
therefore
Probability of interval
between 0.50 and 1.00 min
0.6771 - 0.4584
0.2187
etc.
The fitting of the negative exponential distribution is
summarised in the following table.
16
'-l
......
0.125
7
6
1.00 - 1. 499
1. SO - 1. 999
0.036
0.018
0.018
1.000
2
1
1
56
4.50 - 4.999
(and over)
0.036
2
2.50 - 2.999
3.00 - 3.499
3.50 - 3.999
0.054
3
4.00 - 4.499
0.071
4
2.00 - 2.499
0.107
0.196
11
0.00 - 0.49
0.50 - 0.999
0.340
Probability
19
Frequency
ACTUAL
1.000
0.0305
0.0145
0.0215
0.0308
0.0464
0.0685
0.1013
0.1449
0.2187
0.3229
Probability
55.9
1.7
0.8
1.2
1.7
2.6
3.8
5.7
8.1
12.2
18.1
Frequency
NEGATIVE EXPONENTIAL
MIMI I / SYSTEMS
3
3.1
INTRODUCTION
The theory of this system and also the other systems
described in later chapters can be found in most textbooks on queueing theory and a list of references is
given at the end of this book.
In the M/M/l/oo system, customers arrive randomly for
service - Poisson stream, service time distribution is
negative exponential, single server, no constraint on
queue size. Customers are served in order of arrival.
L
Counter
0
~;~'~'~Lo~
iJ.
= average service rate
Customer being served
Departure on
completion of service
A = average arrival rate
Figure 3.1 M/M/l/oo Systems
3.2
RESUME OF BASIC THEORY AND FOR:r-mLAE
The basic formulae of these systems are given below.
Probability of no customers in the system Po
Probability of n customers in the system Pn
Average numler of customers in the system n
=
=
1 - P
(1 _ p)pn
-pI - p
Average queue length q
Probability that there are more than r
customers in the system
P(>r)
Probability that there are more than r
customers in the queue
p
p
r+l
r+2
Waiting Time
Waiting time distribution wet)
18
p(~
- A)e
-t
(~-A)
dt
Average waiting time
__P_
w
Probability of waiting time
greater than t
1
)l
pe -tell-A) dt
Distribution of total time
in the system
det)
Average time in the system
x
1 - P
a
=
= e)l - A)e
-tell-A)
1
~
Probability of spending longer
than t in the system
3.3
PROBLEMS
1. In a single-channel queueing situation, an activitysampling analysis of the system gave the following details
of the number of persons in the system.
No. of Persons
Times Observed
o
900
821
658
621
503
1
2
3
4
420
5
302
252
181
152
6
7
8
9
Note: The data for more than 9 in the system is not
included.
Do these results support the hypothesis that the queueing
situation is an M/M/l/oo system?
2. A television repairman finds that the time spent on
his jobs has an exponential distribution with mean 30
minutes. If he repairs sets in the order in which they
corne in, and if the arrival of sets is distributed as
Poisson with an average rate of 10 per 8-hour day, what
is the repairman's expected idle time each day?
What is the average number of sets ahead of a set that
has just been brought in?
3. In the design of the layout of handling equipment for
an unloading bay at a factory, three schemes (A, B, C)
are being considered, relevant details being as follows.
19
Schelfle
A
B
C
Variable Of·
Cost/Day
Fixed Cost
/Day
(f)
(£c\l)
60
130
250
100
ISO
200
Handling
Rate/Hour
(NO. of sacks)
1000
2000
6000
Average arrival rate of trucks = IS per 10-hour day;
average truck load = 500 sacks.
If the cost of a truck waiting is given to be flO per
hour and the queueing is M/M/l/oo, which scheme gives
overall minimum cost?
4. In a supermarket the average arrival rate of customers
is 5 every 30 minutes. The average time it takes to
list and calculate the customer's purchases at the cash
desk is 4.5 minutes, and this time is exponentially distributed.
(a) How long will the customer expect to wait for service
at the cash desk?
(b) What is the chance that the queue length will exceed
S?
(c) What is the probability that the cashier is working?
S. In question 4, if by the application of work study
techniques the average service time is reduced to 4
minutes, how long will customers have to wait on average
under this system?
What is the probability that the customer will have
to wait more than 10 minutes for service?
6. At what average rate must a clerk at a supermarket
work in order to ensure a probability of 0.90 that the
customers will not have to wait longer than 12 minutes?
It is assumed that there is only one counter, at which
the customers arrive in a Poisson fashion, at an average
rate of 15 per hour. The length of service by the clerk
has an exponential distribution.
7. During certain weekdays the average arrival of customers for service at the deposit counter in a bank is
4 per hour. The average time to complete the deposit
is 6 minutes with a negative exponential distribution.
Calculate the average queue size and the probability of
having more than 4 persons in the queue.
At Friday lunch time the arrival rate goes up to 8
20
per hour. What must the service time be reduced to in
order to ensure only a 1 in 100 chance of any person
queueing for more than 12 minutes?
8. A repairman is to be hired to repair machines that
break down at an average rate of 3 per hour. Breakdowns
are distributed in time in a manner that may be regarded
as Poisson. Non-productive time on anyone machine is
considered to cost the company £5 per hour. The company
has narrowed the choice down to two repairmen, one slow
but cheap, the other fast but expensive. The slow cheap
repairman asks £3 per hour; in return he will service
breakdown machines exponentially at an average rate of
4 per hour. The fast expellsive repairman demands £5 per
hour, and will repair machines exponentially at an average rate of 6 per hour. Which repairman should be hired?
9. A company is considering installing a tool-grinding
machine for use by its operators. The following proposals
are under review.
Machine A cost £2000
Average grinding time- = 10 minutes
(negative exponential distribution)
Machine B cost £6000 Average grinding time = 8 minutes
(negative exponential distribution)
Machine C cost £10000
Average grinding time = 4 minutes
(All machines have to be written off in 2 years' time).
If the demand rate of operators for grinding is 5 per
hour, and the cost of the operators' non-productive time
(including wages and lost production) is £2 per hour,
which machine should be installed by the company? (Assume
SO weeks/year, 40 hours/week.)
10. Ships arrive randomly at a harbour, and the unloading
time is 1 day on the average. It is negatively exponentially distributed. Given a S-day working week, calculate
the distribution of ships' waiting time for
(a) average arrival of 3 ships/week
(b) average arrival of 4 ships/week.
11. In a supermarket, the company's policy is that customers should wait only an average of 2 minutes for service.
Given that customers arrive randomly at an average rate
of twenty per hour and the average service time is 2.2
minutes (negatively exponentially distributed), what will
be the actual average waiting time?
By how much must the average service time be reduced
to give an average waiting time of 2 minutes?
12. Patients arrive at the casualty department of a hospital at random with an average arrival rate of 3 per
21
hour. The department is served by one doctor who spends
on average 15 minutes with each patient, actual consulting times being exponentially distributed.
(a) What proportion of the time is the doctor idle (that
is, has no patients to examine)?
(b) How many patients are, on average, waiting to see the
doctor?
ec) What is the probability of there being more than 3
patients waiting to see the doctor?
(d) What is the average waiting time of patients?
(e) What is the probability of a patient having to wait
longer than one hour?
13. Before parts are assembled into a vacuum tube, they
must be cleaned in a degreaser. Batcpes of parts are
brought in randomly at an average rate of A batches per
hour, and are cleaned at an average rate of ~ batches
per hour. The cost of delay is £C 1 per batch per hour,
and the cost of owning and operating a degreaser that
works at an average rate of u is £uC2 per hour. Prove
that at minimum total cost
3.4
SOLUTIONS
1. For an M/M/l/oo system, the probability that there are
n persons in the system is
P
n
." pn (1 - p)
Taking logarithms
log Pn = n log p
+
log (1 - p)
Therefore, if log Pn is plotted against n, the points
should fall on a straight line with slope log p and
intercept log (1 - pl. Now
fn
Pn
"fr].
i
Thus log Pn
therefore
log fn
n log p
+
log (1 - p)
+
log rf i
Therefore, if log fn is plotted against n, the points Sh01Jld
22
fallon a straight line with slope log
(1 - p)+ log rf i .
p
and intercept log
Frequency
No. of Persons
in
System
Log fn
(f )
n
(n)
o
2.9542
2.9143
2.8182
2.7931
2.7033
2.6232
2.4800
2.4041
2.2577
2.1818
900
821
658
621
50S
420
302
252
181
152
4810
1
2
3
4
5
6
7
8
9
Total
Plot log f against n (see figure 3.2). For accuracy,
a regression ~ine should be fitted to the points, but for
the purpose of this question it is sufficient to fit a
line by visual inspection. The scatter of the points about
this line is small, and it is safe to assume that within
the error of sampling,these points fallon a straight
line. Therefore the survey does indicate that the queueing process conforms to an M/M/l/oo system.
3.0
2.8
c:
2.6
C>
..Q
2.4
2
3
4
5
n
6
7
8
9
Figure 3.2
Estimation of Traffic Intensity (p)
From the graph, the slope is measured as -0.0833, that
is 1.9167. Thus log p =1.9167
traffic intensity p = 0.825
23
10 sets per day
16 sets per day
2. Arrival rate A
Service rate
A
S
P = - = "8
II
~
Expected idle time per day
Po = 1 -
3
p
proportion of time
system empty
"8
that is,3 hours per day, on average, the repairman is
idle. Average number of television sets in the system is
-n
5
3"
Therefore average number of sets ahead of new arrival= 5/3.
3. Here average arrival rate A = 1.5 trucks per hour.
The service rate II depends on the system. Average unloading tjrne per truck is
of sacks/truck
s = no.
(h)
unload1ng rate/h
Average time a truck is in the system is
CI =
1
~
(h)
therefore
1
~
delay cost/truck arrival
and
delay cost/day
1
~x
no
x
no
x
15
Actual variable operating
cost/day
= variable cost/day
x
utilisation of system
£ C x p
V
Scheme
Average
Unloading
Rate/Hour
(~)
A
B
C
2
4
12
Average
Utilisation Average
Arrival
(p)
Time in
Rate/Hour
£.ystem
0.)
1.5
1.5
1.S
24
d
0.75
0.3i5
0.125
(h)
2.0
0.40
0.095
The costs of each system are giveL
Scheme
Fixed
cost/day
varial'le
A
B
C
Delay
Cosr/day
(S)
cost/day
(S)
60
130
250
belo~.
(S)
75
Total
Cost/day
(S)
30(1
6<'
56
-~
4 ':'v
246
289
14
25
Thus scheme B has the lowest total cost.
4. Assume random arrivals
A = 5
every 30 min =
1
0"
Average service time
per min
1
9
2"
\.l
thus p = 92 per min and p
1 x 9
"6
(a) Average waiting time w
"2
min
3
"4
(p - A)lJ min
1
=
"6
("92 -"61) "9"2
13.5 min
(b) Probability of n customers in the system = Pn
Probability of n customers queueing
Probability of more than n customers
queueing
= Pn +l
E Pi + 1
i=n+l
n+2
=
therefore
Probability of more than 5 customers
queueing
p
0.133
(c) Probability that cashier is working
probability of
one or more customers in the system = 1 - probability of
no customers in system
25
1 - Po
=
also
Po = 1 - P
thus
5.
probability that cashier is working
A = 6
1/.
mln
3
p
'4
2
and p = '3
11 =·l./min
1
4
6
Average waiting time w
min
(i - i) {
8 min
Waiting time distribution
w(t)dt = P(ll - A)e -t C"-A)
~
dt
r
Probability of waiting longer
than T minutes
with T
eowet) dt
Jt=T
-T(ll-A)
pe
10 min
therefore
Probability of waiting
longer than 10 minutes
0.29
6. Let service rate of clerk
II
probability of waiting
longer than T
!reo
T
= 15
=
Then
wet) dt
!e-(ll-A)T
II
=
Given A
per hour.
per hour
0.20 h
therefore
probability of waiting
longer than 12 minutes
26
15 e-Cll-15)0.20
1.1
O.7S
This expression must be less than or equal to 0.1, therefore
0.1 ..._ 12e - (ll -1 5) a . 20
II
Giving II
25 per hour.
to nearest whole number.)
24 gives 0.1033, 25 correct
(ll
7. Average queue length is
q
with A
L
n=l
(n - l)P
4/h and
22
5
q
2
1- 5
II
2
=
n
p
~
10/h
0.267
00
L Pn
n=6
Probability of more than 4 in queue
p6
=
Now A = 8/h
1
Probability of waiting time> 5 h
1
0.004
00
w(t)dt
1/5
This must be less than or equal to 0.01, that is
log
e
pe-()J-A)/5< log 0.01
e
On substituting various values for ll, the value (to the
nearest whole number) II = 26/h satisfies the above inequality.
8. Breakdowns random, A
3 per hour.
Slow Man
Repairs machines exponentially II = 4 per hour.
waiting time of a broken machine is
w
3
(4-3) 4
27
Average
3
'" "4
h
Average service time '" 1/4 h, therefore
Average time to complete repair '" 1 h
therefore
cost per hour of machine breakdo~~s
labour cost per hour
£3
total cost per hour
£18
1 x 5 x£3
£IS
Fast Man
~
6 per hour.
