M/M/3/3 AND M/M/4/4 RETRIAL QUEUES. Tuan Phung-Duc, Hiroyuki Masuyama, Shoji Kasahara and Yutaka Takahashi

JOURNAL OF INDUSTRIAL AND doi:10.3934/imo.2009.5.431 MANAGEMENT OPTIMIZATION Volume 5, Number 3, August 2009 pp. 431 451 M/M/3/3 AND M/M/4/4 RETRIAL QUEUES Tuan Phung-Duc, Hiroyuki Masuyama, Shoi Kasahara and Yutaka Takahashi Graduate School of Informatics, Kyoto University Yoshida-Honmachi, Sakyo-ku, Kyoto 606-8501, Japan Abstract. This paper studies M/M/c/c retrial queues, where c servers are all identical. In the retrial queues, an arriving customer is served immediately if it finds an idle server upon arrival, otherwise the customer tries to enter the system after an exponentially distributed time independently of other customers. As is well known, it is a challenging problem to obtain an analytical solution for the stationary oint distribution of the numbers of retrial customers and busy servers in the M/M/c/c retrial queue especially for c 3. Under some technical assumptions, a few analytical solutions have been presented for c 3. This paper derives analytical solutions for M/M/3/3 and M/M/4/4 retrial queues without such technical assumptions. Through many numerical examples, we show that the derived analytical solutions can be computed by a numerically stable algorithm. 1. Introduction. This paper considers M/M/c/c retrial queues, where c servers are all identical. Customers arrive at the system according to a Poisson process with a rate λ > 0, and their service times are independent and identically distributed (i.i.d. according to an exponential distribution with mean one. Each arriving customer receives service immediately if it finds an idle server on arrival. Otherwise the arriving customer oins a virtual pool, say orbit. A customer in the orbit is called as a retrial customer hereafter. Each retrial customer stays in the orbit for an exponentially distributed time with finite positive mean 1/µ independently of other customers. After the soourn time in the orbit, a retrial customer tries to enter the system and is served immediately if at least one server is idle on its retrial, otherwise it returns to the orbit again. Let C(t and N(t (t 0 denote the numbers of busy servers and retrial customers, respectively, at time t. It is easy to see that X(t = (C(t, N(t (t 0 forms an irreducible level-dependent quasi-birth-and-death process with the state space {0, 1,..., c} Z +, where Z + = {0, 1, 2,... }. It is known that X(t (t 0 is ergodic if and only if λ < c (see, e.g., pp. 97 100 in [3]. Throughout the paper, we assume λ < c and then define i, (i = 0, 1,...,c, Z + as i, = Pr[C(t = i, N(t = ], t where we use superscript c to emphasize the number c of servers in the system. The ergodicity of X(t shows that i, > 0 for all i = 0, 1,...,c and Z +. 2000 Mathematics Subect Classification. Primary: 68M20, 90B22; Secondary: 60K25. Key words and phrases. Multiserver retrial queue, analytical solution, minimal solution, threeterm recurrence relations, continued fractions. 431

432 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI In this paper, we study the stationary distribution { i, ; i = 0, 1,...,c, Z +}. Our M/M/c/c retrial queues and their variants have been studied extensively by many researchers (see a book [3], survey papers [2, 10, 13], and a list of related publications in [4]. However analytical solutions for the stationary distribution } have been obtained only in a few special cases. For the case of c = 1 (i.e., the single-server case, an explicit solution of the stationary distribution is derived in [3]. For the case of c = 2, the stationary distribution is expressed in terms of hypergeometric functions [6, 11]. Compared with the cases of c = 1, 2, the cases of c 3 are much more difficult. Under some technical assumption, Kim [9] and Gomez-Corral and Ramalhoto [5] derive analytical solutions for the M/M/3/3 retrial queue. Through the numerical experiment, Gomez-Corral and Ramalhoto [5] confirm that ergodic M/M/3/3 retrial queues satisfy the technical assumption. For any value of c = 1, 2,..., Pearce { i, [11] constructs an analytical solution for { i, continued fractions, assuming those its are all nonzero. } in terms of the its of extended The main contribution in this paper is to prove, by the method of continued fractions, that analytical solutions for { i, } in M/M/3/3 and M/M/4/4 retrial queues do not require any technical assumptions such as made in [5, 9, 11]. The main results are presented in Section 3, which follow some preinary results in Section 2. In Section 4, we provide a backward algorithm to compute the stationary distribution. Section 5 shows some numerical examples. Finally, Section 6 is devoted to some conclusions and future works. 2. Preinary results. This section provides some preinary results on the stationary distribution { i, ; i = 0, 1,..., c, Z +}. It is easy to see that { i, } satisfies the following balance equations. (λ + i + µ i, = λπ[c] i 1, + ( + 1µπ[c] i 1,+1 + (i + 1π[c] i+1,, (1 i = 0, 1,..., c 1, (λ + c c, = λπ[c] c 1, + ( + 1µπ[c] c 1,+1 + λπ[c] c, 1, (2 i = c, for all = 0, 1, 2,.... Let P [c] i (z (i = 0, 1,..., c denote P [c] i (z = It then follows from (1, (2 and (3 that (λ + ip [c] i (z + µz d dz P [c] i (z = λp [c] i 1 (z + µ d dz P [c] i, z. (3 i 1 [c] (z + (i + 1P i+1 (z, (4 i = 0, 1,..., c 1, (λ + cp c [c] [c] (z = λp c 1 (z + µ d dz P [c] [c] c 1 (z + λzp c (z, (5 i = c, where P [c] [c] 1 (z = (d/dzp 1 (z = 0. Using (4, we can readily prove the following result by mathematical induction.

