Journal of Probability and Statistics

Journal of Probability and Statistics / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 812726 | 11 pages | https://doi.org/10.1155/2011/812726

On a Batch Arrival Queuing System Equipped with a Stand-by Server during Vacation Periods or the Repairs Times of the Main Server

Academic Editor: Rongling Wu
Received25 Jan 2011
Accepted20 May 2011
Published14 Jul 2011

Abstract

We study a queuing system which is equipped with a stand-by server in addition to the main server. The stand-by server provides service to customers only during the period of absence of the main server when either the main server is on a vacation or it is in the state of repairs due to a sudden failure from time to time. The service times, vacation times, and repair times are assumed to follow general arbitrary distributions while the stand-by service times follow exponential distribution. Supplementary variables technique has been used to 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 while the MathCad software has been used to illustrate the numerical results in this work.

1. Introduction

Due to their wide applications in flexible manufacturing or computer communication systems, 𝑀[𝑋]/𝐺/1 queueing system with vacations and 𝑀[𝑋]/𝐺/1 queueing system with breakdowns have been studied by several authors including [1–8]. Recently the authors of [9] have studied some queueing systems with vacations and breakdowns.

In this work, we study an 𝑀[𝑋]/𝐺/1 queueing system with Bernoulli schedule vacations and random breakdowns with an additional significant assumption that the system deploys a stand-by server during the vacation periods and the repair periods of the main server.

Madan [10] studied the steady state behavior of a queuing system with a stand-by server which provides service to costumers only during the repair period. In that work, repair times were assumed to follow an exponential distribution. The present paper considers both vacations and breakdowns with additional assumptions of deployment of a standby during the vacation periods and repair periods. We generalize results obtained not only by Madan [10] but also the results obtained by Maraghi et al. [9]. Most importantly, we assume that the service times, vacation times, and repair times have different general (arbitrary) distributions while the stand-by service times follow exponential distribution. Out of five distributions in this model we assume that four are a generally distributed, this is very important because all the other distributions such as exponential, deterministic, and Erlang-k distributions will be included.

The rest of this paper is arranged as follow. Section 2 gives the assumptions underlying the considered queueing system. Related definitions and used notations are given in Section 3. Equations governing the system are formulated in Section 4. In Section 5, we give the solution of the equations formulated in the previous section to find the queue size distribution at a random epoch. The average queue size and the average waiting time are given in Section 6. In Section 7, we consider a numerical example to illustrate application of our results.

2. Assumptions

Customers arrive at the system in batches of variable size in a compound Poisson process. Let 𝜆𝑐𝑖Δ𝑡(𝑖=1,2,3,…) be the first order probability that a batch of i customers arrives at the system during a short interval of time (𝑡,𝑡+Δ𝑡), where 0≤𝑐𝑖≤1 and âˆ‘âˆžğ‘–=1𝑐𝑖=1 and 𝜆>0 is the mean arrival rate of batches. The customers provided service one by one on a “first come- first served basis.”

The service times of the main server follow a general (arbitrary) distribution with distribution function 𝐺(𝑠) and density function 𝑔(𝑠). Let 𝜇(𝑥)Δ𝑥 be the conditional probability density of service completion during the interval (𝑥,𝑥+Δ𝑥], given that the elapsed service time is 𝑥, so that 𝜇(𝑥)=𝑔(𝑥)1−𝐺(𝑥),(2.1) and, therefore 𝑔(𝑠)=𝜇(𝑠)𝑒−∫𝑠0𝜇(𝑥)𝑑𝑥.(2.2)

On completion of a service, the server may take a vacation of random length with probability 𝑃, or may stay in the system providing service with probability 1−𝑃, where 0≤𝑃≤1.

The server’s vacation times follow a general (arbitrary) distribution with distribution function 𝐵(𝑣) and density function 𝑏(𝑣). Let 𝛽(𝑥)Δ𝑥 be the conditional probability of a completion of a vacation during the interval (𝑥,𝑥+𝑑𝑥) given that the elapsed vacation time is 𝑥, so that 𝛽(𝑥)=𝑏(𝑥)1−𝐵(𝑥),(2.3) and, therefore 𝑏(𝑣)=𝛽(𝑣)𝑒−∫𝑣0𝛽(𝑥)𝑑𝑥.(2.4)

The system may break down at random, and breakdowns are assumed to occur according to a Poisson stream with mean breakdown rate 𝛼>0. Further we assume that once the system breaks down, the customer whose service is interrupted comes back to the head of the queue but it is instantly taken up for service by the stand-by server.

