A repairable retrial queue under Bernoulli schedule and general retrial policy

Abstract

This paper considers a repairable M/G/1 retrial queue with Bernoulli schedule and a general retrial policy, which is motivated by a contention problem in the downlink direction of wireless base stations in cognitive radio networks. Arriving Customers (called primary arrivals) who cannot receive service upon arrival either join the infinite waiting space in front of the server (called as the normal queue) with probability $q$, or enter the orbit with probability $1-q$ according to the FCFS discipline. If the server breaks down in the process of the service of a customer, the customer in service either joins the orbit queue or leaves the system forever. First, we study the ergodicity of two related embedded Markov chains and derive stationary distributions. Second, we find the steady-state joint generating function of the number of customers in both queues. Some important performance measures of the system are obtained. Third, the reliability analysis of the system is also given. Finally, numerical examples are given to illustrate the impact of system parameters on the system performance measures.

Appendix

To use the parabolic method conveniently, a step-by-step procedure is described in this Appendix:

• Step 1 (Initialization): Take a starting 3-point pattern $\mathbf {x}_0=\{x^{(l)}, x^{(m)}, x^{(r)}\}, i=0$ and the tolerance $\varepsilon =10^{-6}$;

• Step 2 (Stopping): If $|\tilde{x}-x^{(m)}|\le \varepsilon$, stop and report approximate optimum solution $x^{(m)}$;

• Step 3 (Quadratic fit): Compute a quadratic fit optimum $\tilde{x}$ according to the formula (7.2). Then if $\tilde{x}\le x^{(m)}$, go to Step 4; and if $\tilde{x}> x^{(m)}$, go to Step 5.

• Step 4 (Left):If $f(x^{(m)})$ is superior to $f(\tilde{x})$ (less for a minimize, greater for a maximize), then update $\tilde{x}\rightarrow x^{(l)}$. Otherwise, replace $x^{(m)}\rightarrow x^{(r)}, \tilde{x}\rightarrow x^{(m)}$. Either way, advance $i = i+1$, and return to Step 2.

• Step 5 (Right): If $f(x^{(m)})$ is superior to $f(\tilde{x})$ (less for a minimize, greater for a maximize), then update $\tilde{x}\rightarrow x^{(r)}$. Otherwise, replace $x^{(m)}\rightarrow x^{(l)}, \tilde{x}\rightarrow x^{(m)}$. Either way, advance $i = i + 1$, and return to Step 2.