Average waiting time is
3
1 h
(6-3)6"'6
w
Average service time '" 1/6 h and average time machine is
out of action
1/3 h, therefore
cost per hour '" £j x 5 x 3
Labour cost per hour
£S
total cost per hour
£10
£5
Hiring the fast man therefore gives a lower cost per hour.
9. Consider a period of 1 hour.
Machine A
Capital depreciation/hour
£2000
2 x SO x 40
£O.S/h
Machine B
Capital depreciation/hour
Machine
£6000
2 x SO x 40
£1. S/h
C
Capital depreciation/hour
£10000
2
x
Consider machine A
Average arrival rate
A
S/h
Average service rate
~
6/h
28
so
x
40
£2.S/h
thus
p
=
5
"6
Utilisation of grinding machine
0.83 or 83%.
5/6
Average delay per call
(includjng service time)
1
iJ=-);"
Total cost of lost time/hour = 5
1
1 h
~
x
1 x £2
= £lO/h
Consider machine B
Average arrival rate A
60
7.5/h
~
8
thus
p
5
7.S
5/h
0.67
Utilisation of grinding machine
Average delay per call
(including service time)
67%.
1
iJ=-);"
Total cost of lost time/hour
=
1 = 0.40 h
2.S
5 x 0.40 x S2 = £4
Consider machine C
Average arrival rate A
l5/h
~
thus
p
5/h
5
15 = 0.33
Utilisation of grinding machine
Average delay per call
=
=
33%.
1
1
iJ=-);"
15 - 5
Total cost of lost time/hour = 5
x
0.1
x
0.1 h
£2 = £l/h
Summary
Total Cost
(£)
Machine A
Machine B
Machine C
Capital depreciation
0.5
1.5
2.5
Cost of lost time
10
4
1
5.5
3.5
Total 10.5
29
Thus machine C (the most expensive) is the best installation, although, as will be seen, all the machines have
the capacity to handle the service.
10. Random arrival and exponential service time mean the
waiting-time distribution is exponential.
w(t)dt
=
p(l.I - A)e -t(l.I- A) dt
with mean
w
=
(1 -
p
pll..l
(a) Average arrival of 3 ships/week, therefore A
day, 1.1 = 1 per day, therefore p = 3/5.
w(t)dt =
~
(1 -
3/5 per
~)e-t(1-3/5)dt
6 e-(2/5)t dt
TI
Mean waiting time w
(1 - ~)l
=~
x
days
~ = 1.5 days
(b) Average arrival of 4 ships/week, therefore A = 4/5
per day, 1.1 = 1 per day, therefore p = A/l.I = 4/5. Waiting
time distribution is
~(l
wet)
_
4
-n
e
~)
e- t (1-4/5)dt
-t/5 dt
4
Mean waiting time -w
11.
1.1 =
(a)
"5
4 days
Random arrival: A = 20/h; exponential service time:
(60/2.2)/h, therefore 1.1
27.3/h.
Average waiting time w
(1 -
P
pJl..l
20/27.3
(1 - 20/27.3)27.3
30
0.733
0.267 x 27.3
6.0 min
(b)
P
w
(1 -
that is
A/ll
: (1 - ~)ll
thus 2]l - 2llA - A
2ll
6ll
II
2
2
0
- 3 ll - 3
1
0
- 2ll - 1
0
2
2
2 min
ph
= 2
= 1(14
12
+
24) per min
One root is negative, thus II = (2 + 5.3)/12 = 0.61 per
minute, that is 60 x 0.61 customers per hour must be
served.
average service rate
Thus average service time
II
=
= 36.6 per hour
60/36.6 = 1.64 min.
12.
(a) Probability (doctor idle)
probability (system empty)
1 - p
therefore doctor is idle for 25 per cent of the time.
(b) Average number of patients
waiting to see the doctor = average number in queue
2
00
q
p
= r-=-p
L (r - l)P r
r=l
(3/4)2
thus average number waiting
1 -
(c) Probability (more than
3 in queue)
00
L Pr
r=5
= p5 =
(i]
=
5
0.24
31
3
'4
2.25
Probability (more
than 4 in system)
Cd) Average waiting time -w
p
(1 - p)]..I
3/4
--3
h
1 - '44
0.7S h
(e)
Probability of a patient having
to wait longer than one hour
:£00w(t)dt
=
1
pe
'43
-(A-J.lL 3 -(4-3)
-
x
'4e
0.3679 = 0.276
13.
Average time batch spends in system
I
~
h
therefore
delay cost per batch
delay cost per hour CD =
CIA
£~
/h
cost of service facility C s = £]..IC 2 /h
total system cost CT = CD
+ Cs
Condition for mlnlmum cost is dCT/d]..l = O.
CT with respect to ]..I
ell - A) 2 + C 2
=0
for minimum
II
32
Differentiating
MjMj I jN SYSTEMS
4
4.1
INTRODUCTION
In this chapter the M/M/l/N system is covered - the
single-channel queueing system given in chapter 3 but
with a constraint that the maximum number in the system
cannot exceed N. Since the arrival rate has to be random
and constant, this special case is generated when the
number of potential customers is infinite, but they only
join the system when n < N; when n = N, the customers
arriving for service go elsewhere, that is, a non-captive
system.
Departure on
completion of service
Arrival of Customers
JI. = Average Service Rate
Maximum number in system = N
Customers go elsewhere if n = N
Figure 4.1
4.2
M/M/1/N System
RESUME OF THEORY AND FORMULAE
The basic formulae for this system are given below.
M/M/1/N System
Arrival
Average rate A Poisson rate distribution
Service
Average rate
Number of servers
Poisson service rate distribution,
neg. expo service time distribution
one
System size limitation
Queue discipline
~
max. size N
first in, first out
Probability of no
customers in the system Po
33
1 - P
1 - p N+l
Probability of n
customers in the system Pn
Proportion of customers not served
(non-captive system)
Average number in system n
Average number in queue q-
p
=
P
N
p
N
(l
-
p)
1 - PN+l
,+ NpN+IJ
[1 - eNp) +(1 l)pN
_ pN+l)
(l
P
2[1 - Np N-l + eN - l)p NJ
N+l
(l-p)(l-p)
NO formulae can be developed for average waiting time
although the general formula
average waiting time w
= ~
II
can be used to obtain the average customer waiticg time.
4.2.1
Optimisation of System under Conditions of Noncaptive Customers, No Constraint on Total Potential
Customers
G = Average profit per service
E = Average cost per service
Note: This condition assumes that the cost of providing
the service is linear, thus doubling the service rate
doubles the total cost of providing the service.
In addition, it is assumed that the service rate II
can be varied. Thus, if all customers were served
gross profit = AG per unit of time
However, only (1 - PN) of customers are served, therefore
gross profit per unit of time is
'G(l - PN l
=
;l_-p~;iN]
,G [1
Net cost of service per unit of time
Net profit per unit time is
Z = ,G ( : -
::+
E"
=
34
AllG
("
Ell
II - ,N
)
llN+l _ AN+l - Ell
If A, G and E are fixed, the problem is to vary u to maximise Z. For maximum Z, dz/d~ = 0, which gives
J~
pN+l[N - eN + l)p + pN+l1=
(1 - pN+l)2
G
For given E/G and N, table 3 gives the optimum value of
p, from which the optimum value of u can be found.
An example of use of this theory is now given.
Example
In this example, the practical difficulty of measuring
the arrival rate A will be demonstrated. Clearly, only
the customers who enter the system can be measured.
In a large department store, at the jewellery counter,
there is only one assistant. Activity sampling studies
of the service gave the following data on the number in
the system.
No. in System
No. of Readings
o
138
286
1
2
576
Work study measurements of the service time gave an average of 10.5 minutes and the distribution was negative
exponential.
Calculate p for the system and show the proportion of
prospective customers lost. Calculate the expected average waiting time.
If the average profit per sale is £5 and the average
cost per service is £2 calculate the optimum service rate
and the increase in net profit.
Solution
This problem illustrates the more practical case where
there is no simple straightforward method of measuring
the average arrival rate A, since all that can be normally measured is the input to the system. Therefore use
is made of activity sampling studies and A is estimated
from the number in the system data.
Estimate of the arrival rate A is as follows.
35
in System
(n)
No.
0
1
2
No.
Total
of Readings
Prorarility
(r n )
0.138
0.286
0.576
1.000
138
286
576
1000
Log
P
n
1.1399
1.4564
1.7604
Since Pn = popn, then p can be calculated in two ways.
The first where pn-r = (Pn/PrJ gives three estimates for
p
p
286
576
or
2.02
286
I~
or
2.07
= 138
2.04
~138
The second method is a graphical one.
then log Pn = n log p + log Po and
log
p
=
Since Pn
(log P n - log PO)
n
If log Pn is plotted against n, the points should fall
on a straight line; the slope of this line gives log p
from which p can be calculated. Figure 4.2 shows this
plot.
0.00
Slope = .310~
1.50
[L
c:
'"
0
...J
1.00
~
~
o
Figure 4.2
Slope
log p
and
,...-
./
V
~
2 n
Log P n against n
0.310, therefore
= 0.310
p = 2.04
36
Average service rate
60
10.5
~
5.71 customers per hour
therefore
A
5.71
=
2.04
Average arrival rate A
2.04
5.71
x
11.65 customers/h
Maximum number in the system N = 2, therefore probability of system being idle is
-
1
1
PI
0.139 x 2.04 = 0.28
P2
0.282 x 2.04
p
p3
1
1
- 2.04
- 2.04 3
Po
-1.04
-7.50
0.14
0.58
(Cross-check: Pc + PI + P 2 = 1.00. Thus P 2 , proportion
of customers who go elsewhere for service, = 0.58 or 58
per cent.)
Optimum Service Rate
Here E = £2, G = 55, N
for optimum
Reference to table 3 gives
1.16
p =
Since A
2.
=
11.62, therefore
~ -- l:T6
11.62 -- 10 customers/h
Compare present service rate of only 5.71 customers per
hour, thus the service rate should be increased.
Increase in Net Profit
Current net profit/t
ss
x
(1 - 0.58)
x
Total profit/h - Cost of
service/h
11.62 - £2
37
x
5.71
£24.4 - 11.4
£13 per hour
With
~
=
10, p
=
1.16, N
1 - p
1 - pN+1
=
2
1 - 1.16
1 - 1.16 3
0.16
0.56
0.286
thus PN = 0.286 x 1.16 2 = 0.385
therefore
Optimum net profit/h
£5 (1 - 0.385) x 11.62
- £2
x
10
US.7 - 20
£1S.7/h
or an increase of £2.7/h in the net profit.
4.3
PROBLEMS
1. Show that for an M/M/l/N system the steady-state probabilities are
1 - p
1 - p
N+l p
n
where p
A/~ and derive the limit values for these probabilities when p+1.
2. A petrol station has a single pump and space for no
more than 3 cars (2 waiting, 1 being served). A car
arriving when the space is filled to capacity goes elsewhere for petrol. Cars arrive according to a Poisson
distribution at a mean rate of one every 8 minutes.
Their service time has an exponential distribution with
a mean of 4 minutes.
The proprietor has the opportunity of renting an adjacent piece of land, which would provide space for an
additional car to wait. (He cannot build another pump.)
The rent would be £10 per week. The expected net profit
from each customer is SOp and the station is open 10
hours every day. Would it be profitable to rent the
additional space?
3. Activity sampling studies at a service counter gave
the following data.
38
No. in System
No. of Readings
o
1605
971
531
342
1
2
3
Average service time
=
10 minutes.
Given that the arrival rate of potential customers is
random, that the service time has a negative exponential
distribution, also that the cost of providing one service
is £1 and that the average gross profit per customer is
£2, what service rate should be adopted and by how much
will the net profit be increased?
What proportion of potential customers will be lost
under the conditions of maximum profit?
4. Customers arrive at random at a single service station
with mean arrival rate A. However, a customer will join
the system to be serviced only if there are less than N
customers already in the system when he arrives, otherwise he goes elsewhere.
Service times are distributed
according to the negative exponential distribution with
mean time l/~, and customers who join the system are
served in the order in which they arrive.
Spow that the expected number of customers passing
through the service station in unit time is given by the
expreSS[i:n _
A
1 _
(~) N
(f)
]
N+l
The above model represents a small petrol filling
station for which N = 2 and ~ = 30 per hour. A sum of
money is available for improving the station and can be
used either to buy more space to increase N to 3 or to
install a faster pump to increase ~ to 40 per hour.
If
A is 10 per hour, which of these two alternatives will
produce the greater increase in profits, assuming the
average profit per customer remains constant?
If A were 30 per hour, would the same answer be valid?
5. Consider a production process with two machines A and
B.
The output from the first machine is Poisson distributed, with the rate of 4 per hour.
The second machine B
processes each item with exponential service time at a
rate of 5 per hour. There is a congestion problem when
39
more than 2 items are waiting to be processed on machine
B.
(a) Calculate the fraction of time that the system is
congested in the steady state.
(b) Suppose machine A is turned off when more than 2
items are waiting to be processed by machine B. Calculate the steady-state probability of the system
being congested in this case. Compare with answer
in (a).
