# On queues with service and interarrival times depending on waiting times

**ABSTRACT** We consider an extension of the standard G/G/1 queue, described by the equation

W= Dmaxmax{0,B-A+YW}W\stackrel{ \mathcal {D}}{=}\max\mathrm{max}\,\{0,B-A+YW\}

, where ℙ[Y=1]=p and ℙ[Y=−1]=1−p. For p=1 this model reduces to the classical Lindley equation for the waiting time in the G/G/1 queue, whereas for p=0 it describes the waiting time of the server in an alternating service model. For all other values of p, this model describes a FCFS queue in which the service times and interarrival times depend linearly and randomly on the

waiting times. We derive the distribution of W when A is generally distributed and B follows a phase-type distribution, and when A is exponentially distributed and B deterministic.

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**ABSTRACT:**In a number of service systems, there can be substantial latitude to vary service rates. However, although speeding up service rate during periods of congestion may address a present congestion issue, it may actually exacerbate the problem by increasing the need for rework. We introduce a state-dependent queuing network where service times and return probabilities depend on the “overloaded” and “underloaded” state of the system. We use a fluid model to examine how different definitions of “overload” affect the long-term behavior of the system and provide insight into the impact of using speedup. We identify scenarios where speedup can be helpful to temporarily alleviate congestion and increase access to service. For such scenarios, we provide approximations for the likelihood of speedup to service. We also identify scenarios where speedup should never be used; moreover, in such a situation, an interesting bi-stability arises, such that the system shifts randomly between two equilibria states. Hence, our analysis sheds light on the potential benefits and pitfalls of using speedup when the subsequent returns may be unavoidable.Operations Research 04/2014; 62(2). · 1.50 Impact Factor - [Show abstract] [Hide abstract]

**ABSTRACT:**The paper studies a general multiserver queue in which the service time of an arriving customer and the next interarrival period may depend on both the current waiting time and the server assigned to the arriving customer. Stability of the system is proved under general assumptions on the predetermined distributions describing the model. The proof exploits a combination of the Markov property of the workload process with a regenerative property of the process. The key idea leading to stability is a characterization of the limit behavior of the forward renewal process generated by regenerations. Extensions of the basic model are also studied. This model generalizes a number of the known state-dependent systems and describes some new systems as particular cases.Journal of Mathematical Sciences 07/2014; 200(4):462-472. - [Show abstract] [Hide abstract]

**ABSTRACT:**We consider a two-station cascade network, where the first station has Poisson input and the second station has renewal input, with i.i.d. service times at both stations. The following partial interaction exists between stations: whenever the second station becomes empty while customers are awaiting service at the first one, one customer can jump to the second station to be served there immediately. However, the first station cannot assist the second one in the opposite case. For this system, we establish necessary and sufficient stability conditions of the basic workload process, using a regenerative method. An extension of the basic model, including a multiserver first station, a different service time distribution for customers jumping from station 1 to station 2, and an arbitrary threshold d 1≥1 on the queue-size at station 1 allowing jumps to station 2, are also treated.Annals of Operations Research 01/2013; 202(1). · 1.10 Impact Factor

Page 1

On queues with service and interarrival times depending on

waiting times

O.J. Boxma?,??, M. Vlasiou???, §

November 26, 2006

?EURANDOM,

P.O. Box 513, 5600 MB Eindhoven, The Netherlands.

??Eindhoven University of Technology,

Department of Mathematics & Computer Science,

P.O. Box 513, 5600 MB Eindhoven, The Netherlands.

???Georgia Institute of Technology,

H. Milton Stewart School of Industrial & Systems Engineering,

765 Ferst Drive, Atlanta GA 30332-0205, USA.

boxma@win.tue.nl, vlasiou@gatech.edu

Abstract

We consider an extension of the standard G/G/1 queue, described by the equation

D= max{0,B − A + Y W}, where P[Y = 1] = p and P[Y = −1] = 1 − p. For p = 1 this

model reduces to the classical Lindley equation for the waiting time in the G/G/1 queue,

whereas for p = 0 it describes the waiting time of the server in an alternating service model.

For all other values of p this model describes a FCFS queue in which the service times

and interarrival times depend linearly and randomly on the waiting times. We derive the

distribution of W when A is generally distributed and B follows a phase-type distribution,

and when A is exponentially distributed and B deterministic.

W

1Introduction

One of the most fundamental relations in queuing and random walk theory is Lindley’s recursion

[11]:

Wn+1= max{0,Bn− An+ Wn}.

In queuing theory, it represents a relation between the waiting times of the n-th and (n + 1)-st

customer in a single server queue, An indicating the interarrival time between the n-th and

(n + 1)-st customer and Bn denoting the service time of the n-th customer. In the applied

probability literature there has been a considerable amount of interest in generalisations of

Lindley’s recursion, namely the class of Markov chains described by the recursion Wn+1 =

g(Wn,Xn). For earlier work on such stochastic recursions see for example Brandt et al. [5]

and Borovkov and Foss [3]. Many structural properties of this recursion have been derived.

(1.1)

§This research has been carried out when this author was affiliated with EURANDOM, The Netherlands.

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For example, Asmussen and Sigman [2] develop a duality theory, relating the steady-state

distribution to a ruin probability associated with a risk process. More references in this domain

can be found for example in Asmussen and Schock Petersen [1] and Seal [16]. An important

assumption which is often made in these studies is that the function g(w,x) is non-decreasing

in its main argument w. For example, in [2] this assumption is crucial for their duality theory

to hold.

