Performance Analysis of Batch Arrival Queue with Feedback,Disasters, and Repairs under a Multiple Vacation Policy

George C. MytalasDept. of Mathematics, CUNY,

Brooklyn, NY 11210, USAand

Michael A. ZazanisDept. of Statistics, Athens University of Economics and Business

Athens 10434, Greece

January 20, 2018

Abstract

We consider a queueing system with batch Poisson arrivals and Bernoulli feedback which is sub-ject to disasters occurring according to an independent Poisson process. Each disaster is followedby a repair period. The server, when idle, takes repeated vacations of random length and, while theserver is under repair or on vacation disasters cannot occur. We analyze this system using the sup-plementary variables technique and we obtain the probability generating function of the stationaryqueue length distribution, the Laplace transform of the busy period distribution, and the probabilitythat a customer completes service. The Laplace transform of the time between two consecutive dis-asters is also obtained. Further, we obtain results regarding the Laplace transform of the system timedistribution for a typical customer completing service. We also analyze the system under a singlevacation policy, as well as a model in which disasters can occur even when the server is not busy andprovide expressions for various performance measures and numerical results.

KEYWORDS: BATCH-ARRIVALS, SINGLE AND MULTIPLE VACATIONS, FEEDBACK, DISASTERS,SUPPLEMENTARY VARIABLES METHOD.

1 Introduction

We study a single server queue with batch Poisson arrivals, general service times, and Bernoulli feed-back. After completing service a customer either joins immediately the tail of the queue as a feedbackcustomer with probability r, or departs forever from the system with probability 1 − r, independentlyof anything else. The system suffers from disasters which occur according to an independent Poissonprocess with rate δ and which instantly remove all customers (both in the queue and in service) fromthe system. Immediately after the occurrence of a disaster the server undergoes a repair with randomduration. Further, the server takes vacations or random length whenever a departing customer leaves thesystem empty. We consider two vacation policies: Multiple vacations that end when an arrival occurs,and single vacations. We analyze in detail the model under the assumption that disasters can occur only

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when the server is busy serving customers. We also examine briefly a system in which disasters can alsooccur when the server is under repair or on vacation.

For background on queues with vacations we refer the reader to Takagi [20] and Tian and Zhang[22]. Numerous researchers have studied queueing models under different vacation policies dependingon the number of possible consecutive vacations server may take every time that system empties. Forexample Baba [2] studied the multiple policy where the number of possible consecutive vacations is un-bounded and there is not an idle state so server either working or is on vacation. Choudhury [7] studiedthe single policy where only one vacation server may take each time that is empty and if returning fromvacation and waiting area is empty then stays idle until first customer arrive. The case where the max-imum number of vacations is a random variable was proposed by Takagi [20] and termed as multipleadaptive vacation (MAV) policy by Tian and Zhang [22].

Disasters, also refereed to as catastrophes, in queueing systems are random events that, when theyoccur, remove all customers from the system (queue and service). Queues with disasters are natural mod-els for communications or manufacturing systems subject to catastrophic failures which result in the lossof all customers currently in service, or waiting to be served. Jain and Sigman [10] analyzed the M/G/1system in the process of disasters and derived the generalization of the Pollaczek-Kinchine formula inthis Boxma et al. [4] considered an M/G/1 queue where disasters occur according to an independentrenewal process and studied the model for different disaster rules. Kyriakidis and Abakuks [13] havestudied such models in the context of birth, death, and immigration processes. Yechiali [26] studiedM/M/c queues with disasters and customers that become impatient during the repair periods followingdisasters. See also Chakravarthy [6] for a generalization of this model. Kumar et al. [12] derived thetransient analysis for the M/G/1 queue with disasters. The busy period of such models has been studiedby Yashkov and Yashkova [25]. Boudali and Economou [5] study the effect of catastrophes on strategiccustomer behavior. We also refer to the papers of Lee et al. [16] and Yang et al. [17] which include repairtimes after disasters occurring for discrete and continuous time models respectively. Recently Mytalasand Zazanis [15] analyzed a batch queueing model, operating under a MAV (Multiple Adapted Vaca-tions) policy and subject to disasters, using the supplementary variables technique.

Queueing models with Bernoulli feedback were first studied by Takács [19]. Further studies on thequeue length, the total sojourn time and waiting time are provided by Disney et al. [8], [9]. Bernoullifeedback queues with vacations were studied by Takagi [20], Takine et al. [21], Wortman and Disney[24]. A batch arrival queueing system with feedback and optional server vacations under single vacationpolicy was studied by Madan and Al-Rawwash [18]. Also, Thangaraj and Vanitha [23] studied a twophase single server queue system with Bernoulli feedbacks and a multiple vacation policy. A generalretrial queue with balking and feedback together with a special case of MAV policy was studied by Keand Chang [11].

Feedback queueing systems with server breakdowns or with vacations on server operations havebeen already studied in the literature. The present paper examines the effect of disasters in such systems,which has not been studied before. The rest of the paper is organized as follows. In section 2 theprincipal model studied, in which the server operates under a multiple vacation policy and disasters donot affect the system when the server is not busy, is described. Sections 3 presents the analysis usingthe supplementary variables technique and section 4 uses these results to obtain various performancecriteria. In section 5 the Laplace transform of the busy period is obtained. In section 6 the Laplacetransform of the length of the fundamental cycles of the system i.e. the time between two consecutivedisasters is evaluated. In section 7 we examine in detail the system time of a tagged customer arriving in

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stationarity. Our analysis follows that of the classic paper of Takàcs [19], taking into account the specialcomplications arising from the presence of disasters, vacations, and the fact that customers arrive inbatches. In section 8 we discuss briefly a model in which the vacation policy consists of single vacations:After the first vacation, even if the server arrives to an empty system he remains present waiting to servecustomers. In section 9 a variation of the original model is analyzed, in which disasters can occur evenwhen the server is in vacation, or under repair. Finally, in section 10 numerical results are presented.

2 Model description and notations

Consider a single server queue with customers arriving in batches according to a Poisson process withrate λ. Let χn denote the size of the nth batch. {χn}, n = 1, 2, . . ., is assumed to be an i.i.d. sequence.Let χ(z) =

∑∞n=1 P(χ1 = n)zn denote the corresponding probability generating function. We do not

assume the mean to be finite. When this assumption is necessary it will be stated explicitly and the meanbatch size will be denoted by mχ. The mean batch size, when finite will be denoted by mχ := Eχ.Service times are assumed to be i.i.d. random variables with common distribution S, correspondingdensity S′ and hazard rate function µ(x) = S

′(x)1−S(x) , x ≥ 0. The Laplace transform of S will be denoted

by Ŝ(s) :=∫∞

0 e−sxdS(x). Again, its mean will not be assumed finite. When this assumption becomes

necessary it will be stated clearly and the mean will be denoted by mS := ES. As soon as service of acustomer is complete, then with probability r he join the tail of the original queue as a feedback customeror else with probability 1− r he may leave the system.

The system is subject to disasters which occur according the Poisson process with rate δ, indepen-dently of all other processes in the system only when server is busy. When a disaster occurs all customerspresent, including the one in service, are removed from the system. Immediately after disaster server ini-tiates a repair period which durations are i.i.d. with distribution function R, density function R′, hazardrate r(x) = R

′(x)1−R(x) , x ≥ 0, and Laplace transform R̂(s) =

∫∞0 e

−sxdR(x) and the mean, which will beassumed finite throughout the paper is mR := ER

Consider the processes

Nt : number of customers in the system at time t

St : elapsed service time at time t (if server is busy, otherwise 0)

Rt : elapsed repair time at time t (if server under repair, otherwise 0)

Ut : elapsed vacation time at time t (if server on vacation otherwise 0)

and the process {ξt} taking values in the set {s, r,v}, whose elements correspond to the server beingbusy serving customers, being under repair, and being on vacation respectively. Due to the presenceof disasters the system has a regenerative structure with the epochs of consecutive disasters acting asregeneration points. In the Appendix the Laplace transform of the time between two consecutive disastersis derived and it is shown to have a finite mean. Therefore a stationary version of the process exists byvirtue of standard results on regenerative processes (e.g. see [1]). Suppose the process is stationary underthe probability measure P and define the densities

Pn(x) := limh↓0

1

hP(N0 = n, ξ0 = s; x < S0 ≤ x+ h),

Wn(x) := limh↓0

1

hP(N0 = n, ξ0 = r; x < R0 ≤ x+ h), (1)

Vn(x) := limh↓0

1

hP(N0 = n, ξ0 = v; x < U0 ≤ x+ h).

The balance equations satisfied by the stationary distribution are

d

dxPn(x) + (λ+ δ + µ(x))Pn(x) = λ

n−1∑k=1

χkPn−k(x), x > 0, n ≥ 1 (2)

d

dxW0(x) + (λ+ r(x))W0(x) = 0 (3)

d

dxWn(x) + (λ+ r(x))Wn(x) = λ

n∑k=1

χkWn−k(x), x > 0, n ≥ 1 (4)

d

dxV0(x) + (λ+ u(x))V0(x) = 0, (5)

d

dxVn(x) + (λ+ u(x))Vn(x) = λ

n∑k=1

χkVn−k(x), x > 0, n ≥ 1 (6)

The boundary conditions of the above system of differential equations are

Pn(0) = (1− r)∫ ∞

0Pn+1(x)µ(x)dx+ r

∫ ∞0

Pn(x)µ(x)dx

+

∫ ∞0

Vn(x)u(x)dx+

∫ ∞0

Wn(x)r(x)dx (7)

V0(0) = (1− r)∫ ∞

0P1(x)µ(x)dx+

∫ ∞0

V0(x)u(x)dx+

∫ ∞0

W0(x)r(x)dx (8)

Vn(0) = 0, n ≥ 1, . . . (9)

W0(0) = δ∞∑n=1

∫ ∞0

Pn(x)dx, (10)

with normalization condition∞∑n=1

∫ ∞0

Pn(x)dx+

∞∑n=0

(∫ ∞0

Wn(x)dx+

∫ ∞0

Vn(x)dx

)= 1. (11)

4

Define the partial probability generating functions (pgf)

P (x; z) :=∞∑n=1

znPn(x), W (x; z) :=∞∑n=0

znWn(x), V (x; z) :=∞∑n=0

znVn(x).

The partial pgf’s for the number of customers in the system in stationarity regardless of the value of thesupplementary variables are then given by

P (z) :=

∫ ∞0

P (x; z)dx, W (z) :=

∫ ∞0

W (x; z)dx, V (z) :=

∫ ∞0

V (x; z)dx. (12)

Proposition 1. The partial pgf for the number of customers in the system when the server is busy underthe multiple vacation policy is given by

P (z) = P (0; z)1− Ŝ(δ + α(z))

δ + α(z)(13)

where

P (0; z) = zδP (1)R̂(α(z))− V0(0)

(1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

. (14)

and α(z) := λ(1− χ(z)).

