# The M/M/N Repairable Queueing System with Variable Breakdown Rates

Authors: Shengli Lv; Jingbo Li
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

This paper considers the M/M/N repairable queuing system. The customers' arrival
is a Poisson process. The servers are subject to breakdown according to Poisson processes with
different rates in idle time and busy time, respectively. The breakdown servers are repaired by repairmen,
and the repair time is an exponential distribution. Using probability generating function
and transform method, we obtain the steady-state probabilities of the system states, the steady-state
availability of the servers, and the mean queueing length of the model.


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## Body

## 1. Introduction

In queuing researches, many researchers have studied the queuing system with repairable servers. Most of the works of the repairable queuing system deal with the single-server models [1–9]. The works about multiserver repairable systems is not sufficient. Mitrany and Avi-Itzhak [10] analyzed the model with N units of servers and the same amount of repairmen, they obtained the steady-state mean queuing length of customers. Neuts and Lucantoni [11] studied the model with N units of repairable servers and c (0≤c≤N) repairmen by matrix analysis method and obtained the steady-state properties of the model.In recent years, many flexible policies have been introduced to the repairable systems. Gray et al. [5] studied the model with a single server which may take a vacation in idle times and may breakdown in busy times; they obtained the mean queue length. Altman and Yechiali [12] presented a comprehensive analysis of the M/M/1 and M/G/1 queues, as well as of the M/M/c queue with server vacations; they obtained various closed-form results for the probability generating function (PGF) of the number of the customers. Zhang and Hou [6] analyzed an M/G/1 queue with working vacations and vacation interruptions; they obtained the queue length distribution and steady-state service status probability. Yang et al. [7] analyzed an M/G/1 queuing system with second optional service, server breakdowns, and general startup times under (N,p)-policy, they obtained the explicit closed-form expression of the joint optimum threshold values of (N,p) at the minimum cost. Chang et al. [8] studied the optimal management problem of a finite capacity M/H2/1 queuing system, where the unreliable server operates F-policy, a cost model is developed to determine the optimal capacity K, the optimal threshold F, the optimal setup rate, and the optimal repair rate at a minimum cost. Wang [9] used a quasi-birth-and-death (QBD) modeling approach to model queuing-inventory systems with a single removable server, performance measures are obtained by using both hybrid and standard procedures; an optimal control policy is proposed and verified through numerical studies.The most works of repairable queuing system assumed that the server breakdown rate is constant, but the breakdown rate of a server may be variable in a real system. It is well known that many kinds of machine are easy to breakdown at their busy times, and some equipments may be easy to fail after a long idle period. For example, the tires of the truck prefer to breakdown when the truck is running on the road. On the other hand, the storage battery in an automobile may not work if the automobile is idle for long period. For the actual demands of the above cases, we study a multiserver repairable queuing system in this paper, and assume that the unreliable servers have different breakdown rates in their busy times and idle times, respectively.The rest of this paper is organized as follows. Section2 describes the model and gives the balance equations. Section 3 presents the equations of PGF. The steady-state availability of the system is derived in Section 4. The steady-state probabilities of the system states and mean queuing length are obtained in Section 5. Case analysis is given in Section 6. Section 7 presents the conclusions.


