Queueing

Queueing Theory

Ivo Adan and Jacques Resing

Department of Mathematics and Computing Science
Eindhoven University of Technology
P.O. Box 513, 5600 MB Eindhoven, The Netherlands

February 28, 2002

Contents

1 Introduction 7
1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Basic concepts from probability theory 11
2.1 Random variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Laplace-Stieltjes transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 Useful probability distributions . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4.1 Geometric distribution . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4.2 Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.4.3 Exponential distribution . . . . . . . . . . . . . . . . . . . . . . . . 13
2.4.4 Erlang distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.4.5 Hyperexponential distribution . . . . . . . . . . . . . . . . . . . . . 15
2.4.6 Phase-type distribution . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5 Fitting distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.6 Poisson process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Queueing models and some fundamental relations 23
3.1 Queueing models and Kendall’s notation . . . . . . . . . . . . . . . . . . . 23
3.2 Occupation rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.3 Performance measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.4 Little’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.5 PASTA property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 M/M/1 queue 29
4.1 Time-dependent behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Limiting behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.2.1 Direct approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.2.2 Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.2.3 Generating function approach . . . . . . . . . . . . . . . . . . . . . 32
4.2.4 Global balance principle . . . . . . . . . . . . . . . . . . . . . . . . 32

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4.3 Mean performance measures . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.4 Distribution of the sojourn time and the waiting time . . . . . . . . . . . . 33
4.5 Priorities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.5.1 Preemptive-resume priority . . . . . . . . . . . . . . . . . . . . . . 36
4.5.2 Non-preemptive priority . . . . . . . . . . . . . . . . . . . . . . . . 37
4.6 Busy period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.6.1 Mean busy period . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.6.2 Distribution of the busy period . . . . . . . . . . . . . . . . . . . . 38
4.7 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 M/M/c queue 43
5.1 Equilibrium probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
5.2 Mean queue length and mean waiting time . . . . . . . . . . . . . . . . . . 44
5.3 Distribution of the waiting time and the sojourn time . . . . . . . . . . . . 46
5.4 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 M/Er /1 queue 49
6.1 Two alternative state descriptions . . . . . . . . . . . . . . . . . . . . . . . 49
6.2 Equilibrium distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
6.3 Mean waiting time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.4 Distribution of the waiting time . . . . . . . . . . . . . . . . . . . . . . . . 53
6.5 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 M/G/1 queue 59
7.1 Which limiting distribution? . . . . . . . . . . . . . . . . . . . . . . . . . . 59
7.2 Departure distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
7.3 Distribution of the sojourn time . . . . . . . . . . . . . . . . . . . . . . . . 64
7.4 Distribution of the waiting time . . . . . . . . . . . . . . . . . . . . . . . . 66
7.5 Lindley’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
7.6 Mean value approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
7.7 Residual service time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
7.8 Variance of the waiting time . . . . . . . . . . . . . . . . . . . . . . . . . . 70
7.9 Distribution of the busy period . . . . . . . . . . . . . . . . . . . . . . . . 71
7.10 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

8 G/M/1 queue 79
8.1 Arrival distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
8.2 Distribution of the sojourn time . . . . . . . . . . . . . . . . . . . . . . . . 83
8.3 Mean sojourn time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

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8.4 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

9 Priorities 87
9.1 Non-preemptive priority . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
9.2 Preemptive-resume priority . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
9.3 Shortest processing time ﬁrst . . . . . . . . . . . . . . . . . . . . . . . . . 90
9.4 A conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
9.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

10 Variations of the M/G/1 model 97
10.1 Machine with setup times . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
10.1.1 Exponential processing and setup times . . . . . . . . . . . . . . . . 97
10.1.2 General processing and setup times . . . . . . . . . . . . . . . . . . 98
10.1.3 Threshold setup policy . . . . . . . . . . . . . . . . . . . . . . . . . 99
10.2 Unreliable machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
10.2.1 Exponential processing and down times . . . . . . . . . . . . . . . . 100
10.2.2 General processing and down times . . . . . . . . . . . . . . . . . . 101
10.3 M/G/1 queue with an exceptional ﬁrst customer in a busy period . . . . . 103
10.4 M/G/1 queue with group arrivals . . . . . . . . . . . . . . . . . . . . . . . 104
10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

11 Insensitive systems 111
11.1 M/G/∞ queue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
11.2 M/G/c/c queue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
11.3 Stable recursion for B(c, ρ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
11.4 Java applet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
11.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Bibliography 119

Index 121

Solutions to Exercises 123

5

Chapter 1

Introduction

In general we do not like to wait. But reduction of the waiting time usually requires extra
investments. To decide whether or not to invest, it is important to know the effect of
the investment on the waiting time. So we need models and techniques to analyse such
situations.
In this course we treat a number of elementary queueing models. Attention is paid
to methods for the analysis of these models, and also to applications of queueing models.
Important application areas of queueing models are production systems, transportation and
stocking systems, communication systems and information processing systems. Queueing
models are particularly useful for the design of these system in terms of layout, capacities
and control.
In these lectures our attention is restricted to models with one queue. Situations with
multiple queues are treated in the course “Networks of queues.” More advanced techniques
for the exact, approximative and numerical analysis of queueing models are the subject of
the course “Algorithmic methods in queueing theory.”
The organization is as follows. Chapter 2 first discusses a number of basic concepts
and results from probability theory that we will use. The most simple interesting queueing
model is treated in chapter 4, and its multi server version is treated in the next chapter.
Models with more general service or interarrival time distributions are analysed in the
chapters 6, 7 and 8. Some simple variations on these models are discussed in chapter 10.
Chapter 9 is devoted to queueing models with priority rules. The last chapter discusses
some insentive systems.
The text contains a lot of exercises and the reader is urged to try these exercises. This
is really necessary to acquire skills to model and analyse new situations.

1.1 Examples
Below we briefly describe some situations in which queueing is important.

Example 1.1.1 Supermarket.
How long do customers have to wait at the checkouts? What happens with the waiting

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time during peak-hours? Are there enough checkouts?

Example 1.1.2 Production system.
A machine produces different types of products.
What is the production lead time of an order? What is the reduction in the lead time
when we have an extra machine? Should we assign priorities to the orders?

Example 1.1.3 Post office.
In a post office there are counters specialized in e.g. stamps, packages, financial transac-
tions, etc.
Are there enough counters? Separate queues or one common queue in front of counters
with the same specialization?

Example 1.1.4 Data communication.
In computer communication networks standard packages called cells are transmitted over
links from one switch to the next. In each switch incoming cells can be buffered when the
incoming demand exceeds the link capacity. Once the buffer is full incoming cells will be
lost.
What is the cell delay at the switches? What is the fraction of cells that will be lost? What
is a good size of the buffer?

Example 1.1.5 Parking place.
They are going to make a new parking place in front of a super market.
How large should it be?

Example 1.1.6 Assembly of printed circuit boards.
Mounting vertical components on printed circuit boards is done in an assembly center
consisting of a number of parallel insertion machines. Each machine has a magazine to
store components.
What is the production lead time of the printed circuit boards? How should the components
necessary for the assembly of printed circuit boards be divided among the machines?

Example 1.1.7 Call centers of an insurance company.
Questions by phone, regarding insurance conditions, are handled by a call center. This call
center has a team structure, where each team helps customers from a specific region only.
How long do customers have to wait before an operator becomes available? Is the number
of incoming telephone lines enough? Are there enough operators? Pooling teams?

Example 1.1.8 Main frame computer.
Many cashomats are connected to a big main frame computer handling all financial trans-
actions.
Is the capacity of the main frame computer sufficient? What happens when the use of
cashomats increases?

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Example 1.1.9 Toll booths.
Motorists have to pay toll in order to pass a bridge. Are there enough toll booths?

Example 1.1.10 Traﬃc lights.
How do we have to regulate traﬃc lights such that the waiting times are acceptable?

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Chapter 2

Basic concepts from probability
theory

This chapter is devoted to some basic concepts from probability theory.

2.1 Random variable
Random variables are denoted by capitals, X, Y , etc. The expected value or mean of X is
denoted by E(X) and its variance by σ 2 (X) where σ(X) is the standard deviation of X.
An important quantity is the coefficient of variation of the positive random variable X
defined as
σ(X)
cX = .
E(X)
The coefficient of variation is a (dimensionless) measure of the variability of the random
variable X.

2.2 Generating function
Let X be a nonnegative discrete random variable with P (X = n) = p(n), n = 0, 1, 2, . . ..
Then the generating function PX (z) of X is defined as
∞
X
PX (z) = E(z ) = p(n)z n .
n=0

Note that |PX (z)| ≤ 1 for all |z| ≤ 1. Further
PX (0) = p(0), PX (1) = 1, PX (1) = E(X),
and, more general,
(k)
PX (1) = E(X(X − 1) · · · (X − k + 1)),

11

where the superscript (k) denotes the kth derivative. For the generating function of the
sum Z = X + Y of two independent discrete random variables X and Y , it holds that
PZ (z) = PX (z) · PY (z).
When Z is with probability q equal to X and with probability 1 − q equal to Y , then
PZ (z) = qPX (z) + (1 − q)PY (z).

2.3 Laplace-Stieltjes transform
The Laplace-Stieltjes transform X(s) of a nonnegative random variable X with distribution
function F (·), is deﬁned as
∞
X(s) = E(e−sX ) = e−sx dF (x), s ≥ 0.
x=0

When the random variable X has a density f (·), then the transform simpliﬁes to
∞
X(s) = e−sx f (x)dx, s ≥ 0.
x=0

Note that |X(s)| ≤ 1 for all s ≥ 0. Further
X(0) = 1, X (0) = −E(X), X (k) (0) = (−1)k E(X k ).
For the transform of the sum Z = X + Y of two independent random variables X and Y ,
it holds that
Z(s) = X(s) · Y (s).
When Z is with probability q equal to X and with probability 1 − q equal to Y , then
Z(s) = q X(s) + (1 − q)Y (s).

2.4 Useful probability distributions
This section discusses a number of important distributions which have been found useful
for describing random variables in many applications.

2.4.1 Geometric distribution
A geometric random variable X with parameter p has probability distribution
P (X = n) = (1 − p)pn , n = 0, 1, 2, . . .
For this distribution we have
1−p p p 1
PX (z) = , E(X) = , σ 2 (X) = , c2 = .
X
1 − pz 1−p (1 − p)2 p

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2.4.2 Poisson distribution
A Poisson random variable X with parameter µ has probability distribution

µn −µ
P (X = n) = e , n = 0, 1, 2, . . .
n!

For the Poisson distribution it holds that

1
PX (z) = e−µ(1−z) , E(X) = σ 2 (X) = µ, c2 =
X .
µ

2.4.3 Exponential distribution
The density of an exponential distribution with parameter µ is given by

f (t) = µe−µt , t > 0.

The distribution function equals

F (t) = 1 − e−µt , t ≥ 0.

For this distribution we have

µ 1 1
X(s) = , E(X) = , σ 2 (X) = , cX = 1.
µ+s µ µ2

An important property of an exponential random variable X with parameter µ is the
memoryless property. This property states that for all x ≥ 0 and t ≥ 0,

P (X > x + t|X > t) = P (X > x) = e−µx .

So the remaining lifetime of X, given that X is still alive at time t, is again exponentially
distributed with the same mean 1/µ. We often use the memoryless property in the form

P (X < t + ∆t|X > t) = 1 − e−µ∆t = µ∆t + o(∆t), (∆t → 0), (2.1)

where o(∆t), (∆t → 0), is a shorthand notation for a function, g(∆t) say, for which
g(∆t)/∆t tends to 0 when ∆t → 0 (see e.g. [4]).
If X1 , . . . , Xn are independent exponential random variables with parameters µ1 , . . . , µn
respectively, then min(X1 , . . . , Xn ) is again an exponential random variable with parameter
µ1 + · · · + µn and the probability that Xi is the smallest one is given by µi /(µ1 + · · · + µn ),
i = 1, . . . , n. (see exercise 1).

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2.4.4 Erlang distribution
A random variable X has an Erlang-k (k = 1, 2, . . .) distribution with mean k/µ if X
is the sum of k independent random variables X1 , . . . , Xk having a common exponential
distribution with mean 1/µ. The common notation is Ek (µ) or briefly Ek . The density of
an Ek (µ) distribution is given by
(µt)k−1 −µt
f (t) = µ e , t > 0.
(k − 1)!
The distribution function equals
k−1
(µt)j −µt
F (t) = 1 − e , t ≥ 0.
j=0 j!

The parameter µ is called the scale parameter, k is the shape parameter. A phase diagram
of the Ek distribution is shown in figure 2.1.

µ µ µ

1 2 k

Figure 2.1: Phase diagram for the Erlang-k distribution with scale parameter µ

In figure 2.2 we display the density of the Erlang-k distribution with mean 1 (so µ = k)
for various values of k.
The mean, variance and squared coefficient of variation are equal to
k k 1
E(X) = , σ 2 (X) = , c2 =
X .
µ µ2 k
The Laplace-Stieltjes transform is given by
k
µ
X(s) = .
µ+s
A convenient distribution arises when we mix an Ek−1 and Ek distribution with the
same scale parameters. The notation used is Ek−1,k . A random variable X has an Ek−1,k (µ)
distribution, if X is with probability p (resp. 1 − p) the sum of k − 1 (resp. k) independent
exponentials with common mean 1/µ. The density of this distribution has the form
(µt)k−2 −µt (µt)k−1 −µt
f (t) = pµ e + (1 − p)µ e , t > 0,
(k − 2)! (k − 1)!
where 0 ≤ p ≤ 1. As p runs from 1 to 0, the squared coefficient of variation of the
mixed Erlang distribution varies from 1/(k − 1) to 1/k. It will appear (later on) that this
distribution is useful for fitting a distribution if only the first two moments of a random
variable are known.

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1.4
k=1
2
4
1.2 10

1

0.8

0.6

0.4

0.2

0
0 0.5 1 1.5 2 2.5 3 3.5 4

Figure 2.2: The density of the Erlang-k distribution with mean 1 for various values of k

2.4.5 Hyperexponential distribution
A random variable X is hyperexponentially distributed if X is with probability pi , i =
1, . . . , k an exponential random variable Xi with mean 1/µi . For this random variable we
use the notation Hk (p1 , . . . , pk ; µ1 , . . . , µk ), or simply Hk . The density is given by

k
f (t) = pi µi e−µi t , t > 0,
i=1

and the mean is equal to
k
pi
E(X) = .
i=1 µi

The Laplace-Stieltjes transform satisfies
k
p i µi
X(s) = .
i=1 µi + s

The coefficient of variation cX of this distribution is always greater than or equal to 1
(see exercise 3). A phase diagram of the Hk distribution is shown in figure 2.3.

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µ1

1
p1

pk
k

µk

Figure 2.3: Phase diagram for the hyperexponential distribution

2.4.6 Phase-type distribution
The preceding distributions are all special cases of the phase-type distribution. The notation
is P H. This distribution is characterized by a Markov chain with states 1, . . . , k (the so-
called phases) and a transition probability matrix P which is transient. This means that
P n tends to zero as n tends to infinity. In words, eventually you will always leave the
Markov chain. The residence time in state i is exponentially distributed with mean 1/µi ,
and the Markov chain is entered with probability pi in state i, i = 1, . . . , k. Then the
random variable X has a phase-type distribution if X is the total residence time in the
preceding Markov chain, i.e. X is the total time elapsing from start in the Markov chain
till departure from the Markov chain.
We mention two important classes of phase-type distributions which are dense in the
class of all non-negative distribution functions. This is meant in the sense that for any
non-negative distribution function F (·) a sequence of phase-type distributions can be found
which pointwise converges at the points of continuity of F (·). The denseness of the two
classes makes them very useful as a practical modelling tool. A proof of the denseness can
be found in [23, 24]. The first class is the class of Coxian distributions, notation Ck , and
the other class consists of mixtures of Erlang distributions with the same scale parameters.
The phase representations of these two classes are shown in the figures 2.4 and 2.5.

µ1 µ2 µk
p1 p2 pk −1
1 2 k

1 − p1 1 − p2 1 − pk −1

Figure 2.4: Phase diagram for the Coxian distribution

A random variable X has a Coxian distribution of order k if it has to go through up to
at most k exponential phases. The mean length of phase n is 1/µn , n = 1, . . . , k. It starts
in phase 1. After phase n it comes to an end with probability 1 − pn and it enters the next
phase with probability pn . Obviously pk = 0. For the Coxian-2 distribution it holds that

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µ

1

p1
1 2

p2 µ µ

pk

1 2 k

µ µ µ

Figure 2.5: Phase diagram for the mixed Erlang distribution

the squared coefficient of variation is greater than or equal to 0.5 (see exercise 8).
A random variable X has a mixed Erlang distribution of order k if it is with probability
pn the sum of n exponentials with the same mean 1/µ, n = 1, . . . , k.

2.5 Fitting distributions
In practice it often occurs that the only information of random variables that is available
is their mean and standard deviation, or if one is lucky, some real data. To obtain an
approximating distribution it is common to fit a phase-type distribution on the mean,
E(X), and the coefficient of variation, cX , of a given positive random variable X, by using
the following simple approach.
In case 0 < cX < 1 one fits an Ek−1,k distribution (see subsection 2.4.4). More specifi-
cally, if
1 1
≤ c2 ≤
X ,
k k−1
for certain k = 2, 3, . . ., then the approximating distribution is with probability p (resp.
1 − p) the sum of k − 1 (resp. k) independent exponentials with common mean 1/µ. By
choosing (see e.g. [28])

1 k−p
p= [kc2 − {k(1 + c2 ) − k 2 c2 }1/2 ], µ= ,
1 + cX X
2 X X
E(X)

the Ek−1,k distribution matches E(X) and cX .
In case cX ≥ 1 one fits a H2 (p1 , p2 ; µ1 , µ2 ) distribution. The hyperexponential distribu-
tion however is not uniquely determined by its first two moments. In applications, the H2

17

distribution with balanced means is often used. This means that the normalization
p1 p2
=
µ1 µ2
is used. The parameters of the H2 distribution with balanced means and fitting E(X) and
cX (≥ 1) are given by
 
1 c2 − 1 
X
p1 = 1 + , p2 = 1 − p1 ,
2 c2 + 1
X

2p1 2p2
µ1 = , µ1 = .
E(X) E(X)

In case c2 ≥ 0.5 one can also use a Coxian-2 distribution for a two-moment fit. The
X
following set is suggested by [18],
µ1 = 2/E(X), p1 = 0.5/c2 ,
X µ2 = µ1 p1 .
It also possible to make a more sophisticated use of phase-type distributions by, e.g.,
trying to match the first three (or even more) moments of X or to approximate the shape
of X (see e.g. [29, 11, 13]).
Phase-type distributions may of course also naturally arise in practical applications.
For example, if the processing of a job involves performing several tasks, where each task
takes an exponential amount of time, then the processing time can be described by an
Erlang distribution.

2.6 Poisson process
Let N (t) be the number of arrivals in [0, t] for a Poisson process with rate λ, i.e. the time
between successive arrivals is exponentially distributed with parameter λ and independent
of the past. Then N (t) has a Poisson distribution with parameter λt, so
(λt)k −λt
P (N (t) = k) = e , k = 0, 1, 2, . . .
k!
The mean, variance and coefficient of variation of N (t) are equal to (see subsection 2.4.2)
1
E(N (t)) = λt, σ 2 (N (t)) = λt, c2 (t) =
N .
λt
From (2.1) it is easily verified that
P (arrival in (t, t + ∆t]) = λ∆t + o(∆t), (∆t → 0).
Hence, for small ∆t,
P (arrival in (t, t + ∆t]) ≈ λ∆t. (2.2)

18

So in each small time interval of length ∆t the occurence of an arrival is equally likely. In
other words, Poisson arrivals occur completely random in time. In ﬁgure 2.6 we show a
realization of a Poisson process and an arrival process with Erlang-10 interarrival times.
Both processes have rate 1. The ﬁgure illustrates that Erlang arrivals are much more
equally spread out over time than Poisson arrivals.

Poisson
t

Erlang-10
t

Figure 2.6: A realization of Poisson arrivals and Erlang-10 arrivals, both with rate 1

The Poisson process is an extremely useful process for modelling purposes in many
practical applications, such as, e.g. to model arrival processes for queueing models or
demand processes for inventory systems. It is empirically found that in many circumstances
the arising stochastic processes can be well approximated by a Poisson process.
Next we mention two important properties of a Poisson process (see e.g. [20]).

(i) Merging.
Suppose that N1 (t) and N2 (t) are two independent Poisson processes with respective
rates λ1 and λ2 . Then the sum N1 (t) + N2 (t) of the two processes is again a Poisson
process with rate λ1 + λ2 .

(ii) Splitting.
Suppose that N (t) is a Poisson process with rate λ and that each arrival is marked
with probability p independent of all other arrivals. Let N1 (t) and N2 (t) denote
respectively the number of marked and unmarked arrivals in [0, t]. Then N1 (t) and
N2 (t) are both Poisson processes with respective rates λp and λ(1 − p). And these
two processes are independent.

So Poisson processes remain Poisson processes under merging and splitting.

19

2.7 Exercises
Exercise 1.
Let X1 , . . . , Xn be independent exponential random variables with mean E(Xi ) = 1/µi ,
i = 1, . . . , n. Define

Yn = min(X1 , . . . , Xn ), Zn = max(X1 , . . . , Xn ).

(i) Determine the distributions of Yn and Zn .

(ii) Show that the probability that Xi is the smallest one among X1 , . . . , Xn is equal to
µi /(µ1 + · · · + µn ), i = 1, . . . , n.

Exercise 2.
Let X1 , X2 , . . . be independent exponential random variables with mean 1/µ and let N be
a discrete random variable with

P (N = k) = (1 − p)pk−1 , k = 1, 2, . . . ,

where 0 ≤ p < 1 (i.e. N is a shifted geometric random variable). Show that S defined as
N
S= Xn
n=1

is again exponentially distributed with parameter (1 − p)µ.
Exercise 3.
Show that the coefficient of variation of a hyperexponential distribution is greater than or
equal to 1.
Exercise 4. (Poisson process)
Suppose that arrivals occur at T1 , T2 , . . .. The interarrival times An = Tn − Tn−1 are
independent and have common exponential distribution with mean 1/λ, where T0 = 0 by
convention. Let N (t) denote the number of arrivals is [0, t] and define for n = 0, 1, 2, . . .

pn (t) = P (N (t) = n), t > 0.

(i) Determine p0 (t).

(ii) Show that for n = 1, 2, . . .

pn (t) = −λpn (t) + λpn−1 (t), t > 0,

with initial condition pn (0) = 0.

(iii) Solve the preceding differential equations for n = 1, 2, . . .

20

Suppose that arrivals occur at T1 , T2 , . . .. The interarrival times An = Tn − Tn−1 are
independent and have common exponential distribution with mean 1/λ, where T0 = 0 by
convention. Let N (t) denote the number of arrivals is [0, t] and deﬁne for n = 0, 1, 2, . . .

pn (t) = P (N (t) = n), t > 0.

(i) Determine p0 (t).

(ii) Show that for n = 1, 2, . . .

t
pn (t) = pn−1 (t − x)λe−λx dx, t > 0.
0

(iii) Solve the preceding integral equations for n = 1, 2, . . .

Prove the properties (i) and (ii) of Poisson processes, formulated in section 2.6.
Exercise 7. (Fitting a distribution)
Suppose that processing a job on a certain machine takes on the average 4 minutes with a
standard deviation of 3 minutes. Show that if we model the processing time as a mixture
of an Erlang-1 (exponential) distribution and an Erlang-2 distribution with density

f (t) = pµe−µt + (1 − p)µ2 te−µt ,

the parameters p and µ can be chosen in such a way that this distribution matches the
mean and standard deviation of the processing times on the machine.
Exercise 8.
Consider a random variable X with a Coxian-2 distribution with parameters µ1 and µ2
and branching probability p1 .

(i) Show that c2 ≥ 0.5.
X

(ii) Show that if µ1 < µ2 , then this Coxian-2 distribution is identical to the Coxian-
2 distribution with parameters µ1 , µ2 and p1 where µ1 = µ2 , µ2 = µ1 and p1 =
ˆ ˆ ˆ ˆ ˆ ˆ
1 − (1 − p1 )µ1 /µ2 .

Part (ii) implies that for any Coxian-2 distribution we may assume without loss of generality
that µ1 ≥ µ2 .
Exercise 9.
Let X and Y be exponentials with parameters µ and λ, respectively. Suppose that λ < µ.
Let Z be equal to X with probability λ/µ and equal to X + Y with probability 1 − λ/µ.
Show that Z is an exponential with parameter λ.

21

Exercise 10.
Consider a H2 distribution with parameters µ1 > µ2 and branching probabilities q1 and
q2 , respectively. Show that the C2 distribution with parameters µ1 and µ2 and branching
probability p1 given by

p1 = 1 − (q1 µ1 + q2 µ2 )/µ1 ,

is equivalent to the H2 distribution.
Exercise 11. (Poisson distribution)
Let X1 , . . . , Xn be independent Poisson random variables with means µ1 , . . . , µn , respec-
tively. Show that the sum X1 + · · · + Xn is Poisson distributed with mean µ1 + · · · + µn .

22

Chapter 3

Queueing models and some
fundamental relations

In this chapter we describe the basic queueing model and we discuss some important fun-
damental relations for this model. These results can be found in every standard textbook
on this topic, see e.g. [14, 20, 28].

3.1 Queueing models and Kendall’s notation
The basic queueing model is shown in ﬁgure 3.1. It can be used to model, e.g., machines
or operators processing orders or communication equipment processing information.

Figure 3.1: Basic queueing model

Among others, a queueing model is characterized by:

• The arrival process of customers.
Usually we assume that the interarrival times are independent and have a common
distribution. In many practical situations customers arrive according to a Poisson
stream (i.e. exponential interarrival times). Customers may arrive one by one, or
in batches. An example of batch arrivals is the customs oﬃce at the border where
travel documents of bus passengers have to be checked.