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 433 Proposition 2.1. P n [c] (z (n = 0, 1,...,c is given by n n!p n [c] (z = e n,i (z di [c] dzip 0 (z, (6 where {e n,i (z; n = 0, 1,...,c, i = 0, 1,..., n} is determined by the following recursion with e 0,0 (z = 1. ( d e n+1,i (z = (λ + ne n,i (z + µz dz e n,i(z + e n,i 1 (z ( d nλe n 1,i (z nµ dz e n,i(z + e n 1,i 1 (z, (7 where e 1,i (z = 0 (i = 0, 1,... and e n, 1 (z = 0 (n = 0, 1,.... It is easy to see from (7 that e n,i (z is written as e n,i (z = i =max(2i n,0 e n,i, z, n = 0, 1,...,c, i = 0, 1,..., n, (8 where e n,i, can be expressed in terms of given parameters λ and µ. Explicit expressions of e n,i, s (n = 0, 1, 2, 3, 4 are given in Appendix C.3. In what follows, for simplicity, we denote following is an immediate consequence of Proposition 2.1. by q[c] for = 0, 1,.... The Proposition 2.2. For the cases of c = 3 and 4, we have ( 1 1, = e 1,i,i ( i + 1 i, c = 3, 4, (9 {( 2 } 2, = 1 2 3, = 1 6 e 2,i,i ( i + 1 i {( 3 e 3,i,i ( i + 1 i c = 3, 4, {( 4 4, = 1 e 4,i,i ( i + 1 i 24 +e 4,2,0 ( + 1( + 2 +2 + + e 2,1,0 ( + 1 +1 + ( 2 e 3,i,i 1 ( i + 2 i i=1 ( 3 e 4,i,i 1 ( i + 2 i i=1, c = 3, 4, (10 +1 } +1, (11 }, c = 4. (12 Proposition 2.2 shows that the stationary distribution { i, } (i = 1, 2,..., c, = 0, 1,... can be obtained through { ; = 0, 1,... }. 3. Main results. This section presents the main results in this paper. 3.1. Analytical solutions. Lemma 3.1. P [c] 0 (z satisfies c s [c] i (z di [c] dzip 0 (z = 0, z < 1, (13

434 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI where for i = 0, 1,..., c, s [c] i (z = λe c,i (z cµ c 1 n=i 1 (c 1! n! Proof. The proof of Lemma 3.1 is given in Appendix A. ( d dz e n,i(z + e n,i 1 (z. (14 Remark 3.1. It follows from (8 and (14 that s [c] i (z (i = 0, 1,...,c is given by i s [c] i (z = ( 1 i+ s [c] i, z, (15 =max(2i c 1,0 where s [c] i, can be expressed in terms of given parameters λ and µ. For c = 3 and 4, explicit expressions of s [c] i, s are written in Appendix C.3. The differential equations for P [3] [4] 0 (z and P 0 (z are given by and (s [3] 3,3 z3 s [3] 3,2 z2 d3 dz 3 P [3] 0 (z + (s[3] 2,2 z2 s [3] 2,1 z + s[3] (s [4] 4,4 z4 s [4] 4,3 z3 d4 dz 4P[4] 0 0 (z 2,0 d2 dz 2P[3] + (s [3] 1,1 z s[3] 1,0 d dz P [3] 0 (z + s[3] (z + (s[4] 3,3 z3 s [4] 3,2 z2 + s [4] 3,1 0,0 P [3] 0 z d3 dz 3 P [4] 0 (z + (s [4] 2,2 z2 s [4] 2,2 z + s[4] 2,0 d2 dz 2 P [4] 0 (z + (s [4] 1,1 z s[4] 1,0 d dz P [4] 0 (z + s[4] 0,0 P [4] 0 (z = 0 (16 (z = 0, (17 respectively. The important point here is that, for c = 3 and 4, s [c] i (z (i = 0, 1..., c is an i-th degree polynomial with less than or equal to 3 terms. This property leads to three terms recurrence relations. In what follows, let (α n (α R denote { 1, n = 0, (α n = α(α + 1 (α + n 1, n = 1, 2,.... From (3, we have P [c] 0 (z = z. (18 Substituting (18 into (16 and (17, we obtain K [c] z = 0, (19 where K [c] K [c] = is given by ( c + ( c 1 s [c] i,i ( i + 1 i i=2 s [c] i,i 2 ( i + 3 i ( c i=1 +2. s [c] i,i 1 ( + 2 i i +1

