REJECTION PROBABILITY IN THE FINITE SIZE ...

REJECTION PROBABILITY IN THE FINITE SIZE SINGLE SERVER QUEUES AHMAD AL HANBALI SCHOOL OF MANAGEMENT AND GOVERNANCE, DEP. INDUSTRIAL ENGINEERING AND BUSINESS INFORMATION SYSTEMS, UNIVERSITY OF TWENTE, P.O. BOX 217, 7500 AE ENSCHEDE, THE NETHERLANDS

Abstract. In this short non-peer reviewed note, we extend the rejection probability of the batch arrival M/G/1/K+1 queue in [5, Sec 9.8.3] by considering the case where the load is larger than one and studying G/M/1/K+1 queue. Rejection probability; M/G/1/K+1; Batch arrivals; G/M/1/K+1;

1. Rejection probability in MX /G/1/K+1 We consider the batch arrival M/G/1/K+1 queue, referred to as MX /G/1/K+1. The total capacity of the queue is K + 1 including the customer in service, if any. The arrival process is a Poisson batch process with a rate λ and a batch size X. The batch size has a discrete probability {βj , j ≥ 1} with mean β and probability generating function β(z), |z| ≤ 1. The service time is generally distributed with LST φ(s), Re(s) ≥ 0, and mean µ. We assume that the partial rejection policy is applied. This is means that an arriving batch whose size exceeds the remaining capacity of the queueing buffer is partially rejected by turning away only those customers in excess of the remaining queueing buffer. Let πj∞ denote the long-run fraction of customers who have j other customers in front of them just after arrival in the MX /G/1/∞ queue (including customers from the same batch). Let us define ρ := λβµ. Let Prej denote the probability that an arriving customer is rejected in MX /G/1/K+1 with partial rejection. It is well known that, see, e.g., [5, Sec 9.8.3] and the references therein, P ∞ (1 − ρ) ∞ j=K+1 πj P Prej = , ρ < 1. (1) ∞ 1−ρ ∞ j=K+1 πj We note that this formula was first discovered in [2]. We note that this formula was also generalized for bulk-service queues in [6]. In all these studies it is assumed that ρ < 1. In the following we shall derive a new formula for the rejection probability that is valid for all values of ρ > 0. The advantage of this new formula is that it gives Prej as function of LST of the service time and the p.g.f. of the batch size and it holds for all values of ρ > 0. 1

REJECTION PROBABILITY IN THE FINITE SIZE SINGLE SERVER QUEUES

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Before reporting our main result, let us introduce some notations. Let E[S] denote the mean number of customers served in a busy period in the MX /G/1/K+1 queue. Let E[L] denote the mean number of rejected customers in a busy period in the queue. Let r denote the root with the smallest absolute value of x − φ λ(1 − β(x)) = 0. (2) Let Dη denote the circle with center at the origin and with radius 0 < η < |r|. It is then known that E[S] gives, see [3], Z 1 (1 − β(x))dx + P(X ≥ K + 1). E[S] = K 2πi Dη x (1 − x) φ λ(1 − β(x)) − x

(3)

where i2 = −1. Moreover, E[L] reads E[L] = β + (ρ − 1)E[S].

(4)

We are now ready to present our result. Lemma 1. The probability that an arriving customer is rejected in MX /G/1/K+1 with partial rejection is given by Prej = 1 −

E[S] , β + ρE[S]

ρ > 0.

(5)

where E[S] is given in (3). Proof. Note that MX /G/1/K+1 is stable for all values of ρ > 0. Therefore, the busy period length of the queue is a proper random variable with finite mean. From the PASTA property we have that the probability that a customer is rejected upon arrival equal to the long-run fraction of customers lost. Applying the renewal reward theorem on the busy period we know that the long-run fraction of customers lost reads E[L] Prej = . E[S] + E[L] Inserting E[S] and E[L] in the latter equation gives the desired result. 2. G/M/1/K+1 Let ψ(s) denote the LST of the inter-arrival times in the G/M/1/K+1 queue. Let o denote the root with the smallest absolute value of (solving according to x) x − ψ s + p + µ(1 − ux) = 0. Let Dη denote the circle with center at the origin and with radius 0 < η < |o|. For the G/M/1/K+1 queue we do not have a closed-form formula of the expected number of customers served during the busy period as in the case of the batch arrival M/G/1/K+1 queue. However, Rosenlund [4] derived a closed-form expression of the joint transform of the busy period B, the idle period I following the busy period,


