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Decay rate for a P H/M/2 queue with shortest queue discipline Yutaka Sakuma, Masakiyo Miyazawa Department of Information Sciences

Yiqiang Q. Zhao School of Mathematics and Statistics

Tokyo University of Science

Carleton University

Abstract In this paper, we consider a P H/M/2 queue in which each server has its own queue and arriving customers join the shortest queue. For this model, it has been conjectured that the decay rate of the tail probabilities for the shortest queue length in the steady state is equal to the square of the decay rate for the queue length in the corresponding P H/M/2 model with a single queue. We prove this fact in the sense that the tail probabilities are asymptotically geometric when the difference of the queue sizes and the arrival phase are fixed. Our proof is based on the matrix analytic approach pioneered by Neuts and recent results on the decay rates.

1. Introduction It is very natural that arriving customers join the shortest queue if there are multiple queues for the same service. We refer to this service rule as shortest queue discipline. Although this queueing discipline is natural, it is very challenging to analyze its stationary characteristics even for Poisson arrivals and exponential services. Partly due to this reason and also because of its own importance, research on stationary tail decay rate with shortest queue discipline has attracted people’s attention. However, studying tail asymptotics is still challenging enough and the only known result is the decay rate for the M/M/2 parallel queues under various assumptions on service rates and arriving streams. Kingman [7] seems to be the first one, who proved the decay rate for the M/M/2 parallel queueing system with two identical servers. Adan, Wessels and Zijm [1, 2] provided expressions for the joint queue length distribution for the symmetric and asymmetric M/M/2 parallel queueing system with shortest queue discipline, based on which tail asymptotics were studied. Takahashi, Fujimoto and Makimoto [18] also obtained the decay rate for the non identical server (asymmetric) model using a quasi-birth-and-death (QBD) process. Foley and McDonald [5] studied a generalized model, in which in addition to a Poisson arrival stream dedicated to each queue there exists a third stream of Poisson arrivals joining the shortest queue. Kurkova and Suhov [9] considered a similar model with identical servers. In both [5] and [9], harmonic functions are extensively used in the analysis. To our knowledge, all existing results on the decay rate for parallel queues with shortest queue discipline were obtained by assuming Poisson arrivals, exponential service and two servers. In this paper, we consider a case, in which arrivals are governed by a renewal process with the phase type interarrival distribution, while still keeping the exponential 1

service and two servers. We also assume that both servers have the same service rate. For this model, Adan and Zhao [3] conjectured that the tail decay rate for the shortest queue is the square of the decay rate for the queue length of the corresponding P H/M/2 model with a single queue. In this paper, intuition to support this conjecture will be presented, and the main objective is to prove this property in the sense that the stationary tail probabilities are exactly asymptotically geometric when the difference of the queue sizes and the arrival phase are fixed. Our proofs are based on the matrix analytic approach for the QBD process and recent results on the tail decay rate; for example, in [13, 18], which also produce some interesting identities particularly on eigenvectors (see Lemmas 4.1, 5.1 and 5.2). The remainder of this paper is composed of four sections. In Section 2, we formally introduce the P H/M/2 queue with shortest queue discipline, and describe it by the QBD process. In Section 3, we give preliminary results on the QBD process, which determine the decay rate of the P H/M/2 queue. In Section 4, the corresponding model with a single queue is considered. In Section 5, we prove the main result on the decay rate. We finally discuss problems on the shortest queue for future research in Section 6.

2. P H/M/2 with shortest queue discipline We formally introduce the P H/M/2 model with shortest queue discipline, depicted in Figure 1. In this model, there are two queues numbered as 1 and 2. Each of the queues L 1 (t )

exp( µ )

PH ( α , T ) w.p. 1

exp( µ )

L 2 (t )

w.p. r

w.p. 1 − r

Figure 1: Shortest queue discipline for different and equal queue sizes has a single exponential server with identical service rate µ. Customers arrive according to a renewal process with a phase type interarrival time distribution, and their service times are independent of their arrival times. We follow Latouche and Ramaswami [10] to introduce the following notation. Let T be a defective transition rate matrix of a continuous time Markov chain with finite state space ST ≡ {1, 2, . . . , m}. That is, T has nonnegative off-diagonal entries and negative diagonal elements such that t ≡ −T 1 ≥ 0,

2

but t ̸= 0, where 1 is the vector of ones and 0 is the null vector. Then, ( ) 0 0 QT ≡ t T

(2.1)

is a transition rate matrix of a continuous time Markov chain with state space {0, 1, . . . , m}, in which 0 is an absorbing state. Let α be the initial distribution concentrated on ST of this Markov chain. Then, the distribution of the time to hit state 0 is said to be of phase type, and denoted by P H(α, T ). When the Markov chain hits 0, it starts over again with the same initial distribution α. In this way, a renewal process, which counts the absorbing times, is defined and adopted as the arrival process in the P H/M/2 model with shortest queue discipline considered in this paper. The state of the above mentioned Markov chain is referred to as the arrival phase. Let J(t) be the continuous time Markov chain with state space ST and the transition rate matrix T + tα. We assume that T + tα is irreducible. Clearly, J(t) presents the state of the arrival phase at time t. Denote the queue length of queue i at time t by Li (t) for i = 1, 2. If an arriving customer finds no difference in length between the two queues, it joins queue 1 with probability r and queue 2 with 1 − r. Otherwise, it joins the shortest queue with probability 1. We describe the P H/M/2 with shortest queue discipline by a Markov chain in the following alternative way. Let X(t) = (min (L1 (t), L2 (t)) , |L1 (t) − L2 (t)|, J(t)) . Then, X(t) is clearly a continuous time Markov chain with the state space S ≡ N × SB , where N = {0, 1, 2, ...} and SB = N × ST . We refer to {min (L1 (t), L2 (t))} as a level process, and {(|L1 (t) − L2 (t)|, J(t))} as a background process. Obviously, the level process is skip free, as depicted in Figure 2. Hence, the transition rate matrix of this Markov

N ×S

T

µI T

T

tα

µI

µI

µI

tα T

µI

tα

µI

tα

µI T

tα

µI 2µ I

T

tα

µI

T

tα

µI

T

tα

µI T

µI

µI 2µ I

T

T

tα tα T

tα

µI 2µ I

tα T

min ( L1 (t ), L 2 (t ) ) Figure 2: Transition diagram of P H/M/2 queue with shortest queue discipline

3

chain X(t) is given by  Q0 Q1   Q−1 Q0 Q1     Q Q Q −1 0 1 Q= , ...   Q Q   −1 0 .. .. . . 

where



T tα  µI T − µI   µI T − µI Q0 =   µI T − µI  ... ... 