23

• The behaviour of customers.
Customers may be patient and willing to wait (for a long time). Or customers may
be impatient and leave after a while. For example, in call centers, customers will
hang up when they have to wait too long before an operator is available, and they
possibly try again after a while.
• The service times.
Usually we assume that the service times are independent and identically distributed,
and that they are independent of the interarrival times. For example, the service
times can be deterministic or exponentially distributed. It can also occur that service
times are dependent of the queue length. For example, the processing rates of the
machines in a production system can be increased once the number of jobs waiting
to be processed becomes too large.
• The service discipline.
Customers can be served one by one or in batches. We have many possibilities for
the order in which they enter service. We mention:
– first come first served, i.e. in order of arrival;
– random order;
– last come first served (e.g. in a computer stack or a shunt buffer in a production
line);
– priorities (e.g. rush orders first, shortest processing time first);
– processor sharing (in computers that equally divide their processing power over
all jobs in the system).
• The service capacity.
There may be a single server or a group of servers helping the customers.
• The waiting room.
There can be limitations with respect to the number of customers in the system. For
example, in a data communication network, only finitely many cells can be buffered
in a switch. The determination of good buffer sizes is an important issue in the design
of these networks.
Kendall introduced a shorthand notation to characterize a range of these queueing mod-
els. It is a three-part code a/b/c. The first letter specifies the interarrival time distribution
and the second one the service time distribution. For example, for a general distribution
the letter G is used, M for the exponential distribution (M stands for Memoryless) and
D for deterministic times. The third and last letter specifies the number of servers. Some
examples are M/M/1, M/M/c, M/G/1, G/M/1 and M/D/1. The notation can be ex-
tended with an extra letter to cover other queueing models. For example, a system with
exponential interarrival and service times, one server and having waiting room only for N
customers (including the one in service) is abbreviated by the four letter code M/M/1/N .

24

In the basic model, customers arrive one by one and they are always allowed to enter
the system, there is always room, there are no priority rules and customers are served in
order of arrival. It will be explicitly indicated (e.g. by additional letters) when one of these
assumptions does not hold.

3.2 Occupation rate
In a single-server system G/G/1 with arrival rate λ and mean service time E(B) the
amount of work arriving per unit time equals λE(B). The server can handle 1 unit work
per unit time. To avoid that the queue eventually grows to inﬁnity, we have to require that
λE(B) < 1. Without going into details, we note that the mean queue length also explodes
when λE(B) = 1, except in the D/D/1 system, i.e., the system with no randomness at all.
It is common to use the notation

ρ = λE(B).

If ρ < 1, then ρ is called the occupation rate or server utilization, because it is the fraction
of time the server is working.
In a multi-server system G/G/c we have to require that λE(B) < c. Here the occupa-
tion rate per server is ρ = λE(B)/c.

3.3 Performance measures
Relevant performance measures in the analysis of queueing models are:

• The distribution of the waiting time and the sojourn time of a customer. The sojourn
time is the waiting time plus the service time.

• The distribution of the number of customers in the system (including or excluding
the one or those in service).

• The distribution of the amount of work in the system. That is the sum of service times
of the waiting customers and the residual service time of the customer in service.

• The distribution of the busy period of the server. This is a period of time during
which the server is working continuously.

In particular, we are interested in mean performance measures, such as the mean waiting
time and the mean sojourn time.
Now consider the G/G/c queue. Let the random variable L(t) denote the number of
customers in the system at time t, and let Sn denote the sojourn time of the nth customer
in the system. Under the assumption that the occupation rate per server is less than one,
it can be shown that these random variables have a limiting distribution as t → ∞ and
n → ∞. These distributions are independent of the initial condition of the system.

25

Let the random variables L and S have the limiting distributions of L(t) and Sn ,
respectively. So

pk = P (L = k) = lim P (L(t) = k), FS (x) = P (S ≤ x) = n→∞ P (Sn ≤ x).
lim
t→∞

The probability pk can be interpreted as the fraction of time that k customers are in the
system, and FS (x) gives the probability that the sojourn time of an arbitrary customer
entering the system is not greater than x units of time. It further holds with probability
1 that
1 t 1 n
lim L(x)dx = E(L), lim Sk = E(S).
t→∞ t x=0 n→∞ n
k=1

So the long-run average number of customers in the system and the long-run average
sojourn time are equal to E(L) and E(S), respectively. A very useful result for queueing
systems relating E(L) and E(S) is presented in the following section.

3.4 Little’s law
Little’s law gives a very important relation between E(L), the mean number of customers
in the system, E(S), the mean sojourn time and λ, the average number of customers
entering the system per unit time. Little’s law states that

E(L) = λE(S). (3.1)

Here it is assumed that the capacity of the system is sufficient to deal with the customers
(i.e. the number of customers in the system does not grow to infinity).
Intuitively, this result can be understood as follows. Suppose that all customers pay 1
dollar per unit time while in the system. This money can be earned in two ways. The first
possibility is to let pay all customers “continuously” in time. Then the average reward
earned by the system equals E(L) dollar per unit time. The second possibility is to let
customers pay 1 dollar per unit time for their residence in the system when they leave. In
equilibrium, the average number of customers leaving the system per unit time is equal
to the average number of customers entering the system. So the system earns an average
reward of λE(S) dollar per unit time. Obviously, the system earns the same in both cases.
For a rigorous proof, see [17, 25].
To demonstrate the use of Little’s law we consider the basic queueing model in figure
3.1 with one server. For this model we can derive relations between several performance
measures by applying Little’s law to suitably defined (sub)systems. Application of Little’s
law to the system consisting of queue plus server yields relation (3.1). Applying Little’s
law to the queue (excluding the server) yields a relation between the queue length Lq and
the waiting time W , namely

E(Lq ) = λE(W ).

26

Finally, when we apply Little’s law to the server only, we obtain (cf. section 3.2)

ρ = λE(B),

where ρ is the mean number of customers at the server (which is the same as the fraction
of time the server is working) and E(B) the mean service time.

3.5 PASTA property
For queueing systems with Poisson arrivals, so for M/·/· systems, the very special property
holds that arriving customers find on average the same situation in the queueing system
as an outside observer looking at the system at an arbitrary point in time. More precisely,
the fraction of customers finding on arrival the system in some state A is exactly the same
as the fraction of time the system is in state A. This property is only true for Poisson
arrivals.
In general this property is not true. For instance, in a D/D/1 system which is empty
at time 0, and with arrivals at 1, 3, 5, . . . and service times 1, every arriving customer finds
an empty system, whereas the fraction of time the system is empty is 1/2.
This property of Poisson arrivals is called PASTA property, which is the acrynom for
Poisson Arrivals See Time Averages. Intuitively, this property can be explained by the
fact that Poisson arrivals occur completely random in time (see (2.2)). A rigorous proof of
the PASTA property can be found in [31, 32].
In the following chapters we will show that in many queueing models it is possible to
determine mean performance measures, such as E(S) and E(L), directly (i.e. not from
the distribution of these measures) by using the PASTA property and Little’s law. This
powerful approach is called the mean value approach.

27

3.6 Exercises
Exercise 12.
In a gas station there is one gas pump. Cars arrive at the gas station according to a Poisson
proces. The arrival rate is 20 cars per hour. An arriving car ﬁnding n cars at the station
immediately leaves with probability qn = n/4, and joins the queue with probability 1 − qn ,
n = 0, 1, 2, 3, 4. Cars are served in order of arrival. The service time (i.e. the time needed
for pumping and paying) is exponential. The mean service time is 3 minutes.

(i) Determine the stationary distribution of the number of cars at the gas station.

(ii) Determine the mean number of cars at the gas station.

(iii) Determine the mean sojourn time (waiting time plus service time) of cars deciding
to take gas at the station.

(iv) Determine the mean sojourn time and the mean waiting time of all cars arriving at
the gas station.

28

Chapter 4

M/M/1 queue

In this chapter we will analyze the model with exponential interarrival times with mean
1/λ, exponential service times with mean 1/µ and a single server. Customers are served
in order of arrival. We require that
λ
ρ= < 1,
µ
since, otherwise, the queue length will explode (see section 3.2). The quantity ρ is the
fraction of time the server is working. In the following section we will first study the
time-dependent behaviour of this system. After that, we consider the limiting behaviour.

4.1 Time-dependent behaviour
The exponential distribution allows for a very simple description of the state of the system
at time t, namely the number of customers in the system (i.e. the customers waiting in
the queue and the one being served). Neither we do have to remember when the last
customer arrived nor we have to register when the last customer entered service. Since the
exponential distribution is memoryless (see 2.1), this information does not yield a better
prediction of the future.
Let pn (t) denote the probability that at time t there are n customers in the system,
n = 0, 1, . . . Based on property (2.1) we get, for ∆t → 0,

p0 (t + ∆t) = (1 − λ∆t)p0 (t) + µ∆tp1 (t) + o(∆t),
pn (t + ∆t) = λ∆tpn−1 (t) + (1 − (λ + µ)∆t)pn (t) + µ∆tpn+1 (t) + o(∆t),
n = 1, 2, . . .

Hence, by letting ∆t → 0, we obtain the following infinite set of differential equations for
the probabilities pn (t).

p0 (t) = −λp0 (t) + µp1 (t),
(4.1)
pn (t) = λpn−1 (t) − (λ + µ)pn (t) + µpn+1 (t), n = 1, 2, . . .

29

It is difficult to solve these differential equations. An explicit solution for the probabilities
pn (t) can be found in [14] (see p. 77). The expression presented there is an infinite sum
of modified Bessel functions. So already one of the simplest interesting queueing models
leads to a difficult expression for the time-dependent behavior of its state probabilities. For
more general systems we can only expect more complexity. Therefore, in the remainder
we will focus on the limiting or equilibrium behavior of this system, which appears to be
much easier to analyse.

4.2 Limiting behavior
One may show that as t → ∞, then pn (t) → 0 and pn (t) → pn (see e.g. [8]). Hence, from
(4.1) it follows that the limiting or equilibrium probabilities pn satisfy the equations

0 = −λp0 + µp1 , (4.2)
0 = λpn−1 − (λ + µ)pn + µpn+1 , n = 1, 2, . . . (4.3)

Clearly, the probabilities pn also satisfy

∞
pn = 1, (4.4)
n=0

which is called the normalization equation. It is also possible to derive the equations (4.2)
and (4.3) directly from a flow diagram, as shown in figure 4.1.

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Figure 4.1: Flow diagram for the M/M/1 model

The arrows indicate possible transitions. The rate at which a transition occurs is λ for
a transition from n to n+1 (an arrival) and µ for a transition from n+1 to n (a departure).
The number of transitions per unit time from n to n + 1, which is also called the flow from
n to n + 1, is equal to pn , the fraction of time the system is in state n, times λ, the rate
at arrivals occur while the system is in state n. The equilibrium equations (4.2) and (4.3)
follow by equating the flow out of state n and the flow into state n.
For this simple model there are many ways to determine the solution of the equations
(4.2)–(4.4). Below we discuss several approaches.

30

4.2.1 Direct approach
The equations (4.3) are a second order recurrence relation with constant coefficients. Its
general solution is of the form

pn = c1 xn + c2 xn ,
1 2 n = 0, 1, 2, . . . (4.5)

where x1 and x2 are roots of the quadratic equation

λ − (λ + µ)x + µx2 = 0.

This equation has two zeros, namely x = 1 and x = λ/µ = ρ. So all solutions to (4.3) are
of the form

pn = c1 + c2 ρn , n = 0, 1, 2, . . .

Equation (4.4), stating that the sum of all probabilities is equal to 1, of course directly
implies that c1 must be equal to 0. That c1 must be equal to 0 also follows from (4.2) by
substituting the solution (4.5) into (4.2).
The coefficient c2 finally follows from the normalization equation (4.4), yielding that
c2 = 1 − ρ. So we can conclude that

pn = (1 − ρ)ρn , n = 0, 1, 2, . . . (4.6)

Apparantly, the equilibrium distribution depends upon λ and µ only through their ratio ρ.

4.2.2 Recursion
One can use (4.2) to express p1 in p0 yielding

p1 = ρp0 .

Substitution of this relation into (4.3) for n = 1 gives

p 2 = ρ2 p 0 .

By substituting the relations above into (4.3) for n = 2 we obtain p3 , and so on. Hence we
can recursively express all probabilities in terms of p0 , yielding

p n = ρn p 0 , n = 0, 1, 2, . . . .

The probability p0 finally follows from the normalization equation (4.4).

31

4.2.3 Generating function approach
The probability generating function of the random variable L, the number of customers in
the system, is given by
∞
PL (z) = pn z n , (4.7)
n=0

which is properly defined for z with |z| ≤ 1. By multiplying the nth equilibrium equation
with z n and then summing the equations over all n, the equilibrium equations for pn can
be transformed into the following single equation for PL (z),
0 = µp0 (1 − z −1 ) + (λz + µz −1 − (λ + µ))PL (z).
The solution of this equation is
∞
p0 1−ρ
PL (z) = = = (1 − ρ)ρn z n , (4.8)
1 − ρz 1 − ρz n=0
where we used that P (1) = 1 to determine p0 = 1 − ρ (cf. section 3.2). Hence, by equating
the coefficients of z n in (4.7) and (4.8) we retrieve the solution (4.6).

4.2.4 Global balance principle
The global balance principle states that for each set of states A, the flow out of set A is
equal to the flow into that set. In fact, the equilibrium equations (4.2)–(4.3) follow by
applying this principle to a single state. But if we apply the balance principle to the set
A = {0, 1, . . . , n − 1} we get the very simple relation
λpn−1 = µpn , n = 1, 2, . . .
Repeated application of this relation yields
p n = ρn p 0 , n = 0, 1, 2, . . .
so that, after normalization, the solution (4.6) follows.

4.3 Mean performance measures
From the equilibrium probabilities we can derive expressions for the mean number of cus-
tomers in the system and the mean time spent in the system. For the first one we get
∞
ρ
E(L) = npn = ,
n=0 1−ρ
and by applying Little’s law,
1/µ
E(S) = . (4.9)
1−ρ

32

If we look at the expressions for E(L) and E(S) we see that both quantities grow to inﬁnity
as ρ approaches unity. The dramatic behavior is caused by the variation in the arrival and
service process. This type of behavior with respect to ρ is characteristic for almost every
queueing system.
In fact, E(L) and E(S) can also be determined directly, i.e. without knowing the
probabilities pn , by combining Little’s law and the PASTA property (see section 3.5).
Based on PASTA we know that the average number of customers in the system seen by an
arriving customer equals E(L) and each of them (also the one in service) has a (residual)
service time with mean 1/µ. The customer further has to wait for its own service time.
Hence
1 1
E(S) = E(L) + .
µ µ
This relation is known as the arrival relation. Together with
E(L) = λE(S)
we ﬁnd expression (4.9). This approach is called the mean value approach.
The mean number of customers in the queue, E(Lq ), can be obtained from E(L) by
subtracting the mean number of customers in service, so
ρ2
E(Lq ) = E(L) − ρ = .
1−ρ
The mean waiting time, E(W ), follows from E(S) by subtracting the mean service time
(or from E(Lq ) by applying Little’s law). This yields
ρ/µ
E(W ) = E(S) − 1/µ = .
1−ρ

4.4 Distribution of the sojourn time and the waiting
time
It is also possible to derive the distribution of the sojourn time. Denote by La the number of
customers in the system just before the arrival of a customer and let Bk be the service time
of the kth customer. Of course, the customer in service has a residual service time instead
of an ordinary service time. But these are the same, since the exponential service time
distribution is memoryless. So the random variables Bk are independent and exponentially
distributed with mean 1/µ. Then we have
La +1
S= Bk . (4.10)
k=1

By conditioning on La and using that La and Bk are independent it follows that
La +1 ∞ n+1
P (S t) = P ( Bk t) = P( Bk t)P (La = n). (4.11)
k=1 n=0 k=1

33

The problem is to find the probability that an arriving customer finds n customers in the
system. PASTA states that the fraction of customers finding on arrival n customers in the
system is equal to the fraction of time there are n customers in the system, so
P (La = n) = pn = (1 − ρ)ρn . (4.12)
n+1
Substituting (4.12) in (4.11) and using that k=1 Bk is Erlang-(n + 1) distributed, yields
(cf. exercise 2)
∞ n
(µt)k −µt
P (S t) = e (1 − ρ)ρn
n=0 k=0 k!
∞ ∞
(µt)k −µt
= e (1 − ρ)ρn
k=0 n=k k!
∞
(µρt)k −µt
= e
k=0 k!
= e−µ(1−ρ)t , t ≥ 0. (4.13)
Hence, S is exponentially distributed with parameter µ(1 − ρ). This result can also be
obtained via the use of transforms. From (4.10) it follows, by conditioning on La , that
S(s) = E(e−sS )
∞
= P (La = n)E(e−s(B1 +...+Bn+1 ) )
n=0
∞
= (1 − ρ)ρn E(e−sB1 ) · · · E(e−sBn+1 ).
n=0

Since Bk is exponentially distributed with parameter µ, we have (see subsection 2.4.3)
µ
E(e−sBk ) = ,
µ+s
so
∞ n+1
n µ µ(1 − ρ)
S(s) = (1 − ρ)ρ = ,
n=0 µ+s µ(1 − ρ) + s
from which we can conclude that S is an exponential random variable with parameter
µ(1 − ρ).
To find the distribution of the waiting time W , note that S = W +B, where the random
variable B is the service time. Since W and B are independent, it follows that
µ
S(s) = W (s) · B(s) = W (s) · .
µ+s
and thus,
(1 − ρ)(µ + s) µ(1 − ρ)
W (s) = = (1 − ρ) · 1 + ρ · .
µ(1 − ρ) + s µ(1 − ρ) + s

34

From the transform of W we conclude (see subsection 2.3) that W is with probability
(1 − ρ) equal to zero, and with probability ρ equal to an exponential random variable with
parameter µ(1 − ρ). Hence

P (W t) = ρe−µ(1−ρ)t , t ≥ 0. (4.14)

The distribution of W can, of course, also be obtained along the same lines as (4.13).
Note that
P (W t)
P (W t|W 0) = = e−µ(1−ρ)t ,
P (W 0)
so the conditional waiting time W |W 0 is exponentially distributed with parameter
µ(1 − ρ).
In table 4.1 we list for increasing values of ρ the mean waiting time and some waiting
time probabilities. From these results we see that randomness in the arrival and service
process leads to (long) waiting times and the waiting times explode as the server utilization
tends to one.

ρ E(W ) P (W t)
t 5 10 20
0.5 1 0.04 0.00 0.00
0.8 4 0.29 0.11 0.02
0.9 9 0.55 0.33 0.12
0.95 19 0.74 0.58 0.35

Table 4.1: Performance characteristics for the M/M/1 with mean service time 1

Remark 4.4.1 (PASTA property)
For the present model we can also derive relation (4.12) directly from the flow diagram
4.1. Namely, the average number of customers per unit time finding on arrival n customers
in the system is equal to λpn . Dividing this number by the average number of customers
arriving per unit time gives the desired fraction, so
λpn
P (La = n) = = pn .
λ

4.5 Priorities
In this section we consider an M/M/1 system serving different types of customers. To keep
it simple we suppose that there are two types only, type 1 and 2 say, but the analysis can
easily be extended the situation with more types of customers (see also chapter 9). Type 1
and type 2 customers arrive according to independent Poisson processes with rate λ1 , and

35

λ2 respectively. The service times of all customers are exponentially distributed with the
same mean 1/µ. We assume that
ρ1 + ρ2 1,
where ρi = λi /µ, i.e. the occupation rate due to type i customers. Type 1 customers are
treated with priority over type 2 jobs. In the following subsections we will consider two
priority rules, preemptive-resume priority and non-preemptive priority.

4.5.1 Preemptive-resume priority
In the preemptive resume priority rule, type 1 customers have absolute priority over type
2 jobs. Absolute priority means that when a type 2 customer is in service and a type 1
customer arrives, the type 2 service is interrupted and the server proceeds with the type 1
customer. Once there are no more type 1 customers in the system, the server resumes the
service of the type 2 customer at the point where it was interrupted.
Let the random variable Li denote the number of type i customers in the system and
Si the sojourn time of a type i customer. Below we will determine E(Li ) and E(Si ) for
i = 1, 2.
For type 1 customers the type 2 customers do not exist. Hence we immediately have
1/µ ρ1
E(S1 ) = , E(L1 ) = . (4.15)
1 − ρ1 1 − ρ1
Since the (residual) service times of all customers are exponentially distributed with the
same mean, the total number of customers in the system does not depend on the order in
which the customers are served. So this number is the same as in the system where all
customers are served in order of arrival. Hence,
ρ1 + ρ2
E(L1 ) + E(L2 ) = , (4.16)
1 − ρ1 − ρ2
and thus, inserting (4.15),
ρ1 + ρ2 ρ1 ρ2
E(L2 ) = − = ,
1 − ρ1 − ρ2 1 − ρ1 (1 − ρ1 )(1 − ρ1 − ρ2 )
and applying Little’s law,
E(L2 ) 1/µ
E(S2 ) = = .
λ2 (1 − ρ1 )(1 − ρ1 − ρ2 )
Example 4.5.1 For λ1 = 0.2, λ2 = 0.6 and µ = 1, we ﬁnd in case all customers are
treated in order of arrival,
1
E(S) = = 5,
1 − 0.8
and in case type 1 customers have absolute priority over type 2 jobs,
1 1
E(S1 ) = = 1.25, E(S2 ) = = 6.25.
1 − 0.2 (1 − 0.2)(1 − 0.8)

36

4.5.2 Non-preemptive priority
We now consider the situation that type 1 customers have nearly absolute priority over
type 2 jobs. The difference with the previous rule is that type 1 customers are not allowed
to interrupt the service of a type 2 customers. This priority rule is therefore called non-
preemptive.
For the mean sojourn time of type 1 customers we find
1 1 1
E(S1 ) = E(L1 ) + + ρ2 .
µ µ µ
The last term reflects that when an arriving type 1 customer finds a type 2 customer
in service, he has to wait until the service of this type 2 customer has been completed.
According to PASTA the probability that he finds a type 2 customer in service is equal
to the fraction of time the server spends on type 2 customers, which is ρ2 . Together with
Little’s law,
E(L1 ) = λ1 E(S1 ),
we obtain
(1 + ρ2 )/µ (1 + ρ2 )ρ1
E(S1 ) = , E(L1 ) = .
1 − ρ1 1 − ρ1
For type 2 customers it follows from (4.16) that
(1 − ρ1 (1 − ρ1 − ρ2 ))ρ2
E(L2 ) = ,
(1 − ρ1 )(1 − ρ1 − ρ2 )
and applying Little’s law,
(1 − ρ1 (1 − ρ1 − ρ2 ))/µ
E(S2 ) = .
(1 − ρ1 )(1 − ρ1 − ρ2 )

Example 4.5.2 For λ1 = 0.2, λ2 = 0.6 and µ = 1, we get
1 + 0.6 1 − 0.2(1 − 0.8)
E(S1 ) = = 2, E(S2 ) = = 6.
1 − 0.2 (1 − 0.2)(1 − 0.8)

4.6 Busy period
In a servers life we can distinguish cycles. A cycle is the time that elapses between two
consecutive arrivals finding an empty system. Clearly, a cycle starts with a busy period BP
during which the server is helping customers, followed by an idle period IP during which
the system is empty.
Due to the memoryless property of the exponential distribution (see subsection 2.4.3),
an idle period IP is exponentially distributed with mean 1/λ. In the following subsections
we determine the mean and the distribution of a busy period BP .

37

4.6.1 Mean busy period
It is clear that the mean busy period divided by the mean cycle length is equal to the
fraction of time the server is working, so
E(BP ) E(BP )
= = ρ.
E(BP ) + E(IP ) E(BP ) + 1/λ
Hence,
1/µ
E(BP ) = .
1−ρ

4.6.2 Distribution of the busy period
Let the random variable Cn be the time till the system is empty again if there are now n
customers present in the system. Clearly, C1 is the length of a busy period, since a busy
period starts when the first customer after an idle period arrives and it ends when the
system is empty again. The random variables Cn satisfy the following recursion relation.
Suppose there are n( 0) customers in the system. Then the next event occurs after an
exponential time with parameter λ + µ: with probability λ/(λ + µ) a new customer arrives,
and with probability µ/(λ + µ) service is completed and a customer leaves the system.
Hence, for n = 1, 2, . . .,
Cn+1 with probability λ/(λ + µ),
Cn = X + (4.17)
Cn−1 with probability µ/(λ + µ),
where X is an exponential random variable with parameter λ + µ. From this relation we
get for the Laplace-Stieltjes transform Cn (s) of Cn that
λ+µ λ µ
Cn (s) = Cn+1 (s) + Cn−1 (s) ,
λ+µ+s λ+µ λ+µ
and thus, after rewriting,
(λ + µ + s)Cn (s) = λCn+1 (s) + µCn−1 (s), n = 1, 2, . . .
For fixed s this equation is a very similar to (4.3). Its general solution is
Cn (s) = c1 xn (s) + c2 xn (s),
1 2 n = 0, 1, 2, . . .
where x1 (s) and x2 (s) are the roots of the quadratic equation
(λ + µ + s)x = λx2 + µ,
satisfying 0 x1 (s) ≤ 1 x2 (s). Since 0 ≤ Cn (s) ≤ 1 it follows that c2 = 0. The
coefficient c1 follows from the fact that C0 = 0 and hence C0 (s) = 1, yielding c1 = 1.
Hence we obtain
Cn (s) = xn (s),
1

38

and in particular, for the Laplace-Stieltjes transform BP (s) of the busy period BP , we
find
1
BP (s) = C1 (s) = x1 (s) = λ+µ+s− (λ + µ + s)2 − 4λµ .
2λ
By inverting this transform (see e.g. [1]) we get for the density fBP (t) of BP ,
1
fBP (t) = √ e−(λ+µ)t I1 (2t λµ), t 0,
t ρ

where I1 (·) denotes the modified Bessel function of the first kind of order one, i.e.
∞
(x/2)2k+1
I1 (x) = .
k=0 k!(k + 1)!

In table 4.2 we list for some values of ρ the probability P (BP t) for a number of t
values. If you think of the situation that 1/µ is one hour, then 10% of the busy periods
lasts longer than 2 days (16 hours) and 5% percent even longer than 1 week, when ρ = 0.9.
Since the mean busy period is 10 hours in this case, it is not unlikely that in a month time
a busy period longer than a week occurs.

ρ P (BP t)
t 1 2 4 8 16 40 80
0.8 0.50 0.34 0.22 0.13 0.07 0.02 0.01
0.9 0.51 0.36 0.25 0.16 0.10 0.05 0.03
0.95 0.52 0.37 0.26 0.18 0.12 0.07 0.04

Table 4.2: Probabilities for the busy period duration for the M/M/1 with mean service
time equal to 1

4.7 Java applet
For the performance evaluation of the M/M/1 queue a JAVA applet is avalaible on the
World Wide Web. The link to this applet is http://www.win.tue.nl/cow/Q2. The applet
can be used to evaluate the mean value as well as the distribution of, e.g., the waiting time
and the number of customers in the system.

39

4.8 Exercises
Exercise 13. (bulk arrivals)
In a work station orders arrive according to a Poisson arrival process with arrival rate λ.
An order consists of N independent jobs. The distribution of N is given by

P (N = k) = (1 − p)pk−1

with k = 1, 2, . . . and 0 ≤ p 1. Each job requires an exponentially distributed amount
of processing time with mean 1/µ.

(i) Derive the distribution of the total processing time of an order.

(ii) Determine the distribution of the number of orders in the system.