Note here that (19 implies that K [c] where a [c] a [c] = Lemma 3.2. and b [c] M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 435 +2 = b[c] q[c] +1 + a[c] = 0 ( = 0, 1,... and therefore q[c] ( = 0, 1,... are given by c s[c] i,i ( i + 1 i c 1 i=2 s[c] i,i 2 ( i + 3 i a [c] =, b [c], b [c] =, = 0, 1, 2,..., (20 =, c i=1 s[c] i,i 1 ( i + 2 i. (21 c 1 i=2 s[c] i,i 2 ( i + 3 i b [c] a [c] = c λ > 1. Proof. It is clear from (21 that the numerators (resp. denominators of both a [c] and b [c] are c-th (resp. (c 1-st degree polynomials of. Thus a [c] = and b [c] =. Furthermore from (21, we have b [c] a [c] = Note here (see Appendix C.3 that s [3] 3,2 s [3] 3,3 i,i 1 ( i + 2 i c s[c] i,i ( i + 1 = s[c] i c i=1 s[c] = 3 λ > 1, s [4] 4,3 = 4 s [4] λ > 1, 4,4 c,c 1 s [c] c,c where the inequalities follow from the ergodic condition λ < c (see, e.g., pp. 97 100 in Falin and Templeton [3] or Choi and Kim [1]. The third equation in Lemma 3.2 implies that there exists some nonnegative integer 0 such that b [c] a[c] + 1, = 0, 0 + 1,.... (22 It follows from (20 that for any k = 0, 1, 2,..., +k = ξ[c] k, q[c] [c] k+1 + ξ k, q[c] k where the two sequences { [c] k, } and { ξ k, } are determined by k,0 = 0, ξ[c] k,1 = 1, ξ[c] k,+2 = b[c] +k ξ[c] k,+1 + a[c], = 2, 3,..., (23 +k ξ[c] k,., = 0, 1,..., (24 k,0 = 1, ξ[c] k,1 = 0, ξ[c] k,+2 = b[c] +k k,+1 + a[c] [c] +k ξ k,, = 0, 1,..., respectively. Remark 3.2. Proposition C.2 shows that If ξ[c] k, /ξ[c] k, exists, See Definition C.2. k, k, = a [c] i+k b [c] i+k 2 a [c] i+k b [c] i+k. = k, k,.

436 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI Lemma 3.3. k, = for all k = 0, 0 + 1,.... Proof. From (24, we have k,+2 = b [c] Thus we obtain k,+2 +k ξ[c] k,+1 + a[c] +k ξ[c] k, k,+1 k,+1 b [c] +k a [c] +k k,, = 0, 1,.... ( b [c] +k 1 k,+1 a [c] +k k, a [c] +k k,+1 a [c] +k k, ( = a [c], = 0, 1,..., (25 +k k,+1 where the second inequality is true because b [c] +k a[c] +k +1 (see (22. It follows from (25 that for i = 1, 2,..., k,i+1 k,i where we use k,0 we have a [c] i 1+k i 1 ( k,i a [c] l+k l=0 k,i 1 ( k,1 k,0 k, i 1 l=i 2 a [c] l+k i 1 = ( a [c] l+k l=0 k,i 1 k,i 2, (26 = 0 and ξ[c] k,1 = 1. Summing both sides of (26 for i = 1, 2,..., 1, k, 1 i 1 1 From (27 and Lemma 3.2, we obtain k, a [c] l+k i=1 l=0. = 2, 3,.... (27 1 i 1 a [c] l+k i=1 l=0 =. Remark 3.3. Gomez-Corral and Ramalhoto [5] derive an analytical solution for { i, }, assuming ξ [3] =. They also say that it is difficult to prove ξ [3] =. Kim [9] asserts that this it always holds for the ergodic M/M/3/3 retrial queue. However his proof contradicts q [3] = 0, which must hold for the ergodic M/M/3/3 retrial queue. Instead of ξ [3] =, we have proved Lemma 3.3. Although Lemma 3.3 does not necessarily show that ξ [3] =, the lemma is sufficient for the derivation of an analytical solution for { i, }, which is shown in Theorem 3.4 below and the subsequent discussion. Theorem 3.4. Let r [c] k r [c] k = q[c] k+1 /q[c] k for any k = 0, 1,.... We then have = a [c] i+k = b [c] i+k ξ[c] k, k,, k = 0, 1,.... (28

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 437 Proof. We first show (28 for k = 0, 0 + 1,.... From Lemma 3.3 and 0 < i < 1 ( i = 0, 1,..., we have +k /ξ[c] k, = 0. It thus follows from (23 that 0 = = ξ[c] q[c] k+1 + k, k = q[c] k+1 + k. which leads to +k k, k, k, = q[c] k+1 k k, Therefore from (29 and Remark 3.2, we obtain a [c] i+k b [c] i+k k, k, = r [c] k, k = 0, 0 + 1,.... (29 = r [c] k, k = 0, 0 + 1,.... Next we show (28 for k = 0, 1,..., 0 1 by mathematical induction method. Since we have proved (28 for k = 0, we suppose that (28 is true for some k = n {1, 2,..., 0 }. From (20, we have n+1 = b[c] n 1 q[c] n + a[c] n 1 q[c] n 1. (30 Because a [c] n, b [c] n 1 0 and q[c] n > 0 for all n = 0, 1,..., dividing both sides of (30 by q n [c] and rearranging the result yields r [c] n 1 = q[c] n n 1 = a[c] n 1 b [c] n 1 q[c] n+1 q n [c] = a[c] n 1 b [c] n 1 r[c] n which shows that (28 is also true for k = n 1. = a [c] n 1 b [c] n 1 + a [c] i+n b [c] i+n = a [c] i+n 1 b [c] i+n 1 From Theorem 3.4, we can obtain the analytical solution for { i, } as follows. We have 1 = 0 k=0 r [c] k, = 1, 2,.... (31 From (31 and (9 (12, { i, } is expressed in terms of only one unknown probability, which is uniquely determined by the normalizing condition. 0 3.2. Asymptotic analysis. This subsection discusses the asymptotic behaviors of the number of customers in the orbit. The following lemma says that for any i = 0, 1,..., c, the probability i, geometrically decreases when is large. Furthermore, this lemma is also useful in the proof of the convergence of the backward algorithm presented in Section 4. Lemma 3.5. We have n,+1 n, where ρ [c] = λ/c denotes the traffic intensity. = r[c] = ρ [c], n = 0, 1...,c, c = 3, 4,,