Once the system breaks down, its repairs start immediately and the duration of repairs follows a general (arbitrary) distribution with distribution function 𝐹(𝑟) and density function 𝑓(𝑟). Let 𝛾(𝑥)Δ𝑥 be the conditional probability density of repair completion during the interval (𝑥,𝑥+Δ𝑥], given that the elapsed repair time is 𝑥, so that 𝛾(𝑥)=𝑓(𝑥)1−𝐹(𝑥),(2.5) and, therefore 𝑓(𝑟)=𝛾(𝑟)𝑒−∫𝑟0𝛾(𝑥)𝑑𝑥.(2.6)

The stand-by server starts serving the customers as soon as the main server breaks down or as soon as the main server leaves for a vacation after completing a service. The stand-by service times follow an exponential distribution with stand-by service rate 𝛿>0 and mean stand-by service time 1/𝛿.

We further assume that the main server joins the system immediately after the completion of its vacation or completion of its repairs, and the customer being served by the stand-by server is immediately transferred to the main server to start a service afresh.

All stochastic processes involved in the system are independent of each other.

3. Notations

We let(i)𝑃𝑛(𝑡,𝑥): probability that at time 𝑡, there are 𝑛≥0 customers in the queue excluding one customer in the service served by the main server, and the elapsed service time of this customer is 𝑥. Accordingly, 𝑃𝑛∫(𝑡)=∞0𝑃𝑛(𝑡,𝑥)𝑑𝑥 denotes the probability that there are 𝑛≥1 customers in the queue excluding one customer in service irrespective of the value of 𝑥;(ii)𝑉𝑛(𝑡,𝑥): probability that at time 𝑡, there are 𝑛≥0 customers in the queue (and one customer is being served by the stand-by server), and the main server is on vacation with elapsed vacation time 𝑥. Accordingly, 𝑉𝑛∫(𝑡)=∞0𝑉𝑛(𝑡,𝑥)𝑑𝑥 denotes the probability that at time 𝑡, there are 𝑛≥0 customers in the queue and the server is on vacation irrespective of the value of 𝑥. As soon as the vacation starts the stand-by server starts serving the customers in the system;(iii)𝑅𝑛(𝑡,𝑥): Probability that at time 𝑡, there are 𝑛 (𝑛≥0) customers in the queue (and one customer is being served by the stand-by server) while the system is under repair with elapsed repair time 𝑥. Accordingly, 𝑅𝑛∫(𝑡)=∞0𝑅𝑛(𝑡,𝑥)𝑑𝑥 denotes the probability that at time 𝑡, there are 𝑛≥0 customers in the queue and the server is under repair irrespective of the value of 𝑥;(iv)𝑄(𝑡): probability that at time 𝑡, there are no customers in the system and the server is idle but available in the system.

Assuming that the steady state exists, we let limğ‘¡â†’âˆžğ´ğ‘›(𝑡,𝑥)=𝐴𝑛(𝑥),limğ‘¡â†’âˆžğ´ğ‘›(𝑡)=limğ‘¡â†’âˆžî€œâˆž0𝐴𝑛(𝑡,𝑥)𝑑𝑥=𝐴𝑛,limğ‘¡â†’âˆžğ‘‘ğ´ğ‘›(𝑡)𝑑𝑡=0,where𝐴=𝑃,𝑉,𝑅,limğ‘¡â†’âˆžğ‘„(𝑡)=𝑄.(3.1)