Exercise 14. (variable production rate)
Consider a work station where jobs arrive according to a Poisson process with arrival rate
λ. The jobs have an exponentially distributed service time with mean 1/µ. So the service
completion rate (the rate at which jobs depart from the system) is equal to µ.
If the queue length drops below the threshold QL the service completion rate is lowered to
µL . If the queue length reaches QH , where QH ≥ QL , the service rate is increased to µH .
(L stands for low, H for high.)
Determine the queue length distribution and the mean time spent in the system.
Exercise 15.
A repair man ﬁxes broken televisions. The repair time is exponentially distributed with
a mean of 30 minutes. Broken televisions arrive at his repair shop according to a Poisson
stream, on average 10 broken televisions per day (8 hours).

(i) What is the fraction of time that the repair man has no work to do?

(ii) How many televisions are, on average, at his repair shop?

(iii) What is the mean throughput time (waiting time plus repair time) of a television?

Exercise 16.
In a gas station there is one gas pump. Cars arrive at the gas station according to a Poisson
process. The arrival rate is 20 cars per hour. Cars are served in order of arrival. The service
time (i.e. the time needed for pumping and paying) is exponentially distributed. The mean
service time is 2 minutes.

(i) Determine the distribution, mean and variance of the number of cars at the gas
station.

(ii) Determine the distribution of the sojourn time and the waiting time.

(iii) What is the fraction of cars that has to wait longer than 2 minutes?

40

An arriving car ﬁnding 2 cars at the station immediately leaves.

(iv) Determine the distribution, mean and variance of the number of cars at the gas
station.

(v) Determine the mean sojourn time and the mean waiting time of all cars (including
the ones that immediately leave the gas station).

Exercise 17.
A gas station has two pumps, one for gas and the other for LPG. For each pump customers
arrive according to a Poisson proces. On average 20 customers per hour for gas and 5
customers for LPG. The service times are exponential. For both pumps the mean service
time is 2 minutes.

(i) Determine the distribution of the number of customers at the gas pump, and at the
LPG pump.

(ii) Determine the distribution of the total number of customers at the gas station.

Exercise 18.
Consider an M/M/1 queue with two types of customers. The mean service time of all
customers is 5 minutes. The arrival rate of type 1 customers is 4 customers per hour and
for type 2 customers it is 5 customers per hour. Type 1 customers are treated with priority
over type 2 customers.

(i) Determine the mean sojourn time of type 1 and 2 customers under the preemptive-
resume priority rule.

(ii) Determine the mean sojourn time of type 1 and 2 customers under the non-preemptive
priority rule.

Exercise 19.
Consider an M/M/1 queue with an arrival rate of 60 customers per hour and a mean
service time of 45 seconds. A period during which there are 5 or more customers in the
system is called crowded, when there are less than 5 customers it is quiet. What is the
mean number of crowded periods per day (8 hours) and how long do they last on average?
Exercise 20.
Consider a machine where jobs arrive according to a Poisson stream with a rate of 20 jobs
per hour. The processing times are exponentially distributd with a mean of 1/µ hours.
The processing cost is 16µ dollar per hour, and the waiting cost is 20 dollar per order per
hour.
Determine the processing speed µ minimizing the average cost per hour.

41

Chapter 5

M/M/c queue

In this chapter we will analyze the model with exponential interarrival times with mean
1/λ, exponential service times with mean 1/µ and c parallel identical servers. Customers
are served in order of arrival. We suppose that the occupation rate per server,
λ
ρ= ,
cµ
is smaller than one.

5.1 Equilibrium probabilities
The state of the system is completely characterized by the number of customers in the
system. Let pn denote the equilibrium probability that there are n customers in the system.
Similar as for the M/M/1 we can derive the equilibrium equations for the probabilities pn
from the flow diagram shown in figure 5.1.
¢ ¢ ¢ ¢ ¢

¡ ¥¥¤
¤ ¤ ©§
¨ ¡ § ¥¥¤
¤ ¤ ¨ ¡ ¥¥¤
¤ ¤

£ £ ¦ £ § £ § £ §

Figure 5.1: Flow diagram for the M/M/c model

Instead of equating the flow into and out of a single state n, we get simpler equations
by equating the flow out of and into the set of states {0, 1, . . . , n − 1}. This amounts to
equating the flow between the two neighboring states n − 1 and n yielding
λpn−1 = min(n, c)µpn , n = 1, 2, . . .
Iterating gives
(cρ)n
pn = p0 , n = 0, . . . , c
n!

43

and
c
n n (cρ)
pc+n = ρ pc = ρ p0 , n = 0, 1, 2, . . .
c!
The probability p0 follows from normalization, yielding
c−1 −1
(cρ)n (cρ)c 1
p0 = + · .
n=0 n! c! 1−ρ

An important quantity is the probability that a job has to wait. Denote this probability
by ΠW . It is usually referred to as the delay probability. By PASTA it follows that

ΠW = pc + pc+1 + pc+2 + · · ·
pc
=
1−ρ
c−1 −1
(cρ)c (cρ)n (cρ)c
= (1 − ρ) + . (5.1)
c! n=0 n! c!

Remark 5.1.1 (Computation of ΠW )
It will be clear that the computation of ΠW by using (5.1) leads to numerical problems
when c is large (due to the terms (cρ)c /c!). In remark 11.3.2 we will formulate a numerically
stable procedure to compute ΠW .

5.2 Mean queue length and mean waiting time
From the equilibrium probabilities we directly obtain for the mean queue length,
∞
E(Lq ) = npc+n
n=0
pc ∞
= n(1 − ρ)ρn
1 − ρ n=0
ρ
= ΠW · , (5.2)
1−ρ
and then from Little’s law,
1 1
E(W ) = ΠW · · . (5.3)
1 − ρ cµ
These formulas for E(Lq ) and E(W ) can also be found by using the mean value technique.
If not all servers are busy on arrival the waiting time is zero. If all servers are busy and
there are zero or more customers waiting, then a new arriving customers ﬁrst has to wait
until the ﬁrst departure and then continues to wait for as many departures as there were
customers waiting upon arrival. An interdeparture time is the minimum of c exponential

44

(residual) service times with mean 1/µ, and thus it is exponential with mean 1/cµ (see
exercise 1). So we obtain

1 1
E(W ) = ΠW + E(Lq ) .
cµ cµ

Together with Little’s law we retrieve the formulas (5.2)–(5.3). Table 5.1 lists the delay
probability and the mean waiting time in an M/M/c with mean service time 1 for ρ = 0.9.

c ΠW E(W )
1 0.90 9.00
2 0.85 4.26
5 0.76 1.53
10 0.67 0.67
20 0.55 0.28

Table 5.1: Performance characteristics for the M/M/c with µ = 1 and ρ = 0.9

We see that the delay probability slowly decreases as c increases. The mean waiting time
however decreases fast (a little faster than 1/c). One can also look somewhat differently at
the performance of the system. We do not look at the occupation rate of a machine, but at
the average number of idle machines. Let us call this the surplus capacity. Table 5.2 shows
for fixed surplus capacity (instead of for fixed occupation rate as in the previous table) and
c varying from 1 to 20 the mean waiting time and the mean number of customers in the
system.

c ρ E(W ) E(L)
1 0.90 9.00 9
2 0.95 9.26 19
5 0.98 9.50 51
10 0.99 9.64 105
20 0.995 9.74 214

Table 5.2: Performance characteristics for the M/M/c with µ = 1 and a fixed surplus
capacity of 0.1 server

Although the mean number of customers in the system sharply increases, the mean
waiting time remains nearly constant.

45

5.3 Distribution of the waiting time and the sojourn
time
The derivation of the distribution of the waiting time is very similar to the one in section
4.4 for the M/M/1. By conditioning on the state seen on arrival we obtain
∞ n+1
P (W t) = P( Dk t)pc+n ,
n=0 k=1
where Dk is the kth interdeparture time. Clearly, the random variables Dk are independent
and exponentially distributed with mean 1/cµ. Analogously to (4.13) we ﬁnd
∞ n
(cµt)k −cµt n
P (W t) = e pc ρ
n=0 k=0 k!
∞ ∞
(cµt)k −cµt n
= e pc ρ
k=0 n=k k!
pc ∞ (cµρt)k −cµt
= e
1 − ρ k=0 k!
= ΠW e−cµ(1−ρ)t , t ≥ 0.
This yields for the conditional waiting time,
P (W t)
P (W t|W 0) = = e−cµ(1−ρ)t , t ≥ 0.
P (W 0)
Hence, the conditional waiting time W |W 0 is exponentially distributed with parameter
cµ(1 − ρ). To determine the distribution of the sojourn time we condition on the length of
the service time, so
P (S t) = P (W + B t)
∞
= P (W + x t)µe−µx dx
x=0
t ∞
= P (W t − x)µe−µx dx + µe−µx dx
x=0 x=t
t
= ΠW e−cµ(1−ρ)(t−x) µe−µx dx + e−µt
x=0
ΠW
= e−cµ(1−ρ)t − e−µt + e−µt
1 − c(1 − ρ)
ΠW ΠW
= e−cµ(1−ρ)t + 1 − e−µt .
1 − c(1 − ρ) 1 − c(1 − ρ)

5.4 Java applet
There is also a JAVA applet is avalaible for the performance evaluation of the M/M/c
queue. The WWW-link to this applet is http://www.win.tue.nl/cow/Q2.

46

5.5 Exercises
Exercise 21. (a fast and a slow machine)
Consider two parallel machines with a common buﬀer where jobs arrive according to a
Poisson stream with rate λ. The processing times are exponentially distributed with mean
1/µ1 on machine 1 and 1/µ2 on machine 2 (µ1 µ2 ). Jobs are processed in order of arrival.
A job arriving when both machines are idle is assigned to the fast machine. We assume
that
λ
ρ=
µ1 + µ2
is less than one.
(i) Determine the distribution of the number of jobs in the system.

(ii) Use this distribution to derive the mean number of jobs in the system.

(iii) When is it better to not use the slower machine at all?

(iv) Calculate for the following two cases the mean number of jobs in the system with
and without the slow machine.

(a) λ = 2, µ1 = 5, µ2 = 1;
(b) λ = 3, µ1 = 5, µ2 = 1.

Note: In [21] it is shown that one should not remove the slow machine if r 0.5 where
r = µ2 /µ1 . When 0 ≤ r 0.5 the slow machine should be removed (and the resulting
system is stable) whenever ρ ≤ ρc , where

2 + r2 − (2 + r2 )2 + 4(1 + r2 )(2r − 1)(1 + r)
ρc = .
2(1 + r2 )

Exercise 22.
One is planning to build new telephone boxes near the railway station. The question is
how many boxes are needed. Measurements showed that approximately 80 persons per
hour want to make a phone call. The duration of a call is approximately exponentially
distributed with mean 1 minute. How many boxes are needed such that the mean waiting
time is less than 2 minutes?
Exercise 23.
An insurance company has a call center handling questions of customers. Nearly 40 calls per
hour have to be handled. The time needed to help a customer is exponentially distributed
with mean 3 minutes. How many operators are needed such that only 5% of the customers
has to wait longer than 2 minutes?
Exercise 24.
In a dairy barn there are two water troughs (i.e. drinking places). From each trough only

47

one cow can drink at the same time. When both troughs are occupied new arriving cows
wait patiently for their turn. It takes an exponential time to drink with mean 3 minutes.
Cows arrive at the water troughs according to a Poisson process with rate 20 cows per
hour.

(i) Determine the probability that there are i cows at the water troughs (waiting or
drinking), i = 0, 1, 2, . . .

(ii) Determine the mean number of cows waiting at the troughs and the mean waiting
time.

(iii) What is the fraction of cows ﬁnding both troughs occupied on arrival?

(iv) How many troughs are needed such that at most 10% of the cows ﬁnd all troughs
occupied on arrival?

Exercise 25.
A computer consists of three processors. Their main task is to execute jobs from users.
These jobs arrive according to a Poisson process with rate 15 jobs per minute. The execu-
tion time is exponentially distributed with mean 10 seconds. When a processor completes
a job and there are no other jobs waiting to be executed, the processor starts to execute
maintenance jobs. These jobs are always available and they take an exponential time with
mean 5 seconds. But as soon as a job from a user arrives, the processor interrupts the
execution of the maintenance job and starts to execute the new job. The execution of the
maintenance job will be resumed later (at the point where it was interrupted).

(i) What is the mean number of processors busy with executing jobs from users?

(ii) How many maintenance jobs are on average completed per minute?

(iii) What is the probability that a job from a user has to wait?

(iv) Determine the mean waiting time of a job from a user.

48

Chapter 6

M/Er /1 queue

Before analyzing the M/G/1 queue, we first study the M/Er /1 queue. The Erlang dis-
tribution can be used to model service times with a low coefficient of variation (less than
one), but it can also arise naturally. For instance, if a job has to pass, stage by stage,
through a series of r independent production stages, where each stage takes a exponen-
tially distributed time. The analysis of the M/Er /1 queue is similar to that of the M/M/1
queue.
We consider a single-server queue. Customers arrive according to a Poisson process with
rate λ and they are treated in order of arrival. The service times are Erlang-r distributed
with mean r/µ (see subsection 2.4.4). For stability we require that the occupation rate
r
ρ=λ· (6.1)
µ
is less than one. In the following section we will explain that there are two ways in which
one can describe the state of the system.

6.1 Two alternative state descriptions
The natural way to describe the state of a nonempty system is by the pair (k, l) where k
denotes the number of customers in the system and l the remaining number of service phases
of the customer in service. Clearly this is a two-dimensional description. An alternative
way to describe the state is by counting the total number of uncompleted phases of work
in the system. Clearly, there is a one to one correspondence between this number and
the pair (k, l). The number of uncompleted phases of work in the system is equal to the
number (k − 1)r + l (for the customer in service we have l phases of work instead of r).
From here on we will work with the one-dimensional phase description.

6.2 Equilibrium distribution
For the one-dimensional phase description we get the flow diagram of figure 6.1.

49

¦ ¦

¡ ¢ ¤¤£
£ £ § ¨ ©§ ¡ ¤¤£
£ £

¥ ¥ ¥

Figure 6.1: One-dimensional flow diagram for the M/Er /1 model

Let pn be the equilibrium probability of n phases work in the system. By equating the
flow out of state n and the flow into state n we obtain the following set of equilibrium
equations for pn .

p0 λ = p1 µ, (6.2)
pn (λ + µ) = pn+1 µ, n = 1, . . . , r − 1, (6.3)
pn (λ + µ) = pn−r λ + pn+1 µ, n = r, r + 1, r + 2, . . . (6.4)

These equations may be solved as follows. We first look for solutions of (6.4) of the form

pn = xn , n = 0, 1, 2, . . . (6.5)

and then we construct a linear combination of these solutions also satisfying the boundary
equations (6.2)–(6.3) and the normalization equation
∞
pn = 1.
n=0

An alternative solution approach is based on generating functions (see exercise 27).
Substitution of (6.5) into (6.4) and then dividing by the common power xn−r yields the
polynomial equation

(λ + µ)xr = λ + µxr+1 . (6.6)

One root is x = 1, but this one is not useful, since we must be able to normalize the
solution of the equilibrium equations. Provided condition (6.1) holds, it can be shown that
equation (6.6) has exactly r distinct roots x with |x| 1, say x1 , . . . , xr (see exercise 26).
We now consider the linear combination
r
pn = ck xn ,
k n = 0, 1, 2, . . . (6.7)
k=1

For each choice of the coefficients ck this linear combination satisfies (6.4). This freedom is
used to also satisfy the equations (6.2)–(6.3) and the normalization equation. Note that,
since the equilibrium equations are dependent, equation (6.2) (or one of the equations in
(6.3)) may be omitted. Substitution of (6.7) into these equations yields a set of r linear

50

equations for r unknown coefficients. It can be shown that this set of equations has a
unique solution, given by
1−ρ
ck = , k = 1, . . . , r,
j=k (1 − xj /xk )

where the subscript j runs from 1 to r. This completes the determination of the equilibrium
probabilities pn . The important conclusion is that for the M/Er /1 queue the equilibrium
probabilities can be expressed as a mixture of r geometric distributions.
From the distribution of the number of phases in the system we can easily find the
distribution of the number of customers in the system. Let qi be the probability of i
customers in the system. Obviously, q0 = p0 and for i ≥ 1 it follows that
ir
qi = pn
n=(i−1)r+1
ir r
= ck xn
k
n=(i−1)r+1 k=1
r ir
= ck xn
k
k=1 n=(i−1)r+1
r
= ck (x−r+1 +
k x−r+2 + · · · + 1)(xr )i .
k k
k=1

Hence, the queue length probabilities can also be expressed as a mixture of r geometric
distributions.
We finally remark that the roots of equation (6.6) can be numerically determined very
efficiently (cf. [2, 3]), and that the results in this section can be easily extended to the case
that the service times are mixtures of Erlang distributions with the same scale parameters
(see subsection 2.4.6).

Example 6.2.1
Consider the M/E2 /1 queue with λ = 1 and µ = 6. The equilibrium equations are given
by
p0 = 6p1 ,
7p1 = 6p2 ,
7pn = pn−2 + 6pn+1 , n = 2, 3, 4, . . .
Substitution of pn = xn into the equations for n ≥ 2 and dividing by xn−2 yields
7x2 = 1 + 6x3 ,
the roots of which are x = 1, x = 1/2 and x = −1/3. Hence, we set
n n
1 1
pn = c1 + c2 − , n = 0, 1, 2, . . .
2 3

51

and determine c1 and c2 from the equilibrium equation in n = 0 and the normalization
equation,
c1 + c2 = 3c1 − 2c2 ,
c1 c2
+ = 1.
1 − 1/2 1 + 1/3
The solution is c1 = 2/5 and c2 = 4/15 (note that the equilibrium equation in n = 1 is
also satisﬁed). So we obtain
n n
2 1 4 1
pn = + − , n = 0, 1, 2, . . .
5 2 15 3
For the queue length probabilities it follows that q0 = p0 = 2/3 and for i ≥ 1,
6 1 i 8 1 i
qi = p2i−1 + p2i = − .
5 4 15 9
Note that the formula above is also valid for i = 0.

6.3 Mean waiting time
Let the random variable Lf denote the number of phases work in the system. Then by
PASTA it follows that
1
E(W ) = E(Lf ) ,
µ
and from (6.7),
∞
E(Lf ) = npn
n=0
∞ r
= ck nxn
k
n=0 k=1
r ∞
= ck nxn
k
k=1 n=0
r
ck xk
= 2
k=1 (1 − xk )

From Little’s law we can also ﬁnd E(Lq ), the mean number of customers waiting in the
queue. A much more direct way to determine E(W ) and E(Lq ) is the mean value approach.
An arriving customer has to wait for the customers in the queue and, if the server
is busy, for the one in service. According to the PASTA property the mean number of
customers waiting in the queue is equal to E(Lq ) and the probability that the server is
busy on arrival is equal to ρ, i.e. the fraction of time the server is busy. Hence,
r
E(W ) = E(Lq ) + ρE(R), (6.8)
µ

52

where the random variable R denotes the residual service time of the customer in service.
If the server is busy on arrival, then with probability 1/r he is busy with the ﬁrst phase
of the service time, also with probability 1/r he is busy with the second phase, and so on.
So the mean residual service time E(R) is equal to
1 r 1 r−1 1 1
E(R) = · + · + ··· + ·
r µ r µ r µ
r+1 1
= · . (6.9)
2 µ
Substitution of this expression into (6.8) yields
r r+1 1
E(W ) = E(Lq ) +ρ· · .
µ 2 µ
Together with Little’s law, stating that

E(Lq ) = λE(W )

we ﬁnd
ρ r+1 1
E(W ) = · · .
1−ρ 2 µ

6.4 Distribution of the waiting time
The waiting time can be expressed as
Lf
W = Bi ,
i=1

where Bi is the amount of work for the kth phase. So the random variables Bi are inde-
pendent and exponentially distributed with mean 1/µ. By conditioning on Lf and using
that Lf and Bi are independent it follows, very similar to (4.13), that
∞ n
P (W t) = P( Bi t)pn
n=1 i=1
∞ n−1
(µt)i −µt r
= e ck xn
k
n=1 i=0 i! k=1
r ∞ ∞
(µt)i −µt n
= ck e xk
k=1 i=0 n=i+1 i!
r ∞
ck xk (µxk t)i −µt
= e
k=1 1 − xk i=0 i!
r
ck
= xk e−µ(1−xk )t , t ≥ 0.
k=1 1 − xk

53

Note that this distribution is a generalization of the one for the M/M/1 model (namely,
not one, but a mixture of exponentials).
In table 6.1 we list for varying values of ρ and r the mean waiting time and some waiting
time probabilities. The squared coeﬃcient of variation of the service time is denoted by c2 .
B
We see that the variation in the service times is important to the behavior of the system.
Less variation in the service times leads to smaller waiting times.

ρ r c2
B E(W ) P (W t)
t 5 10 20
0.8 1 1 4 0.29 0.11 0.02
2 0.5 3 0.21 0.05 0.00
4 0.25 2.5 0.16 0.03 0.00
10 0.1 2.2 0.12 0.02 0.00
0.9 1 1 9 0.55 0.33 0.12
2 0.5 6.75 0.46 0.24 0.06
4 0.25 5.625 0.41 0.18 0.04
10 0.1 4.95 0.36 0.14 0.02

Table 6.1: Performance characteristics for the M/Er /1 with mean service time equal to 1

6.5 Java applet
The link to the Java applet for the performance evaluation of the M/Er /1 queue is
http://www.win.tue.nl/cow/Q2. This applet can be used to evaluate the performance
as a function of, e.g., the occupation rate and the shape parameter r.

54

6.6 Exercises
Exercise 26. (Rouch´’s theorem)
e
Rouch´’s theorem reads as follows.
e
Let the bounded region D have as its boundary a contour C. Let the functions f (z)
and g(z) be analytic both in D and on C, and assume that |f (z)| |g(z)| on C. Then
f (z) + g(z) has in D the same number of zeros as g(z), all zeros counted according to their
multiplicity.

(i) Use Rouch´’s theorem to prove that the polynomial equation (6.6) has exactly r
e
(possibly complex) roots x with |x| 1.
(Hint: take f (z) = µz r+1 and g(z) = −(λ + µ)z r + λ, and take as contour C the
circle with center 0 and radius 1 − with small and positive.)

(ii) Show that all roots are simple.
(Hint: Show that there are no z for which both f (z) + g(z) and f (z) + g (z) vanish
at the same time.)

Exercise 27. (Generating function approach)
Consider the M/Er /1 queue with arrival rate λ and mean service time r/µ. Let P (z) be
the generating function of the probabilities pn , so
∞
P (z) = pn z n , |z| ≤ 1.
n=0

(i) Show, by multiplying the equilibrium equations (6.2)–(6.4) with z n and adding over
all states n, that

P (z)(λ + µ) − p0 µ = P (z)λz r + (P (z) − p0 )µz −1 .

(ii) Show that P (z) is given by

(1 − ρ)µ
P (z) = . (6.10)
µ− λ(z r + z r−1 + · · · + z)

(iii) Show, by partial fraction decomposition of P (z), that the probabilities pn can be
written as
r n
1
pn = ck , n = 0, 1, 2, . . .
k=1 zk

where z1 , . . . , zr are the zeros of the numerator in (6.10).

55

Exercise 28.
Orders arrive according to a Poisson process, so the interarrival times are independent and
exponentially distributed. Each order has to pass, stage by stage, through a series of r
independent production stages. Each stage takes a exponentially distributed time. The
total production time, the workload, of the order is the sum of r independent, identically
distributed random variables. So the distribution of the workload is the r-stage Erlang
distribution.
(i) The state of this system (i.e., the work station together with its queue) can be
characterized by the number of orders in the system and by the number of not yet
completed stages for the order in the work station.
Describe the flow diagram and give the set of equations for the equilibrium state
probabilities.

(ii) The state of the system at a certain time can also be described by the total number
of production stages that have to be completed before all the orders in the system
are ready.
Give the flow diagram and formulate the equations for the equilibrium state proba-
bilities.

(iii) Define

pn = P (total number of orders in the system = n),
qk = P (total number of production stages in the system = k).

Give the relation between these two probabilities.

Exercise 29.
Orders arrive in so called bulks. Each bulk consists of r independent orders. The bulks
themselves arrive according to a Poisson process. The sequence in which orders of one bulk
are processed in the work station is unimportant. All orders of a bulk have to wait until
all orders of bulks that have arrived earlier are completed. The workload of each order is
exponentially distributed. The state of the system is simply characterized by the number
of orders in the system.
Describe the flow diagram for this situation. Compare the result with that of the previous
exercise, part (ii).
Exercise 30.
Jobs arrive at a machine according to a Poisson process with a rate of 16 jobs per hour.
Each job consists of 2 tasks. Each task has an exponentially distributed processing time
with a mean of 1 minute. Jobs are processed in order of arrival.

(i) Determine the distribution of the number of uncompleted tasks in the system.

(ii) Determine the distribution of the number of jobs in the system.

56

(iii) What is the mean number of jobs in the system?

(iv) What is the mean waiting time of a job?

Exercise 31.
Consider an M/E2 /1/2 queue with arrival rate 1 and mean service time 2. At time t = 0
the system is empty. Determine the mean time till the ﬁrst customer is rejected on arrival.
Exercise 32.
Consider the M/E2 /1 queue with an arrival rate of 4 customers per hour and a mean
service time of 8 minutes.

(i) Determine the distribution of the waiting time.

(i) What is the fraction of customers that has to wait longer than 5 minutes?

Exercise 33.
Customers arrive in groups at a single-server queue. These groups arrive according to a
Poisson process with a rate of 3 groups per hour. With probability 1/3 a group consists of
2 customers, and with probability 2/3 it consists of 1 customer only. All customers have an
exponential service time with a mean of 6 minutes. Groups are served in order of arrival.
Within a group, customers are served in random order.

(i) Determine the distribution of the number of customers in the system.

(ii) Determine the mean sojourn time of an arbitrary customer.

Exercise 34.
In a repair shop of an airline company defective airplane engines are repaired. Defects occur
according to a Poisson with a rate of 1 defective engine per 2 weeks. The mean repair time
of an engine is 2/3 week. The repair time distribution can be well approximated by an
Erlang-2 distribution. In the repair shop only one engine can be in repair at the same time.

(i) Show that qn , the probability that there are n engines in the repair shop, is given by
n n
6 1 8 1
qn = − .
5 4 15 9

(ii) Determine the mean sojourn time (waiting time plus repair time) of an engine in the
repair shop.

The airline company has several spare engines in a depot. When a defect occurs, then the
defective engine is immediately replaced by a spare engine (if there is one available) and the
defective engine is send to the repair shop. After repair the engine is as good as new, and
it is transported to the depot. When a defect occurs and no spare engine is available, the
airplane has to stay on the ground and it has to wait till a spare engine becomes available.

57

(iii) Determine the minimal number of spare engines needed such that for 99% of the
defects there is a spare engine available.

