The Israeli queue with retrials

Abstract

The so-called “Israeli queue” (Boxma et al. in Stoch Model 24(4):604–625, 2008; Perel and Yechiali in Probab Eng Inf Sci, 2013; Perel and Yechiali in Stoch Model 29(3):353–379, 2013) is a multi-queue polling-type system with a single server. Service is given in batches, where the batch sizes are unlimited and the service time of a batch does not depend on its size. After completing service, the next queue to be visited by the server is the one with the most senior customer. In this paper, we study the Israeli queue with retrials, where the system is comprised of a “main” queue and an orbit queue. The main queue consists of at most $M$ groups, where a new arrival enters the main queue either by joining one of the existing groups, or by creating a new group. If an arrival cannot join one of the groups in the main queue, he goes to a retrial (orbit) queue. The orbit queue dispatches orbiting customers back to the main queue at a constant rate. We analyze the system via both probability generating functions and matrix geometric methods, and calculate analytically various performance measures and present numerical results.

Appendix

Proposition 8.1

For a given $1\le M <\infty$, the polynomials $q_{m}(z)$ are of the following form:

$$\begin{aligned}&q_m(z)=z^{m-1}\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}+(1-z)h_m^{(M)}(z), \quad m=0,1,\ldots ,M,\\&q_{M+1}(z)=(1-z)h_{M+1}^{(M)}(z), \end{aligned}$$

where

$$\begin{aligned} h_0^{(M)}(z)&= h_1^{(M)}(z)=0,\\ h_m^{(M)}(z)&= h_{m-1}^{(M)}(z)(\lambda (1-p)^{m-1}+\mu +\gamma )z-h_{m-1}^{(M)}(z)\mu z(\lambda (1-p)^{m-2}z+\gamma )\\&-\,\mu \gamma z^{m-2}\prod \limits _{k=0}^{m-3}{(\lambda (1-p)^k+\gamma )}, \quad m=2,3,\ldots ,M, \end{aligned}$$

and

$$\begin{aligned} h_{M+1}^{(M)}(z)&= h_{M}^{(M)}(z)(\lambda (1-p)^M(1-z)+\mu )z-h_{M-1}^{(M)}(z)\mu z(\lambda (1-p)^{M-1}z+\gamma )\\&+\,z^M\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )}-\mu \gamma z^{M-1}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}. \end{aligned}$$

Proof

The proof is conducted by induction over $m$. First, $q_0(z)=1$ and $q_1(z)=\lambda +\gamma$. Therefore, $h_0^{(M)}(z)=h_1^{(M)}(z)=0$. Next,

$$\begin{aligned} q_2(z)&= \alpha _{1}(z)q_{1}(z)-\mu z(\lambda z+\gamma )q_{0}(z)\\&= (\lambda (1-p)+\gamma )z(\lambda +\gamma )-\mu (\lambda z+\gamma )\\&= z\prod \limits _{k=0}^{1}{(\lambda (1-p)^k+\gamma )}-\mu \gamma (1-z), \end{aligned}$$

so that $h_{2}^{(M)}(z)=-\mu \gamma$. Assume now that the proposition holds for any $m=2,3,\ldots ,M-1$. For $m+1$, we have

$$\begin{aligned} q_{m+1}(z)&= \alpha _m(z)q_m(z)-\mu z(\lambda (1-p)^{m-1}z+\gamma )q_{m-1}(z) \nonumber \\&= (\lambda (1-p)^m+\mu +\gamma )z\left( z^{m-1}\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}+(1-z)h_m^{(M)}(z)\right) \nonumber \\&\!-\,\mu z(\lambda (1-p)^{m-1}z\!+\!\gamma )\left( z^{m-2}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k\!+\!\gamma )}\!+\!(1-z)h_{m-1}^{(M)}(z)\right) \nonumber \\&= z^m\prod \limits _{k=0}^{m}{(\lambda (1-p)^k+\gamma )}+\mu z^m\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}\nonumber \\&+\,h_m^{(M)}(z)(\lambda (1-p)^m+\mu +\gamma )z(1-z) \nonumber \\&\!-\,\mu z^m\lambda (1-p)^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k\!+\!\gamma )}\!-\!\mu z^{m-1}\gamma \prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&-\,h_{m-1}^{(M)}(z)\mu z(\lambda (1-p)^{m-1}z+\gamma )(1-z). \end{aligned}$$

(8.1)

Note that

$$\begin{aligned}&\mu z^m\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}-\mu z^m\lambda (1-p)^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )}\nonumber \\&\quad -\,\mu z^{m-1}\gamma \prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&\quad =\mu z^m\lambda (1-p)^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )}+\mu z^m\gamma \prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&\qquad -\,\mu z^m\lambda (1-p)^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )}-\mu z^{m-1}\gamma \prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&\quad =-\mu \gamma z^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )}(1-z). \end{aligned}$$

(8.2)

Substituting (8.2) in (8.1) results in

$$\begin{aligned} q_{m+1}(z)&= z^m\prod \limits _{k=0}^{m}{(\lambda (1-p)^k+\gamma )}+(1-z) \Bigg (h_m^{(M)}(z)(\lambda (1-p)^m+\mu +\gamma )z \nonumber \\&-\,h_{m-1}^{(M)}(z)\mu z(\lambda (1-p)^{m-1}z+\gamma ) -\mu \gamma z^{m-1}\prod \limits _{k=0}^{m-2}{(\lambda (1-p)^k+\gamma )}\Bigg ).\nonumber \\ \end{aligned}$$

