# On theGI/M/1 Queue with Vacations and Multiple Service Phases

Authors: Jianjun Li; Liwei Liu
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

This paper considers aGI/M/1 queue with vacations and multiple service phases. Whenever the system becomes empty, the server takes a vacation, causing the system to move to vacation phase 0. If the server returns from a vacation to find no customer waiting, another vacation begins. Otherwise, the system jumps from phase 0 to some service phase i with probability qi,i=1,2,…,N. Using the matrix geometric solution method and semi-Markov process, we obtain the distributions of the stationary system size at both arrival and arbitrary epochs. The distribution of the stationary waiting time of an arbitrary customer is also derived. In addition, we present some performance measures such as mean waiting time of an arbitrary customer, mean length of the type-i cycle, and mean number of customers in the system at the end of phase 0. Finally, some numerical examples are presented.


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## Body

## 1. Introduction

Queueing system with vacations has been well studied in the past three decades and widely used to model many problems in computer networks, communication, service systems, and manufacturing systems. Levy and Yechiali [1] were first to investigate the queueing model with vacations. Later, many researchers have made an extensive study of such a model under the assumption that the server takes vacations between two successive busy periods. Excellent surveys on queueing models with vacations can be found in Doshi [2], Takagi [3], Tian and Zhang [4], Ke et al. [5], and so on.We consider aGI/M/1 type vacation queue in which the interarrival times form an independent identically distributed sequence of random variables having a general distribution function. Some literatures on this type of vacation queue have emerged. In 1989, Tian et al. [6] applied the matrix analytic method to analyze a GI/M/1 queue with exhaustive service and multiple exponential vacations. Chatterjee and Mukherjee [7] analyzed the same model as that in Tian et al. [6]. But in [7], the vacation times are independently and identically distributed with a general distribution. Chae et al. [8] studied the GI/M/1 queue with single vacation and obtained the stochastic decomposition structures of queue length and the sojourn time. Baba [9] considered a GI/M/1 queue with multiple working vacations. Banik et at. [10] provided a study on a finite buffer GI/M/1 queue with multiple working vacations. The GI/M/1 queue with multiple working vacations and vacation interruption was analyzed by Li et al. [11]. Recently, a GI/M/1 queue with single working vacation can be seen in the survey of Li and Tian [12]. Tao et al. [13] analyzed the GI/M/1 queue with Bernoulli-schedule-controlled vacation and vacation interruption. Ye and Liu [14] investigated the GI/M/1 with single working vacation and vacations.In this paper, we extend the excellent works of Tian et al. [6] and Li and Liu [15] to a GI/M/1 vacation queue with multiple service phases. In [6], the service rate is a constant. In fact, in most of the literature on queueing theory, the service rate is homogeneous; that is, it is a constant. But, it is not reasonable in some cases. In real life, the service rate may change with the change of the environment. Several authors have contributed to the investigation of the queueing systems whose service rates are variable, including Yechiali and Naor [16], Neuts [17], Boxma and Kurkova [18], Cordeiro and Kharoufeh [19], and B. Kim and J. Kim [20]. Most recently, Paz and Yechiali [21] considered an M/M/1 queue in a multiphase random environment with disasters. In their model, the system moves from repair phase to some service environment with a certain probability after being repaired. Jiang et al. [22, 23] extended Paz and Yechiali’s work to an M/G/1 and GI/M/1 queue. In [15], Li and Liu studied a discrete-time Geo/G/1 queue with vacations in random environment in which the system jumps from vacation phase to a service phase with a certain probability when vacation phase ends. Inspired by [6, 15], we investigate the GI/M/1 vacation queue with multiple service phases. When vacation phase ends, the system switches to a service environment with a certain probability, where the service rate in each service phase may be different.The queueing system we discuss has potential application in several situations. For example, in manufacturing systems, a flexible manufacturing facility can be considered to be a server, which is mainly used to produce customer-specified products. When there are no customer backorders, the facility keeps on taking vacations. If, on return from a vacation, there is at least one customer order waiting, the facility then resumes service normally. But the service rate may change since the operator may be different.The rest of this paper is organized as follows. Section2 presents the model description. In Section 3, we provide the transition analysis for an embedded Markov chain. Section 4 is devoted to derive the stationary system size distribution at arrival epochs. Section 5 gives the stationary system size distribution at arbitrary epochs. Sections 6, 7, and 8 present the waiting time distribution, mean number of customers in the system at the end of phase 0, and results for cycle analysis, respectively. Section 9 provides some numerical examples and Section 10 concludes the paper.