Exercise 35.
Jobs arrive according to a Poisson process at a machine. The arrival rate is 25 jobs per
hour. With probability 2/5 a job consists of 1 task, and with probability 3/5 it consists of
2 tasks. Tasks have an exponentially distributed processing time with a mean of 1 minute.
Jobs are processed in order of arrival.

(i) Determine the distribution of the number of uncompleted tasks at the machine.

(ii) Determine the mean waiting time of a job.

(iii) Determine the mean number of jobs in the system.

Exercise 36.
Customers arrive in groups at a server. The group size is 2 and groups arrive according to
a Poisson stream with a rate of 2 groups per hour. Customers are served one by one, and
they require an exponential service time with a mean of 5 minutes.

(i) Determine the distribution of the number of customers in the system.

(ii) What is the mean waiting time of the ﬁrst customer in a group?

(iii) And what is the mean waiting time of the second one?

58

Chapter 7

M/G/1 queue

In the M/G/1 queue customers arrive according to a Poisson process with rate λ and
they are treated in order of arrival. The service times are independent and identically
distributed with distribution function FB (·) and density fB (·). For stability we have to
require that the occupation rate
ρ = λE(B) (7.1)
is less than one. In this chapter we will derive the limiting or equilibrium distribution
of the number of customers in the system and the distributions of the sojourn time, the
waiting time and the busy period duration. It is further shown how the means of these
quatities can be obtained by using the mean value approach.

7.1 Which limiting distribution?
The state of the M/G/1 queue can be described by the pair (n, x) where n denotes the
number of customers in the system and x the service time already received by the customer
in service. We thus need a two-dimensional state description. The first dimension is still
discrete, but the other one is continuous and this essentially complicates the analysis.
However, if we look at the system just after departures, then the state description can be
simplified to n only, because x = 0 for the new customer (if any) in service. Denote by Ldk
the number of customers left behind by the kth departing customer. In the next section
we will determine the limiting distribution
dn = lim P (Ld = n).
k
k→∞

The probability dn can be interpreted as the fraction of customers that leaves behind n
customers. But in fact we are more interested in the limiting distribution pn defined as
pn = lim P (L(t) = n),
t→∞

where L(t) is the number of customers in the system at time t. The probability pn can
be interpreted as the fraction of time there are n customers in the system. From this

59

distribution we can compute, e.g., the mean number of customers in the system. Another
perhaps even more important distribution is the limiting distribution of the number of
customers in the system seen by an arriving customer, i.e.,

an = lim P (La = n),
k
k→∞

where La is the number of customers in the system just before the kth arriving customer.
k
From this distribution we can compute, e.g., the distribution of the sojourn time. What is
the relation between these three distributions? It appears that they all are the same.
Of course, from the PASTA property we already know that an = pn for all n. We will
now explain why also an = dn for all n. Taking the state of the system as the number of
customers therein, the changes in state are of a nearest-neighbour type: if the system is in
state n, then an arrival leads to a transition from n to n + 1 and a departure from n to
n − 1. Hence, in equilibrium, the number of transitions per unit time from state n to n + 1
will be the same as the number of transitions per unit time from n + 1 to n. The former
transitions correspond to arrivals finding n customers in the system, the frequency of which
is equal to the total number of arrivals per unit time, λ, multiplied with the fraction of
customers finding n customers in the system, an . The latter transitions correspond to
departures leaving behind n customers. The frequency of these transitions is equal to the
total number of departures per unit time, λ, multiplied with the fraction of customers
leaving behind n customers, dn . Equating both frequencies yields an = dn . Note that
this equality is valid for any system where customers arrive and leave one by one (see also
exercise 48). Thus it also holds for, e.g., the G/G/c queue.
Summarizing, for the M/G/1 queue, arrivals, departures and outside observers all see
the same distribution of number of customers in the system, i.e., for all n,

an = dn = pn .

7.2 Departure distribution
In this section we will determine the distribution of the number of customers left behind
by a departing customer when the system is in equilibrium.
Denote by Ld the number of customers left behind by the kth departing customer. We
k
first derive an equation relating the random variable Ld to Ld . The number of customers
k+1 k
left behind by the k + 1th customer is clearly equal to the number of customers present
when the kth customer departed minus one (since the k + 1th customer departs himself)
plus the number of customers that arrives during his service time. This last number is
denoted by the random variable Ak+1 . Thus we have

Ld = Ld − 1 + Ak+1 ,
k+1 k

which is valid if Ld 0. In the special case Ld = 0, it is readily seen that
k k

Ld = Ak+1 .
k+1

60

From the two equations above it is immediately clear that the sequence {Ld }∞ forms a
k k=0
Markov chain. This Markov chain is usually called the imbedded Markov chain, since we
look at imbedded points on the time axis, i.e., at departure instants.
We now specify the transition probabilities
pi,j = P (Ld = j|Ld = i).
k+1 k

Clearly pi,j = 0 for all j i − 1 and pi,j for j ≥ i − 1 gives the probability that exactly
j − i + 1 customers arrived during the service time of the k + 1th customer. This holds
for i 0. In state 0 the kth customer leaves behind an empty system and then p0,j gives
the probability that during the service time of the k + 1th customer exactly j customers
arrived. Hence the matrix P of transition probabilities takes the form
 
α0 α1 α2 α3 ···

 α0 α1 α2 α3 ··· 

···
 
 0 α0 α1 α2 
P =
 0 0 α0 α1 ··· ,

 
0 0 0 α0 ···
 
 
. . . . ..
 
.
. .
. .
. .
. .
where αn denotes the probability that during a service time exactly n customers arrive.
To calculate αn we note that given the duration of the service time, t say, the number of
customers that arrive during this service time is Poisson distributed with parameter λt.
Hence, we have
(λt)n −λt
∞
αn = e fB (t)dt. (7.2)
t=0 n!

The transition probability diagram is shown in ﬁgure 7.1.
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Figure 7.1: Transition probability diagram for the M/G/1 imbedded Markov chain

This completes the speciﬁcation of the imbedded Markov chain. We now wish to
determine its limiting distribution. Denote the limiting distribution of Ld by {dn }∞ and
k n=0
the limiting random variable by Ld . So
dn = P (Ld = n) = lim P (Ld = n).
k
k→∞

61

The limiting probabilities dn , which we know are equal to pn , satisfy the equilibrium
equations
dn = dn+1 α0 + dn α1 + · · · + d1 αn + d0 αn
n
= dn+1−k αk + d0 αn , n = 0, 1, . . . (7.3)
k=0

To solve the equilibrium equations we will use the generating function approach. Let us
introduce the probability generating functions
∞ ∞
PLd (z) = dn z n , PA (z) = αn z n ,
n=0 n=0

which are deﬁned for all z ≤ 1. Multiplying (7.3) by z n and summing over all n leads to
∞ n
PLd (z) = dn+1−k αk + d0 αn z n
n=0 k=0
∞ n ∞
−1 n+1−k k
= z dn+1−k z αk z + d0 αn z n
n=0 k=0 n=0
∞ ∞
= z −1 dn+1−k z n+1−k αk z k + d0 PA (z)
k=0 n=k
∞ ∞
= z −1 αk z k dn+1−k z n+1−k + d0 PA (z)
k=0 n=k
−1
= z PA (z)(PLd (z) − d0 ) + d0 PA (z).
Hence we ﬁnd
d0 PA (z)(1 − z −1 )
PLd (z) = .
1 − z −1 PA (z)
To determine the probability d0 we note that d0 is equal to p0 , which is the fraction of time
the system is empty. Hence d0 = p0 = 1 − ρ ( alternatively, d0 follows from the requirement
PLd (1) = 1). So, by multiplying numerator and denominator by −z we obtain
(1 − ρ)PA (z)(1 − z)
PLd (z) = . (7.4)
PA (z) − z
By using (7.2), the generating function PA (z) can be rewritten as
∞
(λt)n −λt
∞
PA (z) = e fB (t)dtz n
n=0 t=0 n!
∞ ∞ (λtz)n
= e−λt fB (t)dt
t=0 n=0 n!
∞
= e−(λ−λz)t fB (t)dt
t=0
= B(λ − λz) (7.5)

62

Substitution of (7.5) into (7.4) finally yields
(1 − ρ)B(λ − λz)(1 − z)
PLd (z) = . (7.6)
B(λ − λz) − z
This formula is one form of the Pollaczek-Khinchin formula. In the following sections
we will derive similar formulas for the sojourn time and waiting time. By differentiating
formula (7.6) we can determine the moments of the queue length (see section 2.2). To find
its distribution, however, we have to invert formula (7.6), which usually is very difficult. In
the special case that B(s) is a quotient of polynomials in s, i.e., a rational function, then in
principle the right-hand side of (7.6) can be decomposed into partial fractions, the inverse
transform of which can be easily determined. The service time has a rational transform
for, e.g., mixtures of Erlang distributions or Hyperexponential distributions (see section
2.4). The inversion of (7.6) is demonstrated below for exponential and Erlang-2 service
times.

Example 7.2.1 (M/M/1)
Suppose the service time is exponentially distributed with mean 1/µ. Then
µ
B(s) = .
µ+s
Thus
µ
(1 − ρ) µ+λ−λz (1 − z) (1 − ρ)µ(1 − z) (1 − ρ)µ(1 − z) 1−ρ
PLd (z) = µ = = = .
µ+λ−λz
−z µ − z(µ + λ − λz) (µ − λz)(1 − z) 1 − ρz
Hence
dn = pn = (1 − ρ)ρn , n = 0, 1, 2, . . .
Example 7.2.2 (M/E2 /1)
Suppose the service time is Erlang-2 distributed with mean 2/µ. Then (see subsection
2.4.4)
2
µ
B(s) = ,
µ+s
so
µ 2
(1 − ρ) µ+λ−λz
(1 − z)
PLd (z) = 2
µ
µ+λ−λz
−z
2
(1 − ρ)µ (1 − z)
=
µ2
− z(µ + λ − λz)2
(1 − ρ)(1 − z)
=
1 − z(1 + ρ(1 − z)/2)2
1−ρ
= .
1 − ρz − ρ2 z(1 − z)/4

63

For ρ = 1/3 we then ﬁnd
2/3 24
PLd (z) = =
1 − z/3 − z(1 − z)/36 36 − 13z + z 2
24 24/5 24/5 6/5 8/15
= = − = − .
(4 − z)(9 − z) 4−z 9−z 1 − z/4 1 − z/9
Hence,
n n
6 1 8 1
dn = pn = − , n = 0, 1, 2, . . .
5 4 15 9
Example 7.2.3 (M/H2 /1)
Suppose that λ = 1 and that the service time is hyperexponentially distributed with
parameters p1 = 1 − p2 = 1/4 and µ1 = 1, µ2 = 2. So the mean service time is equal to
1/4 · 1 + 3/4 · 1/2 = 5/8. The Laplace-Stieltjes transform of the service time is given by
(see subsection 2.4.5)
1 1 3 2 1 8 + 7s
B(s) = · + · = · .
4 1+s 4 2+s 4 (1 + s)(2 + s)
Thus we have
31 15−7z
8 4 (2−z)(3−z)
(1 − z)
PLd (z) = 1 15−7z
4 (2−z)(3−z)
−z
3 (15 − 7z)(1 − z)
= ·
8 (15 − 7z) − 4z(2 − z)(3 − z)
3 15 − 7z
= ·
8 (3 − 2z)(5 − 2z)
3 9/4 3 5/4
= · + ·
8 3 − 2z 8 5 − 2z/5
9/32 3/32
= + .
1 − 2z/3 1 − 2z/5
So
n n
9 2 3 2
dn = pn = + , n = 0, 1, 2, . . .
32 3 32 5

7.3 Distribution of the sojourn time
We now turn to the calculation of how long a customer spends in the system. We will show
that there is a nice relationship between the transforms of the time spent in the system
and the departure distribution.
Let us consider a customer arriving at the system in equilibrium. Denote the total
time spent in the system for this customer by the random variable S with distribution

64

function FS (·) and density fS (·). The distribution of the number of customers left behind
upon departure of our customer is equal to {dn }∞ (since the system is in equilibrium).
n=0
In considering a first-come first-served system it is clear that all customers left behind are
precisely those who arrived during his stay in the system. Thus we have (cf. (7.2))
(λt)n −λt
∞
dn = e fS (t)dt.
t=0 n!

Hence, we find similarly to (7.5) that
PLd (z) = S(λ − λz).
Substitution of this relation into (7.6) yields
(1 − ρ)B(λ − λz)(1 − z)
S(λ − λz) = .
B(λ − λz) − z
Making the change of variable s = λ − λz we finally arrive at
(1 − ρ)B(s)s
S(s) = . (7.7)
λB(s) + s − λ
This formula is also known as the Pollaczek-Khinchin formula.

For exponential service times with mean 1/µ we have
µ
B(s) = .
µ+s
Thus
µ
(1 − ρ) µ+s s (1 − ρ)µs (1 − ρ)µs µ(1 − ρ)
S(s) = µ = = 2
= .
λ µ+s + s − λ λµ + (s − λ)(µ + s) (µ − λ)s + s µ(1 − ρ) + s
Hence, S is exponentially distributed with parameter µ(1 − ρ), i.e.,
FS (t) = P (S ≤ t) = 1 − e−µ(1−ρ)t , t ≥ 0.
Example 7.3.2 (M/E2 /1)
Suppose that λ = 1 and that the service time is Erlang-2 distributed with mean 1/3, so
2
6
B(s) = .
6+s
Then it follows that (verify)
8 3
FS (t) = (1 − e−3t ) − (1 − e−8t ), t ≥ 0.
5 5
Example 7.3.3 (M/H2 /1)
Consider example 7.2.3 again. From (7.7) we obtain (verify)
27 5
FS (t) = (1 − e−t/2 ) + (1 − e−3t/2 ), t ≥ 0.
32 32

65

7.4 Distribution of the waiting time
We have that S, the time spent in the system by a customer, is the sum of W (his waiting
time) and B (his service time), where W and B are independent. Since the transform of
the sum of two independent random variables is the product of the transforms of these two
random variables (see section 2.3), it holds that
S(s) = W (s) · B(s). (7.8)
Together with (7.7) it follows that
(1 − ρ)s
W (s) = , (7.9)
λB(s) + s − λ
which is the third form of the Pollaczek-Khinchin formula.

For exponential service times with mean 1/µ we have
µ
B(s) = .
µ+s
Then from (7.9) it follows that (verify)
µ(1 − ρ)
W (s) = (1 − ρ) + ρ · .
µ(1 − ρ) + s
The inverse transform yields
FW (t) = P (W ≤ t) = (1 − ρ) + ρ(1 − e−µ(1−ρ)t ), t ≥ 0.
Hence, with probability (1−ρ) the waiting time is zero (i.e., the system is empty on arrival)
and, given that the waiting time is positive (i.e., the system is not empty on arrival), the
waiting time is exponentially distributed with parameter µ(1 − ρ) (see also section 4.4).

7.5 Lindley’s equation
In the literature there are several alternative approaches to derive the Pollaczek-Khinchin
formulas. In this section we present one that is based on a fundamental relationship
between the waiting time of the nth customer and the n + 1th customer.
Denote by An the interarrival time between the nth and n+1th customer arriving at the
system. Further let Wn and Bn denote the waiting time and the service time, respectively,
for the nth customer. Clearly, if An ≤ Wn + Bn , then the waiting time of the n + 1th
customer is equal to Wn + Bn − An . If on the other hand An Wn + Bn , then his waiting
time is equal to zero. Summarizing, we have
Wn+1 = max(Wn + Bn − An , 0). (7.10)

66

This equation is commonly referred to as Lindley’s equation. We denote the limit of Wn
as n goes to infinity by W (which exists if ρ is less than one). Then, by letting n go to
infinity in (7.10) we obtain the limiting form

W = max(W + B − A, 0) (7.11)

(where we also dropped the subscripts of An and Bn , since we are considering the limiting
situation). With S = W + B, this equation may also be written as

W = max(S − A, 0) (7.12)

Note that the interarrival time A is exponentially distributed with parameter λ and that
the random variables S and A are independent. We now calculate the transform of W
directly from equation (7.12). This gives

W (s) = E(e−sW )
= E(e−s max(S−A,0) )
∞ ∞
= e−s max(x−y,0) fS (x)λe−λy dxdy
x=0 y=0
∞ x ∞ ∞
= e−s(x−y) fS (x)λe−λy dxdy + fS (x)λe−λy dxdy
x=0 y=0 x=0 y=x
λ ∞ ∞
= (e−λx − e−sx )fS (x)dx + e−λx fS (x)dx
s − λ x=0 x=0
λ
= (S(λ) − S(s)) + S(λ)
s−λ
s λ
= S(λ) − S(s).
s−λ s−λ

Substitution of (7.8) then yields

S(λ)s
W (s) = . (7.13)
λB(s) + s − λ

It remains to find S(λ). Realizing that
∞ ∞
S(λ) = e−λx fS (x)dx = P (A x)fS (x)dx = P (A S),
x=0 x=0

it follows that S(λ) is precisely the probability that the system is empty on arrival, so

S(λ) = 1 − ρ.

Substitution of this relation into (7.13) finally yields formula (7.9).

67

7.6 Mean value approach
The mean waiting time can of course be calculated from the Laplace-Stieltjes transform
(7.9) by differentiating and substituting s = 0 (see section 2.3). In this section we show
that the mean waiting time can also be determined directly (i.e., without transforms) with
the mean value approach.
A new arriving customer first has to wait for the residual service time of the customer
in service (if there is one) and then continues to wait for the servicing of all customers who
were already waiting in the queue on arrival. By PASTA we know that with probability ρ
the server is busy on arrival. Let the random variable R denote the residual service time
and let Lq denote the number of customers waiting in the queue. Hence,

E(W ) = E(Lq )E(B) + ρE(R),

and by Little’s law (applied to the queue),

E(Lq ) = λE(W ).

So we find
ρE(R)
E(W ) = . (7.14)
1−ρ
Formula (7.14) is commonly referred to as the Pollaczek-Khinchin mean value formula. It
remains to calculate the mean residual service time. In the following section we will show
that
E(B 2 )
E(R) = , (7.15)
2E(B)
which may also be written in the form
E(B 2 ) σ 2 + E(B)2 1
E(R) = = B = (c2 + 1)E(B). (7.16)
2E(B) 2E(B) 2 B
An important observation is that, clearly, the mean waiting time only depends upon the
first two moments of service time (and not upon its distribution). So in practice it is
sufficient to know the mean and standard deviation of the service time in order to estimate
the mean waiting time.

7.7 Residual service time
Suppose that our customer arrives when the server is busy and denote the total service
time of the customer in service by X. Further let fX (·) denote the density of X. The basic
observation to find fX (·) is that it is more likely that our customer arrives in a long service
time than in a short one. So the probability that X is of length x should be proportional

68

to the length x as well as the frequency of such service times, which is fB (x)dx. Thus we
may write
P (x ≤ X ≤ x + dx) = fX (x)dx = CxfB (x)dx,
where C is a constant to normalize this density. So
∞
C −1 = xfB (x)dx = E(B).
x=0
Hence
xfB (x)
fX (x) = .
E(B)
Given that our customer arrives in a service time of length x, the arrival instant will be
a random point within this service time, i.e., it will be uniformly distributed within the
service time interval (0, x). So
dt
P (t ≤ R ≤ t + dt|X = x) = , t ≤ x.
x
Of course, this conditional probability is zero when t x. Thus we have
∞ dt ∞ fB (x) 1 − FB (t)
P (t ≤ R ≤ t + dt) = fR (t)dt = fX (x)dx = dxdt = dt.
x=t x x=t E(B) E(B)
This gives the ﬁnal result
1 − FB (t)
fR (t) = ,
E(B)
from which we immediately obtain, by partial integration,
∞ 1 ∞ 1 ∞ 1 2 E(B 2 )
E(R) = tfR (t)dt = t(1 − FB (t))dt = t fB (t)dt = .
t=0 E(B) t=0 E(B) t=0 2 2E(B)
This computation can be repeated to obtain all moments of R, yielding
E(B n+1 )
E(Rn ) = .
(n + 1)E(B)

Example 7.7.1 (Erlang service times)
For an Erlang-r service time with mean r/µ we have
r r
E(B) = , σ 2 (B) = 2 ,
µ µ
so
r(1 + r)
E(B 2 ) = σ 2 (B) + (E(B))2 = .
µ2
Hence (see also (6.9))
1+r
E(R) =
2µ

69

7.8 Variance of the waiting time
The mean value approach fails to give more information than the mean of the waiting time
(or other performance characteristics). But to obtain information on the variance or higher
moments of the waiting time we can use formula (7.9) for the Laplace-Stieltjes transform
of the waiting time. This is demonstrated below.
We first rewrite formula (7.9) as
1−ρ
W (s) = ,
1 − ρ 1−B(s)
sE(B)

The term between brackets can be recognised as the transform of R, since by partial
integration we have
∞ 1 ∞
R(s) = E(e−sR ) = e−st fR (t)dt = e−st (1 − FB (t))dt
t=0 E(B) t=0
1 1 ∞ 1 −st 1 − B(s)
= − e fB (t)dt = .
E(B) s t=0 s sE(B)
Hence
1−ρ
W (s) = . (7.17)
1 − ρR(s)
By differentiating (7.17) and substituting s = 0 we retrieve formula (7.14), i.e.,
ρE(R)
E(W ) = .
1−ρ
By differentiating (7.17) twice and substituting s = 0 we find
2 ρE(R2 )
2
E(W ) = 2(E(W )) + .
1−ρ
Hence, for the variance of the waiting time we now obtain
ρ
σ 2 (W ) = E(W 2 ) − (E(W ))2 = ρ(E(R))2 + (1 − ρ)E(R2 ) .
(1 − ρ)2
The first two moments of the conditional waiting time W |W 0 can be calculated
from
E(W ) E(W 2 )
E(W |W 0) = , E(W 2 |W 0) = .
ρ ρ
It then follows, after some algebra, that the squared coefficient of variation of W |W 0
is given by the simple equation
c2 |W 0 = ρ + (1 − ρ)c2 = 1 + (1 − ρ)(c2 − 1).
W R R

This implies that for ρ near to 1 the squared coefficient of variation of the conditional
waiting time is close to 1. Hence, as a rule of thumb, we may think of the conditional
waiting time as an exponential random variable (which has a coefficient of variation equal
to 1) and thus obtain a rough estimate for its distribution.

70

7.9 Distribution of the busy period
The mean duration of a busy period in the M/G/1 queue can be determined in exactly
the same way as for the M/M/1 queue in subsection 4.6.1. Thus we have
E(B)
E(BP ) = . (7.18)
1−ρ
The calculation of the distribution of a busy period in the M/G/1 queue is more compli-
cated.
We first derive an important relation for the duration of a busy period. What happens
in a busy period? We start with the service of the first customer. His service time is
denoted by B1 . During his service new customers may arrive to the system. Denote this
(random) number by N and label these customers by C1 , . . . , CN . At the departure of
the first customer we take customer C1 into service. However, instead of letting the other
customers C2 , . . . , CN wait for their turn, we choose to take them temporarily out of the
system. Customer C2 will be put into the system again and taken into service as soon as
the system is empty again. So customers arriving during the service of C1 will be served
first (as well as the ones arriving during their service, and so on). Thus it is as if C1 initiates
a new busy period for which C2 has to wait. This busy period will be called a sub-busy
period. In the same way C3 has to wait for the sub-busy period initiated by C2 , and so
on. Finally the sub-busy period due to CN completes the major busy period. Note that
our customers C2 , . . . , CN are not treated first-come first-served anymore. But this does
not affect the duration of the (major) busy period, because its duration is independent of
the order in which customers are served. Denote the busy period by BP and the sub-busy
period due to Ci by BPi . Then we have the important relation
BP = B1 + BP1 + · · · + BPN (7.19)
where the random variables BP, BP1 , BP2 , . . . are independent and all have the same dis-
tribution, and further, they are independent of N . The number N of course depends on
the service time B1 .
We will now use relation (7.19) to derive the Laplace-Stieltjes transform of the busy
period BP . By conditioning on the length of B1 we get
∞
BP (s) = E(e−sBP ) = E(e−sBP |B1 = t)fB (t)dt
t=0

and by also conditioning on N ,
∞
E(e−sBP |B1 = t) = E(e−sBP |B1 = t, N = n)P (N = n|B1 = t)
n=0
∞
(λt)n −λt
= E(e−s(B1 +BP1 +···+BPN ) |B1 = t, N = n) e
n=0 n!
∞
(λt)n −λt
= E(e−s(t+BP1 +···+BPn ) ) e
n=0 n!

71

∞
(λt)n −λt
= E(e−st · e−sBP1 · · · e−sBPn ) e
n=0 n!
∞
(λt)n −λt
= e−st (E(e−sBP ))n e
n=0 n!
= e−(s+λ−λBP (s))t

Thus we have
∞
BP (s) = e−(s+λ−λBP (s))t fB (t)dt,
t=0

which can be recognised as

BP (s) = B(s + λ − λBP (s)). (7.20)

In the example below it is shown that even for the simple M/M/1 queue it is not easy to
determine the distribution of BP from this equation.

For the M/M/1 we have
µ
B(s) = .
µ+s
In this case equation (7.20) reduces to
µ
BP (s) = .
µ + s + λ − λBP (s)

So BP (s) is a root of the quadratic equation

λ(BP (s))2 − (λ + µ + s)BP (s) + µ = 0.

Solving this equation and restricting our solution to the case for which 0 ≤ BP (s) ≤ 1 for
all s ≥ 0 yields
1
BP (s) = λ+µ+s− (λ + µ + s)2 − 4λµ .
2λ
This equation may be inverted to obtain the density of the busy period. This gives an
expression involving a Bessel function (see subsection 4.6.2).

It is, however, straightforward to determine the moments of BP from equation (7.20)
by diﬀerentiating and substituting s = 0 (see section 2.3). This will be demonstrated
below.
Diﬀerentiating (7.20) and substituting s = 0 yields
(1) (1)
−E(BP ) = BP (0) = B (1) (0) · (1 − λBP (0)) = −E(B)(1 + λE(BP )).

72

Hence, with ρ = λE(B), we retrieve formula (7.18), i.e.,
E(B)
E(BP ) = . (7.21)
1−ρ
By differentiating (7.20) twice and substituting s = 0 we get
(2)
E(BP 2 ) = BP (0)
(1) (2)
= B (2) (0) · (1 − λBP (0))2 + B (1) (0) · (−λBP (0))
= E(B 2 )(1 + λE(BP ))2 + λE(B)E(BP 2 ).

From this we obtain
E(B 2 )
E(BP 2 ) = . (7.22)
(1 − ρ)3
From (7.21) and (7.22) it follows that the squared coefficient of variation, c2 , is given by
BP

c2
B ρ
c2 =
BP + . (7.23)
1−ρ 1−ρ
Clearly, as ρ tends to one, the variation in BP explodes.