In this paper we consider a generalisation of Lindley’s recursion, for which the monotonicity

assumption does not hold. In particular, we study the Lindley-type recursion

Wn+1= max{0,Bn− An+ YnWn},

(1.2)

where for every n, the random variable Yn is equal to plus or minus one according to the

probabilities P[Yn= 1] = p and P[Yn= −1] = 1 − p, 0 ? p ? 1. The sequences {An} and {Bn}

are assumed to be independent sequences of i.i.d. non-negative random variables. Our main

goal is to derive the steady-state distribution of {Wn, n = 1,2,...}, when it exists.

Equation (1.2) reduces to the classical Lindley recursion [11] when P[Yn= 1] = 1 for every

n. Furthermore, if P[Yn= −1] = 1, then (1.2) describes the waiting time of the server in an

alternating service model with two service points. For a description of the model for this special

case and related results see Park et al. [13] and Vlasiou et al. [20, 21, 22, 23].

Studying a recursion that contains both Lindley’s classical recursion and the recursion in

[13, 20, 21, 22, 23] as special cases seems of interest in its own right. Additional motivation for

studying the recursion is supplied by the fact that, for 0 < p < 1, the resulting model can be

interpreted as a special case of a queuing model in which service and interarrival times depend

on waiting times. We shall now discuss the latter model.

Consider an extension of the standard G/G/1 queue in which the service times and the

interarrival times depend linearly and randomly on the waiting times. Namely, the model is

specified by a stationary and ergodic sequence of four-tuples of nonnegative random variables

{(An,Bn,? An,?Bn)}, n ? 0. The sequence {Wn} is defined recursively by

Wn+1= max{0,Bn− An+ Wn},

where

An= An+? AnWn,

Bn= Bn+?BnWn.

We interpret Wnas the waiting time and Bnas the service time of customer n. Furthermore,

we take Anto be the interarrival time between customers n and n+1. We call Bnthe nominal

service time of customer n and An the nominal interarrival time between customers n and

n + 1, because these would be the actual times if the additional shift were omitted, that is, if

P[? An=?Bn= 0] = 1.

written Yn= 1 +?Bn−? An. This model – for generally distributed random variables Yn– has

proper steady-state limit, and on approximations for this limit. There are very few exact results

known for queuing models in which interarrival and/or service times depend on waiting times;

we refer to Whitt [24] for some references.

Whitt [24] builds upon previous results by Vervaat [19] and Brandt [4] for the unrestricted

recursion Wn+1= YnWn+ Xn, where Xn= Bn− An. There has been considerable previous

work on this unrestricted recursion, due to its close connection to the problem of the ruin of

Evidently, the waiting times satisfy the generalised Lindley recursion (1.2), where we have

been introduced in Whitt [24], where the focus is on conditions for the process to converge to a

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an insurer who is exposed to a stochastic economic environment. Such an environment has two

kinds of risk, which were called by Norberg [12] insurance risk and financial risk. Indicatively, we

mention the work by Tang and Tsitsiashvili [17], and by Kalashnikov and Norberg [10]. In the

more general framework, Wnmay represent an inventory in time period n (e.g. cash), Ynmay

represent a multiplicative, possibly random, decay or growth factor between times n and n + 1

(e.g. interest rate) and Bn− Anmay represent a quantity that is added or subtracted between

times n and n + 1 (e.g. deposit minus withdrawal). Obviously, the positive-part operator is

appropriate for many applications [24].

This paper presents an exact analysis of the steady-state distribution of {Wn, n = 1,2,...}

as given by (1.2) with P[Yn= 1] = p and P[Yn= −1] = 1 − p. For 0 < p < 1, this amounts

to analysing the above-described G/G/1 extension where? An =?Bn with probability p, and

general, is connected to LaPalice queuing models, introduced by Jacquet [9], where customers

are scheduled in such a way that the period between two consecutively scheduled customers is

greater than or equal to the service time of the first customer.

This paper is organised in the following way. In Section 2 we comment on the stability of the

process {Wn}, as it is defined by Recursion (1.2). In the remainder of the paper it is assumed

that the steady-state distribution of {Wn} exists. Section 3 is devoted to the determination

of the distribution of W when A is generally distributed and B has a phase-type distribution.

In Section 4 we determine the distribution of W when A is exponentially distributed and B

is deterministic. At the end of each section we compare the results that we derive to the

already known results for Lindley’s recursion (i.e. for p = 1) and to the equivalent results for

the Lindley-type recursion arising for p = 0.

At the end of this introduction we mention a few notational conventions. For a random

variable X we denote its distribution by FX and its density by fX. Furthermore, we shall

denote by f(i)the i-th derivative of the function f. The Laplace-Stieltjes transforms (LST) of

A and W are respectively denoted by α and ω. To keep expressions simple, we also use the

function φ defined as φ(s) = ω(s)α(s).

? An= 2 +?Bnwith probability 1 − p. This problem, and state-dependent queuing processes in

2Stability

The following result on the convergence of the process {Wn} to a proper limit W is shown in

Whitt [24]. It is included here only for completeness.

