Analysis of the $M^X/G/1$ retrial queue

Abstract

In this paper, we are concerned with the analysis of the queue length and waiting time distributions in a batch arrival $M^X/G/1$ retrial queue. Necessary and sufficient conditions are obtained for the existence of the moments of the queue length and waiting time distributions. We also provide recursive formulas for the higher order moments of the queue length and waiting time distributions.

Appendix: Proof of Lemma 4

When the batch to which the tagged customer belongs, arrives from outside the system, there are the following three cases to consider: (i) the server is busy, (ii) the server is idle and another customer of that batch starts service, not the tagged customer, (iii) the server is idle and the tagged customer starts service. Note that cases (i) and (ii) correspond to $G=1$ (see Fig. 2) and case (iii) corresponds to $G=0$. Let us denote the three events (i), (ii) and (iii) by $A_1, A_2$ and $A_3$, respectively. By the PASTA property,

$$\begin{aligned} {\mathbb {P}}(A_1)=\rho , ~ {\mathbb {P}}(A_2)=(1-\rho )\left( 1-\frac{1}{{\mathbb {E}}B}\right) , ~ {\mathbb {P}}(A_3)=\frac{1-\rho }{{\mathbb {E}}B}. \end{aligned}$$

By noting $\{G=1\}=A_1 \cup A_2$, we have

$$\begin{aligned} {\mathbb {P}}(A_1\mid G=1)&= \frac{\rho {\mathbb {E}}B}{{\mathbb {E}}B-1+\rho }, \\ {\mathbb {P}}(A_2\mid G=1)&= \frac{(1-\rho )({\mathbb {E}}B-1)}{{\mathbb {E}}B-1+\rho }. \end{aligned}$$

Fig. 2

The cases when $G=1$. The arrows represent the arrival epoch of the batch to which the tagged customer belongs. The left-hand side corresponds to case (i) and the right-hand side corresponds to case (ii)

Since $\tilde{W}^{(1)}(z,s)={\mathbb {E}}[z^{N(\tilde{\tau }^{(1)})} e^{-s(\tilde{\tau }^{(1)}-t^*)}{\mathbb {1}}_{\{\sigma >\tilde{\tau }^{(1)}\}}\mid G=1]$, we have

$$\begin{aligned} \tilde{W}^{(1)}(z,s)&= {\mathbb {P}}(A_1 \mid G=1) {\mathbb {E}}[z^{N(\tilde{\tau }^{(1)})} e^{-s(\tilde{\tau }^{(1)}-t^*)}{\mathbb {1}}_{\{\sigma >\tilde{\tau }^{(1)}\}}\mid A_1] \\&\quad +\, {\mathbb {P}}(A_2 \mid G=1) {\mathbb {E}}[z^{N(\tilde{\tau }^{(1)})} e^{-s(\tilde{\tau }^{(1)}-t^*)}{\mathbb {1}}_{\{\sigma >\tilde{\tau }^{(1)}\}}\mid A_2]. \end{aligned}$$

For convenience we set

$$\begin{aligned} \tilde{W}^{(1)}_1(z,s)&\equiv {\mathbb {E}}[z^{N(\tilde{\tau }^{(1)})} e^{-s(\tilde{\tau }^{(1)}-t^*)}{\mathbb {1}}_{\{\sigma >\tilde{\tau }^{(1)}\}}\mid A_1] \\ \tilde{W}^{(1)}_2(z,s)&\equiv {\mathbb {E}}[z^{N(\tilde{\tau }^{(1)})} e^{-s(\tilde{\tau }^{(1)}-t^*)}{\mathbb {1}}_{\{\sigma >\tilde{\tau }^{(1)}\}}\mid A_2]. \end{aligned}$$

Thus

$$\begin{aligned} \tilde{W}^{(1)}(z,s) = \frac{\rho {\mathbb {E}}B}{{\mathbb {E}}B-1+\rho } \tilde{W}^{(1)}_1(z,s)+ \frac{(1-\rho )({\mathbb {E}}B-1)}{{\mathbb {E}}B-1+\rho } \tilde{W}^{(1)}_2(z,s). \end{aligned}$$