Since c2 = 1 for exponential service times it follows from (7.23) that
B

1+ρ
c2 =
BP .
1−ρ
In table 7.1 we list c2 for increasing values of ρ.
BP

ρ 0.5 0.6 0.7 0.8 0.9 0.95
c2
BP 3 4 5.7 9 19 39

Table 7.1: The squared coefficient of the busy period for the M/M/1 queue

7.10 Java applet
There is a JAVA applet avalaible for the evaluation of mean performance characteristics
of the M/G/1 queue. The link to this applet is http://www.win.tue.nl/cow/Q2.

73

7.11 Exercises
Exercise 37. (Post office)
At a post office customers arrive according to a Poisson process with a rate of 30 customers
per hour. A quarter of the customers wants to cash a cheque. Their service time is
exponentially distributed with a mean of 2 minutes. The other customers want to buy
stamps and their service times are exponentially distributed with a mean of 1 minute. In
the post office there is only one server.

(i) Determine the generating function PL (z) of the number of customers in the system.

(ii) Determine the distribution of the number of customers in the system.

(iii) Determine the mean number of customers in the system.

(iv) Determine S(s), the Laplace-Stieltjes transform of the sojourn time.

(v) Determine the mean and the distribution of the sojourn time.

(vi) Determine the mean busy period duration.

(vii) Determine the mean number of customers in the system and the mean sojourn time
in case all customers have an exponentially distributed service time with a mean of
75 seconds.

Exercise 38.
Consider a single machine where jobs arrive according to a Poisson stream with a rate of
10 jobs per hour. The processing time of a job consists of two phases. Each phase takes
an exponential time with a mean of 1 minute.

(i) Determine the Lapace-Stieltjes transform of the processing time.

(ii) Determine the distribution of the number of jobs in the system.

(iii) Determine the mean number of jobs in the system and the mean production lead
time (waiting time plus processing time).

Exercise 39. (Post office)
At a post office customers arrive according to a Poisson process with a rate of 60 customers
per hour. Half of the customers have a service time that is the sum of a fixed time of 15
seconds and an exponentially distributed time with a mean of 15 seconds. The other half
have an exponentially distributed service time with a mean of 1 minute.
Determine the mean waiting time and the mean number of customers waiting in the queue.
Exercise 40. (Two phase production)
A machine produces products in two phases. The first phase is standard and the same for
all products. The second phase is customer specific (the finishing touch). The first (resp.

74

second) phase takes an exponential time with a mean of 10 (resp. 2) minutes. Orders for
the production of one product arrive according to a Poisson stream with a rate of 3 orders
per hour. Orders are processed in order of arrival.
Determine the mean production lead time of an order.
Exercise 41.
Consider a machine where jobs are being processed. The mean production time is 4 minutes
and the standard deviation is 3 minutes. The mean number of jobs arriving per hour is
10. Suppose that the interarrival times are exponentially distributed.
Determine the mean waiting time of the jobs.
Exercise 42.
Consider an M/G/1 queue, where the server successfully completes a service time with
probability p. If a service time is not completed successfully, it has to be repeated until it
is successful. Determine the mean sojourn time of a customer in the following two cases:
(i) The repeated service times are identical.
(ii) The repeated service times are independent, and thus (possibly) different.

Exercise 43.
At the end of a production process an operator manually performs a quality check. The
time to check a product takes on average 2 minutes with a standard deviation of 1 minute.
Products arrive according to a Poisson stream. One considers to buy a machine that is able
to automatically check the products. The machine needs exactly 84 seconds to perform a
quality check. Since this machine is expensive, one decides to buy the machine only if it
is able to reduce lead time for a quality check (i.e. the time that elapses from the arrival
of a product till the completion of its quality check) to one third of the lead time in the
present situation.
So only if the arrival rate of products exceeds a certain threshold, one will decide to buy
the machine. Calculate the value of this threshold.
Exercise 44. (Robotic dairy barn)
In a robotic dairy barn cows are automatically milked by a robot. The cows are lured into
the robot by a feeder with nice food that can only be reached by first passing through
the robot. When a cow is in the robot, the robot first detects whether the cow has to
be milked. If so, then the cow will be milked, and otherwise, the cow can immediately
leave the robot and walk to the feeder. A visit to the robot with (resp. without) milking
takes an exponential time with a mean of 6 (resp. 3) minutes. Cows arrive at the robot
according to a Poisson process with a rate of 10 cows per hour, a quarter of which will be
milked.
(i) Show that the Laplace-Stieltjes transform of the service time in minutes of an arbi-
trary cow at the robot is given by
1 4 + 21s
B(s) = · .
4 (1 + 6s)(1 + 3s)

75

(ii) Show that the Laplace-Stieltjes transform of the waiting time in minutes of an arbi-
trary cow in front of the robot is given by

3 9 1 1 1
W (s) = + · + · .
8 16 1 + 12s 16 1 + 4s

(iii) Determine the fraction of cows for which the waiting time in front of the robot is less
than 3 minutes.

(iv) Determine the mean waiting time in front of the robot.

Exercise 45.
Consider a machine processing parts. These parts arrive according to a Poisson stream
with a rate of 1 product per hour. The machine processes one part at a time. Each part
receives two operations. The ﬁrst operation takes an exponential time with a mean of 15
minutes. The second operation is done immediately after the ﬁrst one and it takes an
exponential time with a mean of 20 minutes.

(i) Show that the Laplace-Stieltjes transform of the production lead time (waiting time
plus processing time) in hours is given by

5 1 1 5
S(s) = · − · .
4 1+s 4 5+s

(ii) Determine the distribution of the production lead time and the mean production lead
time.

One uses 3 hours as a norm for the production lead time. When the production lead time
of a part exceeds this norm, it costs 100 dollar.

(iii) Calculate the mean cost per hour.

Exercise 46. (Warehouse)
In a warehouse for small items orders arrive according to a Poisson stream with a rate of 6
orders per hour. An order is a list with the quantities of products requested by a customer.
The orders are picked one at a time by one order picker. For a quarter of the orders the
pick time is exponentially distributed with a mean of 10 minutes and for the other orders
the pick time is exponentially distributed with a mean of 5 minutes.

(i) Show that the Laplace-Stieltjes transform of the pick time in minutes of an arbitrary
order is given by

1 4 + 35s
B(s) = · .
4 (1 + 10s)(1 + 5s)

76

(ii) Show that the Laplace-Stieltjes transform of the lead time (waiting time plus pick
time) in minutes of an arbitrary order is given by

5 3 27 1
S(s) = · + · .
32 3 + 20s 32 1 + 20s
(iii) Determine the fraction of orders for which the lead time is longer than half an hour.
(iv) Determine the mean lead time.

Exercise 47. (Machine with breakdowns)
Consider a machine processing parts. Per hour arrive according to a Poisson process on
average 5 orders for the production of a part. The processing time of a part is exactly
10 minutes. During processing, however, tools can break down. When this occurs, the
machine immediately stops, broken tools are replaced by new ones and then the machine
resumes production again. Replacing tools takes exactly 2 minutes. The time that elapses
between two breakdowns is exponentially distributed with a mean of 20 minutes. Hence,
the total production time of a part, B, consists of the processing time of 10 minutes plus
a random number of interruptions of 2 minutes to replace broken tools.
(i) Determine the mean and variance of the production time B.
(ii) Determine the mean lead time (waiting time plus production time) of an order.

Exercise 48. (Arrival and departure distribution)
Consider a queueing system in which customers arrive one by one and leave one by one.
So the number of customers in the system only changes by +1 or −1. We wish to prove
that an = dn . Suppose that the system is empty at time t = 0.
(i) Show that if Ld ≤ n, then La
k+1 k+n+1 ≤ n.

(ii) Show that if La d
k+n+1 ≤ n, then Lk ≤ n.

(iii) Show that (i) and (ii) give, for any n ≥ 0,

lim P (Ld ≤ n) = lim P (La ≤ n).
k k
k→∞ k→∞

Exercise 49. (Number served in a busy period)
Let the random variable Nbp denote the number of customers served in a busy period.
Deﬁne the probabilities fn by
fn = P (Nbp = n).
We now wish to ﬁnd its generating function
∞
FNbp (z) = fn z n .
n=1

77

(i) Show that FNbp (z) satisﬁes

FNbp (z) = z B(λ − λFNbp (z)). (7.24)

(ii) Show that the mean number of customers served in a busy period is given by

1
E(Nbp ) = .
1−ρ

(iii) Solve equation (7.24) for the M/M/1 queue.

(iv) Solve equation (7.24) for the M/D/1 queue (Hint: Use Lagrange inversion formula,
see, e.g., [30]).

78

Chapter 8

G/M/1 queue

In this chapter we study the G/M/1 queue, which forms the dual of the M/G/1 queue. In
this system customers arrive one by one with interarrival times identically and indepen-
dently distributed according to an arbitrary distribution function FA (·) with density fA (·).
The mean interarrival time is equal to 1/λ. The service times are exponentially distributed
with mean 1/µ. For stability we again require that the occupation rate ρ = λ/µ is less
than one.
The state of the G/M/1 queue can be described by the pair (n, x) where n denotes the
number of customers in the system and x the elapsed time since the last arrival. So we
need a complicated two-dimensional state description. However, like for the M/G/1 queue,
the state description is much easier at special points in time. If we look at the system on
arrival instants, then the state description can be simplified to n only, because x = 0 at an
arrival. Denote by La the number of customers in the system just before the kth arriving
k
customer. In the next section we will determine the limiting distribution

an = lim P (La = n).
k
k→∞

From this distribution we will be able to calculate the distribution of the sojourn time.

8.1 Arrival distribution
In this section we will determine the distribution of the number of customers found in the
system just before an arriving customer when the system is in equilibrium.
We first derive a relation between the random variables La and La . Defining the
k+1 k
random variable Dk+1 as the number of customers served between the arrival of the kth
and k + 1th customer, it follows that

La = La + 1 − Dk+1 .
k+1 k

From this equation it is immediately clear that the sequence {La }∞ forms a Markov chain.
k k=0
This Markov chain is called the G/M/1 imbedded Markov chain.

79

We must now calculate the associated transition probabilities

pi,j = P (La = j|La = i).
k+1 k

Clearly pi,j = 0 for all j i + 1 and pi,j for j ≤ i + 1 is equal to the probability that exactly
i + 1 − j customers are served during the interarrival time of the k + 1th customer. Hence
the matrix P of transition probabilities takes the form
 
p0,0 β0 0 · · ·
p1,0 β1 β0 0 · · · 
 

 
P =
 p2,0 β2 β1 β0 0  , 

 p3,0 β3 β2 β1 β0  

.
. ..

. .

where βi denotes the probability of serving i customers during an interarrival time given
that the server remains busy during this interval (thus there are more than i customers
present). To calculate βi we note that given the duration of the interarrival time, t say,
the number of customers served during this interval is Poisson distributed with parameter
µt. Hence, we have

∞ (µt)i −µt
βi = e fA (t)dt. (8.1)
t=0 i!
Since the transition probabilities from state j should add up to one, it follows that
i ∞
pi,0 = 1 − βj = βj .
j=0 j=i+1

The transition probability diagram is shown in ﬁgure 8.1.
¥
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£ ©©¨
¨¨ ¢
¡ ¤¢
£ ¡ £ ©©¨
¨¨
§¥
! §¥
¦ $
# ¥

Figure 8.1: Transition probability diagram for the G/M/1 imbedded Markov chain

This completes the speciﬁcation of the imbedded Markov chain. We now wish to deter-
mine its limiting distribution {an }∞ . The limiting probabilities an satisfy the equilibrium
n=0

80

equations

a0 = a0 p0,0 + a1 p1,0 + a2 p2,0 + · · ·
∞
= ai pi,0 (8.2)
i=0
an = an−1 β0 + an β1 + an+1 β2 + · · ·
∞
= an−1+i βi , n = 1, 2, . . . (8.3)
i=0

To find the solution of the equilibrium equations it appears that the generating function
approach does not work here (verify). Instead we adopt the direct approach by trying to
find solutions of the form

an = σ n , n = 0, 1, 2, . . . (8.4)

Substitution of this form into equation (8.3) and dividing by the common power σ n−1 yields
∞
σ= σ i βi .
i=0

Of course we know that βi is given by (8.1). Hence we have
∞ ∞ (µt)i −µt
σ = σi e fA (t)dt
i=0 t=0 i!
∞
= e−(µ−µσ)t fA (t)dt.
t=0

The last integral can be recognised as the Laplace-Stieltjes transform of the interarrival
time. Thus we arrive at the following equation

σ = A(µ − µσ). (8.5)

We immediately see that σ = 1 is a root of equation (8.5), since A(0) = 1. But this root is
not useful, because we must be able to normalize the solution of the equilibrium equations.
It can be shown that (see exercise 50) as long as ρ 1 equation (8.5) has a unique root
σ in the range 0 σ 1, and this is the root which we seek. Note that the remaining
equilibrium equation (8.2) is also satisfied by (8.4) since the equilibrium equations are
dependent. We finally have to normalize solution (8.4) yielding

an = (1 − σ)σ n , n = 0, 1, 2, . . . (8.6)

Thus we can conclude that the queue length distribution found just before an arriving
customer is geometric with parameter σ, where σ is the unique root of equation (8.5) in
the interval (0, 1).

81

For exponentially distributed interarrival times we have
λ
A(s) = .
λ+s
Hence equation (8.5) reduces to
λ
σ= ,
λ + µ − µσ
so
σ(λ + µ − µσ) − λ = (σ − 1)(λ − µσ) = 0.
Thus the desired root is σ = ρ and the arrival distribution is given by
an = (1 − ρ)ρn , n = 0, 1, 2, . . .
Note that this distribution is exactly the same as the equilibrium distribution of the
M/M/1; see (4.6). This is of course no surprise, because here we have Poisson arrivals.

Example 8.1.2 (E2 /M/1)
Suppose that the interarrival times are Erlang-2 distributed with mean 2/3, so
2
3
A(s) = .
3+s
Further assume that µ = 4 (so ρ = 3/2 · 1/4 = 3/8 1). Then equation (8.5) reduces to
2
3
σ= .
7 − 4σ
Thus
σ(7 − 4σ)2 − 9 = (σ − 1)(4σ − 9)(4σ − 1) = 0.
Hence the desired root is σ = 1/4 and
n
3 1
an = , n = 0, 1, 2, . . .
4 4
Example 8.1.3
Suppose that the interarrival time consist of two exponential phases, the ﬁrst phase with
parameter µ and the second one with parameter 2µ (so it is slightly more complicated than
Erlang-2 where both phases have the same parameter), where µ is also the parameter of
the exponential service time. The Laplace-Stieltjes transform of the interarrival time is
given by
2µ2
A(s) = .
(µ + s)(2µ + s)

82

For this transform equation (8.5) reduces to
2µ2 2
σ= = .
(2µ − µσ)(3µ − µσ) (2 − σ)(3 − σ)
This leads directly to
√ √
σ 3 − 5σ 2 + 6σ − 2 = (σ − 1)(σ − 2 − 2)(σ − 2 + 2) = 0.
√
Clearly only the root σ = 2 − 2 is acceptable. Therefore we have
√ √
an = ( 2 − 1)(2 − 2)n , n = 0, 1, 2, . . .

8.2 Distribution of the sojourn time
Since the arrival distribution is geometric, it is easy to determine the distribution of the
sojourn time. In fact, the analysis is similar to the one for for the M/M/1 queue (see
section 4.4). With probability an an arriving customer ﬁnds n customers in the system.
Then his sojourn time is the sum of n + 1 exponentially distributed service times, each
with mean 1/µ. Hence,
S(s) = E(e−sS )
∞ n+1
µ
= an
n=0 µ+s
∞ n+1
n µ
= (1 − σ)σ
n=0 µ+s
n
µ(1 − σ) ∞ µσ
=
µ + s n=0 µ + s
µ(1 − σ)
= .
µ(1 − σ) + s
From this we can conclude that the sojourn time S is exponentially distributed with pa-
rameter µ(1 − σ), i.e.,
P (S ≤ t) = 1 − e−µ(1−σ)t , t ≥ 0.
Clearly the sojourn time distribution for the G/M/1 is of the same form as for the M/M/1,
the only diﬀerence being that ρ is replaced by σ.
Along the same lines it can be shown that the distributon of the waiting time W is
given by (cf. (4.14))
P (W ≤ t) = 1 − σe−µ(1−σ)t , t ≥ 0.
Note that the probability that a customer does not have to wait is given by 1 − σ (and not
by 1 − ρ).

83

8.3 Mean sojourn time
It is tempting to determine the mean sojourn time directly by the mean value approach.
For an arriving customer we have
1 1
E(S) = E(La ) + , (8.7)
µ µ
where the random variable La denotes the number of customers in the system found on
arrival. According to Little’s law it holds that

E(L) = λE(S). (8.8)

Unfortunately, we do not have Poisson arrivals, so

E(La ) = E(L).

Hence the mean value approach does not work here, since we end up with only two equations
for three unknowns. Additional information is needed in the form of (8.6), yielding
∞ ∞
σ
E(La ) = nan = n(1 − σ)σ n = .
n=0 n=0 1−σ

Then it follows from (8.7) and (8.8) that
σ 1 1 λ ρ
E(S) = + = , E(L) = = .
(1 − σ)µ µ (1 − σ)µ (1 − σ)µ (1 − σ)

8.4 Java applet
There is a JAVA applet avalaible for the evaluation of the G/M/1 queue for several distribu-
tions of the interarrival times. In fact, the applet evaluates the G/Er /1 queue. This queue
has been analyzed in, e.g., [2]. The link to the applet is http://www.win.tue.nl/cow/Q2.

84

8.5 Exercises
Exercise 50.
We want to show that as long as ρ 1 equation (8.5) has a unique root σ in the range
0 σ 1. Set

f (σ) = A(µ − µσ).

(i) Prove that f (σ) is stricly convex on the interval [0, 1], i.e., its derivative is increasing
on this interval.

(ii) Show that f (0) 0 and that f (1) 1 provided ρ 1.

(iii) Show that (i) and (ii) imply that the equation σ = f (σ) has exactly one root in the
interval (0, 1).

Exercise 51.
In a record shop customers arrive according to a hyperexponential arrival process. The
interarrival time is with probability 1/3 exponentially distributed with a mean of 1 minute
and with probability 2/3 it is exponentially distributed with a mean of 3 minutes. The
service times are exponentially distributed with a mean of 1 minute.

(i) Calculate the distribution of the number of customers found in the record shop by
an arriving customer.

(ii) Calculate the mean number of customers in the record shop found on arrival.

(iii) Determine S(s).

(iv) Determine the mean time a customer spends in the record shop.

(v) Calculate the mean number of customers in the record shop (now at an arbitrary
point in time).

Exercise 52.
The distribution of the interarrival time is given by
13 11
FA (t) = 1 − e−3t + 1 − e−2t , t ≥ 0.
24 24
The service times are exponentially distributed with a mean of 1/6.

(i) Determine the distribution of the number of customers in the system just before an
arrival.

(ii) Determine the distribution of the waiting time.

85

Exercise 53.
Determine the distribution of the sojourn time in case of exponentially distributed service
times with mean 1 and hyperexponentially distributed interarrival times with distribution
function
1 1
FA (t) = 1 − e−t/2 + 1 − e−t/4 , t ≥ 0.
2 2

Exercise 54.
Consider a queueing system where the interarrival times are exactly 4 minutes. The service
times are exponentially distributed with a mean of 2 minutes.

(i) Compute σ.

(ii) Determine the distribution of the sojourn time.

Exercise 55.
At a small river cars are brought from the left side to the right side of the river by a ferry.
On average 15 cars per hour arrive according to a Poisson process. It takes the ferry an
exponentially distributed time with a mean of 3 minutes to cross the river and return. The
capacity of the ferry is equal to 2 cars. The ferry only takes oﬀ when there are two or more
cars waiting.

(i) What is the fraction of time that the ferry is on its way between the two river sides?

(ii) Determine the distribution of the number of cars that are waiting for the ferry.

(iii) Determine the mean waiting time of a car.

86

Chapter 9

Priorities

In this chapter we analyse queueing models with diﬀerent types of customers, where one
or more types of customers have priority over other types. More precisely we consider
an M/G/1 queue with r types of customers. The type i customers arrive according to a
Poisson stream with rate λi , i = 1, . . . , r. The service time and residual service of a type
i customer is denoted by Bi and Ri , respectively. The type 1 customers have the highest
priority, type 2 customers the second highest priority and so on. We consider two kinds of
priorities. For the non-preemptive priority rule higher priority customers may not interrupt
the service time of a lower priority customer, but they have to wait till the service time
of the low priority customer has been completed. For the preemptive-resume priority rule
interruptions are allowed and after the interruption the service time of the lower priority
customer resumes at the point where it was interrupted. In the following two sections we
show how the mean waiting times can be found for these two kinds of priorities.

9.1 Non-preemptive priority
The mean waiting time of a type i customer is denoted by E(Wi ) and E(Lq ) is the number
i
of type i customers waiting in the queue. Further deﬁne ρi = λi E(Bi ). For the highest
priority customers it holds that
r
E(W1 ) = E(Lq )E(B1 ) +
1 ρj E(Rj ).
j=1

According to Little’s law we have

E(Lq ) = λ1 E(W1 )
1

Combining the two equations yields
r
j=1 ρj E(Rj )
E(W1 ) = . (9.1)
1 − ρ1

87

The determination of the mean waiting time for the lower priority customers is more
complicated. Consider type i customers with i 1. The waiting time of a type i customer
can be divided in a number of portions. The first portion is the amount of work associated
with the customer in service and all customers with the same or higher priority present
in the queue upon his arrival. Call this portion X1 . The second portion, say X2 , is the
amount of higher priority work arriving during X1 . Subsequently the third portion X3 is
the amount of higher priority work arriving during X2 , and so on. A realization of the
waiting time for a type 2 customer is shown in figure 9.1. The increments of the amount
of work are the service times of the arriving type 1 customers.
amount of work

X1 X2
X3

t

arrivals of type 1 customers

Figure 9.1: Realization of the waiting time of a type 2 customer

Hence the mean waiting time is given by
∞
E(Wi ) = E(X1 + X2 + X3 + · · ·) = E(Xk ).
k=1

As mentioned above, the first portion of work an arriving type i customer has to wait for
is the sum of the service times of all customers with the same or higher priority present in
the queue plus the remaining service time of the customer in service. So
i r
E(X1 ) = E(Lq )E(Bj ) +
j ρj E(Rj ).
j=1 j=1

To determine E(Xk+1 ) note that Xk+1 depends on Xk . We therefore condition on the
length of Xk . Denote the density of Xk by fk (x). Then it follows that
∞
E(Xk+1 ) = E(Xk+1 |Xk = x)fk (x)dx
x=0
∞
= (λ1 xE(B1 ) + · · · + λi−1 xE(Bi−1 ))fk (x)dx
x=0
= (ρ1 + · · · + ρi−1 )E(Xk ).

88

Repeated application of the relation above yields

E(Xk+1 ) = (ρ1 + · · · + ρi−1 )k E(X1 ), k = 0, 1, 2, . . .

Hence we find for i = 2, . . . , r

E(X1 ) i
j=1 E(Lq )E(Bj ) + r ρj E(Rj )
j j=1
E(Wi ) = = , (9.2)
1 − (ρ1 + · · · + ρi−1 ) 1 − (ρ1 + · · · + ρi−1 )

An intuitive argument (which can be made rigorous) to directly obtain the above equation
is by observing that the waiting time of a type i customer is equal to the first portion of
work plus all the higher priority work arriving during his waiting time. So
i−1
E(Wi ) = E(X1 ) + λj E(Wi )E(Bj ),
j=1

from which equation (9.2) immediately follows. Substitution of Little’s law

E(Lq ) = λi E(Wi )
i

into equation (9.2) yields
i−1 r
(1 − (ρ1 + . . . + ρi ))E(Wi ) = E(Lq )E(Bj )
j + ρj E(Rj )
j=1 j=1
= (1 − (ρ1 + . . . + ρi−2 ))E(Wi−1 ).

By multiplying both sides of this equality with 1 − (ρ1 + . . . + ρi−1 ) we get the simple
recursive relation
i i−1 i−1 i−2
(1 − ρj )(1 − ρj )E(Wi ) = (1 − ρj )(1 − ρj )E(Wi−1 ).
j=1 j=1 j=1 j=1

Repeatedly applying this relation and using (9.1) finally leads to
r
ρj E(Rj )
j=1
E(Wi ) = , i = 1, · · · , r. (9.3)
(1 − (ρ1 + . . . + ρi ))(1 − (ρ1 + . . . + ρi−1 ))

The mean sojourn time E(Si ) of a type i customer follows from E(Si ) = E(Wi ) + E(Bi ),
yielding
r
ρj E(Rj )
j=1
E(Si ) = + E(Bi ) , (9.4)
(1 − (ρ1 + . . . + ρi ))(1 − (ρ1 + . . . + ρi−1 ))

for i = 1, · · · , r.

89

9.2 Preemptive-resume priority
We will show that the results in case the service times may be interrupted easily follow
from the ones in the previous section.
Consider a type i customer. For a type i customer there do not exist lower priority
customers due to the preemption rule. So we henceforth assume that λi+1 = · · · = λr = 0.
The waiting time of a type i customer can again be divided into portions X1 , X2 , . . ..
Now X1 is equal to the total amount of work in the system upon arrival, since we assumed
that there are no lower priority customers. Observe that the total amount of work in
the system does not depend on the order in which the customers are served. Hence, at
each point in time, it is exactly the same as in the system where the customers are served
according to the non-preemptive priority rule. So X1 , X2 , . . ., and thus also Wi have the
same distribution as in the system with non-preemptive priorities and, of course, with
λi+1 = · · · = λr = 0. From (9.3) we then obtain
i
j=1 ρj E(Rj )
E(Wi ) = , i = 1, · · · , r.
(1 − (ρ1 + . . . + ρi ))(1 − (ρ1 + . . . + ρi−1 ))
For the mean sojourn time we have to add the service time plus all the interruptions of
higher priority customers during the service time. The mean of such a generalized service
time can be found along the same lines as (9.2), yielding
E(Bi )
.
1 − (ρ1 + · · · + ρi−1 )
So the mean sojourn time of a type i customer is given by
i
ρj E(Rj )
j=1 E(Bi )
E(Si ) = + , (9.5)
(1 − (ρ1 + . . . + ρi ))(1 − (ρ1 + . . . + ρi−1 )) 1 − (ρ1 + · · · + ρi−1 )
for i = 1, · · · , r.

