K retrial queue as K tends to infinity

K retrial queue as K tends to infinity

Available online at www.sciencedirect.com European Journal of Operational Research 189 (2008) 1104–1117 www.elsevier.com/locate/ejor Convergence of ...

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European Journal of Operational Research 189 (2008) 1104–1117 www.elsevier.com/locate/ejor

Convergence of the stationary distributions of M/M/s/K retrial queue as K tends to infinity Yang Woo Shin Department of Statistics, Changwon National University, Changwon, Gyeongnam 641-773, Republic of Korea Received 28 March 2006; accepted 8 March 2007 Available online 26 May 2007

Abstract We investigate the convergence of the stationary distributions xðKÞ of M/M/s/K retrial queue to the stationary distribution p of M/M/s queue as K tends to infinity. It is showed that xðKÞ converges geometrically to p in l1-sense and the convergence rate is characterized by the traffic intensity q ¼ slk , where k and l are the arrival rate and service rate, respectively. Ó 2007 Elsevier B.V. All rights reserved. Keywords: M/M/s/K retrial queue; M/M/s queue; Convergence rate; Censored Markov chain

1. Introduction We consider an M/M/s/K retrial queue with s identical servers and K–s waiting positions. Service times of customers are independent of each other and have a common exponential distribution with parameter l. Customers arrive according to a Poisson process with rate k. When an arriving customer finds that all the servers are busy and no waiting position is available, the customer joins a virtual pool of blocked customers called orbit and repeats its request after a random amount of time until the customer gets into the service facility. The capacity of orbit is infinite. The access from orbit to the service facility is governed by the exponential distribution with rate cn which may depend on the current number n, n P 0 of customers in orbit. That is, the probability of repeated attempt during the interval ðt; t þ DtÞ, given that n customers are in orbit at time t, is cn Dt þ oðDtÞ. We assume that cn P c1 ¼ c > 0, n P 1 and c0 ¼ 0. From the retrial phenomena, one can intuitively expect that the behavior of the M/M/s/K retrial queue is similar to that of ordinary M/M/s queue while there are no customers in orbit and this situation can occur more frequently as the system is less congested. To reduce the system congestion, at least three ways can be considered. One way is to increase the retrial rate. Some works have been done from this perspective by

E-mail address: [email protected] 0377-2217/$ - see front matter Ó 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2007.03.052

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

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showing that higher retrial rates result in less system congestion (e.g. Liang and Kulkarni, 1993; Shin, 2006) or getting closer to the corresponding ordinary queues (Falin and Templeton, 1997). Falin and Templeton (1997) showed that the stationary distribution of M/M/s/s retrial queue (resp. M/G/1 retrial queue) with retrial rate cn ¼ nc converges to that of the ordinary M/M/s queue (resp. M=G=1=1 queue) as c tends to infinity and the convergence rate is Oð1cÞ. Liang (1999) showed that stochastically longer service times or less servers will result in more customers in the system. From this perspective, one can consider the second way to reduce the system congestion in retrial queue that is to increase the service rates so that the traffic intensity becomes small and hence arriving customers get less chance to be rejected and join orbit. The third way is to increase the capacity K in the service facility. By doing so, arriving customers get more chance to get into the service facility and the system performance gets closer to the ordinary queue. In this paper, we investigate the third case, that is, we compare the M/M/s/K retrial queue with the ordinary M/M/s queue as K tends to infinity. The convergence rate of the stationary distributions of the number of customers in M/M/s/K retrial queues is the main concern of this paper. Convergence of the stationary distribution of finite-capacity systems to that of infinite-capacity system in the class of ordinary queueing systems without retrial phenomena has been investigated by several authors (Choi and Kim, 2000; Choi et al., 2003a; Heyman and Whitt, 1989; Simonot, 1997, 1998). The preceding results in ordinary queues are concerned with the convergence of the systems with finite states, while the convergence of the systems with infinite states are considered in this paper. This paper is organized as follows. In Section 2, the main result is presented. The main result is proved by comparing with the stationary distributions of some censored Markov chains. The queue length process in M/ M/s/K retrial queue and its censored Markov chain are investigated in Sections 3 and 4, respectively. In Section 5, M/M/s/K retrial queue with constant retrial rate is considered for an upper bound of the probability that there are more than one customers in orbit in the M/M/s/K retrial queue. In Section 6, some numerical results are presented and the effects of retrial rates to the speed of convergence are numerically investigated. Finally, concluding remarks are given in Section 7. 2. Main result Let X K0 ðtÞ and X K1 ðtÞ represent the number of customers in orbit and in the service facility at time t, respectively in M/M/s/K retrial queue and X K ¼ fX K ðtÞ; t P 0g with X K ðtÞ ¼ ðX K0 ðtÞ; X K1 ðtÞÞ. Then the stochastic process X K is a Markov chain on the state space S ¼ fði; jÞ : i ¼ 0; 1; 2; . . . ; j ¼ 0; 1; 2; . . . ; Kg. We assume that q ¼ slk < 1 which guarantees the existence of the stationary distribution of X K (He et al., 2000). Let xij ðKÞ ¼ lim P ðX K0 ðtÞ ¼ i; X K1 ðtÞ ¼ jÞ; t!1