From Recursion (1.2), it is obvious that if we replace Ynby max{0,Yn} and Bn− Anby

max{0,Bn− An}, then the resulting waiting times will be at least as large as the ones given

by (1.2). Moreover, when we make this change, the positive-part operator is not necessary

anymore.

Lemma 1 (Whitt [24, Lemma 1]). If Wnsatisfies (1.2), then with probability 1, Wn? Znfor

all n, where

Zn+1= max{0,Yn}Zn+ max{0,Bn− An},

and Z0= W0? 0.

n ? 0,

(2.1)

So if Wnsatisfies (1.2), Znsatisfies (2.1), and Znconverges to the proper limit Z, then {Wn}

is tight and P[W > x] ? P[Z > x] for all x, where W is the limit in distribution of any convergent

subsequence of {Wn}. This observation, combined with Theorem 1 of Brandt [4], which implies

that Znsatisfying (2.1) converges to a proper limit if P[max{0,Yn} = 0] = P[Yn? 0] > 0, leads

to the following theorem.

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Page 4

Theorem 1 (Whitt [24, Theorem 1]). The series {Wn} is tight for all ρ = E[B0]/E[A0] and

W0. If, in addition, 0 ? p < 1 and {(Yn,Bn− An)} is a sequence of independent vectors with

P[Y0? 0,B0− A0? 0] > 0,

then the events {Wn= 0} are regeneration points with finite mean time and {Wn} converges in

distribution to a proper limit W as n → ∞ for all ρ and W0.

Naturally, for p = 1, i.e. for the classical Lindley recursion, we need the additional condition

that ρ < 1.

Therefore, assume that the sequences?Bn−? An and Bn− An are independent stationary

negative. Then the conditions of Theorem 1 hold, so there exists a proper limit W, and for the

system in steady-state we write

sequences, that are also independent of one another, and that for all n, Anand Bnare non-

W

D= max{0,B − A + Y W},

(2.2)

where “D=” denotes equality in distribution, where A, B are generic random variables distributed

like An, Bn, and where P[Y = 1] = p and P[Y = −1] = 1 − p.

Remark 1. For x ? 0 Equation (2.2) yields that

FW(x) = P[W ? x] = pP[X + W ? x] + (1 − p)P[X − W ? x],

where X = B −A (note that P[X < 0] > 0). Assuming that the distribution FXof the random

variable X is continuous, the last term is equal to 1 − P[X − W ? x], which gives us that

?x

This means that the limiting distribution of W, provided that FXis continuous, satisfies the

functional equation

?x

Therefore, there exists at least one function that is a solution to (2.3). It can be shown that

in fact there exists a unique measurable bounded function F : [0,∞) → R that satisfies this

functional equation.

To show this, consider the space L∞([0,∞)), i.e. the space of measurable and bounded

functions on the real line with the norm

FW(x) = p

−∞

FW(x − y)dFX(y) + (1 − p)

?

1 −

?∞

x

FW(y − x)dFX(y)

?

.

F(x) = p

−∞

F(x − y)dFX(y) + (1 − p)

?

1 −

?∞

x

F(y − x)dFX(y)

?

.

(2.3)

?F? = sup

t?0|F(t)|.

In this space define the mapping

(T F)(x) = p

?x

−∞

F(x − y)dFX(y) + (1 − p)

?

1 −

?∞

x

F(y − x)dFX(y)

?

.

4

Page 5

Note that T F : L∞?[0,∞)?

?(T F1) − (T F2)? = sup

????p

? sup

x?0

−∞

? sup

x?0

−∞

= ?F1− F2? sup

→ L∞?[0,∞)?, i.e., T F is measurable and bounded. For two

arbitrary functions F1and F2in this space we have

x?0|(T F1)(x) − (T F2)(x)|

= sup

x?0

?x

?x

?x

−∞

[F1(x − y) − F2(x − y)] dFX(y) + (1 − p)

?∞

?∞

sup

x

[F2(y − x) − F1(y − x)] dFX(y)

????

?

?

p

|F1(x − y) − F2(x − y)|dFX(y) + (1 − p)

x

|F2(y − x) − F1(y − x)|dFX(y)

?

p

sup

t?0|F1(t) − F2(t)|dFX(y) + (1 − p)

?pFX(x) + (1 − p)?1 − FX(x)??.

Note that the supremum appearing above is less than or equal to max{p,1−p} for p ∈ (0,1) and

equal to 1−FX(0) for p = 0. Therefore, since for p ?= 1 it holds that supx?0

FX(x)??< 1, for these values of the parameter p we have a contraction mapping. Furthermore,

that (2.3) has a unique solution; for a related discussion, see also [20].

In the next two sections we determine the distribution of W for two cases in which {An}

and {Bn} are independent and i.i.d. sequences of non-negative random variables.

?∞

x

t?0|F2(t) − F1(t)|dFX(y)

?

x?0

?pFX(x)+(1−p)?1−

we know that L∞([0,∞)) is a Banach space, therefore by the Fixed Point Theorem we have

3The GI/PH case

In this section we assume that A is generally distributed, while B follows a particular phase-

type distribution. Specifically, we assume that with probability κnthe nominal service time B

follows an Erlang distribution with parameter µ and n phases, i.e.,

FB(x) =

N

?

