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Journal of Probability and Statistics
Volume 2011, Article ID 812726, 11 pages
http://dx.doi.org/10.1155/2011/812726
Research Article

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

1School of Information Systems Computing and Mathematics, Brunel University, Middlesex UB83PH, UK
2College of Information Technology, Ahlia University, P.O. Box 10878, Bahrain

Received 25 January 2011; Accepted 20 May 2011

Academic Editor: Rongling Wu

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.

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 [18]. 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 0𝑐𝑖1 and 𝑖=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)1𝐺(𝑥),(2.1) and, therefore 𝑔(𝑠)=𝜇(𝑠)𝑒𝑠0𝜇(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 1𝑃, where 0𝑃1.

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)1𝐵(𝑥),(2.3) and, therefore 𝑏(𝑣)=𝛽(𝑣)𝑒𝑣0𝛽(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)1𝐹(𝑥),(2.5) and, therefore 𝑓(𝑟)=𝛾(𝑟)𝑒𝑟0𝛾(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 𝑛0 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 𝑛1 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 𝑛0 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 𝑛0 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 (𝑛0) 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 𝑛0 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 lim𝑡𝐴𝑛(𝑡,𝑥)=𝐴𝑛(𝑥),lim𝑡𝐴𝑛(𝑡)=lim𝑡0𝐴𝑛(𝑡,𝑥)𝑑𝑥=𝐴𝑛,lim𝑡𝑑𝐴𝑛(𝑡)𝑑𝑡=0,where𝐴=𝑃,𝑉,𝑅,lim𝑡𝑄(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 Δ𝑡0, we obtain the following set of differential-difference equations 𝜕𝑃𝜕𝑥𝑛(𝑥)=(𝜆+𝜇(𝑥)+𝛼)𝑃𝑛(𝑥)+𝜆𝑛1𝑖=1𝑐𝑖𝑃𝑛𝑖𝜕(𝑥),𝑛1,(4.1)𝑃𝜕𝑥0(𝑥)=(𝜆+𝜇(𝑥)+𝛼)𝑃0𝜕(𝑥),(4.2)𝑉𝜕𝑥𝑛(𝑥)=(𝜆+𝛽(𝑥)+𝛿)𝑉𝑛(𝑥)+𝜆𝑛𝑖=1𝑐𝑖𝑉𝑛𝑖(𝑥)+𝛿𝑉𝑛+1𝜕(𝑥),𝑛1,(4.3)𝑉𝜕𝑥0(𝑥)=(𝜆+𝛽(𝑥)+𝛿)𝑉0(𝑥)+𝛿𝑉1(𝑥),(4.4)𝜕𝑅𝑛(𝑥)𝜕𝑥=(𝜆+𝛾(𝑥)+𝛿)𝑅𝑛(𝑥)+𝜆𝑛𝑖=1𝑐𝑖𝑅𝑛𝑖(𝑥)+𝛿𝑅𝑛+1(𝑥),𝑛1,(4.5)𝜕𝑅0(𝑥)𝜕𝑥=(𝜆+𝛾(𝑥)+𝛿)𝑅0(𝑥)+𝛿𝑅1(𝑥),(4.6)𝜆𝑄=0𝑅0(𝑥)𝛾(𝑥)𝑑𝑥+(1𝑝)0𝑃0(𝑥)𝜇(𝑥)𝑑𝑥+(1𝑟)0𝑉0(𝑥)𝛽(𝑥)𝑑𝑥.(4.7) The following boundary conditions will be used to solve the above equations: 𝑃𝑛(0)=(1𝑝)0𝑃𝑛+1(𝑥)𝜇(𝑥)𝑑𝑥+0𝑉𝑛+1+(𝑥)𝛽(𝑥)𝑑𝑥0𝑅𝑛+1(𝑥)𝛾(𝑥)𝑑𝑥+𝜆𝑐𝑛+1𝑉𝑄,𝑛0,(4.8)𝑛(0)=𝑝0𝑃𝑛𝑅(𝑥)𝜇(𝑥)𝑑𝑥,𝑛0,(4.9)𝑛(0)=𝛼0𝑃𝑛1𝑅(𝑥)𝑑𝑥,𝑛1,(4.10)0(0)=0.(4.11)