4.4
SOLUTIONS
1. Steady-state equations are
-)'P O
+
llP I
0
o
>.P n - l - (>. + ll)P n + llP n + 1
=
n
(p
Pn + l
PN
+
I)P
(p
Pn
n
p Po
+
=0
n = N
pP N- I
P2
I)P I - pP O
=
p
2
Po
N
I: Pn .. I
n=O
1
Therefore
pn
P .. I - p
n
I - pN+l
lim Pn = lim pn (1
p+l
n ,
n - pP n-l
These may be solved by recursion
From
I,
)./ll
p Po
PI
0
n = N
>.P N- I - llP N = 0
Setting p
n
p+1
-
P~
(1 - p +1)
40
N - I
lim pn lim 1 - P
p+ 1
p+ 1 1 - pN+1
1 lim 1 - p
p+ 1 1 - pN+1
The latter limit can be obtained by applying L'Hospita1's
rule
lim 1 - p
p+ 1 1 - pN+1
lim d (1 _ p) J1im ~ (1 _ pN+1)
p+1 dp
J p+1 dp
-1
1
- (N+1)
N+1
2. Present System
Average arrival rate A
60
8"
7.5 cars/h
Average service rate
4
60
15 cars/h
p
=
A
~
~
7.5 = 0 5
15
.
Maximum number in system, N
being empty
Po =
thus Po
=
3.
Probability of system
1 - P
N+1
1 - p
1 - 0.5
1 - (0.5)4
= 0.533
Probability of system in maximum state, N
P3
0.533
x
3, is
(0.5)3
0.067
Therefore proportion of lost customers
0.067
Proposed System
If maximum number in system increased to N
p
o
=1
- 0.5
1 _ (0.5)5
=
4
0.516
Probability of system being in maximum state is
41
P4 = 0.516
(0.5)4
x
0.032
=
0.032
thus proportion of lost customers
theyefore
Increase in cars
served per hour
\(0.067 - 0.032)
7.5
x
0.035
0.262 cars/h
Increase in cars
served/week
0.262
x
10
x
7
l8.34/week
Thus increase in profit per week
SOp
x
18.34
£9.17
Since rent for additional space would be £10 per week,
it is not economical to increase the existing space.
3. Determine arrival rate (\) by finding average traffic
intensity (p). This is an M/M/l/N process. Thus
P
p)
(1 -
n
1
where N
-
p
N+l
pn
maximum number in system.
Pn
=
Pn - l
Therefore
p
let xn = number of activity sample readings and Pn
Since N = 3
p
=
1
xn
3
1:
"3 n=l x n - l
(see also graphical method, p.
0.644
p
+
0.547
3
+
0.605
0.6, but V = 6/h, therefore
\ = PV =
3.6/h
42
)
0.60
«
xn .
The optimum values of p as a function of Nand E/G are
given in table 3. For N = 3 and E/G = 0.5, p = 1.21.
Hence
A
p
3.6 =
r:zr
3/hour
To determine the increase in profit, compute the present
and the proposed profits.
Present profit
= s.z
x 3.6 x [1 - 1 - 0.6 x 0.6 3] _ 1 x 6
4
1 - 0.6
£O.46/h
s.z
Proposed profit
J-
x 3.6 [1 - 1 - 1. 21 x 1.21 3
4
1 - 1.21
= £1.86
Increase in profit = £1.86 - £0.46
= £1.40/h
Proportion of customers not served is PN
P
N
=
(1 _ p)pN
1 - p N+1
= 0.6, N
original customer loss PN for
p
(1 - 0.6)
0.87
0.10
PN
=
x
0.22
=3
with new service rate
P
N
=
(1 - 1.21) x 1.77 = 0.32
-1.14
(32\)
Note that the solution maximises profit but, as can be
seen, about one-third of the potential customers are lost.
4. Average number passing through service station =
A(l - PN), but
therefore
43
1 x 3
[
1_(~)N ]
, -(;t'
Therefore average number through service station per unit
time Sen) is
Sen)
• A
lJ
[ 1 _(~)N ]
,-(;r
(a) Consider installing a faster pump, N
40/h.
Sen)
10
x
1 - (1/4)2
1 - (1/4) 3
10
x
IS
64
T6 )( 63 = 9.5
10/h,
2, A
that is, an average 9.5 customers will be passing through
the service station per hour.
(b) Consider buying extra space, N
30/h.
10
Sen)
x
3, A = 10/h,
lJ
1 - (1/3)3
1 - (1/3)4
10 x 26 x 81 = 9.75
T!
80
that is, on average 9.75 customers will be passing through
the service station per hour.
Therefore, choose to purchase the extra space. Suppose
A 30/h. For case (a), the faster pump, N = 2, A = 30/h,
lJ
40/h.
(3/4) 2 = 22.7
Sen) = 30 x 1 1 - (3/4)3
For case (b), the extra space, N
30/h. Therefore
p
A
lJ
=
3, A
30/h,
1
Here it is not possible to use the formulae Pn
44
lJ
to calculate the probabilities.
However, since
therefore
thus Po
1
4"
and it follows that PI = P 2 = P 3 = Po = 4"1
With A = 30/h the number of customers who go elsewhere
is 30 x P3 = 7.S/h. Therefore
Sen)
=
22.5
Hence it would be marginally better to install a faster
pump.
S. (a) M/M/l/~ system, with the input (A) the output
from machine (A), the service rate (~), the production
rate of machine (B).
A
4/h
II
S/h
p
4/5 = O.S
Prob. (system congested) = Prob. (more than 2 in
queue)
Prob. (more than 3 in system)
~
0.410
that is, the system is congested 41% of the time.
(b) Machine (A) is turned off whenever there are 4
items in the system. Therefore this is an M/M/l/N system,
with N = 4.
Prob. (4 in the system)
Prob. (system congested)
(1 - p)
p4
(1 _ pS)
0.2
5 x
(1 - O.S )
a • s4
0.122
Now the system is congested for 12.2% of the time.
4S
5
MIMICI
5.1
INTRODUCTION
SYSTEMS
Again the theory of these systems is to be found in most
textbooks. See references at the end of book.
In the M/M/C/oo system, customers arrive randomly for
service, service time distribution is negative exponential, C servers, no constraint on queue size and customers
are served in order of arrival.
'---..,..........-
-
~-
.... - - - - - - - - - - - - - .........---:,.....~
.......
' •• ~
IJ.
,....
=
Average service rate
of each counter
'4.-
Arrival of
customers
'. :::::"
...... ,'.' ..
:.......•..
~
Departure on
completion of service
Figure 5.1
5.2
M/M/C/oo System
RESUME OF BASIC THEORY AND FORMULAE
The basic formulae for these systems are giver. below.
Probability of no customers in the system is
P
o
j 1 + 1 A c c~ ] -1
= [ C-l(A)
L
-...,.....,..
'C! (i1 ) C~-A
j =0 ~ J ~
Probability of n customers in the system is
P
n
Po(~rir
c
n ,
n-c
Pn P0(~) (cA~)
x
ir
c
n ) c
Average number in the queue is
c
POA~(A/~)
q
(c-l) ~ (C~-A) 2
46
Average number in the system is
-
>..
-
n=q+"ij
Average waiting time is
w
Po).l(A/).I)C
')
(c-l): (cw>") '-
Average time in the system is
a = -w +
1
).I
Waitir.g time distribution is
w(t)dt
Prubability of a customer waiting longer than time t is
P(w>t)
s. 3
Poc(>,,/).I)c -(C).l->")t
c! Ccw>") e
PROBLE~!S
1. Arrivals at a telephone booth are considered to be
Poisson, with an average time of 10 minutes between one
arrival and the next. The length of a 'Fhone call is
assumed to be distributed exponentially, with an average
time of three minutes.
(a)
What is the probability that a person arriving at
the booth will have to wait?
(b)
What is the average length of the queue that will
form from time to time?
(c)
The G.P.O. will install a second telephone when
convinced that a customer would expect to have to
wait at least 3 minutes for the 'phone. By how much
must the arrival rate increase in order to justify
the installation of a second 'phone booth?
Cd)
What is the probability that a customer will have
to wait more than 10 minutes for the 'phone?
(e)
If the second telephone is installed what will the
average waiting time now be per customer?
47
2. A small ship-building company has s slipways on which
it builds cabin cruisers and yachts, of a wide range of
tonnages. The average building time is l/~ and may be
considered to be negative exponentially distributed. Orders
for ships have an average inter-arrival time of l/A, which
is also negative exponentially distributed. The orders are
dealt with in order of arrival.
Write down the steady-state probability difference
equations, and show that the expected number of slipways
in use at any moment in time is independent of the number
of slipways (provided that the system is not overloaded).
If there are 4 slipways, calculate the probability that
an order has to wait for a slipway to become available,
given that the average time on the slipway is one calendar
month and there are an average of 12 orders per year.
3. A single server in a supermarket has a service rate
of 30 customers per hour, and the service time distribution is negative exponential.
Given that an acceptable average waiting time per
customer is 2 minutes, calculate at what customer random
arrival rates more assistants should be put on to serving
duties.
[raw a graph to show, for random arrivals, random service time, several servers, the relationship between
waiting time, arrival rate and number of servers.
4. A company manufacture~ ex~ensive custom-built precision instruments. Before de11very the instruments are
inspected, and instruments arrive for inspection according to a Poisson distribution at a mean rate A of one
every 100 minutes. At present the company employs one
inspector and the inspection is distributed negative
exponentially with mean rate ~ of one every 90 minutes.
The average value of the instruments is £5000, and under
the present system the value of work in progress awaiting
future inspection is high. The management consider they
can obtain 25 per cent rate of return in alternative
investments, if capital tied up in this stock can be
reduced.
If an additional inspector costs the company £6000 per
year, would it be economical to employ a second inspector?
If a second inspector is employed what is the proportion
of time they would be working on average?
5. A machine-process has an exponentially distributed
service time with mean l/~. Items arrive randomly for
processing at a rate A; {A/~<l}. At present, management
think there is much congestion and two proposals have been
suggested.
48
(a) To buy a second similar machine-process station to
be operated in parallel;
(b) To replace the present machine-process station with
a machine-process station of twice the service rate
(but still exponentially distributed).
Given that the probability of an item having to wait at
present is equal to the traffic intensity, by setting up
the steady-state probability difference equations for the
two situations, determine which of the two alternatives
will produce the greater reduction in the steady-state
average queue length.
If jobs were to arrive at the rate of 15 per hour, and
the present machine-process can handle 16 per hour, what
would be the expected average queue length for the present
system and its expected value under each of the proposed
alternatives?
6. In a self-service store the arrival process is Poisson
with 9 customers arriving every 5 minutes on average.
A single cashier can serve 10 customers every 5 minutes
on average with time distributed negative exponentially.
The store manager wishes to reduce the probability
that customers have to wait more than 1 minute. In
order to do this he can either double the service rate
by providing an additional server to pack the customers
goods or by providing an additional cash desk. Which
scheme is preferable?
7. In a machine shop the following study was carried out
on a grinding machine installed for the use of operators
(number of operators = 100): the average time taken to
grind tools was 10 minutes and this time was negatively
exponentially distributed; the average number of calls
per week for grinding was 200 in a 44-hour week.
What is the total time lost by operators, in one week,
in waiting in the queue for the grinding machine?
If a second machine is installed, by how much will the
total waiting time be reduced?
8. A company is studying a proposed operation involving
a new fleet of trucks and an unloading depot, at which
it is estimated they will arrive in a random fashion at
an average rate of 3 per hour. It is proposed to construct either one or two unloading bays. The average
time taken to unload a truck at an unloading bay is
estimated to be 15 minutes. A queue of lorries for unloading will form from time to time. The capital cost
of each truck will be £8,000. The cost of constructing
49
the second unloading bay would be £10,000. Consider the
question whether the construction of the second unloading
bay would be worthwhile, given that the unloading time is
negatively exponentially distributed.
5.4
SOLUT IONS
1. A = 0.10 arrivals per minute,
therefore
p
= 0.33
~
calls per minute
= ~ = 0.3
II
(a) Probability that an arrival has to wait = probability
that the telephone is in use = probability that there is
at least one person in the system = 1 - probability that
there is no one in the systen
1 - Po
1 -
(1 - p)
=
p
= 0.3
(b) Average length of queue
2
1-p
q
0.3 2
1-0.3
----- =
0.13 persons
(c) Average or expected waiting time
with
0.33(O.33-A)
II =
0.33
Equating 3 = A/{0.33(0.33-A)} gives A = 1/6 arrivals
per minute. Thus a second booth is only justified,
according to the stated criterion, if the arrival rate
per hour is increased from 6 to 10 or more.
(d) pew > t) = probability of waiting more than time t
in queue
pe -tell-A)
0.3e- 10 (0.33-0.1)
(e) This is an
c by 2
w =
M/M/2/~
= 0.3e- 2 . 3 = 0.03
System.
PO~(A/ll)2
1!(2~-A)2
and
so
From section 5.2, replacing
P
a
=
[1 +
!).I + TI).Ij
1 (!\2 2).1]-1
Z).I-X
With the substitutes X
Po = [1
+
0.3
= 0.1, ).I = 0.33 and X/).I
i
(0 3) 2
+
0.66
0.66-0.1
0.3
]-1
= (1.35) -1 = 0.74
Average waiting time is
2
W= 0.74 0.33(0.3)2
= g:~~~ = 0.07 min = 4.2 sec
(0.66-0.1)
Or Average waiting time, with two booths, is 4.2 sec.