?µ/(µ + s)?n. These distributions, i.e. mixtures of Erlang distributions, are

since it may be used to approximate any given continuous distribution on [0,∞) arbitrarily close;

see Schassberger [15]. Following the proof in [22], one can show that for such an approximation

of FB, the error in the resulting waiting time approximation can be bounded.

We are interested in the distribution of W. In order to derive the distribution of W, we

shall first derive the LST of FW. We follow a method based on Wiener-Hopf decomposition. A

straightforward calculation yields for values of s such that Re(s) = 0:

n=1

κn

?

1 − e−µx

n−1

?

j=0

(µx)j

j!

?

=

N

?

n=1

κn

∞

?

j=n

e−µx(µx)j

j!

,x ? 0,

(3.1)

with LST?N

n=1κn

special cases of Coxian or phase-type distributions. It is sufficient to consider only this class,

ω(s) = E[e−sW] = pE[e−smax{0,B−A+W}] + (1 − p)E[e−smax{0,B−A−W}]

= pP[W + B ? A] + pE[e−s(B−A+W)] − pE[e−s(B−A+W);W + B ? A]+

+ (1 − p)P[B ? W + A] + (1 − p)E[e−s(B−A−W);B ? W + A];(3.2)

here A, B and W are independent random variables. The Lindley-type equation W

A−W} for A generally distributed and B phase-type has already been analysed in Vlasiou and

Adan [21], and the LST of the corresponding W is given there. From Equation (3.8) of [21]

D= max{0,B−

5

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we can readily copy an expression for the last two terms appearing in (3.2), so ω can now be

written as

ω(s) = pP[W + B ? A] + pα(−s)ω(s)

?

N

?

n=1

κn

?

?

µ

µ + s

?n

− pE[e−s(B−A+W);W + B ? A]+

?n−i??

+ (1 − p)1 −

N

?

n=1

n−1

?

i=0

κn(−µ)i

i!

φ(i)(µ)1 −

?

µ

µ + s

.

So for Re(s) = 0 we have that

ω(s)

?

1 − pα(−s)

N

?

n=1

κn

?

µ

µ + s

?n?

= pP[W + B ? A] − pE[e−s(B−A+W);W + B ? A]+

?

n=1

i=0

+ (1 − p)1 −

N

?

n−1

?

κn(−µ)i

i!

φ(i)(µ)

?

1 −

?

µ

µ + s

?n−i??

.

(3.3)

Cohen [6, p. 322–323] shows by applying Rouch´ e’s theorem that the function

1 − pα(−s)

N

?

n=1

κn

?

µ

µ + s

?n

≡

1

(µ + s)N

?

(µ + s)N− pα(−s)

N

?

n=1

κnµn(µ + s)N−n

?

has exactly N zeros ξi(p) in the left-half plane if 0 < p < 1 (it is assumed that α(µ) ?= 0, which

is not an essential restriction) or if p = 1 and E[B] < E[A]. Naturally, this statement is not

valid if p = 0; therefore, this case needs to be excluded from this point on. So we rewrite (3.3)

as follows

ω(s)

N

?

i=1

?s − ξi(p)?=

×

?N

i=1

?s − ξi(p)?

(µ + s)N− pα(−s)?N

p(µ + s)NP[W + B ? A] − p(µ + s)NE[e−s(B−A+W);W + B ? A]+

?

n=1

i=0

n=1κnµn(µ + s)N−n×

?

+ (1 − p)(µ + s)N−

N

?

n−1

?

κn(−µ)i

i!

φ(i)(µ)?(µ + s)N− µn−i(µ + s)N−n+i???

.

(3.4)

The left-hand side of (3.4) is analytic for Re(s) > 0 and continuous for Re(s) ? 0, and the

right-hand side of (3.4) is analytic for Re(s) < 0 and continuous for Re(s) ? 0. So from

Liouville’s theorem [18] we have that both sides of (3.4) are the same N-th degree polynomial,

say,?N

ω(s) =

?N

In the expression above, the constants qiare not determined so far, while the roots ξi(p) are

known. In order to obtain the transform, observe that ω is a fraction of two polynomials of

degree N. So, ignoring the special case of multiple zeros ξi(p), partial fraction decomposition

yields that (3.5) can be rewritten as

i=0qisi. Hence,

?N

i=1

i=0qisi

?s − ξi(p)?.

(3.5)

ω(s) = c0+

N

?

i=1

ci

?s − ξi(p)?,

(3.6)

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Page 7

which implies that the waiting time distribution has a mass at the origin that is given by

P[W = 0] = lim

s→∞E[e−sW] = c0

and has a density that is given by

fW(x) =

N

?

i=1

cieξi(p)x.

All that remains is to determine the N + 1 constants ci. To do so, we work as follows.

We shall substitute (3.6) in the left-hand side of (3.4), and express the terms P[W +B ? A]

and E[e−s(B−A+W);W + B ? A] that appear at the right-hand side of (3.4) in terms of the

constants ci. Note that the terms φ(i)(µ) that appear at the right-hand side of (3.4) can also be

expressed in terms of the constants ci. Thus we obtain a new equation that we shall differentiate

a total of N times. We shall evaluate each of these derivatives for s = 0 and thus we obtain

a linear system of N equations for the constants ci, i = 0,...,N. The last equation that is

necessary to uniquely determine the constants ciis the normalisation equation

?∞

To begin with, note that

c0+

0

fW(x)dx = 1.

