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Steady State Analysis of an M/D/2 Queue with Bernoulli Schedule Server Vacations Kailash C Journal of Modern Applied Statistical Methods Volume 2 | Issue 1 Article 19 5-1-2003 Steady State Analysis Of An M/D/2 Queue With Bernoulli Schedule Server Vacations Kailash C. Madan Yarmouk University, [email protected] Walid Abu-Dayyeh Yarmouk University Firas Tayyan Yarmouk University Follow this and additional works at: http://digitalcommons.wayne.edu/jmasm Part of the Applied Statistics Commons, Social and Behavioral Sciences Commons, and the Statistical Theory Commons Recommended Citation Madan, Kailash C.; Abu-Dayyeh, Walid; and Tayyan, Firas (2003) "Steady State Analysis Of An M/D/2 Queue With Bernoulli Schedule Server Vacations," Journal of Modern Applied Statistical Methods: Vol. 2 : Iss. 1 , Article 19. DOI: 10.22237/jmasm/1051748340 Available at: http://digitalcommons.wayne.edu/jmasm/vol2/iss1/19 This Regular Article is brought to you for free and open access by the Open Access Journals at DigitalCommons@WayneState. It has been accepted for inclusion in Journal of Modern Applied Statistical Methods by an authorized editor of DigitalCommons@WayneState. Journal of Modern Applied Statistical Methods Copyright Ó 2003 JMASM, Inc. May, 2003, Vol. 2, No. 1, 202-209 1538 – 9472/03/$30.00 Steady State Analysis Of An M/D/2 Queue With Bernoulli Schedule Server Vacations Kailash C. Madan Walid Abu-Dayyeh Firas Tayyan Department of Statistics, Faculty of Science Yarmouk University, Irbid, Jordan We examine an M/D/2 queue with Bernoulli schedules and a single vacation policy. We have assumed Poisson arrivals waiting in a single queue and two parallel servers who provide identical deterministic service to customers on first-come, first-served basis. We consider two models; in one we assume that after completion of a service both servers can take a vacation while in the other we assume that only one may take a vacation. The vacation periods in both models are assumed to be exponential. We obtain steady state probability generating functions of system size for various states of the servers. Key words: Two parallel servers, Bernoulli schedules, single vacation policy, deterministic service Introduction We consider two models. In model A we assume that after a service completion both servers Vacation Queues have been studied by numerous may take a vacation of identical exponential researchers including Kleinrock (1983) , Keilson duration and in model B, we assume that only one and Servi (1986), Baba (1986), Doshi of the servers can take a vacation of exponential (1986,1990), Cramer (1989), Choi & Park (1990), duration. In both models, we assume a single Borthakur & Choudhury 1997), Madan (1992, vacation policy which means that whenever a 1999, 2001), to mention a few. Most of these vacation period of a server ends, then he must join authors have investigated single server queues the system irrespective of whether there are assuming Bernoulli schedules or exhaustive customers waiting for service or not. That is, the service or generalized vacations among several server must join the system even if he finds the other vacation policies with a single or multiple system empty on return. The following vacations. Madan and Saleh (2001, 2001, 2001) assumptions briefly describe our models: have studied a single server queue with exponential service and deterministic vacatio ns, Model A: Both Servers Can Take A Vacation. deterministic service with exponential vacations The Underlying Assumptions: and deterministic service with deterministic vacations, assuming Bernoulli schedules. A - Customers arrive at the system one by one and Those articles considered single server their arrivals follow a Poisson distribution with vacation models. Here, we study a queueing mean arrival rate 8, (8 > 0). system with two parallel servers providing identical deterministic service assuming Bernoulli B - Both servers provide identical deterministic schedule server vacations with a single vacation (constant) service with constant service time of policy. length b, (b > 0). C - After every service, both servers together may Send correspondence to Kailash C. Madan, take a vacation with probability p or continue to Department of Statistics, Faculty of Science, stay in the system with probability 1- p. The Yarmouk University, Irbid, Jordan. E-mail him at vacation times follow an exponential distribution [email protected]. with mean vacation time 1/$, ( > 0). 