4. Equations Governing the System

According to the assumptions mentioned above, we have four possible states of our system during a short time interval (𝑡,𝑡+Δ𝑡): the first state is that the main server is providing service, the second is that the main server is on vacation and the stand-by server is providing service, the third state is that the main server is inactive due to a system breakdown and is under repair and the stand-by server is providing service, the last possible state is that the server is idle (there are no customers in the system) but available in the system. By discussing the probabilities of every situation and finding the limit as Δ𝑡→0, we obtain the following set of differential-difference equations 𝜕𝑃𝜕𝑥𝑛(𝑥)=−(𝜆+𝜇(𝑥)+𝛼)𝑃𝑛(𝑥)+𝜆𝑛−1𝑖=1𝑐𝑖𝑃𝑛−𝑖𝜕(𝑥),𝑛≥1,(4.1)𝑃𝜕𝑥0(𝑥)=−(𝜆+𝜇(𝑥)+𝛼)𝑃0𝜕(𝑥),(4.2)𝑉𝜕𝑥𝑛(𝑥)=−(𝜆+𝛽(𝑥)+𝛿)𝑉𝑛(𝑥)+𝜆𝑛𝑖=1𝑐𝑖𝑉𝑛−𝑖(𝑥)+𝛿𝑉𝑛+1𝜕(𝑥),𝑛≥1,(4.3)𝑉𝜕𝑥0(𝑥)=−(𝜆+𝛽(𝑥)+𝛿)𝑉0(𝑥)+𝛿𝑉1(𝑥),(4.4)𝜕𝑅𝑛(𝑥)𝜕𝑥=−(𝜆+𝛾(𝑥)+𝛿)𝑅𝑛(𝑥)+𝜆𝑛𝑖=1𝑐𝑖𝑅𝑛−𝑖(𝑥)+𝛿𝑅𝑛+1(𝑥),𝑛≥1,(4.5)𝜕𝑅0(𝑥)𝜕𝑥=−(𝜆+𝛾(𝑥)+𝛿)𝑅0(𝑥)+𝛿𝑅1(𝑥),(4.6)𝜆𝑄=∞0𝑅0(𝑥)𝛾(𝑥)𝑑𝑥+(1−𝑝)∞0𝑃0(𝑥)𝜇(𝑥)𝑑𝑥+(1−𝑟)∞0𝑉0(𝑥)𝛽(𝑥)𝑑𝑥.(4.7) The following boundary conditions will be used to solve the above equations: 𝑃𝑛(0)=(1−𝑝)∞0𝑃𝑛+1(𝑥)𝜇(𝑥)𝑑𝑥+∞0𝑉𝑛+1+(𝑥)𝛽(𝑥)ğ‘‘ğ‘¥âˆž0𝑅𝑛+1(𝑥)𝛾(𝑥)𝑑𝑥+𝜆𝑐𝑛+1𝑉𝑄,𝑛≥0,(4.8)𝑛(0)=ğ‘âˆž0𝑃𝑛𝑅(𝑥)𝜇(𝑥)𝑑𝑥,𝑛≥0,(4.9)𝑛(0)=ğ›¼âˆž0𝑃𝑛−1𝑅(𝑥)𝑑𝑥,𝑛≥1,(4.10)0(0)=0.(4.11)

