An M/G/1 queue with optional deterministic server vacations

KAILASH C. MADAN (*)

An M/G/1 queue with optional deterministic server vacations Contents: 1. Introduction. — 2. Definitions concerning the states of the system. — 3. The steady state equations. — 4. The steady state probability generating functions. — 5. The expected number in the queue and in the system. — 6. The expected waiting time in the queue and in the system. — 7. Special cases. Acknowledgment. References. Summary. Riassunto. Key words.

1. Introduction Gaver (1959), Levy and Yechilai (1976), Fuhrman (1981), Doshi (1986), Keilson and Servi (1986), Shanthikumar (1988), Cramer (1989) and Madan (1991, 1992), are a few among many authors who have studied queues with server vacations with varying vacation policies including the Bernoulli schedule vacations. In the present paper, we study the steady state behaviour of an M/G/1 queue with deterministic server vacations, using the supplementary variable technique. We assume that at the completion of each service, the server may decide to take a vacation of fixed length d (> 0) with probability p or may continue to be available in the system for the next service with probability 1 − p. One may encounter numerous examples in real life where the server takes deterministic vacations. For convenience, we will designate our system as the M/G/D/1 queueing system which is briefly described by the following assumptions. • Customers arrive at the system one by one in a Poisson stream with mean arrival rate λ (> 0). • The service to customers is provided one by one on a first come, first served basis and their service times follow a general (arbi(*) Department of Statistics, Faculty of Science, Yarmouk University, Irbid, Jordan

84 trary) distribution with distribution function B(v) and the density function b(v). Let µ(x)d x be the conditional probability of completion of a service during the interval (x, x + d x) given that the elapsed time is x, so that b(x) µ(x) = (1) 1 − B(x) and, therefore,




b(v) = µ(v) exp −

 0

v



µ(x)d x

(2)

• As soon as the service of a customer is complete, then with probability p the server decides to take a vacation and with probability 1 − p, he continues to be available for the next service. • We assume that whenever the server takes a vacation, it is of constant duration d (> 0). • Various stochastic processes involved in the system are independent of each other.

2. Definitions concerning the states of the system Let Wn (t, x) be the probability that at time t, there are n (≥ 0) customers in the queue excluding the one in service and the elapsed service time of this customer is x. Correspondingly, Wn (t) = 0∞ Wn (x, t) is the probability that at time t, there are n (≥ 0) customers in the queue excluding the one in service irrespective of the elapsed time x. Let Vn (t) denote the probability that at time t, there are n (≥ 0) customers in the queue and the server is away on vacation. Next, let Q(t) be the probability that at time t, there are no customers in the system and the server is available but idle. Finally, we suppose that K r is the probability of r arrivals during a vacation period so that −λd r K r = e r(λd) , r = 0, 1, 2 . . . ! 3. The steady state equations Assuming that the steady state exists, we let Limt→∞ Wn (t, x) = Wn (x), Limt→∞ Wn (t) = Wn , Limt→∞ Vn (t) = Vn and Limt→∞ Q(t) =Q.

85 Then, we proceed to obtain the steady state equations of the system using the following probability reasoning: Wn (x +d x) = Pr{there are n customers in the queue excluding one in service and the elapsed service time of the customer being served is x + d x} = Pr{there are n customers in the queue excluding one in service with elapsed service time x and no additional customer arrives during the period [x, x + d x) and the service is not completed in this interval} + Pr{there are n−1 customers in the queue excluding one in service with elapsed service time x, one customer arrives during the period [x, x + d x) and the service is not completed in this interval} = Wn (x)(1−λd x)(1−µ(x)d x)+Wn−1 (x)λd x(1−µ(x)d x)(A). From (A) we obtain d Wn (x) + (λ + µ(x))Wn (x) = λWn−1 (x) dx

n ≥ 1.