Proof. Multiplying (2), (4), (6), and (7) by zn and adding we obtain the linear first order PDE’s

∂P (x; z)

∂x+ (α(z) + δ + µ(x))P (x; z) = 0,

∂W (x; z)

∂x+ (α(z) + r(x))W (x; z) = 0, (15)

∂V (x; z)

∂x+ (α(z) + u(x))V (x; z) = 0,

and the equation

∞∑n=1

znPn(0) =

∫ ∞0

∞∑n=1

znVn(x)u(x)dx+ (1− r)∫ ∞

0

∞∑n=1

znPn+1(x)µ(x)dx

+

∫ ∞0

∞∑n=1

znWn(x)r(x)dx+ r

∫ ∞0

∞∑n=1

znPn(x)µ(x)dx (16)

which, taking into account that∑∞

n=1 Vn(x)zn = V (x; z) − V0(x),

∑∞n=1Wn(x)z

n = W (x; z) −W0(x), and

∑∞n=1 Pn+1(x)z

n = z−1P (x; z)− P1(x), gives

P (0; z) = z−1(1− r)∫ ∞

0P (x; z)µ(x)dx− (1− r)

∫ ∞0

P1(x)µ(x)dx+ r

∫ ∞0

P (x; z)µ(x)dx

+

∫ ∞0

V (x; z)u(x)dx−∫ ∞

0V0(x)u(x)dx+

∫ ∞0

W (x; z)r(x)dx−∫ ∞

0W0(x)r(x)dx. (17)

Solving (15) we obtain

P (x; z) = P (0; z)(1− S(x))e−(δ+α(z))x,W (x; z) = W (0; z)(1−R(x))e−α(z)x, (18)V (x; z) = V (0; z)(1− U(x))e−α(z)x.

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Note that since no customers are present when a vacation or a repair period starts,

V (0; z) = V0(0) and W (0; z) = W0(0). (19)

Also, from (3) and (5) we obtain

W0(x) = W0(0)(1−R(x))e−λx, V0(x) = V0(0)(1− U(x))e−λx. (20)

From (18) we obtain∫ ∞0

P (x; z)µ(x)dx =

∫ ∞0

P (0; z)(1− S(x))e−(δ+α(z))xµ(x)dx = P (0; z)Ŝ(δ + α(z)), (21)∫ ∞0

W (x; z)r(x)dx =

∫ ∞0

W (0; z)(1−R(x))e−α(z)xr(x)dx = W (0; z)R̂(α(z)), (22)∫ ∞0

V (x; z)u(x)dx =

∫ ∞0

V (0; z)(1− U(x))e−α(z)xu(x)dx = V (0; z)Û(α(z)). (23)

Similarly from (20) we have∫ ∞0

W0(x)r(x)dx = W0(0)

∫ ∞0

e−λxdR(x) = W0(0)R̂(λ), (24)∫ ∞0

V0(x)u(x)dx = V0(0)

∫ ∞0

e−λxdU(x) = V0(0)Û(λ). (25)

Using (21)–(25) in (17) we obtain

P (0; z) = V (0; z)Û(α(z)) + z−1(1− r)P (0; z)Ŝ(δ + α(z)) +W (0; z)R̂(α(z))−V0(0) + rP (0; z)Ŝ(δ + α(z)).

Using (18) we write (10) as

W0(0) = δ

∞∑n=1

∫ ∞0

Pn(x)dx = δ

∫ ∞0

P (x; 1)dx = δP (1) (26)

whence we obtain (14).

We can prove using Rouche’s theorem (see the Appendix) that the denominator of P (0; z) (given in(14)) has a unique root, z0, in the open unit disk |z| < 1. However, for δ > 0, the system is stable forall values of the other parameters due to the presence of disasters. This implies that the power series thatdefinesP (0; z) converges uniformly on the closed unit disk |z| ≤ 1 and defines an analytic function there.Therefore, the numerator of P (0; z) must also vanish at z0 which implies that the, as of yet, unknownconstants P (0) and V0(0) are connected by the relationship V0(0)

(1− Û(α(z0))

)= δP (1)R̂(α(z0)).

Setting

γ :=R̂(α(z0))

1− Û(α(z0))(27)

and taking into account (26) we thus have

V0(0) = γW0(0). (28)

It will be shown in section 4.2 (see (45)) that z0 can be interpreted as the probability that a busy periodof the system, starting with a single customer, will be completed without the occurrence of a disaster.Also, as can be readily seen from (28), γ is the relative frequency of occurrence of vacation periodsrelative to that of repairs. A probabilistic explanation as to why the right hand side of (28) should havethis interpretation is given in the Appendix.

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3.1 Partial pgf’s

Here we derive the stationary partial pgf’s for the number of customers in the system according to thestate of the server (working, under repair, or on vacation).

Server is working. From proposition 1

P (z) = δP (1) zR̂(α(z))− γ

(1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

1− Ŝ(δ + α(z))δ + α(z)

. (29)

Server is under repair. Taking into account (12), (18), and (26) we obtain

W (z) = δ1− R̂(α(z))

α(z)P (1). (30)

The probability that the server is under repair using the de l’ Hôpital rule is W (1) = P (1)δmR.

Server is on vacation. Taking into account (12), (18), and (28) we obtain

V (z) = δγ1− Û(α(z))

α(z)P (1). (31)

The probability that the server is on vacation using the de l’ Hôpital rule is V (1) = δP (1)mUγ.

The probability that the server is busy. The probability that the server is busy, P (1), is determined bythe normalization condition P (1) +W (1) + V (1) = 1 which gives

P (1) = (1 + γδmU + δmR)−1 . (32)

(The dependence of P (1) on the service time distribution and λ comes through z0 and γ which dependson z0.)

The pgf of the number of customers in the system in stationarity. The pgf of the number of customersin the system in stationarity, Φ(z) :=

∑∞n=0 z

nP(N0 = n), is obtained by adding the above marginalpgf’s. Setting

K(z) := R̂(α(z))− γ(

1− Û(α(z)))

(33)

and using (29), (30), and (31), we have Φ(z) = P (z) +W (z) + V (z) which gives

Φ(z) =δ

1 + γδmU + δmR

×

[1− Ŝ(δ + α(z))

z − (1− r + rz)Ŝ(δ + α(z))zK(z)

δ + α(z)+mRR̂e(α(z)) + γmU Ûe(α(z))

]. (34)

In the above, Ûe(s) :=1−Û(s)smU

and R̂e(s) :=1−R̂(s)smR

denote the Laplace transform of the correspondingequilibrium (integrated tail) distribution of the vacation and repair periods. If the mean of the batch sizedistribution, mχ, and the second moments of the repair and vacation distributions are finite, the expectednumber of customers in stationarity is finite and is given by

Φ′(1) =λmχδ

+1

1 + δmR + γδmU

[− (1− r)Ŝ(δ)

1− Ŝ(δ)+λmχδ

2(ER2 + γEU2)

]. (35)

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3.2 The pgf of the system size at a departure epoch

A departing customer will leave behind l customers in the system at a departure epoch if and only if thereare l + 1 customers in the system just before the departure. Denote by φ+l the probability that, in steadystate, a departing customer leaves behind l customers in the system. Here we consider departures thatleave permanently from the system and are not fed back. From a stochastic intensity argument we obtain

φ+l = C0(1− r)∫ ∞

0Pl+1(x)µ(x)dx, l = 0, 1, 2, . . . , (36)

where C0 is a normalizing constant. If Φ+(z) :=∑∞

l=0 φ+l z

l denotes the corresponding pgf, then

Φ+(z) = C0(1− r)∫ ∞

0

∞∑l=0

zlPl+1(x)µ(x)dx

= C0(1− r)z−1∫ ∞

0P (z;x)µ(x)dx = C0(1− r)z−1P (z; 0)Ŝ(δ + α(z))

= C0(1− r)Ŝ(δ + α(z))δP (1)R̂(α(z))− γ

(1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

where, in the last equation, we have used (14). The normalization constant C0 can be determined fromthe condition Φ+(1) = 1 whence we obtain C0 =

1−Ŝ(δ)δ(1−r)(γ−1)P (1)Ŝ(δ)

. Thus, taking into account (33),

Φ+(z) =Ŝ(δ + α(z))

z − (1− r + rz)Ŝ(δ + α(z))1− Ŝ(δ)Ŝ(δ)

K(z). (37)

It is interesting that if we consider all departures, regardless of whether they are permanent or corre-spond to feedbacks, we obtain the same result. Indeed, if φ̃+l denotes the probability that a departingcustomer leaves behind l customers in steady state, then the stochastic intensity argument would giveφ̃+l = C̃0

∫∞0 Pl+1(x)µ(x)dx.

3.3 The system without repairs

Here we suppose that the repair time following a disaster is negligible and thus following the occurrenceof a disaster the server immediately takes a vacation immediately. The stationary pgf of the number ofcustomers in the system can be obtained by setting R̂(s) = 1 and mR = 0 in (34). Thus

Φ(z) =δγ

1 + δγmU

[z

1− Ŝ(δ + α(z))(1− r + rz) Ŝ(δ + α(z))− z

Û(α(z0))− Û(α(z))δ + α(z)

+1− Û(α(z))

α(z)

]. (38)

In the above expression, and in (39) and (40) as well, γ =(

1− Û(α(z0)))−1

. When mχ < ∞ andEU2

thus recovering the well-known decomposition formula.

4 The Busy Period

4.1 The pgf of the System Size at a Busy Period Initiation Epoch

Let {tl}, l = 1, 2, . . ., be the busy period initiation epochs and Ntl the number of customers in the systemat the initiation epoch of the lth busy period. Let the probability that the typical busy period in stationaritystarts with n customers be denoted by ψn := limk→∞ 1k

∑kl=1 1(Ntl = n) and Ψ(z) :=

∑∞n=1 ψnz

n

denote the corresponding pgf. By a ratio of rates argument we obtain

ψn = D

(∫ ∞0

Wn(x)r(x)dx+

∫ ∞0

Vn(x)u(x)dx

), n = 1, 2, . . . (41)

where D is a normalization constant given by

D−1 =∞∑n=1

(∫ ∞0

Wn(x)r(x)dx+

∫ ∞0

Vn(x)u(x)dx

).

Since∑∞

n=1 znWn(x) = W (x; z)−W0(x), and

∑∞n=1 z

nVn(x) = V (x; z)− V0(x), we conclude that

Ψ(z) =W0(0)

(R̂(α(z))− R̂(λ)

)+ V0(0)

(Û(α(z))− Û(λ)

)W0(0)

(1− R̂(λ)

)+ V0(0)

(1− Û(λ)

)and, taking into account (28) and (33),

Ψ(z) =R̂(α(z))− R̂(λ) + γ

(Û(α(z))− Û(λ)

)1− R̂(λ) + γ

(1− Û(λ)

) = K(z)−K(0)1−K(0)

. (42)

4.2 The Laplace Transform of the Busy Period Length

The basis of our analysis is the remark that the length of the busy period of a queue with a work-conserving priority policy does not depend on the order in which customers are served. Therefore, wemay suppose here that each customer, upon service completion, returns immediately to the server withprobability 1 − r for additional service instead of joining the end of the queue. Thus the length of thebusy period in our system is the same as the length of the busy period in a system without feedback inwhich each customer has a service time requirement whose Laplace transform is

Ŝr(s) :=∞∑n=1

(1− r)n−1rŜ(s)n = (1− r)Ŝ(s)1− rŜ(s)

. (43)

In this framework, for a system operating without disasters, a standard argument gives the Laplacetransform of the length of a busy period starting with a single customer, Γ̂0(s), as the unique solution ofthe Takács equation Γ̂0(s) = Ŝr(s+λ−λχ(Γ̂0(s))) satisfying |Γ̂0(s)| < 1 for s ∈ (0,∞). Equivalently,taking into account (43),

Γ̂0(s) =(

1− r + rΓ̂0(s))Ŝ(s+ λ− λχ(Γ̂0(s))

). (44)

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In particular, comparing (44) with the fixed point problem of Proposition 8 of the Appendix, we see that

z0 = Γ̂0(δ). (45)

Theorem 2. The Laplace transform of the length of a busy period, B̂(s), in our system is given by

B̂(s) = 1− ss+ δ

1− R̂(α(Γ̂0(s+ δ))) + γ[1− Û(α(Γ̂0(s+ δ)))

]1− R̂(λ) + γ

[1− Û(λ)

] = 1− ss+ δ

1−K(Γ̂0(s+ δ))1−K(0)

.