(2.2)



O  tα O      tα O  , Q1 =    tα  

T − 2µI tα  µI T − 2µI   µI T − 2µI Q0 =   µI T − 2µI  .. .. . .

     , Q−1  



O .. .. . .

   ,  



O 2µI  O µI   O µI = .  O ..  .. .

This form of Q is said to be a block tri-diagonal matrix with repeating blocks Q−1 , Q0 and Q1 . Thus, Q represents a QBD process. Let λ be the mean arrival rate of customers, then we have λ = 1/(−αT −1 1). Throughout the paper, we assume that λ < 2µ, which is equivalent to that Q has the stationary distribution. This is intuitively obvious, and can be formally verified through truncation arguments (see Remark 5.3). We denote this distribution by vector π ≡ (π 0 , π 1 , . . .), where π n is the stationary probability vector at level n. We refer to such partitioning of a vector as block partitioning by level.

3. Preliminary results on a QBD process In this section, we present some results for general QBD processes with transition matrices of the form (2.2). These results will be referred to in our proofs. By Q−1 , Q0 and Q1 , we generate an integer-valued Markov additive process (Y (t), B(t)) in such a way that the additive component Y (t) increases by i according to the state transitions of background process B(t) due to Qi for i = 0, ±1. We define SB × SB matrix R as the minimal nonnegative solution of the following equation: R2 Q−1 + RQ0 + Q1 = O.

(3.1)

This R exists if the diagonal entries of Q0 are bounded and the background process, i.e., Q−1 + Q0 + Q1 , has the stationary distribution. Note that (3.1) is equivalent to that, for any real number z ̸= 0, zQ−1 + Q0 + z −1 Q1 = (I − z −1 R)(Q0 + RQ−1 + zQ−1 ). 4

(3.2)

    .  

This equation can be considered as the Wiener-Hopf factorization for the Markov additive process (Y (t), B(t)), and also characterizes nonnegative R (see, e.g., [4, 13]). Assume that Q has the stationary distribution. Denote this distribution by π ≡ (π 0 , π 1 , . . .). Then, n ≥ 0.

π n = π 0 Rn ,

(3.3)

This result is well known for the case that SB is finite (see, e.g., [10, 14]), and this finiteness can be obviously removed. Remark 3.1 For a general QBD process, the background state space for level 0 may be different. In this case, Q−1 , Q0 and Q1 in the first column and row blocks of Q are needed to be replaced by matrices of appropriate sizes, and (3.3) is changed to π n = π 1 Rn−1 ,

n ≥ 1.

(3.4)

Results in this section can be also obtained for this general case, but we employ the simpler situation. In the view of (3.3), the decay rate of π n , which will be specified in Proposition 3.1 below, can be obtained as the maximum eigenvalue of R if SB is finite. If it is infinite, this is not always the case (see, e.g., [8]). We need certain conditions on vector π 0 and matrix R for the maximum eigenvalue of R to be the decay rate. Fortunately, they can be verified for the present model. We also reduce the eigenvalue problem of R to the one of the left-hand side of (3.2). This scenario is indeed realized in the following proposition, which is easily obtained from Theorem 4.1 and Lemma 4.2 of [13] by uniformization since the diagonal entries of Q are assumed to be bounded. Similar but less informative results were also obtained in Theorems 1 and 2 of [18], where extra conditions are assumed. Proposition 3.1 Let SB be a countable set, and let Q0 and Qi be SB × SB matrices for i = 0, ±1 such that the matrix Q of the form (2.2) is a transition rate matrix. Assume that the Q is irreducible, the diagonal entries of Q are bounded below, Q has the stationary distribution π ≡ (π 0 , π 1 , . . .), and the Markov additive process (Y (t), B(t)) generated by Q−1 , Q0 and Q1 is 1-arithmetic. Under these conditions, if there exist positive vectors p, q and η ∈ (0, 1) such that ) ( p η −1 Q1 + Q0 + ηQ−1 = 0, (3.5) ) ( −1 η Q1 + Q0 + ηQ−1 q = 0, (3.6) pq < ∞, (3.7) then R has a nonnegative right eigenvector r with eigenvalue η, which is given by r = −(Q0 + (ηI + R)Q−1 )q,

(3.8)

and unique up to constant multiplication. Furthermore, if π 0 q < ∞, 5

(3.9)

then lim η −n π n (k) =

n→∞

π0r p(k), pr

k ∈ SB .

(3.10)

Thus, the tail of the stationary distribution π geometrically decays with the rate η and a constant prefactor in the increasing direction of the level for each fixed background state. This η is said to be the decay rate of π. Remark 3.2 The constant η satisfying (3.5), (3.6) and (3.7) for some positive vectors p and q is unique since η is a Perron Frobenius eigenvalue of R (see also (3.2) for z = η). That is, we do not need (3.9) for the uniqueness of η. Remark 3.3 If (3.9) does not hold, i.e., π 0 q = ∞, while the other conditions in Proposition 3.1 are satisfied, then we can see from the proof of Theorem 4.1 of [13] that lim inf η −n π n (k) = ∞, n→∞

k ∈ SB .

Hence, the decay rate of π n (k) for each k is not less than η. Remark 3.4 The matrix R is not necessarily irreducible, but has only one irreducible class. The i-th entry of the eigenvector r is positive if and only if i belongs to this irreducible class. These facts easily follow from Lemma 4.2 of [13]. To verify condition (3.9), we truncate the QBD process in such a way that its background state space is finite. For this, we consider eigenvectors with eigenvalue 0 of the following matrix. For i = 0, ±1, let Ai , Ai and Ai be finite dimensional square matrices of the same size. For an integer M ≥ 3, let A(M ) be the block tri-diagonal matrix of (M + 1) × (M + 1) blocks having the following form:   A0 A1  A   −1 A0 A1    A A A −1 0 1     .. .. .. A(M ) =  . . . .     A A A −1 0 1    A−1 A0 A1  A−1 A0 Introduce the following conditions: (i) A(M ) is irreducible, and an ML-matrix; i.e., all of its off-diagonal entries are nonnegative. (ii) A(M ) has positive left and right eigenvectors corresponding to eigenvalue 0. Denote (M ) (M ) these eigenvectors by p(M ) ≡ (pℓ ; 0 ≤ ℓ ≤ M ) and q (M ) ≡ (q ℓ ; 0 ≤ ℓ ≤ M ), respectively. (iii) A−1 + A0 + A1 is subrate, that is, the maximum real part of its eigenvalues is not greater than 0. 6

