An M/M/2 Queueing System with Heterogeneous Servers Including One with Working Vacation

Abstract

This paper analyzes an 𝑀/𝑀/2 queueing system with two heterogeneous servers, one of which is always available but the other goes on vacation in the absence of customers waiting for service. The vacationing server, however, returns to serve at a low rate as an arrival finds the other server busy. The system is analyzed in the steady state using matrix geometric method. Busy period of the system is analyzed and mean waiting time in the stationary regime computed. Conditional stochastic decomposition of stationary queue length is obtained. An illustrative example is also provided.

1. Introduction

Queueing models with vacation have gained significance in the last three decades due to their wide range of applications, especially in the communication and the manufacturing systems. Doshi [1] provides an excellent survey of related works prior to 1986. Takagi [2] and Tian and Zhang [3] provide a good account of developments in this field since then. The literature on the vacation queueing models is growing rapidly.

In multiserver queueing models, we come across two classes of vacation mechanisms: station vacation and server vacation. In the first case, all servers take vacation simultaneously whenever the system becomes empty and they return to the system all together. Thus, station vacation is group vacation for all servers. For example, when a system consists of a number of machines operated by a single individual this scenario occurs. In such a situation, the whole station needs to be treated as a single unit for vacation, when the system is utilized for a secondary task. In the second case, each server takes its own vacation whenever it completes a service and finds no customers waiting for service. This phenomenon also occurs in practice. For example, in a post office or bank, when a clerk completes a service and finds no customer waiting, he or she might go to attend another task. This is what we refer to as the server vacation model. Analysis of a server vacation model is more complicated than that of a station vacation model. This is because at any time point, in the latter, we may have any number of servers between 0 and 𝑐 on vacation. We need to track individual servers going on vacation and completing their vacation. Upon returning from a vacation some servers may find no customers waiting for service. These servers take another vacation. But if any server finds a waiting customer on returning from a vacation, it immediately starts service. For further details on queues with station and server vacations, we refer the reader to Chao and Zhao [4].

Most of the earlier work on multiserver queueing models deal with homogeneous servers; that is, the individual service rates are same for all servers in the system. But this assumption may be valid only when the service process is mechanically or electronically controlled. In a queueing system with human servers the above assumption is highly unrealistic. Often servers providing identical service, serves at different rates. This motivated the researchers to study multiserver queueing system with heterogeneous servers. Several authors have analyzed multiserver queues with vacation. Levy and Yechiali [5], Vinod [6], and Kao and Narayanan [7] discuss asynchronous multiple vacation models with exponentially distributed vacation times. They study the scenario, where any server goes on vacation whenever there are no customers waiting in the system at a service completion epoch. At a vacation termination instant, if there are no waiting customers, the server takes another vacation; if there is a customer waiting for service, the server resumes service. It should be remarked that Vinod [6] was the first to use matrix geometric solutions method to analyze multiserver vacation models. But none of the above models deal with heterogeneous servers.

Servi and Finn [8] introduced a working vacation model with the idea of offering services but at a lower rate, whenever the server is on vacation. Their model was generalized to the case of 𝑀/𝐺/1 in [9, 10] and to 𝐺𝐼/𝑀/1 model in [11]. A survey of working vacation models with emphasis on the use of matrix analytic methods is given in Tian et al. [12]. Working vacation models have a number of applications in practice. Two such examples are given in [12]. Recently, Li and Tian [13] studied an 𝑀/𝑀/1 queue with working vacations in which vacationing server offers services at a lower rate, for the first customer arriving during a vacation. Very recently, Zhang and Hou [14] studied a 𝑀𝐴𝑃/𝐺/1 queue with working vacations and vacation interruptions using supplementary variable method. In this model, the authors assume that the vacation times are exponentially distributed and that the server gets back to normal service mode, when at a service (offered during a vacation) completion, the system has at least one customer waiting in the queue. The server is allowed to take multiple vacations. However, no work on multiserver working vacation model has come to our notice.

Neuts and Takahashi [15] observed that for queueing systems with more than two heterogeneous servers analytical results are intractable and only algorithmic approach could be used to study the steady state behavior of the system. Based on this observation, Krishna Kumar and Pavai Madheswari [16] analyzed 𝑀/𝑀/2 queueing system with heterogeneous servers, where the servers go on vacation in the absence of customers waiting for service. In this paper, we discuss an 𝑀/𝑀/2 queueing model with heterogeneous servers where one server remains idle but the other goes on working vacation in the absence of waiting customers.

This paper is organized as follows. In Section 2, model description is provided. In Section 3, the steady state analysis of the model is presented. In Section 4, we discuss an illustrative example.

