Mathematical Problems in Engineering

Volume 2015, Article ID 165219, 6 pages

http://dx.doi.org/10.1155/2015/165219

## The Analysis, Optimization, and Simulation of a Two-Stage Tandem Queueing Model with Hyperexponential Service Time at Second Stage

^{1}Department of Statistics, Faculty of Science and Arts, Ondokuz Mayıs University, 55200 Samsun, Turkey^{2}Department of Statistics, Faculty of Science and Arts, Amasya University, 5000 Amasya, Turkey

Received 15 May 2015; Accepted 28 October 2015

Academic Editor: Purushothaman Damodaran

Copyright © 2015 Vedat Sağlam 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

The aim of this paper is to analyze a tandem queueing model with two stages. The arrivals to the first stage are Poisson stream and the service time at this stage is exponential. There is no waiting room at first stage. The service time is hyperexponential and no waiting is allowed at second stage. The transition probabilities and loss probabilities of this model are obtained. In addition, the loss probability at second stage is optimized. Performance measures and the variance of the numbers of customers of this tandem queueing model are found. It is seen that the numbers of customers in first stage and second stage are dependent. Finally we have simulated this queueing model. For different values of parameters, exact values, simulated values, and optimal values of obtained performance measures of this model are numerically shown in tables and graphs.

#### 1. Introduction

Queueing systems and modeling of these systems have many uses in production lines, service facilities, telecommunication, computer sciences, transportation, and so forth. The main interest in queueing modeling is that it allows us to process mathematical analysis. Hence the analysis gives us the ability to obtain important properties such as measures of performances of considered models. There follows then optimization of these measures. One of important study areas in Queueing Theory is the tandem queues. Many various important studies are done on tandem queues. Niu gave an upper bound for the stationary expected delay at second server for a sequence of two queues in tandem [1]. Some moment results for certain tandem and multiple-server queues are studied by Wolfson [2]. Ohno and Ichiki discussed the use of modified policy iteration algorithms to find the optimal control for the service rate for a tandem queueing system in [3]. Analysis for the steady state of a two-stage tandem queueing with single server is given by [4]. Scheduling service time in tandem queues is studied in [5]. Ziedins showed that, for some simple service time distributions with support on two points, the throughput can be calculated exactly and that it is always optimal to allocate the capacity as uniformly as possible, even when blocking occurs, in [6]. The moments in tandem queues are widely studied in [7]. A study, on control of a single-server tandem queueing system, is given by [8]. Knessel and Tier considered a diffusion model for two tandem queues with general renewal input [9]. Marin and Bulo studied explicit solutions for queues with hypoexponential service time and they gave applications to product form analysis in [10]. The queues with phase-type distributions are widely investigated by [11]. Throughout maximization for tandem lines is studied by [12]. Recently, performance analysis of tandem queues with small buffers is studied [13] and a paper on a tandem queueing model with parallel phases is given by [14] in which a numerical example is performed. In this paper we analyze a tandem queueing model in which the second stage has hyperexponential distribution. The arrivals to the first stage of the queue are Poisson-distributed and the service time at this stage has exponential distribution. There is no waiting room at first stage of the queue. The service time is hyperexponentially distributed and no waiting is allowed at second stage of the queue. The transition probabilities and loss probabilities of this model are obtained. The loss probability at second stage is optimized. The measures of performances are obtained. We also show that the numbers of customers in first and second stages are dependent. Furthermore this queueing model is simulated. For different values of parameters, exact values, simulated values, and optimal values of obtained performance measures of this model are numerically shown in tables and graphs. In simulation, it is seen that the simulation results tend to exact results after 10000 iteration steps. This queueing model is an extended model of the one studied in [14]. Choosing and gives the model given in [14].

#### 2. The Definition of Model

A new tandem queueing discipline is investigated in which the customers arrive to system with parameter Poisson stream. In the first stage, there is a single server with exponential distributed service time having rate No waiting is allowed in front of this single server. Hence the first loss occurs. The service time at second stage is hyperexponentially distributed. After completing service in first stage, the customers proceed to second stage. At second stage, the customers choose the first server or second server with probabilities and , respectively. The service time parameters of servers at second stage are and . After having service at first stage if any of two servers in second stage is busy then the customers leave the system. In this case second loss occurs.

Now we give a formal mathematical definition of this queueing discipline as below.

Let , , and be the numbers of customers in first stage, in the first server of second stage, and in the second server of second stage, respectively, where and . We define a 3D Markov chain . And the state space of this chain is where Assuming that the limit distribution exists, we need to find the steady-state probabilities

##### 2.1. Difference Equations of the 3D Markov Chain

We have the difference equations of this Markov chain as given in

#### 3. Transition Probabilities

By solving the equation system above, we have the transition probabilities as follows:

##### 3.1. The Dependence of Number of Customers at First and Second Stages

Considering the first stage, in queue if we take , then are obtained in [15]. Where is the probability that the system is idle and is the probability the system is busy. Because there is a probability of loss at first stage, the customers can not arrive to second stage with rate. Therefore the transition probabilities at second stage are found with rate. Then,In this manner, for example, since , the transition probability is written asAs a result, the numbers of customers at first and second stages are dependent on each other.

##### 3.2. Loss Probabilities

There are two loss probabilities in this model. Now we go on by calculating these probabilities:(a)Let denote the loss probability at first stage. Then,

It is easy to see that this probability depends on and parameters only. Hence the loss probability at first stage is independent of the parameters and of second stage.(b)Let denote the loss probability at second stage. This probability is showed as follows.

To be loss at second stage, the second stage must be busy under condition that the first stage is busy. Considering this, the following events are defined:

is the event that first stage is busy.

is the event that second stage is busy.

Considerwhere .

The following theorem on optimization of is given.

Theorem 1. *Fixing and (c is constant), the minimum is obtained when .*

*Proof. * is rewritten as follows: Taking in (11), is written as below:To simplify, we write To make minimum, must be minimum:Equation (13) is written under condition as follows: The values which make minimum are .

Therefore the probability has its minimum value when the product is maximum, that is, when

#### 4. Measures of Performances

##### 4.1. The Mean Number of Customers in the System

If we denote the number of customers in the system by , then we calculate the mean number of customers as follows:

##### 4.2. The Variance of Customers Number in the System

Consider

##### 4.3. Mean Waiting Time in the System

Let be the waiting time in the system. Using total expectation in [16], we havewhere is the event of the loss at first stage. So,and the conditional expectations are written as It follows then that Now the mean waiting time of a customer in the system is calculated as given in

#### 5. The Simulation of the Model

In this section we give the simulation of the queueing model which we analyze. The mean waiting time in the system and the loss probabilities are obtained by performing some number of iterations and the simulation results are given together with exact values in tables. Furthermore, it is seen that the simulation results tend to exact values.

The exact and simulation values of in Tables 1, 2, 3, and 4 are given in Figure 1. Also the exact values of are the initial values of this figure.