Proposition 3.2 Let (Y (t), B(t)) be the integer-valued Markov additive process derived (M ) by the repeating blocks Ai for i = 0, ±1. Let Nk,ℓ (i, j) be the mean sojourn time of the Markov additive process at state (ℓ, j) before the additive component hits level 0 or M given that it starts in (k, i). If conditions (i), (ii) and (iii) are satisfied, then, p(M ) and q (M ) are obtained by (M )

pℓ

(M )

qℓ (M )

where pi

(M )

(M )

(M )

(M )

= p0 A1 N1,ℓ + pM A−1 N(M −1),ℓ , (M )

(M )

= Nℓ,1 A−1 q 0 (M )

and q i

(M )

(M )

+ Nℓ,(M −1) A1 q M ,

1 ≤ ℓ ≤ M − 1,

(3.11)

1 ≤ ℓ ≤ M − 1,

(3.12)

for i = 0 and M are determined by (M )

(M )

(M )

(M )

p0 (A0 + A1 N1,1 A−1 ) + pM A−1 N(M −1),1 A−1 = 0, (M )

(M )

(M )

(M )

p0 A1 N1,(M −1) A1 + pM (A0 + A−1 N(M −1),(M −1) A1 ) = 0, (M )

(M )

(A0 + A1 N1,1 A−1 )q 0 (M )

(M )

A−1 N(M −1),1 A−1 q 0

(M )

(3.14)

(M )

(3.15)

(M )

(3.16)

+ A1 N1,(M −1) A1 q M = 0, (M )

(3.13)

+ (A0 + A−1 N(M −1),(M −1) A1 )q M = 0.

Remark 3.5 The Markov additive process (Y (t), B(t)) may be defective; i.e., it may terminate in finite time because of the assumption (iii). This proposition is known for the left eigenvector when A(M ) is a transition rate matrix; for example, see Theorem 5 of [6]. The proof there can be extended to the present case. For completeness, we provide a detailed proof in Appendix A. Related results can be also found in Chapters 6 and 10 of [10] and references therein.

4. The corresponding P H/M/2 model with a single queue Let us compare the P H/M/2 model with shortest queue discipline with the corresponding two server system with a single queue. Under the stability condition λ < 2µ, the stationary distribution exists for the both systems. Let L1 and L2 be the length of queues 1 and 2, respectively, in the system with shortest queue discipline under the steady state. And let L be the queue length of the corresponding two server system with a single queue. Then, it is expected that the decay rate of P (L1 + L2 = n) is the same as that of P (L = n) as n goes to infinity. Denote this rate by γ. Since the two queues in the shortest queue system are balanced, we may conjecture that for some positive constant c, P (min(L1 , L2 ) = n) ∼ P (L1 + L2 = 2n) ∼ cγ 2n ,

(4.1)

as n → ∞. Indeed, this holds in the sense of (3.10), which will be proved in the next section. In this section, we first consider the decay rate of a GI/M/2 model with a single queue. Let F be the interarrival time distribution of the GI/M/2 system with a single queue, and let F ∗ (s) be its Laplace-Stieltjes transform. The mean of F is assumed to be λ−1 . The

7

service rates of the two servers are identical and denoted by µ. The stability condition that λ < 2µ is assumed. It is easy to see that the equation s = F ∗ (2µ(1 − s))

(4.2)

has a unique root s ∈ (0, 1). Denote this root by σ. Then, it is well known that the queue length distribution of this system decays geometrically with rate σ (see e.g., [16]). If F is a phase type distribution, more specific results can be obtained as follows. Lemma 4.1 Let P H(α, T ) be the interarrival time distribution of the stable P H/M/2 system with a single queue, and let F be the distribution function for P H(α, T ). Then, the root σ of (4.2) is the largest eigenvalue of the matrix U defined as U = tα (2µ(1 − σ)I − T )−1 ,

(4.3)

and t is its unique right eigenvector with eigenvalue σ. Proof. Since the Laplace-Stieltjes transform F ∗ (s) of the interarrival time of customers is F ∗ (s) = α (sI − T )−1 t for the P H(α, T ) distribution, (4.2) with s = σ becomes σ = α (2µ(1 − σ)I − T )−1 t. Premultiplying the vector t in the both sides of this equation gives σt = tα (2µ(1 − σ)I − T )−1 t. Hence, the matrix U in (4.3) has t as an eigenvector corresponding to eigenvalue σ. It remains to prove the uniqueness of t. We prove this by using Theorem 1.6 in [15]. Since U is not necessarily irreducible, we consider a submatrix of U . Let ST+ = {i ∈ ST ; ti > 0} and let U + be the ST+ × ST+ submatrix of U by deleting all rows and columns i, which are not in ST+ . Similarly, we define t+ to be the ST+ subvector of t. We show that U + is positive, therefore irreducible. Let δ be a positive number such that δ > max{|Tii |; i ∈ ST }, then we have −1

U = (2µ(1 − σ) + δ) tα

∞ ∑

1

ℓ=0

(2µ(1 − σ) + δ)

ℓ

(T + δI)ℓ .

For fixed i, j ∈ ST+ , if αj > 0, then Uij > 0 since ti αj > 0. Otherwise, the irreducibility of T + tα implies that there exist k ̸= j and n ≥ 1 such that αk > 0 and {(T + δI)n }kj > 0. This shows that {tα}ik · {(T + δI)n }kj > 0, which implies Uij > 0. Thus, U + is positive. Hence, by Theorem 1.6 in [15], t+ is the unique Perron Frobenius eigenvector of U + , so t is unique. It is easy to see that σ is the largest eigenvalue for U since t is a nonnegative and non null vector. A similar result but for a left eigenvector is obtained in Proposition 2.1 of [17].