## 2. Model Description

The model characteristics are as follows.(1)
There areN units of identical servers in the system. The servers are subject to breakdown according to Poisson processes with different rates which are ξ1 in idle times and ξ2 in busy times, respectively.(2)
Customers arrive according to a Poisson process with rateλ. The service discipline is first come first served (FCFS). The service which is interrupted by a server breakdown will become the first one of the queue of customers. The service time distribution is an exponential distribution with parameter μ.(3)
There arec(1≤c≤N) reliable repairmen to maintain the unreliable servers. The repair discipline is first come first repaired (FCFR). The repair time distribution is an exponential distributions with parameter η. A server is as good as a new one after repair.We defineX(t)≡
the number of available servers in the system at the momentt(0⩽X(t)⩽N),Y(t)≡ the number of customers in the system at the moment t(0⩽Y(t)).The stochastic process {X(t),Y(t),t≥0} is a two-dimensional Markov process which is called quasi-birth-and-death (QBD) process [11] with state space {(i,j),0≤i≤N,j≥0}.LetPi,j(t) denote the probability that the system is in a state of (i,j) at the moment t, and Pi,j denote the steady-state probability of Pi,j(t), then we have
(1)Pi,j={limt→∞Pi,j(t),i=0,1,2,…,N,j=0,1,2,…,0,other.
Assuming that the system is positive recurrent, the balance equations are as follows:
(2)(λ+cη)P0,0=ξ1P1,0,(λ+cη)P0,j=λP0,j-1+ξ2P1,j,j>0,(λ+cη+iξ1)Pi,0=cηPi-1,0+μPi,1+(i+1)ξ1Pi+1,0,0<i≤N-c,j=0,Pi,j[λ+cη+jμ+(i-j)ξ1+jξ2]=λPi,j-1+cηPi-1,j+(j+1)μPi,j+1+[(i+1-j)ξ1+jξ2]Pi+1,j,0<i≤N-c,0<j<i,Pi,j[λ+cη+iμ+iξ2]=λPi,j-1+cηPi-1,j+iμPi,j+1+(ξ1+jξ2)Pi+1,j,0<i≤N-c,j=i,Pi,j[λ+cη+iμ+iξ2]=λPi,j-1+cηPi-1,j+iμPi,j+1+(i+1)ξ2Pi+1,j,0<i≤N-c,j>i,Pi,0[λ+(N-i)η+iξ1]=(N-i+1)ηPi-1,0+μPi,1+(i+1)ξ1Pi+1,0,N-c<i<N,j=0,Pi,j[λ+(N-i)η+jμ+(i-j)ξ1+jξ2]=(N-i+1)ηPi-1,j+λPi,j-1+(j+1)μPi,j+1+[(i+1-j)ξ1+jξ2]Pi+1,j,N-c<i<N,0<j<i,Pi,j[λ+(N-i)η+iμ+iξ2]=λPi,j-1+(N-i+1)ηPi-1,j+iμPi,j+1+(ξ1+jξ2)Pi+1,j,N-c<i<N,j=i,Pi,j[λ+(N-i)η+iμ+iξ2]=λPi,j-1+(N-i+1)ηPi-1,j+iμPi,j+1+(i+1)ξ2Pi+1,j,N-c<i<N,j>i,PN,0(λ+Nξ1)=ηPN-1,0+μPN,1,i=N,j=0,PN,j[λ+jμ+(N-j)ξ1+jξ2]=λPN,j-1+ηPN-1,j+(j+1)μPN,j+1,i=N,0<j<N,PN,j(λ+Nμ+Nξ2)=λPN,j-1+ηPN-1,j+NμPN,j+1,i=N,j≥N.Here, we give the derivation of the second equation of (2). Since the process {X(t),Y(t),t≥0} is a vector Markov process of continuous time, we write the equations of the state of (0,j) by considering the transitions occurring between the moments t and t+Δt(Δt>0) as follows:
(3)P0,j(t+Δt)=P0,j-1(t)λΔt+P1,j(t)ξ2Δt+P0,j(t)[1-(λ+cη)Δt]+o(Δt),
then we have
(4)P0,j(t+Δt)-P0,j(t)=P0,j-1(t)λΔt+P1,j(t)ξ2Δt-P0,j(t)(λ+cη)Δt+o(Δt),(5)P0,j(t+Δt)-P0,j(t)Δt=P0,j-1(t)λ+P1,j(t)ξ2-P0,j(t)(λ+cη)+o(Δt)Δt.
Letting Δt→0 in (5), we have
(6)P0,j(t)′=P0,j-1(t)λ+P1,j(t)ξ2-P0,j(t)(λ+cη).
If the system is positive recurrent, we have the formulas limt→∞P0,j(t)′=0 [13]. Letting t→0 in (6), we obtain the second equation of (2). The derivations of other formulas in (2) are similar.


## 3. Equations of Probability Generating Functions

The PGFs of the number of customers are defined as follows:(7)Gi(z)≡∑j=0∞‍zjPi,j,G(z)≡∑i=0N‍Gi(z),0≤i≤N,|z|≤1.
Then
(8)Gi(1)=∑j=0∞‍Pi,j(i=0,1,2,…,N),
where Gi(1) is the steady-state probability that the number of the available servers of the system is i. Hence,
(9)∑i=0N‍Gi(1)=1.Multiplying the two sides of every equation of (2) by zj+1, and summing over j(j=0,1,2,…) for every i, we obtain
(10)z(λ-λz+cη)G0(z)-zξ2G1(z)=z(ξ1-ξ2)P1,0,-cηzGi-1(z)+[z(iμ+iξ2+λ+cη)-λz2-iμ]Gi(z)-(i+1)zξ2Gi+1(z)=∑m=0i-1‍(ξ2-ξ1)(i-m)Pi,mzm+1+∑m=0i‍(ξ1-ξ2)(i+1-m)Pi+1,mzm+1+(z-1)∑m=0i-1‍μ(i-m)Pi,mzm,0<i≤N-c,-(N-i+1)ηzGi-1(z)+{z[iμ+iξ2+λ+(N-i)η]-λz2-iμ}Gi(z)-(i+1)zξ2Gi+1(z)=∑m=0i-1‍(ξ2-ξ1)(i-m)Pi,mzm+1+∑m=0i‍(ξ1-ξ2)(i+1-m)Pi+1,mzm+1+(z-1)∑m=0i-1‍μ(i-m)Pi,mzm,N-c<i<N,-ηzGN-1(z)+[z(Nμ+Nξ2+λ)-λz2-Nμ]GN(z)=∑m=0N-1‍(ξ2-ξ1)(N-m)PN,mzm+1+(z-1)∑m=0N-1‍μ(N-m)PN,mzm.We give some explanations of (10), the first equation of (2) multiplied by z, we get
(11)(λ+cη)P0,0z=ξ1P1,0z.
The second equation of (2) multiplied by zj+1, we get
(12)(λ+cη)P0,jzj+1=λP0,j-1zj+1+ξ2P1,jzj+1,j>0.
Summing (11) and (12) over j and using (7), we obtain the first equation of (10). The other equations of (10) are obtained in the same way.In order to simplify (10), we defined the following notations:
(13)fi(z)≡(iμ+iξ2+λ+cη)z-λz2-iμ,i=0,1,2,…,N-c,fi(z)≡[iμ+iξ2+λ+(N-i)η]z-λz2-iμ,i=N-c+1,N-c+2,…,N,(14)A(z)≡[f0(z)-ξ2z-cηzf1(z)-2ξ2z⋱⋱⋱⋱⋱⋱-cηzfN-c+1(z)-(N-c+2)ξ2z(c-1)ηzfN-c+2(z)-(N-c+3)ξ2z⋱⋱⋱⋱⋱⋱-2ηzfN-1(z)-Nξ2z-ηzfN(z)],(15)b0(z)≡z(ξ1-ξ2)P1,0,bi(z)≡∑m=0i-1‍(ξ2-ξ1)(i-m)Pi,mzm+1+∑m=0i‍(ξ1-ξ2)(i+1-m)Pi+1,mzm+1+(z-1)∑m=0i-1‍μ(i-m)Pi,mzm,0<i<N,bN(z)≡∑m=0N-1‍(ξ2-ξ1)(N-m)PN,mzm+1+(z-1)∑m=0N-1‍μ(N-m)PN,mzm,b-(z)≡[b0(z)b1(z)⋮bN(z)],g-(z)≡[G0(z)G1(z)⋮GN(z)].
Using the above notations, (10) is rewritten as follows:
(16)A(z)g-(z)=b-(z).
Using Cramer’s rule we obtain
(17)|A(z)|Gi(z)=|Ai(z)|,i=0,1,2,…,N,
where |A(z)| denotes the determinant of A(z), and Ai(z) is a matrix obtained by replacing the (i+1)th column of A(z) with b-(z). In (17), the functions of z are continuous and bounded in the interval [0,1], so the equations in (17) are valid in the interval [0,1] no matter |A(z)|=0 or not.