## 2. Model Description

In this paper, we provide a detailed analysis of the GI/M/1 queue with vacations and multiple service phases. Our queueing system is described as follows. Whenever the system becomes empty, the server begins a vacation of random lengthV; that is to say, the system moves to vacation phase 0. If the server returns from a vacation to find no customer waiting, another vacation begins. Otherwise, the system jumps from phase 0 to some service phase i with probability qi,i=1,2,…,N, where qi>0 and ∑i=1Nqi=1. Let τl represent the arrival epoch of lth customer with τ0=0. The interarrival times {Al,l≥1} are independent and identically distributed (i.i.d.) with a general distribution function, denoted by A(t), a mean, denoted by 1/λ, and a Laplace Stieltjes transform (LST), denoted by A~(s). When the system is in phase 0, the server stops serving customers, and the vacation times are i.i.d. exponentially distributed with rate θ. When the system is in service phase i, the service times are i.i.d. exponential random variables with rate of μi,i=1,2,…,N. Note that the system can not directly move from one service phase to another service phase. That is, if the system becomes empty, it must move to phase 0 first. In addition, once the system moves to service phase i,i=1,2,…,N, the customers are served with the fixed rate μi. Finally, the interarrival times, service times, and vacation times are assumed to be mutually independent, and customers are served according to the discipline of first-come-first-served.


## 3. Embedded Markov Chain

LetL(t) denote the number of customers in the system at time t, and let Ll=L(τl-0) represent the number of customers before the lth arrival. Define(1)Jl=0,thelth  arrival  occurs  during  phase0,i,thelth  arrival  occurs  during  phasei,where i=1,2,…,N.Because both vacation and service times have memoryless property, the process{(Ll,Jl),l≥1} is an embedded Markov chain with the state space(2)Ω=m,0,m≥0∪m,i,m≥1,i=1,2,…,N.In order to express the transition matrix of{(Ll,Jl),l≥1}, let us define(3)Pn,i,m,j=PLl+1=m,Jl+1=j∣Ll=n,Jl=i,bk,i=∫0∞μitkk!e-μitdAt,k=0,1,2,…,i=1,2,…,N,vk,i=∫0∞∫0tμit-xkk!e-μit-xθe-θxdxdAt,k=0,1,2,…,i=1,2,…,N.Now, we consider the transition probabilities of(Ll,Jl). First, considering the case that the system is in service phase i,i=1,2,…,N, the transition from (n,i) to (m,i) occurs if there are n+1-m service completions during an interarrival time. Then, we have(4)Pn,i,m,i=∫0∞μitn+1-mn+1-m!e-μitdAt=bn+1-m,i,1≤m≤n+1,0,m≥n+2.The transition from (n,i) to (0,0) arises if n+1 customers are served and then the server enters into the vacation phase. From (4), we obtain(5)Pn,i,0,0=∫0∞∫0tμiμixnn!e-μixdxdAt=1-∑k=0nbk,i,n=1,2,….Second, the transition from (n,0) to (m,0) is possible, only if m=n+1. This transition occurs if the remaining vacation time is greater than an interarrival time. Then, we have(6)Pn,0,n+1,0=∫0∞e-θtdAt=A~θ.Third, the transition from (n,0) to (m,i) occurs if there are n+1-m service completions between the end of a vacation and a customer arriving. Therefore we have(7)Pn,0,m,i=qi∫0∞∫0tμit-xn+1-mn+1-m!e-μit-xθe-θxdxdAt=qivn+1-m,i,for m=1,2,…,n+1,i=1,2,…,N. For m=n+2,n+3,…, i=1,2,…,N, we have(8)Pn,0,m,i=0.From (6)–(8), we get(9)Pn,0,0,0=1-A~θ-∑i=1Nqi∑k=0nvk,i.If we sequence the states in the lexicographic order,0,0,k,0,k,1,…,k,N,k=1,2,…, the transition probability matrix of {(Ll,Jl),l≥1} can be written as the Block-Jacobi matrix(10)P=B00A0100⋯B10A1A00⋯B20A2A1A0⋯B30A3A2A1⋯⋮⋮⋮⋮,where(11)B00=1-A~θ-∑i=1Nqiv0,i,A01=A~θ,q1v0,1,q2v0,2,…,qNv0,N,Bk0=1-A~θ-∑i=1Nqi∑h=0kvh,i,1-∑h=0kbh,1,1-∑h=0kbh,2,…,1-∑h=0kbh,NT,A0=A~θq1v0,1q2v0,2⋯qNv0,N0b0,100000b0,200⋮⋮⋮⋱⋮0000b0,N,Ak=0q1vk,1q2vk,2⋯qNvk,N0bk,100000bk,2⋯0⋮⋮⋮⋱⋮0000bk,N,k≥1.It is clear that P is a stochastic matrix and the process {(Ll,Jl),l≥1} is irreducible and aperiodic.