(46)The probability q1 that a busy period ends normally (by the departure of a customer leaving the systemempty), and q2, that it ends by a disaster, are given by

q1 =−R̂(λ) + γ

(1− Û(λ)

)1− R̂(λ) + γ

(1− Û(λ)

) = −K(0)1−K(0)

, (47)

q2 =1

1− R̂(λ) + γ(

1− Û(λ)) = 1

1−K(0). (48)

Finally, the conditional Laplace transform of a busy period given that it ends normally, B̂1(s), or giventhat it ends by a disaster, B̂2(s) are

B̂1(s) =R̂(α(Γ̂0(s+ δ)))− R̂(λ) + γ

[Û(α(Γ̂0(s+ δ)))− Û(λ)

]−R̂(λ) + γ

[1− Û(λ)

]=

K(Γ̂0(s+ δ)))−K(0)−K(0)

, (49)

B̂2(s) =[1− R̂(α(Γ̂0(s+ δ))) + γ

[1− Û(α(Γ̂0(s+ δ)))

]] δδ + s

=δ

δ + s

[1−K(Γ̂0(s+ δ))

]. (50)

whereΓ̂(s) = Ψ

(Γ̂0(s)

)(51)

and Ψ(z) is given by (42).

Proof. Let us first imagine that the system operates in stationarity and that at the moment when a busyperiod is initiated, the disaster mechanism becomes inactive. Let Γ denote the duration of the busyperiod in such a situation. At an initiation epoch of a typical busy period the distribution of the numberof customers present has pgf given by (42). Then the Laplace transform of Γ is given by

Γ̂(s) =∞∑n=1

(Γ̂0(s)

)nψn = Ψ

(Γ̂0(s)

).

Let ∆ be an exponential random variable with rate δ, independent of Γ. Then, q1, the probability that abusy period terminates normally is given by q1 := P(∆ > Γ) = Γ̂(δ) and the complementary probabilityq2 that it terminates by a disaster is q2 := P(∆ ≤ Γ) = 1 − Γ̂(δ). This establishes (47) and (48). A

10

simple conditioning argument gives the Laplace transforms of typical busy period given that it terminatesnormally, B̂1(s), or by a disaster, B̂1(s),

B̂1(s) = E[e−s(Γ∧∆) | Γ < ∆] =Γ̂(s+ δ)

Γ̂(δ),

B̂2(s) = E[e−s(Γ∧∆) | Γ > ∆] =1− Γ̂(s+ δ)

1− Γ̂(δ)δ

s+ δ.

Taking into account (27), (33), (42), (45), and (51) we obtain, after some algebraic manipulations, andtaking also into account that K(z0) = 0, (49) and (50). Finally, (46) follows from B̂(s) = q1B̂1(s) +q2B̂2(s).

5 Performance Measures

P1. Actual rate of disasters. The rate of busy period terminations due to disasters is equal to the rateof repair initiations given by W0(0) = δP (1) = δ1+γδmU+δmR where P (1) is given by (32). This is ofcourse the rate of actual disaster occurrence.

P2. Expected length of a vacation string. A vacation string consists of a number of vacations duringwhich there were no arrivals and a final vacations during which at least one batch arrived. Since vacationsare i.i.d. and batches arrive according to an independent Poisson process, if U denotes a generic vacationduration and Λ an independent exponential random variable with rate λ then β := Û(λ) = P(Λ > U).The expected number of vacations without arrivals is β/(1 − β) and the expected length of each suchvacation is E[U |U < Λ]. Thus the total expected length of the vacation string is

β

1− βE[U |U < Λ] + E[U |U ≥ Λ],

the second term above corresponding to the length of the final vacation during which arrivals occur.Taking into account that βE[U |U < Λ] + (1 − β)E[U |U ≥ Λ] = mU , we see that the expected lengthof the vacation string is

mU (1− β)−1.

P3. Rate of initiation of vacations. This is clearly given by V0(0) = γW0(0) (see equation 28). Eachvacation string following a busy period, or a repair period during which there were no arrivals, consistson the average of (1 − β)−1 vacations, each starting with no customers present. (Of course the last oneends with customers present.) Hence the rate of vacation string initiations is V0(0)(1− β).

P4. Rate of normal busy period terminations, θ1 and busy period terminations by disasters, θ2. Here,normal is meant to signify busy period terminations caused by a customer completing service and de-parting for good, leaving the system empty. Thus

θ1 = (1− r)∫ ∞

0P1(x)µ(x)dx.

Setting b := R̂(λ) and using (8) together with (24), (25),(26), and (28) we obtain

θ1 = (1− r)∫ ∞

0P1(x)µ(x)dx = V0(0)−

∫ ∞0

V0(x)u(x)dx−∫ ∞

0W0(x)r(x)dx

= V0(0)− V0(0)Û(λ)−W0(0)R̂(λ) = V0(0)(1− β)−W0(0)b= δP (1)(γ(1− β)− b).

11

This makes perfect sense and can also be obtained by a different argument: Each normal busy periodending initiates a string of vacations. From the rate of vacation string initiations however we have tosubtract the rate of repair initiations corresponding to repairs with no arrivals.

By a conditional PASTA argument we see that the rate of busy periods terminated by disasters is

θ2 = P (1)δ =δ

1 + δmR + δγmU.

From these two rates we can obtain the probabilities q1, q2, obtained in (47), (48) by a ratio of ratesargument:

q1 :=θ1

θ1 + θ2=

γ(1− β)− b1− b+ γ(1− β)

, q2 :=θ2

θ1 + θ2=

1

1− b+ γ(1− β).

P5. Expected length of an inactive period, EI . If the inactive period follows a busy period that terminatesnormally then its expected length is simply the expected length of a vacation string obtained in P6. Ifon the other hand the inactive perod follows a busy period that terminates as a result of a disaster thenits expected length is mR + b mU1−β , then the second term is the sum corresponding to the probability thatthere are no arrivals during the repair period, b, multiplied by the expected length of a vacation string.Thus

EI = q1mU

1− β+ q2

(mR + b

mU1− β

)=

γmU +mR1− b+ γ(1− β)

.

P6. Expected length of a busy period, EB. This could of course be obtained from (46). However it iseasier to use instead renewal theoretic argument from which it follows that P (1) = EBEB+EI whence weget

EB = EI1

P (1)−1 − 1=

γmU +mR1− b+ γ(1− β)

1

δ(γmU +mR)=

1

δ

1

1− b+ γ(1− β). (52)

This expression is striking enough to merit some commentary. Note that the service time distribution, orthe feedback probability r for that matter, do not appear explicitly in the formula. They are buried in theparameter γ = R̂(z0)

1−Û(z0)through its dependence on z0(δ) which is the unique solution of modulus less

than 1 ofz0 = Ŝ(δ + λ− λχ(z0))(1− r + rz0). (53)

In particular, z0(0) = 1. Let us examine for simplicity the case where the repair time is negligible (i.e.R̂(s) ≡ 1). Then b = R̂(λ) = 1 and denoting the mean busy period EB by mB(δ) we have

mB(δ) =1− Û(λ− λz0(δ))

δ

1

1− Û(λ). (54)

Thus

mB(0) =1

1− Û(λ)limδ→0

1− Û(λ− λz0(δ))δ

=1

1− Û(λ)

(−Û ′(0)

) (−λz′0(0)

)= − λmU

1− βz′0(0). (55)

In the above we have taken into account that z0(0) = 1 and Û ′(0) = mU . z′0(δ) can be obtained by dif-ferentiating the implicit equation (53) with respect to δ: Thus z′0 = Ŝ

′(δ+λ−λχ(z0)) (1− λχ′(z0)z′0)+Ŝ(δ + λ− λχ(z0))rz′0 and setting δ = 0 in this equation we obtain

z′0(0) = −mS

1− r − λmχmS.

12

Substitute this into (55) to obtain

mB(0) =λmU1− β

mS1− r − λmχmS

.

This is indeed the mean duration of the busy period in a queue with no disasters, non-terminating vaca-tions, and Bernoulli feedback.

P7. The number of customers removed by a typical disaster. Denote the corresponding pgf by φd(z).Then

φd(z) =P (z)

P (1), (56)

i.e. equal to the stationary number of customers in the system, conditional on the server being busy, andthus subject to disasters. A rigorous proof of this can be given by appealing to Papangelou’s theorem(see [3]). Heuristically, this follows from a PASTA type of argument, taking into account the fact that theintensity of the disaster process is δ, when the server is busy, and equal to zero when the server is underrepair or on vacation. Thus the number of customers seen by a disaster is the stationary distribution ofthe number of customers conditional on the server being busy. Therefore, from (29),

φd(z) = δ zR̂(α(z))− γ

(1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

1− Ŝ(δ + α(z))δ + α(z)

. (57)

The mean number of customers eliminated by a disaster occurrence is given by

φ′d(1) =λmχδ

(1 + γδmU + δmR)− (1− r)Ŝ(δ)

1− Ŝ(δ). (58)

P8. Rate of customer departures after completing service. This is the rate of service completions thatcorrespond to customers who leave the system for good (as opposed to being fed back). This is given by

(1− r)∫ ∞

0

∞∑n=1

Pn(x)µ(x)dx = (1− r)∫ ∞

0P (x; 1)µ(x)dx = (1− r)P (0; 1)Ŝ(δ)

= (1− r)δP (1) Ŝ(δ)1− Ŝ(δ)

. (59)

P9. Rate of customer removals by disasters. This is obtained by subtracting (59) from the customerarrival rate, λmχ, which gives

λmχ − (1− r)δP (1)Ŝ(δ)

1− Ŝ(δ).

P10. Fraction of customers which complete service. This is obtained by a ratio of rates argument bydividing the rate of service completions that result to departures from the system, given by (59) by therate of customer arrivals, λmχ. We obtain

p =1

λmχ

((1− r)δP (1) Ŝ(δ)

1− Ŝ(δ)

)=

δ(1− r)λmχ(1 + γδmU + δmR)

Ŝ(δ)

1− Ŝ(δ). (60)

13

6 The Laplace transform of the time between two consecutive disasters inthe system with multiple vacations

Here we provide a renewal theoretic analysis of the cycles of the system with multiple vacations. Thesample path of the system consists of cycles which we define as the segments of the sample path of thesystem between consecutive disasters. These cycles consist in turn of sub-cycles: There is an initialinactive-active sub-cycle which consists of a repair time, possibly followed by a string of vacations (ifthere are no arrivals during the repair period). The active part of this sub-cycle is the busy period thatensues. If a disaster does not happen during this busy period it ends by a customer departure leaving thesystem empty. Then an ordinary inactive-active sub-cycle begins. It consists of a string of vacations andthe ensuing busy period. Again, the busy period either ends as a result of a disaster, in which case thewhole cycle ends, or as a result of a customer departure leaving the system empty. In this second case,a new ordinary inactive-active sub-cycle begins. Thus, a cycle consists of an initial active-inactive sub-cycle plus an number (possible zero) of ordinary active-inactive sub-cycles. By analyzing the structureof these sub-cycles we obtain the following

Proposition 3. The Laplace Transform of the time between two consecutive disasters is given by

Ω̂(s) =δ

s+ δ

(R̂(s)− R̂(s+ η(s+ δ)) 1− Û(s)

1− Û(s+ η(s+ δ))

). (61)

whereη(s) := α(Γ̂0(s)) = λ− λχ(Γ̂0(s)). (62)

In particular, when mU and mR are finite, so is the mean time between two consecutive disasters, and itis given by

−Ω̂′(0) = 1δ

+mR +mUγ. (63)

This establishes the claim that a stationary version of the process exists, made in the beginning ofsection 3, on which the analysis there is based. Also it is worth pointing out that (63) has an intuitiveexplanation in view of the meaning of γ discussed in the Appendix. Each cycle consists of a singlerepair period, γ vacation periods, and a number of busy periods, the last of which ends by a disaster.Thus the total length of time in a cycle during which the server is busy serving customers is exponentialwith mean 1δ . A renewal-reward argument then also shows that the fraction of time the server is busy is

P (0) =1δ

1δ

+mR+mUγwhich gives an intuitive explanation of (32).

