The Israeli Queue with a general group-joining policy

Abstract

We consider a single-server multi-queue system with unlimited-size batch service where the next queue to be served is the one with the most senior customer (the so called ‘Israeli Queue’). We study a Markovian system with state-dependent group-joining policy and derive results for various performance measures, such as steady-state distribution of the number of groups in the system, sojourn times, group sizes, and lengths of busy periods. Closed-form expressions are obtained for both the Uniform and the Geometric joining policies. Numerical results are presented.

Appendix

1.1 Proof of Proposition 3.1

Proof

We need to show that

$$\begin{aligned} \frac{1}{\mu }+\frac{\lambda }{2\mu ^2} \le \frac{\lambda }{\mu ^2(1-e^{-\frac{\lambda }{\mu }})}. \end{aligned}$$

(6.1)

By setting $a=\frac{\lambda }{\mu }$, and some straightforward algebra, Eq. (6.1) is equivalent to

$$\begin{aligned} \frac{2-a}{2+a} \le e^{-a}. \end{aligned}$$

(6.2)

Equation (6.2) clearly holds for $a\ge 2$. We will prove that it also holds for $0\le a<2$. Note that (6.2) can be written as

$$\begin{aligned} a+2+(a+2)e^a-4e^a\ge 0. \end{aligned}$$

Define $f(a)=a+2+(a+2)e^a-4e^a$. We need to prove that $f(a)\ge 0$ for all $0\le a<2$. Note that $f'(a)=1+e^a(a-1)$, and $f''(a)=ae^a\ge 0$. Therefore, $f'(a)$ is non-decreasing, and with $f'(0)=0$ we have that $f'(a)\ge 0$ for $0\le a<2$. This implies that f(a) is also non-decreasing for $0\le a<2$, and with $f(0)=0$, the proof is completed. $\square$