## 4. Stationary Distribution at Arrival Epochs

In this section, we will derive the stationary distribution of{(Ll,Jl),l≥1} at the arrival epochs by using the matrix geometric approach. First, we introduce the following lemmas.Lemma 1.
The Markov chain{(Ll,Jl),l≥1} is stable if and only if ρi=λ/μi<1 and θ>0, i=1,2,…,N.Proof.
It is easy to see that our system acts as the classicalGI/M/1 queue with arrival rate λ and service rate μi during service phase i,i=1,2,…,N. So as long as ρi=λ/μi<1 and θ>0, i=1,2,…,N, the Markov chain {(Ll,Jl),l≥1} is stable. That is, ρi=λ/μi<1 and θ>0, i=1,2,…,N, are a sufficient condition for stability of {(Ll,Jl),l≥1}. On the other hand, obviously, if θ=0, the Markov chain {(Ll,Jl),l≥1} will not be stable, and if there is an i, ρi=λ/μi≥1,i=1,2,…,N, since qi>0, then when the system moves to service phase i, the Markov chain {(Ll,Jl),l≥1} will not be stable, either. Thus, ρi=λ/μi<1 and θ>0, i=1,2,…,N, are necessary for stability of {(Ll,Jl),l≥1}.Lemma 2.
Ifρi=λ/μi<1, then the equation z=A~(μi(1-z)) has a unique root ri,i in the range 0<z<1, i=1,2,…,N.Proof.
We setφ(z)=z-A~(μi(1-z)). Evidently, φ(0)=-A~(μi)<0, and φ(1)=1-A~(0)=0. Further, for 0<z<1, we have(12)φ′′z=-μi2∫0∞t2e-μi1-ztdAt<0.Since φ′(z)=1-μi∫0∞te-μi(1-z)tdA(t), ρi=λ/μi<1, then φ′(1)=1-1/ρi<0. Thus, the equation z=A~(μi(1-z)) has a unique root in the range 0<z<1, denoted by ri,i.Lemma 3.
Ifρi=λ/μi<1, i=1,2,…,N, then the matrix equation R=∑k=0∞‍RkAk has the minimal nonnegative solution(13)R=r0,0r0,1r0,2⋯r0,N0r1,100000r2,200⋮⋮⋮⋱⋮000⋯rN,N,where r0,0=A~(θ), ri,i is the unique root of equation z=A~(μi(1-z)) in the range 0<z<1, i=1,2,…,N, and(14)r0,i=qiθθ-μi1-A~θri,i-A~θ,i=1,2,…,N.Proof.
Since eachAk is a special upper diagonal matrix, if there exists the solution R of the matrix equation R=∑k=0∞RkAk, then R can be assumed having the same upper diagonal structure as(15)R=r0,0r0,1r0,2⋯r0,N0r1,100000r2,200⋮⋮⋮⋱⋮000⋯rN,N.Then(16)Rk=r0,0kr0,1∑j=0k-1r0,0jr1,1k-1-jr0,2∑j=0k-1r0,0jr2,2k-1-j⋯r0,N∑j=0k-1r0,0jrN,Nk-1-j0r1,1k00000r2,2k00⋮⋮⋮⋱⋮000⋯rN,Nk,k≥1.Substituting Rk into the matrix equation, we have(17)r0,0=A~θ,ri,i=∑k=0∞ri,ikbk,i=A~μi1-ri,i,i≥1,r0,i=qi∑k=0∞r0,0kvk,i+r0,i∑k=1∞bk,i∑j=0k-1ri,ijr0,0k-1-j,i=1,2,…,N.Thus, r0,0=A~(θ), and from Lemma 2, ri,i is the unique root of equation z=A~(μi(1-z)) in the range 0<z<1, i=1,2,…,N. Substituting r0,0=A~(θ) in the third equation of (17), we have(18)qi∑k=0∞r0,0kvk,i=qi∑k=0∞A~θkvk,i=qi∫0∞∫0t∑k=0∞A~θkμit-xkk!e-μit-xθe-θxdxdAt=qi∫0∞∫0te-μit-x1-A~θ-θxθdxdAt=qiθθ-μi1-A~θA~μi1-A~θ-A~θ,1-∑k=1∞bk,i∑j=0k-1ri,ijA~θk-1-j=1-1ri,i-A~θ∑k=1∞bk,iri,ik-A~θk=1-1ri,i-A~θri,i-A~μi1-A~θ=1ri,i-A~θA~μi1-A~θ-A~θ.By