## 4. Steady-State Availability

In this section, we discuss the steady-state availabilitiesGi(1)(i=0,1,2,…,N).Lettingz=1 in (10) we obtain
(18)cηG0(1)-ξ2G1(1)=(ξ1-ξ2)P1,0,cηGi-1(1)-iξ2Gi(1)-[cηGi(1)-(i+1)ξ2Gi+1(1)]=(ξ1-ξ2)∑m=1i‍mPi,i-m-(ξ1-ξ2)∑m=1i+1‍mPi+1,i+1-m,0<i≤N-c,[N-(i-1)]ηGi-1(1)-iξ2Gi(1)-[(N-i)ηGi(1)-(i+1)ξ2Gi+1(1)]=(ξ1-ξ2)∑m=1i‍mPi,i-m-(ξ1-ξ2)∑m=1i+1‍mPi+1,i+1-m,N-c<i<N,ηGN-1(1)-Nξ2GN(1)=(ξ1-ξ2)∑m=1N‍mPN,N-m.
The (N+1) equations of (18) are simplified to N independent equations, joined with (9), we have
(19)cηGi(1)-(i+1)ξ2Gi+1(1)=(ξ1-ξ2)∑m=1i+1‍mPi+1,i+1-m,0≤i≤N-c,(N-i)ηGi(1)-(i+1)ξ2Gi+1(1)=(ξ1-ξ2)∑m=1i+1‍mPi+1,i+1-m,N-c≤i≤N-1,∑i=0N‍Gi(1)=1.
Using (2), Pi,j(0≤j≤i-1,1≤i≤N) are reduced to  Pi,0(1≤i≤N) which will be solved in Section 5. Given  Pi,0(1≤i≤N), we obtain Gi(1)(i=0,1,2,…,N) by solving (19), then the steady-state availability of the system is as follows:
(20)A=1-G0(1).