8

5. Decay rate for the model with shortest queue discipline We now prove the main result of this paper. In what follows, Q and Qi for i − 0, ±1 are matrices given by (2.2), and σ is the decay rate of the stationary queue length distribution for the corresponding P H/M/2 model with a single queue, which is obtained as the solution of (4.2). We first define an SB × SB matrix K by K = σ 2 Q−1 + Q0 + σ −2 Q1 . Theorem 5.1 Suppose that the P H/M/2 queue with shortest queue discipline is stable; i.e., λ < 2µ, where λ is the mean arrival rate and µ is the service rate of each of the two identical servers. Then, there exist positive left and right eigenvectors p and q, respectively, of K corresponding to eigenvalue 0 such that pq < ∞, and the nonnegative minimal solution R of (3.1) has a positive right eigenvector r corresponding to the eigenvalue σ 2 , which yield lim σ −2n P (min (L1 , L2 ) = n, |L1 − L2 | = ℓ, J = j) =

n→∞

π0r p(ℓ, j) pr

(5.1)

for each fixed (ℓ, j) ∈ SB , where Li and J are the length of queue i, i = 1, 2 and the arrival phase under steady state, respectively. Remark 5.1 From Proposition 3.1 and Remark 3.2, vectors p, q and r are unique up to constant multiplication. We prove this theorem by applying Proposition 3.1 with η = σ 2 . In other words, we need to verify the conditions (3.5)–(3.7) and (3.9), which will be done in terms of a series of lemmas. Note that K is a block tri-diagonal matrix of the following form:   K0 K 1  K−1 K0 K1      K−1 K0 K1  , ..   .  K−1 K0  ... ... where K 1 = tα + 2σ 2 µI,

K0 = T − 2µI,

K1 = σ 2 µI,

K−1 = σ −2 tα + µI.

Because of the block tri-diagonal structure of K, we first try to find a positive vector q = (q ℓ ; ℓ ≥ 0), partitioned according to the level of K, satisfying Kq = 0 which corresponds to (3.6). Lemma 5.1 Let v = (2µ(1 − σ)I − T )−1 t and q = (σ −ℓ v; ℓ ≥ 0). Then, q is positive, and Kq = 0. 9

Proof. From Lemma 4.1, we have K0 v + K 1 (σ −1 v) = = = = =

(K0 + σ −1 K 1 )v σ −1 (tα + σ(T − 2µI) + 2σ 2 µI)v σ −1 (tα − σ(2µ(1 − σ)I − T )) (2µ(1 − σ)I − T )−1 t σ −1 (tα (2µ(1 − σ)I − T )−1 − σI)t 0.

Similarly, we have, for ℓ ≥ 1, { } σ −ℓ σK−1 + K0 + σ −1 K1 v = σ −ℓ−1 (tα + σ(T − 2µI) + 2σ 2 µI)v = 0. Thus, Kq = 0. It remains to prove that q, equivalently, v, is positive. This is obvious since (−T )−1 t = 1 > 0. We next try to construct a positive vector p = (pℓ ; ℓ ≥ 0) satisfying (3.5). For this purpose, we convert K into a transition rate matrix of a Markov chain. By ∆q we denote the diagonal matrix whose diagonal entries are the corresponding entries of vector q. Similarly, ∆v is defined for vector v. Let D = ∆q , and KD = D−1 KD. Then,   ∆−1 σ −1 ∆−1 v K0 ∆ v v K 1 ∆v  σ∆−1  ∆−1 σ −1 ∆v−1 K1 ∆v v K−1 ∆v v K0 ∆v   −1 −1 −1 −1   σ∆v K−1 ∆v ∆v K0 ∆v σ ∆v K1 ∆v KD =  . .  −1 −1 . .  σ∆v K−1 ∆v ∆v K0 ∆v   .. .. . . Since KD 1 = D−1 Kq = 0, KD is a transition rate matrix of a QBD process with finite number of phases in each level. Let u be the stationary probability vector of −1 ∆−1 v (σK−1 + K0 + σ K1 )∆v .

The mean drift at off-boundary states is computed as ( ) ( −2 ) −1 −1 u σ −1 ∆−1 σ K1 − K−1 v v K1 ∆v − σ∆v K−1 ∆v 1 = σu∆v = −σ −1 u∆−1 v · tα · v < 0. Hence, KD is ergodic by Theorem 3.1.1 of [14]. Thus, there exists a stationary distribution ξ = (ξ ℓ ; ℓ ≥ 0) satisfying ξD−1 KD = 0, i.e., ξD−1 K = 0. This leads to the following result. ) ( Lemma 5.2 Let p = σ ℓ ξ ℓ ∆−1 v ; ℓ ≥ 0 . Then, p is a positive vector satisfying pK = 0 and pq < ∞, which correspond to (3.5) and (3.7). Proof. We only need to verify that pq < ∞, which is immediate from pq = ξ1 = 1 < ∞.

10

It remains to verify the condition corresponding to (3.9) in order to apply Proposition 3.1. This may look intuitively obvious. However, it is very challenging and requires non-trivial proofs. For verifying this condition for an M/M/2 parallel system with shortest queue discipline, Foley and McDonald [5] employed a Lyapunov function, while Takahashi et al. [18] used a state truncation method. Both proofs are complicated. Moreover, key results were used without proofs in the latter paper (see the two lines above equation (B1) on page 21 of [18]). In the following, we use truncation arguments similar to that used in [18], but detailed proofs are provided. For the stationary distribution π = (π n ; n ≥ 0) of Q, we further partition it as π ≡ (π n,ℓ ; n, ℓ ≥ 0), where for fixed n and ℓ, π n,ℓ is an ST -vector given by π n,ℓ (j) = P (min(L1 , L2 ) = n, |L1 − L2 | = ℓ, J = j),

(n, ℓ, j) ∈ S.

To prove (3.9), we need to show that the tail of π 0 is “lighter” than that of q, intuitively. We will not directly prove this result. Instead, we establish a relationship between π 0,ℓ and π n,0 , and then estimate the decay rate of π n,0 by considering a truncated model of Q and adopting max(L1 , L2 ) as its level. To this end, for each integer n ≥ 0, define Mn = {(m, ℓ, j) ∈ S; m + ℓ ≤ n}, which is the set of all states such that the longer queue is not greater than n. Consider the flow balance equation between Mn and S \ Mn for n ≥ 0. Notice that the only state set through which the process could leave Mn for S \ Mn is {(n, 0, j); j ∈ ST } due to an arrival, and the state sets through which the process could leave S \ Mn for Mn are {(ℓ, n + 1 − ℓ, j); j ∈ ST } for ℓ = 0, 1, . . . , n due to service completions. Therefore, we have π n,0 tα 1 =

n ∑

µπ k,n+1−k 1,

k=0

which implies that π 0,n+1 1 ≤

1 π n,0 t. µ

This yields π0q =

∞ ∑

σ −ℓ π 0,ℓ v ≤ π 0,0 v +

ℓ=0

d ∑ −n σ π n−1,0 t, µ n=1 ∞

(5.2)

where d is a positive constant satisfying v ≤ d1. Since the phase space ST for arrivals is finite, π 0,0 v is finite. So, the condition (3.9) is obtained if we can show that the remaining sum in (5.2) is finite. In the following lemma, we show a stronger result, from which the finiteness of (5.2) is a direct consequence. Lemma 5.3 For any small ϵ > 0, we have lim sup σ (−2+ϵ)n π n,0 = 0. n→∞