i ¼ 0; 1; 2; . . . ; j ¼ 0; 1; 2; . . . ; K

and xðKÞ ¼ ðx0 ðKÞ; x1 ðKÞ; x2 ðKÞ; . . .Þ with xn ðKÞ ¼ ðxn0 ðKÞ; . . . ; xnK ðKÞÞ, n P 0. In this section, we present the comparison result between xðKÞ and the stationary distribution p ¼ fpj ; j P 0g of M/M/s queue with arrival rate k and service rate l. In order to map from the state space f0; 1; 2; . . .g of M/M/s queue to the state space S of X K , we consider M/M/s queue as the M/M/s/K retrial queue with infinite retrial rate cn ¼ 1, n P 1. Thus if there are n > K customers in M/M/s queue, then we consider there are n  K customers in orbit and K customers in service facility. From this perspective, we ~ ¼ f~ define p pij ; ði; jÞ 2 Sg as 8 i ¼ 0; 0 6 j 6 K; > < pj ; ~ij ¼ piþK ; i P 1; j ¼ K; p > : 0; i P 1; 0 6 j 6 K  1 ~ by the following formula: and compare xðKÞ with p using p kp  xðKÞk,

1 X K X i¼0

j¼0

j~ pij  xij ðKÞj ¼

K X j¼0

jpj  x0j ðKÞj þ

1 X i¼1

jpiþK  xiK ðKÞj þ

1 X K 1 X i¼1

j¼0

xij ðKÞ:

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The main objective of this paper is to show the following theorem, that is, the difference kp  xðKÞk converges geometrically to zero as K ! 1 and the convergence rate is closely related to the traffic intensity q. Theorem 2.1. There exist positive constants C0 and C1 such that for large K kp  xðKÞk 6 ðC 0 þ C 1 KÞqK :

ð1Þ

For the proof of Theorem 2.1, we need a notion of censored (or restricted) Markov chain. The censored Markov chain Y ðEÞ ¼ fY ðEÞ ðtÞ; t P 0g with censoring set E  E of a Markov chain Y ¼ fY ðtÞ; t P 0g on the state space E is the stochastic process which records transitions of Y during visits E. For more about the censored Markov chains, please see Latouche and Ramaswami (1981) and Freedman (1972). Here, we present well known results about censored Markov chain that will be used later. The stationary probability distribuðEÞ tions yðEÞ ¼ ðy j ; j 2 EÞ and y ¼ ðy j ; j 2 EÞ of Y ðEÞ and Y, respectively, are related as follows: 1 y ; j 2 E; yðEÞ j X ðEÞ kyðEÞ  yk, jy j  y j j ¼ 2ð1  yðEÞÞ; ðEÞ

yj ¼

ð2Þ ð3Þ

j2E

P ðEÞ where yðEÞ ¼ j2E y j and y j ¼ 0, j 2 E  E. ð0Þ Now we sketch the main steps of the proof of Theorem 2.1. Let xð0Þ ðKÞ ¼ ðx0j ðKÞ; j ¼ 0; 1; . . . ; KÞ be the ð0Þ stationary distribution of censored Markov chain X K of X K with censoring set 0 ¼ fð0; 0Þ; . . . ; ð0; KÞg. Idenð0Þ ð0Þ tifying the state ð0; jÞ by j, we write x0j ðKÞ  xj ðKÞ, j ¼ 0; 1; 2; . . . ; K and Theorem 2.1 is proved through the following inequality: kp  xðKÞk 6 kp  pðKÞ k þ kpðKÞ  xð0Þ ðKÞk þ kxð0Þ ðKÞ  xðKÞk;

ð4Þ

where pðKÞ is the stationary distribution of the queue length process in M/M/s/K queue that is a censored Markov chain with censoring set f0; 1; 2; . . . ; Kg of that in M/M/s queue. It is immediate from (3) and the well known results for p (e.g. see Gross and Harris, 1998) pn ¼ p0

sminðn;sÞ n q; minðn; sÞ!