## 5. Steady-State Probabilities of the System States and Mean Queuing Length

### 5.1. The Roots of|A(z)| in the Interval (0,1)

In order to get the steady-state probabilitiesPi,0(1≤i≤N), we need all N independent linear equations. Further, for getting the linear equations of Pi,0(1≤i≤N) we need the roots of |A(z)| in the interval (0,1), so we discuss the roots of |A(z)| as follows.We define the following notations:(21)Q0(z)≡1,Q1(z)≡fN(z),(22)(z)≡|fN-1(z)-Nξ2z-ηzfN(z)|,⋮⋮QN(z)≡|f1(z)-2ξ2z-cηzf2(z)-3ξ2z⋱⋱⋱-cηzfN-c+1(z)-(N-c+2)ξ2z(c-1)ηzfN-c+2(z)-(N-c+3)ξ2z⋱⋱⋱-2ηzfN-1(z)-Nξ2z-ηzfN(z)|,(23)QN+1(z)≡|A(z)|,(0≤z≤∞),
where Qi(z)(i=1,2,…,N) is formed by the last i rows and last i columns of |A(z)|. According to (14) we have
(24)Qk+1(z)=fN-k(z)Qk(z)-(N-k+1)ξ2kηz2Qk-1(z),1≤k≤c-1,Qk+1(z)=fN-k(z)Qk(z)-(N-k+1)ξ2cηz2Qk-1(z),c≤k≤N.
The properties of Qi(z)(i=1,2,…,N) and the necessary proofs are as follows. (a)
Q0(z) has no roots.(b)
Qk(z) and Qk+1(z) have no common roots in the interval (0,∞)(k=1,2,…,N).Proof.
Suppose that (b) is not true andz0(>0) is a common roots of Qk(z) and Qk+1(z), then Qk-1(z0)=0 due to (24). Similarly, Qk-2(z0)=0, so we get Q0(z0)=0 which contradicts the statement (a).(c)
Ifz0 is a positive root of Qk(z)(k=1,2,…,N), then Qk-1(z0) and Qk+1(z0) are opposite in sign. This property readily follows from (24).
(d)
Q
k
(
1
)
>
0, k=0,1,2,…,N.Proof.
Substitutingz=1 into Qk(z), we get Qk(1)=(N-k+1)(N-k+2)⋯(N-1)Nξ2k>0, k=1,2,…,N, and Q0(1)=1.(e)
Q
N
+
1
(
1
)
=
|
A
(
1
)
|
=
0, and QN+1(0)=|A(0)|=0.Proof.
The first row of|A(z)| has a common factor z, so QN+1(0)=|A(0)|=0. The sum of every column of |A(z)| has a common factor (z-1). Replacing every element of the last row of |A(z)| with the sum of the corresponding column and extracting the common factor (z-1), |A(z)| is written as follows: (25)|A(z)|=z(z-1)×[λ+cη-λz-ξ20⋯00-cηzf1(z)-2ξ2z⋯000-cηf2(z)⋯00⋮000⋯fN-1(z)-Nξ2z-λz-λz+μ-λz+2μ⋯-λz+(N-1)μ-λz+Nμ].
If we define
(26)D(z)≡|λ+cη-λz-ξ20⋯00-cηzf1(z)-2ξ2z⋯000-cηf2(z)⋯00⋮000⋯fN-1(z)-Nξ2z-λz-λz+μ-λz+2μ⋯-λz+(N-1)μ-λz+Nμ|,
then(27)|A(z)|=z(z-1)D(z).(f)
Sign
[
Q
k
(
0
)
]
=
(
-
1
)
k
(
k
=
0,1
,
2
,
…
,
N
).Proof.
From the definitions offi(z)(i=0,1,2,…,N), we got f0(0)=0 and fk(0)<0(k=1,2,…,N),  so we get this property from (24).(g)
Sign
[
Q
k
(
+
∞
)
]
=
(
-
1
)
k
(
k
=
0,1
,
2
,
…
,
N
+
1
).Proof.
It is since the highest power term ofQk(z) is (-λz2)k(k=0,1,2,…,N+1) and the sign of Qk(+∞) is determined by its highest power term.Theorem 1.
IfD(1)>0, the polynomial |A(z)| has exactly (N-1) distinct roots in the interval (0,1).Proof.
SinceQ1(z)=fN(z)=[N(μ+ξ2)+λ]z-λz2-Nμ, Q1(z) is a 2-power polynomial of z. Further, we find that Q1(1)=ξ2>0 and Q1(0)=-μ<0, so Q1(z) has two distinct roots which are denoted by  z1,1(0<z1,1<1) and  z1,2(>1).
With the fact thatz1,1 and z1,2 are roots of Q1(z), and Q0(z)=1>0, according to the property (c) or (24), we get Q2(z1,1)<0 and Q2(z1,2)<0.
Q
2
(
z
) is a 4-power polynomial of z. From the properties (c), (d), (f), and (g), we find that Q2(z) has one and only one root in each interval of (0,z1,1), (z1,1,1), (1,z1,2), and (z1,2,+∞).
So on, we find thatQN(z) is a 2N-power polynomial of z, it has N distinct roots in the interval (0,1) and N distinct roots in the interval (1,∞). We denote the 2N roots of QN(z) by  zN,i(i=1,2,…,2N) orderly.
From the properties (c), (d), (e), and (f), we find that|A(z)| has one and only one root in each interval (zN,i,zN,i+1)(i=1,2,…,N-1,N+1,…,2N-1), all of them are 2(N-1) distinct roots of |A(z)|.
Since|A(1)|=0 and D(1)=(|A(z)|)z=1′>0, it has a real number ε (>0) satisfies 1+ε<zN,N+1 and |A(1+ε)|>0. On the other hand, from (c) and (d) we get
(28)Sign[QN+1(zN,i)]=(-1)N+i,i=N+1,N+2,…,2N,
then Sign[QN+1(zN,N+1)]=(-1)N+N+1 or |A(zN,N+1)|=QN+1(zN,N+1)<0, so |A(z)| has at least one root in the interval (1,zN,N+1).
From (28), we get Sign[|A(zN,2N)|]=Sign[QN+1(zN,2N)]=(-1)N+2N=(-1)3N=(-1)N. From the property (g), we get Sign[|A(+∞)|]=Sign[QN+1(+∞)]=(-1)N+1. So we know that |A(z)| has at least one root in the interval (zN,2N,∞).
From the properties (e) and (f), we know that 0 and 1 are roots of|A(z)|.
In conclusion,|A(z)| is a 2(N+1)-power polynomial of z, it has 2(N+1) distinct roots at most. Now we make certain all roots of |A(z)| and find that it has N-1 distinct roots in the interval (0,1).From the proof, we find that the(N-1) distinct roots in the interval (0,1) of |A(z)| are also the roots of D(z).