5. Queue Size Distribution at a Random Epoch

Defining the following probability generating functions 𝐴𝑞(𝑥,𝑧)=𝑛=0𝑧𝑛𝐴𝑛(𝑥),𝐴𝑞(𝑧)=𝑛=0𝑧𝑛𝐴𝑛,𝐶𝐴=𝑃,𝑉,𝑅,(𝑧)=𝑖=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 𝜕𝑃𝜕𝑥𝑞(𝑥,𝑧)+(𝜆𝜆𝐶(𝑧)+𝜇(𝑥)+𝛼)𝑃𝑞(𝑥,𝑧)=0.(5.2) Using the same process, from (4.3), (4.4) and (4.5), (4.6) we get, respectively, 𝜕𝑉𝜕𝑥𝑞𝛿(𝑥,𝑧)+𝜆𝜆𝐶(𝑧)+𝛽(𝑥)+𝛿𝑧𝑉𝑞𝜕(𝑥,𝑧)=0,(5.3)𝑅𝜕𝑥𝑞𝛿(𝑥,𝑧)+𝜆𝜆𝐶(𝑧)+𝛾(𝑥)+𝛿𝑧𝑅𝑞(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 𝑧𝑃𝑞(0,𝑧)=(1𝑝)0𝑃𝑞(𝑥,𝑧)𝜇(𝑥)𝑑𝑥+0𝑉𝑞(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+0𝑅𝑞(𝑥,𝑧)𝛾(𝑥)𝑑𝑥+𝜆𝐶(𝑧)𝑄(1𝑝)0𝑃0(𝑥)𝜇(𝑥)𝑑𝑥+(1𝑟)0𝑉0(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+0𝑅0.(𝑥,𝑧)𝛾(𝑥)𝑑𝑥(5.5) From (4.7), we have 𝑧𝑃𝑞(0,𝑧)=(1𝑝)0𝑃𝑞(𝑥,𝑧)𝜇(𝑥)𝑑𝑥+0𝑉𝑞(𝑥,𝑧)𝛽(𝑥)𝑑𝑥+0𝑅𝑞(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 𝑉𝑞(0,𝑧)=𝑝0𝑃𝑞(𝑥,𝑧)𝜇(𝑥)𝑑𝑥.(5.7) Similarly, from (4.10) we get 𝑅𝑞(0,𝑧)=𝛼𝑧0𝑃𝑞(𝑥,𝑧)𝑑𝑥=𝛼𝑧𝑃𝑞(𝑧),𝑛0.(5.8) Integrating (5.2) from 0 to 𝑥 yields 𝑃𝑞(𝑥,𝑧)=𝑃𝑞(0,𝑧)𝑒(𝜆𝜆𝐶(𝑧)+𝛼)𝑥𝑥0𝜇(𝑡)𝑑𝑡,(5.9) where 𝑃𝑞(0,𝑧) is given by (5.6). Let us consider 𝑎=𝜆𝜆𝐶(𝑧)+𝛼.

Integrating equation (5.9) by parts with respect to 𝑥 yields 𝑃𝑞(𝑧)=𝑃𝑞(0,𝑧)1𝐺(𝑎)𝑎,(5.10) where 𝐺(𝑎)=0𝑒𝑎𝑥𝑑𝐺(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 0𝑃𝑞(𝑥,𝑧)𝜇(𝑥)𝑑𝑥=𝑃𝑞(0,𝑧)𝐺(𝑎).(5.11) Using (5.11), from (5.7) we get 𝑉𝑞(0,𝑧)=𝑝𝑃𝑞(0,𝑧)𝐺(𝑎).(5.12) Similarly, we integrate (5.3) from 0 to 𝑥, we get 𝑉𝑞(𝑥,𝑧)=𝑉𝑞(0,𝑧)𝑒(𝜆𝜆𝐶(𝑧)+𝛿𝛿/𝑥)𝑥𝑥0𝛽(𝑡)𝑑𝑡.(5.13) Substituting by the value of 𝑉𝑞(0,𝑧) from (5.12) in (5.13) we get 𝑉𝑞(𝑥,𝑧)=𝑝𝑃𝑞(0,𝑧)𝐺(𝑎)𝑒(𝜆𝜆𝐶(𝑧)+𝛿𝛿/𝑧)𝑥𝑥0𝛽(𝑡)𝑑𝑡.(5.14)