Proof. 1. Analysis of the structure of an ordinary active-inactive sub-cycle.Let τ1 denote the length of a string of vacations until an arrival occurs. If {Ui} is an i.i.d. sequence orrandom variables with the vacation distribution then the length of a vacation string is τ1 = U1 + U2 +· · ·+UM whereM = min{n : U1 + · · ·+Un > Λ} and Λ is an exponential random variable with rate λ,independent of the vacation lengths, corresponding to the next arrival time. The probability that during avacation period there are no arrivals is given by

∫∞0 e

−λtdU(t) = Û(λ). The length of a vacation period,

given that there are no arrivals in its duration, has Laplace transform given by Û(s+λ)Û(λ)

. Finally, the jointdistribution of the length of a vacation, given that there were arrivals in its duration, and the number ofcustomers present in the system when the vacation ends (and the busy period begins) is given by

1

1− Û(λ)

∫ ∞0

∞∑n=1

χ(z)n(λt)n

n!e−λte−stdU(t) =

Û(s+ λ− λχ(z))− Û(λ+ s)1− Û(λ)

. (64)

14

The probability that a vacation string consists of n vacations, the first n− 1 without arrivals, and the lastcontaining arrivals, is Û(λ)n−1(1 − Û(λ)). Thus the joint distribution of the length, τ1, of the string ofvacations constituting the inactive period and the number of customers, ν1, present in the system when itends is

E[e−sτ1zν1 ] =∞∑n=1

(1− Û(λ))Û(λ)n−1(Û(s+ λ)

Û(λ)

)n−1Û(s+ λ− λχ(z))− Û(λ+ s)

1− Û(λ)

=Û(s+ λ− λχ(z))− Û(λ+ s)

1− Û(s+ λ). (65)

Next we consider the second part of the sub-cycle, namely the busy period that ensues. To this end,we imagine first that the disaster mechanism has been deactivated. Denote by τ2 the length of the busyperiod that follows. Its Laplace transform, conditional on the fact that it starts with ν1 customers, is givenby E[e−s2τ2 | ν1] = Γ̂0(s2)ν1 where Γ̂0(s2) is the Laplace transform of the length of a busy period ofthe system without disasters and starting with a single customer. (See section 4.2 for further discussion.)Thus we obtain the joint Laplace transform of the duration of an inactive period τ1 and the busy periodthat follows it, τ2, (without disasters) as

E[e−s1τ1−s2τ2 ] =Û(s1 + η(s2))− Û(λ+ s1)

1− Û(s1 + λ). (66)

At this point we can take into account the possible occurrence of disasters. Let ∆ be an exponential ran-dom variable with rate δ (the rate of occurrence of disasters), independent of all other random variables.Then the joint distribution of the inactive and the busy part of the sub-cycle, given that a disaster doesnot occur during the busy part, is

E[e−s1τ1−s2τ2 | ∆ > τ2] =E[e−s1τ1−s2τ21(∆ > τ2)]

E[1(∆ > τ2)]=

E[e−s1τ1−(s2+δ)τ2 ]E[e−δτ2 ]

=Û(s1 + η(s2 + δ)))− Û(λ+ s1)

1− Û(s1 + λ)1− Û(λ)

Û(η(δ)))− Û(λ).

Thence, the length of a sub-cycle starting with a vacation, given that a disaster does not occur during itsbusy period is

ψ1(s) :=Û(s+ η(s+ δ)))− Û(λ+ s)

1− Û(s+ λ)1− Û(λ)

Û(η(δ)))− Û(λ). (67)

The probability that such a sub-cycle is completed without a disaster happening is

a1 := P(∆ > τ2) = E[e−δτ2 ] =1− Û(η(δ))

1− Û(λ)(68)

and the probability that a disaster occurs during such a sub-cycle is

1− a1 =Û(η(δ))− Û(λ)

1− Û(λ). (69)

Using (66) we can also obtain the joint distribution of the inactive and busy period given that the latterends by a disaster. In that case the length of the busy period is equal to ∆ and the corresponding jointLaplace transform is

E[e−s1τ1−s2∆ | ∆ ≤ τ2] =1

1− E[e−δτ2 ]δ

δ + s2

(E[e−s1τ1 ]− E[e−s1τ1−(δ+s2)τ2 ]

). (70)

15

Therefore, the Laplace transform of the length of the sub-cycle, given that its busy period ends by adisaster, which we will denote by ψ1(s), is obtained by setting s1 = s2 = s in the above to obtain

ψ1(s) =δ

δ + s

1− Û(λ)1− Û(η(δ))

Û(s)− Û(s+ η(s+ δ))1− Û(s+ λ)

. (71)

2. The initial active-inactive sub-cycle.The inactive period here consist of a repair period, followed possibly by a string of vacations if there areno arrivals during the repair period. The length of the initial inactive period is τ01 := R+U1 + · · ·+UMwhere M = inf{n ≥ 0 : R + U1 + · · ·+ Un ≤ ∆}. We denote the number of customers present at theend of this inactive period by ν01 . There are two cases: If there are customer arrivals during the repairperiod, an event occurring with probability 1− R̂(λ), then the joint distribution of (τ01 , ν01), conditionalon this event, is

R̂(s1 + λ− λχ(z))− R̂(s1 + λ)1− R̂(λ)

. (72)

If there are no arrivals during the repair period, which happens with probability R̂(λ), then a stringof vacations begins and, by the same arguments that lead to (65), the conditional joint distribution of(τ01 , ν

01) is

R̂(s1 + λ)

R̂(λ)

∞∑n=0

(Û(s1 + λ)

Û(λ)

)nÛ(s1 + λ− λχ(z))− Û(s1 + λ)

1− Û(λ)

(1− Û(λ)

)Û(λ)n (73)

=R̂(s1 + λ)

R̂(λ)

(Û(s1 + λ− λχ(z))− Û(s1 + λ)

) 11− Û(s1 + λ)

. (74)

Be multiplying (72) by 1− R̂(λ) and the right hand side of (73) by R̂(λ) and adding we obtain

E[e−s1τ01 zν

01 ] = R̂(s1 + λ− λχ(z))− R̂(s1 + λ)

1− Û(s1 + λ− λχ(z))1− Û(s1 + λ)

.

Arguing as in (66), the joint Laplace transform of the duration of the initial inactive period τ01 and thebusy period that follows it, τ02 , (without disasters) is

E[e−s1τ01−s2τ02 ] = R̂(s1 + η(s2))− R̂(s1 + λ)

1− Û(s1 + η(s2))1− Û(s1 + λ)

and the Laplace transform of the initial inactive-active sub-cycle, given that its busy period does notsuffer a disaster, ψ0(s), is given by

ψ0(s) =E[e−s1τ01−(s2+δ)τ02 ]

E[e−δτ02 ]=

R̂(s+ η(s+ δ))− R̂(s+ λ)1−Û(s+η(s+δ))1−Û(s+λ)

R̂(η(δ))− R̂(λ)1−Û(η(δ))1−Û(λ)

. (75)

The probability that the initial sub-cycle is completed without a disaster happening is

a0 := R̂(η(δ))− R̂(λ)1− Û(η(δ))

1− Û(λ). (76)

16

The complementary probability, 1 − a0, is of course the probability that during the busy period of theinitial sub-cycle a disaster occurs. Again, using (70) in this case, the Laplace transform of the length ofthe initial sub-cycle, given that a disaster occurs, ψ0(s), can be obtained as follows:

E[e−s1τ01−s2∆ | ∆ ≤ τ02 ] (77)

=δ

δ + s2

1

1− a0

(R̂(s1)− R̂(s1 + η(s2 + δ))− R̂(s1 + λ)

Û(s1 + η(s2 + δ))− Û(s1)1− Û(s1 + λ)

)Setting s1 = s2 = s in (77) we obtain

ψ0(s) =δ

δ + s

1

1− a0

(R̂(s)− R̂(s+ η(s+ δ))− R̂(s+ λ) Û(s+ η(s+ δ))− Û(s)

1− Û(s+ λ)

)(78)

The Laplace Transform of the time between two consecutive disasters is

Ω̂(s) := (1− a0)ψ0(s) + a0(1− a1)ψ0(s)ψ1(s)∞∑n=0

an1ψ1(s)n

= (1− a0)ψ0(s) + a0ψ0(s)(1− a1)ψ1(s)1− a1ψ1(s)

.

Substituting in the above (68), (67), (71), (76), (75), (78), we obtain (61).

Differentiating (61) and evaluating at s = 0 we obtain Ω̂′(0) = 1δ − R̂′(0)− Û ′(0) R̂(η(δ))

1−Û(η(δ)). Since

η(δ) = α(Γ̂0(δ)) and Γ0(δ) = z0, Ω̂′(0) = 1δ +mR +mUR̂(α(z0))

1−Û(α(z0)). This, together with the definition

(27), establishes (63).

Corollary 4. When the repair time is negligible it suffices to take R = U in order to obtain the corre-sponding Laplace transform which is

Ω̂(s) =δ

s+ δ

Û(s)− Û(s+ η(s+ δ))1− Û(s+ η(s+ δ))

.

In particular, when the duration of each vacation is exponentially distributed, i.e. Û(s) = ωω+s , theLaplace transform becomes

Ω̂(s) =δ

δ + s

ω

ω + s

λ− λχ(Γ̂0(s+ δ))s+ λ− λχ(Γ̂0(s+ δ))

.

7 The System Time

Here we obtain the Laplace transform of the system time for a customer arriving in steady state, i.e.the amount of time from the epoch of arrival until the epoch the customer departs for good from thesystem. The basic approach in our analysis goes back to Takàcs [19]. However the vacation policy, theoccurrence of disasters, and the fact that customers arrive in batches require important modifications. (Inregard to this see [21].)

We consider a tagged (typical) customer arriving at time t = 0 in a stationary system. This customerbelongs then to a batch distributed according to the size-biased distribution, i.e. one containing k cus-tomers with probability m−1χ kχk, k = 1, 2, . . .. Denoting by K0 the number of customers that precede

17

the tagged customer in the batch and by L0 the number of customers that follow him, and taking intoaccount the fact that the tagged customer may occupy any position in the batch equally likely, the jointpgf of K0 and L0 is

E[zK01 zL02 ] =

∞∑k=1

kχkmχ

k−1∑l=0

zl1zk−l−12

k=

χ(z2)− χ(z1)mχ(z2 − z1)

. (79)

The total sojourn time of the tagged customer through the system depends, among other things, onthe state of the server at the t = 0 which, in the notation of §3, is described by the value of the process{ξt} at zero. If ξ0 = s then the server is busy serving a customer and we denote by S0 the residualservice time of the customer in service at time zero. If ξ0 is equal to u or r then the server is either onvacation or under repair and we denote by U0 and R0 the residual vacation time and the residual repairtime respectively. Note that, since at t = 0 a batch arrives, if the server is on vacation or under repair,after time U0 or R0 respectively, the server will start serving customers since there will be customerspresent in the system.