### 5.2. Steady-State Probabilities

Assuming that the system parameters meetD(1)>0. Letting zk(k=1,2,…,N-1) denote the roots of |A(z)| in the interval (0,1). Substituting z1 in (17), we obtain a set of linear equations about the steady-state probabilities of Pi,0(i=1,2,…,N), but these equations are similar to each other. However, the equations belong to different zk(k=1,2,…,N-1) are independent mutually, so we obtain (N-1) independent equations by the (N-1) different roots of zk, respectively.In the following, we discuss about theNth-independent linear equation of Pi,0(i=1,2,…,N). Similar to (27), |Ai(z)| is written as follows:
(29)|Ai(z)|=z(z-1)Di(z),i=0,1,2,…,N,
where
(30)Di(z)=|cη+λ(1-z)-ξ2⋯(ξ1-ξ2)P1,0⋯0-cηzf1(z)⋯b1(z)⋯00-cηz⋯b2(z)⋯0⋮⋮00⋯bN-1(z)⋯-Nξ2z-λz-λz+μ⋯∑i=1N‍∑m=1i‍mμpi,i-mzi-m(i+1)thcolumn⋯-λz+Nμ|.
Substituting (27) and (29) into (17), we obtain
(31)D(z)Gi(z)=Di(z),i=0,1,2,…,N.
Substituting z=1 in (31), we obtain
(32)D(1)Gi(1)=Di(1),i=0,1,2,…,N.
From (19), we know that Gi(1)(i=0,1,2,…,N) can be expressed by Pi,0(i=1,2,…,N), so (32) are linear equations of Pi,0(i=1,2,…,N), but they are similar to each other. However, every equation of (32) is independent with the (N-1) linear equations obtained by the roots of |Ai(z)| in the interval (0,1). Then all independent linear equations of Pi,0(i=1,2,…,N) are as follows:
(33)|A0(zk)|=0,k=1,2,…,N-1,D(1)G0(1)=D0(1).
Further, from (29), (33) is equivalent to the follows:
(34)D0(zk)=0,k=1,2,…,N-1,D(1)G0(1)=D0(1).The steady-state probabilities ofPi,0(i=0,1,2,…,N) are obtained by solving (34). Using Pi,0(i=0,1,2,…,N) and (2), we obtain the other steady-state probabilities of Pi,j(i=0,1,2,…,N,j=1,2,…).


### 5.3. Mean Queuing Length

After getting the probabilitiesPi,j(0≤j≤i-1,1≤i≤N) in (19) we obtain Gi(1)(i=0,1,2,…,N) by solving (19) and obtain the steady-state availability (A) of the model by (20).From (31), we obtain
(35)Gi(z)=Di(z)D(z),D(z)≠0,|z|≤1,i=0,1,2,…,N.
The PGF of G(z) is obtained by (7). Using the property of PGF [13] we obtain the steady-state mean queuing length is as follows:
(36)L=dG(z)dz|z=1.


## 5.1. The Roots of|A(z)| in the Interval (0,1)

In order to get the steady-state probabilitiesPi,0(1≤i≤N), we need all N independent linear equations. Further, for getting the linear equations of Pi,0(1≤i≤N) we need the roots of |A(z)| in the interval (0,1), so we discuss the roots of |A(z)| as follows.We define the following notations:(21)Q0(z)≡1,Q1(z)≡fN(z),(22)(z)≡|fN-1(z)-Nξ2z-ηzfN(z)|,⋮⋮QN(z)≡|f1(z)-2ξ2z-cηzf2(z)-3ξ2z⋱⋱⋱-cηzfN-c+1(z)-(N-c+2)ξ2z(c-1)ηzfN-c+2(z)-(N-c+3)ξ2z⋱⋱⋱-2ηzfN-1(z)-Nξ2z-ηzfN(z)|,(23)QN+1(z)≡|A(z)|,(0≤z≤∞),
where Qi(z)(i=1,2,…,N) is formed by the last i rows and last i columns of |A(z)|. According to (14) we have
(24)Qk+1(z)=fN-k(z)Qk(z)-(N-k+1)ξ2kηz2Qk-1(z),1≤k≤c-1,Qk+1(z)=fN-k(z)Qk(z)-(N-k+1)ξ2cηz2Qk-1(z),c≤k≤N.
The properties of Qi(z)(i=1,2,…,N) and the necessary proofs are as follows. (a)
Q0(z) has no roots.(b)
Qk(z) and Qk+1(z) have no common roots in the interval (0,∞)(k=1,2,…,N).Proof.
Suppose that (b) is not true andz0(>0) is a common roots of Qk(z) and Qk+1(z), then Qk-1(z0)=0 due to (24). Similarly, Qk-2(z0)=0, so we get Q0(z0)=0 which contradicts the statement (a).(c)
Ifz0 is a positive root of Qk(z)(k=1,2,…,N), then Qk-1(z0) and Qk+1(z0) are opposite in sign. This property readily follows from (24).
(d)
Q
k
(
1
)
>
0, k=0,1,2,…,N.Proof.
Substitutingz=1 into Qk(z), we get Qk(1)=(N-k+1)(N-k+2)⋯(N-1)Nξ2k>0, k=1,2,…,N, and Q0(1)=1.(e)
Q
N
+
1
(
1
)
=
|
A
(
1
)
|
=
0, and QN+1(0)=|A(0)|=0.Proof.
The first row of|A(z)| has a common factor z, so QN+1(0)=|A(0)|=0. The sum of every column of |A(z)| has a common factor (z-1). Replacing every element of the last row of |A(z)| with the sum of the corresponding column and extracting the common factor (z-1), |A(z)| is written as follows: (25)|A(z)|=z(z-1)×[λ+cη-λz-ξ20⋯00-cηzf1(z)-2ξ2z⋯000-cηf2(z)⋯00⋮000⋯fN-1(z)-Nξ2z-λz-λz+μ-λz+2μ⋯-λz+(N-1)μ-λz+Nμ].
If we define
(26)D(z)≡|λ+cη-λz-ξ20⋯00-cηzf1(z)-2ξ2z⋯000-cηf2(z)⋯00⋮000⋯fN-1(z)-Nξ2z-λz-λz+μ-λz+2μ⋯-λz+(N-1)μ-λz+Nμ|,
then(27)|A(z)|=z(z-1)D(z).(f)
Sign
[
Q
k
(
0
)
]
=
(
-
1
)
k
(
k
=
0,1
,
2
,
…
,
N
).Proof.
From the definitions offi(z)(i=0,1,2,…,N), we got f0(0)=0 and fk(0)<0(k=1,2,…,N),  so we get this property from (24).(g)
Sign
[
Q
k
(
+
∞
)
]
=
(
-
1
)
k
(
k
=
0,1
,
2
,
…
,
N
+
1
).Proof.
It is since the highest power term ofQk(z) is (-λz2)k(k=0,1,2,…,N+1) and the sign of Qk(+∞) is determined by its highest power term.Theorem 1.
IfD(1)>0, the polynomial |A(z)| has exactly (N-1) distinct roots in the interval (0,1).Proof.
SinceQ1(z)=fN(z)=[N(μ+ξ2)+λ]z-λz2-Nμ, Q1(z) is a 2-power polynomial of z. Further, we find that Q1(1)=ξ2>0 and Q1(0)=-μ<0, so Q1(z) has two distinct roots which are denoted by  z1,1(0<z1,1<1) and  z1,2(>1).
With the fact thatz1,1 and z1,2 are roots of Q1(z), and Q0(z)=1>0, according to the property (c) or (24), we get Q2(z1,1)<0 and Q2(z1,2)<0.
Q
2
(
z
) is a 4-power polynomial of z. From the properties (c), (d), (f), and (g), we find that Q2(z) has one and only one root in each interval of (0,z1,1), (z1,1,1), (1,z1,2), and (z1,2,+∞).
So on, we find thatQN(z) is a 2N-power polynomial of z, it has N distinct roots in the interval (0,1) and N distinct roots in the interval (1,∞). We denote the 2N roots of QN(z) by  zN,i(i=1,2,…,2N) orderly.
From the properties (c), (d), (e), and (f), we find that|A(z)| has one and only one root in each interval (zN,i,zN,i+1)(i=1,2,…,N-1,N+1,…,2N-1), all of them are 2(N-1) distinct roots of |A(z)|.
Since|A(1)|=0 and D(1)=(|A(z)|)z=1′>0, it has a real number ε (>0) satisfies 1+ε<zN,N+1 and |A(1+ε)|>0. On the other hand, from (c) and (d) we get
(28)Sign[QN+1(zN,i)]=(-1)N+i,i=N+1,N+2,…,2N,
then Sign[QN+1(zN,N+1)]=(-1)N+N+1 or |A(zN,N+1)|=QN+1(zN,N+1)<0, so |A(z)| has at least one root in the interval (1,zN,N+1).
From (28), we get Sign[|A(zN,2N)|]=Sign[QN+1(zN,2N)]=(-1)N+2N=(-1)3N=(-1)N. From the property (g), we get Sign[|A(+∞)|]=Sign[QN+1(+∞)]=(-1)N+1. So we know that |A(z)| has at least one root in the interval (zN,2N,∞).
From the properties (e) and (f), we know that 0 and 1 are roots of|A(z)|.
In conclusion,|A(z)| is a 2(N+1)-power polynomial of z, it has 2(N+1) distinct roots at most. Now we make certain all roots of |A(z)| and find that it has N-1 distinct roots in the interval (0,1).From the proof, we find that the(N-1) distinct roots in the interval (0,1) of |A(z)| are also the roots of D(z).


