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

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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

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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 bz 2 [ p(z 2 - 1)B e - lb(1- z) + pz (z - 1)B e - lb(1- z ) ] ù 0 1 , (9) + 2 -lb(1- z) 2 2 -lb(1- z) ú [z - (1 - p)e ][( l - lz + b ) z ] - pbz e û

é[ p(z 2 -1)B e -lb(1- z) + pz(z -1)B e -lb(1-z) ][z 2 - (1- p)e -lb(1-z) ] N(z) 0 1 = ê 2 -lb(1-z) 2 2 -lb(1- z) ë [z - (1 - p)e ][(l - lz + b)z ] - pbz e pe -lb(1-z) [(1- p)(z 2 -1)B e -lb(1-z) + (1- p)z(z -1)B e -lb(1- z) ]ù 0 1 . (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 ) .

Because f ( z) ³ g( z ) , therefore by Rouche’s theorem f ( z) + g( z) has the same of zeros inside or on z =1 as that of f ( z) . Now, it is easy to show that f ( z) has four zeros on or inside z =1. Therefore, f ( z) + g (z) , i.e., the denominator of the right hand side of (11) has four zeros on or inside z =1. For each of these four zeros the numerator of the right hand side of (11) must vanish, thus giving us four linear equations in the two unknowns B 0 and B 1 . Then two of these four equations are sufficient to determine the two unknowns, whereas the other two may just be redundant. Hence, the probability generating functions B (z), N (z) and P (z) obtained in (9), (10) and (11) can be completely determined.

Next, we shall use normalizing condition P(1) = B(1) + N(1) = 1. (12) zero At z = 1, P(1) = , therefore using L'Hopital's rule we have from (11) zero

( p + b )(2B + B ) P(1) = lim P(z) = 0 1 = 1, (13) z®1 2b - lbb - pl which gives

( p+ b )(2B0 + B1 ) = 2b -lbb - pl . (14)

Equation (14) will hold only if 2b - lbb - pl > 0 which gives the steady state condition

l (bb + p) < 1. (15) 2b

Note that when there are no server vacations, then with p = 0, N n =0 for all n³ 0 , equation (10) yields N(z)=0 as it should be. Further, equations (9) and (15) respectively give

é[(z 2 -1)B e -lb(1-z) + z(z -1)B e-lb(1-z) ] B(z)= 0 1 , (16) = ê 2 -lb(1-z) ë [z -e ]

lb < 2 . (17) MADAN, ABU-DAYYEH, & TAYYAN 206

Note that (16) and (17) are the known the current service started. results for the M/D/c queue for c=2. (See Kashyap and Chaudhury, 1988, p. 60-61.) On (t): as the probability that at time t only one server is available in the system providing service Model B: Only One Server At A Time Can Take A and there were n ( ³ 0) customers when the current Vacation service started. The Underlying Assumptions. In this case, the assumptions (a), (b) and Pn (t): as the probability that at time t there are n (d) in section 2.1 for the previous case are the ( ³ 0) customers in the system regardless of the same. However, assumption (c) is different in this state of the servers. case under which we assume that after every service completion, only one server may take a Steady State Forward Equations of the System vacation with probability p or continue to stay in Assuming that steady state exists, we let the system with probability 1- p. The vacation limBn (t) = Bn , lim On (t) = On and times follow an exponential distribution with mean t®¥ t®¥ vacation time 1/ b , ( b > 0). limPn (t) = Pn . Thus Bn , On and Pn denote the t®¥ Definitions and Notations corresponding steady state probabilities. Then we We define: obtain the following set of steady state equations: Bn (t): as the probability that at time t both servers are available in the system providing service and there were n ( ³ 0) customers in the system when

n+2 B n = (1 - p)k n [B0 + B1 + B2 ]+ (1 - p)å Bi k n+ 2-i + bOn , n ³ 0, (18) i=3

n+1 n+2 (1+ b)On =O0kn +O1kn +åOj kn+1- j + pkn [B0 + B1 +B2 ] + på Bik n+2-i , n ³ 0. (19) j=2 i=3

Steady State Probability Generating Functions for the System Size In addition to the probabilit y generating functions defined in (4a) and 4b) in section 2.4, we define the following probability generation function: ¥ n O(z) = åOn z , | z | £ 1. (20) n=0

n+2 We multiply both sides of equation (16) by z and add for all n = 0, 1, 2… Thus 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 k n+2-i z + bz åOn z , n ³ 0. (21) n=0 i=3 n=0