(8.3)

Now, for $m=M+1$, we get

$$\begin{aligned} q_{M+1}(z)&= \alpha _M(z)q_M(z)-\mu z(\lambda (1-p)^{M-1}z+\gamma )q_{M-1}(z) \nonumber \\&= (\lambda (1-p)^M(1-z)+\mu )z\left( z^{M-1}\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )}+(1-z)h_M^{(M)}(z)\right) \nonumber \\&-\,\mu z(\lambda (1-p)^{M-1}z+\gamma )\left( z^{M-2}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}+(1-z)h_{M-1}^{(M)}(z)\right) \nonumber \\&= z^M\lambda (1-p)^M(1-z)\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )}+\mu z^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )} \nonumber \\&+\,h_M^{(M)}(z)(\lambda (1-p)^M(1-z)+\mu )z(1-z)\nonumber \\&-\,z^M\lambda (1-p)^{M-1}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&-\,\mu z^{M-1}\gamma \prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}-h_{M-1}^{(M)}(z)\mu z(\lambda (1-p)^{M-1}z\!+\!\gamma )(1-z),\qquad \qquad \end{aligned}$$

(8.4)

which after some algebra leads to

$$\begin{aligned} q_{M+1}(z)&= (1-z)\Bigg (h_M^{(M)}(z)(\lambda (1-p)^M(1-z)+\mu )z\nonumber \\&-\,h_{M-1}^{(M)}(z)\mu z(\lambda (1-p)^{M-1}z+\gamma )\nonumber \\&+\,z^M\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )}\nonumber \\&-\,\mu \gamma z^{M-1}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}\Bigg ). \end{aligned}$$

(8.5)

This completes the proof.

Proposition 8.2

For all $1\le M < \infty$,

$$\begin{aligned} h_{M+1}^{(M)}(1)=\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )} -\gamma \mu ^M\sum \limits _{m=0}^{M-1}{\frac{1}{\mu ^m}\prod \limits _{k=0}^{m-1} {(\lambda (1-p)^k+\gamma )}}. \end{aligned}$$

(8.6)

Proof

By induction over $M$. First, assume $M=1$. Recall from Proposition 8.1 that for all $1\le M <\infty$, $h_0^{(M)}(z)=h_1^{(M)}(z)=0$. Then, from Proposition 8.1, we get

$$\begin{aligned} h_2^{(1)}(1)=\lambda (1-p)(\lambda +\gamma )-\mu \gamma , \end{aligned}$$

which coincides with Eq. (8.6) when $M=1$.

Assume now that the proposition holds for $M-1$. Using Proposition 8.1 for $M$, we get

$$\begin{aligned} h_{M+1}^{(M)}(1)&= h_{M}^{(M)}(1)\mu -h_{M-1}^{(M)}(1)\mu (\lambda (1-p)^{M-1}+\gamma ) \nonumber \\&+\,\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k\!+\!\gamma )}\!-\! \mu \gamma \prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}.\qquad \qquad \end{aligned}$$

(8.7)

In addition, we use the following two relations

($a$):

$h_{M-1}^{(M)}(z)=h_{M-1}^{(M-1)}(z)$,

($b$):

$h_{M}^{(M-1)}(1)=h_{M}^{(M)}(1)-h_{M-1}^{(M-1)}(1)(\lambda (1-p)^{M-1}+\gamma ) + \lambda (1-p)^{M-1}\prod _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}$,

$\square$

so that Eq. (8.7) translates to

$$\begin{aligned} h_{M+1}^{(M)}(1)&= \mu \left( h_{M}^{(M-1)}(1)+h_{M-1}^{(M)}(1)(\lambda (1-p)^{M-1}+\gamma )-\, \lambda (1-p)^{M-1}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}\right) \nonumber \\&-h_{M-1}^{(M)}(1)\mu (\lambda (1-p)^{M-1}+\gamma )+\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )} \nonumber \\&-\,\mu \gamma \prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}. \end{aligned}$$

(8.8)

Using the validity for $M-1$ yields

$$\begin{aligned} h_{M+1}^{(M)}(1)&= \mu \Bigg (\lambda (1-p)^{M-1}\prod \limits _{k=0}^{M-2} {(\lambda (1-p)^k+\gamma )} -\gamma \mu ^{M-1}\sum \limits _{m=0}^{M-2}{\frac{1}{\mu ^m}\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}} \nonumber \\&-\, \lambda (1-p)^{M-1}\prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )}\Bigg )+\lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )} \nonumber \\&-\,\mu \gamma \prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&= \lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )} -\gamma \mu ^M\sum \limits _{m=0}^{M-2}{\frac{1}{\mu ^m}\prod \limits _{k=0}^{m-1}{(\lambda (1-p)^k+\gamma )}} \nonumber \\&-\,\mu \gamma \prod \limits _{k=0}^{M-2}{(\lambda (1-p)^k+\gamma )} \nonumber \\&= \lambda (1-p)^M\prod \limits _{k=0}^{M-1}{(\lambda (1-p)^k+\gamma )}- \gamma \mu ^M\sum \limits _{m=0}^{M-1}{\frac{1}{\mu ^m}\prod \limits _{k=0}^{m-1} {(\lambda (1-p)^k+\gamma )}}.\qquad \end{aligned}$$

(8.9)

This completes the proof.