## 5.2. Steady-State Probabilities

Assuming that the system parameters meetD(1)>0. Letting zk(k=1,2,…,N-1) denote the roots of |A(z)| in the interval (0,1). Substituting z1 in (17), we obtain a set of linear equations about the steady-state probabilities of Pi,0(i=1,2,…,N), but these equations are similar to each other. However, the equations belong to different zk(k=1,2,…,N-1) are independent mutually, so we obtain (N-1) independent equations by the (N-1) different roots of zk, respectively.In the following, we discuss about theNth-independent linear equation of Pi,0(i=1,2,…,N). Similar to (27), |Ai(z)| is written as follows:
(29)|Ai(z)|=z(z-1)Di(z),i=0,1,2,…,N,
where
(30)Di(z)=|cη+λ(1-z)-ξ2⋯(ξ1-ξ2)P1,0⋯0-cηzf1(z)⋯b1(z)⋯00-cηz⋯b2(z)⋯0⋮⋮00⋯bN-1(z)⋯-Nξ2z-λz-λz+μ⋯∑i=1N‍∑m=1i‍mμpi,i-mzi-m(i+1)thcolumn⋯-λz+Nμ|.
Substituting (27) and (29) into (17), we obtain
(31)D(z)Gi(z)=Di(z),i=0,1,2,…,N.
Substituting z=1 in (31), we obtain
(32)D(1)Gi(1)=Di(1),i=0,1,2,…,N.
From (19), we know that Gi(1)(i=0,1,2,…,N) can be expressed by Pi,0(i=1,2,…,N), so (32) are linear equations of Pi,0(i=1,2,…,N), but they are similar to each other. However, every equation of (32) is independent with the (N-1) linear equations obtained by the roots of |Ai(z)| in the interval (0,1). Then all independent linear equations of Pi,0(i=1,2,…,N) are as follows:
(33)|A0(zk)|=0,k=1,2,…,N-1,D(1)G0(1)=D0(1).
Further, from (29), (33) is equivalent to the follows:
(34)D0(zk)=0,k=1,2,…,N-1,D(1)G0(1)=D0(1).The steady-state probabilities ofPi,0(i=0,1,2,…,N) are obtained by solving (34). Using Pi,0(i=0,1,2,…,N) and (2), we obtain the other steady-state probabilities of Pi,j(i=0,1,2,…,N,j=1,2,…).