2. Mathematical Model

We consider an 𝑀/𝑀/2 queueing model with heterogeneous servers, server 1 and server 2. Server 1 is always available, whereas server 2 goes on vacation whenever there are no customers waiting for service. Let the service rates of servers 1 and 2 be 𝜇₁ and 𝜇₂, respectively, where 𝜇₁≠𝜇₂. Customers arrive to the system according to a Poisson process with parameter 𝜆. The duration of vacation is exponentially distributed with parameter 𝜂. At the end of a vacation, service commences if there is a customer waiting for service. Otherwise the server goes on another vacation. During vacation, if an arrival finds server 1 busy, server 2 returns to serve the customer but at a lower rate. To be precise, server 2 serves this customer at the rate 𝜃𝜇₂, 0<𝜃≤1. As this vacation gets over, server 2 instantaneously switches over to the normal service rate 𝜇₂, if there is at least one customer waiting for service. Upon completion of a service at low rate, the server will (a) continue the current vacation if it is not finished and no customer is waiting for service; (b) continue the slow service if the vacation has not expired and if there is at least one customer waiting for service. For clarity, we make it clear that if an arriving customer finds a free server, he enters service immediately. Else he joins the queue.

2.1. The QBD Process

The model discussed in Section 2 can be studied as a quasi-birth-and-death (QBD) process. First, we set up the necessary notations.

At time 𝑡, let 𝑁(𝑡) be the number of customers in the system and 𝐽(𝑡)=0,iftheserver2isonvacation,1,iftheserver2isworkinginvacationmode,2,iftheserver2isworkinginnormalmode, (2.1) Let 𝑋(𝑡)=(𝑁(𝑡),𝐽(𝑡)). Then (𝑋(𝑡)∶𝑡≥0) is a continuous time Markov chain (CTMC) with states space Ω={(0,0),(1,0),(1,1),(1,2)}^∞_𝑖=2𝑙(𝑖), (2.2) where 𝑙(𝑖)=𝑖,𝑗∶𝑖≥2,𝑗=1or2. (2.3) Note that when 𝑁(𝑡)=0, the only possible value of 𝐽(𝑡) is 0 and when 𝑁(𝑡)=1, 𝐽(𝑡) has three possible values 0, 1, and 2.

The infinitesimal generator matrix 𝑄 of this Markov chain is given by 𝐵𝑄=⁰⁰𝐵₀₁𝐵₁₀𝐵₁₁𝐵₁₂𝐵₂₁𝐴₁𝐴₀𝐴₂𝐴₁𝐴₀⋱⋱⋱, (2.4) where the block matrices appearing in 𝑄 are as follows: 𝐵₀₀=−𝜆,𝐵₀₁=,𝐵𝜆00₁₀=𝜇₁𝜃𝜇₂𝜇₂,𝐵₁₁=−𝜆−𝜇₁000−𝜆−𝜃𝜇₂−𝜂𝜂00−𝜆−𝜇₂,𝐵₁₂=,𝐵𝜆0𝜆00𝜆₂₁=𝜃𝜇₂𝜇₁0𝜇₂0𝜇₁,𝐴₀=,𝐴𝜆00𝜆₁=−𝜆−𝜇₁−𝜃𝜇₂−𝜂𝜂0−𝜆−𝜇₁−𝜇₂,𝐴₂=𝜇₁+𝜃𝜇₂00𝜇₁+𝜇₂. (2.5)

3. Steady State Analysis

In this section, we will discuss the steady state analysis of the model under study.

3.1. Stability Condition

Theorem 3.1. The queueing system described above is stable if and only if 𝜌<1 where 𝜌=𝜆/(𝜇₁+𝜇₂).

Proof. To establish the stability condition we use Pakes' lemma (see [17]). Let 𝑁_𝑖 be the number of customers in the system immediately after the departure of the 𝑖th customer. Then {𝑁_𝑖∶𝑖∈ℕ} satisfies the equation 𝑁_𝑖=𝑁_𝑖−1−1+𝑉_𝑖if𝑁_𝑖−1𝑉≥1_𝑖if𝑁_𝑖−1=0, (3.1) where 𝑉_𝑖 is the number of arrivals during the service of 𝑖th customer. Clearly {𝑁_𝑖∶𝑖∈ℕ} is an irreducible aperiodic Markov chain. Pakes' lemma asserts that an aperiodic irreducible Markov chain is ergodic, if there exists an 𝜖>0 such that the mean drift 𝜙_𝑗=𝐸𝑁_𝑖+1−𝑁_𝑖∣𝑁_𝑖=𝑗 (3.2) is finite for all 𝑗∈ℕ and 𝜙_𝑗≤−𝜖 for all 𝑗∈ℕ except perhaps for a finite number. In the present model, value of the mean drift is 𝜙_𝑗=−1+𝜌if𝑗≥1𝜌if𝑗=0. (3.3) Thus, if 𝜌<1, the Markov chain {𝑁_𝑖∶𝑖∈ℕ} is ergodic and hence the condition is sufficient.
To prove the necessity of the condition, assume that 𝜌≥1. We use Theorem 3.1 in Sennot et al. [18], which states that {𝑁_𝑖∶𝑖∈ℕ} is nonergodic if it satisfies Kaplan's condition, 𝜙_𝑗<∞, for 𝑗≥0 and there is a 𝑗₀ such that 𝜙_𝑗≥0, for 𝑗≥𝑗₀. When 𝜌≥1 Kaplan's condition is readily satisfied. Hence, the Markov chain is not ergodic.