(3)

Similarly for n = 0 we have d W0 (x) + (λ + µ(x))W0 (x) = 0 . dx

(4)

To develop the next equation, we shall first consider its timedependent version and then pass to the limit as t → ∞ to get the corresponding steady state equation. Therefore Q(t +dt) = Pr{at time t + dt there is no customer in queue or service, the server is idle but available in the system} = Pr{at time t there is no customer in queue and one customer is in service irrespective of elapsed service time x and the service of this customer gets completed during the interval [t, t +dt) and with probability 1− p the server chooses not to take a vacation}

86 + Pr{at time t there is no customer in queue or in service and the server is on vacation, the vacation terminates during the interval [t, t +dt) and there is no new arrival during this interval} + Pr{at time t there is no customer in queue or in service and the server is available but idle and there is no new arrival during this interval} 

= (1 − p)

∞

0



W0 (t, x)µ(x)d x dt+

+ V0 (t)k0 dt + Q(t)(1 − λ dt) (B) From the right side of (B), we transfer Q(t) to its left side and then divide both sides by dt and take the limit as t → ∞. Thus we have  ∞  d Q(t) = (1 − p) W0 (t, x)µ(x)d x + V0 (t)k0 − λQ(t) . (C) dt 0 For the steady state, we let dtd Q(t) = 0 in (C) and drop the argument t and simplify. Thus we obtain λQ = (1 − p)



0

∞

W0 (x)µ(x)d x + V0 K 0 .

(5)

Using the similar reasoning as above, we obtain 

Vn = p

∞

0

Wn (x)µ(x)d x

n ≥ 0.

(6)

The above equations are to be solved subject to the following boundary conditions: Wn (0) = (1 − p)



∞

0

Wn+1 (x)µ(x)d x + V0 K n+1 + V1 K n

+ . . . + Vn K 1 + Vn+1 K 0 W0 (0) = (1 − p)



0

∞

(7)

n ≥ 1;

W1 (x)µ(x)d x + V0 K 1 + V1 K 0 + λQ .

(8)

We note that the reasoning in obtaining the boundary conditions (7) and (8) is that the terms in the right hand sides of these equations indicate either the completion of a service with the server choosing to stay in the system with probability 1 − p or the end of a vacation period. At such an instant each of the right hand terms in equations (7) and (8) leads to the left hand term with zero initial value of the service time x, because the service of the next customer starts immediately at such an instant.

87 4. The steady state probability generating functions We define the following probability generating functions: W (x, z) =

∞ 

Wn (x, z)z n ;

W (z) =

n=0

V (z) =

∞ 

∞ 

Wn z n ;

(9a)

n=0

Vn z n .

(9b)

n=0

On multiplying equations (3) and (4) by z n , summing over n and using (9a), we have d W (x, z) + (λ − λz + µ(x))W (x, z) = 0 . dx

(10)

Similarly, multiplying equation (6) by z n , summing over n and using (9a) and (9b), we get V (z) = p



∞

W (x, z)µ(x)d x .

0

(11)

In the same manner as above, equations (7) and (8), on using (9a), yield zW (0, z) = (1 − p)

 0

− (1 − p) + V (z)e

∞



W (x, z)µ(x)d x ∞

W0 (x)µ(x)d x

0 −λd(1−z)

− V0 K 0 + λQ(z)

(12a)

wherein we have used the fact that ∞ 

Kn zn =

0

∞  0

e−λd

(λd)n n z = e−λd(1−z) . n!

Using equation (5) we can re-write equation (12a) as W (0, z) =

(1− p)

∞ 0

w(x, z)µ(x)d x +V (z)e−λd(1−z) + λQ(z −1) (12b) z

88 Now, we integrate equation (10) between the limits 0 and x, and obtain x −(λ−λz)− µ(t)dt 0 W (x, z) = W (0, z)e (13) where W (0, z) is given by (12b). Then we integrate equation (13) with respect to x and get 

¯ − λz) 1 − b(λ W (z) = W (0, z) λ − λz



(14)



¯ where b(λ−λz) = 0∞ e−(λ−λz)x b(x)d x is the Laplace transform of b(x). Then, by virtue of equations (2) and (13), we have 

∞

0

¯ − λz) . W (x, z)µ(x)d x = W (0, z)b(λ

(15)

Using (15) in equation (12b) and simplifying, we have W (0, z) =

V (z)e−λd(1−z) + λQ(z − 1) . ¯ − λz) z − (1 − p)b(λ

(16)

Using equation (16), equation (14) becomes 

¯ − λz) 1 − b(λ W (z) = λ − λz





V (z)e−λd(1−z) + λQ(z − 1) . ¯ − λz) z − (1 − p)b(λ

(17)