substituting (18) and in the third equation of (17), we can obtain (19)r0,i=qiθθ-μi1-A~θri,i-A~θ,i=1,2,…,N.Then the proof is completed.Letπ=π0,π1,π2,…,πk,… be the stationary probability vector of {(Ll,Jl),l≥1}, where(20)π0=π0,0=liml→∞⁡PLl=0,Jl=0,πk=πk,0,πk,1,…,πk,N,k≥1,πk,i=liml→∞⁡PLl=k,Jl=i,k≥1,i=0,1,2,…,N.Theorem 4.
Ifρi=λ/μi<1 and θ>0, i=1,2,…,N, the stationary probability vector π0,0,π1,0,π1,1,…,π1,N exists and is given by(21)π0,0,π1,0,π1,1,…,π1,NBR=π0,0,π1,0,π1,1,…,π1,N,π0,0+π1,0,π1,1,…,π1,NI-R-1e=1,and for k≥2, πk can be obtained by(22)πk=π1Rk-1,where e is a column vector with all its components equal to one, I is the identity matrix, and(23)BR=Δ0β0β1β2⋯βNγ00δ1δ2⋯δNγ10Δ10⋯0γ200Δ2⋯0⋮⋮⋮⋮⋱⋮γN000⋯ΔNwith(24)Δ0=1-r0,0-∑i=1Nqiθθ-μiA~μi-r0,0,β0=r0,0,βi=qiθθ-μiA~μi-r0,0,i=1,2,…,N,δi=qir0,0θθ-μi1-r0,0A~μi1-r0,0-r0,0-θθ-μiA~μi-r0,0+r0,ir0,0-ri,i1r0,0A~μi1-r0,0-A~μi-1ri,iri,i-A~μi,i=1,2,…,N,γ0=1-∑i=1Nδi,γi=A~μiri,i,i=1,2,…,N,Δi=1-A~μiri,i=1-γi,i=1,2,…,N.Proof.
Based on Theorem  1.5.1 of Neuts [24], if the Markov chain {(Ll,Jl),l≥1} is positive recurrent, then the matrix(25)BR=B00A01∑k=1∞Rk-1Bk0∑k=1∞Rk-1Akhas positive left invariant vector, and π0,0,π1,0,π1,1,…,π1,N is given by the positive left invariant vector and the normalizing condition(26)π0,0+π1,0,π1,1,…,π1,NI-R-1e=1,where e is a column vector with all its components equal to one and I is the identity matrix. From Lemma 1, if ρi=λ/μi<1 and θ>0, i=1,2,…,N, then {(Ll,Jl),l≥1} is positive recurrent; thus we have (21) and (22).
Fork≥2, using equation (1.5.4)  of Neuts [24], we obtain(27)πk=π1Rk-1.
Now, we are ready to derive the expression ofB[R] in the following.
Substituting the expressions forB00, A01, R, Ak, and Bk0 in (25), we have(28)BR=Δ0β0β1β2⋯βNγ00δ1δ2⋯δNγ10Δ10⋯0γ200Δ2⋯0⋮⋮⋮⋮⋱⋮γN000⋯ΔN,where(29)Δ0=B00=1-A~θ-∑i=1Nqiv0,i,β0,β1,…,βN=A01=A~θ,q1v0,1,q2v0,2,…,qNv0,N,γ0,γ1,…,γNT=∑k=1∞Rk-1Bk0=∑k=1∞r0,0k-11-A~θ-∑i=1Nqi∑h=0kvh,i+∑i=1Nr0,i∑j=0k-2r0,0jri,ik-2-j1-∑h=0kbh,i∑k=1∞r1,1k-11-∑h=0kbh,1⋮∑k=1∞rN,Nk-11-∑h=0kbh,N,0δ1⋯δN0Δ1⋯0⋮⋮⋱⋮00⋯ΔN=∑k=1∞Rk-1Ak=∑k=1∞0q1r0,0k-1vk,1+r0,1bk,1∑j=0k-2r0,0jr1,1k-2-j⋯qNr0,0k-1vk,N+r0,Nbk,N∑j=0k-2r0,0jrN,Nk-2-j0r1,1k-1bk,1⋯0⋮⋮⋱⋮00⋯rN,Nk-1bk,N.
ForΔ0, we first have(30)v0,i=∫0∞∫0te-μit-xθe-θxdxdAt=θθ-μiA~μi-A~θ,i=1,2,…,N.From r0,0=A~(θ) (see Lemma 3), then Δ0 can be obtained by(31)Δ0=1-A~θ-∑i=1Nqiv0,i=1-r0,0-∑i=1Nqiθθ-μiA~μi-r0,0.Obviously, we have β0=A~(θ)=r0,0, and from (30), we obtain(32)βi=qiv0,i=qiθθ-μiA~μi-r0,0,i=1,2,…,N.