In what follows we will denote by Es the conditional expectation given that ξ0 = s and similarly forEv,Er. We begin with the following

Lemma 5. Let N0 denote the number of customers in the system and S0, U0, R0, the ages of the service,vacation, or repair process, according to whether the server is working, on vacation, or under repair attime 0. Then

Es[zN0e−sS0 ] =δzK(z)

z − (1− r + rz)Ŝ(δ + α(z))Ŝ(s)− Ŝ(δ + α(z))

δ + α(z)− s, (80)

Ev[zN0e−sU0 ] =1

mU

Û(s)− Û(α(z))α(z)− s

, (81)

Er[zN0e−sR0 ] =1

mR

R̂(s)− R̂(α(z))α(z)− s

. (82)

Proof. The conditional joint distribution of N0 and S0 given that ξ0 = s is obtained readily from (18)and (14), using the fact that

P (0; z)

P (1)=

zδK(z)

z − (1− r + rz)Ŝ(δ + α(z)). (83)

Then

Es[zN01(S0 ∈ dx)] := E[zN01(S0 ∈ dx)|ξ0 = s] =P (0; z)

P (1)(1− S(x))e−(δ+α(z))xdx

=zδ K(z)

z − (1− r + rz)Ŝ(δ + α(z))(1− S(x))e−(δ+α(z))dx. (84)

Obviously,

Es[zN0e−sS0 |N0, S0] = zN0∫ ∞

0e−sy

S′(S0 + y)

1− S(S0)dy,

18

and hence, taking into account (84), we obtain

Es[zN0e−sS0 ] =∫ ∞

0Es[zN01(S0 ∈ dx)]

∫ ∞0

e−syS′(x+ y)

1− S(x)dy

=P (0; z)

P (1)

∫ ∞x=0

(1− S(x))e−(δ+α(z))x∫ ∞

0e−ys

S′(x+ y)

1− S(x)dydx

=P (0; z)

P (1)

∫ ∞0

∫ ∞0

e−ys−(δ+α(z))xS′(x+ y)dydx

=P (0; z)

P (1)

Ŝ(s)− Ŝ(δ + α(z))δ + α(z)− s

.

Using (83) in the last equation above we obtain (80).

The proofs of (81) and (82) are entirely similar so we will only discuss the first. From (18) and (19)we easily deduce that

Ev[zN01(U0 ∈ dx)] =1

mU(1− U(x))e−α(z)xdx. (85)

From this we obtain the joint transform of the number of customers in the system at time 0, right beforethe arrival of the tagged customer, and the residual vacation time at time 0 (conditional on the event thatthe server is on vacation at time 0):

Ev[zN0e−sU0 ] =∫ ∞

0Ev[zN01(U0 ∈ dx)]

∫ ∞0

e−syU ′(x+ y)

1− U(x)dy

=1

mU

∫ ∞0

∫ ∞0

e−ys−α(z)xU ′(x+ y)dydx.

From this (81) follows immediately.

Let ν be a geometric random variable independent of everything else, with P(ν = n) = (1−r)rn−1,n = 1, 2, . . ., representing the number of passes through the system for the tagged customer. Let us nowassume for the moment that at t = 0, when the batch containing the tagged customer arrives, the disastermechanism is shut off. This will allow us to obtain the Laplace transform of the total sojourn time of thetagged customer assuming that no disaster occur during this time. Once this is accomplished, we willbe able to determine the corresponding conditional Laplace transform, given that the tagged customerdoes not suffer a disaster during his stay in the system. The total sojourn time of the tagged customer,H, consists of a possible initial delay, W0, which represents the time that elapses from the moment thetagged customer arrives to the system (t = 0) to the moment the server becomes operational, plus thetime Tν , from the moment the server becomes operational (t = W0) to the moment the tagged customercompletes his νth service. Thus H = W0 + Tν with W0 = 1(ξ0 = u)U0 + 1(ξ0 = r)R0.

If ∆ is an exponential random variable with rate δ, independent of everything else, the Laplacetransform of the total sojourn time of the tagged customer, given that he departs without suffering adisaster, is given by

Ĥ(s) := E[e−sH|Tν < ∆] =E[e−s(W0+Tν)1(Tν < ∆)]

P(Tν < ∆). (86)

The above equation accounts for the fact that the disaster mechanism will not affect the system duringthe initial period W0. Denoting by Tn the amount of time that elapses from the epoch the server becomes

19

operational until the epoch the tagged customer completes its nth cycle through the system, n = 1, 2, . . . ,we see through a conditioning argument that

E[e−s(W0+Tν)1(Tν < ∆)] = E[e−sW0−(s+δ)Tν ] =∞∑n=1

(1− r)rn−1E[e−sW0−(s+δ)Tn ]

=

∞∑n=1

(1− r)rn−1Gn(s, s+ δ) (87)

P(Tν < ∆) = E[e−δTν ] =∞∑n=1

(1− r)rn−1E[e−δTn ] =∞∑n=1

(1− r)rn−1Gn(0, δ) (88)

whereGn(s0, s) := E[e−s0W0−sTn ], n = 1, 2, . . . . (89)

Denote by T1, T2, . . . , Tn the successive sojourn times of the tagged customer through the system.In particular, T1 is the amount of time from the moment the server becomes operational, i.e. W0, (alter-natively: from t = 0) to the completion of the first service time of the tagged customer. Similarly, Tk,k = 2, 3, . . . , n is the amount of time that elapses from the epoch of the (k−1)th to the kth service com-pletion for the tagged customer. Thus, conditioning on the number of passes ν = n, Tn = T1 + · · ·+Tn.

Denote by Q1 the number of customers ahead of the tagged customer at time t = 0, excluding theone in service if ξ0 = s, i.e.

Q1 = N0 − 1(ξ0 = s) +K0. (90)

Then

T1 = 1(ξ0 = s)S0 + 1(ξ0 = u)U0 + 1(ξ0 = r)R0 +

Q1+1∑i=1

σ1,i . (91)

Here {σl,i}, l = 1, . . . , n, i = 1, 2, . . . are i.i.d. random variables distributed according to the servicetime distribution S. Since the tagged customer will go n times through the system, the total system timeof this customer is Tn := T1 + T2 + · · · + Tn. During his n consecutive arrivals at the system he seesQ1, Q2, . . . , Qn customers ahead of him. Thus,

Q2 = L0 +Br(Q1) +Nλ,χ(T1) (92)

Tk =

Qk+1∑i=1

σk,i, k = 2, 3, . . . , n, (93)

Qk = Br(Qk−1) +Nλ,χ(Tk−1), k = 3, 4, . . . , n, (94)

where Br(m) is a binomial random variable with probability of success r and m trials and Nλ,χ(x) is acompound Poisson random variable with pgf e−λx(1−χ(z)), independent from all other random variables.Equation (94) states that, at the epoch when the tagged customer completes the (k − 1)th pass throughthe system and is fed back he sees in front of him Qk customers. These are the customers who havearrived during the (k − 1)th pass, which has duration Tk−1, i.e. the term Nλ,χ(Tk−1) in (94), plus thoseof the Qk−1 that where present at the beginning of the (k − 1)th pass and were fed back to the system.Since each of the customers returns for additional service with probability r, the total number of theserecycled customers is Br(Qk−1), independent of everything else.

The main result of this section is the following

20

Theorem 6. The Laplace transform of the overall system time of a customer arriving in stationarity,conditional on the event that the customer departs from the system without being affected by a disasteris given by

Ĥ(s) =

∑∞n=1 r

n−1Gn(s, s+ δ)∑∞n=1 r

n−1Gn(0, δ)(95)

where

Gn(s, s+ δ) :=δP (1)

λmχ

α(ws+δn−1)− α(ws+δn )s+ α(ws+δn−1)− α(w

s+δn )

1

ws+δn−1 − ws+δn

n−1∏k=0

Ŝ(s+ δ + α(ws+δk ))

×

K̂(s+ α(ws+δn−1)) + K̂(α(ws+δn )) r(ws+δn − ws+δn−1)Ŝ(δ + α(ws+δn ))ws+δn −

(1− r + rws+δn

)Ŝ(δ + α(ws+δn ))

(96)

Gn(0, δ) =δP (1)

λmχ

1

wδn−1 − wδn

n−1∏k=0

Ŝ(δ + α(wδk))

×

(K̂(α(wδn−1)) + K̂(α(w

δn))

r(wδn − wδn−1)Ŝ(δ + α(wδn))wδn − (1− r + rwδn) Ŝ(δ + α(wδn))

)(97)

In the above relationships, K̂(s) := R̂(s)− γ(1− Û(s)) and the sequence of functions {wsn} appearingin (96) and (97) is given by the recursion

ws0 := 1, ws1 := Ŝ(s), w

sk :=

(1− r + rwsk−1

)Ŝ(s+ α(wsk−1)), k = 2, 3, . . . , s ≥ 0. (98)

Proof. The proof of this theorem is long and will be presented in three stages. The analysis will becarried out by conditioning first on the state of the system at the arrival epoch of the tagged customer.

1. The tagged customer arrives during a working period

Here we assume that at time 0 a tagged customer arrives to a stationary system and finds it in workingcondition (as opposed to a vacation or repair state). T1, the time for the tagged customer to go throughthe system and complete his first service, is given by T1 = S0 +

∑Q1i=1 σ1,i (see (91)). From the form of

the equations (90)–(94) it becomes clear that we need to distinguish two cases:

(i) The tagged customer passes only once through the system (n = 1). Here, taking into account (90) and(91), we have

Es[e−sT1 |N0,K0, S0] = e−sS0Ŝ(s)N0+K0

and using (79) we obtain

Es[e−sT1 |N0, S0] = e−sS0Ŝ(s)N01− χ(Ŝ(s))mχ(1− Ŝ(s))

.

Thus,

Es[e−sT1 |N0, S0] = Ŝ(s)N01− χ(Ŝ(s))mχ(1− Ŝ(s))

∫ ∞0

e−syS′(y + S0)

1− S(S0)dy.

21

Taking into account (80) and using the fact that ws1 = Ŝ(s) (see the recursive definition in (98)) weobtain

Es[e−sT1 ] =δ

mχ

1− χ(ws1)1− ws1

ws1K(ws1)

ws1 − (1− r + rws1)Ŝ(δ + α(ws1))ws1 − Ŝ(δ + α(ws1))δ + α(ws1)− s

. (99)

(ii) The tagged customer passes more than once through the system (n > 1). Recall that Tk := T1+· · ·+Tk, k = 1, . . . , n. Our goal here is to evaluate Es[e−sTn ]. This is accomplished through the evaluationof a series of conditional expectations through the following recursive argument: From (93)

Es[e−sTn |Q1, T1, . . . , Qn−1, Tn−1, Qn] = e−sTn−1 Ŝ(s)Qn+1.