438 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI Proof. Theorem 3.4 yields r [c] = b [c] a [c] r [c] +1, = 0, 0 + 1,.... (32 It follows from (22, Theorem 3.4 and Proposition C.3 that r [c] 1 for all = 0, 0 + 1,.... Thus since b [c] = (see Lemma 3.2, we have From (32 and (33, we obtain r[c] = r [c] +1 b [c] a [c] / b[c] 1 r [c] +1 /b[c] = = 0. (33 a [c] while the second equality follows from Lemma 3.2. Thus since r [c] for all = 0, 1,..., we have r [c] = ρ [c] and therefore From (9 (12 and (35, we have n, n As a result, (35 and (36 lead to n,+1 n, for all n = 1, 2,..., c. = +1 b [c] = ρ[c], (34 = +1 /π[c] > 0 = ρ [c]. (35 = e n,n,n, n = 1, 2,...,c, (36 n! n,+1 +1 ( + 1n +1 ( + 1n n n n, = ρ [c], 4. Numerical algorithm. Based on Theorem 3.4, { } can be computed using the sequence of continued fractions {r [c] k }. However, in the numerical computation (k = 0, 1,... in order to obtain {q[c] }. The following lemma says that { } is a minimal solution of (20 which can be computed by several recursive formulae. point of view, we have to compute each r [c] k Lemma 4.1. The sequence {, = 0, 1,... } is a minimal solution of the threeterm recurrence relations (20. Proof. From (23, we have Theorem 3.4 yields = ξ[c] 1 + 0, = 0, 1,.... (37 = r [c] q[c] 0 = 1 0,

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 439 from which and (37 it follows that Note here that { ; = 0, 1,... } is a solution of (20 due to (24. Thus {q[c] } is a minimal solution of (20 (see Definition C.1. = 0. The normalization condition for { i, } is given by c i, = 1. From (9 (12 in Proposition 2.2, this normalizing condition is rewritten as β [c] q[c] = 1, (38 where β [c] is some c-th polynomial of which is determined from (9 (12. Remark 4.1. We can compute the minimal solution { } by starting with the two initial terms 0 and 1 = q[c] 0 r[c] 0, and then computing the terms q[c] 2, q[c] 3,..., with the recursive use of (20. However, since rounding error is inevitable in numerical calculation, we need to take this error in consideration, in particular for the calculation of the infinite sequence of { }. Suppose we start the recursive calculation with ˆ 0 and ˆ 1 instead of 0 and 1 due to the rounding error. We then obtain a solution {ˆ } of (20 which is not a minimal solution. Since {q[c] } is a minimal solution, thus /ˆq[c] = 0, according to Proposition C.1. Hence the it of the relative error is given by ˆ = ˆ 1 =. This shows that the forward algorithm using the recurrence relation (20 is numerically unstable. Gautschi [8] reports this phenomenon and suggests that the backward algorithm is numerically stable. Next, we present in detail, a backward algorithm and discuss its convergence. 4.1. Backward algorithm for minimal solution. The problem of computing a minimal solution of (20 for which (38 holds is extensively studied by Gautschi [8], who proposes a backward algorithm to compute the minimal solution satisfying (38 as follows.

440 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI For some nonnegative integer number ν, the sequences r [c] ( = ν, ν 1,...,1, 0 are computed by the following formulae. r ν [c] (ν = 0, r[c] η [c] ν (ν = 1, (ν = Note here that r [c] The sequence {η [c] 1 (ν = η [c] 1 η [c] (ν ν m=0 β[c] m η m [c] (ν b [c] a [c] (ν = η[c] (ν (ν, η[c] (ν and q[c] (ν = ν, ν 1,...,1, (39 r [c] (ν, = ν, ν 1,..., 1, (40 r [c] 1 (ν,, = 0, 1, 2,..., ν. (41 (ν and q[c] (ν are approximations to r[c] and, respectively. (ν} is a solution of (20 which is determined by the two consecutive terms, η ν [c] (ν = 1 and η [c] (ν = 0 (see pp. 36 39 in Gautschi [8]. Theorem 4.2. The backward algorithm (39 (41 converges in the sense ν q[c] (ν = q[c], = 0, 1,.... Proof. Proof of Theorem 4.2 is presented in Appendix B. We now provide a detailed algorithm to compute a finite sequence { i, }, which is an approximation to the stationary distribution { i, }. This algorithm is based on the backward recursive formulae, (39 (41 and Theorem 4.2. Input: c, λ, µ, ε. Output: ν, { i, [Begin Algorithm] ; i = 0, 1,...,c, = 0, 1,...,ν} and { p[c] Step 1. Set the initial value of ν. Step 2. Compute (ν and p[c] ; = 0, 1,..., ν}. (ν ( = 0, 1..., ν by the backward recursive formulae, (39 (41 and (9 (12, respectively, where p [c] (ν is an approximation to which denotes the probability that there are customers in the orbit in steady p [c] state. If ν ( p [c] 2 (ν + 1 p[c] (ν < ε, then go to Step 3. Otherwise, increase ν by ν := ν + 1 and return to Step 2. Step 3. Set = q[c] (ν ( = 0, 1..., ν and compute π[c] i, (i = 1, 2,..., c by formulae (9 (12 where 0, = 0 and π[c] 0,ν+2 = 0 and set [End Algorithm] p [c] = p [c] (ν, = 0, 1,..., ν.