Forδi,i=1,2,…,N, we have(33)δi=∑k=1∞qir0,0k-1vk,i+r0,ibk,i∑j=0k-2r0,0jri,ik-2-j=qi∑k=1∞r0,0k-1∫0∞∫0tμit-xkk!e-μit-xθe-θxdxdAt+r0,i∑k=1∞bk,ir0,0k-1-ri,ik-1r0,0-ri,i=qir0,0∫0∞∫0teμir0,0t-x-1e-μit-xθe-θxdxdAt+r0,ir0,0-ri,i∫0∞∑k=1∞μir0,0tkr0,0k!e-μitdAt-∫0∞∑k=1∞μiri,itkri,ik!e-μitdAt=qir0,0θθ-μi1-r0,0A~μi1-r0,0-A~θ-θθ-μiA~μi-A~θ+r0,ir0,0-ri,i1r0,0A~μi1-r0,0-A~μi-1ri,iri,i-A~μi.
It is easy to obtain the expressions forΔi,i=1,2,…,N,(34)Δi=∑k=1∞ri,ik-1bk,i=∑k=1∞ri,ik-1∫0∞μitkk!e-μitdAt=1ri,iA~μi1-ri,i-A~μi=1ri,iri,i-A~μi=1-A~μiri,i.To compute the expression for δ0, we first have(35)∑k=1∞r0,0k-11-A~θ-∑i=1Nqi∑h=0kvh,i=11-r0,0-A~θ1-r0,0-∑i=1Nqi∑h=0∞∑k=h∞r0,0k-1∫0∞∫0tμit-xhh!e-μit-xθe-θxdxdAt=11-r0,0-A~θ1-r0,0-∑i=1Nqi1-r0,0r0,0∫0∞e-μi1-r0,0t∫0tθe-θ-μi1-r0,0xdxdAt+∑i=1Nqi1-r0,0r0,0∫0∞e-μit∫0tθe-θ-μixdxdAt-∑i=1Nqi1-r0,0∫0∞e-μit∫0tθe-θ-μixdxdAt=11-r0,0-A~θ1-r0,0-∑i=1Nqiθ1-r0,0r0,0θ-μi1-r0,0A~μi1-r0,0-A~θ+∑i=1Nqiθr0,0θ-μiA~μi-A~θ,∑k=1∞∑i=1Nr0,i∑j=0k-2r0,0jri,ik-2-j1-∑h=0kbh,i=∑k=1∞∑i=1Nr0,ir0,0k-1-ri,ik-1r0,0-ri,i1-∑h=0kbh,i=∑i=1Nr0,ir0,0-ri,i11-r0,0-11-ri,i-∑k=1∞r0,0k-1∑h=0kbh,i+∑k=1∞ri,ik-1∑h=0kbh,i=∑i=1Nr0,ir0,0-ri,i11-r0,0-11-ri,i-∑i=1Nr0,ir0,0-ri,i11-r0,0∫0∞∑h=1∞μir0,0thr0,0h!+1e-μitdAt+∑i=1Nr0,ir0,0-ri,i11-ri,i∫0∞∑h=1∞μiri,ithri,ih!+1e-μitdAt=∑i=1Nr0,ir0,0-ri,i11-r0,0-11-ri,i-∑i=1Nr0,ir0,0-ri,i11-r0,0r0,0A~μi1-r0,0-A~μi+11-r0,0A~μi+∑i=1Nr0,ir0,0-ri,i11-ri,iri,iA~μi1-ri,i-A~μi+11-ri,iA~μi=∑i=1Nr0,ir0,0-ri,i11-r0,0+1r0,0-1ri,iA~μi-11-r0,0r0,0A~μi1-r0,0.
Therefore, from (14),(36)δ0=11-r0,0-A~θ1-r0,0-∑i=1Nqiθ1-r0,0r0,0θ-μi1-r0,0A~μi1-r0,0-A~θ+∑i=1Nqiθr0,0θ-μiA~μi-A~θ+∑i=1Nr0,ir0,0-ri,i11-r0,0+1r0,0-1ri,iA~μi-11-r0,0r0,0A~μi1-r0,0=1-∑i=1Nr0,ir0,0-ri,i1r0,0A~μi1-r0,0-A~μi+1ri,iA~μi-1+∑i=1Nqiθr0,0θ-μiA~μi-A~θ-∑i=1Nqiθr0,0θ-μi1-r0,0A~μi1-r0,0-A~θ=1-∑i=1Nδi.
Finally, the expressions forγi,i=1,2,…,N, can be easily obtained by(37)γi=∑k=1∞ri,ik-11-∑h=0kbh,i=11-ri,i-11-ri,iri,i∫0∞∑h=1∞μiri,ithh!dAt-11-ri,iA~μi=11-ri,i-11-ri,iri,iA~μi1-ri,i-A~μi-11-ri,iA~μi=A~μiri,i.Then the proof is completed.Remark 5.
IfN=1, let μ1=μ; we have(38)R=r0,0r0,10r1,1,where r0,0=A~(θ), r11 is the unique root in the range 0<z<1 of equation z=A~(μ(1-z)), and r0,1=θ/(θ-μ(1-r0,0))(r1,1-r0,0), which are derived by Tian et al. [6].Remark 6.
IfN=1, let μ1=μ; then(39)BR=Δ0β0β1γ00δ1γ10Δ1,where(40)Δ0=1-r0,0-θθ-μA~μ-r0,0,β0=r0,0,β1=θθ-μA~μ-r0,0,γ0=1-δ1,δ1=θθ-μ1-r0,0A~μr0,0-A~μr1,1-θr0,0θ-μA~μ-r0,0,γ1=A~μr1,1,Δ1=1-A~μr1,1.From (21) and (22), we have(41)πk,0=1-r1,1σr0,0k,k≥0,πk,1=1-r1,1σβr1,1k-r0,0k,k≥1,where(42)σ=θ-μ1-r0,0θ-μ1-r1,1,β=θθ-μ1-r0,0.All the above results coincide with the results derived in [6].