(We will often write wi instead of wsi in what follows to make the notation simpler.) Taking into account(98) and (94), we have

Es[e−sTn |Q1, T1, . . . , Qn−1, Tn−1] = w1 E[e−sTn−1 wQn1 |Q1, T1, . . . , Qn−1, Tn−1]= w1 e

−sTn−2(1− r + rw1)Qn−1 e−Tn−1(s+α(w1)).

Using (93) and (98) we obtain

Es[e−sTn |Q1, T1, . . . , Qn−1] = w1 e−sTn−2(1− r + rw1)Qn−1 Ŝ(s+ α(w1))Qn−1+1

= w1 Ŝ(s+ α(w1)) e−sTn−2w

Qn−12 .

Working recursively we arrive at

Es[e−sTn |Q1, T1, Q2] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−2))e−sT1wQ2n−1. (100)

From the above and (92),

Es[e−sTn |Q1, T1,K0, L0] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−2))wL0n−1(1− r + rwn−1)Q1e−T1(s+α(wn−1)). (101)

Since T1 = S0 +∑Q1

i=1 when ξ0 = s (see (91))

Es[e−sTn |N0, S0,K0, L0] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−2))wL0n−1e−S0(s+α(wn−1))

×(

(1− r + rwn−1)Ŝ(s+ α(wn−1)))N0 (

(1− r + rwn−1)Ŝ(s+ α(wn−1)))K0

= w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−2))wL0n−1e−S0(s+α(wn−1))wN0n w

K0n .

where in the last equation we have used again (98). Then, from (79),

Es[e−sTn |N0, S0] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−2))χ(wn)− χ(wn−1)mχ(wn − wn−1)

wN0n e−S0(s+α(wn−1)).

Hence, from (80), taking into account that wn = (1− r + rwn−1)Ŝ(s+ α(wn−1)), it follows that

Gs,n(s) := Es[e−sTn ] = w1Ŝ(s+ α(w1))Ŝ(s+ α(w2)) · · · Ŝ(s+ α(wn−1))χ(wn)− χ(wn−1)mχ(wn − wn−1)

× δ(1− r + rwn−1)K(wn)wn − (1− r + rwn)Ŝ(δ + α(wn))

Ŝ(s+ α(wn−1))− Ŝ(δ + α(wn))δ + α(wn)− s− α(wn−1)

. (102)

22

2. The tagged customer arrives during a vacation period

Here we must take into account the residual vacation time that elapses before the server becomes opera-tional. We distinguish again two cases:

(i) The tagged customer passes only once through the system (n = 1). We shall write T1 = U0 +T ∗1 withT ∗1 =

∑Q1+1i=1 σ1,i. It is necessary to split T1 into these two parts because the first is not subject to

disasters whereas the second is. In order to obtain the joint Laplace transform of U0 and T ∗1 note that

Ev[e−s0U0−sT∗1 |U0,N0,K0] = e−s0U0Ŝ(s)N0+K0+1

from which, together with (79), we have

Ev[e−s0U0−sT∗1 |U0,N0] = e−s0U0Ŝ(s)N0+1

1− χ(Ŝ(s))mχ(1− Ŝ(s))

.

From this last equation, together with (81) we obtain

Gv,1(s0, s) := Ev[e−s0U0−sT∗1 ] = Ŝ(s)

1− χ(Ŝ(s))mχ(1− Ŝ(s))

Ev[e−s0U0Ŝ(s)N0+1]

= Ŝ(s)1− χ(Ŝ(s))mχ(1− Ŝ(s))

1

mU

Û(s0)− Û(α(Ŝ(s)))α(Ŝ(s))− s0

. (103)

(ii) The tagged customer passes more than once through the system (n > 1). The analysis remains the sameup to (101). Similarly we define T∗n := T

∗1 + T2 + · · ·+ Tn, Tn := U0 + T∗n.

Ev[e−s0U0−sT∗n |N0,U0,K0, L0] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−1))wL0n−1e

−U0(s0+α(wn−1))

(wn)N0 (wn)

K0 .

Taking into account (79)

Ev[e−s0U0−sT∗n |N0,U0] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−1))

χ(wn)− χ(wn−1)mχ(wn − wn−1)

wN0n e−U0(s0+α(wn−1)).

Hence, from (80),

Gv,n(s0, s) := Ev[e−s0U0−sTn ] = w1Ŝ(s+ α(w1)) · · · Ŝ(s+ α(wn−1))χ(wn)− χ(wn−1)mχ(wn − wn−1)

× 1mU

Û(s0 + α(wn−1))− Û(α(wn))α(wn)− s0 − α(wn−1)

. (104)

The situation is entirely similar if, at time 0, the server is under repair.

3. Computation of the joint transform of W0 and Tn

From (89) Gn(s0, s) := Es[e−sTn ]P(ξ0 = s) + Ev[e−s0U0−sTn ]P(ξ0 = v) + Er[e−s0R0−sTn ]P(ξ0 = r)and hence

Gn(s0, s) = P (1)Gs,n(s) + P (1)δγmUGv,n(s0, s) + P (1)δmRGr,n(s0, s).

23

From (102), (104), and the corresponding relationship for Gr,n(s0, s).

Gn(s0, s) =δP (1)

λmχ

α(wn−1)− α(wn)wn − wn−1

n−1∏k=0

Ŝ(s+ α(wk))

×

((1− r + rwn−1)K(wn)

wn − (1− r + rwn)Ŝ(δ + α(wn))Ŝ(s+ α(wn−1))− Ŝ(δ + α(wn))

δ + α(wn)− s− α(wn−1)

+R̂(s0 + α(wn−1))− R̂(α(wn))

α(wn)− s0 − α(wn−1)+ γ

Û(s0 + α(wn−1))− Û(α(wn))α(wn)− s0 − α(wn−1)

)Define

K̂(s) := R̂(s)− γ[1− Û(s)] (105)and note that K(z), as defined in (33), is related to it via K(z) = K̂(α(z)). Using this notation,

Gn(s0, s) =δP (1)

λmχ

α(wn−1)− α(wn)wn − wn−1

n−1∏k=0

Ŝ(s+ α(wk))

×

(wn − (1− r + rwn−1)Ŝ(δ + α(wn))wn − (1− r + rwn)Ŝ(δ + α(wn))

K̂(α(wn))

δ + α(wn)− s− α(wn−1)+K̂(s0 + α(wn−1))− K̂(α(wn))

α(wn)− s0 − α(wn−1)

)From the above equation we obtain the expressions (96) and (97).

8 The Single Vacation Model

Here we discuss briefly a related model in which the server, at the end of a busy period that is notterminated by a disaster, takes only a single vacation. At the end of this vacation, if no customers arepresent, the server remains idle, waiting to serve immediately the first customer to arrive. The maindifference with the model analyzed thus far is that, in this case, there is an additional possible serverstate, namely the idle state. Thus the process {ξt} defined in section 3 takes now values in the set{i, s, r,v}, state i corresponding to the server being idle. In addition to the probability densities definedin (1) we also define the probability P0 = P(N0 = 0, ξ0 = i) which corresponds to the server being idle(and of course the number of customers in the system being zero). The balance equations (2), (3), (4),(5), (6), still hold for the stationary densities, together with

λP0 =

∫ ∞0

V0(x)u(x)dx. (106)

The boundary conditions (9) and (10) also hold while the boundary conditions for Pn(0) and V0(0), (7)and (8) respectively, are replaced by

Pn(0) = (1− r)∫ ∞

0Pn+1(x)µ(x)dx+

∫ ∞0

Vn(x)u(x)dx+ r

∫ ∞0

Pn(x)µ(x)dx

+

∫ ∞0

Wn(x)r(x)dx+ λχnP0, n ≥ 1, (107)

V0(0) = (1− r)∫ ∞

0P1(x)µ(x)dx+

∫ ∞0

W0(x)r(x)dx. (108)

The normalization condition for this system becomes

P0 +

∞∑n=1

∫ ∞0

Pn(x)dx+

∞∑n=0

(∫ ∞0

Wn(x)dx+

∫ ∞0

Vn(x)dx

)= 1. (109)

24

Proposition 7. The partial pgf for the number of customers in the system when the server is busy underthe single vacation policy is again given by (13) with (14) replaced by

P (0; z) = zδP (1)R̂(α(z))− V0(0)

(Û(λ)(1− χ(z)) + 1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

. (110)

Proof. The linear first order PDE’s described by relations (15) are the same for the single vacation modeland thus relations (18) still hold, as do (21), (22), and (23). Multiplying the boundary conditions (107)by zn and adding for all n,

P (0; z) = z−1(1− r)∫ ∞

0P (x; z)µ(x)dx− (1− r)

∫ ∞0

P1(x)µ(x)dx+ r

∫ ∞0

P (x; z)µ(x)dx

+

∫ ∞0

V (x; z)u(x)dx−∫ ∞

0V0(x)u(x)dx+

∫ ∞0

W (x; z)r(x)dx

−∫ ∞

0W0(x)r(x)dx+ λP0χ(z). (111)

Equation (20) still holds and hence (106) yields

λP0 =

∫ ∞0

V0(0)(1− U(x))e−λxU ′(x)

1− U(x)dx = V0(0)Û(λ). (112)

Using (21), (22), and (112) in (111) we obtain (110).

Rouché’s theorem can be again used to establish a relationship between V0(0) and P (1). In fact thedenominator of P (0; z), given in this case by (110), is the same as that of (14) and hence z0 is again theunique root of equation (159) in the unit disk. (see the Appendix). Thus

V0(0)(Û(λ)(1− χ(z0)) + 1− Û(α(z0))

)= δP (1)R̂(α(z0)). (113)

Equation (26) still holds, as does (28), but with γ replaced by

γs :=R̂(α(z0))

β(1− χ(z0)) + 1− Û(α(z0)). (114)

In view of the above,

P (z) = zδP (1)R̂(α(z))− γs

((1− χ(z))β + 1− Û(α(z))

)z − (1− r + rz)Ŝ(δ + α(z))

1− Ŝ(δ + α(z))δ + α(z)

. (115)

while the expressions (31), (30), for the partial generating functions V (z),W (z), remain the same. Also,from (112), (113), and (114), P0 = λ−1βδγsP (1). The normalization condition gives P (1) + V (1) +W (1) + P0 = 1 in this case, whence we have

P (1) =(1 + γsδmU + γsδλ

−1β + δmR)−1

. (116)

Thus the pgf of the number of customers in the system in stationarity is

Φ(z) = δP (1)

(zKs(z)

δ + α(z)

1− Ŝ(δ + α(z))z − (1− r + rz)Ŝ(δ + α(z))

− Ks(z)− 1α(z)

)(117)

25

withKs(z) := R̂(α(z))− γs

(β(1− χ(z)) + 1− Û(α(z))

).

The mean number of customers in stationarity is

Φ′(1) =λmχδ

+1

1 + δ γs(mU + λ−1β) + δmR

(δλmχ

2

(ER2 + γsEU2

)− (1− r)Ŝ(δ)

1− Ŝ(δ)

).

Also, the probability that the server is idle in stationarity is

P0 =λ−1βδγs

1 + γsδmU + γsδλ−1β + δmR. (118)

S1. Rate of disasters.

W0(0) = δP (1) = δ(1 + γsδmU + γsδλ

−1β + δmR)−1

. (119)

S2. Rate of initiation of vacations.

V0(0) = γsW0(0) = δγs(1 + γsδmU + γsδλ

−1β + δmR)−1 (120)

S3. Rate of normal busy period terminations, θ1 and busy period terminations by disasters, θ2.