5. Queue Size Distribution at a Random Epoch

Defining the following probability generating functions ğ´ğ‘ž(𝑥,𝑧)=âˆžî“ğ‘›=0𝑧𝑛𝐴𝑛(𝑥),ğ´ğ‘ž(𝑧)=âˆžî“ğ‘›=0𝑧𝑛𝐴𝑛,𝐶𝐴=𝑃,𝑉,𝑅,(𝑧)=âˆžî“ğ‘–=1𝑧𝑖𝑐𝑖,(5.1) we multiply (4.1) by 𝑧𝑛, take summation over 𝑛 from 1 to ∞, adding to (4.2) then by simplifying and using (5.1) we get ğœ•ğ‘ƒğœ•ğ‘¥ğ‘ž(𝑥,𝑧)+(𝜆−𝜆𝐶(𝑧)+𝜇(𝑥)+𝛼)ğ‘ƒğ‘ž(𝑥,𝑧)=0.(5.2) Using the same process, from (4.3), (4.4) and (4.5), (4.6) we get, respectively, ğœ•ğ‘‰ğœ•ğ‘¥ğ‘žî‚€ğ›¿(𝑥,𝑧)+𝜆−𝜆𝐶(𝑧)+𝛽(𝑥)+ğ›¿âˆ’ğ‘§î‚ğ‘‰ğ‘žğœ•(𝑥,𝑧)=0,(5.3)ğ‘…ğœ•ğ‘¥ğ‘žî‚€ğ›¿(𝑥,𝑧)+𝜆−𝜆𝐶(𝑧)+𝛾(𝑥)+ğ›¿âˆ’ğ‘§î‚ğ‘…ğ‘ž(𝑥,𝑧)=0.(5.4) Multiply (4.8) by 𝑧𝑛+1, sum over 𝑛 from 0 to ∞, and use the generating functions defined in (5.1), we get ğ‘§ğ‘ƒğ‘žî€œ(0,𝑧)=(1−𝑝)∞0ğ‘ƒğ‘žî€œ(𝑥,𝑧)𝜇(𝑥)𝑑𝑥+∞0ğ‘‰ğ‘žî€œ(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+∞0ğ‘…ğ‘žâˆ’î‚¸î€œ(𝑥,𝑧)𝛾(𝑥)𝑑𝑥+𝜆𝐶(𝑧)𝑄(1−𝑝)∞0𝑃0(𝑥)𝜇(𝑥)𝑑𝑥+(1−𝑟)∞0𝑉0(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+∞0𝑅0.(𝑥,𝑧)𝛾(𝑥)𝑑𝑥(5.5) From (4.7), we have ğ‘§ğ‘ƒğ‘žî€œ(0,𝑧)=(1−𝑝)∞0ğ‘ƒğ‘žî€œ(𝑥,𝑧)𝜇(𝑥)𝑑𝑥+∞0ğ‘‰ğ‘žî€œ(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+∞0ğ‘…ğ‘ž(𝑥,𝑧)𝛾(𝑥)𝑑𝑥+𝜆𝑄(𝐶(𝑧)−1).(5.6) Multiply (4.9) by 𝑧𝑛 and sum over 𝑛 from 0 to ∞, we get ğ‘‰ğ‘žî€œ(0,𝑧)=ğ‘âˆž0ğ‘ƒğ‘ž(𝑥,𝑧)𝜇(𝑥)𝑑𝑥.(5.7) Similarly, from (4.10) we get ğ‘…ğ‘žî€œ(0,𝑧)=ğ›¼ğ‘§âˆž0ğ‘ƒğ‘ž(𝑥,𝑧)𝑑𝑥=ğ›¼ğ‘§ğ‘ƒğ‘ž(𝑧),𝑛≥0.(5.8) Integrating (5.2) from 0 to 𝑥 yields ğ‘ƒğ‘ž(𝑥,𝑧)=ğ‘ƒğ‘ž(0,𝑧)𝑒∫−(𝜆−𝜆𝐶(𝑧)+𝛼)𝑥−𝑥0𝜇(𝑡)𝑑𝑡,(5.9) where ğ‘ƒğ‘ž(0,𝑧) is given by (5.6). Let us consider ğ‘Ž=𝜆−𝜆𝐶(𝑧)+𝛼.

Integrating equation (5.9) by parts with respect to 𝑥 yields ğ‘ƒğ‘ž(𝑧)=ğ‘ƒğ‘žî‚µ(0,𝑧)1−𝐺∗(ğ‘Ž)ğ‘Žî‚¶,(5.10) where 𝐺∗∫(ğ‘Ž)=∞0ğ‘’âˆ’ğ‘Žğ‘¥ğ‘‘ğº(𝑥) is the Laplace-Stieltjes transform of the service time 𝐺(𝑥).