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and the number of customers served S and the number of losses L during the busy period as follows, see [4], R µu(1−vz) ψ(w)−ψ(s+p+µ(1−uz)) dz 1 2πi Dη z K+1 (s+µ(1−uz)) ψ(s+p+µ(1−uz))−z −sB−wI S L E e u v = , K ≥ 1. (6) R (1−vz)dz 1 2πi D η

z K+1 ψ(s+p+µ(1−uz))−z

Substituting s = 0, w = 0 and u = 1 into the latter equation gives E v L . Taking the limit of (E v L − 1)/(v − 1) when v → 1 gives that E[L] =

1 1 2πi

(1−z)dz Dη z K+1 ψ(µ(1−z))−z

R

.

We derive now E S . Substituting s = 0, w = 0 and v = 1 into (6) gives that R (1−z) 1−ψ(µ(1−z)) dz 1 2πi Dη z K+1 (1−uz) ψ(µ(1−uz))−z S−1 . E u = R (1−z)dz 1 2πi D η

z K+1 ψ(µ(1−uz))−z

Taking the limit of (E[uS−1 ] − 1)/(u − 1) when u → 1 yields that R dz −2πi + Dη K z ψ(µ(1−z))−z E[S] = 1 + R (1−z)dz D η

z K+1 ψ(µ(1−z))−z

1 = E[L] − 1 + 2πi

Z Dη

z K+1

dz . ψ(µ(1 − z)) − z

(7)

Lemma 2. The long-run fraction of customers lost in G/M/1/K+1 is given by 1 ∞ πrej = 1 R , ρ > 0, K ≥ 1. (8) dz 2πi D η

z K+1 ψ(µ(1−z))−z

Proof. The proof follows right away from the renewal reward theorem, which gives ∞ that πrej = E[L]/(E[S] + E[L]). 3. Discussion First, note that one is a root (2). Second, by Tak´acs’ lemma, see, e.g., [1, Appendix], we have that r, the root with the smallest absolute value in (2), satisfies the following property: r = 1 for ρ ≤ 1, and 0 < r < 1 for ρ > 1. Taking advantage of the latter property the contour integration, Int, in (3) can be transformed to the definite integration as follows: Z 2π 1 (1 − β(ηeiθ ))dθ , Int = 2π(η)K−1 0 eiθ(K−1) (1 − ηeiθ ) φ λ(1 − β(ηeiθ )) − ηeiθ


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where 0 < η < 1 for ρ ≤ 1 and 0 < η < r < 1 for ρ > 1. Note that for ρ > 1 it is sufficient to just have a lower bound of the root r. Moreover, since r is real a lower bound can be simply derived using the dichotomic search algorithm. Advantages of the new formula in (8) compared to (1): (1) The new formula is valid for all values of ρ > 0. As opposed to Tijms’ Formula that it is restricted to ρ < 1. (2) It is not necessary to compute the state probability πj∞ , j = 0, 1, . . ., in order to find Prej . However, we just need to compute a definite integration for which to solve it there are many numerical tools available. 4. Possible extensions • The result can be extended to the renewal batch arrival queue with exponential service. • The result can be extended to complete rejection policy. • What is the heavy traffic asymptotic of the loss probability? How to define heavy traffic regim for finite queue? References [1] J. W. Cohen. The single server queue. North-Holland, 1982. [2] J. Keilson and L. Servi. Blocking probability for M/G/1 vacation systems with occupancy level dependent schedules. Operations Research, 37(1):134–140, 1989. [3] S. I. Rosenlund. Busy periods in time-dependent M/G/1 queue. Adv. Appl. Prob., 8:195–208, 1976. [4] S. I. Rosenlund. The queue G/M/m/N: busy period and cycle, waiting time under reverse and random order service. Naval Research Logistics Quarterly, 25(1):107–119, 1978. [5] H. Tijms. A first course in stochastic models. John Wiley & Sons Inc, 2003. [6] J. Van Ommeren. Loss probabilities in batch-arrival bulk-service queues with finite capacity. Stochastic Models, 13(2):371–379, 1997. E-mail address: {A.ALHANBALI}@UTWENTE.NL