## 5. The Stationary Distribution at Arbitrary Epochs

In this section, we will derive the limiting distribution ofL(t) by using the method of the semi-Markov process (SMP). To this end, we consider another stochastic process {(Z(t),K(t)),t≥0}, where Z(t)=Ll,τl≤t<τl+1, and K(t)=Jl,τl≤t<τl+1. Clearly, (Z(t),K(t)) is a SMP, and (Ll,Jl) is its embedded Markov chain.Letωk,i denote the time that SMP (Z(t),K(t)) resides in state (k,i), where (k,i)∈Ω. By the definition of SMP, we have P{ωk,i≤t}=A(t), and E(ωk,i)=λ-1, for all (k,i)∈Ω. Let(43)fk,i=limt→∞⁡PZt=k,Kt=i.Then(44)fk,i=πk,iEωk,i∑h,i∈Ωπh,iEωh,isee  Gross  et  al.25,P.299.Evidently, we have fk,i=πk,i; that is, the SMP and its embedded Markov chain have an identical limiting distribution.Letpk=limt→∞⁡PL(t)=k,k=0,1,…. From the theory of SMP (see Gross et al. [25], P. 299), we obtain(45)pk=∑h,i∈Ωfh,i∑j=0N∫0∞Prequired  changes  intto  bring  state  fromh,itok,jλ1-Atdt.For k≥2, from (22), we have(46)pk=∑h=k-1∞πh,0∑j=1Nqj∫0∞∫0tμjt-xh+1-kh+1-k!e-μjt-xθe-θxdxλ1-Atdt+πk-1,0∫0∞e-θtλ1-Atdt+∑i=1N∑h=k-1∞πh,i∫0∞μith+1-kh+1-k!e-μitλ1-Atdt=π1,0r0,0k-2∑j=1Nqj∫0∞∫0te-μj1-r0,0t-xθe-θxdxλ1-Atdt+π1,0r0,0k-2λθ1-A~θ+∑i=1N∑h=k-1∞π1,0r0,ir0,0h-1-ri,ih-1r0,0-ri,i+π1,iri,ih-1∫0∞μith+1-kh+1-k!e-μitλ1-Atdt=π1,0r0,0k-2θλ∑j=1Nqjθ-μj1-r0,01-A~μj1-r0,0μj1-r0,0-1-A~θθ+π1,0r0,0k-2λ1-A~θθ+π1,0r0,0k-2λ∑i=1Nr0,ir0,0-ri,iμi1-r0,01-A~μi1-r0,0-π1,0λ∑i=1Nr0,iri,ik-2r0,0-ri,iμi+λ∑i=1Nπ1,iri,ik-2μi.Similarly, fork=1, we have(47)p1=π0,0θλ∑j=1Nqjθ-μj1-A~μjμj-1-A~θθ+π1,0r0,0-1θλ∑j=1Nqjθ-μj1-r0,01-A~μj1-r0,0μj1-r0,0-1-A~θθ-π1,0r0,0-1θλ∑j=1Nqjθ-μj1-A~μjμj-1-A~θθ+π0,0λθ1-A~θ+π1,0λ∑i=1Nr0,ir0,0-ri,ir0,01-A~μi1-r0,0μi1-r0,0-1-A~μiμi-π1,0λ∑i=1Nr0,iA~μir0,0-ri,iri,iμi+λ∑i=1Nπ1,iA~μiri,iμi.Finally, for k=0, we can obtain p0=1-∑k=1∞‍pk.Then, the expected stationary system size at an arbitrary epoch, denoted byE(L), can be given by(48)EL=∑k=0∞kpk.Remark 7.
IfN=1, let μ1=μ; substituting the results in Remarks 5 and 6 into (46) and (47), we obtain(49)p1=1-r1,1σλμβ+λθ1-β1-r0,0,pk=1-r1,1σλμβr1,1k-1+λθr0,0k-11-β1-r0,0,k≥2;then(50)p0=1-σλμβ+λθ1-β1-r1,1,which coincide with the result derived in [6].