n ¼ 0; 1; 2; . . .

that kp  p

ðKÞ

k¼2

1 X

! pn

n¼Kþ1

¼

 p0

 2ss qKþ1 : s!ð1  qÞ

Thus it is sufficient to show that there exist constants C i , i ¼ 0; 1; 2 such that kpðKÞ  xð0Þ ðKÞk 6 ðC 0 þ C 1 KÞqK ; ! 1 X ð0Þ xi ðKÞ 1 6 C 2 qK ; kx ðKÞ  xðKÞk ¼ 2

ð5Þ ð6Þ

i¼1

where 1 is the column vector of corresponding size whose components are all 1 and the relation (3) is used in (6). The proofs of (5) and (6) are presented in Sections 4 and 5, respectively. 3. The Markov chain X K In this section, we briefly describe the generator and the stationary distribution of X K for later use. Setting n ¼ fðn; 0Þ; ðn; 1Þ; . . . ; ðn; KÞg, n P 0, the generator of the Markov chain X K is of the form

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

ðnÞ

1107

ðnÞ

where A0, A1 and A2 are square matrices of order K + 1 defined by 1 0 k Kn;0 C B Kn;1 k C B l C B C B 2l Kn;2 k C B C B .. .. .. C B C B . . . ðnÞ C; A1 ¼ B C B sl Kn;s k C B C B .. .. .. C B C B . . . C B B sl Kn;s k C A @ sl K0;s with Kn;j ¼ k þ cn þ jl, n P 0, and 0 

O

A0 ¼

k

 ;

0

B B B ¼B B @

ðnÞ A2

cn 0

n P 0;

ð7Þ

1 C C C C; .. C . cn A 0 ..

.

n P 1:

Let fRn ; n P 0g be the sequence of ðK þ 1Þ  ðK þ 1Þ matrices which are the minimal nonnegative solutions to the system of equations ðnþ1Þ

A0 þ R n A1

ðnþ2Þ

þ Rn ðRnþ1 A2

Þ ¼ 0;

n P 0:

Then we can see from the special structure of the matrix A0 that ðnþ1Þ

Rn ¼ A0 ðA1

ðnþ2Þ 1

 Rnþ1 A2

Þ

ð8Þ

has the following formula:   OKðKþ1Þ Rn ¼ ; n P 0; rn

ð9Þ

where OKðKþ1Þ is the zero matrix of size K  ðK þ 1Þ and rn ¼ ðrn0 ; rn1 ; . . . ; rnK Þ is a ðK þ 1Þ-row vector. Thus it follows from (9) and Bright and Taylor (1995) that the stationary distribution xðKÞ ¼ ðxn ðKÞ; n P 0Þ of X K is given by ! ! n1 n2 Y Y xn ðKÞ ¼ x0 ðKÞ Rj ¼ x0K ðKÞ rjK rn1 ; n P 1 j¼0

j¼0

and x0 ðKÞ is the unique solution of the equation ð0Þ

ð1Þ

x0 ðKÞðA1 þ R0 A2 Þ ¼ 0;

ð10Þ

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with the normalizing condition " !# 1 n1 Y X x0 ðKÞ I þ Rj 1 ¼ 1: n¼1

j¼0

ð0Þ

4. The Markov chain X K

ð0Þ

In order to prove (5), we first derive the generator of the censored Markov chain X K with censoring set 0 ¼ fð0; 0Þ; ð0; 1Þ; . . . ; ð0; KÞg of X K and then derive the upper bound for kxð0Þ ðKÞ  pðKÞ k. ð0Þ

Lemma 4.1. The generator of X K is of the form ð0Þ e 0; Qxð0Þ ðKÞ ¼ A1 þ c R

ð11Þ

where ~0 ¼ R



OKðKþ1Þ ~r0



and ~r0 ¼ ð0; r00 ; r01 ; . . . ; r0;K1 Þ is the ðK þ 1Þ-row vector. Proof. Write the matrix Qx ðKÞ in a partitioned form

where E ¼ 0 and Ec ¼ f1; 2; . . .g and let Aði; jÞ be the square matrix of order K + 1 that is the ði; jÞ-block com1 ponent of the fundamental matrix ðQ Þ ¼ ðAði; jÞÞi;jP1 . Then the first row blocks Að1; jÞ, j P 1 satisfy the following equations: ð1Þ

ð2Þ

Að1; 1ÞA1 þ Að1; 2ÞA2 ¼ I; ðjþ1Þ

Að1; jÞA0 þ Að1; j þ 1ÞA1

ð12Þ ðjþ2Þ

þ Að1; j þ 2ÞA2

¼ 0;

j P 1:

It can be seen that Að1; jÞ, j P 1 defined by  1 ð1Þ ð2Þ Að1; 1Þ ¼  A1 þ R1 A2 ; Að1; jÞ ¼ Að1; 1ÞR1 R2 . . . Rj1 ;

ð13Þ

ð14Þ ð15Þ

jP2

satisfy (12) and (13). We can see from (8) and (14) that A0 Að1; 1Þ ¼ R0 . It follows from Latouche and Ramaswð0Þ ami (1981, Theorem 5.5.3) that the generator of X K is given by ð0Þ

ð1Þ

ð0Þ

ð1Þ

Qxð0Þ ðKÞ ¼ A1 þ A0 Að1; 1ÞA2 ¼ A1 þ R0 A2 ð1Þ

~0. and the lemma is proved by setting R0 A2 ¼ cR

h

Proposition 4.2. There exist positive constants C 0 and C 1 such that for large K kxð0Þ ðKÞ  pðKÞ k 6 ðC 0 þ C 1 KÞqK : Proof. Expanding xð0Þ ðKÞQð0Þ x ðKÞ ¼ 0, we have that ð0Þ

ð0Þ

ð0Þ

k þ ln xð0Þ n ðKÞ ¼ kxn1 ðKÞ þ ln xnþ1 ðKÞ þ cr 0;n1 xK ðKÞ;

0 6 n 6 K  1;

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

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ð0Þ

where ln ¼ minðs; nÞl, 0 6 n 6 K  1 and xj ðKÞ ¼ r0j ¼ 0 for j < 0. Solving these equations yields that ! n2 X p c n ð0Þ ð0Þ r0j xK ðKÞ; 0 6 n 6 K; ð16Þ xnð0Þ ðKÞ ¼ x0 ðKÞ  ln j¼0 p0 P1 where the convention j¼0 aj ¼ 0 is used for n = 1. Summing over n ¼ 0; 1; 2; . . . ; K in (16) and using xð0Þ ðKÞ1 ¼ 1, we have that ð0Þ

ðKÞ

ðKÞ

ð0Þ

x0 ðKÞ ¼ p0 þ p0 w0 ðKÞxK ðKÞ;

ð17Þ

where w0 ðKÞ ¼ c

K X r00 þ    þ r0;n2 ln n¼2

ð18Þ

and hence xnð0Þ ðKÞ, 1 6 n 6 K are given by xnð0Þ ðKÞ ¼ pðKÞ n þ

! n2 X c ð0Þ pðKÞ r0n xK ðKÞ; n w0 ðKÞ  ln j¼0

1 6 n 6 K:

ð19Þ

It follows from (19) for n ¼ K that: ðKÞ

pK  : ðKÞ c 1  pK w0 ðKÞ  sl ðr00 þ r01 þ    þ r0;K2 Þ

ð0Þ

xK ðKÞ ¼

ð20Þ

Thus we have from (17), (19) and (20) that ðKÞ

ð0Þ

kxð0Þ ðKÞ  pðKÞ k 6 2w0 ðKÞxK ðKÞ 6

2w0 ðKÞpK

ðKÞ

1  w0 ðKÞpK

:

ð21Þ

It suffices to show that for some constants c0 and c1, ðKÞ

w0 ðKÞpK 6 ðc0 þ c1 KÞqK : Noting from k¼c

Qxð0Þ ðKÞ1

K 1 X

ð22Þ

¼ 0 that

r0j ;

j¼0

we have

  k 1 1 K sþ1 þ  þ þ w0 ðKÞ 6 6 ðs lnðs  1Þ  s þ 1 þ KÞq: l 2 s1 s

Thus the inequality (22) is immediate from  s  s s ðKÞ ðKÞ s K pK ¼ p0 q 6 qK s! s! and the proposition is proved.

h

5. M/M/s/K retrial queue with constant retrial rate We consider the M/M/s/K retrial queue with constant retrial rate ~cn ¼ c, n P 1 and let Y K ¼ fY K ðtÞ; t P 0g with Y K ðtÞ ¼ ðY K0 ðtÞ; Y K1 ðtÞÞ, where Y K0 ðtÞ and Y K1 ðtÞ are the number of customers in orbit and in the service facility, respectively, at time t. Then the generator Qy ðKÞ of the Markov chain Y K is obtained from Qx ðKÞ in ðnÞ ð1Þ ðnÞ ð1Þ ð1Þ ð1Þ (7) by letting A1 ¼ A1 and A2 ¼ A2 for n P 1. We denote A1 ¼ A1 and A2 ¼ A2 . Let h be the stationary distribution of A0 þ A1 þ A2 . It can be easily seen that if q < 1, then