2. This is an M/M/C/= system with C = s.
state equations are
- APo
+ ).IP 1 =
The steady-
a
- (X+).I)P 1 + XP O + 2).1P 2 =
- (X+:).I)P 2 + X:1 +
3).1~3 =
a
a
a
- (X+S).l)P s + XP s - 1 + S).lP s +1
- (X+S).l)P n + Pn- 1 + S).lP n+ 1
=
=
a
a
n < S
n = s
n
>
s
Solving the equations recursively gives
n
~P
n. a
n , s
n
~
s
where p = X/(S).l).
NOW, expected number of slipways in use = expected
number of customers in service = expected number in system
expected number in queue, that is
n - q =
51
5
l:nP + l: nP n
1 n
5+1
00
l: nP + 5 l: Pn
5+1
5+1 n
00
5
l:nP + 5 l: Pn
1 n
s+l
SS p n P
L.
-s'
0
s+l'
s~n
~ sp
p
00
+ s~
L.
l.
' 0
n1I.
E(s)
5
f(,
PP0 L~ + Sp +
~
2!
+ •••
+
(sp) S-l) + (S
~ s
s-l!
s!p
+
Ss Ps+l
sr
+
•••
)11~
therefore
E(s)
=
spP O
1
~
o
= sp
since p = A/(S~), then E(s) = A/~, which is independent
of the number of slipways. With s = 4
probability that an order has to wait
44 (A In
Pn = 4~ ~J Po
n"} 4
=(ir- 4 t:Po
since
L~o(»Ji: l(~)' 4:-'
L:oi \ ('~l)
"0
+
+ ,;
=e66 +
is)
-1 =
(g)
r
~
r
-1
and
00
therefore, probability of wait
= Po
l! n=4l: i )n-4
co
(
S2
1
C-i74)
Po 4l!
4
3x4!PO
Po
1
IS
49
3. In order to obtain the arrival rate gIvIng an average
waiting time w of 2 minutes, we must compute the waiting
times and derive the result by approximation. The simplest
method for c > 1 (that is, more than one server) is to
dEtermine values of l/i the reciprocal of average waiting
time.
Let
For an M/M/C/oo process
A/~.
£
c
+ (c-l) ~ (c-£)
Therefore
1
].I( C-E)
w
[1
+
(C_E)C~l
]-1
(c-l)!
J!
j =0
xl]
~J
E
+ •••
.!.£:~) ')rc-,) ]
+
For c
>
+
5.1
1, this expression reduces to
C
1
- = ll(C-E; £"
W
and for c = 1
1
ll(l-E)
+
(i - 1)
c-2 c-l
x
L
II
j=O k=j +1
-kE
5.2
from 5.1
E
The following results were calculated on the Olivetti
(101) and give the graph shown in figure 5.2.
S3
No.
of Servers
1
Upper limit of A
2
15
Increase in A before
change
42.4
27.4
71.1
28.7
5
4
3
100.2
29.1
129.5
29.3
4.
(a) Present System
1
Rate of arrival
100 /min
1
= mJ
/min
Rate of inspection
thus
p
= -IIA = 0.9
Average number of instruments in the system
n =-P--9
1-p (b) Proposed System (Two Inspectors)
From section 5.2, average number of instruments in inspection system is
n
where q
and Po
=
-
q +
IIA
POAll(A/ll)
2
with n = 2
(211-A)2 1 !
= [1
+
~1J
+
..l..(~)2
2! II
that is
+ (0.9)
1
+ Z(0.9)
2
~
2
90
= [1
+ (0. 9)
+
i
(0 • 81 )
= (2.64) -1 = 0.379
54
]-1
1
TOO
-1
J
2-~. 9
U1
U1
l\)
I.ro
C1)
J::
11
I.Q
...."l
W=
AVERAGE
WAITING
TIME IN
QUEUE,
MINUTES
0.0
0.5
1.0
1.5
2.0
2.5
o
I !
I
20
/
60
BO
V
I
A= ARRIVAL RATE, CUSTOMERS PER HOUR
40
r/
I
100
7
/
I •
,.. = SERVICE RATE PER SERVER C = NUMBER OF SERVERS
30 PER HOUR (POTENTIAL)
120
I
140
~
A = 129.5
I
t-
~
1::~
,...-...
p;
2
3to
'""C
a
'"iil3
:::l
"
1::
0::!.
a3
>'
~
"
l>
x+-
0 ....
:-'~'"
§",t.>
-....11: ~
>'
I
0
-I
n
..n
r
1
1
0.379 x 100
x 90 (0.9) 2
q
l!
- TOO
1
(0.9)2x90X100
0.379
0.379
(9~
(200-90)2
7290
12100
0.379 x 0.602
= 0.23
So that the average number in the system is
-n
0.23 + 0.9
1.13
Reduction in number of units in work in progress
=
9 - 1.13
=
7.87
Average value of this reduction = 7.87 x £5000 = £39350
Annual reduction in stockholding costs
£39350 x
25
TOO
£9837.5
Net annual saving by employing second inspector
= £9837.5 - £6000 = £3837.5
For a system with c servers the average utilisation per
server is
o
p
·i
O = 0.45
Therefore, on average, the inspectors will be working
45 per cent of the time.
5.
(a) Second Machine
Installing an additional machine.
The steady-state difference equations are
-AP
o
+)1P
1
=
0
56
-
H'O
=
(A + ~)Pl + 2~P2
(I
HI
2~P3
0
+ 2~)P3 + 2~P4
0
+ 2~)PZ
P 2 - (A
0
etc.
Sol,,-ing this equations recursively gives
A
PI
~
l(~)
2P
2 ~
0
Pz
(~r Po
1
Pn
or
Po
Zen-I)
Let ~ = £ ;
j.J
To
obtain Po
Po
+
thus PO (1
Po
thus
£
+
= 1
+
1£2
+
;Z£3
+
••• )
2[i +(1)+ (7)2 + (~/
PO{-1
Po
+
Pz
+
PI
= 1
+ ... J = 1
+ Z [1 +(1)+ (1)2 + (1)3 + ...
(1-~72
-1 ) =1
Z-£
Po = Z+£
Average number in the queue is
57
J}=
1
(2+E) (2-d
5.3
(b) One Machine Twice as Fast
q ;
l~:
where p ;
(~)
5.4
2 (2-d
Subtract equation in 5.3 from equation 5.4
5.5
Equation 5.5 is positive, hence two machines in parallel
give a shorter queue. Original A ; 15, ~ = 16, therefore
E = 15/16.
One Old Machine
- _
q -
(15/16)2
(1-15/16)
14
One Faster Machine
f15/16i 2
- _
225
2x17x16
q - 2 2-15/ 6)
0.413
Two Machines
-_
715 / 16)3
q - (2+15 16)(2-15/16)
16x47x17
3375
12784 = 0.264
6. Double Service Rate
A = 1.8,
~
=
4,
p =
0.45.
58
pe -().I-A)t
P(w>t)
0.45 e -2.2
Thus P(w>l)
0.45 x 0.1108
0.05
'Two
c
=
Cashiers
2; A = 1.8; ].1
pew > t)
2;
£
= A/].1 = 0.9;
CE
~
1)
£/c
0.45
c
(cp)
-c).l(l-p)t
Po c! (l_p)e
C
Poc!(c-£)
P (I<:
p
e-(C]1-:qt
2£2
-(2p-A)
2(2-E) e
2-£
2+£
-- x
e
-2.2
0.81
2.9
x
0.1108
0.031
Therefore 1 customer in 32 ~ill have to wait longer than
1 minute. Thus, if it is the store manager's objective
to minimise the probability of waiting more than 1 minute
for service he should install 2 cash desks.
7.
Single Grinding Machine
Requirement for grinding machine A = 200/week, rate of
grinding ].1 = 6/h = 264/week, therefore
p
A
].1200
264 = 0.76
Average waiting time;
p
v(l-p) weeks
0.0118 weel<s
Thus 200 x 0.118
per week.
2.36 man-weeks are lost due to waiting
Two Grinding Machines
This is an
M/~/2/oo
system.
From sect jon 5.2
S0
w
and
Po
1
+zr
=
zoo,
'll = 264
(0.76)
Substituting A
AI'll
Po
[\
+
0.76 + tr(0.76)2
ill ] -1
[ 1 . 76 + 2\ x O. 57 xl. 61 ] -1
[ 1.76 + 0.46
l -1
J
(2.22)
-1
therefore
Po = 0.45
therefore
264 x (0.76)2
0.0006 weeks
(2 x 264 - 200)Z
Man-weeks lost in waiting: ZOO x ~ = 0.12 weeks
-w
0.45 x
Reduction in time lost
2.36 - 0.12 = 2.24 weeks
8.
One Unloading Bay
Average arrival rate
'll = 4/h.
p
A
3/h, average unloading rate
= ~'ll = 0.75
Average number of trucks in the system is
n
=
p- -- 0.75
-1-p
1-0.75
3 t
k
ruc s
Average waiting time per truck is
w =
P
(l-p)'ll
= 3 x3" l=I h
Two Unloading Bays
From Section 5.2
60
Po
[1
= j ~O
(A)
j 1
iJ IT
[1 + 0.75
+
1
+ 2T
(A)
2
iJ
1
2! (0.75)
2
2ll ]
2ll-A
-1
2x4 ] -1
2x4-3
0.45
Average number in the queue is
q
(C-l) ! (Cll-A) 2
0.45 x3x4 x (0.75)2
lx(2 x3-4)2
0.12
Thus
Average number of trucks in the system is
-n
0.12 + 0.75
= 0.87
Average waitjng time is
w
~ = 0 312 = 0.04 h
The construction of an additional loading bay reduces the
number of trucks in the system by (3-0.87) or 2.13 trucks.
Thus for an investment of flO,OOO, two trucks (2.13 exactly)
are released for haulage duties. Since the trucks are worth
£8,000 each, it seems a worthwhile investment (providing the
two extra trucks do not lie idle).
6
6.1
SYSTEMS WITH ARRIVAL RATE AN D lOR
SERVICE RATE DEPENDENT ON THE
NUMBER IN THE SYSTEM (Mn/Mn/-I- SYSTEMS)
INTRODUCTION
There are a large number of practical problems that are
systems with either the arrival rate and/or service rate
dependent on the number in the system. For example, consider a maintenance gang responsible for five large machines each of which breaks down randomly, at the average
rate of once per week. Clearly the total arrival rate
depends on the number already broken down (number in the
system); again problems of telephone switchboards, car
hire, incentives to service personnel, some stock control
models are similar typical systems.
6.2
RESUME OF BASIC THEORY AND FORMULAE
These models have basic random arrival and negative exponential distribution of service time, with the additional
condition that the average arrival rate and/or the average
service rate are a function of the number in the system.
Thus, let An = arrival rate with n in the system and ~n
= service rate with n in the system.
The systems are otherwise general, that is, they can be finite or infinite
systems and either single- or mUlti-channel systems. These
systems are represented by
Mn/Mn/-/the dashes indicating that there is no constraint on these
conditions.
These problems are solved using the following formula:
probability of n customers in the system is
The average waiting time can be obtained by using the
general formulae
w
w
n
~
62
The wide area of practical application of this model is
clearly illustrated in the problems for solution in
section 6.4, and the special applications in section 6.3.
6.3
SPECIAL APPLICATIOKS OF THEORY
The use of this theory will now be demonstrated for four
special cases.
1. Derivation of Basic Formulae for M/M/l/N Systems
Here An
also
J.l n
p
n
A
o , n
0
n ) N
J.l
all n
AO
An _l
III
J.l n
< N
Po
therefore
A
PI
j.l
(~) 2Po
Pz
P
N
Po
(~)Np
J.l
0
=
N
thus 1 = r P.
i=O 1
therefore
1
thus Po
and Pn
=
1-0
1 -0 N+I
l-p
1 -p N+l
on
(see chapter 4.)
63
2. Derivation of Basic Formulae for M/M/C/oo Systems
If number in system n , c then there are no customers
waiting. However, if n > c, then c customers are being
served and (n-c) are queueing for service. Thus
An
A
all n
~n
n~
0
c~
n ) c
=
,n
<
c
therefore
An-I)
Pn =(AO
~l
Thus PI
Po
~P
~ 0
A
Pn
(~) 7iJP O
1
(~)n -,PO
n.
Pn
(~)n
P2
and
~n
n
c
<
1
P
c! c n-c 0
n ) c
(see chapter S.)
3. Machine Interference
This general problem is stated as follows: A single
operator is minding m machines; this system is shown
diagramatically in figure 6.1.
Machines
!~--------l~
r-------~:
operator
f
~
SoN;""
Rot. .
Figure 6.1
The system can be regarded as a single channel (one operator) with n customers. Thus
64
all n
Il
(m-n) A
a
n
=
m
etc.
Thus 1
m
L P.
i=O 1
=
Po [1
+
(~
m )
+
m(m -1)
(~) 2 ...
m!
(~) m]
Solving for Po enables a solution to be obtained for all
Pn . The following can then be evaluated
Operator utilisation = 1 - Po
Machine utilisation
where p = A/Il. Average time a machine is in the system
(broken down) is
m
1
d = Ile l - PO) -
r
6.4
PROBLEMS
1. A skilled setter controls 5 transfer machines. Calls
for resetting occur randomly on each machine with an
average per machine of 1 per hour of actual running time
(a machine is stopped as soon as resetting is required).