9.3 Shortest processing time first
In production systems one often processes jobs according to the shortest processing time
first rule (SPTF). The mean production lead time in a single machine system operating
according to the SPTF rule can be found using the results in section 9.1.
Consider an M/G/1 queue with arrival rate λ and service times B with density fB (x).
Assume that ρ = λE(B) 1. The server works according to the SPTF rule. That is, after
a service completion, the next customer to be served is the one with the shortest service
time.
Define type x customers as the ones with a service time between x and x + dx. The
mean waiting time of a type x customer is denoted by E(W (x)) and ρ(x)dx is the fraction
of time the server helps type x customers, so
ρ(x)dx = (λfB (x)dx)x = λxfB (x)dx. (9.6)

90

From (9.3) and by observing that the numerator in (9.3) corresponds to the mean amount
of work at the server, which in the present situation is simply given by ρE(R), we obtain

ρE(R)
E(W (x)) = y=x
(1 − y=0 ρ(y)dy)2
ρE(R)
= y=x .
(1 − λ y=0 yfB (y)dy)2

Hence the mean overall waiting time is given by
∞
E(W ) = E(W (x))fB (x)dx
x=0
∞ fB (x)dx
= ρE(R) y=x . (9.7)
x=0 (1 − λ y=0 yfB (y)dy)2

In table 9.1 we compare the SPTF rule with the usual first come first served (FCFS) rule
for an M/M/1 system with mean service time 1. The mean waiting time for the SPTF
rule is given by
∞ e−x dx
E(W ) = ρ
x=0 (1 − ρ(1 − e−x − xe−x ))2
and for the FCFS rule it satisfies
ρ
E(W ) = .
1−ρ
The results in table 9.1 show that considerable reductions in the mean waiting time are
possible.

E(W )
ρ FCFS SPTF
0.5 1 0.713
0.8 4 1.883
0.9 9 3.198
0.95 19 5.265

Table 9.1: The mean waiting time for FCFS and SPTF in an M/M/1 with E(B) = 1

9.4 A conservation law
In this section we consider a single-server queue with r types of customers. The type i
customers arrive according to a general arrival stream with rate λi , i = 1, . . . , r. The mean

91

service time and mean residual service time of a type i customer is denoted by E(Bi ) and
E(Ri ), respectively. Define ρi = λi E(Bi ). We assume that
r
λi E(Bi ) 1,
i=1

so that the server can handle the amount of work offered per unit of time. Customers enter
service in an order independent of their service times and they may not be interrupted
during their service. So, for example, the customers may be served according to FCFS,
random or a non-preemptive priority rule. Below we derive a conservation law for the
mean waiting times of the r types of customers, which expresses that a weighted sum of
these mean waiting times is independent of the service discipline. This implies that an
improvement in the mean waiting of one customer type owing to a service discipline will
always degrade the mean waiting time of another customer type.
Let E(V (P )) and E(Lq (P )) denote the mean amount of work in the system and the
i
mean number of type i customers waiting in the queue, respectively, for discipline P . The
mean amount of work in the system is given by
r r
E(V (P )) = E(Lq (P ))E(Bi ) +
i ρi E(Ri ). (9.8)
i=1 i=1

The first sum at the right-hand side is the mean amount of work in the queue, and the
second one is the mean amount of work at the server. Clearly the latter does not depend
on the discipline P .
The crucial observation is that the amount of work in the system does not depend
on the order in which the customers are served. The amount of work decreases with one
unit per unit of time independent of the customer being served and when a new customer
arrives the amount of work is increased by the service time of the new customer. Hence,
the amount of work does not depend on P . Thus from equation (9.8) and Little’s law

E(Lq ) = λi E(Wi (P )),
i

we obtain the following conservation law for the mean waiting times,
r
ρi E(Wi (P )) = constant with respect to service discipline P .
i=1

Below we present two examples where this law is used.

Example 9.4.1 (M/G/1 with FCFS and SPTF)
The SPTF rule selects customers in a way that is dependent of their service times. Nev-
ertheless, the law above also applies to this rule. The reason is that the SPTF rule can be
translated into a non-preemptive rule as explained in section 9.3. Below we check whether
for the M/G/1 the weighted sum of the mean waiting times for SPTF is indeed the same
as for FCFS.

92

In case the customers are served in order of arrival it holds that (see (7.14))
ρ2 E(R)
ρE(W ) = .
1−ρ
When the server works according to the SPTF rule we have (see (9.6) and (9.7))
∞ ∞ ρE(R)λxfB (x)dx
E(W (x))ρ(x)dx = y=x
x=0 x=0 (1 − λ y=0 yfB (y)dy)2
∞
ρE(R)
= y=x
1 − λ y=0 yfB (y)dy x=0
ρ2 E(R)
= ,
1−ρ
which indeed is the same as for the FCFS rule.

Example 9.4.2 (M/G/1 with non-preemptive priority)
Consider an M/G/1 queue with two types of customers. The type 1 customers have
non-preemptive priority over the type 2 customers. The mean waiting time for the type 1
customers is given by (9.1). We now derive the mean waiting time for the type 2 customers
by using the conservation law. According to this law it holds that

ρ1 E(W1 ) + ρ2 E(W2 ) = C, (9.9)

where C is some constant independent of the service discipline. To determine C consider
the FCFS discipline. For FCFS it follows from (7.14) that
ρ1 E(R1 ) + ρ2 E(R2 )
E(W1 ) = E(W2 ) = .
1 − ρ1 − ρ2
Hence,
ρ1 E(R1 ) + ρ2 E(R2 )
C = (ρ1 + ρ2 ) . (9.10)
1 − ρ1 − ρ2
By substituting (9.1) and (9.10) into equation (9.9) we retrieve formula (9.3) for the mean
waiting time of the type 2 customers under the non-preemptive priority rule.

93

9.5 Exercises
Exercise 56.
Customers arrive to a single-server queue according to a Poisson process with a rate of 10
customers per hour. Half of the customers has an exponential service time with a mean of
3 minutes. The other half has an exponential service time with a mean of 6 minutes. The
ﬁrst half are called type 1 customers, the second half type 2 customers. Determine in each
of the following three cases the mean waiting time:
(i) Service in order of arrival (FCFS).
(ii) Non-preemptive priority for type 1 customers.
(ii) Non-preemptive priority for type 2 customers.

Exercise 57.
Consider an M/D/1 queue with three types of customers. The customers arrive according
to a Poisson process. Per hour arrive on average 1 type 1 customer, 2 type 2 customers
and also 2 type 3 customers. Type 1 customers have preemptive resume priority over
type 2 and 3 customers and type 2 customers have preemptive resume priority over type 3
customers. The service time of each customer is 10 minutes.
Calculate the mean sojourn time for each of the three types of customers.
Exercise 58.
Consider a machine where jobs arrive according to a Poisson stream with a rate of 4 jobs
per hour. Half of the jobs have a processing time of exactly 10 minutes, a quarter of
the jobs have a processing time of exactly 15 minutes and the remaining quarter have a
processing time of 20 minutes. The jobs with a processing time of 10 minutes are called
type 1 jobs, the ones with a processing time of 15 minutes type 2 jobs and the rest type 3
jobs. The jobs are processed in order of arrival.
(i) Determine the mean sojourn time (waiting time plus processing time) of a type 1, 2
and 3 job and also of an arbitrary job.
One decides to process smaller jobs with priority. So type 1 orders have highest priority,
type 2 orders second highest priority and type 3 orders lowest priority.
Answer question (i) for the following two cases:
(ii) Jobs processed at the machine may not be interrupted.
(iii) Type 1 and type 2 jobs may interrupt the processing of a type 3 job. Type 1 jobs
may not interrupt the processing of a type 2 job.

Exercise 59.
A machine produces a speciﬁc part type. Orders for the production of these parts arrive
according to a Poisson stream with a rate of 10 orders per hour. The number of parts
that has to be produced for an order is equal to n with probability 0.5n , n = 1, 2, . . . The
production of one part takes exactly 2 minutes. Orders are processed in order of arrival.

94

(i) Denote by B the production time in minutes of an arbitrary order. Show that

E(B) = 4, σ 2 (B) = 8.

(ii) Determine the mean production lead time (waiting time plus production time) of an
arbitrary order.
The management decides to give orders for the production of 1 part priority over the other
orders. But the production of an order may not be interrupted.
(iii) Determine the mean production lead of an order for the production of 1 part, and of
an order for the production of at least 2 parts.

(iv) Determine the mean production lead time of an arbitrary order.

Exercise 60.
Consider a machine for mounting electronic components on printed circuit boards. Per
hour arrive according to a Poisson process on average 30 printed circuit boards. The
time required to mount all components on a printed circuit board is uniformly distributed
between 1 and 2 minutes.
(i) Calculate the mean sojourn time of an arbitrary printed circuit board.
One decides to give printed circuit boards with a mounting time between 1 and x (1 ≤
x ≤ 2) minutes non-preemptive priority over printed circuit boards the mounting time of
which is greater than x minutes.
(ii) Determine the mean sojourn time of an arbitrary printed circuit board as a function
of x.

(iii) Determine the value of x for which the mean sojourn time of an arbitrary printed
circuit board is minimized, and calculate for this speciﬁc x the relative improvement
with respect to the mean sojourn time calculated in (i).

Exercise 61.
A machine produces 2 types of products, type A and type B. Production orders (for one
product) arrive according to a Poisson process. For the production of a type A product
arrive on average 105 orders per day (8 hours), and for a type B product 135 orders per
day. The processing times are exponentially distributed. The mean processing time for a
type A order is 1 minute and for a type B order it is 2 minutes. Orders are processed in
order of arrival.
(i) Calculate the mean production lead time for a type A product and for a type B
product.

(ii) Determine the Laplace-Stieltjes transform of the waiting time.

95

(iii) Determine the Laplace-Stieltjes transform of the production lead time of a type A
product and determine the distribution of the production lead time of a type A
product.

For the production lead time of a type A product one uses a norm of 10 minutes. Each
time the production lead time of a type A product exceeds this norm, a cost of 100 dollar
is charged.

(iv) Calculate the average cost per day.

To reduce the cost one decides to give type A products preemptive resume priority over
type B products.

(v) Determine the mean production lead time for a type A product and for a type B
product.

(vi) Calculate the average cost per day.

Exercise 62.
A machine mounts electronic components on three diﬀerent types of printed circuit boards,
type A, B and C boards say. Per hour arrive on average 60 type A boards, 18 type B
boards and 48 type C boards. The arrival streams are Poisson. The mounting times are
exactly 20 seconds for type A, 40 seconds for type B and 30 seconds for type C. The
boards are processed in order of arrival.

(i) Calculate for each type of printed circuit board the mean waiting time and also
calculate the mean overall waiting time.

Now suppose that the printed circuit boards are processed according to the SPTF rule.

(ii) Calculate for each type of printed circuit board the mean waiting time and also
calculate the mean overall waiting time.

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Chapter 10

Variations of the M/G/1 model

In this chapter we treat some variations of the M/G/1 model and we demonstrate that the
mean value technique is a powerful technique to evaluate mean performance characteristics
in these models.

10.1 Machine with setup times
Consider a single machine where jobs are being processed in order of arrival and suppose
that it is expensive to keep the machine in operation while there are no jobs. Therefore the
machine is turned oﬀ as soon as the system is empty. When a new job arrives the machine
is turned on again, but it takes some setup time till the machine is ready for processing.
So turning oﬀ the machine leads to longer production leadtimes. But how much longer?
This will be investigated for some simple models in the following subsections.

10.1.1 Exponential processing and setup times
Suppose that the jobs arrive according to a Poisson stream with rate λ and that the
processing times are exponentially distributed with mean 1/µ. For stability we have to
require that ρ = λ/µ 1. The setup time of the machine is also exponentially distributed
with mean 1/θ. We now wish to determine the mean production lead time E(S) and the
mean number of jobs in the system. These means can be determined by using the mean
value technique.
To derive an equation for the mean production lead time, i.e. the arrival relation, we
evaluate what is seen by an arriving job. We know that the mean number of jobs in the
system found by an arriving job is equal to E(L) and each of them (also the one being
processed) has an exponential (residual) processing time with mean 1/µ. With probability
1 − ρ the machine is not in operation on arrival, in which case the job also has to wait
for the (residual) setup phase with mean 1/θ. Further the job has to wait for its own

97

processing time. Hence
1 1 1
E(S) = (1 − ρ) + E(L) +
θ µ µ
and together with Little’s law,

E(L) = λE(S)

we immediately find
1/µ 1
E(S) = + .
1−ρ θ
So the mean production lead time is equal to the one in the system where the machine is
always on, plus an extra delay caused by turning off the machine when there is no work.
In fact, it can be shown that the extra delay is exponentially distributed with mean 1/θ.

10.1.2 General processing and setup times
We now consider the model with generally distributed processing times and generally
distributed setup times. The arrivals are still Poisson with rate λ. The first and second
moment of the processing time are denoted by E(B) and E(B 2 ) respectively, E(T ) and
E(T 2 ) are the first and second moment of the setup time. For stability we require that
ρ = λE(B) 1. Below we demonstrate that also in this more general setting the mean
value technique can be used to find the mean production lead time. We first determine the
mean waiting time. Then the mean production lead time is found afterwards by adding
the mean processing time.
The mean waiting time E(W ) of a job satisfies

E(W ) = E(Lq )E(B) + ρE(RB )
+ P (Machine is off on arrival)E(T )
+ P (Machine is in setup phase on arrival)E(RT ), (10.1)

where E(RB ) and E(RT ) denote the mean residual processing and residual setup time, so
(see (7.15))
E(B 2 ) E(T 2 )
E(RB ) = , E(RT ) = .
2E(B) 2E(T )
To find the probability that on arrival the machine is off (i.e. not working and not in the
setup phase), note that by PASTA, this probability is equal to the fraction of time that
the machine is off. Since a period in which the machine is not processing jobs consists of
an interarrival time followed by a setup time, we have
1/λ
P (Machine is off on arrival) = (1 − ρ) .
1/λ + E(T )

98

Similarly we find
E(T )
P (Machine is in setup phase on arrival) = (1 − ρ) .
1/λ + E(T )
Substituting these relations into (10.1) and using Little’s law stating that
E(Lq ) = λE(W )
we finally obtain that
ρE(RB ) 1/λ E(T )
E(W ) = + E(T ) + E(RT ).
1−ρ 1/λ + E(T ) 1/λ + E(T )
Note that the first term at the right-hand side is equal to the mean waiting time in the
M/G/1 without setup times, the other terms express the extra delay due to the setup
times. Finally, the mean production lead time E(S) follows by simply adding the mean
processing time E(B) to E(W ).

10.1.3 Threshold setup policy
A natural extension to the setup policy in the previous section is the one in which the
machine is switched on when the number of jobs in the system reaches some threshold
value, N say. This situation can be analyzed along the same lines as in the previous
section. The mean waiting time now satisfies
E(W ) = E(Lq )E(B) + ρE(RB )
N
N −i
+ P (Arriving job is number i) + E(T )
i=1 λ
+ P (Machine is in setup phase on arrival)E(RT ), (10.2)
The probability that an arriving job is the i-th one in a cycle can be determined as follows.
A typical production cycle is displayed in figure 10.1.
Non-processing period Processing period

A1 A2 ... AN Setup B1 B2

Figure 10.1: Production cycle in case the machine is switched on when there are N jobs

The probability that a job arrives in a non-processing period is equal to 1 − ρ. Such a
period now consists of N interarrival times followed by a setup time. Hence, the probability
that a job is the i-th one, given that the job arrives in a non-processing period, is equal to
1/λ divided by the mean length of a non-processing period. So
1/λ
P (Arriving job is number i) = (1 − ρ) , i = 1, . . . , N,
N/λ + E(T )

99

and similarly,
E(T )
P (Machine is in setup phase on arrival) = (1 − ρ) .
N/λ + E(T )
Substituting these relations into (10.2) we obtain, together with Little’s law, that
ρE(RB ) N/λ N −1 E(T )
E(W ) = + + E(T ) + E(RT ).
1−ρ N/λ + E(T ) 2λ N/λ + E(T )

10.2 Unreliable machine
In this section we consider an unreliable machine processing jobs. The machine can break
down at any time it is operational, even though it is not processing jobs. What is the impact
of these breakdowns on the production leadtimes? To obtain some insight in the effects of
breakdowns we formulate and study some simple models in the following subsections.

10.2.1 Exponential processing and down times
Suppose that jobs arrive according to a Poisson stream with rate λ. The processing times
are exponentially distributed with mean 1/µ. The machine is successively up and down.
The up and down times of the machine are also exponentially distributed with means 1/η
and 1/θ, respectively.
We begin with formulating the condition under which the machine can handle the
amount of work offered per unit time. Let ρU and ρD denote the fraction of time the
machine is up and down, respectively. So
1/η 1 1
ρU = = , ρD = 1 − ρU = .
1/η + 1/θ 1 + η/θ 1 + θ/η
Then we have to require that
λ
ρU . (10.3)
µ
We now proceed to derive an equation for the mean production lead time. An arriving
job finds on average E(L) jobs in the system and each of them has an exponential processing
time with mean 1/µ. So in case of a perfect machine his mean sojourn time is equal to
(E(L) + 1)/µ. However, it is not perfect. Breakdowns occur according to a Poisson
process with rate η. So the mean number of breakdowns experienced by our job is equal to
η(E(L) + 1)/µ, and the mean duration of each breakdown is 1/θ. Finally, with probability
ρD the machine is already down on arrival, in which case our job has an extra mean delay
of 1/θ. Summarizing we have
1 1 1 1
E(S) = (E(L) + 1) + η(E(L) + 1) · + ρD
µ µ θ θ

100

η 1 ρD
= (1 + )(E(L) + 1) +
θ µ θ
1 ρD
= (E(L) + 1) + .
µρU θ

Then, with Little’s law stating that

E(L) = λE(S),

we immediately obtain

1/(µρU ) + ρD /θ
E(S) = .
1 − λ/(µρU )

In table 10.1 we investigate the impact of the variability in the availability of the
machine on the mean production leadtime. The mean production time is 1 hour (µ = 1).
The average number of jobs that arrive in a week (40 hours) is 32, so λ = 0.8 jobs per
hour. The fraction of time the machine is available is kept constant at 90%, so ρU = 0.9
(and thus η/θ = 1/9). The rate η at which the machine breaks down is varied from (on
the average) every 10 minutes till once a week. In the former case the mean down time is
1.1 minute, in the latter case it is more dramatic, namely nearly half a day (4.4 hours).
The results show that the variation in the availability of the machine is essential to the
behavior of the system. Note that as η and θ both tend to infinity such that η/θ = 1/9,
then E(S) tends to 10, which is the mean sojourn time in an M/M/1 with arrival rate 0.8
and service rate 0.9.

η θ E(S)
6 54 10.02
3 27 10.03
1 9 10.1
0.125 1.125 10.8
0.0625 0.5625 11.6
0.025 0.225 14

Table 10.1: The mean production leadtime in hours as a function of the break-down rate
η per hour for fixed availability of 90%

10.2.2 General processing and down times
We now consider the model with general processing times and general down times. The
arrivals are still Poisson with rate λ. The first and second moment of the processing
time are denoted by E(B) and E(B 2 ) respectively. The time between two breakdowns is

101

exponentially distributed with mean 1/η and E(D) and E(D2 ) are the first and second
moment of the down time. For stability we require that (cf. (10.3))
1
λE(B) .
1 + ηE(D)
Below we first determine the mean waiting time by using the mean value technique. The
mean production leadtime is determined afterwards.
We start with introducing the generalized processing time, which is defined as the pro-
cessing time plus the down times occurring in that processing time. Denote the generalized
processing time by G. Then we have
N (B)
G=B+ Di ,
i=1

where N (B) is the number of break-downs during the processing time B and Di is the i-th
down time. For the mean of G we get by conditioning on B and N (B) that
∞
∞ (ηx)n
E(G) = E(G|B = x, N (B) = n)e−ηx fB (x)dx
x=0 n=0 n!
∞ ∞ (ηx)n
= (x + nE(D))e−ηx
fB (x)dx
x=0 n=0 n!
∞ ∞ (ηx)n
= (x + xηE(D))e−ηx fB (x)dx
x=0 n=0 n!
= E(B) + E(B)ηE(D),
and similarly,
E(G2 ) = E(B 2 )(1 + ηE(D))2 + E(B)ηE(D2 ).
A typical production cycle is shown in figure 10.2. The non-processing period consists of
cycles which start with an up time. When the machine is up two things can happen: a job
arrives, in which case the machine starts to work, or the machine goes down and has to
be repaired. Hence, an up time is exponentially distributed with mean 1/(λ + η) and it is
followed by a processing period (i.e. a job arrives) with probability λ/(λ + η) or otherwise,
it is followed by a down time. During a processing period the machine works on jobs with
generalized processing times (until all jobs are cleared). For the mean waiting time it holds
that
E(W ) = E(Lq )E(G) + ρG E(RG )
+P [Arrival in a down time in a non-processing period]E(RD ), (10.4)
where ρG is the fraction of time the machine works on generalized jobs and E(RG ) and
E(RD ) denote the mean residual generalized processing time and the mean residual down
time, so
E(G2 ) E(D2 )
ρG = λE(G), E(RG ) = , E(RD ) = .
2E(G) 2E(D)

102

Non-processing period Processing period

up down up down G1 G2

first arrival

Figure 10.2: Production cycle of an unreliable machine

The probability that a job arriving in a non-processing period finds that the machine is
down, is given by
E(D)η/(λ + η) E(D)η
= ,
1/(λ + η) + E(D)η/(λ + η) 1 + E(D)η
Hence, since a job arrives with probability 1 − ρG in a non-processing period, we have
E(D)η
P (Arrival in a down time in a non-processing period) = (1 − ρG ) .
1 + E(D)η
Substitution of this relation into (10.4) and using Little’s law yields
ρG E(RG ) E(D)η
E(W ) = + E(RD ) ,
1 − ρG 1 + E(D)η
and the mean production lead time finally follows from

E(S) = E(W ) + E(G).

10.3 M/G/1 queue with an exceptional first customer
in a busy period
A servers’ life in an M/G/1 queue is an alternating sequence of periods during which
no work is done, the so-called idle periods, and periods during which the server helps
customers, the so-called busy periods. In some applications the service time of the first
customer in a busy period is different from the service times of the other customers served
in the busy period. For example, consider the setup problem in section 10.1 again. This
problem may, alternatively, be formulated as an M/G/1 queue in which the first job in
a busy period has an exceptional service time, namely the setup time plus his actual
processing time. The problem in section 10.2 can also be described in this way. Here we
can take the generalized processing times as the service times. However, on arrival of the
first job in a busy period the machine can be down. Then the (generalized) servicing of
that job cannot immediately start, but one has to wait till the machine is repaired. This
repair time can be included in the service time of the first job. In this case, however, the
determination of the distribution, or the moments of the service time of the first job is
more difficult then in the setup problem. Below we show how, in the general setting of an

103

M/G/1 with an exceptional first customer, the mean sojourn time can be found by using
the mean-value approach.
Let Bf and Rf denote the service time and residual service time, respectively, of the
first customers in a busy period. For the mean waiting time we have

E(W ) = E(Lq )E(B) + ρf E(Rf ) + ρE(R), (10.5)

where ρf is the fraction of time the server works on first customers, and ρ is the fraction of
time the server works on other (ordinary) customers. Together with Little’s law we then
obtain from (10.5) that
ρf E(Rf ) + ρE(R)
E(W ) = ,
1 − λE(B)
and for the mean sojourn time we get

E(S) = E(W ) + (1 − ρf − ρ)E(Bf ) + (ρf + ρ)E(B).

It remains to determine ρf and ρ. The probability that an arriving customer is the first
one in a busy cycle is given by 1 − ρf − ρ. Hence, the number of first customers arriving
per unit of time is equal to λ(1 − ρf − ρ). Thus ρf and ρ satisfy

ρf = λ(1 − ρf − ρ)E(Bf ),
ρ = λ(ρf + ρ)E(B),

from which it follows that
λE(Bf )(1 − λE(B)) λE(Bf )λE(B)
ρf = , ρ= .
1 + λE(Bf ) − λE(B) 1 + λE(Bf ) − λE(B)

10.4 M/G/1 queue with group arrivals
In this section we consider the M/G/1 queue where customers do not arrive one by one,
but in groups. These groups arrive according to a Poisson process with rate λ. The group
size is denoted by the random variable G with probability distribution

gk = P (G = k), k = 0, 1, 2, . . .

Note that we also admit zero-size groups to arrive. Our interest lies in the mean waiting
time of a customer, for which we can write down the following equation.
∞
E(W ) = E(Lq )E(B) + ρE(R) + rk (k − 1)E(B) , (10.6)
k=1

where ρ is the server utilization, so

ρ = λE(G)E(B),

104

and rk is the probability that our customer is the kth customer served in his group. The
first two terms at the right-hand side of (10.6) correspond to the mean waiting time of
the whole group. The last one indicates the mean waiting time due to the servicing of
members in his own group.
To find rk we first determine the probability hn that our customer is a member of a
group of size n (cf. section 7.7). Since it is more likely that our customer belongs to a
large group than to a small one, it follows that hn is proportional to the group size n as
well as the frequency of such groups. Thus we can write
hn = Cngn ,
where C is a constant to normalize this distribution. So
∞
C −1 = ngn = E(G).
n=1

Hence
ngn
hn = , n = 1, 2, . . .
E(G)
Given that our customer is a member of a group of size n, he will be with probability 1/n
the kth customer in his group going into service (of course, n ≥ k). So we obtain
∞ ∞
1 1
rk = hn · = gn ,
n=k n E(G) n=k
and for the last term in (10.6) it immediately follows that
∞
1 ∞ ∞
rk (k − 1)E(B) = gn (k − 1)E(B)
k=1 E(G) k=1 n=k
1 ∞ n
= gn (k − 1)E(B)
E(G) n=1 k=1
1 ∞ 1
= n(n − 1)gn E(B)
E(G) n=1 2
E(G2 ) − E(G)
= E(B). (10.7)
2E(G)
From (10.6) and (10.7) and Little’s law stating that
E(Lq ) = λE(G)E(W ),
we finally obtain
ρE(R) (E(G2 ) − E(G))E(B)
E(W ) = + .
1−ρ 2E(G)(1 − ρ)
The first term at the right-hand side is equal to the mean waiting time in the system where
customers arrive one by one according to a Poisson process with rate λE(G). Clearly, the
second term indicates the extra mean delay due to clustering of arrivals.