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Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

lim

K!1

hA0 1 <1 hA2 1

and q < 1 is a sufficient condition for Qy ðKÞ to be positive recurrent for sufficiently large K (e.g. see Neuts, 1981, Theorem 3.1.1). Let R be the minimal nonnegative solution of the matrix equation A0 þ RA1 þ R2 A2 ¼ 0:

ð23Þ

It follows from the special structure of the matrix A0 that R has the following formula:   OKðKþ1Þ R¼ ; r

ð24Þ

where r ¼ ðr0 ; r1 ; . . . ; rK Þ is a ðK þ 1Þ-row vector. Thus Rn ¼ ðrK Þ

n1

R;

ð25Þ

nP1

and hence Eq. (23) is equivalent to R ¼ A0 ðA1  rK A2 Þ1 :

ð26Þ

It follows from (25) and Neuts (1981, Theorem 3.1.1) that the stationary distribution y ¼ ðy0 ; y1 ; . . .Þ with yn ¼ ðy n0 ; . . . ; y nK Þ, n P 0 of Y K is given by yn ¼ y 0K rKn1 r;

n ¼ 1; 2; 3; . . .

ð27Þ

Since the retrial rates of Y K are less than those of X K , that is, ~cn ¼ c 6 cn for all n P 1, fY K0 ðtÞ; t P 0g is stochastically greater than fX K0 ðtÞ; t P 0g (Shin, 2006). Thus we have the following: ! ! 1 1 X X r1 xn ðKÞ 1 ¼ lim P ðX K0 ðtÞ P 1Þ 6 lim P ðY K0 ðtÞ P 1Þ ¼ yn 1 ¼ y : ð28Þ t!1 t!1 1  rK 0K n¼1 n¼1 Now we prove the inequality (6) from (28) by showing that rK is far away from 1 and r1 is bounded, and y0K is bounded from above by a constant times of qK. Lemma 5.1 lim rK ¼ q:

ð29Þ

K!1

Proof. See Appendix A.

h

Proposition 5.2. There is a constant C such that for large K, ! 1 X xn ðKÞ 1 6 CqK :

ð30Þ

n¼1

Proof. We can see from (28) that it is sufficient to show r1 y 6 CqK : 1  rK 0K Simplifying ðA0 þ RA1 þ R2 A2 Þ1 ¼ 0 with (24) yields for each K, K 1 X j¼0

k rj ¼ : c

It follows from (29) and (31) that

ð31Þ r1 1rK

is bounded for large K.

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

1111 ð0Þ

It remains to show that y 0K 6 C  qK for a constant C*. We consider the censored Markov chain Y K of Y K ð0Þ ð0Þ with the censoring set 0 ¼ fð0; 0Þ; ð0; 1Þ; . . . ; ð0; KÞg and let yð0Þ ¼ ðy 0 ; . . . ; y K Þ be the stationary distribution ð0Þ ð0Þ ð0Þ of Y K , where ð0; jÞ is identified by j, 0 6 j 6 K. Since y 0K 6 y K , it suffices to show that y K 6 C  qK for a ð0Þ * constant C . Following the same procedure used in previous section for x ðKÞ, we have that:  s ðKÞ pK s 1 ð0Þ  6 yK ¼ qK ; ðKÞ ðKÞ c s! 1  p wðKÞ 1  pK wðKÞ  sl ðr0 þ r1 þ    þ rK2 Þ K where wðKÞ ¼ c

K X r0 þ    þ rn2 : ln n¼2

We can easily see from (31) that wðKÞ 6 ðs lnðs  1Þ  s þ 1 þ KÞq

ð32Þ

ðKÞ

and hence pK wðKÞ tends to zero as K ! 1. Therefore, we have for large K ð0Þ

yK 6

 2

 ss K q s!

and the proposition is proved.

h

Remark. It follows from (32) that wðKÞ is bounded by a linear function of K. In fact, we can show that wðKÞ is a bounded function, that is, for some constant Cw: ð33Þ

wðKÞ 6 C w and hence following the same procedure in Section 4, it can be seen the inequality: ð0Þ

kyð0Þ  pðKÞ k 6 2wðKÞy K 6 C w qK :