## 6. The Stationary Waiting Time Distribution

LetW and W~(s) denote the stationary waiting time of an arbitrary customer and its LST, respectively. Due to the memoryless property of exponential distribution, the waiting time distribution of an arbitrary customer is given by(51)Wt=PW≤t=π0,0Vt+∑k=1∞πk,0∑i=1NqiV∗Bik∗t+∑i=1N∑k=1∞πk,iBik∗t,where V(t) denotes the distribution of the vacation time, Bi(t) denotes the distribution of the service time in service phase i, star ∗ denotes convolution, and Bik∗(t) denotes the k-fold convolution of Bi(t).Then, from (22) and (51), we obtain the LST of the waiting time distribution as follows:(52)W~s=π0,0V~s+∑k=1∞πk,0∑i=1NqiV~sB~isk+∑i=1N∑k=1∞πk,iB~isk=π0,0θθ+s+π1,0∑k=1∞r0,0k-1∑i=1Nθθ+sμiμi+sk+∑i=1N∑k=1∞π1,0r0,ir0,0k-1-ri,ik-1r0,0-ri,i+π1,iri,ik-1μiμi+sk=θθ+sπ0,0+∑i=1Nπ1,0μiμi1-r0,0+s+∑i=1Nπ1,iμiμi1-ri,i+s+∑i=1Nπ1,0r0,iμir0,0-ri,i1μi1-r0,0+s-1μi1-ri,i+s,where V~(s) and B~i(s) denote the LSTs of V(t) and Bi(t), respectively.By (52), the mean waiting time of an arbitrary customer can be given by(53)EW=-dW~sdss=0=π0,0θ+Nπ1,0θ1-r0,0+∑i=1Nπ1,0μi1-r0,02+∑i=1Nπ1,iμi1-ri,i2+∑i=1Nπ1,0r0,ir0,0-ri,iμi11-r0,02-11-ri,i2.Remark 8.
IfN=1, let μ1=μ; substituting the results in Remarks 5 and 6 into (52) and (53), we have(54)W~s=1-r1,1σθθ+s+μr0,0μ1-r0,0+sθθ+s-β+βμr1,1μ1-r0,0+s,EW=1θ+r1,1μ1-r1,1,which are derived in [6].


## 7. Mean Number of Customers in the System at the End of Phase 0

LetL0 denote the mean number of customers in the system at the end of phase 0, and let A+ represent the remaining interarrival time. The probability that no customer arrives during a vacation period is given by(55)PA+>V=∫0∞1-e-θtλ1-Atdt=1-λθ1-A~θ.That is, the probability that at least one customer arrives during a vacation period is λ(1-A~(θ))/θ.In our model, if a vacation ends, the server finds no customer waiting, and a new vacation begins; otherwise, the system moves from phase0 to some service phase i. Thus, the number of vacations in phase 0 is distributed according to geometric distribution with parameter λ(1-A~(θ))/θ. Because the mean time of each vacation is 1/θ, the mean time that the system spends in phase 0, denoted by T0, can be obtained by(56)T0=∑k=1∞1-λ1-A~θθk-1λ1-A~θθkθ=1λ1-A~θ,which is the product of the mean number of vacations that the system resides in phase 0 and the mean time of each vacation.From (56), the mean number of customers in the system at the end of phase 0 is(57)L0=1λ1-A~θ·λ=11-A~θ,which is just the product of the mean time that the system spends in phase 0 and the arrival rate of customers in phase 0.


## 8. Cycle Analysis

We define a cycle as the time interval between two successive epochs at which phasei commences, i=0,1,…,N. Then there are (N+1) types of cycles. The length of type-i cycle is denoted by Ci.First, we derive the mean length of the type-0 cycle.During the type-0 cycle, the system visits vacation phase 0 and certain service phase i one time, i=1,2,…,N, respectively. We know that the mean time that the system spends in phase 0 is T0 (see (56)). And when the system jumps from phase 0 into certain service phase i with probability qi,i=1,2,…,N, from (57), the mean time that the system spends in service phase i should be equal to the product of the mean number of customers in the system at the end of phase 0 and the mean busy period that one customer induces in service phase i. Let Ti represent the mean time that the system resides in service phase i. Then we have(58)Ti=11-A~θ·1μi-λ,i=1,2,…,N,where 1/(μi-λ) is the mean busy period that one customer induces in service phase i.Therefore, the mean length of the type-0 cycle is(59)EC0=T0+∑i=1NqiTi=1λ1-A~θ+11-A~θ∑i=1Nqiμi-λ.Next, we consider the type-i cycle, i=1,2,…,N.During the type-i cycle, the time that the system visits phase 0 follows a geometric distribution with parameter qi. And it is easy to find that the mean length of type-i cycle during which the system visits phase 0k times is (60)Ti+T0+k-1T0+∑j=1,j≠iNqj1-qiTj.Then we can obtain the mean length of type-i cycle as follows:(61)ECi=∑k=1∞1-qik-1qiTi+T0+k-1T0+∑j=1,j≠inqj1-qiTj=Ti+T0+∑k=1∞1-qik-1qik-1T0+11-qiα-T0-qiTi=Ti+T0+1-qiqiT0+11-qiα-T0-qiTi=αqi,where α=T0+∑i=1NqiTi.