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 441 5. Performance measures and numerical results. 5.1. Performance measures. The probability that there are customers in the orbit at steady state p [c] is given by c p [c] = n,, = 0, 1, 2,.... (42 n=0 From Lemma 3.5 and (9 (12, we also have p [c] +1 = ρ [c]. (43 p [c] Let P B denote the stationary blocking probability that an arriving customer finds all the servers are occupied, which is given by P B = c,. Let E[C] and E[N] denote the average number of busy servers and the average number of retrial customers in the orbit, respectively, in steady state. We then have E[C] = c i=1 i i,, E[N] = According to Falin and Templeton [3], E[C] is explicitly given by p [c]. (44 E[C] = cρ [c] = λ. (45 5.2. Numerical examples. In this section, we show some numerical examples. In all the numerical examples, we set ε = 10 10. This is equivalent to increasing ν by one until at least the first ten digits of p [c] (ν ( = 0, 1...ν are the same as those of p [c] (ν + 1 ( = 0, 1...ν. Firstly, we investigate the accuracy of our algorithm. According to Falin [2, 3], the average number of busy servers is given by (45. On the other hand, the average number of busy servers can be approximately computed using (44 truncated at ν. Figure 1 presents the absolute error between the approximation of the average number of busy servers and its exact value, against the traffic intensity, with various retrial rates. It is observed that the absolute error is bounded by 10 8. This shows that the algorithm can compute the average number of busy servers with high accuracy. Figure 2 describes the truncation point ν at which the algorithm stops, against the traffic intensity. It is observed that, at low traffic intensity, our algorithm stops in a few steps. However, ν is likely to increase exponentially as traffic intensity increases. In particular, when the traffic intensity approaches to 1, ν increases rapidly. Figure 3 presents the distribution of the number of customers in the orbit for the case of three servers with traffic intensities ρ [3] = 0.7, 0.9, 0.95, 0.97 and 0.99, respectively, while the retrial rate is kept constant µ = 1. It is observed that p [c] geometrically decreases when is large, according to (43. Figure 4 shows the relation between the blocking probability P B against the traffic intensity for the cases of c = 3 and 4 while keeping µ = 1. It is observed that the blocking probability increases with the traffic intensity, as expected. It is also observed that P B of the case c = 3 is larger than that of the case c = 4.

442 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI Absolute error 1 0.01 0.0001 1e-006 1e-008 Three servers, Retrial rate = 0.01 Four servers, Retrial rate = 0.01 Three Servers, Retrial rate = 0.1 Four Servers, Retrial rate = 0.1 Three Servers, Retrial rate = 1 Four Servers, Retrial rate = 1 Three Servers, Retrial rate = 10 Four Servers, Retrial rate = 10 Three Servers, Retrial rate = 100 Four Servers, Retrial rate = 100 10-8 1e-010 1e-012 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Traffic intensity Figure 1. Absolute error E[C] λ. 10000 1000 Three servers, Retrial rate = 0.01 Four servers, Retrial rate = 0.01 Three servers, Retrial rate = 0.1 Four servers, Retrial rate = 0.1 Three servers, Retrial rate = 1 Four servers, Retrial rate = 1 Three servers, Retrial rate = 100 Four servers, Retrial rate = 100 Truncation point 100 10 1 0 0.2 0.4 0.6 0.8 1 Traffic intensity Figure 2. Truncation point ν. Figure 5 shows the average number of customers in the orbit E[N], against traffic intensity for four cases: µ = 0.01, 0.1, 1 and 100. The average queue length of the conventional M/M/3/ and M/M/4/ queues are also plotted. It is observed that E[N] s for the four cases increase as traffic intensity increases and E[N] for a larger µ is smaller, as expected. We observe that when µ = 100, E[N] s of M/M/3/3 and M/M/4/4 retrial queues are very close to the average numbers of customers in the waiting lines of the conventional M/M/3/ and M/M/4/ queues, respectively. This implies that the average number of customers in the orbit of a retrial queue is

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 443 1 0.01 Traffic Intensity: 0.7 Traffic Intensity: 0.9 Traffic Intensity: 0.95 Traffic Intensity: 0.98 Traffic Intensity: 0.99 Probability 0.0001 1e-006 1e-008 1e-010 0 500 1000 1500 2000 Number of customers in the orbit Figure 3. Distribution of the number of customers in the orbit. 1 0.1 0.01 0.001 Probability 0.0001 1e-005 1e-006 1e-007 1e-008 Three Servers, Retrial rate = 1 Four Servers, Retrial rate = 1 1e-009 0 0.2 0.4 0.6 0.8 1 Traffic intensity Figure 4. Blocking probability. asymptotic to that of customers in the waiting line of the corresponding conventional queue, when µ tends to infinity. This agrees with the intuition and the theoretical result (see Theorem 2.5 in Falin and Templeton [3]. Figure 6 presents the standard deviation of the number of customers in the orbit against the traffic intensity keeping µ = 0.01, 0.1, 1 and 100. We observe that the standard deviation monotonically increases as traffic intensity increases. Figure 7 illustrates the impact of the retrial rate µ on the blocking probability P B, keeping ρ [c] = 0.2, 0.4 and 0.6. Note that, the horizontal lines are the probabilities