3.2. Steady State Probability Vector

Let 𝐱, partitioned as 𝐱=(𝑥₀,𝐱₁,𝐱₂,…), be the steady state probability vector of 𝑄. Note that 𝑥₀ is a scalar, 𝐱₁=(𝑥₁₀,𝑥₁₁,𝑥₁₂) and 𝐱_𝑖=(𝑥_𝑖1,𝑥_𝑖2) for 𝑖≥2. The vector 𝐱 satisfies the condition 𝐱𝑄=𝟎 and 𝐱𝐞=1, where 𝐞 is a column vector of 1's of appropriate dimension. Apparently when the stability condition is satisfied, the subvectors of 𝐱, corresponding to the different levels, are given by the equation 𝐱_𝑗=𝐱₂𝑅^𝑗−2,𝑗≥3, where 𝑅 is the minimal nonnegative solution of the matrix quadratic equation (see [20]) 𝑅²𝐴₂+𝑅𝐴₁+𝐴₀=0. (3.4) Knowing the matrix 𝑅, 𝑥₀, 𝐱₁ and 𝐱₂ are obtained by solving the equations 𝑥₀𝐵₀₀+𝐱₁𝐵₁₀𝑥=0,₀𝐵₀₁+𝐱₁𝐵₁₁+𝐱₂𝐵₂₁𝐱=0,₁𝐵₁₂+𝐱₂𝐴₁+𝑅𝐴₂=0, (3.5)(3.6)(3.7) subject to the normalizing condition 𝑥₀+𝐱₁𝐞+𝐱₂(𝐼−𝑅)⁻¹𝐞=1, (3.8)

Theorem 3.2. The matrix R of (3.4) is given by 𝑅=^𝑅11𝑅12_0𝑅22, where 𝑅₁₁=(𝜆+𝜇₁+𝜃𝜇₂+𝜂−(𝜆+𝜇₁+𝜃𝜇₂+𝜂)²−4𝜆(𝜇₁+𝜃𝜇₂))/2(𝜇₁+𝜃𝜇₂), 𝑅₁₂=𝜌−(𝜇₁+𝜃𝜇₂)𝑅₁₁/(𝜇₁+𝜇₂) and 𝑅₂₂=𝜌.

Proof. Since 𝐴₀, 𝐴₁, and 𝐴₂ are upper triangular, 𝑅 is essentially an upper triangular matrix. The value of 𝑅₁₁, follows from the assertion that 𝑅 is the minimal nonnegative solution of (3.4). The rest of the proof is an easy consequence of the condition 𝑅𝐴₂𝐞=𝐴₀𝐞.

Remark 3.3. Though 𝑅 has a nice structure which enables us to make use of the properties like 𝑅^𝑘=^{𝑅𝑘11𝑅12∑𝑘−1𝑗=0𝑅𝑗11𝑅𝑘−𝑗−122}_0𝑅𝑘22, for 𝑘≥1, due to the form of the expression for 𝑅₁₁ it may not be easy to carry out the computations required in the forthcoming discussions. Hence, we explore the possibility of algorithmic computation of 𝑅. The computation of 𝑅 matrix can be carried out using a number of well-known methods such as logarithmic reduction. We list here only the main steps involved in logarithmic reduction algorithm for computation of 𝑅. For full details on the logarithmic reduction algorithm we refer the reader to [19].

Logarithmic Reduction Algorithm for 𝑅:
Step  0. 𝐻←(−𝐴₁)⁻¹𝐴₀, 𝐿←(−𝐴₁)⁻¹𝐴₂, 𝐺=𝐿, and 𝑇=𝐻.
Step  1. 𝑈=𝐻𝐿+𝐿𝐻𝑀=𝐻²𝐻⟵(𝐼−𝑈)⁻¹𝑀𝑀⟵𝐿²𝐿⟵(𝐼−𝑈)⁻¹𝑀𝐺⟵𝐺+𝑇𝐿𝑇⟵𝑇𝐻, (3.9) continue Step 1 until ‖𝐞−𝐺𝐞‖_∞<𝜖.
Step  2. 𝑅=−𝐴₀(𝐴₁+𝐴₀𝐺)⁻¹.

