On the Mx/G/1 queue with feedback and optional server vacations based on a single vacation policy

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Abstract

We study a single server queue with batch arrivals and general (arbitrary) service time distribution. The server provides service to customers, one by one, on a first come, first served basis. Just after completion of his service, a customer may leave the system or may opt to repeat his service, in which case this customer rejoins the queue. Further, just after completion of a customer's service the server may take a vacation of random length or may opt to continue staying in the system to serve the next customer. We obtain steady state results in explicit and closed form in terms of the probability generating functions for the number of customers in the queue, the average number of customers and the average waiting time in the queue. Some special cases of interest are discussed and some known results have been derived. A numerical illustration is provided.

Introduction

The Mx|G|1 queue has been studied by numerous authors including Medhi [13], Scholl and Kleinrock [14], Bhat [1], Trivedi [18], Gross and Harris [7], Doshi [4], Kashyap and Chaudhry [8], Shanthikumar [16], Choi and Park [3] and Madan [11], [12], to mention a few. In this paper, we consider an extended Mx|G|1 queue with feedback and optional server vacations and designate this queue as Mx|G|G|1 feedback queue. It is assumed that the system receives bulk input in batches of variable size and that there is a single server who provides one by one service to customers on a first come, first service basis. We assume that just after completion of his service a customer either leaves the system or he may demand re-service, in which case this customer has to rejoin the queue. In queuing literature, one finds some types of queuing networks including some cyclic queuing systems with feedback. For example, see Finch [5], Scharge [15], Simon [17], Glenbe and Pujolle [6] and Madan [9], [10]. In most of such systems, a fraction of customers may re-enter the system by joining the same queue again or the queue in front of another server. However, this kind of feedback has been considered mainly in papers dealing with multi-server serial queuing networks. In our present model, we introduce feedback in a vacation queue. One may encounter many such queuing situations where a re-service may be necessary. If we are dealing with a human customer, then on completion of a service, such a customer may find his service unsatisfactory and consequently may demand a re-service, or analogous to a human customer, if we are dealing with production units then, on completion of a service, a quick on-the-spot inspection of such a unit may reveal that the produced unit is defective and consequently a re-service may be necessary. The following assumptions briefly describe the mathematical model of our problem.

Section snippets

Model description and assumptions

Customers arrive at the system in batches of variable size in a compound Poisson process. Let λcidt (i=1, 2, 3, …) be the first order probability that a batch of i units (customers) arrives at the system during a short interval of time (t,t+dt], where 0⩽ci⩽1 and ∑i=1ci=1 and λ>0 is the mean arrival rate of batches. Customers are provided one by one service on a “first come, first served” basis and their service time S follows a general (arbitrary) distribution with distribution function G(s)

Definitions, notations and equations governing the system

We define Wn(x,t) to be probability that at time t, there are n⩾1 customers in the system including one customer in service and the elapsed service time of this customer is x. Accordingly, Wn(t)=∫0Wn(x,t)dx denotes the probability that there are n⩾1 customers in the queue including one customer in service irrespective of the value of x. Next, we define Vn(x,t) to be probability that at time t, there are n⩾0 customers in the queue and the server is on vacation with elapsed vacation time x.

Queue size distribution at a random epoch

We integrate Eqs. , with respect to x and obtainWq(x,z)=Wq(0,z)[1−μ(x)]exp[−λ(1−C(z))x],Vq(x,z)=Vq(0,z)[1−β(x)]exp[−λ(1−C(z))x].

Next, we multiply both sides of Eq. (3.7) zn+1, take summation over n from 1 to ∞ and simplify the result using , , . Thus we getzWq(0,z)=(1−r)(1−p)Wq(0,z)G*[λ−λC(z)]+r(1−p)zWq(0,z)G*[λ−λC(z)]−(1−r)(1−p)z∫0W1(x)μ(x)dx−z∫0V0(x)β(x)dx+λzC(z)Q+zVq(0,z)B*[λ−λC(z)],where G*[λ−λC(z)]=∫0exp(−[λ−λC(z)]x)dG(x) and B*[λ−λC(z)]=∫0exp(−[λ−λC(z)]x)dB(x) are the

The average queue size and the average waiting time

Let Lq denote the mean queue size under the steady state. Then, we have from (4.10), Lq=ddzPq(z)|z=1. Next, we write Pq(z) in (4.10) as Pq(z)=N(z)D(z), where N(z) and D(z) are respectively the numerator and denominator of the right side of (4.10). ThenLq=limz→1ddzPq(z)=limz→1D(z)N(z)−N(z)D(z)2(D(z))2=D(1)N(1)−N(1)D(1)2(D(1))2,where primes and double primes in (5.1) denote first and second derivatives at z=1, respectively. Then, carrying out the derivatives and after some algebra we get

Special cases

Case 1

Single Poisson arrivals

When customers arrive at the system one by one, then c1=1 and ci=0 for i≠1. Consequently, C(z)=z, E(I)=1 and E(I(I−1))=0. Hence, using these substitutions in the main results of 4 Queue size distribution at a random epoch, 5 The average queue size and the average waiting time, we obtainPq(z)=(1−r)(z−1)G*[λ−λz]Qz−(1−r+rz)(1−p+pB*[λ−λz])G*[λ−λz],Q=1−r−λE(S)−pλE(V)1−r,ρ=λE(S)+pλE(V)1−r,r≠1,r+λE(S)+pλE(V)<1.

Further, Lq and Wq are given by , , respectively. Thus, we haveN

A numerical illustration

For the purpose of a numerical illustration, we choose the following arbitrary values: E(S)=0.02, E(S2)=0.015, E(V)=0.02, E(V2)=0.05 and λ=2. Also, we vary values of r, p to be 0.25, 0.50 and 0.75 such that the steady state condition (4.13) is always satisfied. Table 1, Table 2, Table 3 below give the computed values of various states of the server, the proportion of idle time, the utilization factor, the mean queue size, and customers' mean waiting time in the queue.

The above Table 1, Table 2

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