## 9. Numerical Examples

In this section, some numerical experiments are shown to explore the effect of the model parameters on the system performance measures. First, we assume that customers arrive according to a Poisson process with rateλ, and N=2; that is, the system we consider translates into an M/M/1 queue with two service phases and a vacation phase. The model parameters are chosen as λ=1.2, μ1=2, μ2=2.5, q1=0.4, q2=0.6, and θ=1.4,1.5,1.6, unless they are considered as variables or their values are given in the respective figures. Using Lemma 3 and Theorem 4, the stationary state probability {πk,k≥0} can be obtained. Then we can derive a variety of performance measures of the system.Figure1 demonstrates the effect of λ on E(W) for different vacation rate θ. As intuitively expected, the mean waiting time of an arbitrary customer increases with the increase of λ for any θ. And for a fixed λ, E(W) decreases as θ increases. This is obvious because as θ increases, the mean time of vacation decreases, and there are fewer customers in the system. The effect of μ2 on E(W) for different values of θ is illustrated in Figure 2. It can be observed that E(W) decreases with the increase of μ2 for different values of θ. And if μ2 is fixed, as intuition tells us, E(W) decreases as θ increases. In Figure 3, we plot the change trend of the mean waiting time of an arbitrary customer with varying values of q1. It is obvious that, for any θ, the larger q1 is, the larger the mean waiting time E(W) becomes. And for a fixed q1, E(W) decreases with the increase of θ, as we expect.Figure 1
The effect ofλ on E(W).Figure 2
The effect ofμ2 on E(W).Figure 3
The effect ofq1 on E(W).Figure4 demonstrates the effect of λ on π0,0 for different vacation rate θ. It can be observed that, with the increase of λ, the probability that the system is empty declines for any θ, and if λ is fixed, π0,0 increases with the increase of θ. Evidently, the change trends agree with the intuitive expectations. The effect of μ2 on π0,0 is illustrated in Figure 5. As intuitively expected, we can see that increased μ2 leads to the larger π0,0 for any given θ. Moreover, if μ2 is fixed, π0,0 increases with the increase of θ. In Figure 6, we plot the change trend of π0,0 with q1 varying. We can find that π0,0 decreases as q1 increases for any given θ. For a given q1, when q1 is relatively small, π0,0 increases with the increase of θ; however, when q1 is relatively large, π0,0 decreases with the increase of θ. That is, when θ is larger, the influence of q1 to π0,0 is greater. This phenomenon can be explained from the fact that μ1<μ2.Figure 4
The effect ofλ on π0,0.Figure 5
The effect ofμ2 on π0,0.Figure 6
The effect ofq1 on π0,0.Then, in Figure7, we present a comparison result of three queueing models: the M/M/1, E2/M/1, and D/M/1 models, where E2 stands for a 2-Erlangian arrival interval and D stands for a deterministic arrival interval. As demonstrated in Figure 7, the mean waiting time E(W) increases with the increase of λ for the three models, and for a fixed λ, the M/M/1 model has the largest value of E(W), and the D/M/1 model has the smallest value.Figure 7
The effect ofλ on E(W) for three interarrival time distributions (θ=1.5).Finally, Figures8–10 demonstrate the effect of (μ1,μ2) and (q1,q2) on E(W) for the above three interarrival distributions when μ¯ is fixed, where μ¯=(μ1+μ2)/2. In Figure 8, we plot the change trend of E(W) as μ2 increases from 2 to 3 under the corresponding three models, given that μ¯=2.5 and q1=q2=0.5. We can observe that E(W) is symmetrical about μ2=2.5 and the system efficiency is best when μ2=2.5 for each model. However, in Figures 9 and 10, we can see that if μ¯=2.5, q1=0.2, and q2=0.8, E(W) is not symmetrical and the value of μ2 that the system efficiency is best is greater than 2.5; if μ¯=2.5, q1=0.8, and q2=0.2, E(W) is not symmetrical as well and the value of μ2 that the system efficiency is best is less than 2.5. Note that when μ1=μ2=2.5, the model with two service phases actually translates into the model with only one service phase. From Figures 8–10, we can see that multiple service phases have important influence on the performance of the queueing system.Figure 8
The effect ofμ1 and μ2 on E(W) (μ¯=2.5, q1=q2=0.5, θ=1.5).Figure 9
The effect ofμ1 and μ2 on E(W) (μ¯=2.5, q1=0.2,q2=0.8, θ=1.5).Figure 10
The effect ofμ1 and μ2 on E(W) (μ¯=2.5, q1=0.8,q2=0.2, θ=1.5).


## 10. Conclusion

In this paper, we have carried out an analysis ofGI/M/1 queue with vacations and multiple service phases. We have obtained the distributions of the stationary system size at arrival and arbitrary epochs. Further, we have obtained the stationary waiting time distribution of an arbitrary customer. Some important performance measures of the system such as the mean waiting time of an arbitrary customer, the mean length of the type-i cycle, and the mean number of customers in the system at the end of phase 0 are obtained. Finally, the influence of the model parameters on several performance characteristics is demonstrated through some numerical experiments.



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*Source: 3246934-2017-08-08.xml*