And by virtue of equation (15), (11) can be re-written as ¯ − λz) V (z) = pW (0, z)b(λ

(18a)

which, on using (16), becomes 



V (z)e−λd(1−z) + λQ(z − 1) ¯ V (z) = p b(λ − λz) . ¯ − λz) z − (1 − p)b(λ

(18b)

Equation (18b), can be further simplified to V (z) =

¯ − λz)Q(z − 1) λp b(λ . ¯ − λz) + p b(λ ¯ − λz)(1 − e−λd(1−z) ) z − b(λ

(19)

89 Then, substituting for V (z) from equation (19) into equation (17) and simplifying, we obtain W (z) =

¯ − λz) − 1]Q [b(λ . ¯ − λz) + p b(λ ¯ − λz)(1 − e−λd(1−z) ) z − b(λ

(20)

We see that at z = 1, the right hand sides of both equations (19) and (20) are indeterminate of the form 0/0. Therefore, applying L’Hopital’s rule we obtain λµp Q ; µ − λ − λµpd λQ W (1) = Lim W (z) = z→1 µ − λ − λµpd V (1) = Lim V (z) =

(21)

z→1

(22)

¯ wherein we have used the facts that b(0) = 1 and −b¯  (0) = µ1 . Now, to determine the only unknown constant Q, we use (21) and (22) in the normalizing condition Q + V (1) + W (1) = 1 and have Q=

λ(1 + pµ) µ(1 − λpd) − λ =1− µ(1 − λpd) + pλµ µ(1 − λpd) + λpµ

(23)

where λ < µ(1 − λpd). Equation (23) gives the steady state probability that there is no customer in the system and the server is idle. It is easy to verify that when there are no server vacations, then with p = 0, equation (23) reduces to Q = 1− µλ , a well-known steady state result of the queueing system M/M/1. Also from equation (23), we obtain ρ, the utilisation factor of the system as ρ =1− Q =

λ(1 + pµ) < 1. µ(1 − λpd) + pλµ

(24)

Substituting for Q found in equation (23), we finally have from equations (19) and (20)


V (z) =



λ(1 + pµ) µ(1 − λpd) + pλµ ; ¯ − λz) + p b(λ ¯ − λz)(1 − e−λd(1−z) ) z − b(λ

¯ − λz)(z − 1) 1 − λp b(λ




λ(1 + pµ) µ(1 − λpd) + pλµ . W (z) = ¯ − λz) + p b(λ ¯ − λz)(1 − e−λd(1−z) ) z − b(λ ¯ − λz) − 1] 1 − [b(λ

(25)

(26)

90 Let Pq = V (z) + W (z) denote the probability generating function of the queue length irrespective of whether the server is away on vacation or is available in the system. Then, adding (25) and (26) and simplifying, we have


Pq (z) =



λ(1+ pµ) µ(1 − λpd)+ pλµ . (27) ¯ ¯ z − b(λ − λz)+ p b(λ − λz)(1 − e−λd(1−z) )

¯ − λz)(λpz − λp+1) − 1] 1 − [b(λ

We note that in the case when there are no server vacations, we let p = 0 in equation (27) and have




¯ − λz) − 1] 1 − λ [b(λ µ . (28) Pq (z) = ¯ − λz) z − b(λ The result in equation (28) is a known result for the M/G/1 queue [see Kashyap & Chaudhry (1988), equation (26), page 58]. Further, let P(z) denote the probability generating function of the number in the system. Then we have from equations (23) and (27) P(z) = Q + z Pq (z) =

¯ − λz)[z(λpz − λp + 1) + p(1 − e−λd(1−z) ) − 1] b(λ · ¯ − λz) + p b(λ ¯ − λz)(1 − e−λd(1−z) ) z − b(λ


· 1−



λ(1 + pµ) . µ(1 − λpd) + pλµ

(29)

We note that the results found above in equations (27) and (29) are the new and more general results. And in the particular case when there are no server vacations, we let p = 0 in equation (29) and get 