Let us consider 𝑏=𝜆𝜆𝐶(𝑧)+𝛿𝛿/𝑧 now integrating (5.14) by parts with respect to 𝑥 we get 𝑉𝑞(𝑧)=𝑝𝑃𝑞(0,𝑧)𝐺(𝑎)1𝐵(𝑏)𝑏,(5.15) where 𝐵(𝑏)=0𝑒𝑏𝑥𝑑𝐵(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 0𝑉𝑞(𝑥,𝑧)𝛽(𝑥)𝑑𝑥=𝑝𝑃𝑞(0,𝑧)𝐺(𝑎)𝐵(𝑏).(5.16)

Now integrating (5.4) from 0 to 𝑥, yields 𝑅𝑞(𝑥,𝑧)=𝑅𝑞(0,𝑧)𝑒(𝜆𝜆𝐶(𝑧)+𝛿𝛿/𝑧)𝑥𝑥0𝛾(𝑡)𝑑𝑡.(5.17)

Substituting by the value of 𝑅𝑞(0,𝑧) from (5.8) in (5.17) we get 𝑅𝑞(𝑥,𝑧)=𝛼𝑧𝑃𝑞(0,𝑧)1𝐺(𝑎)𝑎𝑒(𝜆𝜆𝐶(𝑧)+𝛿𝛿/𝑧)𝑥𝑥0𝛾(𝑡)𝑑𝑡,(5.18) integrating (5.18) by parts with respect to 𝑥 we get 𝑅𝑞(𝑧)=𝛼𝑧𝑃𝑞(0,𝑧)1𝐺(𝑎)1𝐹(𝑏)𝑎𝑏,(5.19) where 𝐹(𝑏)=0𝑒(𝜆𝜆𝐶(𝑧)+𝛿𝛿/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 0𝑅𝑞(𝑥,𝑧)𝛾(𝑥)𝑑𝑥=𝛼𝑧𝑃𝑞(0,𝑧)1𝐺(𝑎)𝑎𝐹(𝑏).(5.20)

Now using (5.11), (5.16) and (5.20), (5.6) becomes 𝑃𝑞(0,𝑧)=𝑎𝑐𝑄𝑎𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧1𝐺𝐹(𝑎)(𝑏),(5.21) where 𝑐=𝜆𝜆𝐶(𝑧), from (5.21) equations (5.10), (5.15) and (5.19) become, respectively, 𝑃𝑞(𝑧)=𝑐𝑄1𝐺(𝑎)𝑎𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧1𝐺(𝐹𝑎)(,𝑉𝑏)𝑞(𝑧)=𝑎𝑐𝑄𝑝𝐺(𝑎)1𝐵(𝑏)𝑎𝑏𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧𝑏1𝐺(𝐹𝑎)(,𝑅𝑏)𝑞(𝑧)=𝛼𝑧𝑐𝑄1𝐺(𝑎)1𝐹(𝑏)𝑎𝑏𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧𝑏1𝐺(𝐹𝑎)(.𝑏)(5.22) Let 𝑆𝑞(𝑧) denote the probability generating function of the queue size irrespective of the state of the system. That is, 𝑆𝑞(𝑧)=𝑃𝑞(𝑧)+𝑉𝑞(𝑧)+𝑅𝑞(𝑧).

Then adding (5.22) we obtain 𝑆𝑞(𝑧)=𝑐𝑄1𝐺(𝑎)𝑏+𝛼𝑧1𝐹(𝑏)𝑎𝑐𝑄𝑝𝐺(𝑎)1𝐵(𝑏)𝑎𝑏𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧𝑏1𝐺(𝐹𝑎)(𝑏).(5.23)