θ1 = (1− r)∫ ∞

0P1(x)µ(x)dx = δP (1)(γs − b), θ2 = W0(0) = δP (1). (121)

S4. Fraction of busy periods terminating normally, q1, or terminating by a disaster, q2. From a ratio ofrates argument, using (119) and (121), one can see that

q1 :=θ1

θ1 + θ2=

γs − b1 + γs − b

, q2 :=θ2

θ1 + θ2=

1

1 + γs − b.

S5. Expected length of an inactive period, EI .

EI = q1 (mU + βλ−1) + q2(mR + b(mU + βλ

−1))

=γsmU +mR + βγsλ

−1

1 + γs − b.

S6. Mean length of busy period. By a regenerative argument, P (1) = EBEB+EI and hence, using theprevious result we obtain

EB =1

δ

1

1− b+ γs. (122)

S6. Rate of customer departures after completing service, f .

f = (1− r)∫ ∞

0

∞∑n=1

Pn(x)µ(x)dx = (1− r)∫ ∞

0P (x; z)µ(x)dx = (1− r)P (0; 1)Ŝ(δ)

= (1− r)δP (1) Ŝ(δ)1− Ŝ(δ)

. (123)

S7. Rate of customer removals by disasters. This is obtained by subtracting (123) from λmχ, the rate ofcustomer arrivals:

λmχ − δP (1)(1− r)Ŝ(δ)

1− Ŝ(δ).

S8. Fraction of customers who complete service. This is obtained by dividing the rate of departures fromthe system obtained in (123) divided by the rate of arrivals to the system, λmχ, to obtain

δ

λmχ (1 + γsδmU + δγsλ−1β + δmR)

(1− r)Ŝ(δ)1− Ŝ(δ)

. (124)

26

9 Disasters Affecting All Aspects of the System’s Operation

In this section we consider a variation of the model in which disasters can occur at any time, regardlessof the state of the server. This includes periods where server is on vacation, under repair or for singlevacation model if it is idle. In all cases, when a disaster occurs, a repair period begins, which may ofcourse be in turn interrupted by a new disaster.

9.1 Supplementary Variables Analysis

The balance equations for the multiple vacation policy model are given by

d

dxPn(x) + (λ+ δ + µ(x))Pn(x) = λ

n−1∑k=1

χkPn−k(x), x > 0, n ≥ 1 (125)

d

dxW0(x) + (λ+ δ + r(x))W0(x) = 0 (126)

d

dxWn(x) + (λ+ δ + r(x))Wn(x) = λ

n∑k=1

χkWn−k(x), x > 0, n ≥ 1 (127)

d

dxV0(x) + (λ+ δ + u(x))V0(x) = 0, (128)

d

dxVn(x) + (λ+ δ + u(x))Vn(x) = λ

n∑k=1

χkVn−k(x), x > 0, n ≥ 1. (129)

The boundary conditions of the above system of differential equations are

Pn(0) = (1− r)∫ ∞

0Pn+1(x)µ(x)dx+ r

∫ ∞0

Pn(x)µ(x)dx

+

∫ ∞0

Vn(x)u(x)dx+

∫ ∞0

Wn(x)r(x)dx, n ≥ 1 (130)

V0(0) = (1− r)∫ ∞

0P1(x)µ(x)dx+

∫ ∞0

V0(x)u(x)dx+

∫ ∞0

W0(x)r(x)dx (131)

Vn(0) = 0, n ≥ 1 (132)W0(0) = δ, (133)

with normalization condition (11). Using the same methodology and arguments as in section 3 we obtain

P (0; z) = zδR̂(α(z) + δ)− V0(0)

(1− Û(α(z) + δ)

)z − (1− r + rz) Ŝ(δ + α(z))

. (134)

Equation (18) of section 3 is modified here in what regards V (x; z) and W (x; z) as follows:

W (x; z) = W0(0)[1−R(x)]e−(δ+α(z))x, V (x; z) = V0(0)[1− V (x)]e−(δ+α(z))x. (135)

The value of V0(0) is again determined using Rouche’s theorem in (134) yielding V0(0) = δγδ where

γδ :=R̂(α(z0) + δ)

1− Û(α(z0) + δ). (136)

27

z0 is again the unique root of the denominator of (134) within the unit disk. Setting now

Kδ(z) := R̂(δ + α(z))− γδ(

1− Û(δ + α(z)))

(137)

we rewrite (134) as

P (0; z) =δz Kδ(z)

z − (1− r + rz) Ŝ(δ + α(z)).

9.2 Probability Generating Functions and Performance Measures

T1. The partial pgf when the server is working.

P (z) =δz Kδ(z)

z − (1− r + rz) Ŝ(δ + α(z))1− Ŝ(δ + α(z))

δ + α(z)(138)

withP (1) = R̂(δ)− γδ

(1− Û(δ)

)the steady-state probability that the server is busy serving customers.

T2. The partial pgf when the server is under repair.

W (z) = δ1− R̂(α(z) + δ)

α(z) + δ. (139)

The probability that the server is under repair is W (1) = 1− R̂(δ).

T3. The partial pgf when the server is on vacation.

V (z) = δγδ1− Û(α(z) + δ)

α(z) + δ. (140)

The probability that the server is on vacation is V (1) = γδ(1− Û(δ)).

T4. The pgf of the number of customers in the system in stationarity.

Φ(z) =zKδ(z)[1− Ŝ(δ + α(z))]

z − (1− r + rz) Ŝ(δ + α(z))δ

δ + α(z)+ δmRR̂e(α(z) + δ) + δγδmU Ûe(α(z) + δ).

The mean number of customers in stationarity is

Φ′(1) =λmχδ−(R̂(δ)− γδ

(1− Û(δ)

)) (1− r)Ŝ(δ)1− Ŝ(δ)

. (141)

T5. The pgf of the system size at a departure epoch.

Φ+(z) = C0(1− r)δKδ(z)

z − (1− r + rz) Ŝ(δ + α(z))(142)

where C0 =1−Ŝ(δ)

δ(1−r)(R̂(δ)−γδ(1−Û(δ))).

28

T6. The pgf of the system size at a busy period initiation epoch. Using the ratio of rates argument thatled to (41) we find in the same way that

Ψ(z) =R̂(α(z) + δ)− R̂(λ+ δ) + γδ

(Û(α(z) + δ)− Û(λ+ δ)

)R̂(δ)− R̂(λ+ δ) + γδ

(Û(δ)− Û(λ+ δ)

) = Kδ(z)−Kδ(0)1−Kδ(0)

(143)

T7. Rate of customer departures after completing service, f . This is given by

f = (1− r)∫ ∞

0

∞∑n=1

Pn(x)µ(x)dx. (144)

Adding up the equations (130), (131), and using (144) we obtain

∞∑n=1

Pn(0) + V0(0) = (1− r)f + rf +∫ ∞

0

∞∑n=0

Vn(x)u(x)dx+

∫ ∞0

∞∑n=0

Wn(x)r(x)dx (145)

∞∑n=1

Pn(0) = P (0; 1) =δKδ(1)

1− Ŝ(δ)=

δ

1− Ŝ(δ)

(R̂(δ)− γδ(1− Û(δ))

).

Also∑∞

n=0Wn(x) = W (x; 1) = W0(0)[1−R(x)]e−δx and hence∫∞

0

∑∞n=0Wn(x)r(x)dx = W0(0)R̂(δ).

Similarly∫∞

0

∑∞n=0 Vn(x)u(x)dx = V0(0)Û(δ). Thus, from (145),

f = (1− r)P (0; 1) + V0(0)[1− Û(δ)]−W0(0)R̂(δ),

and hence,

f = (1− r) δŜ(δ)1− Ŝ(δ)

[R̂(δ)− γδ(1− Û(δ))

]. (146)

T8. Fraction of customers that complete service without suffering a disaster, p. The rate of departuresfrom the system after a service completion is (1− r)f where f is given by (146). The rate of arrivals tothe system on the other hand is λmχ. Hence, by a ratio of rates argument, the probability that a customercompletes service and departs from the system without suffering a disaster is

p =δ

λmχ

(1− r)Ŝ(δ)1− Ŝ(δ)

[R̂(δ)− γδ(1− Û(δ))

]. (147)

It is worth noting that the above expression, together with Φ′(1) gives the following interesting formulafor the average number of customers in the system:

Φ′(1) =λmχδ

(1− p). (148)

This expression has a simple explanation: Written as δΦ′(1) = λmχ(1 − p), the left hand side is therate at which customers are removed due to disasters, as a result of PASTA. The right rand side is therate at which customers arrive multiplied by the probability that a customer will suffer a disaster. Thesetwo rates must of course be equal.

T9. Rate of ordinary busy period terminations, θ1:

θ1 = (1− r)∫ ∞

0P1(x)µ(x)dx

29

From (131) we have

V0(0) = θ1 +

∫ ∞0

V0(x)u(x)dx+

∫ ∞0

W0(x)r(x)dx

hence

V0(0) = θ1 +

∫ ∞0

V0(0)e−(λ+δ)x[1− U(x)]u(x)dx+

∫ ∞0

W0(0)e−(λ+δ)x[1−R(x)]r(x)dx

orθ1 = δ

(γδ[1− Û(λ+ δ)]− R̂(λ+ δ)

).

T10. Rate of busy period terminations by disasters, θ2:

θ2 = δP (1) = δδKδ(1)

1− Ŝ(δ)1− Ŝ(δ)

δ= δKδ(1) = δ

(R̂(δ)− γδ[1− Û(δ)]

).

T11. Probability that a busy period terminates by a departure, q1 and that it terminates by a disaster, q2.From a ratio of rates argument and the above results we have

q1 =θ1

θ1 + θ2=

γδ

(1− Û(δ + λ)

)− R̂(δ + λ)

R̂(δ)− R̂(λ+ δ) + γδ(Û(δ)− Û(λ+ δ)

) , (149)q2 =

θ2θ1 + θ2

=R̂(δ)− γδ

(1− Û(δ)

)R̂(δ)− R̂(λ+ δ) + γδ

(Û(δ)− Û(λ+ δ)

) . (150)T12. Rate of disasters. The rate of busy period terminations due to disasters is equal to the rate of repairinitiations given by W0(0) = δ.

T13. Rate of initiation of vacations. V0(0) = γδW0(0) = δγδ.

9.3 Laplace transform of an inactive period.

We define as inactive period the time that elapses from the moment the server ceases serving customers(either because of a service completion leaving the system empty or because of a disaster) to the momentthe next busy period begins (and thus the server begins serving customers again). We distinguish betweentwo types of inactive periods, those that begin with a disaster, and those that begin with a vacation. Ifwe denote by Î(s) the Laplace transform of the duration of a typical inactive period, by Î1(s) that ofan inactive period that begins with a vacation, and by Î2(s) that of an inactive period that begins with adisaster, then we have

Î(s) = q1Î1(s) + q2Î2(s) (151)

where the probabilities q1, q2, are given by (149), (150). We evaluate first Î1(s), taking into account thefact that, in the model considered in this section, disasters may occur during vacation or repair periods.Let An, n = 1, 2, . . ., denote the event that such an inactive period consists of a string of n− 1 vacationsduring which neither a disaster nor an arrival occurs, ending by a vacation during which either a disasteror an arrival occurs. Suppose that {Ui}, {Λi}, {∆i}, are three independent sequences of i.i.d. randomvariables. The elements of the first sequence are distributed according to the vacation distribution, while

30

those of the second and the third are exponentially distributed with rates λ and δ respectively. Then,taking into account the memoryless property of the exponential distribution,

P(An) = P {U1 < (Λ1 ∧∆1), . . . , Un−1 < (Λn−1 ∧∆n−1), ∆n < Un} .