ð34Þ

The proof of (33) is sketched in Appendix B. 6. Numerical results In order to investigate the behavior of kp  xðKÞk, we depict ErrðKÞ ¼ logq kp  xðKÞk as a function of K in Fig. 1 for the M=M=3=K retrial queue with l ¼ 1:0, cn ¼ 5n, n P 0 and three cases of q ¼ 0:5, 0:75 and 0:9. For computing xðKÞ, we use the algorithm in Shin and Moon (2006) in which the generalized truncation method proposed by Neuts and Rao (1990) is used and effective algorithms for computing the rate matrices RN ¼ R and Rk, 0 6 k 6 N  1 for each truncation level N are presented. In order to reduce the effect of approximation, we slightly modify the way of choosing N in Neuts and Rao (1990). The truncation level N is selected such that N satisfies one of the following conditions: ðiÞ jrK ðN Þ  qj < 0:001

and

krðN Þ  rðN  1Þk1 < 0:01;

ðiiÞ

and

krðN Þ  rðN  1Þk1 < 0:001:

jrK ðN Þ  qj < 0:01

The truncation levels used in Fig. 1 are N = 6, 7 and 24 for q ¼ 0:5; 0:75 and 0:9, respectively and N does not depend on K. Fig. 1 shows that logq kp  xðKÞk is almost linear and that the behavior of kp  xðKÞk is similar to that of the function CqK . To investigate the effects of retrial rates to the speed of convergence, Err(K) with q ¼ 0:75 is depicted for three cases of c ¼ 0:1; 1:0 and 10:0 in Fig. 2. The values N in Fig. 2 are chosen as N = 13 and 5 for c ¼ 1:0 and 10:0, respectively and N depends on K for c ¼ 0:1 as follows N 5 ¼ 99, N 6 ¼ 55, N 7 ¼ 36 and N K ¼ 26 for 8 6 K 6 55. As one expected, we can see that kp  xðKÞk is a decreasing function of c. We can also see in

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Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117 Err K 60 rho =0.9 50 40 rho =0.75 30 20 rho =0.5 10 10

20

30

40

50

K

Fig. 1. Behavior of ErrðKÞ ¼ logq ðkp  xðKÞkÞ.

Err K gamma = 1.0 40 gamma =10.0 30 gamma =0.1 20 10

10

20

30

40

50

K

Fig. 2. Effects of c to ErrðKÞ ¼ log0:75 ðkp  xðKÞkÞ.

the computing procedure that the truncation level N becomes sensitive to the capacity K as the retrial rate c decreases. 7. Conclusions Considering the ordinary M/M/s queue as an M/M/s/K retrial queue with infinite retrial capacity, we mapped the one dimensional state space of M/M/s queue to the two dimensional state space of M/M/s/K retrial queue with finite retrial rate and then showed that the stationary distribution xðKÞ of M/M/s/K retrial queue converges geometrically to the stationary distribution p of ordinary M/M/s queue. More specifically speaking, it has been showed that qK kp  xðKÞk is dominated by a linear function of K for large K. The difference kp  xðKÞk is divided into three parts: the difference kp  pðKÞ k between p and the stationary distribution pðKÞ of ordinary M/M/s/K queue, the difference kpðKÞ  xð0Þ ðKÞk between pðKÞ and the stationary distribution xð0Þ ðKÞ of the censored Markov chain with the level 0 as censoring set, and the difference kxð0Þ ðKÞ  xðKÞk. We showed that the first and third differences are bounded by a constant times of qK, but the second difference between pðKÞ and xð0Þ ðKÞ is bounded by a linear function of K times of qK. A tighter bound of the difference kp  xðKÞk could be obtained if w0 ðKÞ in (18) has a tighter bound. That is, if one can show that w0 ðKÞ is bounded by a constant, then Proposition 4.2 becomes kxð0Þ ðKÞ  pðKÞ k 6 C  qK and the main result (1) becomes kp  xðKÞk 6 CqK :

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

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It follows from the numerical results and Remark for wðKÞ in Section 5 that the boundedness of w0 ðKÞ is conjectured, but it has not been proved yet. Acknowledgements This research was supported by the MIC (Ministry of Information and Communication), Korea, under the ITRC (information Technology Research Center) support program supervised by the IITA (Institute of Information Technology Assessment). The author is grateful to the referees for their valuable comments and suggestions that improved the presentation of the paper. Appendix A. Proof of Lemma 5.1 Recall the formula R ¼ A0 ðA1  rK A2 Þ form   B11 B12 B¼ ; B21 B22

1

in (26) and write the matrix B ¼ A1  rK A2 in the partitioned

where B11 and B22 are the square matrices of order s and M ¼ K þ 1  s, respectively and are of the form 1 0 k þ c ðk þ crK Þ C B k þ c þ l ðk þ crK Þ C B l C B C B . . . C; B .. .. .. B11 ¼B C C B B ðs  2Þl k þ c þ ðs  2Þl ðk þ crK Þ C A @ 0 B B B B B22 ¼B B B B @

ðs  1Þl k þ c þ sl

ðk þ crK Þ

sl

k þ c þ sl

ðk þ crK Þ

..

