Scaling limits for single server retrial queues with two-way communication

Abstract

This paper studies M/G/1 retrial queues in which there are two arrival flows, i.e., incoming calls made by regular customers and outgoing calls made by the server in idle time. The stationary analysis of this system has been carried out in a recent paper by Artalejo and Phung-Duc (Appl Math Model 37(4):1811–1822, 2013). In this paper, we obtain a decomposition property where we prove that the queue length is decomposed into the sum of three independent random variables with clear physical meaning. We then derive scaling limits for the queue length distribution under some extreme conditions (i) heavy traffic, (ii) slow retrials and (iii) instantaneous connection to outgoing calls. Furthermore, we also investigate the convergence of our model to that without outgoing calls.

Appendix: Uniform convergence

In order for the interchange between limit and integration uniform convergence of the integrand is required. To this end, we prove the following two lemmas.

Lemma 1

(Page 30 in Falin and Templeton 1997) If $\beta _1^2 < \infty$ then we have

$$\begin{aligned} \beta _1 (\varepsilon s) = 1- \beta _1^1 \varepsilon s + \frac{\beta _1^2}{2} \varepsilon ^2 s^2 + o (\varepsilon ^2), \end{aligned}$$

uniformly with respect to $s \in [0,M]$.

Proof

A proof is given in page 30 of Falin and Templeton (1997). $\square$

Lemma 2

If $\beta _2^1 < \infty$ then we have

$$\begin{aligned} \beta _2 (\varepsilon s) = 1- \beta _2^1 \varepsilon s + o (\varepsilon ) \end{aligned}$$

uniformly with respect to $s \in [0,M]$.

Proof

Letting $g(t) = \beta _2 (t) -1 + \beta _2^1 t$, because $\beta _2^1$ exists, we have

$$\begin{aligned} g(0) = 0, \quad g^\prime (0) = 0. \end{aligned}$$

We have $g^{\prime \prime } (t) = \beta _2^{\prime \prime } (t)> 0$ for any $t > 0$. Thus, $g^\prime (t) \ge g^\prime (0) = 0$ for $t \ge 0$. Therefore, we have

$$\begin{aligned} 0 \le \frac{g(\varepsilon s)}{\varepsilon } \le M \frac{g(\varepsilon M)}{\varepsilon M}, \quad s \in [0,M], \end{aligned}$$

where $M$ is an arbitrary fixed positive number. Since $\frac{g(\varepsilon M)}{\varepsilon M} \rightarrow 0$ as $\varepsilon \rightarrow 0$, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \frac{g(\varepsilon s)}{\varepsilon } = 0, \end{aligned}$$

uniformly with respect to $s \in [0,M]$ for any $M>0$. $\square$