444 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI Average number of customers in the orbit 10000 1000 100 10 1 Three servers, Retrial rate = 0.01 Four servers, Retrial rate = 0.01 Three servers, Retrial rate = 0.1 Four servers, Retrial rate = 0.1 Four servers, Retrial rate = 1 Four servers, Retrial rate = 1 Three servers, Retrial rate = 100 Four servers, Retrial rate = 100 Conventional M/M/3/Infinity Conventional M/M/4/Infinity 0.1 0 0.2 0.4 0.6 0.8 1 Traffic intensity Figure 5. Average number of customers in the orbit. 10000 1000 Three servers, Retrial rate = 0.01 Four servers, Retrial rate = 0.01 Three servers, Retrial rate = 0.1 Four servers, Retrial rate = 0.1 Three servers, Retrial rate = 1 Four servers, Retrial rate = 1 Three servers, Retrial rate = 100 Four servers, Retrial rate = 100 Standard deviation 100 10 1 0.1 0 0.2 0.4 0.6 0.8 1 Traffic intensity Figure 6. Standard deviation of the number of customers in the orbit. that arriving customers have to wait for service in the conventional M/M/3/ queueing system with the same traffic intensities 0.4 and 0.6. These probabilities are computed using Erlang C formula. It is observed that when 0.01 µ 0.1, the blocking probability is almost insensitive to the retrial rate. In the interval, 0.1 µ 100, the blocking probability slightly increases as µ increases. When the retrial rate is extremely high, 100 µ 10000, the blocking probability is insensitive to the retrial rate again. This implies that when the retrial rate µ tends to infinity, the blocking probability in the M/M/c/c retrial queue tends to the

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 445 1 0.1 Blocking probability 0.01 Three Servers, Traffic intensity = 0.2 Four Servers, Traffic intensity = 0.2 Three Servers, Traffic intensity = 0.4 Four Servers, Traffic intensity = 0.4 Three Servers, Traffic intensity = 0.6 Four Servers, Traffic intensity = 0.6 M/M/3/Infinity, Traffic intensity = 0.4 M/M/3/Infinity, Traffic intensity = 0.6 0.001 0.001 0.01 0.1 1 10 100 1000 10000 Retrial rate Figure 7. Blocking probability against retrial rate. 10000 1000 Three Servers, Traffic intensity = 0.2 Four Servers, Traffic intensity = 0.2 Three Servers, Traffic intensity = 0.4 Four Servers, Traffic intensity = 0.4 Three Servers, Traffic intensity = 0.6 Four Servers, Traffic intensity = 0.6 Average number of customers in the orbit 100 10 1 0.1 0.01 0.001 0.001 0.01 0.1 1 10 100 1000 10000 Retrial rate Figure 8. Average number of retrial customers against retrial rate. waiting probability in conventional M/M/c/ queue (see Theorem 2.5 in Falin and Templeton [3]. In Figure 8, we investigate the impact of retrial rate µ on the average number of customers in the orbit E[N], while keeping the traffic intensity constant, ρ [c] = 0.2, 0.4, 0.6, respectively. It is observed that E[N] monotonically decreases in the interval 0.001 µ 1. This is explained as follows. When the retrial rate µ is large, the time between two consecutive retrials of a customer is short. Therefore, each retrial customer stays in the orbit for a short time, resulting in a small number

446 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI 1 Three servers, Retrial rate = 0.1 Three servers, Retrial rate = 1 Three servers, Retrial rate = 10 Three servers, Retrial rate = 100 0.01 0.0001 Probability 1e-006 1e-008 1e-010 1 10 100 1000 Number of customers in the orbit Figure 9. Probability distribution of the number of customers in the orbit. of customers in the orbit. It is also observed that when µ is large, E[N] is also insensitive to µ. Figure 9 illustrates the asymptotic behavior of the distribution of the number of customers in the orbit when the retrial rate tends to infinity, keeping the traffic intensity constant, ρ [3] = 0.8. We observe that not only P B and E[N], but also the probability distribution of the number of customers in the orbit is likely to have a it when the retrial rate µ tends to infinity. This agrees with the theoretical results (see Lemma 2.1 in Falin and Templeton [3]. 6. Conclusions and future works. In this paper, we have derived analytical solutions for M/M/3/3 and M/M/4/4 retrial queues without any technical assumptions which were made in previous literature. First, we have reduced the problem of finding the oint distribution of the numbers of customers in the orbit and those in service { i, } to find {π[c] } which is probability that all the servers are idle. We have derived a system of three-term recurrence relations for { } and have proved that } is a minimal solution of the three-term recurrence relations which is expressed { in terms of continued fractions. We also have presented a backward algorithm to compute the stationary distribution { i, } and have proved its convergence. Through many numerical examples, we have confirmed that the algorithm performs well even with a variety of critical situations such as, very high traffic intensities and very low retrial rates. The numerical examples have also shown that the numerical results computed by the algorithm agreed well with theoretical results. For the future works, we will find analytical solutions for retrial queueing systems with more than four servers. When the retrial queueing system has more than four servers, unfortunately, a system of three-term recurrence relations no longer exists but some higher order system of recurrence relations does. Further studies and efforts are needed to resolve this problem.