11

(5.3)

{0,1,..., M } × ST µI

T

µI

tα

T

µI

tα tα

2µ I

T

tα

T

T

tα

µI

µI

tα

T

µI

T

tα

T

tα

tα 2µ I

T

µI

µI

tα 2µ I

T

tα

µI µI

µI

µI

2µ I

T

tα T

max ( L1( M ) (t ), L 2( M ) (t ) )

Level M

Figure 3: A truncated model Clearly, Lemma 5.3 implies the finiteness of π 0 q and completes the proof of Theorem 5.1. In what follows, we prove Lemma 5.3 employing truncation arguments. Since these arguments are lengthy, they are divided into three steps. Step 1 For each integer M ≥ 3, we modify the original system with shortest queue discipline in such a way that the service of the shorter queue is stopped when the difference of the two queues attains M . Namely, the state (n, ℓ, j) ∈ S is truncated as ℓ ≤ M by removing the state transitions from (n, M, j) to (n − 1, M + 1, j). For this truncated model, we define a new level by n + ℓ for the state (n, ℓ, j), which means that the longer queue length is chosen for the level instead of the shorter queue length. The state of this truncated model is (max(L1 (t), L2 (t)), |L1 (t) − L2 (t)|, J(t)) with the limitation that |L1 (t) − L2 (t)| ≤ M . See Figure 3 for the transition diagram of this model. The truncated model can be considered as a QBD process with transition rate matrix:   (M ) (M ) Q0,0 Q0,1   (M ) (M ) (M ) Q1,2   Q1,0 Q1,1   .. .. ..   . . .     (M ) (M ) (M ) (M ) Q = QM −1,M −2 QM −1,M −1 QM −1,M ,   (M ) (M ) (M )   QM,M −1 QM,M Q1   (M ) (M ) (M )   Q Q Q   −1 0 1 .. .. .. . . . (M )

(M )

(M )

where Qi,i , Qi,i+1 and Qi+1,i for 0 ≤ i ≤ M −1 are m(i+1)×m(i+1), m(i+1)×m(i+2) (M ) and m(i + 2) × m(i + 1) matrices, respectively, QM,M is m(M + 1) × m(M + 1) matrix,

12

(M )

(M )

and Q−1 , Q0

(M )

and Q1



(M ) Q−1

O  µI O  = ... ...  µI O

are m(M + 1) × m(M + 1)   T − 2µI  tα T    (M )   , Q0 =     

(M )

2µI − 2µI µI .. .. .. . . . tα T − 2µI µI tα T − µI

    ,  



O tα  .. = .

Q1

matrices such that

 . O (M )

(M )

(M )

Step 2 In this step, we show that the QBD process that has Q−1 , Q0 , Q1 as its repeating blocks is stable for a sufficiently large M . By Theorem 3.1.1 of [14], this is equivalent to show that (M )

(M )

− Q−1 )1 < 0,

p(M ) (Q1

(5.4) (M )

(M )

(M )

for a sufficiently large M , where p(M ) is the stationary distribution of Q−1 +Q0 +Q1 . To prove this, we consider the continuous time Markov chain with transition rate matrix (∞) (∞) (∞) (∞) (M ) Q−1 + Q0 + Q1 , where Qi are the matrices obtained from Qi by letting M → ∞ for i = 0, ±1. Clearly, this Markov chain is a QBD process, whose transition rate matrix (∞) (∞) (∞) Q−1 + Q0 + Q1 has µI + tα, T − 2µI, µI as its repeating blocks. Consider the Markov additive process generated by this repeating blocks. Its background process has the unique stationary distribution because T + tα is a finite dimensional and irreducible rate matrix. Denote this distribution by row vector ν. Then, its mean drift is given by ν(−(µI + tα) + µI)1 = −νt < 0. (∞)

(∞)

(∞)

Hence, Q−1 + Q0 + Q1 has the stationary distribution, which is denoted by vector (∞) p(∞) . As usual, we partition this vector in blocks as p(∞) = (pℓ ; ℓ ≥ 0). Then, we have the following result, whose proof is deferred to Appendix B since it involves just computations. Lemma 5.4 Under the above settings, the stability condition λ < 2µ implies (∞)

p(∞) (Q1

1 (∞) − Q−1 )1 = λ − µ < 0. 2

(5.5)

We are now back to the proof of (5.4). Similar to (B.4) in Appendix B, we have (M )

p(M ) (Q1

(M )

(M )

− Q−1 )1 = p0 (µ1 + t) − µ. (M )

(M )

(M )

Applying Theorem 3.4 in [11] to Q−1 + Q0

+ Q1

(M )

= p0 .

lim p0

M →∞

(∞)

(∞)

and Q−1 + Q0

(∞)

Hence, (5.6) and Lemma 5.4 yield (5.4) for a sufficiently large M . 13

(5.6) (∞)

+ Q1 , we have (5.7)

Remark 5.2 The assumption of finiteness of the phase assumed in Theorem 3.4 of [11] is not essential, and the conclusion in the theorem is still valid for infinitely many phases. We also notice that (5.7) can be obtained by using (3.13) in Proposition 3.2. Roughly (M ) (M ) speaking, this is because N1,1 and pM converge to N1,1 and the null vector, respectively, as M goes to infinity, where N1,1 (i, j) is the mean sojourn time of the corresponding Markov additive process at state (1, j) before hitting level 0 when it starts at (1, i). (M )

Remark 5.3 For the truncated model by Q(M ) , let Li (t) be the size of queue i at time t for i = 1, 2. Since the difference between the two queue lengths is restricted by the service blocking in this model, we can construct sample paths such that, for i = 1, 2, (M ) (M ) Li (t) ≤ Li (t) for all t ≥ 0 with probability one when Li (0) = Li (0) = 0 and the same arrival phase process is given. This observation concludes that Q has the stationary distribution if λ < 2µ since Q(M ) has the stationary distribution for a sufficiently large M under the same condition. Step 3 Choose a sufficiently large M such that (5.4) holds. Then, Q(M ) has the (M ) stationary distribution, which is denoted by vector π (M ) = (π n ; n ≥ 0) partitioned (M ) according to the level. We consider the decay rate of π n as n goes to infinity. Since Q(M ) represents the QBD process with finitely many background states, we can find a positive constant ηM < 1 and positive vector p(M ) and q (M ) such that p(M ) K (M ) (ηM ) = 0, where

K (M ) (ηM )q (M ) = 0,

K (M ) (η) = η −1 Q1

(M )

(M )

+ Q0

(5.8)

(M )

+ ηQ−1 .