(3.7)

P[W + B ? A] = P[W = 0]P[B ? A] +

?∞

0

P[B ? A − x]

N

?

i=1

cieξi(p)xdx,

(3.8)

with

P[B ? A] =

?∞

0

N

?

n=1

κn

?

e−µx

∞

?

j=n

(µx)j

j!

?

dFA(x) =

N

?

n=1

∞

?

i=n

κn(−µ)i

i!

α(i)(µ),

(3.9)

and

?∞

0

P[B ? A − x]

N

?

i=1

cieξi(p)xdx =

?∞

0

?∞

N

?

0

P[B ? y − x]

?µ(y − x)?j

µj?µ + ξi(p)?k−j−1

N

?

i=1

cieξi(p)xdxdFA(y)

=

?∞

N

?

0

?y

∞

?

0

e−µ(y−x)

n=1

∞

?

j=n

κn

j!

N

?

i=1

cieξi(p)xdxdFA(y)

=

n=1

j=n

N

?

i=1

∞

?

k=j+1

κnci

k!(−1)k

α(k)(µ).

(3.10)

Likewise, we have that

E[e−s(B−A+W);W + B ? A] = P[W = 0]E[e−s(B−A);B ? A]+

+

?∞

0

E[e−s(B−A+x);x + B ? A]

N

?

i=1

cieξi(p)xdx,

(3.11)

7

Page 8

with

E[e−s(B−A);B ? A] =

?∞

?∞

N

?

0

?x

0

e−s(y−x)

N

?

µ

µ + s

n=1

κnµe−µy(µy)n−1

(n − 1)!dydFA(x)

e−x(µ+s)xi(µ + s)i

=

0

exs

N

?

κnµn(−1)i

n=1

κn

?

?n ∞

?

i=n

i!

dFA(x)

=

n=1

∞

?

i=n

i!

(µ + s)i−nα(i)(µ),

(3.12)

and

?∞

0

E[e−s(B−A+x);x + B ? A]

N

?

i=1

cieξi(p)xdx

=

?∞

?∞

N

?

0

?y

?y

∞

?

0

?y−x

N

?

N

?

0

e−s(z−y+x)

N

?

n=1

κnµe−µz(µz)n−1

(n − 1)!

e−(µ+s)(y−x)(µ + s)j(y − x)j

N

?

i=1

cieξi(p)xdzdxdFA(y)

=

00

n=1

κn

?

µ

µ + s

?n

?

e−s(x−y)

∞

?

j=n

j!

N

?

i=1

cieξi(p)xdxdFA(y)

=

n=1

j=ni=1

∞

?

k=j+1

κnci

µ

µ + s

?n(µ + s)j?µ + ξi(p)?k−j−1

k!(−1)k

α(k)(µ).

(3.13)

So, using (3.9) and (3.10), substitute (3.8) in the right-hand side of (3.4), and similarly for

(3.11). Furthermore, as mentioned before, substitute (3.6) into the left-hand side of (3.4) to

obtain an expression, where both sides can be reduced to an N-th degree polynomial in s. By

evaluating this polynomial and all its derivatives for s = 0 we obtain N equations binding the

constants ci. These equations, and the normalisation equation (3.7), form a linear system for

the constants ci, i = 0,...,N, that uniquely determines them (see also Remark 2 below). For

example, the first equation, evaluated at s = 0, yields that

c0−

N

?

i=1

ci

ξi(p)=

1 − p

1 − pα(0)= 1,

since α(0) = 1. We summarise the above in the following theorem.

Theorem 2. Consider the recursion given by (1.2), and assume that 0 < p < 1. Let (3.1) be

the distribution of the random variable B. Then the limiting distribution of the waiting time

has mass c0at the origin and a density on [0,∞) that is given by

N

?

In the above equation, the constants ξi(p), with Re(ξi(p)) < 0, are the N roots of

fW(x) =

i=1

cieξi(p)x.

(µ + s)N− pα(−s)

N

?

n=1

κnµn(µ + s)N−n= 0,

and the N + 1 constants ciare the unique solution to the linear system described above.

8

Page 9

Remark 2. Although the roots ξi(p) and coefficients cimay be complex, the density and the

mass c0at zero will be positive. This follows from the fact that there is a unique equilibrium

distribution and thus a unique solution to the linear system for the coefficients ci. Of course,

it is also clear that each root ξi(p) and coefficient ci have a companion conjugate root and

conjugate coefficient, which implies that the imaginary parts appearing in the density cancel.

Remark 3. In case that ξi(p) has multiplicity greater than one for one or more values of i, the

analysis proceeds in essentially the same way. For example, if ξ1(p) = ξ2(p), then the partial

fraction decomposition of ω becomes

ω(s) = c0+

c1

?s − ξ1(p)?2+

N

?

i=2

ci

s − ξi(p),

the inverse of which is given by

fW(x) = c1xeξ1(p)x+

N

?

i=2

cieξi(p)x.

Remark 4. For the nominal service time B we have considered only mixtures of Erlang distrib-

utions, mainly because this class approximates well any continuous distribution on [0,∞) and

because we can illustrate the techniques we use without complicating the analysis. However,

we can extend this class by considering distributions with a rational Laplace transform. The

analysis in [21] can be extended to such distributions, and the analysis in Cohen [6, Section

II.5.10] is already given for such distributions, so the results given there can be implemented

directly.