3.3. Busy Period Analysis

For the system under study, busy period is the interval between arrival of a customer to the empty system and the first epoch thereafter when the system becomes empty again. Thus, it is precisely the first passage time from the state (1,0) to the state (0,0). For the working vacation model, busy cycle for the system is the time interval between two successive departures, which leave the system empty. Thus, the busy cycle is the first return time to state (0,0) with at least one visit to any other state. But before analyzing the busy period structure we need to introduce the notion of fundamental period. For the QBD process under consideration, it is the first passage time from level 𝑖, where 𝑖≥3, to the level 𝑖−1. The cases 𝑖=2, 𝑖=1, and 𝑖=0, corresponding to the boundary states, need to be discussed separately. It should be noted that due to the structure of the QBD process, the distribution of the first passage time is invariant in 𝑖.

Let 𝐺_𝑗𝑗′(𝑘,𝑥) denote the conditional probability that a QBD process starting in the state (𝑖,𝑗) at time 𝑡=0 reaches the level 𝑖−1 for the first time no later than time 𝑥, after exactly 𝑘 transitions to the left and does so by entering the state (𝑖−1,𝑗′). For convenience we introduce the joint transform 𝐺_𝑗𝑗′(𝑧,𝑠)=^∞_𝑘=1𝑧^𝑘∞₀𝑒^−𝑠𝑥d𝐺_𝑗𝑗′(𝑘,𝑥);|𝑧|≤1,Re(𝑠)≥0, (3.10) and the matrix 𝐺(𝑧,𝑠)=𝐺_𝑗𝑗′(𝑧,𝑠). (3.11) The matrix 𝐺(𝑧,𝑠) is the unique solution to the equation (see [20]) 𝐺(𝑧,𝑠)=𝑧𝑠𝐼−𝐴₁⁻¹𝐴₂+𝑠𝐼−𝐴₁⁻¹𝐴₀𝐺²(𝑧,𝑠). (3.12) The matrix 𝐺=𝐺(1,0) takes care of the first passage times, except for the boundary states. If we know the 𝑅 matrix then 𝐺 matrix can be computed using the result (see [19]) 𝐺=−𝐴₁+𝑅𝐴₂⁻¹𝐴₂. (3.13) Otherwise, we may use logarithmic reduction method to compute 𝐺. For the boundary level states 2, 1, and 0, let 𝐺^(2,1)_𝑗𝑗′(𝑘,𝑥), 𝐺^(1,0)_𝑗𝑗′(𝑘,𝑥), and 𝐺^(0,0)_𝑗𝑗′(𝑘,𝑥) be the conditional probability discussed above for the first passage times from level 2 to level 1, level 1 to level 0, and the first return time to the level 0, respectively. Then as in (3.12) we get 𝐺^(2,1)(𝑧,𝑠)=𝑧𝑠𝐼−𝐴₁⁻¹𝐵₂₁+𝑠𝐼−𝐴₁⁻¹𝐴₀𝐺(𝑧,𝑠)𝐺^(2,1)(),𝐺𝑧,𝑠_(1,0)(𝑧,𝑠)=𝑧𝑠𝐼−𝐵₁₁⁻¹𝐵₁₀+𝑠𝐼−𝐵₁₁⁻¹𝐵₁₂𝐺^(2,1)(𝑧,𝑠)𝐺^(1,0)(),𝐺𝑧,𝑠_(0,0)()=𝜆𝑧,𝑠𝐺𝑠+𝜆00^(1,0)(𝑧,𝑠). (3.14)(3.15)(3.16) Note that 𝐺^(1,0)(𝑧,𝑠) is a 3×1 matrix. Thus, the Laplace Stieltjes transform (LST) of the busy period is the first element of 𝐺^(1,0)(1,𝑠). For convenience, we use the notations 𝐺₂₁=𝐺^(2,1)(1,0),𝐺₁₀=𝐺^(1,0)(1,0),𝐺₀₀=𝐺^(0,0)(1,0). (3.17) Due to the positive recurrence of the QBD process, matrices 𝐺, 𝐺₂₁, 𝐺₁₀, and 𝐺₀₀ are all stochastic. If we let 𝐶₀=−𝐴₁⁻¹𝐴₂,𝐶₂=−𝐴₁⁻¹𝐴₀, (3.18) then 𝐺 is the minimal nonnegative solution (see [20]) to the matrix equation 𝐺=𝐶₀+𝐶₂𝐺². (3.19) From (3.14), (3.15), and (3.16), we get 𝐺₂₁=−𝐴₁+𝐴₀𝐺⁻¹𝐵₂₁,𝐺₁₀=−𝐵₁₁+𝐵₁₂𝐺₂₁⁻¹𝐵₁₀,𝐺₀₀=100𝐺₁₀ (3.20) respectively. Equation (3.12) is equivalent to 𝑧𝐴₂−𝑠𝐼−𝐴₁𝐺(𝑧,𝑠)+𝐴₀𝐺²(𝑧,𝑠)=0. (3.21)