(M )

This ηM is the decay rate of π n , and p(M ) and q (M ) are unique. Partition p(M ) = (M ) (M ) (pℓ ; 0 ≤ ℓ ≤ M ) and q (M ) = (q ℓ ; 0 ≤ ℓ ≤ M ) in ST blocks. We note the following facts. (a) ηM , as the decay rate for the longer queue of the truncated model, is also the decay rate for the shorter queue of the truncated model. (b) ηM is not less than the decay rate for the shorter queue of the original model as described in Proposition 3.1. (c) ηM is decreasing in M . (d) ηM ≥ σ 2 . Here, (a) is immediate since we have (M )

(M )

(M )

(M )

(M )

(M )

max(L1 (t), L2 (t)) = min(L1 (t), L2 (t)) + |L1 (t) − L2 (t)|, (M )

(M )

where |L1 (t) − L2 (t)| is bounded by M . Furthermore, (b) and (c) follow from the observation in Remark 5.3, while (d) is obtained from (b), Remark 3.3 and Lemmas 5.1 and 5.2. 14

In the following, we show that ηM can arbitrarily approach σ 2 . Therefore, (5.3) follows according to (b). Due to (c), we can define η∞ as η∞ = lim ηM , M →∞

which is not less than σ 2 by (d). The corresponding limiting vectors p = (pℓ ; ℓ ≥ 0) and q = (q ℓ ; ℓ ≥ 0) are defined as, for ℓ ≥ 0, (M )

pℓ = lim inf pℓ M →∞

,

(M )

q ℓ = lim inf q ℓ M →∞

.

We also define generating function matrix K (∞) (η) as K (∞) (η) = η −1 Q1

(∞)

(∞)

+ Q0

(∞)

+ ηQ−1 ,

η > 0.

Lemma 5.5 The vectors p and q are positive, p q < ∞, and pK (∞) (η∞ ) = 0,

K (∞) (η∞ )q = 0.

Proof. Let A(M ) = K (M ) (η) in Proposition 3.2. Since this A(M ) also depends on η, we (M ) denote it by A(M ) (η), and its block entries by Ai (η) if they depend on η. Namely, A−1 (η) = tα + ηµI, A0 = T − 2µI, A1 = µI, −1 A−1 = A−1 , A0 = A0 , A1 (η) = η tα + 2µI, A−1 = A−1 , A0 = T − µI, A1 = A 1 . Then, it is easily seen that all the conditions of Proposition 3.2 are satisfied for η = ηM . Since A−1 (η) + A0 + A1 is a defective transition rate matrix for 0 < η < 1, we have, for each fixed ℓ ≥ 1, (M )

(M )

lim N(M −1),ℓ (η) = lim Nℓ,(M −1) (η) = 0,

M →∞ (M )

M →∞

(M )

where Ni,j (η) are the Ni,j

(M )

(M )

for A(M ) (η). Since N1,ℓ (η) and Nℓ,1 (η) are increasing and (∞)

(∞)

bounded in M for each fixed η, there exist finite matrices N1,ℓ (η) and Nℓ,1 (η) such that (∞)

(M )

N1,ℓ (η) = lim N1,ℓ (η), M →∞

(∞)

(M )

Nℓ,1 (η) = lim Nℓ,1 (η). M →∞

(M )

Note that Ni,j (η) is increasing in η, since A−1 (η) is increasing in η. So, we have, for any ϵ > 0 satisfying ϵ + η∞ < 1, (∞)

(M )

(M )

(∞)

N1,ℓ (η∞ ) ≤ lim inf N1,ℓ (ηM ) ≤ lim sup N1,ℓ (ηM ) ≤ N1,ℓ (η∞ + ϵ). M →∞

(∞)

M →∞

(∞)

Since limϵ↓0 N1,ℓ (η∞ + ϵ) = N1,ℓ (η∞ ), we have (M )

(∞)

lim N1,ℓ (ηM ) = N1,ℓ (η∞ ).

M →∞

15

Similarly, we have (M )

(∞)

lim Nℓ,1 (ηM ) = Nℓ,1 (η∞ ).

M →∞

Hence, from (3.11), (3.12), (3.13) and (3.15), we have (∞)

(∞)

pℓ = p0 A1 N1,ℓ (η∞ ),

q ℓ = Nℓ,1 (η∞ )A−1 q 0

where (∞)

(∞)

p0 (A0 + A1 N1,1 (η∞ )A−1 ) = 0, (M )

(A0 + A1 N1,1 (η∞ )A−1 )q 0 = 0.

(M )

(M )

(M )

We can choose p0 and q 0 in such a way that p0 (0) = q 0 (0) = 1 for all M . This implies that p0 (0) = q 0 (0) = 1. Hence, from the above equations and the irreducibility (∞)

of rate matrix A0 + A1 N1,1 A−1 , p0 and q 0 must be positive, so p and q are positive. (∞)

(∞)

Furthermore, N1,ℓ (η∞ ) and Nℓ,1 (η∞ ) geometrically decay entry-wise as ℓ goes to infinity since A−1 (η) + A0 + A1 is a defective transition rate matrix such that (A−1 (η) + A0 + A1 )1 < 0. This implies that p q < ∞. Thus, we have proved the first half of the lemma. The remaining part is easily obtained since K (M ) (ηM ) converges to K (∞) (η∞ ) entry-wise as M goes to infinity and these matrices have the tri-diagonal structure. (∞)

(∞)

Lemma 5.6 Let q ℓ = σ ℓ v for ℓ ≥ 0 and q (∞) = (q ℓ ; ℓ ≥ 0) for v of Lemma 5.1, then K (∞) (σ 2 )q (∞) = 0, and there exists a unique positive vector p(∞) such that p(∞) q (∞) < ∞.

p(∞) K (∞) (σ 2 ) = 0,

Proof. Similarly to the proof of Lemma 5.1, we have (∞)