Now multiplying both sides of (5.9) by 𝜇(𝑥) and integrating over 𝑥 we get ∞0ğ‘ƒğ‘ž(𝑥,𝑧)𝜇(𝑥)𝑑𝑥=ğ‘ƒğ‘ž(0,𝑧)𝐺∗(ğ‘Ž).(5.11) Using (5.11), from (5.7) we get ğ‘‰ğ‘ž(0,𝑧)=ğ‘ğ‘ƒğ‘ž(0,𝑧)𝐺∗(ğ‘Ž).(5.12) Similarly, we integrate (5.3) from 0 to 𝑥, we get ğ‘‰ğ‘ž(𝑥,𝑧)=ğ‘‰ğ‘ž(0,𝑧)𝑒∫−(𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑥)𝑥−𝑥0𝛽(𝑡)𝑑𝑡.(5.13) Substituting by the value of ğ‘‰ğ‘ž(0,𝑧) from (5.12) in (5.13) we get ğ‘‰ğ‘ž(𝑥,𝑧)=ğ‘ğ‘ƒğ‘ž(0,𝑧)𝐺∗(ğ‘Ž)𝑒∫−(𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑧)𝑥−𝑥0𝛽(𝑡)𝑑𝑡.(5.14)

Let us consider 𝑏=𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑧 now integrating (5.14) by parts with respect to 𝑥 we get ğ‘‰ğ‘ž(𝑧)=ğ‘ğ‘ƒğ‘ž(0,𝑧)𝐺∗(ğ‘Ž)1−𝐵∗(𝑏)𝑏,(5.15) where 𝐵∗∫(𝑏)=∞0𝑒−𝑏𝑥𝑑𝐵(𝑥) is the Laplace-Stieltjes transform of the vacation time 𝐵(𝑥).

Now multiplying both sides of (5.14) by 𝛽(𝑥) and integrating over 𝑥 we get ∞0ğ‘‰ğ‘ž(𝑥,𝑧)𝛽(𝑥)𝑑𝑥=ğ‘ğ‘ƒğ‘ž(0,𝑧)𝐺∗(ğ‘Ž)𝐵∗(𝑏).(5.16)

Now integrating (5.4) from 0 to 𝑥, yields ğ‘…ğ‘ž(𝑥,𝑧)=ğ‘…ğ‘ž(0,𝑧)𝑒∫−(𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑧)𝑥−𝑥0𝛾(𝑡)𝑑𝑡.(5.17)

Substituting by the value of ğ‘…ğ‘ž(0,𝑧) from (5.8) in (5.17) we get ğ‘…ğ‘ž(𝑥,𝑧)=ğ›¼ğ‘§ğ‘ƒğ‘žî€·(0,𝑧)1−𝐺∗(ğ‘Ž)ğ‘Žğ‘’âˆ«âˆ’(𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑧)𝑥−𝑥0𝛾(𝑡)𝑑𝑡,(5.18) integrating (5.18) by parts with respect to 𝑥 we get ğ‘…ğ‘ž(𝑧)=ğ›¼ğ‘§ğ‘ƒğ‘žîƒ©î€·(0,𝑧)1−𝐺∗(ğ‘Ž)1−𝐹∗(𝑏)îƒªğ‘Žğ‘,(5.19) where 𝐹∗∫(𝑏)=∞0𝑒−(𝜆−𝜆𝐶(𝑧)+𝛿−𝛿/𝑧)𝑥𝑑𝐹(𝑥) is the Laplace-Stieltjes transform of the repair time 𝐹(𝑥).

Now multiplying both sides of (5.18) by 𝛾(𝑥) and integrating over 𝑥 we get ∞0ğ‘…ğ‘ž(𝑥,𝑧)𝛾(𝑥)𝑑𝑥=ğ›¼ğ‘§ğ‘ƒğ‘žî‚µ(0,𝑧)1−𝐺∗(ğ‘Ž)ğ‘Žî‚¶ğ¹âˆ—(𝑏).(5.20)

