Analysis of an M^[X]/G/1 unreliable retrial G-queue with orbital search and feedback under Bernoulli vacation schedule

Abstract

In this paper, we consider a batch arrival retrial queue with feedback under Bernoulli vacation schedule, where the busy server is subjected to breakdown due to the arrival of negative customers. Any arriving batch of positive customers finds the server free, one of the customers from the batch enters into the service area and the rest of them join into the orbit. Arriving positive customers may balk (or renege) the system at particular times. After completion of service the unsatisfied positive customer may rejoin into the orbit to get another regular service as feedback customer. The server takes Bernoulli vacation after service completion of positive customers. After completion of service (if the server is not taking vacation), repair or vacation the server searches for the customers in the orbit or remains idle. The steady state probability generating function for the system size is obtained by using the supplementary variable method. Some system performance measures, reliability measures and stochastic decomposition law are discussed. Finally, some numerical examples and cost optimization analysis are presented.

Appendix A

The embedded Markov chain {Z _n ; n ∈ N} is ergodic if and only if ρ < 1 for our system to be stable, where $\rho =\overline{\theta}\left(1-{R}^{\ast}\left(\lambda \right)\right)\left(r+{X}_{\left[1\right]}-1\right)-p{S}^{*}\left(\delta \right)-\lambda b{X}_{\left[1\right]}\omega$ and ω = qν ⁽¹⁾ S*(δ) + ((1 − S*(δ))/δ)(1 + δg ⁽¹⁾)

• Proof To prove the sufficient condition of ergodicity, it is very convenient to use Foster’s criterion (see Pakes [26]), which states that the Markov chain {Z _n ; n ∈ N} is an irreducible and aperiodic Markov chain is ergodic if there exists a nonnegative function f(j), j ∈ N and ε > 0,, such that mean drift ψ _j  = E[f(Z _n + 1) − f(Z _n )/Z _n  = j] is finite for all j∈ N and ψ _j  ≤ − ε for all j∈ N, except perhaps for a finite number j’s. In our case, we consider the function f(j) = j. then we have

$$ {\psi}_j=\left\{\begin{array}{l}p{S}^{*}\left(\delta \right)+\lambda b{X}_{\left[1\right]}\omega -1,\kern13.25em j=0,\\ {}\\ {}\overline{\theta}\left(1-{R}^{\ast}\left(\lambda \right)\right)\left(r+{X}_{\left[1\right]}-1\right)-p{S}^{*}\left(\delta \right)-\lambda b{X}_{\left[1\right]}\omega -1,\kern2.75em j=1,2\dots \end{array}\right. $$

Clearly the inequality $\overline{\theta}\left(1-{R}^{\ast}\left(\lambda \right)\right)\left(r+{X}_{\left[1\right]}-1\right)+p{S}^{*}\left(\delta \right)+\lambda b{X}_{\left[1\right]}\left(q{\nu}^{(1)}{S}^{*}\left(\delta \right)+\left(\left(1-{S}^{*}\left(\delta \right)\right)/\delta \right)\left(1+\delta {g}^{(1)}\right)\right)<1$ is a sufficient condition for ergodicity.

To prove the necessary condition, As noted in Sennot et al. [31], if the Markov chain {Z _n ; n ≥ 1} satisfies Kaplan’s condition, namely, ψ _j  < ∞ for all j ≥ 0 and there exits j ₀ ∈ N such that ψ _j  ≥ 0 for j ≥ j ₀. Notice that, in our case, Kaplan’s condition is satisfied because there is a k such that m _ij  = 0 for j < i − k and i > 0, where M = (m _ij ) is the one step transition matrix of {Z _n ; n ∈ N}. Then $\overline{\theta}\left(1-{R}^{\ast}\left(\lambda \right)\right)\left(r+{X}_{\left[1\right]}-1\right)+p{S}^{*}\left(\delta \right)+\lambda b{X}_{\left[1\right]}\left(q{\nu}^{(1)}{S}^{*}\left(\delta \right)+\left(\left(1-{S}^{*}\left(\delta \right)\right)/\delta \right)\left(1+\delta {g}^{(1)}\right)\right)\ge 1$ implies the non-ergodicity of the Markov chain.