## 5.3. Mean Queuing Length

After getting the probabilitiesPi,j(0≤j≤i-1,1≤i≤N) in (19) we obtain Gi(1)(i=0,1,2,…,N) by solving (19) and obtain the steady-state availability (A) of the model by (20).From (31), we obtain
(35)Gi(z)=Di(z)D(z),D(z)≠0,|z|≤1,i=0,1,2,…,N.
The PGF of G(z) is obtained by (7). Using the property of PGF [13] we obtain the steady-state mean queuing length is as follows:
(36)L=dG(z)dz|z=1.


## 6. Case Analysis

We analyze the case ofN=2 and c=1 in this section. According to the above discussion, the determinant |A(z)| (or D(z)) of this case has only one root z1 in the interval (0,1). The notations of this case are as follows:
(37)f0(z)=(λ+η)z-λz2,f1(z)=(μ+ξ2+λ+η)z-λz2-μ,f2(z)=(2μ+2ξ2+λ)z-λz2-2μ,b0(z)=(ξ1-ξ2)P1,0z,b1(z)=(μ+ξ2-ξ1)P1,0z+∑m=12‍m(ξ1-ξ2)P2,2-mz3-m-μP1,0=μP1,0(z-1)+z(ξ1-ξ2)(P2,1z+2P2,0-P1,0),b2(z)=∑m=12‍m(μ+ξ2-ξ1)P2,2-mz3-m-∑m=12‍mμP2,2-mz2-m,A(z)=[f0(z)-ξ2z0-ηzf1(z)-2ξ2z0-ηzf2(z)],|A(z)|=z(z-1)×|λ+η-λz-ξ20-ηz(μ+ξ2+λ+η)z-λz2-μ-2ξ2z-λz-λz+μ-λz+2μ|,D(z)=|λ+η-λz-ξ20-ηz(μ+ξ2+λ+η)z-λz2-μ-2ξ2z-λz-λz+μ-λz+2μ|=-(η+λ-zλ)(zλ-2μ)[-μ+z(η+λ-zλ+μ)]+zξ2[zλ(-2η-3λ+3zλ)+2(η-2zλ+2λ)μ-2zλξ2],
then
(38)D(1)=|η-ξ20-ηη+ξ2-2ξ2-λ-λ+μ-λ+2μ|=2μ(ξ2η+η2)-λ(2ξ22+2ξ2η+η2),
and D(1)>0 is equivalent to
(39)2μ(ξ2η+η2)-λ(2ξ22+2ξ2η+η2)>0,
or
(40)λμ<2ξ2η+2η22ξ22+2ξ2η+η2=2η/(ξ2+η)1+(ξ2/(ξ2+η))2.
The left of (40) is the mean service quantities that all customers need per unit time. The right of (40) is the mean service quantities that the two servers provide per unit time. So (40) is the necessary and sufficient condition of recurrence of the system.Equations (19) in this case are as follows:
(41)ηG0(1)-ξ2G1(1)=(ξ1-ξ2)P1,0,ηG1(1)-2ξ2G2(1)=∑m=12‍m(ξ1-ξ2)P2,2-m,∑i=02‍Gi(1)=1.Equations (2) in this case are as follows:
(42)(λ+η)P0,0=ξ1P1,0,i=0,j=0,(λ+η)P0,j=λP0,j-1+ξ2P1,j,i=0,j>0,(λ+η+ξ1)P1,0=ηP0,0+μP1,1+2ξ1P2,0,i=1,j=0,(λ+η+ξ2+μ)P1,1=λP1,0+ηP0,1+μP1,2+(ξ1+ξ2)P2,1,i=1,j=1,(λ+η+ξ2+μ)P1,j=λP1,j-1+ηP0,j+μP1,j+1+2ξ2P2,j,i=1,j>1,(λ+2ξ1)P2,0=ηP1,0+μP2,1,i=2,j=0,(λ+ξ1+ξ2+μ)P2,1=λP2,0+ηP1,1+2μP2,2,i=2,j=1,(λ+2μ+2ξ2)P2,j=λP2,j-1+ηP1,j+2μP2,j+1,i=2,j≥2.
Using (42), we obtain
(43)P2,1=(2ξ1+λ)P2,0-ηP1,0μ.
Using (41) and (43), G0(1), G1(1), and G2(1) are expressed in an algebraic expressions of P1,0 and P2,0.If (40) is satisfied, we obtain z1 by solving
(44)D(z)=0.
Equations (34) in this case are as follows:
(45)D0(z1)=0,D(1)G0(1)=D0(1).
Solving (45), we obtain P1,0 and P2,0.Using (42), we obtain Pi,j (i=0,1,2, j=1,2,…).Using (20) and (41), we obtain the steady-state availability A.For the mean queuing lengths, we have(46)D0(z)=|(ξ1-ξ2)P1,0-ξ20b1(z)(μ+ξ2+λ+η)z-λz2-μ-2ξ2μ(P2,1z+2P2,0+P1,0)-λz+μλz+2μ|,D1(z)=|λ+η-λz(ξ1-ξ2)P1,00-ηzb1(z)-2ξ2-λzμ(P2,1z+2P2,0+P1,0)λz+2μ|,D2(z)=|λ+η-λz-ξ2(ξ1-ξ2)P1,0-ηz(μ+ξ2+λ+η)z-λz2-μb1(z)-λz-λz+μμ(P2,1z+2P2,0+P1,0)|,(47)G(z)=D0(z)+D1(z)+D2(z)D(z).
Using (36), we obtain the mean queuing lengths L is as follows:
(48)L=dG(z)dz|z=1=1[λη2-2μη2+2ξ2(ηλ-ημ+λξ2)]2×{-[η(-2ηλ+2λ2+2ημ-5λμ+2μ2)+ξ2(-4ηλ+3λ2+2ημ-4λμ-4λξ2)]×[(2P2,0+P2,1)μ(η2+2ξ1η+2ξ1ξ2)P1,0+μ(η2+ηξ2+ηξ1+2ξ1ξ2)(2P2,0+P2,1)]+μ[-η2λ+2μη2-2ξ2(ηλ-ημ+λξ2)]×[(2P2,0+P2,1)(η2-2ηλ+2ημ-2λξ2+2ξ1η-λξ1+2μξ1+2ξ1ξ2)P1,0+(η2+ηξ2+ηξ1+2ξ1ξ2)P2,1+(η2-2ηλ+ημ+ηξ2-2λξ2+ηξ1-λξ1+2ξ1ξ2)(2P2,0+P2,1)(2P2,0+P2,1)(-2ηλ+2λ2+2ημ-5λμ+2μ2)]}.Numerical Example. Letting N=2, c=1, λ=1, ξ1=0.5, ξ2=1, η=1, and μ=2, we have