Now using (5.11), (5.16) and (5.20), (5.6) becomes ğ‘ƒğ‘ž(0,𝑧)=âˆ’ğ‘Žğ‘ğ‘„ğ‘Žî€·ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧1−𝐺∗𝐹(ğ‘Ž)∗(𝑏),(5.21) where 𝑐=𝜆−𝜆𝐶(𝑧), from (5.21) equations (5.10), (5.15) and (5.19) become, respectively, ğ‘ƒğ‘žî€·(𝑧)=−𝑐𝑄1−𝐺∗(ğ‘Ž)ğ‘Žî€·ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧1−𝐺∗(î€¸ğ¹ğ‘Ž)∗(,𝑉𝑏)ğ‘ž(𝑧)=âˆ’ğ‘Žğ‘ğ‘„ğ‘ğºâˆ—î€·(ğ‘Ž)1−𝐵∗(𝑏)î€·ğ‘Žğ‘ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧𝑏1−𝐺∗(î€¸ğ¹ğ‘Ž)∗(,𝑅𝑏)ğ‘žî€·(𝑧)=−𝛼𝑧𝑐𝑄1−𝐺∗(ğ‘Ž)1−𝐹∗(𝑏)î€·ğ‘Žğ‘ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧𝑏1−𝐺∗(î€¸ğ¹ğ‘Ž)∗(.𝑏)(5.22) Let ğ‘†ğ‘ž(𝑧) denote the probability generating function of the queue size irrespective of the state of the system. That is, ğ‘†ğ‘ž(𝑧)=ğ‘ƒğ‘ž(𝑧)+ğ‘‰ğ‘ž(𝑧)+ğ‘…ğ‘ž(𝑧).

Then adding (5.22) we obtain ğ‘†ğ‘žî€·(𝑧)=−𝑐𝑄1−𝐺∗(ğ‘Ž)𝑏+𝛼𝑧1−𝐹∗(𝑏)î€¸î€¾âˆ’ğ‘Žğ‘ğ‘„ğ‘ğºâˆ—î€·(ğ‘Ž)1−𝐵∗(𝑏)î€·ğ‘Žğ‘ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧𝑏1−𝐺∗(î€¸ğ¹ğ‘Ž)∗(𝑏).(5.23)

In order to find 𝑄, we use the normalization condition ğ‘†ğ‘ž(1)+𝑄=1.(5.24) Note that if 𝑧=1 then 𝑏=0 and 𝑐=0, so ğ‘†ğ‘ž(1) is indeterminate of 0/0 form. Therefore, we apply L’Hopitals Rule twice on (5.23), we get ğ‘†ğ‘ž(1)=lim𝑧→1ğ‘î…žî…ž(𝑧)ğ·î…žî…ž(𝑧),(5.25) where 𝑁(𝑧) and 𝐷(𝑧) are the numerator and denominator of the right hand side of (5.23) respectively. Double primes in (5.25) denote the second derivative at 𝑧=1. Carrying out the derivatives at 𝑧=1 we have ğ‘î…žî…ž(1)=−2𝑄𝜆𝐸(𝐼)(𝜆𝐸(𝐼)−𝛿)1−𝐺∗(𝛼){1+𝛼𝐸(𝑅)}+𝛼𝑝𝐺∗𝐷(ğ‘Ž)𝐸(𝑉),(5.26)(1)=2(𝜆𝐸(𝐼)−𝛿)1−𝐺∗((î€·ğ‘Ž)𝜆𝐸(𝐼)+𝛼{1+(𝜆𝐸(𝐼)−𝛿)𝐸(𝑅)})−𝛼1−𝑝(𝜆𝐸(𝐼)−𝛿)𝐺∗,(ğ‘Ž)𝐸(𝑉)(5.27) where 𝐶(1)=1,ğ¶î…ž(1)=𝐸(𝐼)is the mean batch size of the arriving customers, 𝐵∗(0)=1, and 𝐵∗′(0)=−𝐸(𝑉) the mean vacation time, and 𝐹∗(0)=1, and 𝐹∗′(0)=−𝐸(𝑅) is the mean repair time.

Therefore, adding 𝑄 to (5.25) and equaling to 1 and simplifying we get 𝛼𝑄=1−𝑝(𝜆𝐸(𝐼)−𝛿)𝐺∗−(ğ‘Ž)𝐸(𝑉)1−𝐺∗(ğ‘Ž)(𝜆𝐸(𝐼)+𝛼{1+(𝜆𝐸(𝐼)−𝛿)𝐸(𝑅)})𝛼𝛿𝐸(𝑅)1−𝐺∗(𝛼)+𝐺∗(𝛼)(1+𝑝𝛿𝐸(𝑉)).(5.28) From (5.28) we can find the utilization factor, 𝜌, where 𝜌=1−𝑄.