(T − 2µI)q 0

+ (σ −2 tα + 2µI)q 1

(∞)

= σ −1 (tα + σ(T − 2µI) + 2σ 2 µI)v = 0,

and for ℓ ≥ 1, (∞)

(∞)

(tα + σ 2 µI)q ℓ−1 + (T − 2µI)q ℓ

(∞)

+ µIq ℓ+1

= σ ℓ−1 (tα + σ(T − 2µI) + 2σ 2 µI)v = 0. Thus, we have that K (∞) (σ 2 )q (∞) = 0. The remaining part is proved similarly to ˜ = ∆−1(∞) K (∞) (σ 2 )∆q(∞) , i.e., Lemma 5.2. Namely, let K q   ˜ 01 ˜ 00 K K   K ˜1 ˜0 K ˜ 10 K  ˜ = K ,  ˜ ˜ ˜ K K K 1 0 −1   .. .. .. . . . ˜ 10 = σ −1 ∆−1 (tα + ˜ 01 = σ∆−1 (σ −2 tα + 2µI)∆v , K ˜ 00 = K ˜ 0 = ∆−1 (T − 2µI)∆v , K where K v v v 2 2 −1 −1 ˜ 1 = σµI and K ˜ −1 = σ ∆ (tα + σ µI)∆v . Then, there is a stationary σ µI)∆v , K v ˜ such that probability vector u ˜ −1 + K ˜0 + K ˜ 1 ) = 0. ˜ (K u 16

Since ˜1 − K ˜ −1 )1 = −σ −1 u ˜ (K ˜ ∆−1 u v tα∆v 1 < 0, ˜ is ergodic by Theorem 3.1.1 of [14]. So there is a distribution ξ˜ such that ξ˜K ˜ = 0. Let K −1 (∞) ˜ p = ξ∆q(∞) , then we have ˜ = 1 < ∞. p(∞) q (∞) = ξ1 This completes the proof. (Proof of Lemma 5.3) By Remark 3.2 and Lemma 5.4, if K (∞) (η) has positive left and right eigenvectors with eigenvalue 0 such that their inner product is finite, then η is unique. Hence, Lemma 5.5 and Lemma 5.6 conclude that η∞ = σ 2 . So (5.3) follows from (b).

6. Further problems This paper seems to be the first one for studying on the shortest queue with a non-Poisson arrival process, and there arise many interesting questions on such a queueing model. The first question is on the arrival process. Our results suggests that a similar result may be obtained for a more general arrival process. For example, Theorem 5.1 may be true for the Markovian arrival process of Neuts, MArP for short, if the service times are exponentially distributed. This may hold even for a general stationary arrival process that allows for the corresponding single queue model to have a light tailed queue length distribution. Of course, the latter would be a much harder problem, and the case of MArP may be more promising for verification. Another related question is on the case that each queue has their own arrival stream as in the model of [5]. It is interesting to see when the two queues are well balanced in the case that there are PH type renewal arrivals dedicated to each queue. However, the present approach may not be useful in this case since the verification of (3.9) would be harder. The second question is on the service time distributions. The exponential service greatly simplifies arguments for deriving the square form of the decay rate. But we do not know what happens for the non-exponential service. So, it could be questioned for which class of the service time distributions we can have the square form of the decay rate. The third question is on the decay rate of the marginal distribution. The arrival phases are obviously aggregated because the number of those phases are finite. The aggregation seems to be true also for the difference of the queue sizes. That is, we may expect to have lim σ −2n P (min (L1 , L2 ) = n) =

n→∞

π0r p1. pr

(6.1)

This may be obtained from (5.1) by the summation over the difference of the two queue sizes. However, it requires to verify the exchange of the limit and summation since the difference of the queue sizes is unbounded. The following corollary may be useful for this verification. 17

Corollary 6.1 Assume the same conditions as in Proposition 3.1. Let SB+ ⊂ SB be the set of irreducible indexes of R. Then, if r(i)−1 is bounded over i ∈ SB+ , we have lim η −n π n 1 =

n→∞

π0r p1. pr

(6.2)

That is, the marginal distribution of the level also decays with rate η. The proof of this corollary is given in Appendix C. From (3.8), r seems to have the same asymptotics as q, so Corollary 6.1 leads to (6.1). However, we have not formally proved this, where the difficulty lies in less information on matrix R. It may be also interesting to see the tail decay rate of the shortest queue length at an arrival instant. This is immediately seen to be also σ 2 from the observation that the stationary probability in state (n, i, j) at the arrival instant is bπ n (i, j)tj for a normalizing constant b, where tj is the j-th entry of t. A formal derivation for this can be obtained by the rate conservation law (see, e.g., [12]). Since the stationary distribution for the arrival phase is proportional to α(−T )−1 , we can see that b = α(−T )−1 1. We also note the case that there are k servers for k ≥ 3. If the service time distributions are identical and exponential, then similar arguments to derive (4.1) yield P (min(L1 , . . . , Lk ) = n) ∼ P (L1 + . . . + Lk = kn) ∼ cσ kn , where Li is the size of the i-th queue in the steady state for i = 1, . . . , k. There seems no verification to have been reported for this asymptotic behavior even for the Poisson arrivals. The present approach might be useful for this verification. Finally, we comment on that when the assumption of an equal service rate for the both exponential servers made in this paper is dropped, a similar characterization on tail asymptotics can be expected by using the same idea as used in this paper.

Acknowledgements The authors are grateful to the anonymous referees for their careful reading and invaluable comments. This research was originated from mutual visits of the second and third authors in 2004 and 2003, respectively. Both of them are grateful to their institutes for supporting those visits. We were also supported in part by JSPS under grant No. 13680532, and by a research grant from NSERC.

References [1] Adan, I.J.B.F., Wessels, J. and Zijm, W.H.M. (1990) Analysis of the symmetric shortest queue problem, Stochastic Models 6, 691–713. [2] Adan, I.J.B.F., Wessels, J. and Zijm, W.H.M. (1991) Analysis of the asymmetric shortest queue problem, Queueing Systems 8, 1–58. [3] Adan, I.J.B.F. and Zhao, Y.Q. (2003) Shortest queue model with Erlangian arrivals, unpublished note.