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 447 Appendix A. Proof of Lemma 3.1. Proof. By summing up (4 (i = 0, 1,...,c 1 and (5 and rearranging it, we obtain ( c 1 λp [c] c (z(1 z = µ n=0 d dz P [c] n (z (1 z. (46 For z < 1, dividing both sides of (46 by (1 z/(c! yields ( c 1 λc!p c [c] c! d [c] (z = µ n!p n (z. (47 n! dz n=0 n=0 From (6 and (47, we have ( c 1 λc!p c [c] (z = µ n c! d e n,i (z di [c] n! dz dzip 0 (z c 1 = µ = µ n=0 c i=1 n+1 c! n! i=1 ( c 1 n=i 1 ( d dz e n,i(z + e n,i 1 (z d i ( c! d n! dz e n,i(z + e n,i 1 (z Using (6, the left side hand of (48 is rewritten as c λc!p c [c] (z = λ e c,i (z di dz From (48 and (49, we obtain (13 and (14. Appendix B. Proof of Theorem 4.2. [c] ip 0 [c] dzip 0 (z d i dz i P [c] 0 (z. (48 (z. (49 Proposition B.1 (Theorem 3.1 in [8]. Suppose the recurrence relation (20 has a nonvanishing minimal solution,, for which (38 holds. Let g[c] be any other solution of (38. Then the backward algorithm (39 and (41 converges in the sense if and only if ν q[c] ν g [c] (ν = q[c] (50 ν β [c] g[c] = 0. (51 Proof of Theorem 4.2. We consider a solution of (20 such that g [c] 0 = 0 and g [c] = 0+1 1. It is clear that g [c] = 0 for all = 0, 0 + 1,.... Therefore, it follows from (26 that g [c] +1 g[c] for all = 0, 0 + 1,.... We prove that the sequence g [c] satisfies (51. We have g [c] ν β [c] g[c] = q[c] g [c] 0 β [c] g[c] + q[c] g [c] Because { } is a minimal solution of (20, we have ν g [c] 0 β [c] g[c] = 0. ν = 0+1 β [c] g[c].

448 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI On the other hand, ν g [c] β [c] g[c] = 0+1 q[c] ν β [c] = 0+1 g [c] g [c] According to the definition of r [c] in Theorem 3.4, = q[c] 0 ν r [c]. q[c] ν = 0+1 β [c]. (52 It follows from Lemma 3.5 that r [c] = ρ [c] < 1. Therefore, there exists some positive number ρ [c] < 1 and some nonzero natural number J( ρ [c] such that r [c] < ρ [c], = J( ρ [c], J( ρ [c] + 1,.... (53 From (52 and (53, we have J( ρ [c] 1 ( ρ [c] ν J( ρ[c] +1, ν = J( ρ [c], J( ρ [c] + 1,.... (54 q[c] 0 Because ν (54 that = 0+1 β[c] r [c] is some polynomial of ν, therefore, it follows from (52 and ν g [c] ν β [c] g[c] = 0+1 = 0. Appendix C. Recurrence relations and continued fractions. C.1. Recurrence relations and minimal solutions. Definition C.1. We consider three-term recurrence relations y n+2 = b n y n+1 + a n y n, a n 0, n = 1, 2,..., (55 where the {a n }, {b n } and {y n } are sequences on R. The set of all solutions {y n } of (55 forms a linear vector space V of dimension two over R. If there exists a non-trivial solution {h n } (i.e. h n 0 for some n and another solution {g n } of (55 such that h n = 0, n g n then {h n } is called a minimal solution of (55 [7]. Proposition C.1. For any solution {y n } of (55 not proportional to {h n }, Proof. See pp. 163 166 in [7]. h n = 0. n y n

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 449 C.2. Basic results on continued fractions. In this section, we use the following notation to describe continued fractions. a 0 = a 0, b 0 b 0 b 0 + a 0 a 1 a 0 b 0 + a 1 b 1 = b 1 + a 2 b 2 +... = a 0 b 0 + a 1 b 1, a b. a 0 b 0 + a 1 b 1 + a 2 b 2 = a 0 b 0 + a 1 b 1 + a 2 b 2, We now consider the infinite continued fraction n=0 a n b n, where the {a n ; n = 0, 1,... } is nonzero and {b n ; n = 0, 1,... } is real. Let A n and B n (n = 0, 1,... denote real numbers determined by A 0 = 1, A 1 = 0, A n+2 = b n A n+1 + a n A n, n = 0, 1,..., B 0 = 0, B 1 = 1, B n+2 = b n B n+1 + a n B n, n = 0, 1,.... Proposition C.2 (Theorem 2.1 in [7]. For any n = 2, 3,..., n 2 A n a =. B n b Definition C.2 (Definition 1.1 in [12]. If A n = γ, n B n for some finite γ, then the continued fraction n=0 a n b n is said to be convergent and its convergent value is equal to γ. Proposition C.3 (Theorem 4.35 in [7]. If b n a n +1 for all n = 0, 1,..., then n=0 where γ is some number such that γ 1. a n b n = γ, C.3. Explicit expressions of e n,i, s and s [c] i, s. This appendix provides explicit expressions of e n,i, s (n = 0, 1, 2, 3, 4 and s [c] i, s (c = 3, 4. Using (7 and (8, we