As a particular case if we assume there is no stand by server this means that 𝛿=0, 𝑏=𝑐=𝜆−𝜆𝐸(𝐼) using this in the main results of this paper, we get, ğ‘†ğ‘žî€·(𝑧)=−𝑄1−𝐺∗(ğ‘Ž)𝑏+𝛼𝑧1−𝐹∗(𝑏)î€¸î€¾âˆ’ğ‘Žğ‘„ğ‘ğºâˆ—î€·(ğ‘Ž)1−𝐵∗(𝑏)ğ‘Žî€·ğ‘§âˆ’ğºâˆ—(ğ‘Ž)(1−𝑝+𝑝𝐵∗(𝑏))−𝛼𝑧1−𝐺∗(î€¸ğ¹ğ‘Ž)∗(1𝑏),(5.29)𝑄=1−𝜆𝐸(𝐼)𝛼𝐺∗−1(ğ‘Ž)𝛼+𝐸(𝑅)𝐺∗(ğ‘Ž)−𝐸(𝑅)+𝑝𝐸(𝑉).(5.30) These results agree with results given by [9].

6. The Average Queue Size and the Average Waiting Time

Let ğ¿ğ‘ž denote the mean number of customers in the queue under the steady state. Then ğ¿ğ‘ž=ğ‘‘ğ‘†ğ‘‘ğ‘§ğ‘ž|||(𝑧)𝑧=1.(6.1) Since this formula gives 0/0 form, then using the L'Hopital's rule four times we obtain ğ¿ğ‘ž=lim𝑧→1ğ·î…žî…ž(𝑧)ğ‘î…žî…žî…ž(𝑧)âˆ’ğ‘î…žî…ž(𝑧)ğ·î…žî…žî…ž(𝑧)3î€·ğ·î…žî…žî€¸(𝑧)2,(6.2) where ğ‘î…žî…ž(1) and ğ·î…žî…ž(1)are given in (5.26) and (5.27), respectively, and ğ‘î…žî…žî…žî€·(1)=−3𝑄𝑛𝑦1−𝐺∗(𝛼)(1+𝛼𝐸(𝑅))−6𝑄𝑦𝑚2ğºâˆ—î…ž[](𝛼)1+𝛼𝐸(𝑅)−3𝑄𝑚1−𝐺∗(ğ‘Ž)𝑥(1+𝛼𝐸(𝑅))+2𝛼𝑦𝐸(𝑅)+𝛼𝑦2𝐸𝑅2+3𝑝𝑄𝐸(𝑉)𝑦𝐺∗(ğ‘Ž)2𝑚2−𝛼𝑛+6𝛼𝑝𝑄𝑚2𝑦𝐸(𝑉)ğºâˆ—î…ž(ğ‘Ž)−3𝛼𝑝𝑄𝑚𝐺∗𝑦(𝛼)2𝐸𝑉2,𝐷+𝑥𝐸(𝑉)(6.3)(1)=31−𝐺∗(ğ‘Ž)𝑝𝐸(𝑉)𝑦{2𝑚𝑦−𝛼𝑥}+6𝑚2ğ‘¦ğºâˆ—î…žî€·(𝛼)+31−𝐺∗(ğ‘Ž){𝑛𝑦+𝑚𝑥}−3𝛼𝑦𝑝2ğºâˆ—î…ž(ğ‘Ž)𝑚𝑦𝐸(𝑉)−𝐺∗𝐸𝑉(ğ‘Ž)2𝑦2+𝑥𝐸(𝑉)+6𝛼𝑦21−𝐺∗𝑥(ğ‘Ž)𝐸(𝑅)+3𝛼{1+𝑦𝐸(𝑅)}1−𝐺∗(ğ‘Ž)+2ğ‘šğ‘¦ğºâˆ—î…žî€¾î€·(𝛼)+3𝛼𝑦1−𝐺∗𝑦(ğ‘Ž)2𝐸𝑅2,+𝑥𝐸(𝑅)(6.4) where (𝜆𝐸(𝐼(𝐼−1))+2𝛿)=𝑥,𝜆𝐸(𝐼)−𝛿=𝑦,𝜆𝐸(𝐼(𝐼−1))=𝑛,and𝜆𝐸(𝐼)=𝑚.