Remark 5. The analysis we have presented so far can be directly extended to the case where Y

takes any finite number of negative values. In other words, let the distribution of Y be given

by P[Y = 1] = p, and for i = 1,...,n, P[Y = −ui] = pi, where ui> 0 and?

?

n=1

ipi= 1−p. Then,

for example, Equation (3.3) becomes

ω(s)1 − pα(−s)

N

?

κn

?

µ

µ + s

?n?

= pP[W + B ? A] − pE[e−s(B−A+W);W + B ? A] +

?N

n=1

n

?

i=1

piP[B ? uiW + A]+

+

n

?

i=1

pi

?

κn

?

µ

µ + s

?n

α(−s)ω(−uis) − E[e−s(B−A−uiW);B ? uiW + A]

?

.

Following the same steps as below (3.3), we can conclude that the waiting time density is again

given by a mixture of exponentials of the form

fW(x) =

N

?

i=1

ˆ cieξi(p)x,

where the new constants ˆ ci(and the mass of the distribution at zero, given by ˆ c0) are to be

determined as the unique solution to a linear system of equations. The only additional remark

necessary when forming this linear system is to observe that both the probability P[B ? uiW+A]

and the expectation E[e−s(B−A−uiW);B ? uiW + A] can be expressed linearly in terms of the

constants ˆ ci.

9

Page 10

The case p = 0

We have seen that the case where Yn= −1 for all n, or in other words the case p = 0, had

to be excluded from the analysis. Equation (3.4) is still valid if we take the constants ξi(0) to

be defined as in Theorem 2. However, one cannot apply Liouville’s theorem to the resulting

equation. The transform can be inverted directly. As it is shown in [21], the terms φ(i)(µ) that

remain to be determined follow by differentiating (3.4) N − 1 times and evaluating ωi(s) at

s = µ for i = 0,...,N − 1. The density in this case is a mixture of Erlang distributions with

the same scale parameter µ for all exponential phases. As we can see, for p = 0 the resulting

density is intrinsically different from the one described in Theorem 2.

The case p = 1

If p = 1 and E[B] < E[A], then we are analysing the steady-state waiting time distribution of a

G/PH/1 queue. Equation (3.4) now reduces to

ω(s)

N

?

i=1

?s − ξi(1)?=

?N

i=1

?s − ξi(1)?

(µ + s)N− α(−s)?N

(µ + s)NP[W + B ? A] − (µ + s)NE[e−s(B−A+W);W + B ? A]

n=1κnµn(µ + s)N−n×

×

??

.

(3.14)

Earlier we have already observed that the right-hand side of (3.14) is equal to an N-th degree

polynomial?N

against zeros of the denominator, and the term

i=0qisi. Inspection of the right-hand side of (3.14) reveals that it has an N-fold

zero in s = −µ. Indeed, all zeros of the numerator of the quotient in the right-hand side cancel

P[W + B ? A] − E[e−s(B−A+W);W + B ? A]

is finite for s = −µ. Hence,

N

?

i=0

qisi= qN(µ + s)N.

(3.15)

Combining (3.14) and (3.15), we conclude that

ω(s)

N

?

i=1

?s − ξi(1)?= qN(µ + s)N,

and since ω(0) = 1, the last equation gives us that

qN=

?N

i=1

?−ξi(1)?

µN

.

Thus, we have that

ω(s) =

?µ + s

µ

?N N

?

i=1

ξi(1)

ξi(1) − s,

which is in agreement with Equation II.5.190 in [6, p. 324].

10

Page 11

4The M/D case

We have examined so far the case where the nominal interarrival time A is generally distributed

and the nominal service time B follows a phase-type distribution. In other words, we have

studied the case which is in a sense analogous to the ordinary G/PH/1 queue. We now would

like to study the reversed situation; namely, the case analogous to the M/G/1 queue.

The M/G/1 queue has been studied in much detail. However, the analogous alternating

service model – i.e., take P(Y = −1) = 1 in (1.2), so p = 0 – seems to be more complicated

to analyse. As shown in [20], if p = 0, the density of W satisfies a generalised Wiener-Hopf

equation, for which no solution is known in general. The presently available results for the

distribution of W with p = 0 are developed in [20], where B is assumed to belong to a class

strictly bigger than the class of functions with rational Laplace transforms, but not completely

general. Moreover, the method developed in [20] breaks down when applied to (2.2) with Y not

identically equal to −1.

We shall refrain from trying to develop an alternative approach for the M/G case with a more

general distribution for B than the one treated in Section 3. Instead, we give a detailed analysis

of the M/D case: A is exponentially distributed and B is deterministic. This case is neither

contained in the G/PH case of the previous section nor has it been treated (for the special

choice of p = 0) in [20]. Its analysis is of interest for various reasons. To start with, the model

generalises the classical M/D/1 queue; additionally, the analysis illustrates the difficulties that

arise when studying (2.2) in case A is exponentially distributed and B is generally distributed;

finally, the different effects of Lindley’s classical recursion and of the Lindley-type recursion

discussed in [20] are clearly exposed. As we shall see in the following, the analysis can be

practically split into two parts, where each part follows the analysis of the corresponding model

with Y ≡ 1, or Y ≡ −1.