¯ − λz) 1 − λ (1 − z)b(λ µ . (30) P(z) = ¯ − λz) − z b(λ Again the result in equation (30) is the well-known PollaczecKhinchine formula [see Medhi (1982), equation 6.7, page 313 or Choi & Park (1990), page 225]. It may be noted that corresponding results for many particular cases of interest such as M/Ek /D/1, M/M/D/1 and M/D/D/1, etc. can be derived from the main results found in equations (27) and (29) by ¯ − λz). substituting the appropriate values of b(λ

91 5. The expected number in the queue and in the system Let L q denote the expected number of customers in the queue. d Pq (z) at z = 1. Since Then, we have from equation (27), L q = dz (z) Pq (z) is indeterminate of the 0/0 form at z = 1, we let Pq (z) = ND(z) , where N (z) and D(z) are respectively the numerator and denominator of the right hand side of equation (27). Then we use the following well-known result in queueing theory [see Madan (1991)] L q = Lim z→1

d D  (z)N  (z) − N  (z)D  (z) Pq (z) = P  (1) = Lim z→1 dz 2[D  (z)]2

(31)

D  (1)N  (1) − N  (1)D  (1) = 2[D  (1)]2

where primes stand for the derivatives with respect to z. We carry out the desired derivatives of the right hand side of equation (27) at z = 1, using the fact that −b¯  (0) = µ1 and b¯  (0) = E(v 2 ), the second moment of the service time. After some algebra and simplification, we have


N  (1) = λp + 




N (1) = λ

2

D  (1) = 1 −

λ µ




1−

2p E(v ) + µ

λ(1 + pµ) µ(1 − λpd) + λpµ




λ(1 + pµ) 1− µ(1 − λpd) + λpµ

2

λ − λpd µ


D  (1) = −λ2 E(v 2 ) +







2 pd + pd 2 . µ

Substituting the above values in equation (31) and simplifying we finally have

Lq =



λ2 E(v 2 )(1+λp−λpd)+ 2µp 1−




· 1−

2 1− λ(1 + λpµ) µ(1 − λpd) + pλµ



λ µ

λ µ



− λpd

+λp 2 d 2 +

2

λpd 2 µ

+

2λpd µ2

· (32)

92 Again, (32) is a new result. We note that the expected number in the system is given by L = L q + ρ, where ρ has already been found in equation (24). 6. The expected waiting time in the queue and in the system The expected waiting time in the queue is given by Lq λ

λ E(v 2 )(1+λp−λpd)+

Wq = =






µ(1 + pµ) µ(1 − λpd) + pλµ





1−

λ µ

2 1−

· 1−



2p µ

λ µ

− λpd

+λp 2 d 2 +

2

λpd 2 µ

+

2λpd µ2

(33)

and the expected waiting time in the system is given by W =

L q +ρ . λ

7. Special cases We will only find L q for the following special cases. The other parameters such as Wq , L and W can be easily obtained similarly. I.

M/Ek /D/1 Queue If the service times are assumed to be k-Erlangian, then we have E(v 2 ) =

 0

∞

k+1 v 2 (kµ)k v k−1 ekµv dv = (k − 1)! kµ2

(34)

Using (34) in equation (32), we have

Lq =

λ2





k+1 kµ2




· 1−

(1+λp−λpd)+

2p µ

2 1− λ(1 + pµ) µ(1 − λpd) + pλµ





λ µ

1−



λ µ

− λpd

+λp 2 d 2 +

2

λpd 2 µ

+

2λpd µ2

(35)

93 II.

M/M/D/1 Queue

In this case the service times have an exponential distribution. Therefore, with k = 1 in (34) gives E(v 2 ) = µ22 and substituting this value of E(v 2 ) in equation (32) or k = 1 in equation (35), we obtain λ2

Lq =



2(1+λp−λpd) µ2




· 1− III.

+

2p µ





λ µ

1−

2 1−

λ µ

λ(1 + pµ) µ(1 − λpd) + pλµ



+ λp 2 d 2 +

− λpd

2

λpd 2 µ

+

2λpd µ2



(36)

M/D/D/1 Queue

If we assume that service time have a constant duration c (> 0), then σv2 = 0 and consequently, E(v 2 ) = σv2 + c2 = c2 and with this value of E(v 2 ), equation (32) yields

λ2 c2 (1+ λp − λpd)+

Lq =



2p µ

λ(1 + pµ) · 1− µ(1 − λpd) + pλµ



1−

2 1−




IV.





