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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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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)
?
.
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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
3 The 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−
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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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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)
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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.
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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
4 The 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)+ λ FW(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)
12
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
Page 16
and from Equation (4.12) we have that
fW(b+) = λ(FW(b) − 1 + p − pπ0).
That is, fW(b−) − fW(b+) = λπ0.
The case p = 0
Observe that if p = 0 then the support of W is the interval [0,b]. To determine the density of
the waiting time, we insert p = 0 into Equation (4.8). Thus, we obtain that
f
??
W(x) = 0,
from which we immediately have that
fW(x) = ν1x + ν2,
for some constants ν1and ν2such that (4.6) is satisfied. The latter condition implies that for
every x ∈ (0,b) the following equation must hold:
ν1= λ(ν1x + ν2) − λ(ν1(b − x) + ν2).
From this we conclude that ν1is equal to zero, i.e. the waiting time has a mass at zero and
is uniformly distributed on (0,b). To determine the mass π0and the constant ν2we evaluate
(4.5) at x = 0 and we use the normalisation equation (4.4), keeping in mind that fW(x) = 0 for
x ∈ [b,∞). These two equations yield that if p = 0, then
λ
1 + λb,
fW(x) =
0 < x < b,
and
π0=
1
1 + λb.
(4.17)
Evidently, the density in this case is quite different from the density for p ?= 0, which is on (0,b)
a mixture of two exponentials; see (4.9).
Another way to see that fW(x) = λπ0, 0 < x < b, is as follows. Recall that for p = 0 and
x ? b we have that fW(x) = 0. Equation (4.5) can now be written as
fW(x) = λπ0+ λP[W ∈ (0,x)] − λP[W ∈ (b − x,b)].
Replacing x by b−x shows that fW(x) = fW(b−x), which implies that P[W ∈ (0,x)] = P[W ∈
(b − x,b)] and finally that fW(x) = λπ0, 0 < x < b. It seems less straightforward to explain
probabilistically that W, given that W > 0, is uniformly distributed. With a view towards the
recursion W
n Poisson arrivals occur in some interval, then they are distributed like the n order statistics of
the uniform distribution on that interval; see Ross [14, Section 2.3].
D= max{0,b − A − W}, we believe that this property is related to the fact that, if
The case p = 1
For the M/D/1 queue, Erlang [7] derived the following expression for the waiting-time distrib-
ution:
i
?
where ρ is the traffic intensity. Recall that for the M/D/1 queue we have that FW(0) = 1−ρ. We
see that for p = 1 Equation (4.5) indeed leads to the waiting-time distribution (1−ρ)eλx, as it is
P[W ? x] = (1 − ρ)
j=0
?−λ(x − jb)?j
j!
eλ(x−jb), ib ? x < (i + 1)b,
16
Page 17
given by Erlang’s expression for the first interval [0,b). For x ? b, one needs to recursively solve
Equation (4.13) in order to obtain Erlang’s expression. However, since the recursive solution
we have obtained for our model makes use of FW(x) as it is given by (4.11), which is not valid
for p = 1, the waiting-time distribution we have obtained in Theorem 3 cannot be extended to
the case for p = 1.
The terms both in Erlang’s expression for the waiting-time distribution of an M/D/1 queue
and in Theorem 3 alternate in sign and in general are much larger than their sum. Thus, the
numerical evaluation of the sum may be hampered by roundoff errors due to the loss of significant
digits, in particular under heavy traffic. For the M/D/1 queue, however, a satisfactory solution
has been given by Franx [8] in a way that only a finite sum of positive terms is involved; thus,
this expression presents no numerical complications, not even for high traffic intensities. For
our model, extending Franx’s approach is a challenging problem as the representation of various
quantities appearing in [8] which are related to the queue length at service initiations is not
straightforward.
As we see, the waiting-time distribution in Theorem 3 is quite similar to Erlang’s expression,
so we expect that eventually the solution will suffer from roundoff errors. However, a significant
difference in the numerical computation between the M/D/1 queue and the model described by
Recursion (1.2) arises when computing π0. For any single server queue we know a priori that
P[W = 0] = 1−ρ. In our model, π0has to be computed from the normalisation equation, where
the numerical complications when calculating the waiting-time distribution become apparent.
In particular, as p tends to 1, i.e. as the system behaves almost like an M/D/1 queue, the
computation of π0becomes more problematic.
As a final observation, we note that the effects of Lindley’s classical recursion and of the
Lindley-type recursion discussed in [20] are quite apparent. The analysis for our model is in a
sense separated into two parts: the derivation of the waiting-time distribution in [0,b) and in
[b,∞). In the first part, we see that Equation (4.6) is quite similar to the differential equation
appearing in [22] for the derivation of the waiting-time distribution in case p = 0 and B follows
a polynomial distribution. Moreover, one could use the same technique to derive a solution,
but Equation (4.6) is too simple to call for such means. In the second part, we see the effects
of the M/D/1 queue, as we eventually derive FWin a recursive manner. Furthermore, this
model inherits all the numerical difficulties appearing in the classical solution for the M/D/1
queue, plus the additional difficulties of computing π0. For Lindley’s recursion, π0 is known
beforehand, while for the Lindley-type recursion described in [13, 20, 21, 22, 23] π0is derived
by the normalisation equation.
Acknowledgements
The authors would like to thank Ivo Adan for his helpful remarks and his assistance in simulating
this model.
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