202 203 STEADY STATE ANALYSIS OF AN M/D/2 QUEUE D- All stochastic processes involved in the system Pn (t): as the probability that at time t there are are independent of each other. n (0) customers in the system without regardless of the state of the servers Definitions and Notations ki : as the probability of i arrivals during a service Define: period of constant length b. Bn (t): as the probability that at time t both servers Steady State Forward Equations of the System are available in the system providing service and Assuming that the steady state exists, let there were n (0) customers in the system when the limBn(t)=Bn, limNn (t) = Nn , and current service started. t®¥ t®¥ limPn (t) = Pn . Thus, Bn , N n and Pn denote the Nn (t): as the probability that at time t there are t®¥ n (0) customers in the system and both servers are corresponding steady state probabilities. Then on vacation. applying the usual probability reasoning we obtain the following set of steady state forward equations: n+ 2 Bn = (1 - p)k n [B0 + B1 + B2 ]+ (1 - p)å Bi k n+2-i + bN n , n ³ 0, (1) i=3 n+2 (l + b)N n = lN n-1 + pkn [B0 + B1 + B2 ] + på Bi kn+2-i , n ³ 1, (2) i=3 (l + b ) N 0 = pk 0 [B 0 + B1 + B 2 ], n =0. (3) Steady State Probability Generating Functions for the System Size We define the following probability generation functions: ¥ ¥ ¥ n n n B(z) = å Bn z , N(z) = å N n z , P(z) = å Pn z , (4a) n=0 n=0 n=0 ¥ ¥ e-lb(lb)n zn ¥ (lbz)n n -lb -lb(1-z) z £ K(z) = åKnz = å = e å = e , 1. (4b) n=0 n=0 n! n=0 n! n+2 We multiply (1) by z and add for all n = 0, 1, 2…Then we have ¥ ¥ ¥ ¥ 2 n 2 n 2 n 2 n z å Bn z = (1- p)B0 z åk n z + (1- p)B1z å kn z + (1- p)B2 z å kn z n=0 n=0 n=0 n=0 ¥ n+2 ¥ n+2 2 n + (1- p)åå Bi kn+2-i z + bz å Nn z , n ³ 0 (5) n=0 i=3 n=0 MADAN, ABU-DAYYEH, & TAYYAN 204 Then using (4) we obtain from (5) bz2N(z)+ (1- p)B (z2 -1)e-lb(1-z) + (1- p)B z(z -1)e-lb(1-z) B(z) = 0 1 (6) z2 - (1- p)e-lb(1-z) n+2 Similarly we multiply (2) by z and (3) by z and add them for all n = 0,1,2… Then we have ¥ ¥ ¥ ¥ 2 n 3 n 2 n 2 n (l + b)z å N n z = lz å N n z +pB0 z å kn z + pB1z åk n z n=0 n=0 n=0 n=0 ¥ ¥ n+2 2 n n+2 + pB2 z å kn z + påå Bik n+2-i z ,n ³ 0 (7) n=0 n=0 i=3 Using (4) we obtain from (7) -lb(1-z) 2 -lb(1-z) -lb(1- z) pB(z)e + pB0 (z -1)e + pB1z(z -1)e N(z) = (8) (l - lz + b)z 2 Then we solve (6) and (8) simultaneously for B(z) and N (z) and obtain on simplifying é[(1- p)(z 2 -1)B e-lb(1-z) + (1- p)z(z -1)B e-lb(1-z) ][(l -lz + b)z 2 ] B(z) 0 1 = ê 2 -lb(1-z) 2 2 -lb(1-z) ë [z - (1- p)e ][(l -lz + b)z ] - pbz e 2 2 - lb(1- z) - lb(1- z ) bz [ p(z - 1)B0 e + pz (z - 1)B1e ] ù , (9) + 2 -lb(1- z) 2 2 -lb(1- z) ú [z - (1 - p)e ][( l - lz + b ) z ] - pbz e û 2 -lb(1- z) -lb(1-z) 2 -lb(1-z) é[ p(z -1)B0e + pz(z -1)B1e ][z - (1- p)e ] N(z) = ê 2 -lb(1-z) 2 2 -lb(1- z) ë [z - (1 - p)e ][(l - lz + b)z ] - pbz e -lb(1-z) 2 -lb(1-z) -lb(1- z) pe [(1- p)(z -1)B0e + (1- p)z(z -1)B1e ]ù . (10) + 2 -lb(1-z) 2 2 -lb(1-z) ú [z - (1- p)e ][(l - lz + b)z ]- pbz e û Hence, adding (9) and (10) we have P(z) = B(z) + N(z) . (11) Now we have to determine the unknown probabilities and which appear in the numerators of the right B 0 B1 hand sides of equations (9), (10) and (11). For this purpose we use Rouche's theorem as follows. Let f ( z) = [z 2 -(1- p)e - lb( 1-z ) ][(l -lz + b ) z 2 ], g( z) = - pbz 2 e -lb (1-z) . 205 STEADY STATE ANALYSIS OF AN M/D/2 QUEUE Note that both f (z) and g (z) are regular on and inside | z | =1. We aim to prove that f (z) ³ g( z) on | z | =1. Now, on | z | =1, f ( z ) = [z 2 - (1 - p)e - lb ( 1- z ) ][(l - lz + b ) z 2 ] = z 2 - (1 - p ) e - lb ( 1- z ) ( l - lz + b ) z 2 é2-l-b(1z)2ùéù ³ëz-(1-p)eûëû(l-lz+b)z = [1- (1- p)e - lb ( 1-z ) ][l - l + b ]= pbe - lb( 1- z ) = g ( z ) .
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