(32)

First we compute $\tilde{W}^{(1)}_1(z,s)$. When the tagged customer arrives while the server is busy, let $\tau ^{(0)}$ be the beginning epoch of the service period in which the tagged customer arrives and ${\mathcal {A}}$ denote the number of customers who arrive from outside the system, excluding the tagged customer, during the time interval $(\tau ^{(0)}, \tilde{\tau }^{(1)})$. Then, by well established techniques (see, for example, Takagi 1991), we have

$$\begin{aligned} {\mathbb {E}}[ z^{{\mathcal {A}}} e^{-s(\tilde{\tau }^{(1)}-t^*)}\mid G=1] = \frac{\beta (\lambda -\lambda b(z))-\beta (s+\lambda -\lambda b(z))}{s {\mathbb {E}}S}. \end{aligned}$$

Since the size of the batch to which the tagged customer belongs has the distribution $\frac{kb_k}{{\mathbb {E}}B}$, $k=1,2,\ldots$, by length biased sampling, we have

$$\begin{aligned} \tilde{W}^{(1)}_1(z,s)&= \pi (z) \sum _{k=1}^\infty \frac{kb_k}{{\mathbb {E}}B}z^{k-1} \frac{\beta (\lambda -\lambda b(z))-\beta (s+\lambda -\lambda b(z))}{s {\mathbb {E}}S} \nonumber \\&= \frac{1}{{\mathbb {E}}B{\mathbb {E}}S}\pi (z)b'(z) \frac{\beta (\lambda -\lambda b(z))-\beta (s+\lambda -\lambda b(z))}{s}. \end{aligned}$$

(33)

Now we compute $\tilde{W}^{(1)}_2(z,s)$. By the PASTA property, given that the batch which contains the tagged customer arrives while the server is idle, the probability generating function of the number of customers in the system immediately before $t^*$ is $\frac{q(z)}{1-\rho }$. The tagged customer belongs to a batch of size k with probability $\frac{kb_k}{{\mathbb {E}}B}$ due to length biased sampling. The tagged customer is not selected with probability $\frac{k-1}{k}$ given that there are k customers in the batch to which the tagged customer belongs. Thus, given $A_2$, the tagged customer belongs to a batch of size k with probability $\frac{\frac{kb_k}{{\mathbb {E}}B}\frac{k-1}{k}}{\sum _{j=1}^\infty \frac{jb_j}{{\mathbb {E}}B}\frac{j-1}{j}}=\frac{(k-1)b_k}{{\mathbb {E}}B-1}$. Therefore, given $A_2$, the probability generating function of the number of customers in the orbit, excluding the tagged customer, immediately after $t^*$ is

$$\begin{aligned} \frac{q(z)}{1-\rho }\sum _{k=2}^\infty \frac{(k-1)b_k}{{\mathbb {E}}B-1} z^{k-2}, \end{aligned}$$

where the exponent $k-2$ of z follows because the two customers (the tagged customer and a customer selected for service) are excluded. Since the joint transform of the service time of a customer and the number of customers who arrive from outside the system during that service time is given by $\beta (s+\lambda -\lambda b(z))$, we have

$$\begin{aligned} \tilde{W}^{(1)}_2(z,s)&=\frac{q(z)}{1-\rho } \sum _{k=2}^\infty \frac{(k-1)b_k}{{\mathbb {E}}B-1}z^{k-2} \beta (s+\lambda -\lambda b(z)) \nonumber \\&= \frac{q(z)}{(1-\rho )({\mathbb {E}}B-1)} \frac{b'(z)-\tilde{b}(z)}{z} \beta (s+\lambda -\lambda b(z)). \end{aligned}$$

(34)

Finally, substituting (33) and (34) into (32), we have (19).