4.1Deterministic nominal service times

As before, consider Equation (2.2), and assume that Y = 1 with probability p and Y = −1 with

probability 1 − p. Let A be exponentially distributed with rate λ and B be equal to b, where

b > 0. Furthermore, we shall denote by π0the mass of the distribution of W at zero; that is,

π0= P[W = 0].

For this setting, we have from (2.2) that for x ? 0,

FW(x) = P[max{0,b − A + Y W} ? x] = P[b − A + Y W ? x]

= pP[b − A + W ? x] + (1 − p)P[b − A − W ? x]

= pπ0P[b − A ? x] + p

?∞

0

P[b − A ? x − y]fW(y)dy + (1 − p)π0P[b − A ? x]+

?∞

P[A ? b − x + y]fW(y)dy+

?∞

+ (1 − p)

0

P[b − A ? x + y]fW(y)dy

= π0P[A ? b − x] + p

?∞

0

+ (1 − p)

0

P[A ? b − x − y]fW(y)dy.

(4.1)

11

Page 12

So, for 0 ? x < b the above equation reduces to

FW(x) = π0e−λ(b−x)+ p

?∞

0

e−λ(b−x+y)fW(y)dy + (1 − p)

?b−x

0

e−λ(b−x−y)fW(y)dy+

?∞

+ (1 − p)

b−x

fW(y)dy,

(4.2)

and for x ? b, Equation (4.1) reduces to

FW(x) = π0 + p

?x−b

0

fW(y)dy + p

?∞

x−b

e−λ(b−x+y)fW(y)dy + (1 − p)(1 − π0),

(4.3)

where we have utilised the normalisation equation

π0+

?∞

0

fW(y)dy = 1.

(4.4)

In the following, we shall derive the distribution on the interval [0,b) and on the interval

[b,∞) separately. At this point though, one should note that from Equation (2.2) it is apparent

that for A exponentially distributed and B = b, the distribution of W is continuous on (0,∞).

Also, one can verify that Equation (4.2) for x = b reduces to Equation (4.3) for x = b. The fact

that FWis continuous on (0,∞) will be used extensively in the sequel. Notice also that from

Equations (4.2) and (4.3) we can immediately see that we can differentiate FW(x) for x ∈ (0,b)

and x ∈ (b,∞); see, for example, Titchmarsh [18, p. 59].

The distribution on [0,b).

In all subsequent equations it is assumed that x ∈ (0,b). In order to derive the distribution

of W on [0,b], we differentiate (4.2) once to obtain

fW(x) = λπ0e−λ(b−x)+ λp

?∞

0

e−λ(b−x+y)fW(y)dy + λ(1 − p)

?b−x

0

e−λ(b−x−y)fW(y)dy−

− (1 − p)e−λ(b−x)eλ(b−x)fW(b − x) + (1 − p)fW(b − x).

We rewrite this equation after noticing that the second line is equal to zero, while the sum of

the integrals in the first line can be rewritten by using (4.2). Thus, we have that

?

= λFW(x) − λ(1 − p)

b−x

In order to obtain a linear differential equation, differentiate (4.5) once more, which leads to

fW(x) = λπ0e−λ(b−x)+ λFW(x) − π0e−λ(b−x)− (1 − p)

?∞

?∞

b−x

fW(y)dy

?

fW(y)dy.

(4.5)

f

?

W(x) = λfW(x) − λ(1 − p)fW(b − x).

(4.6)

Equation (4.6) is a homogeneous linear differential equation, not of a standard form because

of the argument b−x that appears at the right-hand side. To solve it, we substitute x for b−x

in (4.6) to obtain

f

?

W(b − x) = λfW(b − x) − λ(1 − p)fW(x).

(4.7)

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Page 13

Then, we differentiate (4.6) once more to obtain

f

??

W(x) = λf

?

W(x) + λ(1 − p)f

?

W(b − x),

and we eliminate the term f

?

W(b − x) by using (4.7). Thus, we conclude that

??

W(x) = λ2p(2 − p)fW(x).

f

(4.8)

For p ?= 0, the solution to this differential equation is given by

fW(x) = d1er1x+ d2er2x,

(4.9)

where r1and r2are given by

r1,2= ±λ

?

p(2 − p),

(4.10)

and the constants d1and d2will be determined by the initial conditions. Namely, the solution

needs to satisfy (4.6) and the condition FW(0) = π0. Thus, for the first equation, substitute

the general solution we have derived into (4.6). For the second equation, first rewrite (4.5) as

follows:

fW(x) = λFW(x) − λ(1 − p)

then substitute fW(x) from (4.9), and finally evaluate the resulting equation for x = 0. This

system uniquely determines d1and d2. Specifically, we have that

λ2(1 − p)?1 − p(1 − π0) − 2π0

ebr1λ(1 − p(1 − π0) − 2π0)r1(λ − r1)

(ebr1− 1)λ2(2 − p)(1 − p) + ebr1r1

where in the process we have assumed that p ?= 1. Up to this point we have that the waiting-time

distribution on [0,b] is given by

?

1 − π0−

?b−x

0

fW(y)dy

?

,

d1=

?r1

(ebr1− 1)λ2(2 − p)(1 − p) + ebr1r1

?r1− λ(2 − p)?,

?r1− λ(2 − p)?,d2=

FW(x) =d1

r1(er1x− 1) +d2

r2(er2x− 1) + π0,

(4.11)

where d1and d2are known up to the probability π0. The cases for p = 0 and p = 1 follow

directly from Equation (4.8) and will be handled separately in the sequel.