450 T. PHUNG-DUC, H. MASUYAMA, S. KASAHARA AND Y. TAKAHASHI can calculate e n,i, s (n = 0, 1, 2, 3, 4 as follows. e 0,0,0 = 1, e 1,0,0 = λ, e 1,1,1 = µ, e 2,0,0 = λ 2, e 2,1,0 = µ, e 2,1,1 = µ (2 λ + µ + 1, e 2,2,2 = µ 2, e 3,0,0 = λ 3, e 3,1,0 = µ (3 λ + 2 µ + 2, e 3,1,1 = µ ( 3 λ 2 + µ 2 + 3 λµ + 3 λ + 3 µ + 2, e 3,2,1 = 3µ 2, e 3,2,2 = 3 µ 2 (λ + µ + 1, e 3,3,3 = µ 3, e 4,0,0 = λ 4, e 4,1,0 = µ (6 λ 2 + 3 µ 2 + 8 λµ + 8 λ + 9 µ + 6, e 4,1,1 = µ (4 λ 3 + µ 3 + 6 λ 2 µ + 4 λµ 2 + 6 λ 2 + 6 µ 2 + 12 λµ + 8 λ + 11 µ + 6, e 4,2,0 = 3 µ 2, e 4,2,1 = 2 µ 2 (6 λ + 7 µ + 7, e 4,2,2 = µ 2 (6 λ 2 + 7 µ 2 + 12λµ + 12 λ + 18 µ + 11, e 4,3,2 = 6µ 3, e 4,3,3 = 2 µ 3 (2 λ + 3 µ + 3, e 4,4,4 = µ 4. Further in terms of (8, (14 and (15, s [c] i, s (c = 3, 4 are calculated as follows. s [3] 0,0 = λ4, s [3] 1,0 = µ (6 λ2 + 3 µ 2 + 8 λµ + 8 λ + 9 µ + 6, s [3] 1,1 = λµ(3 λ2 + µ 2 + 3 λµ + 3 λ + 3 µ + 2, s [3] 2,0 = 3µ2, s [3] 2,1 = 9 µ2 (λ + µ + 1, s [3] 2,2 = 3 λµ2 (λ + µ + 1, s [3] 3,2 = 3µ3, s [3] 3,3 = λµ3, s [4] 0,0 = λ5, s [4] 1,0 = µ (10 λ3 + 4 µ 3 + 20 λ 2 µ + 15 λµ 2 + 20 λ 2 + 24 µ 2 + 45 λµ + 30 λ + 44 µ + 24, s [4] 1,1 = λµ(4 λ3 + µ 3 + 6 λ 2 µ + 4 λµ 2 + 6 λ 2 + 6 µ 2 + 12 λµ + 8 λ + 11 µ + 6, s [4] 2,0 = 5µ2 (3 λ + 4 µ + 4, s [4] 2,1 = 2 µ2 (12 λ 2 + 14 µ 2 + 25 λµ + 25 λ + 36 µ + 22, s [4] 2,2 = λµ2 ( 6 λ 2 + 7 µ 2 + 12 λµ + 12 λ + 18 µ + 11, s [4] 3,1 = 12µ3, s [4] 3,2 = 6 µ3 (3 λ + 4 µ + 4, s [4] 3,3 = 2 λµ3 (2 λ + 3 µ + 3, s [4] 4,3 = 4µ4, s [4] 4,4 = λµ4. Acknowledgments. We would like to thank the referee very much for valuable comments and suggestions. REFERENCES [1] B. D. Choi and B. Kim, Non-ergodicity criteria for denumerable continuous time Markov processes, Operations Research Letters, 32 (2004, 574 580. [2] G. I. Falin, A survey of retrial queues, Queueing Systems, 7 (1990, 127 168. [3] G. I. Falin and J. G. C. Templeton, Retrial Queues, Chapman & Hall, London, 1997. [4] A. Gomez-Corral, A bibliographical guide to the analysis of retrial queues through matrix analytic techniques, Annals of Operations Research, 141 (2006, 163 191.

M/M/3/3 AND M/M/4/4 RETRIAL QUEUES 451 [5] A. Gomez-Corral and M. F. Ramalhoto, The stationary distribution of a Markovian process arising in the theory of multiserver retrial queueing systems, Mathematical and Computer Modelling, 30 (1999, 141 158. [6] T. Hanschke, Explicit formulas for the characteristics of the M/M/2/2 queue with repeated attempts, Journal of Applied Probability, 24 (1987, 486 494. [7] W. B. Jones and W. J. Thron, Continued Fractions: Analytic Theory and Applications, Addison-Wesley, Massachusetts, 1980. [8] W. Gautschi, Computational aspects of three-term recurrence relation, SIAM Review, 9 (1967, 24 82. [9] Y. C. Kim, On M/M/3/3 retrial queueing system, Honam Mathematical Journal, 17 (1995, 141 147. [10] M. F. Neuts, Numerical investigation of a multiserver retrial model, Queueing Systems, 7 (1990, 169 190. [11] C. E. M. Pearce, Extended continued fractions, recurrence relations and two-dimentional Markov processes, Advances in Applied Probability, 21 (1989, 357 375. [12] H. S. Wall, Analytic Theory of Continued Fractions, AMS Chelsea Publishing, Providence, Rhode Island, 2000. [13] T. Yang and J. G. C. Templeton, A survey on retrial queues, Queueing Systems, 2 (1987, 201 233. Received September 2008; revised December 2008. E-mail address: tuan@sys.i.kyoto-u.ac.p E-mail address: masuyama@sys.i.kyoto-u.ac.p E-mail address: shoi@i.kyoto-u.ac.p E-mail address: takahashi@i.kyoto-u.ac.p