Let 𝜕𝐺()𝑀=−𝑧,𝑠𝜕𝑠_{𝑧=1,𝑠=0},𝜕𝐺()𝑀=𝑧,𝑠𝜕𝑧_{𝑧=1,𝑠=0}. (3.22) Differentiation of (3.21) with respect to 𝑠 and 𝑧 followed by setting 𝑧=1 and 𝑠=0 leads to (see [20]) 𝑀=−𝐴₁⁻¹𝐺+𝐶₂(𝐺𝑀+𝑀𝐺),𝑀=𝐶₀+𝐶₂𝐺𝑀+𝑀𝐺, (3.23) with 𝟎 as starting value for 𝑀 and 𝑀, successive substitutions in the above equations yield the values of 𝑀 and 𝑀. Applying an exactly similar reasoning to (3.14), (3.15), and (3.16), we get 𝑀₂₁=−𝐴₁+𝐴₀𝐺⁻¹𝐼+𝐴₀𝑀𝐺₂₁,𝑀₁₀=𝐵₁₁+𝐵₁₂𝐺₂₁⁻¹𝐼+𝐵₁₂𝑀₂₁𝐺₁₀,𝑀₀₀=1𝜆00𝐺₁₀+100𝑀₁₀, (3.24) where 𝑀₂₁𝜕𝐺=−^(2,1)(𝑧,𝑠)𝜕𝑠_{𝑧=1,𝑠=0},𝑀₁₀𝜕𝐺=−^(1,0)(𝑧,𝑠)𝜕𝑠_{𝑧=1,𝑠=0},𝑀₀₀𝜕𝐺=−^(0,0)(𝑧,𝑠)𝜕𝑠_{𝑧=1,𝑠=0}. (3.25) Note that 𝑀₁₀ is a 3×1 matrix and 𝑀₀₀ is a scalar. The first element of the matrix 𝑀₁₀ and 𝑀₀₀ are mean lengths of a busy period and a busy cycle, respectively. The second and third elements of the matrix 𝑀₁₀ are the first passage times to the state (0,0) from (1,1) and (1,2), respectively. With the notations 𝑀₂₁=𝜕𝐺^(2,1)(𝑧,𝑠)𝜕𝑧_{𝑧=1,𝑠=0},𝑀₁₀=𝜕𝐺^(1,0)(𝑧,𝑠)𝜕𝑧_{𝑧=1,𝑠=0}, (3.26) it follows from (3.14) and (3.15) that 𝑀₂₁=−𝐴₁+𝐴₀𝐺⁻¹𝐵₂₁+𝐴₀𝑀𝐺₂₁,𝑀₁₀=−𝐵₁₁+𝐵₁₂𝐺₂₁⁻¹𝐵₁₀+𝐵₁₂𝑀₂₁𝐺₁₀. (3.27) The first component of the vector 𝑀₁₀ is the mean number of service completions in a busy period.

3.4. Stationary Waiting Time Distribution in the Queue

Let 𝑊(𝑡) be the distribution function for the waiting time in the queue of an arriving (tagged) customer. Note that if there is no customer in the system, the arrival receives service immediately. If either of the two servers is not busy then also there would be no delay in getting service. Thus, the probability that the customer gets his service without waiting is 𝑥₀+𝑥₁₀+𝑥₁₁+𝑥₁₂. Hence, with probability 1−𝑥₀−𝑥₁₀−𝑥₁₁−𝑥₁₂, the customer has to wait before getting the service. The waiting time may be viewed as the time until absorption in a Markov chain with state space Ω₁={∗}{𝟐,𝟑,…}. (3.28) Here * is the absorbing state, which corresponds to taking the tagged customer into service and is obtained by lumping together the level states 𝟎={(0,0)} and 𝟏={(1,0),(1,1),(1,2)}. For 𝑖≥2, the level 𝐢 is given by 𝐢={(𝑖,𝑗),𝑗=1or2}. The states other than the absorbing state correspond to the number of customers present in the system as the tagged customer arrives. Once the tagged customer joins the queue, the subsequent arrivals will not affect his waiting time in the queue. Hence the parameter 𝜆 does not show up in the generator matrix 𝑄 of this Markov process, given by ∗23⋮𝑄=𝐴∗23…₂𝐴𝐞𝐷₂𝐷,⋱⋱where𝐷=−𝜇₁−𝜃𝜇₂−𝜂𝜂0−𝜇₁−𝜇₂. (3.29)

