STOCHASTIC DECOMPOSITION OF THE M/M/1 QUEUE WITH ENVIRONMENT DEPENDENT WORKING VACATION

We consider an M/M/1 queue with n types of working vacation. After a non zero busy period if the server finds the system empty, it opts for one of the n types of working vacation depending on the environment. On vacation completion epoch finding an empty system, it remains in the respective vacation. We demonstrate the stochastic decomposition structure of the queue length and waiting time of this M/M/1 queue and obtain the distribution of the additional queue length and additional delay.


INTRODUCTION
If a queue is empty the server remains idle. The idle time of the server can be utilized for supplementary jobs. This gives rise to extensive research work in the field of vacation queueing models. The details regarding the research on queueing models can be found in the survey of Doshi [1], the monograph of Takagi [2] and Tian and Zhang [3]. If the number of customers in the queue is less, the functioning of the server at a slow rate will reduce the operating cost, energy consumption, and start-up cost. These advantages are pointing towards The rest of the paper is organized as follows. In section 2 the model is described in detail. Section 3 discusses the stochastic decomposition structures of the number of customers in the system and waiting time and obtains the distributions of additional queue length and additional delay.

MODEL DESCRIPTION
Consider a single server queueing system with working vacation in which arrival occurs according to a Poisson process with parameter λ . The service time is exponentially distributed with parameter µ. On completion of a service if the server finds the system empty it goes for a working vacation. There are n types of working vacations. Depending on the environment, after a busy period, the server goes for i th type of vacation with probability p i , 1 ≤ i ≤ n. The duration of i th type of vacation is exponentially distributed with parameter γ i , 1 ≤ i ≤ n. During vacation, if customers arrive, the server provides service at a lower rate µ i , provided the server is in i th type of vacation, 1 ≤ i ≤ n. On completion of service during vacation, if there is no customer in the system the server continues to stay on vacation. Otherwise, the vacation is interrupted, i.e. the server returns to normal service without completing the vacation and starts service at the normal rate µ. On completion of vacation, if the server finds the system empty, it remains on the corresponding vacations. Figure 1 is a diagrammatic representation of the model.

MATHEMATICAL DESCRIPTION
We establish the stochastic decomposition of the state space by induction on the number of environmental factors.
Case.1 First we consider the case of n = 2.
Let N(t) be the number of customers in the system and S(t) be the status of the server at time t: if the server is serving in normal mode; 1, if server is in the type I working vacation; 2, if server is in the type II working vacation; The infinitesimal generator associated with the Markov chain is Then A is the infinitesimal generator of a Markov chain with state space {0, 1, 2} which represents the status of the server. Let y = (y 0 , y 1 , y 2 ) be the invariant probability vector of A. Then yA = 0 and ye = 1. The left drift rate of the original Markov chain is yA 2 e and that for right drift is yA 0 e. Left drift indicates a service completion and right drift represents arrival of customer.
Thus the system is stable if and only if yA 0 e < yA 2 e. Here yA 0 e = λ and yA 2 e = µ.
Hence we have Theorem: The system is stable if and only if λ < µ.

Steady State Analysis.
For the analysis of the model it is necessary to solve for the minimal non-negative solution R 1 of the matrix quadratic equation Since the Matrices A 2 , A 1 , A 0 are lower triangular R 1 is also lower triangular. Solving (1) we ob- and r 2 = λ (λ +µ 2 +θ 2 ) . Let x = (x 0 , x 1 , x 2 , . . . , ) be the steady state probability vector associated with the Markov process X. Here x 0 = (x 01 , x 02 ) and . ., then x can be obtained by solving xQ = 0 using the boundary condition (2) x 0 e + x 1 (I − R 1 ) −1 e = 1.
From xQ = 0 we get From (3) and (4) we will get Assume x 01 = k 1 and x 02 = k 2 , then from (8) and (9), Substituting the values of x 11 and x 01 in (5) we will get x 10 = k 1 r 1 p 1 . Also To find the value of k 1 we use the normalizing condition Substituting k 2 in (10) Let Q v (z) be the PGF associated with the number of customers in the system. Then Case.2 Now consider the case of n = 3. Then S(t) has four states.
if the server is serving in normal mode; 1, if server is in the type I working vacation; 2, if server is in the type II working vacation; 3, if server is in the type III working vacation; The state space of X is {(0, k)|k = 1, 2, 3}∪{( j, k), j = 1, 2, . . . ; k = 0, 1, 2, 3}. The infinitesimal generator associated with the Markov chain is We get λ < µ as the condition for stability.

As in the earlier sections
Let y = (y 0 , y 1 , y 2 , . . . , y n ) be the invariant probability vector of A satisfying yA = 0 and ye = 1. The system is stable if and only if yA 0 e < yA 2 e. Here yA 0 e = λ and yA 2 e = µ.
Then Q v (z) = ∑ ∞ n=0 x n z n = n ∑ j=1 x 0 j + x 10 z 1 − r 0 z + n ∑ j=1 x 1 j z 1 − r j z + n ∑ j=1 x 1 j r j z r 0 − r j

CONFLICT OF INTERESTS
The author(s) declare that there is no conflict of interests.