The distribution of setting time is negative exponential
with an average of 6 minutes.
(a) What is the probability of no machines being set
up at any given time?
(b)
What is the average delay for any machine in waiting
for the start of resetting?
65
2. With reference to a process characterised by random
arrivals and random departures, and using An for the parameter associated with an arrival when the system is in
state n and ~n for that associated with a departure,
derive the expression for the equilibrium probability
that the system is in state n.
Explain what type of system corresponds to the following expression for An' ~n'
(a)
~.
n
~n
(b) )'n
~n
n~
=
(n , c),
n~
~n
c~
(n > c)
Cc) An
ACn, k-l), An
o
Cn ;, k)
Ce) An = A(n , k-l), An
o
(n
(d) An
~
k)
~n
nll(n , k)
Prove that in case (e)
p
n
1
Cl/n!)pk
2
1 + p + l!P
+
.••
+
1
I! p
k
where Pn = probability that the system is in state nand
p = A/~ = traffic offering.
3. A company has a spares-replacement policy, which is
to order a spare when one is used from stock. The usage
rate is random with mean ~ per unit time and the replacement time has an exponential distribution with mean l/A
units of time. When there is no stock, required spares
are purchased immediately from a local supplier at a
premium.
If the maximum stock is N, show that the system can
be represented by a queueing process with random arrivals
and exponential service time distribution in which the
mean arrival is
An
(N-n)A
0
0
n >
,n
, N
N
and the service rate is
lIn =
~
n
~
1
66
where n is the number of spares in stock.
Hence show that the steady-state probability of having
n spares in stock is
A
~ +
=0
n >
N! (A) 2 +
(n-2)! ~
•••
N
If (a) the mean usage rate is 2 spares per week, and
the mean replacement time is 0.5 per week, (b) the cost
of a spare is normally £100, but costs £115 if bought
specially when no stocks are available, and (c) the cost
of stock holding is 0.5% per week, what is the maximum
stock level that minimises average total weekly cost?
4. A single-server queueing process is such that customers
arrive at a mean rate A per unit time, but only join the
system at a rate An where n is the number in the system.
Given that the service time distribution is negative
exponential with mean l/~, queue discipline is first-comefirst-served and that A = A/(n+l), derive an expression
for the expected numbernof customers in the system in the
steady state, and show that the expected queue length is
+
e -A/~
5. A shipping company has a single unloading berth with
ships arriving in a Poisson fashion at an average rate of
three per day. The unloading time distribution for a
ship with n unloading crews is found to be exponential
with average unloading time 1/2n days. The company has
a large labour supply without regular working hours, and
to avoid long waiting lines, the company has a policy of
using as many unloading crews on a ship as there are
ships waiting in line or being unloaded.
(a)
Under these conditions, what will be the average
number of unloading crews working at any time?
(b)
What is the probability that more than 4 crews
will be needed?
6.5
SOLUTIONS
1. The system is said to be empty when all the machines
are running, that is, no machines are being set up, and
67
the system is full when one machine is being set up and
four machines are waiting to be set up.
The arrival rate is dependent on queue size, therefore
let An = (S-n)A; n = 0, 1, Z, 3, 4, S. The steady-state
equations are
AOP O
(AI
+
}..l)P I
+
llP2
0
Al PI
(A Z
+
ll)P Z
+
llP 3
0
AZP Z
- (A3
+
1l)P 3
+
llP 4
0
"3 P 3
(A4
+
1l)P 4
+
llP S
0
A4 P4 - llP S
0
Solving these equations recursively gives
AO
PI
ilPo
Pz
-Z-P O
"0"1
].l
AOAl"2"3 A4
II
5
Po
(a) The probability that the system is empty = PO.
before
5
L Pn
(S-n)A, therefore
n=O
68
As
A
=
1 per hour,
~
=
10 per hour, therefore p
A/~
1/10.
Thus
POC1
+ 0.5 + 0.2 + 0.06 + 0.013 + 0.0012)
Po
1
1. 7732
PI
5p
x
0.564
0.282
P2
4p
x
0.282
0.113
P3
3p
.x
0.113
0.0.34
P4
2p
x
0.034
0.0068
P5
p
x
0.0068
=
0.564
=
0.0007
Cb) To find the average delay for any machine waiting for
the start of resetting.
(i)
Short Method Based on w
lilA Relationship
n
q
Pn
qP n
X
X P
0
1
2
3
4
5
0
0
1
2
3
4
0.564
0.282
0.113
0.034
0.0068
0.0007
0.000
0.000
0.113
0.068
2.820X
1.128X
0.339X
0.068X
0.003
SA
4X
3A
2A
A
0
0.204
X
O.OZO
q-
Therefore waiting time in queue is
- =
w
- =
~
~
0.204 x 60
4.362
=
2.8 min
69
n
n n
o.oon
0.0000
4.362A
(ii)
Short Method Based on w =
n/v
Relationship
Any machine arriving finds n in the system. When all n
have been serviced then the new machine will have finished
queueing and just be starting service. n
5
=n:gPn but the
machines arriving depend upon Pn , therefore
-
(lhl)
W
and l/l.!
6 min.
Pn
n
0
1
2
3
4
5
5
E (nP n An/X)
n=O
.564
.282
.113
.034
.0068
.0007
An/X
nP n
0.917
0.688
0.459
0.229
0.000
0.000
0.282
0.226
0.102
0.027
0.003
0.000
0.259
0.155
0.047
0.006
0.000
0.640
0.467
An
SA
4A
3A
2A
H
0
n =
Therefore w = 6 x 0.467
(iii)
nPnAn/ X
2.8 min.
Full Method Based on waiting Time Distribution
Assume any machine arriving finds n in the system. The
new arrival will wait until all the n machines have been
served (in time t+dt). Then the waiting time distribution
w(t)dt is based on
(a)
(b)
(c)
(d)
probability of an arrival (depends on system size)
system size on arrival
n - 1 machines being served in time t
the nth machine being served in interval (t+dt)-t
Now An
(5 - n)A and X =
5
E AnPn = (5 - n)A= 4.36A.
n=O
70
So
but
-
v;
~ (S-n) P
~n
(00 t n e-~t dt
n=1 ~ n (n-I)! Jo
Integrating term by term
n = I
n =
~
'-
,10
00
Ioc,
,
te-~t
;1=
dt =
~!ooo
t2e-~t dt
~
0
1
e- l1t cIt
0
11
te-~t dt
2
3
~
00
~fo
i!ooo t3e-~t
t2e-~t dt
n = 3,10 00 t3e-~t dt
10
n = 4, 0
00
t 4 e -~t dt
~
6
4"
~
dt
0
24
5
~
therefore
w
\.in
Cn-I)!
5 S-n
l: - - P
n=14.36 n
x
n!
-n+l
~
This has now reduced to the format in (ii)
ahove.
2. The steady-state equations are
"n-lPn-l
-
(,\
n
+
~n)Pn + \In+lPn+l
Solving recursively
PI
Pz
Pn
"0
-P
~l
1.. 0 "1
~l~Z
0
Po
"01..1. •. "n-l
~1~2···~n
Po
71
0
n > 0
where Po is given by
+
PO· (
l:
0
Pn
1, that is
AO
AOAI
+
+
]11
]11]12
+
AOAI
]11]12
...
...
An-l
+
]1n
.. )-1
(a) An
A;]1n = n]1. Arrival rate independent of
number in system with parameter A. Service rate varies
according to number in system. This model describes a
system with ample servers so that a queue never forms.
(b) An
A,]1n = n]1 (n , c), ]1n = C]1 (n ~ c). Arrival
rate independent of number in system with paramete~ A.
c servers so that for n , c no queue forms and the service
rate has parameter n]1. For n > c, all the servers are
busy and a queue of n-c customers is formed.
(c) An = A/(n+l),]1 n =]1. Arrival rate dependent on
number in system, service rate independent with parameter
]1. This describes a queueing system with discouragement,
that is, the sight of a long queue discourages fresh customers from joining it.
(d) An
A(n, k - 1), An = 0 (n ~ k); ]1n =]1. Arrival
rate dependent on queue size, service rate independent
with parameter]1.
This describes a queue with limited
waiting room (k only in system). Thus customers who arrive
to find the system full go elsewhere for service.
(e) An = A(n l> k - 1), An =0 (n ~ k), ]1n = n]..1. This
describes a system with k servers and waiting room for
only k customers, for example a telephone exchange with
just k lines and no facility for holding subscribers who
require a line but cannot be supplied with one. It is
assumed that such calls are lost.
For case (e)
An
--P
n
,0
]1
n.
where
1
72
1 n
n!p
Pn
2
1 + p + L +
2!
3. If n
(;
An
k
k!
N, N-n spares are on order.
IN-n)A
and An
P
+
=0
n
n
~
Thus
N
N
~
Since the maximum inventory is N
one item used
one completion of service
Probability (1 completion of service in t, t+ct)
~ot
where
the difference differential equations are
Pn (t+Ct)
PO(t)(l-Aoot) +
Pl(t)~ot
Pn_lCt)An_lot +
PnCt)[l-(An+~)otJ
+
o
Pn+l(t)~ot
< n < N
n
=
N
In the steady state
Thus the steady-state equations are
-AOP O
+ ~Pl
=0
n =0
o
An-lP n - l - CAn+~) Pn + ~Pn+l
-~PN
thus PI
Pn+l
PN
=
+ AN-l PN- l
AO
V Po
-A n-l
~
=
n
0
Pn-l +
{An+~2
~
AN-l
-~- PN-l
From equation 6.1, with n
=
1
73
o
< n <
N
=N
Pn
(6.1)
By recursion
1
thus
Total cost CT = average stock x stockholding cost +
premium x average usage x probability of no stock
N
100 L nP n + 15
n=l
x
2P O
We require the value of N that minimises this total cost.
N
1
2
1/2
1/5
1/16
1/65
1/326
1/2
2/5
3/16
4/65
5/326
2/5
6/16
12/65
20/326
24/65
60/326
24/65
120/326
3
4
5
120/326
15.25
6.6
2.9
74
2.0
4.2
Thus N
4 gives maximum stock level for minimum cost.
=
= Aj(n+1)
4. Here An
Thus
1 P
( -~A)n -n!
0
P
n 1
-,PO
n.
1
+ + ••• )
The average number in the queue is given by
= L (n-l)P
q
n=l
= e - P(
~~
+
n
zh 3h
+
+ ••• )
= e-PS
Now e p
= 1
+
P
+
~Z
TI
p3
TI
+
+
•••
therefore
eP
~
=
1 + 1 + P + PZ +
TI
TI
P
...
Z
3 Z
d (e P)_ -1 + 1 + TIP
+ 4'!P
P - 2
TI
P
dp
..,
pZ app
d (le P)
= -1
thus P Z (-1
;Ze P + %e P )
L-
+
Zp3
P
TI
+
3!
-1
+
S
7S
+
+
...
3p4
if!
+
...
and
S
-
e- P S = e- P - 1 + P
q
= e-A/~
- 1 + ~
~
5. For this system, A = 3/day, ~n = n~/day, where ~ =
number of ships that may be unloaded per day with one
A/~ = 1.5.
Then steady-state equations
crew ~ = 2. Let P
are
-pP O + PI = 0
=
n
0
o
PP n - l - (n + p)P n + (n + l)P n + l
n > 0
Solving recursively gives
pn
nrPO
1
+ ••• )
P
o = e-
P
Average number of unloading crews
ships in system, therefore
n-
average number of
00
!:
n=O
nP n
2
+l£+
2!
3p3
+
3T
... )
p = 1. 5 crews
00
Prob. (more than 4 crews)
n~sPn
.019
7
7.1
SINGLE-CHANNEL SYSTEMS WITH GENERAL
SERVICE TIME DISTRIBUTIONS
(M/GI I I SYSTEMS)
INTRODUCTION
The general formulae for these systems were obtained
by Pollaczeh-Khintchine, while Fry obtained a solution
to systems with constant service times, that is, M/D/l/oo
systems.
7.2
,
,
RESUME OF BASIC THEORY
The formulae of Pollaczeh-Khintchine are general and
also cover the M/D/l/oo system that Fry studied.
Probability of no customers in system Po
(For Pn for M/D/l systems only see Fry.)
(l+io;)
Average queue length q = p2 2Cl-p)
Average number in system -n
=1
- P
q + p
2
p(l+ll 2cr s)
211 (l-p)
Average time in system d = w + 1
Average waiting time -w
II
7.3
PROBLEMS
1. The time to process a claim in an insurance office is
exactly 25 minutes for each claim submitted. If claimants arrive in random stream at the average rate of one
every 40 minutes, how long on average must a claimant wait
for service given that there is only one insurance clerk
processing the claims.
2. In a heavy machine shop, the overhead crane is 75 per
cent utilised. Time study observations gave the average
slinging time as 10.5 min with a standard deviation of 8.8
min. What is the average calling rate for the services of
the crane, and what is the average delay in getting service? If the average service time is cut to 8.0 min, with
standard deviation of 6.0 min, how much reduction will
occur, on average, in the delay of getting served?
3. A firm employs a team of skilled fitters to repair
breakdowns that occur to its machines. Consider the
following alternative methods of operation.