Define vector 𝐲(𝑡)=𝑦_∗(𝑡),𝐲₂(𝑡),𝐲₃(𝑡),…, (3.30) where 𝐲_𝑖(𝑡)=𝑦_𝑖1(𝑡),𝑦_𝑖2(𝑡),for𝑖≥2. (3.31) The components of the 𝐲_𝑖(𝑡) are the probabilities that at time 𝑡, the CTMC with generator 𝑄 is in the respective states of level 𝑖. Note that the scalar 𝑦_∗(𝑡) is the probability that the process is in the absorbing state at time 𝑡. By the PASTA property, we get 𝐲(0)=𝑥₀+𝑥₁₁+𝑥₁₀+𝑥₁₂,𝐱₂,𝐱₃,…. (3.32) Clearly 𝑊(𝑡)=𝑦_∗(𝑡),for𝑡≥0. (3.33) The LST of 𝑊(𝑡) is given by (see [20]) 𝑊(𝑠)=^∞_𝑖=2𝐲_𝑖(0)(𝑠𝐼−𝐷)⁻¹𝐴₂^𝑖−2(𝑠𝐼−𝐷)⁻¹𝐴₂𝐞. (3.34) The mean waiting time can be obtained from 𝑊(𝑠) as 𝐸(𝑊)=−𝑊^′(0)=^∞_𝑖=1𝐱_2+𝑖^𝑖−1_𝑗=0𝑈^𝑗(−𝐷)⁻¹𝑈^𝑖−𝑗𝑈𝐞+^∞_𝑖=0𝐱_2+𝑖𝑈^𝑖(−𝐷)⁻²𝐴₂𝐞, (3.35) where 𝑈=(−𝐷)⁻¹𝐴₂ is a stochastic matrix. Hence, (3.34) can be simplified as 𝐸(𝑊)=−𝑊^′(0)=^∞_𝑖=1𝐱_2+𝑖^𝑖−1_𝑗=0𝑈^𝑗(−𝐷)⁻¹𝐞+^∞_𝑖=0𝐱_2+𝑖𝑈^𝑖(−𝐷)⁻¹𝐞. (3.36) Let 𝐻=^∞_𝑖=0𝐱_2+𝑖𝑈^𝑖, (3.37) Since 𝑈 is stochastic, we get 𝐻𝐞=𝐱₂(𝐼−𝑅)⁻¹𝐞=1−𝑥₀−𝑥₁₀−𝑥₁₁−𝑥₁₂ (3.38) This result can be used to find an approximate value of 𝐻 and hence that of the second term in (3.36) to any desired degree of accuracy. Thus, only the first term in (3.36) demands serious computation. For this we make use of the ideas in [15, 16, 21].

Now consider the matrix 𝑈₂=0101, (3.39) which has the property that 𝑈𝑈₂=𝑈₂𝑈=𝑈₂. (3.40) Then we get ^𝑖−1_𝑗=0𝑈^𝑗𝐼−𝑈+𝑈₂=𝐼−𝑈^𝑖+𝑖𝑈₂,for𝑖≥1. (3.41) By the classical theorem on finite Markov chains, the matrix (𝐼−𝑈+𝑈₂) is nonsingular (see [22]). In view of the last equation, the first term in (3.36) becomes ^∞_𝑖=1𝐱_2+𝑖𝐼−𝑈^𝑖+𝑖𝑈₂𝐼−𝑈+𝑈₂⁻¹(−𝐷)⁻¹𝐞. (3.42) With this simplification, we get 𝐸(𝑊)=𝐱₂𝑅(𝐼−𝑅)⁻¹+𝐼+𝑅(𝐼−𝑅)⁻²𝑈₂−𝐻𝐼−𝑈+𝑈₂⁻¹(−𝐷)⁻¹𝐞+𝐻(−𝐷)⁻¹𝐞. (3.43)

3.5. Conditional Stochastic Decomposition of Queue Length

In this section, we provide a stochastic decomposition of queue length in the stationary regime, subject to the condition that both servers are busy. Note that from (3.5)–(3.8) we get 𝑥₀, 𝑥₁₀, 𝑥₁₁, 𝑥₁₂, 𝑥₂₁, and 𝑥₂₂. Let 𝑄_𝑣 be the queue length of the vacation model under study, subject to the condition that both servers are busy. Then we have the following.