Similarly, let Bn be the event that we have n− 1 consecutive vacations during which neither a vacationnor an arrival occurs, followed by a vacation during at least one arrival and no disaster occur. Then

P(Bn) = P {U1 < Λ1 ∧∆1, . . . , Un−1 < Λn−1 ∧∆n−1 , Λn < Un < ∆n} .

Clearly the events {An, Bn;n = 1, 2, . . .} are disjoint and their union is the whole sample space so that

E[e−sI1 ] =∞∑n=1

E[e−sI11(An)] + E[e−sI11(Bn)] (152)

On An it holds that I1 = U1 + · · ·+Un−1 + ∆n+ Ĩ2 since, during the nth vacation a disaster occurs andafter this an inactive period starting with a disaster begins. Here Ĩ2 is an independent random variablewith Laplace transform Î2(s). Thus

E[e−sI11(An)] = E[e−s(U1+···+Un−1)1 (U1 < Λ1 ∧∆1, . . . , Un−1 < Λn−1 ∧∆n−1)]×E[e−s∆n1(∆n < Un)] Î2(s). (153)

On the other hand, on Bn it holds that I1 = U1 + · · ·+ Un−1 + Un and therefore

E[e−sI11(Bn)] = E[e−s(U1+···+Un−1)1 (U1 < Λ1 ∧∆1, . . . , Un−1 < Λn−1 ∧∆n−1)]×{E[e−sUn1(Λn < Un < ∆n)]

}. (154)

Given that

E[e−s∆n1(∆n < Un)] = E[∫ Un

0e−sxδe−δxdx

]=

δ

δ + s

(1− Û(s+ δ)

),

E[e−sUn1(Λn < Un < ∆n)] = Û(s+ δ)− Û(s+ λ+ δ), (155)

and

E[e−s(U1+···+Un−1)1 (U1 < Λ1 ∧∆1, . . . , Un−1 < Λn−1 ∧∆n−1)] =n−1∏i=1

E[e−sUi1 (Ui < Λi ∧∆i)]

= Û(s+ λ+ δ)n−1 (156)

we have

Î1(s) =

∞∑n=1

Û(s+ λ+ δ)n−1[

δ

δ + s[1− Û(s+ δ)]Î2(s) + Û(s+ δ)− Û(s+ λ+ δ)

]whence we obtain

Î1(s) =δδ+s [1− Û(s+ δ)]Î2(s) + Û(s+ δ)− Û(s+ δ + λ)

1− Û(s+ δ + λ). (157)

The corresponding computation for the Laplace transform of an inactive period starting with a disas-ter can be obtained by a similar argument, which we will not present at the same level of detail. We will

31

consider again a partition of the sample space according to the number, n = 1, 2, . . . of consecutive re-pair times at the beginning of this inactive period. Of these, the first n−1 are all interrupted by disasters,whereas the n is completed without the occurrence of a disaster. If during this nth repair time customerarrivals occur then the end of the repair period is also the end of the inactive period, otherwise at the endof the repair period there follows an independent inactive period starting with a vacation with Laplacetransform given by (157). With {Ri} being an i.i.d. sequence of random variables with the distributionof the repair period we have the counterpart of (152), (153),

Î2(s) =

∞∑n=1

(E[e−s(∆1+···+∆n−1)1(R1 > ∆1, . . . , Rn−1 > ∆n−1)]×{E[e−sRn1(Rn < ∆n, Rn < Λ)] Î1(s) + E[e−sRn1(Λ < Rn < ∆)]

} ) .Evaluating the above quantities we obtain

Î2(s) =∞∑n=1

(δ

δ + s[1− R̂(s+ δ)]

)n−1 [R̂(s+ λ+ δ)Î1(s) + R̂(s+ δ)− R̂(s+ λ+ δ)

]whence we obtain

Î2(s) =R̂(s+ λ+ δ)Î1(s) + R̂(s+ δ)− R̂(s+ λ+ δ)

1− δδ+s [1− R̂(s+ δ)]. (158)

From (157) and (158) we obtain

Î1(s) =

[s+ δR̂(s+ δ)

] [1− Û(s+ λ+ δ)

]−[s+ δR̂(s+ λ+ δ)

] [1− Û(s+ δ)

][s+ δR̂(s+ δ)

] [1− Û(s+ λ+ δ)

]− δR̂(s+ λ+ δ)

[1− Û(s+ δ)

] ,Î2(s) =

(s+ δ)R̂(s+ δ)[1− Û(s+ λ+ δ)

]− (s+ δ)R̂(s+ λ+ δ)

[1− Û(s+ δ)

][s+ δR̂(s+ δ)

] [1− Û(s+ λ+ δ)

]− δR̂(s+ λ+ δ)

[1− Û(s+ δ)

] .From the above, together with (151) and (149), (150), we obtain the Laplace transform of the typicalinactive period

Î(s) =

[(δ + q2s)R̂(s+ δ) + q1s

] [1− Û(s+ λ+ δ)

]−[(δ + q2s)R̂(s+ λ+ δ) + q1s

] [1− Û(s+ δ)

][s+ δR̂(s+ δ)

] [1− Û(s+ λ+ δ)

]− δR̂(s+ λ+ δ)

[1− Û(s+ δ)

] .The corresponding mean values are given by

EI1 =1

δ

1− Û(δ)R̂(δ)[1− Û(λ+ δ)]− R̂(λ+ δ)[1− Û(δ)]

EI2 =1

δ

[1− R̂(δ)][1− Û(λ+ δ)] + R̂(λ+ δ)[1− Û(δ)]R̂(δ)[1− Û(λ+ δ)]− R̂(λ+ δ)[1− Û(δ)]

EI =1

δ

1− R̂(δ) + γδ[1− Û(δ)]R̂(δ)− R̂(λ+ δ) + γδ(Û(δ)− Û(λ+ δ))

10 Numerical Results

We present results for two types of systems. In the first, disasters cannot occur when the system is underrepair or on vacation. This is the system analyzed in sections 3 through 7 and we will refer to it in brief

32

as the partial disaster model. In the second system, analyzed in section 9, disasters occur even when theserver is on vacation or under repair and we will refer to it here as the total disaster model. Numericalresults are plotted for two performance criteria, the average number of customers in the system (35) and(141) and the probability of a customer completing service and departing without suffering a disaster (60)and (147). In all cases presented the Poisson arrival rate is λ = 1, the batch size is constant and equalto 1, and the feedback probability is r = 0.5. Regarding the distributions for the service time, repair,and vacation durations, two types of systems are also considered. In the first all these distributions areexponential and the in the second deterministic.

In figures 1 and 2 an exponential partial disaster system is considered. The mean number of cus-tomers and the probability of not suffering a disaster is plotted against the total mean service time forfour cases (small and large mean repair and vacation times). Total mean system time is mean servicetime divided by 1− r, i.e. the total expected time the customer spends in the server.

Figures 3 and 4 repeat the same for the total disaster system. Note that when the total mean servicetime exceeds 1 the systems would be instable in the absence of disasters. In each plot the disaster rateδ takes values from 0.01 (red) to 0.1 (blue). Notice in particular the behavior of the mean queue lengthwhen the mean vacation time is small and the mean repair time is large.

Figures 5–8 display graphs of the mean queue length and the probability of not suffering a disasterfor the exponential and deterministic system with partial disasters. Here the independent variable is thedisaster rate δ and several curves in each plot are given for values of the total mean service time rangingfrom 0.12 (red) to 1.2 (blue). Again the behavior when the mean vacation time is small (0.2) and meanrepair time is large (5) is particularly interesting. Also note the interesting non-monotonic behavior ofthe probability of not suffering a disaster when utilization is low (0.12) mean vacation time is large (5)and mean repair time small (0.2) clearly displayed for the deterministic system in figure 12.

11 Appendix

11.1 Determination of z0

Proposition 8. For δ > 0 the equation

z = (1− r + rz)Ŝ(δ + α(z)) (159)

has a single root, z0, inside the unit disk {z ∈ C : |z| < 1} which is real and positive.

Proof. Let h(z) := (1 − r + rz)Ŝ(δ + α(z)) − z. Since h(0) = (1 − r)Ŝ(δ + λ) > 0 and h(1) =Ŝ(δ)− 1 < 0 there exists z0 ∈ (0, 1) such that h(z0) = 0. This shows that (159) has a real and positiveroot within the unit disk. To complete the proof we must show that h has no other roots there. Letf(z) := −z and g(z) := (1 − r + rz)Ŝ(δ + α(z)) which are both analytic in |z| ≤ 1. When |z| ≤ 1,|1− r + rz| ≤ |1− r|+ |rz| = 1− r + r|z| ≤ 1. Thus

|g(z)| ≤ |1− r + rz|∫ ∞

0| e−(δ+α(z))x | dS(x) ≤

∫ ∞0

e−δx e−x

and thus |g(z)| ≤∫∞

0 e−δxdS(x) < 1 when |z| = 1. It follows by Rouché’s theorem that f(z) and

f(z) + g(z) will have the same number of zeros in the unit disk |z| < 1. Since f(z) has only one zero inthe unit disk (namely z = 0) h(z) also has a single zero, z0, there.

11.2 A probabilistic interpretation for γ

Consider the succession of repair and vacation periods, disregarding any busy periods that may intervene,as a discrete time process {Yn}, with state space {u, r}. We argue that this process is a Markov chainand determine its transition probability matrix. Indeed if Yn = r, then Yn+1 = u in two cases. Thefirst is that there are no arrivals during the repair period so that a vacation period starts immediately afterit, an event that happens with probability R̂(λ). The second is when there are arrivals and no disasterhappens during the ensuing busy period so that a vacation period follows after it. The probability of thisis R̂(α(z0)) − R̂(λ) and hence P(Yn+1 = u|Yn = r) = R̂(α(z0)). This argument also shows that theevent that Yn+1 = u, conditional on Yn = r, does not depend on Yk, k < n.

Similarly, suppose that Yn = u. Then the event that Yn+1 = u happens also in two ways: The firstcase is that there are no arrivals during the current vacation period and thus another vacation followsimmediately, and event that happens with probability Û(λ). In the second case there are arrivals, whichhappens with probability 1 − Û(λ). The pgf of the number of customers that arrive in such a period,given that there are arrivals is Û(α(z))−Û(λ)

1−Û(λ). A busy period ensues which, in the absence of vacations,

would have duration with Laplace transform Û(α(Γ̂(s)))−Û(λ)1−Û(λ)

. The probability that such a busy period isnot interrupted by a disaster is then

Û(α(Γ̂(δ)))− Û(λ)1− Û(λ)

Thus, using the fact that z0 = Γ̂(δ),

P(Yn+1 = u|Yn) = Û(λ) + (1− Û(λ))Û(α(Γ̂(δ)))− Û(λ)

1− Û(λ)= Û(α(z0)).

Hence the transition probability matrix of {Yn} is

( r ur 1− R̂(α(z0)) R̂(α(z0))u 1− Û(α(z0)) Û(α(z0))

)The stationary distribution (πr, πu) satisfies then the equation πu = πr R̂(α(z0)) + πu Û(α(z0)) andhence

πuπr

=R̂(α(z0))

1− Û(α(z0))= γ.

Therefore γ gives the relative frequency of the occurrence of vacation periods relative to repair periodsin steady state and provides a probabilistic interpretation for (28).

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