77
(~)
N6 specialisation, all fitters tackle all repairs.
This method gives a standard deviation of repaIr
time equal to the mean repair rate (coefficient of
variation = 1.0)
(b) Partial specialisation, which reduces the standard
deviation of repair time to half the mean repair
rate
ec) Full specialisation, which ensures that all repairs
are completed in the same time.
Given that the mean repair rate is the same for all
methods of operation and that repairs arrive at random
~ith respect to time show that
(i) Method (b) gives a 37.5 per cent reduction on average
time for repair over method (a)
~aiting
(ii) Method (c) gives a 50 per cent reduction on average
waiting time over method (a).
7.4
SOLUT IONS
1. Here
0
2
s
=
0, A
1.5/h,
\1
Average waiting time is w
2.4/h,
p
= 1.5/2.4
0.625
(1+\12 0 ;)
p 2lJ (l-p)
2(1-p)JJ
0.625
2(l-0.625)x2.4
0.35 h
21 min
2. This is a M/GLI/oo process.
getting service w is given by
Initial situation
p
0.75
~:
60
T"Q."!
A
p x
0.75
S.71/h
Jj
x
5.71
4.29/h
7.8
Thus the average delay in
Average waiting time is
w
P
2 (I-p)
w
0.75
2 (1-0. 75)
w
0.75
o:T""
(1
x 1.
+
(1
70
~20;) x 1~
5.71 2 x 8.8 2) 60
min
60 2 x"5":iT
+
x
60
"5":II min
w = 26.8 min
I f service time is cut to 8 minutes
60
).I
7.5/h
"8
p = 4.29 = 0.571
~
or utilisation of the crane reduced to 57.1 per cent
w= 0.571
2 (I-O. 571)
(1
+
0.571
2xO.429 x 1.562
7 52 x 6.0 2)x
.
602
x
60 min
"'1.'5"
8
8.3 min
a reduction of 18.S min or approximately 70 per cent.
3. All these are M/G/l/oo processes with equal service
rate ~ and arrival rate A. The average waiting time for
an M/G/l/oo process is given by
w =
2
where as is the variance of the service time.
(a) 0 s2
wa
(b) 0 s2
1/).12
~
p
(l-p)
= 1/4).12
P(l +
wb
~
2
2~(1-p)
/t) =
).I
5p
8~(1-p)
79
Therefore
o
Z\.l (1-p)
The cost arisicg from the waitirg in the queue for
jobs is proportional to A~. The percentage savings achieved are
(i) saving in waiting time
wa - wb
wa
5
p
(:.
"8 \.Ill-p)
\.l(l-p)
300/8
100 per cent
x
=
x
100 per cent
37.5 per cent
(ii) , saving in waiting time
wa - wc
x 100 per cent
wa
r
p
j.J::..!(~l:....-....!::p:.L)_-=-21J~(l==--.....tp~)
p
\.lel-p)
x
100 per c e n t
so
per cent
STATISTICAL TABLES
POISSON DISTRIBUTION
TABLE 1
The table gives the probability that r or more random
events are contained in an interval when the average
number of such events per interval is m, that is
~
x=r
e
x
-m m
x!
Where there is no entry for a particular pair of values
of rand m, this indic?tes that the appropriate probability is less than 0.000 05. Similarly, except for the
case r = 0 when the entry is exact, a tabulated value of
1.0000 represents a probability greater than 0.999 95.
0.1
m=
r = 0 1. 0000
1
.0952
2 .0047
3
.0002
4
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
.0001
.0002
.0004
.0008
.0001
.0014
.0002
.0023
.0003
.0037
.0006
.0001
.1813
.0175
.0011
.0001
.2592
.0369
.0036
.0003
5
.3297
.0616
.0079
.0008
.3935
.0902
.0144
.0018
.4512
.1219
.0231
0034
6
7
m
1.0
0.2
1. 0000
.5034
.1558
.0341
.0058
.5507
.1912
.0474
.0091
.5934
.2275
.0629
.0135
.6321
.2642
.0803
.0190
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
r = 0
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
5
6
7
8
9
.0054
.0010
.0001
.0077
.0015
.0003
.0107
.0022
.0004
.0001
.0143
.0032
.0006
.0001
.0186
.0045
.0009
.0002
.0237
.0060
.0013
.0003
.0296
.0080
.0019
.0004
.0001
.0364
.0104
.0026
.0006
.0001
.0441
.0132
.0034
.0008
.0002
.0527
.0166
. "0045
.0011
.0002
=
1
2
3
4
.6671
.3010
.0996
.0257
2.1
m=
r = 0 1. 0000
1 .8775
2
.6204
.3504
3
4
.1614
5
6
7
8
9
10
11
12
.6988
.3374
.1205
.0338
.7275
.3732
.1429
.0431
.7534
.4082
.1665
.0537
.7769
.4422
.1912
.0656
.7981
.4751
.2166
.0788
.8173
.5068
.2428
.0932
.8347
'.5372
.2694
.1087
.8504
.5663
.2963
.1253
.8647
.5940
.3233
.1429
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
.8892
.3773
.1806
.8997
.6691
.4040
.2007
.0621
.0204
.0059
.0015
.0003
.0725
.0249
.0075
.0020
.0005
.0838
.0300
.0094
.0026
.0006
.0959
.0357
.0116
.0033
.0009
.1088
.0420
.0142
.0042
.0011
.1226
.0490
.0172
.0053
.0015
.1371
.0567
.0206
.0066
.0019
.1523
.0651
.0244
.0081
.0024
.1682
.0742
.0287
.0099
.0031
.1647
.0839
.0335
.0119
.0038
.0001
.0001
.0001
.0002
.0003
.0001
.0004
.0001
.0005
.0001
.0007
.0002
.0009
.0002
.0001
.0011
.0003
.0001
.6454
.9093
.6916
.4303
.2213
.9179
.7127
.4562
.2424
81
.9257
.7326
.4816
.2640
.9328
.7513
.5064
.2859
.9392
.7689
.5305
.3081
.9450
.7854
.5540
.3304
.9502
.8009
.5768
.3528
m
r
=
3.1
3.3
3.5
3.6
3.7
3.e
3.9
4.0
1. 0000
1. 0000
1
2
3
4
.9550
.8153
.5988
.3752
.9592
.8288
.6201
.3975
.9631
.8414
.6406
.4197
.9666
.8532
.6603
.4416
.9698
.8641
.6792
.4634
.9727
.8743
.6973
.4848
.9753
.8838
.7146
.5058
.9776
.8926
.7311
.5265
5
6
7
8
9
.2018
.0943
.0388
.0142
.0047
.2194
.1054
.0446
.0168
.0057
.2374
.1171
.0510
.0198
.0069
.2558
.1295
.0579
.0231
.0083
.2746
.1424
.0653
.0267
.0099
.2936
.1559
.0733
.0308
.0117
.3128
.1699
.0818
.0352
.0137
.3322
.1844
.0909
.0401
.0160
. :;516
.1994
.1005
.0454
.0185
.3712
.2149
.1107
.0511
.0214
10
11
12
13
14
.0014
.0004
.0001
.0018
.0005
.0001
.0022
.0006
.0002
.0027
.0008
.0002
.0001
.0033
.0010
.0003
.0001
.0040
.0013
.0004
.0001
.0048
.0016
.0005
.0001
.0058
.0019
.OOOG
.0069
.0023
.0007
.0002
.0001
.0081
.0028
.0009
.0003
.0001
4. 1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
.0002
.9798
.900a
.7469
.5468
.9817
.9084
.7619
.5665
= 0 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1.0000 1. 0000
1
2
3
4
.9834
.9155
.7762
.5858
.9850
.9220
. 78~8
.6046
.9864
.9281
.8026
.6228
.9877
.9337
.8149
.6406
.9889
.9389
.8264
.6577
.9899
.9437
.8374
.6743
.9909
.9482
.8477
.6903
.9918
.9523
.8575
.7058
.9926
.9561
.8667
.7207
.993:;
.9596
.8753
.7350
5
6
7
8
9
.3907
.2307
.1214
.0573
.0245
.4102
.2469
.1325
.0639
.0279
.4296
.2633.1442
.0710
.0317
.4488
.2801
.1564
.0786
.0358
.4679
.2971
.1689
.0866
.0403
.4868
.3142
.1820
.0951
.0451
.5054
.3316
.1954
.1040
.0503
.523"7
.3490
.2092
.1133
.0558
.5418
.3665
.2233
.1231
.0618
.5595
.3840
.2378
.1334
.0681
10
11
12
13
14
.0095
.0034
.0011
.0003
.0001
.0111
.0041
.0014
.0004
.0001
.0129
.0048
.0017
.0005
.0002
.0149
.0057
.0020
.0007
.0002
.0171
.0067
.0024
.0008
.0003
.0195
.0078
.0029
.0010
.0003
.0222
.0090
.0034
.0012
.0004
.0251
.0104
.0040
.0014
.0005
.0283
.0120
.0047
.0017
.0006
.0318
.0137
.0055
.0020
.0007
.0001
.0001
.0001
.0001
.0001
.0002
.0001
.0001
5.8
6.0
6.2
6.4
6.6
6.8
7.0
15
16
m : -:
r
3.4
= 0 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1.0000 1. 0000 1. 0000
m
r
3.2
5.2
5.4
5.6
. a002
= 0 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000 1. 0000
1
2
3
4
.9945
.9658
.8912
.7619
.9955
.9711
.9052
.7867
.9963
.9756
.9176
.8094
.9970
.9794
.9285
.8300
.9975
.9826
.9380
.8488
.9980
.9854
.9464
.8658
.998:;
.9877
.9537
.8811
.9986
.9897
.9600
.8948
.9989
.9913
.9656
.9072
.9991
.9927
.9704
.9182
5
6
7
8
9
.5939
.4191
.2676
.1551
.0819
.6267
.4539
.2983
.1783
.0974
.6579
.4881
.3297
.2030
.1143
.6873
.5217
.3616
.2290
.1328
.7149
.5543
.3937
.2560
.1528
.7408
.5859
.4258
.2840
.1741
.7649
.6163
.4577
.3127
.1967
.7873
.6453
.4892
. ;'419
.2204
.8080
.6730
.5201
.3715
.2452
.8270
.6993
.5503
.4013
.2709
10
11
12
13
14
.0397
.0177
.0073
.0028
.0010
.0488
.0225
.0096
.0038
.0014
.0591
.0282
.0125
.0051
.0020
.0708
.0349
.0160
.0068
.0027
.0839
.0426
.0201
.0088
.0036
.0984
.0514
.0250
.0113
.0048
.1142
.0614
.0307
.0143
.0063
.1314
.0726
.0373
.0179
.0080
. 1498
.0849
.0448
.0221
.0102
.1695
.0985
.0534
.0270
.0128
15
16
17
18
19
.0003
.0001
.0005
.0002
.0001
.0007
.0002
.0001
.0010
.0004
.0001
.0014
.0005
.0002
'.0001
.0019
.0007
.0003
.0001
.0026
.0010
.0004
.0001
.0034
.0014
.0005
.0002
.0001
.0044
.0018
.0007
.0003
.0001
.0057
.0024
.0010
.0004
.0001
82
m
7.2
0
7.4
7.6
7.8
8.0
8.2
8.4
8.6
8.8
9.0
r = 0
1
2
3
4
1. 0000
.9993
.9939
.9745
.9281
1. 0000
.9994
.9949
.9781
.9368
1. 0000
.9995
.9957
.9812
.9446
1. 0000
.9996
.9964
.9839
.9515
1. 0000
.9997
.9970
.9862
.9576
1. 0000
.9997
.9975
.9882
.9630
1. 0000
.9998
.9979
.9900
.9677
1. 0000
.9998
.9982
.9914
.9719
1. 0000
.9998
.9985
.9927
.9756
1. 0000
.9999
.9988
.9938
.9788
5
6
7
8
9
.8445
.7241
.5796
.4311
.2973
.8605
.7474
.6080
.4607
.3243
.8751
.7693
.6354
.4900
.3518
.8883
.7897
.6616
.5188
.3796
.9004
.8088
.6866
.5470
.4075
.9113
.8264
.7104
.5746
.4353
.9211
.8427
.7330
.6013
.4631
.9299
.8578
.7543
.6272
.4906
.9379
.8716
.7744
.6522
.5177
.9450
.8843
.7932
.6761
.5443
10
11
12
13
14
.1904
.1133
.0629
.0327
.0159
.2123
.1293
.0735
.0391
.0195
.2351
.1465
.0852
.0464
.0238
.2589
.1648
.0980
.0546
.0286
.2834
.1841
.1119
.0638
.0342
.3085
.2045
.1269
.0739
.0405
.3341
.2257
.1429
.0850
.0476
.3600
.2478
.1600