Theorem 3.4. If 𝜌<1, then 𝑄_𝑣=𝑄₀+𝑄_𝑑, where 𝑄₀ and 𝑄_𝑑 are two independent random variables. 𝑄₀ is the queue length of the M/M/2 queueing model with heterogeneous servers without vacation and 𝑄_𝑑 can be interpreted as the additional queue length due to vacation and slow service, subject to the condition that both servers are busy.

Proof. Let 𝑃_𝑏 denote the Probability that both servers are busy. Then 𝑃_𝑏=^∞_𝑛=2𝑥_𝑛2=^∞_𝑛=2𝑥₂₂𝜌^𝑛−2+^∞_𝑛=3𝑥₂₁𝑅₁₂^𝑛−3_𝑗=0𝑅^𝑗₁₁𝜌^{𝑛−𝑗−3}=𝑥₂₂𝜌^∞_𝑘=0𝜌^𝑘+𝑥₂₁𝑅^∞_12𝑘=0𝑅^𝑘∞_11𝑘=0𝜌^𝑘=;𝑘=𝑛−31−𝜌⁻¹𝑥₂₂𝜌+𝑥₂₁𝑅₁₂1−𝑅₁₁⁻¹, (3.44) so that 1𝑃_𝑏=1−𝜌𝑥₂₂𝜌+𝑥₂₁𝑅₁₂1−𝑅₁₁⁻¹⁻¹=1−𝜌𝛿, (3.45) where 𝛿=𝑥₂₂𝜌+𝑥₂₁𝑅₁₂1−𝑅₁₁⁻¹⁻¹, (3.46)𝑄_𝑣(𝑧), the generating function of the queue length subject to the condition that both servers are busy, is given by 𝑄_𝑣(𝑧)=1𝑃_𝑏^∞_𝑛=2𝑥_𝑛1𝑧^𝑛−2=1𝑃_𝑏^∞_𝑛=2𝑥₂₂𝜌^𝑛−2𝑧^𝑛−2+1𝑃_𝑏^∞_𝑛=3𝑥₂₁𝑅₁₂^𝑛−3_𝑗=0𝑅^𝑗₁₁𝜌^{𝑛−𝑗−3}𝑧^𝑛−3. (3.47) By following a computational procedure similar to that of 𝑃_𝑏, we arrive at 𝑄_𝑣(𝑧)=1−𝜌𝛿𝑥1−𝜌𝑧²²𝑥𝜌𝑧+²¹R₁₂1−𝑅₁₁𝑧=𝑄₀(𝑧)𝑄_𝑑(𝑧), (3.48) where 𝑄₀(𝑧)=1−𝜌,𝑄1−𝜌𝑧_𝑑(𝑧)=𝛿𝑥₂₂𝑥𝜌𝑧+²¹R₁₂1−𝑅₁₁𝑧. (3.49)(3.50) From (3.49) it follows that 𝑄₀(𝑧) is the generating function of an 𝑀/𝑀/2 heterogeneous queuing model without vacations, which is precisely the case 𝛽=1 in [23]. Equation (3.50) suggests that 𝑄_𝑑 has a geometric distribution with parameter 1−𝑅₁₁.

Remark 3.5. Due to the algorithmic approach used in the derivation stationary waiting time distribution, a similar decomposition result for the waiting time distribution is far from reality.

3.6. Key System Performance Measures

In this section, we list a number of key system performance measures along with their formulae in addition to the busy period structure and the mean waiting time discussed above.  (1) The probability that the system is empty: 𝑃_EMP=𝑥₀.  (2) The probability that the server 1 is idle: 𝑃_IDL=𝑥₀+𝑥₁₁+𝑥₁₂.  (3) The probability that the server 2 is on vacation: 𝑃_VAC=𝑥₀+𝑥₁₀.  (4) The probability that the server 2 is working in vacation mode: 𝑃_SLOW=∑^∞_𝑗=1𝑥_𝑗1=𝑥₁₁+𝑥₂₁/(1−𝑅₁₁).  (5) The probability that the server 2 is working in normal mode: 𝑃_NORM=1−𝑥₀−𝑃_SLOW.  (6) The mean number of customers in the system: 𝜇_𝑁𝑆=∑^∞_𝑗=1𝑗𝐱_𝑗𝐞=𝑥₁₀+𝑥₁₁+𝑥₁₂+𝐱₂(𝐼−𝑅)⁻²𝑅⁻¹𝐞−𝐱₂𝑅⁻¹𝐞.

4. Numerical Results

4.1. Illustrative Example

We analyze the effect of the parameters 𝜆, 𝜂 and 𝜃 on the key performance measures. Table 1 analyzes the effect of 𝜆, Table 2 explains the effect of 𝜂, and Table 3 examines the effect of 𝜃 on the performance measures. To this end, we use the following abbreviations in addition to the notations used in Section 3.6.