..

.

..

.

sl

.

k þ c þ sl sl

k þ c þ ðs  1Þl 1 C C C C C C C ðk þ crK Þ C A k þ sl

and B12 and B21 are matrices of size s  M and M  s, respectively and are of the form B12 ¼ ðk þ crK Þes ðsÞeT1 ðMÞ; B21 ¼ sle1 ðMÞeTs ðsÞ; where ek ðnÞ ¼ ð0; . . . ; 0; 1; 0; . . . ; 0ÞT is the column n-vector whose kth component is 1 and others are all zeros and vT denotes the transpose of v. It is clear that B11 and B22 are invertible for each 0 6 rK 6 1 and all the 1 entries of B1 11 and B22 are positive. It follows from (26) that: rj1 ¼ k½B1 Kþ1;j ;

j ¼ 1; 2; . . . ; K þ 1;

ð35Þ

where [A]ij denotes the ði; jÞ component of the matrix A. Using the inverse formula for the block partitioned matrix, e.g. see Horn and Johnson (1985, p. 18), we have that 1 ½B1 Kþ1;sþj ¼ ½ðB22  B21 B1 11 B12 Þ Mj ;

1 6 j 6 M;

with 1 T B21 B1 11 B12 ¼ slðk þ cr K Þ½B11 ss e1 ðMÞe1 ðMÞ

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and 1 1 ½B1 Kþ1;j ¼ slð1 þ bðKÞ½B1 22 11 Þ½B22 M1 ½B11 sj ; 1 1 ½B1 Kþ1;sþj ¼ ½B1 22 Mj þ bðKÞ½B22 M1 ½B22 1j ;

1 6 j 6 s;

1 6 j 6 M;

ð36Þ ð37Þ

where bðKÞ ¼

slðk þ crK Þ½B1 11 ss : 1 1  slðk þ crK Þ½B1 11 ss ½B22 11

The following lemma is immediate from Choi et al. (2003b, Theorem 6) and we describe the results without proof. Lemma A.1. The first and last rows of B1 22 are given by j1 þ h12 ðKÞg2M1 g1Mj ; ½B1 22 1j ¼ h11 ðKÞg 2

½B1 22 Mj

¼

j1 h21 ðKÞgM 1 g2

þ

h22 ðKÞg1Mj ;

1 6 j 6 M; 1 6 j 6 M;

ð38Þ ð39Þ

where g1 ¼ g2 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 k þ c þ sl  ðk þ c þ slÞ  4slðk þ crK Þ 2ðk þ crK Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 k þ c þ sl  ðk þ c þ slÞ  4slðk þ crK Þ

;

2sl

ð40Þ

ð41Þ

and the quantities hij, i; j ¼ 1; 2 are as follows: 1 1 ðk þ slð1  g2 ÞÞ; h12 ðKÞ ¼  ðslg2  cÞ; D D 1 1 h21 ðKÞ ¼  ðk þ crK Þ; h22 ðKÞ ¼ ðk þ c þ slð1  g2 ÞÞ D D h11 ðKÞ ¼

with M1 : D ¼ ðk þ c þ slð1  g2 ÞÞðk þ slð1  g2 ÞÞ  ðslg2  cÞðk þ crK ÞgM 1 g2

Now we are ready to prove Lemma 5.1. Proof of Lemma 5.1. Noting from (24) that rK is the spectral radius of R, we have that 0 6 rK < 1 for sufficiently large K. Let lim sup rK ¼ g K!1

and frK j ; j ¼ 1; 2; . . .g be the subsequence of frK ; K ¼ 1; 2; . . .g such that limK j !1 rK j ¼ g. Then it is clear that 0 6 g 6 1. To prove the lemma, we first show that 0 6 g < 1. It is immediate from (31) that rK1 6

K 1 X j¼0

k rj ¼ : c

ð42Þ

We have from (23) with (24) that ðcrK  kÞrK1 ¼ ðk þ slÞrK  k:

ð43Þ

If g = 1, then we have from (42) and (43) that g1 ¼ limK j !1 rK j 1 satisfies the equation sl ¼ ðc  kÞg1 6 cg1 6 k which violates the stability condition k < sl and hence g < 1.