.0971
.0555
.3863
.2706
.1780
.1102
.0642
.4126
.2940
.1970
.1242
.0739
15
16
17
18
19
.0073
.0031
.0013
.0005
.0002
.0092
.0041
.0017
.0007
.0003
.0114
.0052
.0022
.0009
.0004
.0141
.0066
.0029
.0012
.0005
.0173
.0082
.0037
.0016
.0006
.0209
.0102
.0047
.0021
.0009
.0251
.0125
.0059
.0027
.0011
.0299
.0152
.0074
.0034
.0015
.0353
.0184
.0091
.0043
.0019
.0415
.0220
.0111
.0053
.0024
20
21
22
23
.0001
.0001
.0001
.0002
.0001
.0003
.0001
.0003
.0001
.0005
.0002
.0001
.0006
.0002
.0001
.0008
.0003
.0001
.0011
.0004
.0002
.0001
m r
=
9.2
0 1. 0000
.9999
1
2
.9990
3
.9947
4
.9816
9.4
9.6
9.8
10.0
11.0
12.0
13.0
14.0
15.0
1. 0000
.9999
.9991
.9955
.9840
1. 0000
.9999
.9993
.9962
.9862
1. 0000
.9999
.9994
.9967
.9880
1. 0000
1. 0000
.9995
.9972
.9897
1. 0000
1. 0000
.9998
.9988
.9951
1. 0000
1. 0000
.9999
.9995
.9977
1. 0000
1. 0000
1. 0000
.9998
.9990
1. 0000
1. 0000
1. 0000
.9999
.9995
1. 0000
1. 0000
1. 0000
1. 0000
.9998
.9707
.9924
.9797
.9542
.9105
.8450
.9963
.9893
.9741
.9460
.9002
.9982
.9945
.9858
.9684
.9379
.9991
.9972
.9924
.9820
.9626
5
6
7
8
9
.9514
.8959
.8108
.6990
.5704
.9571
.9065
.8273
.7208
.5958
.9622
.9162
.8426
.7416
.6204
.9667
.9250
.8567
.7612
.6442
.8699
.7798
.6672
.9849
.9625
.9214
.8568
.7680
10
11
12
13
14
.4389
.3180
.2168
.4911
.3671
.2588
.1721
.1081
.5168
.3920
.2807
.1899
.1214
.5421
.4170
.3032
.2084
.1355
.6595
.5401
.4207
.3113
.2187
.7576
.6528
.5:;84
.4240
.3185
.8342
.7483
.6468
.5369
.4270
.8906
.8243
.7400
.6415
.5356
.9301
.8815
.8152
.7324
.6:;68
.9~29
.0844
.4651
.3424
.2374
.1552
.0958
15
16
17
18
19
.0483
.0262
.0135
.0066
.0031
.0559
.0309
.0162
.0081
.0038
.0643
.0362
.0194
.0098
.0048
.0735
.0421
.0230
.0119
.0059
.0835
.0487
.0270
.0143
.0072
.1460
.0926
.0559
.0322
.0177
.2280
.1556
.1013
.0630
.0374
.3249
.2364
.1645
.1095
.0698
.4296
.3306
.2441
.1728
.1174
.5343
.4319
.3359
.2511
.1805
20
21
22
23
24
.0014
.0006
.0002
.0001
.0017
.0008
.0003
.0001
.0022
.0010
.0004
.0002
.0001
.0028
.0012
.0005
.0002
.0001
.0035
.0016
.0007
.0003
.0001
.0093
.0047
.0023
.0010
.0005
.0213
.0116
.0061
.0030
.0015
.0427
.0250
.0141
.0076
.0040
.0765
.0479
.0288
.0167
.0093
.1248
.0830
.0531
.0327
.0195
.0002
.0001
.0007
.0003
.0001
.0001
.0020
.0010
.0005
.0002
.0001
.0050
.0026
.0013
.0006
.0003
.0112
.0062
.0033
.0017
.0009
.0001
.0001
.0004
.0002
.0001
.139~
25
26
27
28
29
30
31
32
83
m=
16.0
17.0
18.0
19.0
20.0
21. 0
22.0
23.0
24.0
25.0
r=
1. 0000
1. 0000
1. 0000
1. 0000
.9999
1. 0000
1. 0000
1.0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
1. 0000
.9996
.9986
.9960
.9900
.9780
.9998
.9993
.9979
.9946
.9874
.9999
.9997
.9990
.9971
.9929
1. 0000
.9998
.9995
.9985
.9961
1. 0000
.9999
.9997
.9992
.9979
1. 0000
1. 0000
.9999
.9996
.9989
1. 0000
1. 0000
.9999
.9998
.9994
1. 0000
1. 0000
1. 0000
.9999
.9997
1. 0000
1. 0000
1. 0000
1. 0000
.9998
1. 0000
1. 0000
10
11
12
13
14
.9567
.9226
.8730
.8069
.7255
.9739
.9509
.9153
.8650
.7991
.9846
.9696
.9451
.9083
.8574
.9911
.9817
.9653
.9394
.9016
.9950
.9892
.9786
.9610
.9339
.9972
.9937
.9871
.9755
.9566
.9985
.9965
.9924
.9849
.9722
.9992
.9980
.9956
.9909
.9826
.9996
.9989
.9975
.9946
.9893
.9998
.9994
.9986
.9969
.9935
15
16
17
18
19
.6325
.5333
.4340
.3407
.2577
.7192
.6285
.5323
.4360
.3450
.7919
.7133
.6249
.5314
.4378
.8503
.7852
.7080
.6216
.5305
.8951
.8435
.7789
.7030
.6186
.9284
.8889
.8371
.7730
.6983
.9523
.9231
.8830
.8310
.7675
.9689
.9480
.9179
.8772
.8252
.9802
.9656
.9437
.9129
.8717
.9876
.9777
.9623
.9395
.9080
20
21
22
23
24
.1878
.1318
.0892
.0582
.0367
.2637
.1945
.1385
.0953
.0633
.3491
.2693
.2009
.1449
.1011
.4394
.3528
.2745
.2069
.1510
.5297
.4409
.3563
.2794
.2125
.6157
.5290
.4423
.3595
.2840
.6940
.6131
.5284
.4436
.3626
.7623
.6899
.6106
.5277
.4449
.8197
.7574
.6861
.6083
.5272
.8664
.8145
.7527
.6825
.6061
25
26
27
28
29
.0223
.0131
.0075
.0041
.0022
.0406
.0252
.0152
.0088
.0050
.0683
.0446
.0282
.0173
.0103
.1067
.0731
.0486
.0313
.0195
.1568
.1122
.0779
.0525
.0343
.2178
.1623
.1174
.0825
.0564
.2883
.2229
.1676
.1225
.0871
.3654
.2923
.2277
.1726
.1274
.4460
.3681
.2962
.2323
.1775
.5266
.4471
.3706
.2998
.2366
30
31
32
33
34
.0011
.0006
.0003
.0001
.0001
.0027
.0014
.0007
.0004
.0002
.0059
.0033
.0018
.0010
.0005
.0118
.0070
.0040
.0022
.0012
.0218
.0135
.0081
.0047
.0027
.0374
.0242
.0152
.0093
.0055
.0602
.0405
.0265
.0169
.0105
.0915
.0640
.0436
.0289
.0187
.1321
.0958
.0678
.0467
.0314
.1821
.1367
.1001
.0715
.0498
.0001
.0002
.0001
.0001
.0006
.0003
.0002
.0001
.0015
.0008
.0004
.0002
.0001
.0032
.0018
.0010
.0005
.0003
.0064
.0038
.0022
.0012
.0007
.0118
.0073
.0044
.0026
.0015
.0206
.0132
.0082
.0050
.0030
.0338
.0225
.0146
.0092
.0057
.0001
.0001
.0001
.0004
.0002
.0001
.0008
.0004
.0002
.0001
.0001
.0017
.0010
.0005
.0003
.0002
.0034
.0020
.0012
.0007
.0004
.0001
.0002
.0001
35
36
37
38
39
40
41
42
43
44
45
46
84
1. 0000
1. 0000
.9999
26.0
27.0
28.0
29.0
30.0
32.0
34.0
36.0
38.0
40.0
9
1. 0000
1.0000
1. 0000
1. 0000
1.0000
1. 0000
1. 0000
1.0000
1. 0000
1.0000
10
11
12
13
14
.9999
.9997
.9992
.9982
.9962
.9999
.9998
.9996
.9990
.9978
1. 0000
.9999
.9998
.9994
.9987
1.0000
1. 0000
.9999
.9997
.9993
1. 0000
1.0000
.9999
.9998
.9996
1.0000
1.0000
1.0000
1.0000
.9999
1.0000
1.0000
1.0000
1.0000
1. 0000
1. 0000
1.0000
1.0000
1. 0000
1. 0000
1.0000
1.0000
1.0000
1.0000
1.0000
1. 0000
1. 0000
1. 0000
1.0000
1.0000
15
16
17
18
19
.9924
.9858
.9752
.9580
.9354
.9954
.9912
.9840
.9726
.9555
.9973
.9946
.9899
.9821
.9700
.9984
.9967
.9937
.9885
.9801
.9991
.9981
.9961
.9927
.9871
.9997
.9993
.9986
.9972
.9948
.9999
.9998
.9995
.9990
.9980
1. 0000
.9999
.9998
.9997
.9993
1.0000
1. 0000
1.0000
.9999
.9998
1. 0000
1. 0000
1.0000
1.0000
.9999
20
21
22
23
24
.9032
.8613
.8095
.7483
.6791
.9313
.8985
.8564
.8048
.7441
.9522
.9273
.8940
.8517
.8002
.9674
.9489
.9233
.8896
.8471
.9781
.9647
.9456
.9194
.8854
.9907
.9841
.9740
.9594
.9390
.9963
.9932
.9884
.9809
.9698
.9986
.9973
.9951
.9915
.9859
.9995
.9990
.9981
.9965
.9938
.9998
.9996
.9993
.9986
.9974
25
26
27
28
29
.6041
.5261
.4481
.3730
.3033
.6758
.6021
.5256
.4491
.3753
.7401
.6728
.6003
.5251
.4500
.7958
.7363
.6699
.5986
.5247
.8428
.7916
.7327
.6671
.5969
.9119
.8772
.8344
.7838
.7259
.9540
.9326
.9047
.8694
.8267
.9776
.9655
.9487
.9264
.8977
.9897
.9834
.9741
.9611
.9435
.9955
.9924
.9877
.9807
.9706
30
31
32
33
34
.2407
.1866
.1411
.1042
.0751
.3065
.2447
.1908
.1454
.1082
.3774
.3097
.2485
.1949
.1495
.4508
.3794
.3126
.2521
.1989
.5243
.4516
.3814
.3155
.2556
.6620
.5939
.5235
.4532
.3850
.7765
.7196
.6573
.5911
.5228
.8621
.8194
.7697
.7139
.6530
.9204
.8911
.8552
.8125
.7635
.9568
.9383
.9145
.8847
.8486
35
36
37
38
39
.0528
.0363
.0244
.0160
.0103
.0787
.0559
.0388
.0263
.0175
.1121
.0822
.0589
.0413
.0283
.1535
.1159
.0856
.0619
.0438
.2027
.1574
.1196
.0890
.0648
.3208
.2621
.2099
.1648
.1268
.4546
.3883
.3256
.2681
.2166
.5885
.5222
.4558
.3913
.3301
.7086
.6490
.5862
.5216
.4570
.8061
.7576
.7037
.6453
.5840
40
41
42
43
44
.0064
.0039
.0024
.0014
.0008
.0113
.0072
.0045
.0027
.0016
.0190
.0125
.0080
.0050
.0031
.0303
.0205
.0136
.0089
.0056
.0463
.0323
.022f
.0148
.0097
.0956
.0707
.0512
.0364
.0253
.1717
.1336
.1019
.0763
.0561
.2737
.2229
.1783
.1401
.1081
.3941
.3343
.2789
.2288
.1845
.5210
.4581
.3967
.3382
.2838
45
46
47
48
49
.0004
.0002
.0001
.0001
.0009
.0005
.0003
.0002
.0001
.0019
.0011
.0006
.0004
.0002
.0035
.0022
.0013
.0008
.0004
.0063
.0040
.0025
.0015
.0009
.0173
.0116
.0076
.0049
.0031
.0404
.0286
.0199
.0136
.0091
.0819
.0609
.0445
.0320
.0225
.1462
.1139
.0872
.0657
.0486
.2343
.1903
.1521
.1196
.0925
.0001
.0001
.0002
.0001
.0001
.0005
.0003
.0002
.0001
.0001
.0019
.0012
.0007
.0004
.0002
.0060
.0039
.0024
.0015
.0009
.0156
.0106
.0071
.0047
.0030
.0353
.0253
.0178
.0123
.0084
.0703
.0526
.0387
.0281
.0200
.0001
.0001
.0006
.0003
.0002
.0001
.0001
.0019
.0012
.0007
.0005
.0003
.0056
.0037
.0024
.0015
.0010
.0140
.0097
.0066
.0044
.0029
.0002
.0001
.0001
.0006
.0004
.0002
.0001
.0001
.0019
.0012
.0008
.0005
.0003
m=
=
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
.0002
.0001
.0001
65
66
67
For values of m greater than 40 use the Normal Distribution curve by setting jJ = m and (J = 1m.
(See J. Murdoch and J.A. Barnes, Statistical Tables,
Macmillan, London and Basingstoke, 1971.)
85
TABLE 2
EXPONENTIAL FUNCTION e -x
For any negative exponential distribution, the tabulated
function may be used to find the proportion of the distribution in excess of x times the mean. As an example,
in random sampling of an exponential variate with a mean
of 8, the probability that a single value will exceed 6
is 0.4724. Further, the 1% point of the distribution is
seen to be 4.61 times the mean.