Then using (4a), (4b) and (20) we obtain from (21)

207 STEADY STATE ANALYSIS OF AN M/D/2 QUEUE

bz 2O(z) + (1 - p)B (z 2 -1)e -lb(1-z) + (1 - p)B z(z -1)e -lb(1-z) B(z) = 0 1 . (22) z 2 - (1 - p)e -lb(1- z)

n +2 Similarly, we multiply both sides of (19) by z and add them for all n = 0, 1, 2… Then we have

¥ ¥ ¥ ¥ n+1 2 n 2 n 2 n n+2 (1+ b )z å On z = O0 z å k n z + O1 z å k n z + å å O jk n+1- j z n=0 n=0 n=0 n=0 j=2 ¥ ¥ ¥ 2 n 2 n 2 n + pB0 z å k n z + pB1 z å k n z + pB2 z å k n z n=0 n=0 n=0 ¥ n+2 n+2 + påå Bi kn+2-i z , n ³ 0 . (23) n=0 i=3

Then using (4a), (4b) and (18) we obtain from (23)

-lb(1-z) 2 -lb(1-z) -lb(1-z) -lb(1-z) pB(z)e + pB0 (z -1)e + pB1z(z-1)e + z(z-1)O0e O(z) = . (24) (1+b)z2 -ze-lb(1-z)

Then, we solve equations (22) and (24) simultaneously for B(z) and O(z) and obtain

é[(1- p)(z2 -1)B e-lb(1-z) + (1- p)z(z -1)B e-lb(1-z)][(1+ b)z2 - ze-lb(1-z) ] B(z) = 0 1 ê 2 -lb(1-z) 2 -lb(1-z) 2 -lb(1-z) ë [z - (1- p)e ][(1+ b)z - ze ]- pbz e

bz 2 [ p(z 2 -1)B e - lb(1- z) + pz(z -1)B e -lb(1-z) + z(z -1)O e -lb(1- z) ]ù 0 1 n , (25) + 2 -lb(1-z) 2 -lb(1- z) 2 -lb(1- z) ú [z - (1 - p)e ][(1+ b )z - ze ] - pbz e û

2 -lb(1-z) -lb(1-z) -lb(1-z) 2 -lb(1-z) é[p(z -1)B0e + pz(z -1)B1e +z(z-1)O0e ][z -(1- p)e ] O(z) = ê 2 -lb(1-z) 2 -lb(1-z) 2 -lb(1-z) ë [z -(1- p)e ][(1+b)z -ze ]- pbz e

pe -lb(1- z) [(1- p)( z 2 -1)B e -lb(1- z) + (1- p)z( z -1)B e -lb(1- z) ]ù 0 1 . (26) + 2 -lb(1- z) 2 -lb(1- z) 2 -lb(1- z) ú [z - (1- p)e ][(1+ b )z - ze ] - pbz e û

Then adding (25) and (26), we obtain

P(z) = B(z) + O(z) . (27)

The unknowns B0 , B1 and O0 can be determined by applying Rouche’s theorem as before. Hence, the probability generating functions B(z), O(z) and P(z) can be completely determined.

MADAN, ABU-DAYYEH, & TAYYAN 208

Further, we use the normalizing condition

P(1) = B(1) + O(1) = 1 . (28)

zero At z = 1, because P(1) = , and hence, using L'Hopital's rule we have from (27) zero

( p + b)(2B + B + O ) P(1) = lim P(z) = 0 1 0 = 1, (29) z®1 2b - lbb + p - plb which gives

( p+ b)(2B0 + B1 +O0 ) = 2b -lbb + p - plb. (30)

Equation (30) will hold only if 2b - lbb + p - plb > 0 which yields the steady state condition

lb(b + p) < 1. (31) 2b + p

Again note that when there are no server vacations, then with p = 0 and On =0 for all n³ 0 , equation (26) yields O(z)=0 as it should be. Further, (25) and (31) respectively give

é[(z 2 -1)B e -lb(1-z) + z(z -1)B e-lb(1-z) ]ù B(z)= 0 1 , (32) = ê 2 -lb(1-z) ú ë [z -e ] û

lb < 2 . (33)

Note that (32) and (33) are the same known results for the M/D/c queue for c=2 as in section 2.4. (See Kashyap & Chaudhury, 1988, p. 60-61.)

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