105

Example 10.4.1 (Uniform group sizes)
In case the group size is uniformly distributed over 1, 2, . . . , n, so
1
gk = , k = 1, . . . , n,
n
we ﬁnd
n n
k n+1 k2 − k (n − 1)(n + 1)
E(G) = = , E(G2 − G) = = .
k=1 n 2 k=1 n 3

Hence
ρE(R) (n − 1)E(B)
E(W ) = + .
1−ρ 3(1 − ρ)

106

10.5 Exercises
Exercise 63.
Consider an M/G/1 queue where at the end of each busy period the server leaves for a
fixed period of T units of time. Determine the mean sojourn time E(S).
Exercise 64.
In a factory a paternoster elevator (see figure 10.3) is used to transport rectangular bins
from the ceiling to the floor. This is a chain-elevator with product carriers (each carrier
can hold at most one bin) at a certain fixed distance. The number of carriers between the
ceiling and the floor is 4. Every time a carrier reaches the ceiling (and, simultaneously,
another one reaches the floor) the elevator stops 2 seconds. This is exactly the time a bin
needs to enter or leave the carrier. Subsequently the elevator starts to move again till the
next carrier reaches the ceiling. This takes exactly 3 seconds. Then it waits for 2 seconds,
moves again, and so on. The process of waiting and moving goes on continuously. Note
that a bin, finding on arrival an empty carrier positioned in front of it, cannot enter the
carrier, because the residual time till the carrier start to move again is less than 2 seconds.
The bin has to wait till the next carrier arrives.

Figure 10.3: Paternoster elevator

Bins arrive according to a Poisson process with a rate of 10 bins per minute.
(i) Determine the mean waiting time of a bin in front of the elevator.
(ii) Determine the mean time that elapses from the arrival of a bin till the moment the
bin leaves the elevator downstairs.

Exercise 65.
Consider a machine which is turned off when there is no work. It is turned on and restarts
work when enough orders, say N , arrived to the machine. The setup times are negligible.
The processing times are exponentially distributed with mean 1/µ and the average number
of orders arriving per unit time is λ( µ).

107

(i) Determine the mean number of orders in the system.

(ii) Determine the mean production lead time.

Suppose that λ = 20 orders per hour, 1/µ = 2 minutes and that the setup cost is 54
dollar. In operation the machine costs 12 dollar per minute. The waiting cost is 1 dollar
per minute per order.

(iii) Determine, for given threshold N , the average cost per unit time.

(iv) Compute the threshold N minimizing the average cost per unit time.

Exercise 66.
At a small river pedestrians are brought from the left side to the right side of the river
by a ferry. On average 40 pedestrians per hour arrive according to a Poisson process. It
takes the ferry exactly 2 minutes to cross the river and return. The capacity of the ferry
is sufficiently big, so it is always possible to take all waiting pedestrians to the other side
of the river. The ferry travels continuously back and forth, also when there are no waiting
pedestrians.

(i) Determine the mean waiting time and the mean number of pedestrians waiting for
the ferry.

The ferry now only takes off when there are one or more pedestrians. In case there are no
pedestrians the ferry waits for the first one to arrive, af ter which it immediately takes off
(the ferry never waits at the other side of the river).

(ii) Determine the probability that an arriving pedestrian finds the ferry waiting to take
off.

(iii) Determine the mean waiting time and the mean number of pedestrians waiting for
the ferry.

Exercise 67.
A machine produces products in two phases. The first phase is standard and the same for
all products. The second phase is customer specific (the finishing touch). The first (resp.
second) phase takes an exponential time with a mean of 10 (resp. 2) minutes. Orders for
the production of one product arrive according to a Poisson stream with a rate of 3 orders
per hour. Orders are processed in order of arrival.

(i) Determine the mean production lead time (waiting time plus production time) of an
order.

The machine is switched off when the system is empty and it is switched on again as soon
as the first order arrives. A fixed cost of 20$ is incurred each time the machine is switched
on (the time needed to switch the machine on or off is negligible).

108

(ii) Determine the average switch-on cost per hour.

To reduce the production lead time one decides to start already with the production of
phase 1 when the system is empty. If upon completion of phase 1 no order has arrived
yet, the production stops and the machine is switched off. When the first order arrives the
machine is switched on again and can directly start with phase 2.

(iii) Determine the reduction in the mean production lead time.

(iv) Determine the average switch-on cost per hour.

Exercise 68.
In a bank there is a special office for insurances. Here arrive according to a Poisson process
on average 8 customers per hour. The service times are exponentially distributed with a
mean of 5 minutes. As soon as all customers are served and the office is empty again, the
clerck also leaves to get some coffee. He returns after an exponential time with a mean of
5 minutes and then starts servicing the waiting customers (if there are any) or patiently
waits for the first customer to arrive.

(i) What is the mean sojourn time and the mean number of customers in the office?

Now suppose that when the clerck finds an empty office upon his return, he immediately
leaves again to get another cup of coffee. This takes again an exponential time with a mean
of 5 minutes. The clerck repeats to leave until he finds a waiting customer upon return.

(ii) How many cups of coffee the clerck drinks on average before he starts servicing again?

(iii) Determine the mean sojourn time and the mean number of customers in the office.

Exercise 69.
Consider a machine processing orders. These orders arrive according to a Poisson process
with a rate of 1 order per hour. When there are no orders, the machine is switched off. As
soon as a new order arrives, the machine is switched on again, which takes exactly T hours.
The processing time of an order is exponentially distributed with a mean of 30 minutes.

(i) Determine as a function of T :

(a) The mean numbers of orders processed in a production cycle.
(b) The mean duration of a production cycle.
(c) The mean production lead time of an order.

Suppose that it costs 17 dollar each time the machine is switched on again and that the
waiting cost per hour per order are 1 dollar.

(ii) Show that the average cost per hour are minimal for T = 3 hour.

109

Exercise 70.
Consider a queueing system where on average 3 groups of customers arrive per hour. The
mean group size is 10 customers. The service time is exactly 1 minute for each customer.
Determine the mean sojourn time of the ﬁrst customer in a group, the last customer in a
group and an arbitrary one in the following two cases:

(i) the group size is Poisson distributed;

(ii) the groups size is geometrically distributed.

Exercise 71.
Customers arrive in groups at a server. The groups consist of 1 or 3 customers, both with
equal probability, and they arrive according to a Poisson stream with a rate of 2 groups
per hour. Customers are served one by one, and they require an exponential service time
with a mean of 10 minutes.

(i) Determine the mean sojourn time of an arbitrary customer.

(ii) Determine the mean number of customers in the system.

Exercise 72.
Passengers are brought with small vans from the airport to hotels nearby. At one of
those hotels on average 6 vans per hour arrive according to a Poisson process. With
probability 1/4 a van brings 2 guests for the hotel, with probability 1/4 only one guest
and with probability 1/2 no guests at all. At the reception of the hotel there is always one
receptionist present. It takes an exponential time with a mean of 5 minutes to check in a
guest.

(i) Determine the distribution of the number of guests at the reception.

(ii) Determine the mean waiting time of an arbitrary guest at the reception.

Exercise 73.
Customers arrive at a server according to a Poisson stream with a rate of 6 customers per
hour. As soon as there are no customers the server leaves to do something else. He returns
when there are 2 customers waiting for service again. It takes exactly 5 minutes for him to
return. The service time of a customer is uniformly distributed between 5 and 10 minutes.

(i) Determine the mean duration of a busy period, i.e., a period during which the server
is servicing customers without interruptions.

(ii) Determine the mean number of customers served in a busy period.

(iii) Determine the mean sojourn time of a customer.

110

Chapter 11

Insensitive systems

In this chapter we study some queueing systems for which the queue length distribution is
insensitive to the distribution of the service time, but only depends on its mean.

11.1 M/G/∞ queue
In this model customers arrive according to a Poisson process with rate λ. Their service
times are independent and identically distributed with some general distribution function.
The number of servers is infinite. So there is always a server available for each arriving
customer. Hence, the waiting time of each customer is zero and the sojourn time is equal
to the service time. Thus by Little’s law we immediately obtain that
E(L) = ρ,
where ρ = λE(B) denotes the mean amount of work that arrives per unit time. In the
remainder of this section we want to also determine the distribution of L, i.e., the proba-
bilities pn that there are n customers in the system.

Example 11.1.1 (M/M/∞)
In this model the service times are exponentially distributed with mean 1/µ.
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Figure 11.1: Flow diagram for the M/M/∞ model

From figure 11.1 we obtain by equating the flow from state n − 1 to n and the flow
from n to n − 1 that
pn−1 λ = pn nµ.

111

Thus
λ ρ ρ2 ρn
pn = pn−1 = pn−1 = pn−2 = · · · = p0 .
nµ n n(n − 1) n!
Since the probabilities pn have to add up to one, it follows that
∞
−1 ρn
p0 = = eρ .
n=0 n!
Summarizing, we have found that
ρn
pn = e−ρ . (11.1)
n!
Thus the number of customers in the system has a Poisson distribution with mean ρ.
Example 11.1.2 (M/D/∞)
Let b denote the constant service time. The probability pn (t) that there are exactly n
customers in the system at time t is equal to the probability that between time t − b and t
exactly n customers arrived. Since the number of customers arriving in a time interval of
length b is Poisson distributed with mean λb, we immediately obtain
(λb)n −λb ρn −ρ
pn (t) = e = e ,
n! n!
which is valid for all t b, and thus also for the limiting distribution.
Example 11.1.3 (Discrete service time distribution)
Suppose that the service time distribution is discrete, i.e., there are nonnegative numbers
bi and probabilities qi such that
P (B = bi ) = qi , i = 1, 2, . . .
In this case it is also easy to determine the probabilities pn . We split the Poisson stream
with rate λ into the countably many independent Poisson streams numbered 1, 2, . . .. The
intensity of stream i is λqi and the customers of this stream have a constant service time
bi . The number of servers is inﬁnite and thus we immediately obtain from the previous
example that the number of type i customers in the system is Poisson distributed with
mean λqi bi and this number is of course independent of the other customers in the system.
Since the sum of independent Poisson random variables is again Poisson (see exercise 11) it
follows that the total number of customers in the system is Poisson distributed with mean
λq1 b1 + λq2 b2 + · · · = ρ.
The examples above suggest that also in the M/G/∞ the number of customers in
the system is Poisson distributed with ρ. In fact, since each distribution function can be
approximated arbitrary close by a discrete distribution function, this follows from example
11.1.3 (see also exercise 74). Summarizing we may conclude that in the M/G/∞ it holds
that
ρn
pn = e−ρ , n = 0, 1, 2, . . . ,
n!
where ρ = λE(B). Note that this is true regardless of the form of the distribution function
FB (·) of the service time.

112

11.2 M/G/c/c queue
In this model customers also arrive according to a Poisson process with rate λ. Their
service times are independent and identically distributed with some general distribution
function. There are c servers available. Each newly arriving customer immediately goes
into service if there is a server available, and that customer is lost if all servers are occupied.
This system is therefore also referred to as the M/G/c loss system.
In this section we want to find the probabilities pn of n customers in the system. Of
special interest is the probability pc , which, according to the PASTA property, describes
the fraction of customers that are lost.

Example 11.2.1 (M/M/c/c)
In this model the service times are exponentially distributed with mean 1/µ.
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Figure 11.2: Flow diagram for the M/M/c/c model

From figure 11.1 we immediately obtain that
pn−1 λ = pn nµ, n = 1, 2, . . . , c.
Hence, it is readily verified that
(λ/µ)n /n! ρn /n!
pn = c n
= c n
, n = 0, 1, . . . , c,
n=0 (λ/µ) /n! n=0 ρ /n!

where ρ = λ/µ.

It can be proved that also for a general service time distribution the probabilities pn
are given by (see e.g. [7])
ρn /n!
pn = c n
, n = 0, 1, . . . , c,
n=0 ρ /n!

where ρ = λE(B). Hence, the so-called blocking probability B(c, ρ) is given by
ρc /c!
B(c, ρ) = pc = c n
. (11.2)
n=0 ρ /n!

The name given to this formula is Erlang’s loss formula. Note that by Little’s law we
obtain that
E(L) = ρ(1 − B(c, ρ))

113

Example 11.2.2 (Parking lot)
Customers arrive according to a Poisson process at a parking lot near a small shopping
center with a rate of 60 cars per hour. The mean parking time is 2.5 hours and the parking
lot offers place to 150 cars. When the parking lot is full, an arriving customer has to park
his car somewhere else. Now we want to know the fraction of customers finding all places
occupied on arrival.
The parking lot can be described by a M/G/150/150 model with ρ = 60 · 2.5 = 150.
Hence the fraction of customers finding all places occupied on arrival is given by
150150 /150!
B(150, 150) = 150 n
.
n=0 150 /n!

It will be clear that the computation of B(150, 150) gives rise to a serious problem!

11.3 Stable recursion for B(c, ρ)
Fortunately it is easy to derive a simple and numerically stable recursion for the blocking
probabilities B(c, ρ). From (11.2) we have
ρc /c!
B(c, ρ) = c−1 n c
.
n=0 ρ /n! + ρ /c!
c−1
Dividing the numerator and denominator of this expression by n=0 ρn /n! yields
ρB(c − 1, ρ)/c ρB(c − 1, ρ)
B(c, ρ) = = . (11.3)
1 + ρB(c − 1, ρ)/c c + ρB(c − 1, ρ)
Starting with B(0, ρ) = 1 we can use relation (11.3) to subsequently compute the blocking
probabilities B(c, ρ) for c = 1, 2, 3, . . ..

c B(c, 150)
150 0.062
155 0.044
160 0.028
165 0.017
170 0.009

Table 11.1: Blocking probability B(c, ρ) for ρ = 150 and several values of c

Example 11.3.1 (Parking lot)
Using the recursion (11.3) it is easy to compute B(c, 150) for c = 150. In table 11.1 we
list B(c, 150) for several values of c. Clearly in the present situation with 150 parking
places 6% of the arriving customers find all places occupied. With 20 additional places
this percentage drops below 1%.

114

Remark 11.3.2 (Delay probability in the M/M/c queue)
It will be clear from (5.1) that the formula for delay probability ΠW in the M/M/c suﬀers
from the same numerical problems as (11.2). Luckely there is a simple relation between
the queueing probability ΠW in the M/M/c with server utilization ρ and the blocking
probability B(c, cρ). From (5.1) we immediately obtain
(cρ)c /c! ρB(c − 1, cρ)
ΠW = c−1 = .
(1 − ρ) n=0 (cρ)
n /n! + (cρ)c /c! 1 − ρ + ρB(c − 1, cρ)
By ﬁrst computing B(c − 1, cρ) from the recursion (11.3) we can use the relation above to
determine ΠW .

11.4 Java applet
There is a JAVA applet is avalaible for the evaluation of the M/G/∞ queue. The WWW-
link to this applet is http://www.win.tue.nl/cow/Q2.

115

11.5 Exercises
Exercise 74.
Prove that the fact that (11.1) holds for discrete service time distributions implies that it
also holds for general service time distributions.
(Hint: Approximate the service time distribution from below and from above by discrete
distributions and then consider the probability that there are n or more customers in the
system.)
Exercise 75.
In a small restaurant there arrive according to a Poisson proces on average 5 groups of
customers. Each group can be accommodated at one table and stays for an Erlang-2
distributed time with a mean of 36 minutes. Arriving groups who find all tables occupied
leave immediately.
How many tables are required such that at most 7% of the arriving groups is lost?
Exercise 76.
For a certain type of article there is a stock of at most 5 articles to satisfy customer
demand directly from the shelf. Customers arrive according to a Poisson process at a rate
of 2 customers per week. Each customer demands 1 article. The ordering policy is as
follows. Each time an article is sold to a customer, an order for 1 article is immediately
placed at the supplier. The lead time (the time that elapses from the moment the order
is placed until the order arrives) is exponentially distributed with a mean of 1 week. If on
arrival of a customer the shelf is empty, the customer demand will be lost. The inventory
costs are 20 guilders per article per week. Each time an article is sold, this yields a reward
of 100 guilders.
(i) Calculate the probability distribution of the number of outstanding orders.

(ii) Determine the mean number of articles on stock.

(iii) What is the average profit (reward - inventory costs) per week?

Exercise 77.
In our library there are 4 VUBIS terminals. These terminals can be used to obtain infor-
mation about the available literature. If all terminals are occupied when someone wants
information, then that person will not wait but leave immediately (to look for the required
information somewhere else). A user session on a VUBIS terminal takes on average 2.5
minutes. Since the number of potential users is large, it is reasonable to assume that users
arrive according to a Poisson stream. On average 72 users arrive per hour.
(i) Determine the probability that i terminals are occupied, i = 0, 1, . . . , 4.

(ii) What is the fraction of arriving users finding all terminals occupied?

(iii) How many VUBIS terminals are required such that at most 5% of the arriving users
find all terminals occupied?

116

Exercise 78.
Consider a machine continuously processing parts (there is always raw material available).
The processing time of a part is exponentially distributed with a mean of 20 seconds. A
finished part is transported immediately to an assembly cell by an automatic conveyor
system. The transportation time is exactly 3 minutes.

(i) Determine the mean and variance of the number of parts on the conveyor.

To prevent that too many parts are simultaneously on the conveyor one decides to stop
the machine as soon as there are N parts on the conveyor. The machine is turned on again
as soon as this number is less than N .

(ii) Determine the throughput of the machine as a function of N .

(iii) Determine the smallest N for which the throughput is at least 100 parts per hour.

Exercise 79.
A small company renting cars has 6 cars available. The costs (depreciation, insurance,
maintenance, etc.) are 60 guilders per car per day. Customers arrive according to a Poisson
process with a rate of 5 customers per day. A customer rents a car for an exponential time
with a mean of 1.5 days. Renting a car costs 110 guilders per day. Arriving customers for
which no car is available are lost (they will go to another company).

(i) Determine the fraction of arriving customers for which no car is available.

(ii) Determine the mean profit per day.

The company is considering to buy extra cars.

(iii) How many cars should be bought to maximize the mean profit per day?

117

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[2] I.Adan, Y.Zhao, Analyzing GI/Er /1 queues, Opns. Res. Lett., 19 (1996), pp. 183–
190.

[3] I.J.B.F. Adan, W.A. van de Waarsenburg, J. Wessels, Analyzing Ek |Er |c
queues, EJOR, 92 (1996), pp. 112–124.

[4] N.G. de Bruijn, Asymptotic methods, Dover, 1981.

[5] B.D. Bunday, An introduction to queueing theory, Arnold, London, 1996.

[6] J.A. Buzacott, J.G. Shanthikumar, Stochastic models of manufacturing systems,
Prentice Hall, Englewood Cliﬀs, 1993.

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CRC Press, Boca Raton, 1995.

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Thesis, Vrije Universiteit, Amsterdam, 1993.

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London, 1982.

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119

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383–387.

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ceedings Performance’ 80, Toronto, (May 28–30, 1980), pp. 117–125.

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(1985), pp. 879–892.

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Math. Statist., 41 (1970), pp. 182–187.

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[25] S. Stidham, A last word on L = λW , Opns. Res., 22 (1974), pp. 417–421.

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[27] H.C. Tijms, Stochastic modelling and analysis: a computational approach, John Wi-
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1989.

120

Index

G/M/1, 79 hyperexponential distribution, 15
M/D/∞, 112
M/Er /1, 49 idle period, 37, 103
M/G/1, 59 imbedded Markov chain, 61, 79
M/G/∞, 111
Kendall’s notation, 24
M/G/c/c, 113
M/M/1, 29 Laplace-Stieltjes transform, 12
M/M/∞, 111 last come first served, 24
M/M/c, 43 Lindley’s equation, 67
M/M/c/c, 113 Little’s law, 26
loss system, 113
arrival distribution, 60, 79
arrival relation, 33, 97 mean, 11
busy period, 25, 37, 71, 103 mean value approach, 27, 33, 68
memoryless property, 13
coefficient of variation, 11 merging Poisson processes, 19
conditional waiting time, 35, 70 mixtures of Erlang distributions, 16
conservation law, 91, 92
Coxian distribution, 16 non-preemptive priority, 37, 87
cycle, 37
occupation rate, 25
delay probability, 44
PASTA property, 27
departure distribution, 59, 60
performance measures, 25
discrete distribution, 112
phase diagram, 14
Erlang distribution, 14 phase representation, 16
Erlang’s loss formula, 113 phase-type distribution, 16
exponential distribution, 13 Poisson distribution, 13, 18
Poisson process, 18, 20, 21
FCFS, 91 Pollaczek-Khinchin formula, 63, 65, 66, 68
first come first served, 24, 91 preemptive-resume priority, 36, 87
flow diagram, 30 processor sharing, 24
generating function, 11 queueing model, 23
geometric distribution, 12
global balance principle, 32 random variable, 11
group arrivals, 104 rational function, 63

121

residual service time, 53, 68
Rouch´’s theorem, 55
e

server utilization, 25
shortest processing time ﬁrst, 90
sojourn time, 25, 33
splitting Poisson processes, 19
SPTF, 90
standard deviation, 11

transient Markov chain, 16

variance, 11

waiting time, 35
work in the system, 25

122

Solutions to Exercises
Exercise 1.

(i) Use that
n
P (Yn x) = P (Xi x)
i=1

and
n
P (Zn ≤ x) = P (Xi ≤ x).
i=1

(ii) It follows that
∞
P (Xi = min (X1 , . . . , Xn )) = P (Xj x)fXi (x)dx
0 j=i
∞ µi
= µi e−(µ1 +···+µn ))x dx = .
0 (µ1 + · · · + µn )

Exercise 1

123

Exercise 2.

Use Laplace-Stieltjes transforms to prove that

∞
E(e−sS ) = P (N = k)E(e−sS | N = k)
k=1
∞ k
µ µ(1 − p)
= (1 − p)pk−1 = .
k=1 µ+s µ(1 − p) + s

Exercise 2

124

Exercise 3.

Use that the random variable

 1/µ1 ,
 with probability p1 ,
. .

X= . . .
.
 1/µ , with probability p ,

k k

has variance ≥ 0, and hence that
k k 2
2
pi (1/µi ) ≥ pi (1/µi ) .
i=1 i=1

Exercise 3

125

Exercise 4.

(i) p0 (t) = P (A1 t) = e−λt .

(ii) Use that

pn (t + ∆t) = λ∆tpn−1 (t) + (1 − λ∆t)pn (t) + o(∆t),

and let ∆t tend to zero.

(iii) Prove by induction that

(λt)n −λt
pn (t) = e
n!

is the solution of the diﬀerential equation in (ii).

Exercise 4

126

Exercise 5.

(i) p0 (t) = P (A1 t) = e−λt .

(ii) Use that
∞
P (N (t) = n) = P (N (t) = n | A1 = x)fA1 (x) dx.
0

(iii) Prove by induction that

(λt)n −λt
pn (t) = e
n!

is the solution of the integral equations in (ii).

Exercise 5

127

Exercise 6.

Merging property: Use that the minimum of independent exponential random variables is
again an exponential random variable. (See also Exercise 1)
Splitting property: Use the result of Exercise 2.
Exercise 6

128

Exercise 7.
√
Choose p = (18 − 4 14)/25 = 0.1213 and µ = (2 − p)/4 = 0.4697.

Exercise 7

129

Exercise 8.

(i) For a Coxian-2 distribution we have

2 2p1 2p1 1 p1
E(X 2 ) = 2
+ + 2, and E(X) = + .
µ1 µ1 µ2 µ2 µ1 µ2

Use this to show that E(X 2 ) ≥ 3 E(X)2 and hence that c2 ≥ 1 .
2 X 2

(ii) Show that both distributions have the same Laplace-Stieltjes transform. Try to
understand why these distributions are equivalent! (cf. Exercise 9)

Exercise 8

130

Exercise 9. Use Laplace-Stieltjes transforms, or use the formula

X1 = min(X1 , X2 ) + (X1 − min(X1 , X2 )) ,

where X1 is an exponential random variable with parameter λ and X2 is an exponential
random variable, independent of X1 , with parameter µ − λ. Exercise 9

131

Exercise 10. Use Exercise 9 with µ = µ1 and λ = µ2 . Exercise 10

132

Exercise 11. Use generating functions and the fact that for the sum Z = X + Y of
independent discrete random variables X and Y , it holds that (see subsection 2.2)

PZ (z) = PX (z) · PY (z).

Exercise 11

133

Exercise 12. As time unit we take 1 minute.

(i) Solve the (global) balance equations

λqn pn = µpn+1 , n = 0, 1, 2, 3,

where λ = µ = 1/3, together with the normalization equation. This gives

32 32 24 12 3
p0 = , p1 = , p2 = , p3 = , p4 = .
103 103 103 103 103

(ii) E(L) = 128/103 ≈ 1.24.

(iii) E(S) = 384/71 ≈ 5.41 minutes.

(iv) E(S) = 384/103 ≈ 3.73 minutes.
E(W ) = 171/103 ≈ 1.66 minutes.

Exercise 12

134

Exercise 13.

(i) Exponential with parameter µ∗ = µ(1 − p) (see Exercise 2).

(ii) P (L = n) = (1 − ρ)ρn for n = 0, 1, 2, . . . , where ρ = λ/µ∗ .

Exercise 13

135

Exercise 14. We have that

 µL ,
 if nr. of customers QL ,
service completion rate =  µ, if QL ≤ nr. of customers QH ,

µH , if nr. of customers ≥ QH .

The (global) balance equations are

λpn = µL pn+1 , if n + 1 QL ,
λpn = µ pn+1 , if QL ≤ n + 1 QH ,
λpn = µH pn+1 , if n + 1 ≥ QH .

The solution of these equations is given by
n
λ

 p0
 µL
, if n QL ,

 QL −1 n−QL +1
λ λ
pn =

p0 µL µ
, if QL ≤ n QH ,
 QL −1 QH −QL n−QH +1
λ λ λ
, if n ≥ QH .


p0 µL µ µH

Finally, p0 follows from the normalization equation. Exercise 14

136

Exercise 15.

(i) 3/8

(ii) 5/3

(iii) 80 minutes

Exercise 15

137

Exercise 16.
n
1 2
(i) P (L = n) = 3 3
, n = 0, 1, 2, . . .
and hence E(L) = 2 and σ 2 (L) = 6.

(ii) P (S ≤ t) = 1 − e−t/6 , t ≥ 0,
P (W ≤ t) = 1 − 2 e−t/6 ,
3
t ≥ 0.
2 −1/3
(iii) 3
e ≈ 0.48.
9 6 4
(iv) p0 = 19
, p1 = 19
, p2 = 19
,
hence E(L) = 14/19 ≈ 0.737 and σ 2 (L) = 22/19 − (14/19)2 ≈ 0.615.

(v) E(S) = 42/19 ≈ 2.21 minutes and E(W ) = 12/19 ≈ 0.63 minutes.

Exercise 16

138

Exercise 17.

(i) It holds that
n
1 2
P (L(Gas) = n) = , n = 0, 1, 2, . . .
3 3

and
n
(LP G) 5 1
P (L = n) = , n = 0, 1, 2, . . . .
6 6

(ii) Use
n
P (L = n) = P (L(Gas) = k, L(LP G) = n − k)
k=0

to show that
n n
20 2 5 1
P (L = n) = − , n = 0, 1, 2, . . .
54 3 54 6

Exercise 17

139

Exercise 18.

(i) E(S1 ) = 7.5 minutes.
E(S2 ) = 30 minutes.

(ii) E(S1 ) = 10.625 minutes.
E(S2 ) = 27.5 minutes.