λ µ

λ µ

+ λp 2 d 2 +

− λpd

2

λpd 2 µ

+

2λpd µ2

(37)

Compulsory Server Vacations

If we let p = 1, this essentially means that after completion of each service, the server always takes a vacation of constant duration d. Thus with p = 1, we have from equation (32)

Lq =

λ2 E(v 2 )(1 + λ − λd) +




2 µ

2 1−

λ(1 + λµ) · 1− µ(1 − λd) + λµ



 λ µ

1−

λ µ

− λd



2

+ λd 2 +

λd 2 µ

+

2λd µ2

(38)

We note that all results derived above in the particular cases I to IV are also new. But the following particular case is a known result:

94 V.

No Server Vacations (M/G/1 Queue) In this case, we let p = 0 in (32) and obtain Lq =

λ2 E(v 2 )
. λ 2 1− µ

(39)

We note that the result in equation (39) agrees with a known result (see Kashyap and Chaudhry (1988), equation 34, page 59). Acknowledgment The author wishes to express his gratitude and sincere thanks to the referee for his valuable suggestions to revise the paper in the present form.

REFERENCES Choi, B. D. and Park, K.K. (1990) The M/G/1 retrial queue with Bernoulli schedule, Queueing Systems, 7, 219-228. Cramer, M. (1989) Stationary distributions in a queueing system with vacation times and limited service, Queueing Systems Theory and Applications, vol. 4, no. 1, 57-68. Doshi, B. T. (1986) Queueing systems with vacations-a survey, Queueing Systems, 1, 29-66. Fuhrman, S. (1981) A note on the M/G/1 queue with server vacations, Oper. Res., 31, 1368. Gaver, D. P. (1959) Imbedded Markov chain analysis of a waiting line process in continuous time, Ann. Math. Stat., 30, 698-720. Kashyap, B. R. K. and Chaudhry, M.L. (1988) An introduction to queueing theory, A & A Publications, Kingston, Ont. Canada. Keilson, J. and Servi, N.D. (1986) Oscillating random walk models for GI/G/1 vacation systems with Bernoulli schedules, Journal of Applied Probability, 23, 790-802. Levy, Y. and Yechilai, U. (1976) An M/M/s queue with server vacations, Infor, 14, no. 2, 153-163. Madan, K. C. (1991) On a queueing system with general vacation times, Internat. J. Man. & Inform. Sc., vol. 1, 51-60. Madan, K. C. (1992) An M/G/1 queueing system with compulsory server vacations, Trabjos de Investigacion Operativa, J. Man. & Inform. Sc., vol. 7, no. 1, 105-115. Medhi, J. (1982) Stochastic Processes, Wiley Eastern. Shanthikumar, J.G. (1988) On stochastic decomposition in the M/G/1 type queues with generalised vacations, Operations Research, 36, 566-569.

95 An M/G/1 queue with optional deterministic server vacations Summary We investigate the steady state behaviour of a single server vacation queue with Poisson arrivals and arbitrary (general) service times. At the completion of each service, the server may take a vacation of a fixed (constant) duration with probability p or may continue to be available in the system for next service with probability 1 − p. The supplementary variable technique is employed to find explicitly the probability generating function of the number in the system and the mean number in the system. Some particular cases of interest are discussed and some known results are derived as special cases.

Una fila d’attesa M/G/1 con assenze opzionali e deterministiche del server Riassunto Si studia il comportamento dell’assenza di un singolo server con arrivi poissoniani e tempi di servizio arbitrari (generali). A compimento di ogni servizio, il server pu`o assentarsi per un periodo di tempo costante con probabilit`a p o pu`o essere ancora disponibile per il successivo servizio con probabilit`a 1 − p. La tecnica della variabile supplementare viene impiegata per trovare la forma esplicita della funzione generatrice dei momenti del numero nel sistema e del numero medio nel sistema. Alcuni casi particolari di interesse vengono discussi ed alcuni risultati noti vengono derivati come casi particolari.

Key words Single server; General service times; Optional deterministic server vacations; Steady state.

[Manuscript received July 1998; final version received March 1999.]