ð44Þ

Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

1115

Now we show that g ¼ q. Noting from (35) and (37) that 1 1 rK j ¼ k½B1 22 M j M j þ kbðK k Þ½B22 M j 1 ½B22 1M j ;

where M j ¼ K j þ 1  s, we have that g satisfies the following equation: 1 1 g ¼ k lim ½B1 22 M j M j þ k lim bðK j Þ½B22 M j 1 ½B22 1M j  I þ II: K j !1

K j !1

Consider gi in Lemma A.1 as a function of rK and write them as gi ðrK Þ, i ¼ 1; 2. Since gi ðgÞ < 1, we have that lim gi ðrK j ÞM j ¼ 0;

K j !1

i ¼ 1; 2

and hence lim ½B1 22 M j 1 ¼ 0:

K j !1

Similarly, we have that limK j !1 ½B1 22 1M j ¼ 0. In order to get II = 0, it suffices to show that bðK j Þ is bounded for sufficiently large Kj which is equivalent to 1 lim slðk þ crK j Þ½B1 11 ss ½B22 11 < 1:

ð45Þ

K j !1

Noting that 1 cð1  rK j Þ C B .. C B 1 1 B . C; 1 ¼ ðB11 B11 Þ1 ¼ B11 B C @ cð1  rK j Þ A 0

kþc we have ½B1 11 ss

s1 X 1 1  cð1  rK j Þ ¼ ½B1 11 si kþc i¼1

! <

1 ; kþc

where we use the fact that ½B1 11 ij > 0; 1 6 i; j 6 s. Since sl lim sl½B1 ¼ g1 ðgÞ < 1; 22 11 ¼ K!1 k þ c þ sl  slg2 ðgÞ we have that 1 lim slðk þ crK j Þ½B1 11 ss ½B22 11 6 lim

K j !1

K j !1

k þ crK j k þ cg g ðgÞ < 1: ðsl½B1 22 11 Þ ¼ kþc 1 kþc

Thus II = 0. It is easily seen that I ¼ k lim ½B1 22 M j M j ¼ k lim h22 ðK j Þ ¼ K j !1

K j !1

k kg2 ðgÞ ¼ : k þ sl  slg2 ðgÞ k þ cg  cg2 ðgÞ

Thus 0 < g < 1 satisfies the following equation: g¼

kg2 ðgÞ : k þ cg  cg2 ðgÞ

ð46Þ

Simplifying (46) yields that g2  ð1 þ qÞg þ q ¼ 0 and we have that g ¼ q. Similarly we can show that lim inf K!1 rK ¼ q.

h

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Y.W. Shin / European Journal of Operational Research 189 (2008) 1104–1117

Appendix B. Proof of (33) Here we just sketch the proof. Rewrite wðKÞ by " ! s  s1 X cX r0 þ    þ rn2  c ðK  sÞ wðKÞ ¼ rn þ þ l n¼2 sl n n¼0

Ks1 X

!# nrK1n

:

n¼1

1 We have showed in the proof of Lemma 5.1 that slð1 þ bðKÞ½B1 22 11 Þ½B11 sj , 1 6 j 6 s are bounded for suffie 1 such that for large K ciently large K. We can easily see that there exists a positive constant C

e M1 : ½B1 22 M1 6 C 1 g 1

ð47Þ

e 0 such that for sufficiently large K It can be seen from (35), (36) and (39) that there exists a positive constant C e 1 gM1 ; e 0 ½B1  6 C e0C rj1 6 C 22 M1 1 and hence lim ðK  sÞ

s1 X

K!1

! ¼ 0:

rj

j¼0

Now we show that lim

16j6s

Ks1 X

K!1

! < 1:

nrK1n

n¼1

e 2 and C e 3 such that for large K We can see from (38) and (47) that there exist positive constants C 1 e M1 gj1 þ C e 3 ðg1 g2 ÞM1 gMj ; ½B1 2 1 22 M1 ½B22 1j 6 C 2 g1

1 6 j 6 M:

e 4 and C e 5 such that for large K We can also see from (39) that there exist positive constants C e M j1 þ C e 5 gMj ; ½B1 1 22 Mj 6 C 4 g 1 g 2 Thus Ks1 X

! nrK1n

¼k

1 6 j 6 M:

M2   X 1 1 n ½B1 22 M;M1n þ bðKÞ½B22 M1 ½B22 1;M1n

n¼1

n¼1

e 6C

M 2 X

M1 nþ1 g1 Þ

Mn2 nðgM þ g1nþ1 þ g1M1 g2Mn2 þ ðg1 g2 Þ 1 g2

n¼1

is bounded for all K. Thus wðKÞ is bounded.

h

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