In order to find 𝑄, we use the normalization condition 𝑆𝑞(1)+𝑄=1.(5.24) Note that if 𝑧=1 then 𝑏=0 and 𝑐=0, so 𝑆𝑞(1) is indeterminate of 0/0 form. Therefore, we apply L’Hopitals Rule twice on (5.23), we get 𝑆𝑞(1)=lim𝑧1𝑁(𝑧)𝐷(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 𝑁(1)=2𝑄𝜆𝐸(𝐼)(𝜆𝐸(𝐼)𝛿)1𝐺(𝛼){1+𝛼𝐸(𝑅)}+𝛼𝑝𝐺𝐷(𝑎)𝐸(𝑉),(5.26)(1)=2(𝜆𝐸(𝐼)𝛿)1𝐺((𝑎)𝜆𝐸(𝐼)+𝛼{1+(𝜆𝐸(𝐼)𝛿)𝐸(𝑅)})𝛼1𝑝(𝜆𝐸(𝐼)𝛿)𝐺,(𝑎)𝐸(𝑉)(5.27) where 𝐶(1)=1,𝐶(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 𝛼𝑄=1𝑝(𝜆𝐸(𝐼)𝛿)𝐺(𝑎)𝐸(𝑉)1𝐺(𝑎)(𝜆𝐸(𝐼)+𝛼{1+(𝜆𝐸(𝐼)𝛿)𝐸(𝑅)})𝛼𝛿𝐸(𝑅)1𝐺(𝛼)+𝐺(𝛼)(1+𝑝𝛿𝐸(𝑉)).(5.28) From (5.28) we can find the utilization factor, 𝜌, where 𝜌=1𝑄.

As a particular case if we assume there is no stand by server this means that 𝛿=0, 𝑏=𝑐=𝜆𝜆𝐸(𝐼) using this in the main results of this paper, we get, 𝑆𝑞(𝑧)=𝑄1𝐺(𝑎)𝑏+𝛼𝑧1𝐹(𝑏)𝑎𝑄𝑝𝐺(𝑎)1𝐵(𝑏)𝑎𝑧𝐺(𝑎)(1𝑝+𝑝𝐵(𝑏))𝛼𝑧1𝐺(𝐹𝑎)(1𝑏),(5.29)𝑄=1𝜆𝐸(𝐼)𝛼𝐺1(𝑎)𝛼+𝐸(𝑅)𝐺(𝑎)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 𝐿𝑞 denote the mean number of customers in the queue under the steady state. Then 𝐿𝑞=𝑑𝑆𝑑𝑧𝑞|||(𝑧)𝑧=1.(6.1) Since this formula gives 0/0 form, then using the L'Hopital's rule four times we obtain 𝐿𝑞=lim𝑧1𝐷(𝑧)𝑁(𝑧)𝑁(𝑧)𝐷(𝑧)3𝐷(𝑧)2,(6.2) where 𝑁(1) and 𝐷(1)are given in (5.26) and (5.27), respectively, and 𝑁(1)=3𝑄𝑛𝑦1𝐺(𝛼)(1+𝛼𝐸(𝑅))6𝑄𝑦𝑚2𝐺[](𝛼)1+𝛼𝐸(𝑅)3𝑄𝑚1𝐺(𝑎)𝑥(1+𝛼𝐸(𝑅))+2𝛼𝑦𝐸(𝑅)+𝛼𝑦2𝐸𝑅2+3𝑝𝑄𝐸(𝑉)𝑦𝐺(𝑎)2𝑚2𝛼𝑛+6𝛼𝑝𝑄𝑚2𝑦𝐸(𝑉)𝐺(𝑎)3𝛼𝑝𝑄𝑚𝐺𝑦(𝛼)2𝐸𝑉2,𝐷+𝑥𝐸(𝑉)(6.3)(1)=31𝐺(𝑎)𝑝𝐸(𝑉)𝑦{2𝑚𝑦𝛼𝑥}+6𝑚2𝑦𝐺(𝛼)+31𝐺(𝑎){𝑛𝑦+𝑚𝑥}3𝛼𝑦𝑝2𝐺(𝑎)𝑚𝑦𝐸(𝑉)𝐺𝐸𝑉(𝑎)2𝑦2+𝑥𝐸(𝑉)+6𝛼𝑦21𝐺𝑥(𝑎)𝐸(𝑅)+3𝛼{1+𝑦𝐸(𝑅)}1𝐺(𝑎)+2𝑚𝑦𝐺(𝛼)+3𝛼𝑦1𝐺𝑦(𝑎)2𝐸𝑅2,+𝑥𝐸(𝑅)(6.4) where (𝜆𝐸(𝐼(𝐼1))+2𝛿)=𝑥,𝜆𝐸(𝐼)𝛿=𝑦,𝜆𝐸(𝐼(𝐼1))=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(𝐼1))=0, we consider that 𝛿 takes the values 0, 1, 4, 6, 8, and 10.

tab1
Table 1: Computed values of various queue performance measures (𝜇=7,𝜆=2,𝛼=2,𝑃=0.5,𝛾=3,𝛽=7).

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.

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