Exercise 18

140

Exercise 19. Fraction of time that it is crowded: ρ5 ≈ 0.24.
Number of crowded periods (per 8 hours = 480 minutes): λp4 = 480(1 − ρ)ρ4 ≈ 38.
E(crowded period) = E(busy period) = 3 minutes. Exercise 19

141

Exercise 20. We have

average costs per hour = 16µ + 20E(Lq )
ρ2
= 16µ + 20
(1 − ρ)
8000
= 16µ + .
µ(µ − 20)
For µ 20, this function is minimal for µ = µ∗ ≈ 25. Exercise 20

142

Exercise 21. As state description we use the number of jobs in the system, and if this
number of jobs is equal to 1, we distinguish between state (1, f ) in which the fast server is
working and state (1, s) in which the slow server is working.

(i) As solution of the balance equations we ﬁnd

1−ρ
p0 = ,
1−ρ+C
p1 = p1,f + p1,s = Cp0 ,
pn = ρn−1 p1 , n 1,

where we used the notation µ = µ1 + µ2 , ρ = λ/µ and

λµ(λ + µ2 )
C= .
µ1 µ2 (2λ + µ)

(ii) For the mean number of jobs in the system we ﬁnd
∞
C
E(L) = npn = .
n=1 (1 − ρ)(1 − ρ + C)

(iii) It is better not to use the slower machine at all if E(Lf ), the expected number of
jobs in the system when you only use the fast server is smaller than E(L). This is
the case if µ1 λ and

λ C
.
µ1 − λ (1 − ρ)(1 − ρ + C)

(iv) In case (a) we have

2 81
E(Lf ) = = E(L).
3 104

In case (b) we have

3 24
E(Lf ) = = E(L).
2 17

Exercise 21

143

Exercise 22. As time unit we choose 1 minute: λ = 4/3 and µ = 1. In order to have
ρ 1 we need that c ≥ 2. Hence, we ﬁrst try c = 2. This gives ΠW = 8/15 ≈ 0.533 and
E(W ) = 24/30 = 0.8 minutes. Hence, we conclude that 2 boxes is enough. Exercise 22

144

Exercise 23. As time unit we choose 1 minute: λ = 2/3 and µ = 1/3. In order to have
ρ 1 we need that c ≥ 3. Hence, we ﬁrst try c = 3. This gives ΠW = 4/9 ≈ 0.444 and

P (W 2) = ΠW · e−2/3 ≈ 0.228 0.05.

Similarly, for c = 4 we ﬁnd ΠW = 4/23 ≈ 0.174 and

P (W 2) = ΠW · e−4/3 ≈ 0.046 0.05.

Hence, we need at least 4 operators. Exercise 23

145

Exercise 24. As time unit we choose 1 minute: λ = 1/3 and µ = 1/3.

(i) We have

n−1
1 1 1
p0 = , pn = , n ≥ 1.
3 3 2

(ii) Using (5.2) and (5.3) we have

ρ 1 1
E(Lq ) = ΠW · = 1/3, E(W ) = ΠW · · = 1 minute.
1−ρ 1 − ρ cµ

(iii) ΠW = 1/3.

(iv) For c = 2 we have ΠW = 1/3 1/10. Similarly, we can ﬁnd for c = 3 that
ΠW = 1/11 1/10. Hence, we need 3 troughs.

Exercise 24

146

Exercise 25. As time unit we choose 1 minute: λ = 15 and µ = 6.

(i) c · ρ = λ/µ = 2.5.

(ii) c · (1 − ρ) · 12 = 6 maintenance jobs per minute.

(iii) We have

n−2
8 20 25 5
p0 = , p1 = , pn = , n ≥ 2,
178 178 178 6

and hence (see (5.1))

125
ΠW = ≈ 0.702.
178

(iv) Using (5.3), we have

1 1 1
E(W ) = ΠW · · = · ΠW ≈ 0.234 minutes.
1 − ρ cµ 3

Exercise 25

147

Exercise 30.

(i) The distribution of the number of uncompleted tasks in the system is given by
n n
7 2 7 2
pn = + − , n = 0, 1, 2, . . . .
24 3 40 5

(ii) The distribution of the number of jobs in the system is given by
n n
35 4 21 4
qn = − , n = 0, 1, 2, . . . .
48 9 80 25
∞
(iii) The mean number of jobs equals E(L) = n=1 nqn = 104/105.

(iv) The mean waiting time of a job equals equals

ρ
E(W ) = E(RB ) = 8/7 · 3/2 = 12/7 minutes.
1−ρ

(Check: Little’s formula E(L) = λE(S) is satisﬁed, with E(L) = 104/105 job, λ = 4/15
job per minute and E(S) = 26/7 minutes.) Exercise 30

148

Exercise 31. Define Ti as the mean time till the first customer is rejected if we start with
i phases work in the system at time t = 0. Then we have

T0 = 1 + T2 ,
1 1 1
T1 = + T0 + T3 ,
2 2 2
1 1 1
T2 = + T1 + T4 ,
2 2 2
1 1
T3 = + T2 ,
2 2
1 1
T4 = + T3 .
2 2
The solution of this set of equations is given by
1 1
(T0 , T1 , T2 , T3 , T4 ) = (4, 3 , 3, 2, 1 ).
2 2
Hence, if at time t = 0 the system is empty, the mean time till the first customer is rejected
is equal to 4. Exercise 31

149

Exercise 32. The distribution of the number of phases work in the system is given by
n n
7 2 7 2
pn = + − , n = 0, 1, 2, . . . .
24 3 40 5
(i) The distribution of the waiting time (in minutes) is given by

7 −1t 1 7
P (W ≤ t) = 1 − e 12 + e− 20 t .
12 20

(ii) The fraction of customers that has to wait longer than 5 minutes is given by

7 −5 1 7
P (W 5) = e 12 − e− 4 ≈ 0.376.
12 20

Exercise 32

150

Exercise 33.

(i) The distribution of the number of customers in the system is given by
n n
3 1 6 1
pn = + − , n = 0, 1, 2, . . . .
7 2 35 5
∞
(ii) The mean number of customers equals E(L) = n=1 npn = 5/6. Now, either use the
PASTA property

E(S) = (5/6) · 6 + (1/4) · 6 + 6

or use Little’s formula

5/6
E(S) =
1/15

to conclude that the mean sojourn time of an arbitrary customer is equal to 12.5
minutes.

Exercise 33

151

Exercise 34.

(i) See Example 6.2.1.

(ii) Use PASTA and/or Little to conclude that E(S) = 11/12 week.

(iii) Because p0 + p1 + p2 0.99 and p0 + p1 + p2 + p3 0.99 we need at least 4 spare
engines.

Exercise 34

152

Exercise 35.

(i) The distribution of the number of uncompleted tasks at the machine is given by
n n
9 3 4 1
pn = + − , n = 0, 1, 2, . . . .
39 4 39 3

(ii) The mean number of uncompleted tasks equals E(Ltask ) = ∞ npn = 11/4. Hence,
n=1
using PASTA, the mean waiting time of a job is E(W ) = E(Ltask )·1 = 11/4 minutes.

(iii) The mean sojourn time of a job equals equals E(S) = E(W ) + E(B) = 11/4 + 8/5 =
87/20 minutes. Hence, using Little’s formula, we have E(Ljob ) = 5/12·87/20 = 29/16.

Exercise 35

153

Exercise 36.

(i) The distribution of the number of customers in the system is given by
n n
2 1 4 1
pn = + − , n = 0, 1, 2, . . . .
5 2 15 3

(ii) For the mean number of customers in the system we have E(L) = ∞ npn = 3/4.
n=1
Hence, using PASTA, the mean waiting time of the ﬁrst customer in a group equals
E(W1 ) = E(L) · 5 = 15/4 minutes.

(iii) The mean waiting time of the second customer in a group equals E(W2 ) = E(W1 ) +
5 = 35/4 minutes.

(Check: Little’s formula E(Lq ) = λE(W ) is satisﬁed, with E(Lq ) = 3/4 − 1/3 = 5/12
customer, λ = 1/15 customer per minute and E(W ) = 25/4 minutes.) Exercise 36

154

Exercise 37. As time unit we take 1 minute. Hence, λ = 1/2 and
1
1 2 3 1
B(s) = · 1 + · .
4 2
+s 4 1+s

(i) From (7.6) we have
9 3
(1 − ρ)B(λ − λz)(1 − z) 3 15 − 7z
PL (z) = = = 32 2 + 32 2 .
B(λ − λz) − z 8 (3 − 2z)(5 − 2z) 1 − 3z 1 − 5z

(ii) The distribution of the number of customers is given by
n n
9 2 3 2
pn = + , n = 0, 1, 2, . . . .
32 3 32 5
∞
(iii) E(L) = n=1 npn = 43/24.
(iv) From (7.7) we have
1 3
(1 − ρ)B(s)s 3 4 + 7s 27 4 5 4
S(s) = = = · 1 + · 3 .
λB(s) + s − λ 4 (1 + 4s)(3 + 4s) 32 4
+s 32 4
+s

(v) The distribution function of the sojourn time is given by

27 1 5 3
FS (x) = (1 − e− 4 x ) + (1 − e− 4 x ),
32 32
and the mean sojourn time by
27 5 4 43
E(S) = ·4+ · = minutes.
32 32 3 12
(vi) From (7.21) we have

E(B) 10
E(BP ) = = minutes.
1−ρ 3

(vii) For the M/M/1 queue we have

ρ 5
E(L) = = ,
1−ρ 3

and
E(L) 10
E(S) = = minutes.
λ 3
Exercise 37

155

Exercise 38. As time unit we take 1 minute, so λ = 1/6.
2
1
(i) B(s) = 1+s
.
n n
6 1 8 1
(ii) pn = 5 4
− 15 9
, n = 0, 1, 2, . . . .

(iii) E(L) = 11/24 and E(S) = 11/4 minutes.

Exercise 38

156

Exercise 39. As time unit we take 1 minute, so λ = 1. Let X be exponential with
parameter 4 and Y be exponential with parameter 1. Then,
1 1 1 3
E(B) = · E( + X) + · E(Y ) = minutes,
2 4 2 4
1 1 1 1 5 1 37
E(B 2 ) = · E([ + X]2 ) + · E(Y 2 ) = · + ·2= minutes,
2 4 2 2 16 2 32
and so E(R) = 37/48 minutes. Hence, E(W ) = 37/16 minutes and E(Lq ) = 37/16
customers. Exercise 39

157

Exercise 40. As time unit we take 1 minute, so λ = 1/20. Furthermore, E(B) = 12
minutes and E(R) = 31/3 minutes. Hence, E(S) = 55/2 = 27.5 minutes. Exercise 40

158

Exercise 41. The mean waiting time of jobs is given by
ρ 25
E(W ) = · E(R) = minutes.
1−ρ 4
Exercise 41

159

Exercise 44. The time unit is 1 minute: λ = 1/6, E(B)=15/4, ρ = 5/8, E(R) =
(2/5) · 6 + (3/5) · 3 = 21/5.

(i) The service time is hyperexponentially distributed with parameters p1 = 1/4, p2 =
3/4, µ1 = 1/6 and µ2 = 1/3.

(ii) From (7.9) we have

(1 − ρ)s 1 + 9s + 18s2 3 9 1 1 1
W (s) = = = + · + · .
λB(s) + s − λ (1 + 12s)(1 + 4s) 8 16 1 + 12s 16 1 + 4s

(iii) The distribution function of the waiting time is given by

3 9 1 1 1
FW (x) = + (1 − e− 12 x ) + (1 − e− 4 x ).
8 16 16

Hence, the fraction of cows for which the waiting time is less than 3 minutes equals

3 9 1 1 3
FW (3) = + (1 − e− 4 ) + (1 − e− 4 ) ≈ 0.532.
8 16 16

(iv) The mean waiting time is given by

9 1
E(W ) = · 12 + · 4 = 7 minutes.
16 16

Alternatively, from a mean value analysis we have

ρ 5 21
E(W ) = E(R) = · = 7 minutes.
1−ρ 3 5

Exercise 44

160

Exercise 45. The time unit is 1 hour: λ = 1, E(B)=7/12, ρ = 7/12, E(R) = (3/7) ·
(7/12) + (4/7) · (1/3) = 37/84.

(i) The Laplace-Stieltjes transform of the processing time is given by

4 3
B(s) = · .
4+s 3+s

From (7.7) we have

(1 − ρ)B(s)s 5 5 1 1 5
S(s) = = = · − · .
λB(s) + s − λ (1 + s)(5 + s) 4 1+s 4 5+s

(ii) The distribution function of the production lead time is given by

5 1
FS (x) = (1 − e−x ) − (1 − e−5x ).
4 4

The mean production lead time is given by

5 1 1 6
E(S) = · 1 − · = hours.
4 4 5 5


ρ 7 37 7 6
E(S) = E(R) + E(B) = · + = hours.
1−ρ 5 84 12 5

(iii) The mean cost per hour equals

5 −3 1 −15
λ · (1 − FS (3)) · 100 = ·e − ·e · 100 ≈ 6.22 dollar.
4 4

Exercise 45

161

Exercise 46. The time unit is 1 minute: λ = 1/10, E(B)=25/4, ρ = 5/8, E(R) =
(2/5) · 10 + (3/5) · 5 = 7.

(i) The pick time is hyperexponentially distributed with parameters p1 = 1/4, p2 = 3/4,
µ1 = 1/10 and µ2 = 1/5.

(ii) From (7.7) we have

(1 − ρ)B(s)s 12 + 105s 5 3 27 1
S(s) = = = · + · .
λB(s) + s − λ 4(3 + 20s)(1 + 20s) 32 3 + 20s 32 1 + 20s

(iii) The distribution function of the sojourn time is given by

5 3 27 1
FS (x) = (1 − e− 20 x ) + (1 − e− 20 x ).
32 32

Hence, the fraction of orders for which the lead time is longer than half an hour is
given by

5 9 27 − 3
1 − FS (30) = · e− 2 + · e 2 ≈ 0.190.
32 32

(iv) The mean lead time is given by

5 20 27 215
E(S) = · + · 20 = minutes.
32 3 32 12


ρ 5 25 215
E(S) = E(R) + E(B) = · 7 + = minutes.
1−ρ 3 4 12

Exercise 46

162

Exercise 51. As time unit we choose 1 minute: µ = 1. The Laplace-Stieltjes transform
of the interarrival time distribution is given by
1 1 2 1
A(s) = · + · .
3 1 + s 3 1 + 3s

(i) an = (1 − σ) σ n , where σ, the solution in (0,1) of σ = A(µ − µσ), is given by
√
21 − 153
σ= ≈ 0.48.
18
σ
(ii) E(La ) = 1−σ
≈ 0.92.
1−σ
(iii) S(s) = 1−σ+s
.
1
(iv) E(S) = 1−σ
≈ 1.92.
3
(v) E(L) = λ · E(S) = 7
· E(S) ≈ 0.82.

Exercise 51

163

Exercise 52. We have µ = 6 and the Laplace-Stieltjes transform of the interarrival time
distribution is given by
13 3 11 2
A(s) = · + · .
24 3 + s 24 2 + s
5
(i) an = (1 − σ) σ n , where σ, the solution in (0,1) of σ = A(µ − µσ), is given by σ = 12
.
7
(ii) FW (t) = 1 − σ e−µ(1−σ)t = 1 − 5
12
e− 2 t .

Exercise 52

164

Exercise 53. The sojourn time is exponentially distributed with parameter µ(1 − σ),
where µ = 1 and
√
7 − 17
σ= ≈ 0.36.
8
Exercise 53

165

Exercise 54.

(i) The solution in (0,1) of σ = e−2(1−σ) , is given by σ ≈ 0.203.

(ii) FS (t) = 1 − e−µ(1−σ)t ≈ 1 − e−(0.4)·t .

Exercise 54

166

Exercise 55.

(i) 3/8.

(ii) Let pn denote the probability that there are n cars waiting for the ferry. Then,

1 3 1 2 7
p0 = + · = ,
4 4 2 16
3 1 1 3 1 3 15
p1 = · + · = ,
4 2 4 2 32
3 1 n+2
pn = · , n ≥ 2.
4 2

∞
(iii) E(Lq ) = n=0 npn = 3 , and hence using Little’s formula we have E(W ) = 3 minutes.
4

Exercise 55

167

Exercise 56. As time unit we choose 1 minute:

λ = 1/6, E(B) = 9/2, ρ = 3/4 and E(R) = 5,

λ1 = 1/12, E(B1 ) = 3, ρ1 = 1/4 and E(R1 ) = 3,

λ2 = 1/12, E(B2 ) = 6, ρ2 = 1/2 and E(R2 ) = 6.
ρ
(i) E(W ) = 1−ρ
E(R) = 15 minutes.

(ii) Use formula (9.3) on page 89:

ρ1 E(R1 ) + ρ2 E(R2 )
E(W1 ) = = 5 minutes ,
1 − ρ1

ρ1 E(R1 ) + ρ2 E(R2 )
E(W2 ) = = 20 minutes ,
(1 − ρ1 )(1 − ρ1 − ρ2 )

1 1
E(W ) = E(W1 ) + E(W2 ) = 12.5 minutes .
2 2

(iii) Similar to formula (9.3), we now have

ρ1 E(R1 ) + ρ2 E(R2 ) 15
E(W2 ) = = = 7.5 minutes ,
1 − ρ2 2

ρ1 E(R1 ) + ρ2 E(R2 )
E(W1 ) = = 30 minutes ,
(1 − ρ2 )(1 − ρ1 − ρ2 )

1 1
E(W ) = E(W1 ) + E(W2 ) = 18.75 minutes .
2 2

Exercise 56

168


λ1 = 1/60, E(B1 ) = 10, ρ1 = 1/6 and E(R1 ) = 5,

λ2 = 1/30, E(B2 ) = 10, ρ2 = 1/3 and E(R2 ) = 5,

λ3 = 1/30, E(B3 ) = 10, ρ3 = 1/3 and E(R3 ) = 5.

Now, use formula (9.5) for E(Si ) on page 90:
ρ1 E(R1 )
E(S1 ) = + E(B1 ) = 11 minutes ,
1 − ρ1

ρ1 E(R1 ) + ρ2 E(R2 ) E(B2 )
E(S2 ) = + = 18 minutes ,
(1 − ρ1 )(1 − ρ1 − ρ2 ) 1 − ρ1

ρ1 E(R1 ) + ρ2 E(R2 ) + ρ3 E(R3 ) E(B3 )
E(S3 ) = + = 70 minutes ,
(1 − ρ1 − ρ2 )(1 − ρ1 − ρ2 − ρ3 ) 1 − ρ1 − ρ2

Exercise 57

169


λ = 1/15, E(B) = 55/4, ρ = 11/12 and E(R) = 15/2,

λ1 = 1/30, E(B1 ) = 10, ρ1 = 1/3 and E(R1 ) = 5,

λ2 = 1/60, E(B2 ) = 15, ρ2 = 1/4 and E(R2 ) = 15/2,

λ3 = 1/60, E(B3 ) = 20, ρ3 = 1/3 and E(R3 ) = 10.
ρ
(i) E(W1 ) = E(W2 ) = E(W3 ) = E(W ) = 1−ρ E(R) = 165/2 = 82.5 minutes. Hence,
E(S1 ) = 92.5 minutes, E(S2 ) = 97.5 minutes, E(S3 ) = 102.5 minutes and E(S) =
96.25 minutes.

(ii) Use formula (9.4) for E(Si ) on page 89:

ρ1 E(R1 ) + ρ2 E(R2 ) + ρ3 E(R3 ) 325
E(S1 ) = + E(B1 ) = ≈ 20.31 minutes ,
1 − ρ1 16

ρ1 E(R1 ) + ρ2 E(R2 ) + ρ3 E(R3 ) 159
E(S2 ) = + E(B2 ) = = 39.75 minutes ,
(1 − ρ1 )(1 − ρ1 − ρ2 ) 4

ρ1 E(R1 ) + ρ2 E(R2 ) + ρ3 E(R3 )
E(S3 ) = + E(B3 ) = 218 minutes ,
(1 − ρ1 − ρ2 )(1 − ρ1 − ρ2 − ρ3 )

1 1 1
E(S) = E(S1 ) + E(S2 ) + E(S3 ) ≈ 74.59 minutes .
2 4 4

(iii) Combine the arguments of Sections 9.1 and 9.2:

ρ1 E(R1 ) + ρ2 E(R2 ) 245
E(S1 ) = + E(B1 ) = ≈ 15.31 minutes ,
1 − ρ1 16

ρ1 E(R1 ) + ρ2 E(R2 ) 111
E(S2 ) = + E(B2 ) = = 27.75 minutes ,
(1 − ρ1 )(1 − ρ1 − ρ2 ) 4

ρ1 E(R1 ) + ρ2 E(R2 ) + ρ3 E(R3 ) E(B3 )
E(S3 ) = + = 246 minute s ,
(1 − ρ1 − ρ2 )(1 − ρ1 − ρ2 − ρ3 ) 1 − ρ1 − ρ2

1 1 1
E(S) = E(S1 ) + E(S2 ) + E(S3 ) ≈ 76.09 minutes .
2 4 4

Exercise 58

170

Exercise 59. As time unit we choose 1 minute: λ = 1/6.

(i) For N , the number of parts that has to be produced for an order, we have
n
1
P (N = n) = , n = 1, 2, 3, . . .
2

and hence E(N ) = 2 and σ 2 (N ) = 2 (see also Section 2.4.1). From B = 2N , it now
follows that E(B) = 4 and σ 2 (B) = 8.

(ii) Using that ρ = 2/3 and E(R) = 3, we have

ρ
E(S) = E(R) + E(B) = 10 minutes .
1−ρ

(iii) We now have

λ1 = 1/12, E(B1 ) = 2, ρ1 = 1/6 and E(R1 ) = 1,

λ2 = 1/12, E(B2 ) = 6, ρ2 = 1/2 and E(R2 ) = 11/3.

Hence,

ρ1 E(R1 ) + ρ2 E(R2 ) 22
E(S1 ) = + E(B1 ) = = 4.4 minutes ,
1 − ρ1 5

ρ1 E(R1 ) + ρ2 E(R2 ) 66
E(S2 ) = + E(B2 ) = = 13.2 minutes ,
(1 − ρ1 )(1 − ρ1 − ρ2 ) 5

1
(iv) E(S) = 2 E(S1 ) + 1 E(S2 ) =
2
44
5
= 8.8 minutes.

Exercise 59

171

Exercise 63.
ρ T T
E(S) = · E(RB ) + 1 −λT · + E(B).
1−ρ T + λe 2
Exercise 63

172

Exercise 64.

As time unit we choose 1 second:

λ = 1/6.

(i) The mean waiting time satisﬁes

E(W ) = 2.5 + E(Lq ) · 5 .

Together with Little’s formula, E(Lq ) = λE(W ), this yields E(W ) = 15 seconds.

(ii) The time elapsing from entering the carrier till the departure of that bin is 4 cycles
(= 4 · 5 = 20 seconds) plus moving out of the carrier (= 2 seconds), so 22 seconds.
Hence, the mean sojourn time is equal to 15 + 22 = 37 seconds.

Exercise 64

173

Exercise 65.

(i) The mean number of orders in the system is given by

ρ N −1
E(L) = + .
1−ρ 2

(ii) From Little’s formula we obtain

1/µ N −1
E(S) = + .
1−ρ 2λ

(iii) The average cost (setup cost + machine cost + waiting cost) per minute equals

6 3 6 21 3N
+ 8 + (4 + (N − 1)) = + + .
N 2 N 2 2

(iv) N = 2.

Exercise 65

174

Exercise 66. See exercise 4 of the exam of June 21, 1999. Exercise 66

175

Exercise 69. As time unit we choose 1 hour:

λ = 1, E(B) = 1/2, ρ = 1/2 and E(RB ) = 1/2.

(i) The fraction of time that the machine processes orders is 1/2. The mean duration of
a period that the machine is switched oﬀ equals 1, the mean duration of a switch-on
period equals T . Hence, the mean duration of a period that the machine processes
orders equals 1+T . Hence, both the mean number of orders processed in a production
cycle and the mean duration of a production cycle equals 2 + 2T . The mean waiting
time of an order equals

1 T T 1+T
E(W ) = E(Lq ) · 1/2 + ·T + · + · 1/2.
2 + 2T 2 + 2T 2 2 + 2T

Together, with Little’s formula E(Lq ) = 1 · E(W ) this gives

T 2 + 3T + 1
E(W ) = .
2 + 2T

Hence, the mean production lead time of an order equals

T 2 + 4T + 2
E(S) = .
2 + 2T

(ii) The average cost per hour equals

17 T 2 + 3T + 1 T 2 + 3T + 18
+ =
2 + 2T 2 + 2T 2 + 2T

which is minimal for T = 3.

Exercise 69

176

Exercise 71.

(i) E(S) = 105/2 = 52.5 minutes .

(ii) E(L) = 21/6.

Exercise 71

177


λ = 1/10, E(B) = 15/2, ρ = 3/4 and E(RB ) = 35/9.

(i) The fraction of time that the server serves customers is 3/4. The mean duration of
a period that the server is away equals 10 + 10 + 5 = 25 minutes. Hence, the mean
duration of a busy period equals 75 minutes.

(ii) 75/7.5 = 10 customers.

(iii) The mean waiting time of a customer equals

E(W ) = E(Lq ) · 15/2 + 1/10 · 5 + 1/20 · 5/2 + 3/4 · 35/9.

Together, with Little’s formula E(Lq ) = 1/10 · E(W ) this gives E(W ) = 121/6 =
20.17 minutes. Hence, the mean sojourn time of a customer equals E(S) = 27.67
minutes.

Exercise 73

178

Exercise 77.

(i) For the probability that i terminals are occupied we have

3i
i! 8 3i
pi = 4 3n = .
n=0 n! 131 i!

Hence,

8 24 36 36 27
(p0 , p1 , p2 , p3 , p4 ) = , , , , .
131 131 131 131 131
27
(ii) B(4, 3) = p4 = 131
= 0.2061.

(iii) Use the recursion (11.3):

B(4, 3) = 0.2061, B(5, 3) = 0.11005, B(6, 3) = 0.05215, B(7, 3) = 0.0219.

So, we need at least 7 terminals.

Exercise 77

179

Exercise 79.

(i) B(6, 7.5) = 0.3615.

(ii) The mean profit per day equals

5 · 110 · 1.5 · (1 − B(6, 7.5)) − 6 · 60 = 166.7 guilders .

(iii) When the company has c cars, the mean profit per day equals

5 · 110 · 1.5 · (1 − B(c, 7.5)) − c · 60 guilders .

So, if the company buys 1 extra car, the mean profit becomes 174.7 guilders, if the
company buys 2 extra cars, it becomes 173.8 guilders, if the company buys 3 extra
cars, it becomes 163.4 guilders, and so on. The mean profit per day is maximized
when the company buys 1 extra car.

Exercise 79

180

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