(49)λμ=12<45=2η/(ξ2+η)1+(ξ2/(ξ2+η))2.
The roots of D(z)=0 are as follows:
(50)z1=0.349123,z2=1.84513,z3=4.46896,z4=8.33679,
and only z1 in the interval (0,1).Solving (45), we obtain
(51)P1,0=0.151375,P2,0=0.297974.
Using (42) we obtain
(52)P0,0=0.037844,P2,1=0.146599.
Solving (41), we obtain
(53)G0(1)=0.280333,G1(1)=0.35602,G2(1)=0.363647.
Finally, the availability and mean queuing lengths of this example are as follows:
(54)A=0.719667,L=6.04039.The other numerical results are shown in Table1. All the system parameters in Table 1 satisfy (40).Table 1
The availabilityA and mean queuing length L (N=2, c=1, and λ=1).
μ
ξ
1
=
0
,
ξ
2
=
0.5
,and η=1.2
ξ
1
=
0.3
,
ξ
2
=
0.5, andη=1
ξ
1
=
0.5
,
ξ
2
=
0.5, and η=1
L
A
L
A
L
A
1.1
3.0797
0.8951
4.6956
0.8209
4.8505
0.8
1.2
2.3810
0.9058
3.5018
0.8266
3.6549
0.8
1.3
1.9425
0.9147
2.8136
0.8315
2.9650
0.8
1.4
1.6411
0.9224
2.3641
0.8354
2.5138
0.8
1.5
1.4207
0.9290
2.0464
0.8383
2.1947
0.8
1.6
1.2524
0.9348
1.8095
0.8429
1.9563
0.8
1.7
1.1195
0.9399
1.6256
0.8458
1.7710
0.8
1.8
1.0119
0.9443
1.4784
0.8484
1.6225
0.8
μ
ξ
1
=
0.5
,
ξ
2
=
0.5
,and η=0.8
ξ
1
=
0.5
,
ξ
2
=
1, andη=1.5
ξ
1
=
1
,
ξ
2
=
0.8, and η=1.5
L
A
L
A
L
A
1.1
9.4410
0.7423
9.9094
0.7412
5.0503
0.7734
1.2
6.1455
0.7423
5.7928
0.7509
3.7017
0.7700
1.3
4.6345
0.7423
4.1374
0.7608
2.9584
0.7671
1.4
3.7636
0.7423
3.2405
0.7687
2.4853
0.7646
1.5
3.1952
0.7423
2.6761
0.7756
2.1565
0.7623
1.6
2.7937
0.7423
2.2873
0.7817
1.9140
0.7603
1.7
2.4944
0.7423
2.0025
0.7872
1.7273
0.7585
1.8
2.2389
0.7423
1.7846
0.7920
1.5788
0.7569We find that the mean queuing length (L) decreases with the increasing of the parameter μ in Table 1, it is because of the greater service rate the less customers in the system. Furthermore, we find that the availability (A) increases with the increasing of the parameter μ, where ξ1<ξ2 (the cases: ξ1=0,ξ2=0.5;ξ1=0.3,ξ2=0.5;ξ1=0.5,ξ2=1); on the contrary, the availability decreases with the increasing of the parameter μ, where ξ1>ξ2 (the case: ξ1=1,ξ2=0.8); otherwise, the availability is constant, where ξ1=ξ2 (the case: ξ1=0.5,ξ2=0.5).


## 7. Conclusions

In Section  5.1, the inequalityD(1)>0 of Theorem 1 is the necessary and sufficient condition for the system to be positive recurrent, and a probability explanation of this condition is given by (40).We find that the idle time breakdown rateξ1 does not appear in (40). This is because the busy time breakdown rate ξ2 is at work when the number of the customers is greater than or equal to the number of the available servers, and the criteria of positive recurrence depends on the busy time breakdown rate.A case analysis is given to illustrate the analysis of this paper, and the numerical results indicate that the variation of breakdown rates has a significant effect on the steady-state availability and steady-state queue length of the system.



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*Source: 173407-2013-02-21.xml*