7. Numerical Example

To illustrate the results of this chapter numerically we consider that the service times, vacation times, stand-by service times, and repair times are exponentially distributed. All the values were chosen so that the steady state condition is satisfied. In Table 1 we will show the effect of the new contribution of this paper, where we will show the influence of the new parameter 𝛿 (stand-by service rate) on the trends. We choose the following values 𝜇=7, 𝛾=3, 𝜆=2, 𝛽=7, 𝛼=2, 𝑃=0.5, 𝐸(𝐼)=1, and 𝐸(𝐼(𝐼−1))=0, we consider that 𝛿 takes the values 0, 1, 4, 6, 8, and 10.


𝛿 𝜌 𝑄 𝐿 ğ‘ž 𝑊 ğ‘ž 𝐿 𝑊

00.6190.3811.40480.70242.02381.0119
10.53060.46940.80640.40321.33710.6685
40.37140.62860.34320.17160.71470.3573
60.30950.69050.28070.14030.59020.2951
80.26530.73470.26170.13090.5270.2635
100.23210.76790.25910.12950.49120.2456

Table 1 shows that increasing the value of 𝛿 decreases the value of utilization factor, the mean queue size, and the mean waiting time of the customers while the server idle time increases. All the trends shown by the table are as expected.

Acknowledgments

The authors would like to express their deep appreciation and thanks to the referee(s) for their comments and suggestions, which led to improvement of the paper in its present form.

References

  1. Y. Levy and U. Yechiali, “An M/M/s queue with servers vacations,” INFOR Journal, vol. 14, no. 2, pp. 153–163, 1976. View at: Google Scholar
  2. S. W. Fuhrmann and R. B. Cooper, “Stochastic decomposition in M/G/1 queue with generalized vacations,” Operations Research, vol. 33, no. 5, pp. 1117–1129, 1985. View at: Google Scholar
  3. B. T. Doshi, “Queueing systems with vacations—a survey,” Queueing Systems, vol. 1, no. 1, pp. 29–66, 1986. View at: Publisher Site | Google Scholar
  4. A. Aissani and J. R. Artalejo, “On the single server retrial queue subject to breakdowns,” Queueing Systems, vol. 30, no. 3-4, pp. 309–321, 1998. View at: Google Scholar
  5. J. Wang, J. Cao, and Q. Li, “Reliability analysis of the retrial queue with server breakdowns and repairs,” Queueing Systems, vol. 38, no. 4, pp. 363–380, 2001. View at: Publisher Site | Google Scholar
  6. W. J. Gray, P. P. Wang, and M. Scott, “A queueing model with multiple types of server breakdowns,” Quality Technology and Quantitative Management, vol. 1, no. 2, pp. 245–255, 2004. View at: Google Scholar
  7. Q. Xu, S. Bao, Z. Ma, and N. Tian, “M[X]/G/1 queue with multiple vacations,” Stochastic Analysis and Applications, vol. 25, no. 1, pp. 127–140, 2007. View at: Publisher Site | M[X]/G/1%20queue%20with%20multiple%20vacations&author=Q. Xu&author=S. Bao&author=Z. Ma&author=&author=N. Tian&publication_year=2007" target="_blank">Google Scholar
  8. K. C. Madan, W. Abu-Dayyeh, and M. Gharaibeh, “Steady state analysis of two Mx/Ma,b/1 queue models with random breakdowns,” International Journal of Information and Management Sciences, vol. 14, no. 3, pp. 37–51, 2003. View at: Google Scholar
  9. F. A. Maraghi, K. C. Madan, and K. Darby-Dowman, “'Batch arrival queueing system with random breakdowns and Bernoulli schedule server vacations having general vacation time distribution,” International Journal of Information and Management Sciences, vol. 20, no. 1, pp. 55–70, 2009. View at: Google Scholar
  10. C. K. Madan, “A bulk input queue with a stand-by,” The South African Statistical Journal, vol. 29, pp. 1–7, 1995. View at: Google Scholar

Copyright © 2011 Rehab F. Khalaf et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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