Delay analysis of a two-class batch-service queue with class-dependent variable server capacity

Abstract

In this paper, we analyse the delay of a random customer in a two-class batch-service queueing model with variable server capacity, where all customers are accommodated in a common single-server first-come-first-served queue. The server can only process customers that belong to the same class, so that the size of a batch is determined by the length of a sequence of same-class customers. This type of batch server can be found in telecommunications systems and production environments. We first determine the steady state partial probability generating function of the queue occupancy at customer arrival epochs. Using a spectral decomposition technique, we obtain the steady state probability generating function of the delay of a random customer. We also show that the distribution of the delay of a random customer corresponds to a phase-type distribution. Finally, some numerical examples are given that provide further insight in the impact of asymmetry and variance in the arrival process on the number of customers in the system and the delay of a random customer.

Appendix A: Eliminating branching points

We analyse the steady-state pgf $D_{A,n}(z)$ for the delay of a random customer with n customers in the queue and the customer at the head of the queue is of class A. From Eq. (6) we obtain that the pgf $D_{A,n}(z)$ is equal to

$$\begin{aligned} D_{A,n}(z)&= z\frac{(1-\sigma )z}{\lambda _1(z)-\sigma }\frac{\lambda _1(z)+\sigma (z-1)}{2\lambda _1(z)-1}\lambda _1(z)^n\\&\quad +\, z\frac{(1-\sigma )z}{\lambda _2(z)-\sigma }\frac{\lambda _2(z)+\sigma (z-1)}{2\lambda _2(z)-1}\lambda _2(z)^n. \end{aligned}$$

We note that for any polynomial f(z) with real coefficients then $f(z)+f(z*)$, with $z^\star$ the complex conjugate of z, gives a real number. This is also the case for the arguments $\lambda _1(z)$ and $\lambda _2(z)$ given by

$$\begin{aligned} \lambda _{1,2}(z) = \frac{1}{2}\Big (1\pm \sqrt{1+4\sigma (1-\sigma )(z^2-1)}\Big ), \end{aligned}$$

(25)

so that $f(\lambda _1(z))+f(\lambda _2(z))$ is a function without roots and also without branching points. We first note that $(2\lambda _1(z)-1)$ and $(2\lambda _2(z)-1)$ can be written as

$$\begin{aligned} 2\lambda _1(z)-1&= \sqrt{1+4\sigma (1-\sigma )(z^2-1)},\\ 2\lambda _2(z)-1&= -\sqrt{1+4\sigma (1-\sigma )(z^2-1)}. \end{aligned}$$

By using Newton’s binomial expansion for the nth powers, we obtain

$$\begin{aligned}&D_{A,n}(z)= \Bigg (\frac{1}{2}\Bigg )^n\frac{(1-\sigma )z^2}{(2\lambda _1(z)-1)(\lambda _1(z)-\sigma )(\lambda _2(z)-\sigma )} \\&\quad \times \, \Bigg [ (\lambda _2(z)-\sigma )(\lambda _1(z)+\sigma (z-1)) \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (\sqrt{1+4\sigma (1-\sigma )(z^2-1)}\Big )^k \\&\quad -\,(\lambda _1(z)-\sigma )(\lambda _2(z)+\sigma (z-1)) \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (-\sqrt{1+4\sigma (1-\sigma )(z^2-1)}\Big )^k \Bigg ]. \end{aligned}$$

Invoking the definitions of $\lambda _1(z)$ and $\lambda _2(z)$ in Eq. (25) results in

$$\begin{aligned} D_{A,n}(z)&= \Bigg (\frac{1}{2}\Bigg )^n\frac{-z}{\sqrt{1+4\sigma (1-\sigma )(z^2-1)}}\\&\quad \,\times \,\left[ \left( \frac{1-2\sigma }{2}-(1-\sigma )z- \frac{\sqrt{1+4\sigma (1-\sigma )(z^2-1)}}{2}\right) \right. \\&\quad \times \,\sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (\sqrt{1+4\sigma (1-\sigma )(z^2-1)}\Big )^k\\&\quad -\,\left( \frac{1-2\sigma }{2}-(1-\sigma )z+\frac{\sqrt{1+4 \sigma (1-\sigma )(z^2-1)}}{2}\right) \\&\quad \times \, \left. \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (-\sqrt{1+4\sigma (1-\sigma )(z^2-1)}\Big )^k \right] . \end{aligned}$$

Rearranging of the summations leads to

$$\begin{aligned} D_{A,n}(z)&= \Bigg (\frac{1}{2}\Bigg )^n\frac{-z}{\sqrt{\cdots }} \Bigg [\Big (\frac{1-2\sigma }{2}-(1-\sigma )z\Big ) \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (\sqrt{\cdots }\Big )^k-\Big (-\sqrt{\cdots }\Big )^k \\&\quad -\,\frac{\sqrt{\cdots }}{2} \sum \limits _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \Big (\sqrt{\cdots }\Big )^k+\Big (-\sqrt{\cdots }\Big )^k\Bigg ], \end{aligned}$$

where we abbreviated $\sqrt{1+4\sigma (1-\sigma )(z^2-1)}$ as $\sqrt{\cdots }$. We clearly see that in the first summation only the terms when k is odd are non-zero and in the second summation only the even values of k remain. As a result we can write $D_{A,n}(z)$ as

and the analogue equation for a class B customer at the head of the queue is

We clearly see that both these functions are polynomials of degree $n+1$.