The distribution on [b,∞).

As before, we obtain a differential equation by differentiating (4.3) once, and substituting

the resulting integrals by using (4.3) once more. Thus, we obtain the equation

?

which can be reduced to

fW(x) = λ?FW(x) − 1 + p − pFW(x − b)?.

Equation (4.12) is a delay differential equation that can be solved recursively. Observe that

for x ∈ (b,2b), the term FW(x − b) has been derived in the previous step, so for x ∈ (b,2b),

fW(x) = λFW(x) − π0− (1 − p)(1 − π0) − p

?x−b

0

fW(y)dy

?

,

(4.12)

13

Page 14

Equation (4.12) reduces to an ordinary linear differential equation from which we can easily

derive the distribution of W in the interval (b,2b).

For simplicity, denote by Fi(x) the distribution of W when x ∈ [ib,(i+1)b], and analogously

denote by fi(x) the density of W, when x ∈ (ib,(i + 1)b). Then (4.12) states that

fi(x) = λ?Fi(x) − 1 + p − pFi−1(x − b)?,

which leads to an expression for Fi that is given in terms of an indefinite integral that is a

function of x, that is,

??

The constants γi can be derived by exploiting the fact that the waiting-time distribution is

continuous. In particular, every γiis determined by the equation

Fi(x) = eλx

λ?−1 + p − pFi−1(x − b)?e−λxdx + γi

?

,i ? 1.

(4.13)

Fi(ib) = Fi−1(ib).

(4.14)

Solving Equation (4.13) recursively, we obtain that

Fi(x) = 1 − pi(1 − π0) − pi

?d1

r1

+d2

r2

?

+

2

?

j=1

?

λp

λ − rj

?idj

(−λp)jγi−j(x − jb)j−1

rjerj(x−ib)+

+ x

i−1

?

j=0

j!

eλ(x−jb).

(4.15)

Observe that for i = 0, if we define the empty sum at the right-hand side to be equal to zero, then

the above expression is satisfied. Notice that, since we have made use of the distribution on [0,b)

as it is given by (4.11), Equation (4.15) is not valid for p = 0 or p = 1. From Equation (4.14)

we now have that for every i ? 1,

?

j=1

i−1

?

where we have assumed that γ0= 0, and that for i = 1, the second sum is equal to zero. These

expressions can be simplified further by observing that

?d1

Recall that d1 and d2, and thus also all constants γi, are known in terms of π0.

probability π0that still remains to be determined will be given by the normalisation equation

(4.4). Notice though, that since the waiting-time distribution is determined recursively for every

interval [ib,(i + 1)b], Equation (4.4) yields an infinite sum. The sum is well defined, since a

unique density exists. The above findings are summarised in the following theorem.

γi=e−λib(1 − p)pi−1

π0− 1 −d1− d2

r1

?

−

2

?

e−λibdj

rj

?

λp

λ − rj

?i?

1 −ebrj(λ − rj)

λp

?

+

+ i

j=1

e−λjb(i − j)j−1(−λpb)j(γi−1−j− γi−j)

j!

+ γi−1,

(4.16)

1 − pi(1 − π0) − pi

r1

+d2

r2

?

= 1 −

pi

2 − p.

The

14

Page 15

Theorem 3. Consider the recursion given by (1.2), and assume that 0 < p < 1. Let A be

exponentially distributed with rate λ and B be equal to b, where b > 0. Then for x ∈ [ib,(i+1)b],

i = 0,1,..., the limiting distribution of the waiting time is given by

FW(x) = 1 −

pi

2 − p+

2

?

j=1

?

λp

λ − rj

?idj

rjerj(x−ib)+ x

i−1

?

j=0

(−λp)jγi−j(x − jb)j−1

j!

eλ(x−jb),

where the constants γi are given by Equation (4.16) and the probability π0 is given by the

normalisation equation (4.4).

One might expect though that Equation (4.4) may not be suitable for numerically determin-

ing π0. However, if the probability p is not too close to one, or in other words, if the system does

not almost behave like an M/D/1 queue, then one can numerically approximate π0from the

normalisation equation. As an example, in Figure 1 we display a typical plot of the waiting-time

distribution. We have chosen b = 1, λ = 2, and p = 1/3.

FW?x?

1

23

45

x

0.2

0.4

0.6

0.8

1

Figure 1: The waiting time distribution for b = 1, λ = 2, and p = 1/3.

For p close to one, we can see from the expressions for d1and d2that both the numerators

and the denominators of these two constants approach zero. Furthermore, the denominators

λ − rj, j = 1,2 that appear in the waiting-time distribution also approach zero, which makes

Theorem 3 unsuitable for numerical computations for values of p close to one. Moreover, we

also see that very large values of the parameter λ may also lead to numerical problems, since

λ is involved in the exponent of almost all exponential terms that appear in the waiting-time

distribution.

As one can observe from Figure 1, and show from Theorem 3, FWis not differentiable

for x = b. This is not surprising, as the waiting-time distribution is defined by two different

equations; namely Equation (4.2) for x < b and Equation (4.3) for x ? b. Furthermore, from

Equation (4.5) we have that

fW(b−) = λFW(b) − λ(1 − p)(1 − π0),

15