Table 1 

Case A: 𝜇₁=10, 𝜇₂=5, 𝜂=1 and 𝜃=0.6. Case B: 𝜇₁=5, 𝜇₂=10, 𝜂=1 and 𝜃=0.6.

Table 2 

Case A: 𝜆=12, 𝜇₁=10, 𝜇₂=5 and 𝜃=0.6. Case B: 𝜆=12, 𝜇₁=5, 𝜇₂=10 and 𝜃=0.6.

Table 3 

Case A: 𝜆=12, 𝜇₁=10, 𝜇₂=5 and 𝜂=1. Case B: 𝜆=12, 𝜇₁=5, 𝜇₂=10 and 𝜂=1.

𝜇_WTQ: Mean waiting time in the queue.

𝜇_LBP: Mean length of a busy period.

𝜇_LBC: Mean length of a busy cycle.

𝜇_NSBP: Mean number of service completions in a busy period.(i)Since 𝜇₁ and 𝜇₂ are fixed, the traffic intensity 𝜌 increases with 𝜆. Due to this 𝑃_NORM, 𝜇_NS and 𝜇_WTQ increase and 𝑃_VAC and 𝑃_IDL decrease as 𝜆 increases. Note that the busy period starts with the Markov chain in the state (1,0); that is, with server 2 on vacation. Hence, initially 𝑃_SLOW increases with 𝜆. For this reason 𝜇_LBP, 𝜇_LBC, and 𝜇_NSBP show an early downward trend. But as 𝜆 further increases 𝑃_SLOW declines as expected due to the high traffic intensity. Hence, 𝜇_LBP and 𝜇_NSBP reverse the direction of change. Due to the effect of 𝑃_VAC and 𝑃_IDL, this reversal occurs only at a later stage for 𝜇_LBC. It is worth comparing the values of the measures in cases A and B. Even though the net service rate 𝜇₁+𝜇₂=15 in both cases, the effect of the vacation parameter 𝜂 becomes more predominant when 𝜇₁<𝜇₂. Due to this the measures 𝑃_VAC and 𝑃_IDL take smaller values and the measures 𝜇_NS, 𝜇_WTQ, 𝜇_LBP, and 𝜇_NSBP take larger values in case B, compared to their values in case A.(ii)As 𝜂 increases, the mean duration of vacation 1/𝜂 decreases. But as the mean duration of vacation decreases, the probability of the expiry of the vacation without initiating the slow service increases. The chance of an early expiry of vacation always results in an increase in 𝑃_NORM and 𝑃_VAC and a decrease in 𝑃_SLOW. Note that 𝑃_VAC+𝑃_SLOW decreases as 𝜂 increases and 𝑃_VAC+𝑃_SLOW<𝑃_NORM for any value of 𝜂 in the given range. So 𝑃_IDL increases with 𝜂. Thus, the proportion of time in which both servers work at the normal rate increases as 𝜂 increases. Hence, the measures 𝜇_NS, 𝜇_WTQ, 𝜇_LBP, 𝜇_LBC, and 𝜇_NSBP decrease as 𝜂 increases. The argument given the last paragraph holds good for the difference in magnitude of the measures in cases A and B.(iii)As 𝜃 increases, the service rate 𝜃𝜇₂ of the second server during vacation mode of service increases. As a result, server 2 clears out customers at an increased rate in slow service mode. This produces an increase in 𝑃_VAC, 𝑃_SLOW, and 𝑃_IDL and a decrease in 𝑃_NORM as expected. Consequently, 𝜇_LBP, 𝜇_LBC and 𝜇_NSBP decrease as 𝜃 increases. The huge difference in the value of net service rate 𝜇₁+𝜃𝜇₂ between cases A and B, during vacation mode of service, is the reason for the pattern of behavior of 𝜇_NS in these two cases. Increase in 𝜃 does not affect 𝜇_WTQ significantly in case A but it affects the measure in case B. This is because the effect of 𝜃 becomes significant only when 𝜇₂ is large compared to 𝜇₁.

5. Concluding Remarks

In this paper, we studied an 𝑀/𝑀/2 queueing model with heterogeneous servers. One server follows multiple vacation policy. But this server offers service at a lower rate during vacation if customers arrive. The other server remains in the system even when it is empty. The busy period of the system was analyzed in the stationary regime. Mean waiting time of a customer has been computed. Conditional stochastic decomposition of queue length has been derived. An illustrative numerical example to bring out the qualitative nature of the model has been presented.

Acknowledgments

The authors would like to thank the anonymous referees for their valuable suggestions and comments which greatly improved the presentation of the paper. C. Sreenivasan's research is supported by the University Grants Commission, Government of India, under Faculty Development Programme (Grant no. F.FIP/11th Plan/KLCA042TF02).