"
.1
.2
.3
.4
.5
.6
.7
.8
.3679
.1353
.0498
.0183
.3329
.1225
.0450
.0166
.3012
.1108
.040B
.0150
.2725
.1003
.0369
. 0136
.2466
.0907
.2019
.0743
.0273
.0101
.1827
.0672
.0247
.0 2910
.1653
.0608
.0224
.0 2823
.1496
.0550
.0202
.0 2745
5.0
6.0
7.0
8.0
".0
10.0
11. 0
.0 2674
.0 2248
.0 3912
.0 3335
.0 3123
.0 2610
.0 2224
.0 3825
.0 3304
.0 3112
.0 2552
.0 2203
.0 3747
.0 3275
.0 3101
.0 2452
.0 2166
.0 3611
.0 3225
.0 4827
.0 2 370
.0 2 136
.0 3500
.0 4411
.0 4151
.0 5556
.0 5205
.0 6752
.0 4372
.0 4137
.0 5503
.0 5185
.0 2335
.0 2123
.0 3453
.0 3167
.0 4613
.0 4225.
.0:>829
.0 5305
.0 5112
.0 6413
.0 2303
.0 2111
.0 3410
.0 3 151
.0 4555
.0 4454
.0 4167
.0 5614
,0 5226
,0 6832
.0 2499
.02}84
.0 3676
.0 3249
.0 4914
,0 4336
.0 4 124
.0 5455
.0 5167
,0 6616
.2231
.0621
.0302
· 0111
.0 2409
.0 2 150
· 0 3553
.0 3203
.0 4749
,0 4275
.0 4101
.0 5373
,0 5137
.0 6504
.0 2274
.0 2 101
.0 3371
.0 3136
.0 4502
.0 4185
.0 5679
.052fiO
.0 6919
.0 6338
15.0
16.0
17.0
18.0
19.0
.0 6306
.0 6 113
.0 7414
.0 1152
.0 6277
.0 6102
.0 7375
.0 1138
,0 6227
.0 1834
.0 1307
.0 1113
.0 6205
.0 7754
20.0
.0 8206
.0
1.0
2.0
3.0
4.0
12.0
13,0
14.0
. 08560
. 0 8 507
. 0 8415
.0 4304
.0 4112
.0 5412
.0 5 15'2
,0 6557
,0 6186
.0 7683
.0 7251
.0 8924
· 0 1278
.0 1102
· 0 8 376
· 0 8 340
· 03184
.0 4677
.0 4249
.0:>917
.0 5337
.0 5 124
.0 6 456
,0 6168
.0 6152
.0 1559
.0 7206
.0 8756
.0 8278
· 0 7618
.0 7227
.0 8836
· 0 8 307
.0 4204
.0:>750
.0 5276
,0 5102
.0 6374
.0 6137
.0 7506
• 07186
.0 8 684
.0 8252
.02
.OJ
.04
.05
.06
.07
.08
.00
.6636
.9802
.886"
.8025
.7261
.6570
.9704
.8781
.7945
.7189
.6505
.9608
.8694
.7866
.7118
.6440
.9512
.8607
.7788
.7047
.6376
.9418
.8521
.7711
.6977
.6313
.9324
.8437
.7634
.6907
.6250
.9231
.8353
.7558
.6839
.6188
.9139
.8270
.7483
.6771
.6126
.6005
.5434
.4916
.4449
.4025
.5945
.5379
.4868
.4404
.3985
.5886
.5326
.4819
.4360
.3946
.5827
.5273
.4771
.4317
.3906
.5770
.5220
.4724
.4274
.3867
.5712
.5169
.4677
.4232
.3829
.5655
.5117
.4630
.4190
.3791
.5599
.5066
.4584
.4148
.3753
.5543
.5016
.45S8
.4107
.3716
.3679
.3329
.3012
.2725
.2466
.3642
.3296
.2892
.2698
.2441
.3606
.326S
.2952
.2671
.2417
.3570
.3230
.2923
.2645
.2393
.3535
.3198
.2894
.2618
.2S69
.3499
.3166
.2865
.2592
.2346
.3465
.3135
.2837
.2567
.2322
.3430
.3104
.2808
.2541
.2299
.3396
.3073
.2780
.2516
.2276
.3362
.3042
.2753
.2491
.2254
1.5
1.6
1.7
1.8
I."
2.0
2.1
2.2
2.3
2.4
.2231
.2019
.1827
.1653
.1496
.2209
.1999
.1809
.1637
.1481
.2187
.1979
.1791
.1620
.1466
.2165
.1959
.1773
.1604
.1451
.2144
.1940
.1755
.1588
.1437
.2122
.1920
.1738
.1572
.1423
.2101
.1901
.1720
.1557
.1409
.2080
.1882
.1703
.1541
.1395
.2060
.1864
.1686
.1526
.1381
.2039
.1845
.1670
.1511
.1367
.1353
.1225
.H08
.1003
.0907
.1340
.1212
.1097
.0993
.0898
.1327
.1200
.1086
.0983
.0889
.1313
.1188
.1075
.0973
.0880
.1300
.1177
.1065
.0063
.0872
.1287
.1165
.1054
.0954
. 0863
.1275
.1153
.1044
.0944
.0854
.1262
.1142
.1035
.0935
.0846
.1249
.1130
.1023
.0926
.0837
.1237
.1119
.1013
· 0916
.0829
2.5
2.6
2.7
2.8
2. "
3.0
3.1
3.2
3.3
3.4
.0821
.0743
.0672
.0608
.0550
. 0813
.0735
.0665
.0602
.0545
. 0805
.0797
.0721
.0652
.0590
.0534
.0789
.0714
.0646
.0584
.0529
.0781
.0707
.063"
.0578
.0523
. 0773
.0699
.0633
.0573
.0518
.0765
.0693
.0627
.0567
.0513
.0758
.0686
.0620
.0561
.0508
.0750
.0679
.0614
.0556
.050S
.0498
.0450
.0408
.036"
. 0334
.0493
.0446
.0404
.0365
.0330
.0483
.0437
.0396
.0358
.0324
.0478
.0433
.OS92
. 0354
.0321
.0474
.0429
.0388
.0351
. 0317
.0469
.0424
.0384
.0347
.0314
.0464
.0420
.0380
.0344
.0311
.0460
.0416
.0376
.OS40
.0308
.0455
.0412
.0373
· 0337
.OS05
3.5
3.6
3.7
3.8
3.9
.0302
.0273
.0247
.0224
.0202
.0299
.0271
.0245
.0221
.0200
.0296
.0268
.0242
.0290
.0263
.0238
.0215
.0194
.0287
.0260
.0235
.0213
.0193
.0284
.0257
.0233
.0211
.0191
.0282
.0255
.0231
.0209
.018"
.0279
.0252
.0228
.0207
.0187
.0276
.0250
.0226
.0219
.0198
.0293
.0265
.0240
.0217
.0196
4.0
4.1
4.2
4.3
4.4
.0183
.0166
.0150
.0136
.0123
.0181
.0164
.0148
.0134
.0122
.0180
.0178
.0176
.0\59
.0146
.0132
.0119
.0144
.0130
.0174
.0158
.0143
.0129
.0117
.0172
.0156
.0\41
.0128
.0116
.0171
.0155
.0140
.0127
.0114
.0169
.0153
.0138
.0125
.0113
.0167
.0151
.0137
· 0124
· 0112
4.5
4.6
4.7
4.8
4."
5.0
.0111
.0101
.0091
.0082
.0074
.0110
.0100
.0090
.0081
.0109
.009"
.008"
.0081
.0073
.0108
.0098
.0107
.0097
.0087
.0079
.0072
.0106
.0105
.0095
.0086
.0078
.0070
.0104
.009.
.0103
.009S
.0084
.0076
.0069
.0102
.0092
.008S
.0075
.0068
.00
.01
.08'681
,0 6250
.0 1921
.0 7339
.0 1125
. 0 8459
· 0334
.0123
.
0
.1
.2
.3
.4
1. 0000
.9048
.8187
.7408
.6703
.9900
.5
.6
.7
.8
.6065
.5488
.4966
.4493
.4066
1.0
1.1
1.2
1.3
1.4
."
.8958
.8106
. 73S4
.0074
.0728
.065"
.0596
.0539
.0488
.0442
.0400
. 0362
.0327
.0162
.0147
.0133
.0120
.0161
.0088
. ooso
.0072
.0118
.0096
.0087
.0078
.0071
.0067
86
.0085
.0077
.0069
.0204
.0\85
,0 6124
.0 7458
.0 1168
.0 8619
. 0 8228
00
"
1. 98
0.65
0.70
0 .. 75
0.80
0.85
0.90
0.95
0.99
1. 43
1. 59
1. 77
1.03
1.16
1. 28
1. 55
1. 69
1. 86
1. 21
1. 31
1. 42
0.96
1.04
1.12
0.73
0.81
0.88
0.44
0.55
0.64
0.36
o. SO
0.61
0.71
0.82
0.93
3
2
1. 73
1. 93
1. 37
1. 4 7
1. 58
1.13
1.20
1. 28
0.94
1.00
1.06
0.75
0.82
0.88
O. SO
0.60
0.68
4
1. 55
1. 69
1. 89
1. 27
1. 35
1. 44
1.09
1.15
1. 21
0.93
0.98
1.04
0.77
0.83
0.88
0.55
0.64
0.72
5
-_._._-
1. 43
1. 55
1. 70
1. 98
1. 22
1. 28
1. 35
1.06
1.11
1.16
0.93
0.97
1.02
0.80
0.84
0.89
0.59
0.68
0.74
6
G
p
1. 36
1. 46
1. 58
1. 79
1.18
1. 24
1. 30
1.05
1.09
1.13
0.93
0.97
1.01
0.81
0.85
0.89
0.63
0.70
0.77
7
1. 31
1. 39
1. SO
1. 64
1.15
1. 20
1. 25
1.04
1.07
1.11
0.94
0.97
1.00
0.83
0.87
0.90
0.66
0.73
0.79
8
1. 28
1. 35
1. 44
1. 5 5
1.13
1.18
1. 22
1.03
1.06
1.10
0.94
0.97
1.00
0.84
0.87
0.90
0.68
0.75
0.80
9
1. 24
1. 30
1. 38
1. 51
1. 80
1.12
1.15
1.19
1.02
LOS
1.09
0.94
0.97
1.00
0.85
0.88
0.91
0.70
0.77
0.81
10
N
FOR M/M/I/N SYSTEMS
average profit per service
OPTIMUM VALUE OF
average cost per service
O. SO
0.55
0.60
1. 45
1. 72
0.81
1.00
1. 21
0.20
0.25
0.30
0.35
0.40
0.45
0.29
0.46
0.63
1
0.05
0.10
0.15
~
Where E
TABLE 3
1. 20
1. 24
1. 31
1. 41
1. 65
1.09
1.12
1.15
1.02
1.04
1.07
0.95
0.97
0.99
0.87
0.90
0.92
0.74
0.80
0.84
12
1.16
1.20
1. 26
1. 35
1. SO
1.08
1.10
1.13
1.01
1.03
LOS
0.95
0.97
0.99
0.88
0.91
0.93
0.77
0.82
0.86
14
1.14
1.16
1. 22
1. 29
1. 45
1.07
1.09
1.11
1.01
1.03
LOS
0.96
0.98
0.99
0.90
0.92
0.94
0.79
0.84
0.87
16
18
1.12
1.16
1.19
1. 26
1.40
1.06
1.08
1.11
1.01
1.02
1.04
0.96
0.98
0.99
0.91
0.93
0.95
0.81
0.85
0.88
maximum system size.
1.11
1.14
1.17
1. 23
1. 35
LOS
1.07
1.09
1.01
1.02
1.04
0.96
0.98
0.99
0.91
0.93
0.95
0.82
0.87
0.89
20
REFERENCES
For an introduction to the theory of random arrival, random service queueing systems (M/M/-/-), readers are
referred to the following general introductory textbooks
on Operational Research.
Sasieni, M.W., et al., Operations Research (Wiley, New
York, 1959) pp. 125-54.
Wagner, H.M., Principles of Operational Research (PrenticeHall, Englewood Cliffs, N.J., 1969) pp. 837-85.
pp. 75-115.
Churchman, C.W., Ackoff, R.L., and Arnoff, E.L., Introduction to Operations Research (Wiley, New York, I957)
pp. 391-416.
Specially recommended for readers wishing to study Queueing Theory in greater depth is
Page, E.G., Queueing Theory in O.R. (Butterworth, London,
1972) •
Other specialistic textbooks on Queueing Theory include
Cohen, J.W., The Single Server Queue (North-Holland,
Amsterdam, 1969) .
Cox, D.R., and Smith, W.L., Queues (Chapman & Hall, London,
1971) .
~~~~~~~~~~~~~~~_U~se~s (Van Nostrand,
formulae, see p.
Lee, A.M., Applied Queueing Theory (Macmillan, London and
Basingstoke, 1966).
Jaiswal, N.K., Priority Queues (Academic Press, London,
1968) •
Khintchine, A.Y., Mathematical Models in the Theory of
Queueing (Hafner, New York, 1968).
Morse, P.M., Queues, Inventories and Maintenance (Wiley,
New York, 1958).
Saaty, R.L., Elements of Queueing Theory (McGraw-Hill,
London, 1961).