18

[4] Arjas, E. and Speed, T. P. (1973) Symmetric Wiener-Hopf factorizations in Markov additive processes, Z. Wahrcheinlich. verw. Geb. 26, 105–118. [5] Foley, R. D. and McDonald, D. R. (2001) Join the shortest queue: stability and exact asymptotics, Annals of Applied Probability 11(3), 569–607. [6] Hajek, B. (1982) Birth-and death processes on the integers with phases and general boundaries, Journal of Applied Probability 19, 488–499. [7] Kingman, J.F.C. (1961) Two similar queues in parallel, Annals of Mathematical Statistics 32, 1314–1323. [8] Kroese, D.P., Scheinhardt, W.C.W. and Taylor, P.G. (2004) Spectral Properties of the Tandem Jackson Network, Seen as a Quasi-Birth-and-Death Process, Annals of Applied Probability 14, 2057-2089. [9] Kurkova, I.A. and Suhov, Y. M. (2003) Malyshev’s theory and JS-queues. Asymptotics of stationary probabilities, Annals of Applied Probability, 13(4), 1313–1354. [10] Latouche, G. and Ramaswami, V. (1999) Introduction to Matrix Analytic Methods in Stochastic Modeling, American Statistical Association and the Society for Industrial and Applied Mathematics, Philadelphia. [11] Li, H. and Zhao, Y.Q. (2000) Stochastic block-monotonicity in the approximation of the stationary distribution of infinite Markov chains, Stochastic Models, 16, 313–333. [12] Miyazawa, M. (1994) Rate conservation laws: a survey, Queueing Systems 15, 1–58. [13] Miyazawa, M. and Zhao, Y.Q. (2004) The stationary tail asymptotics in the GI/G/1 type queue with countably many background states, Adv. in Appl. Probab. 36(4), 1231–1251. [14] Neuts, M.F. (1981) Matrix-Geometric Solutions in Stochastic Models, Johns Hopkins University Press, Baltimore. [15] Seneta, E. (1981) Nonnegative Matrices and Markov Chains, Springer-Verlag, New York. [16] Tak´acs, L. (1962) Introduction to the Theory of Queues, Oxford University Press, New York. [17] Takahashi, Y. (1981) Asymptotic exponentiality of the tail of the waiting-time distribution in a P H/P H/c queue. Adv. Appl. Prob., 13, 619–630. [18] Takahashi, Y., Fujimoto, K. and Makimoto, N. (2001) Geometric decay of the steadystate probabilities in a quasi-birth-and-death process with a countable number of phases, Stochastic Models 17, 1–24.

Appendix 19

A Proof of Proposition 3.2 We prove this proposition only for q (M ) since the proof for p(M ) is similar. Delete the first and the last block rows and columns from (M + 1) × (M + 1) block matrix A(M ) . Denote this (M − 1) × (M − 1) block matrix by A∗ . Then, the vector equation A(M ) q (M ) = 0 can be written as  (M )      (M ) q1 0 A−1 q 0 ..    (M )    0      . ∗  q2 + A +  = 0,      . . ..     ..   0 (M )

A0 q 0

(M )

(M )

0

A1 q M

q M −1 (M )

+ A1 q 1

(M )

= 0,

(M )

A−1 q M −1 + A0 q M = 0.

By the assumption (iii), (−A∗ )−1 exists, and its kℓ block is Nkℓ . These equations immediately conclude (3.12), (3.15) and (3.16). (M )

B Proof of Lemma 5.4 (∞)

(∞)

(∞)

Since Q−1 + Q0 + Q1 is a transition rate matrix of the QBD process, we can apply Neuts’ matrix geometric solution. Namely, there exists the minimal nonnegative and nonzero matrix R of the following matrix equation: µI + R(T − 2µI) + R2 (tα + µI) = O.

(B.1)

Then, we have the expression: (∞)

(∞)

where p1

(∞)

ℓ ≥ 1,

(B.2)

p0 (T − 2µI) + p1 (tα + µI) = 0,

(B.3)

pℓ

= p1 Rℓ−1 ,

is determined by (∞)

(∞)

(∞) p0 (tα

+ 2µI) +

(∞) p1 (T

− 2µI + R(tα + µI)) = 0.

We start with computing the left-hand side of (5.5) as (∞)

p(∞) (Q1

(∞)

(∞)

− Q−1 )1 = p0 t − µ

∞ ∑

(∞)

pℓ

1

ℓ=1

=

(∞) p0 (µ1

+ t) − µ.

(B.4)

Postmultiplying (B.3) by 1 leads to (∞)

(∞)

p0 (2µ1 + t) = p1 (t + µ1). From this and (B.2), (B.4) becomes (∞)

p

(∞) (Q1

−

(∞) Q−1 )1

∞ ( ) ∑ (∞) + t) − µ 2 − pℓ 1

=

(∞) p1 (µ1

=

ℓ=1 (∞) p1 (µ1 + t + µ(I − R)−1 1) − 2µ (∞) p1 (I − R)−1 ((I − R)(µ1 + t) + µ1)

=

20

− 2µ.

(B.5)

Note that (B.1) yields (I − R)(µ(I − R)1 − Rt) = 0. Since I − R is invertible, we have that µ(I − R)1 = Rt. Using this and the fact that ∞ ∑

(∞)

pℓ

t = λ,

ℓ=0

(B.5) becomes (∞)

p(∞) (Q1

− Q−1 )1 = p1 (I − R)−1 (t + µ1) − 2µ (∞)

(∞)

(∞)

= λ − p0 (µ1 + t) − µ. Comparing this with (B.4), we have 1 (∞) p0 (µ1 + t) = λ. 2 This and (B.4) lead to the equality of (5.5). Thus, the proof of Lemma 5.4 is completed by the stability condition λ < 2µ.

C Proof of Corollary 6.1 For SB+ -dimensional vector u = (u(i); i ∈ SB ), define ∆u to be the SB ×SB diagonal matrix whose i-th diagonal entry is u(i) if i ∈ SB+ and 0 otherwise. Let r + = (r(i); i ∈ SB+ ) and (r + )−1 = (r(i)−1 ; i ∈ SB+ ). Then, from (3.3), we have ( )n−1 π n = η n−1 π 1 ∆r+ η −1 ∆(r+ )−1 R∆r+ ∆(r+ )−1 . From the proof of Theorem 4.1 of [13], we can see that η −1 ∆(r+ )−1 R∆r+ is a transition probability matrix which has the recurrent class SB+ , and (3.9) implies that π 1 ∆r+ is a finite measure. Furthermore, ∆(r+ )−1 1 is bounded by the assumption. Hence, the standard theory of a Markov chain concludes (6.2).

21