A general workload conservation law with applications to queueing systems

Abstract

In the spirit of Little’s law $L=\lambda W$ and its extension $H=\lambda G$ we use sample-path analysis to give a general conservation law. For queueing models the law relates the asymptotic average workload in the system to the conditional asymptotic average sojourn time and service times distribution function. This law generalizes previously obtained conservation laws for both single- and multi-server systems, and anticipating and non-anticipating scheduling disciplines. Applications to single- and multi-class queueing and other systems that illustrate the versatility of this law are given. In particular, we show that, for anticipative and non-anticipative scheduling rules, the unconditional delay in a queue is related to the covariance of service times and queueing delays.

Appendix

Our first result is a bound on the busy period for a single-server queue. Let $\{A_k= T_k-T_{k-1}, k\ge 1\}$ be the sequence of inter-arrival times, and define $A'_k=\min \{A_k,M\}$, where M is chosen such that $\lambda ' ES <1$; i.e., $\lambda<\lambda '<\mu$, ($\mu =1/ES$). Here $\lambda '$ is the arrival rate of the new modified system. Also, let EB be the long-run average length of a busy period.

Lemma 4.1

Let $\rho <1$, then for a G / G / 1 model where relevant limits are well-defined, we have

$$\begin{aligned} EB \le \rho M\Big /\left( \frac{\lambda }{\lambda '}-\rho \right) . \end{aligned}$$

Proof

We follow the notation used in El-Taha and Stidham [7, pp. 25–26]. Let A(0, t), Y(0, t) ($A'(0,t)$, and $Y'(0,t)$) be the original (modified) system arrivals that find the system in state 0 and total time in state 0 during [0, t), respectively. Note that $A'(0,t) \le A(0,t)$ and $Y'(0,t)\le Y(0,t)$. This follows by noting the effect of reducing inter-arrival times on, possibly, joining neighboring busy periods. Using Theorem 1.12 and Example 1.7 in El-Taha and Stidham [7, p. 26], the long-run average idle period for the modified system $I'=\lambda '(0)^{-1} \le M$; moreover,

$$\begin{aligned} EB:= \lim _{t\rightarrow \infty }\frac{t-Y(0,t)}{A(0,t)}\le \lim _{t\rightarrow \infty }\frac{t-Y'(0,t)}{A'(0,t)}:= EB'. \end{aligned}$$

Therefore,

$$\begin{aligned} EB\le EB'= & {} \frac{\lambda ' ES}{\lambda '(0)(1-\lambda 'ES)}\\\le & {} \frac{\lambda ' ES M}{(1-\lambda 'ES)}\\= & {} \rho M/\left( \frac{\lambda }{\lambda '}-\rho \right) , \end{aligned}$$

which completes the proof. $\square$

Ayesta [1] gives an upper bound for the busy period of a single-server queue using a different approach. Our bound is given using a sample-path approach consistent with the article analysis. No stochastic assumptions are used. In the same spirit, we prove the second result that is needed in the proof of Corollary 2.3.

Lemma 4.2

Suppose that $E[S^2] = \int _{0}^{\infty }x^2 \mathrm{d}F(x) <\infty$, then

$$\begin{aligned} \lim _{x\rightarrow \infty }x^2 F^c(x)=0. \end{aligned}$$

Proof

Note that

$$\begin{aligned} E[S^2]=\int _{0}^{x}t^2\mathrm{d}F(t) + \int _{x}^{\infty }t^2\mathrm{d}F(t). \end{aligned}$$

Now, taking limits of both sides as $x \rightarrow \infty$ leads to

$$\begin{aligned} E[S^2]=\int _{0}^{\infty }t^2\mathrm{d}F(t) + \lim _{x\rightarrow \infty }\int _{x}^{\infty }t^2\mathrm{d}F(t). \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{x\rightarrow \infty }\int _{x}^{\infty }t^2\mathrm{d}F(t)=0. \end{aligned}$$

(27)

Now,

$$\begin{aligned} \lim _{x\rightarrow \infty }x^2 F^c(x)= & {} \lim _{x\rightarrow \infty }\int _{x}^{\infty }x^2 \mathrm{d}F(t)\\\le & {} \lim _{x\rightarrow \infty }\int _{x}^{\infty }t^2 \mathrm{d}F(t)\\= & {} 0. \end{aligned}$$

The following known result does not seem to be well-reported in the literature. $\square$

Lemma 4.3

Let f(x) be a monotone continuous function, then

$$\begin{aligned} \int _a^b f(x)\mathrm{d}x+ \int _{f(a)}^{f(b)} f^{-1}(y)\mathrm{d}y =bf(b)-af(a). \end{aligned}$$

Proof

We start by computing the integral $\int _{f(a)}^{f(b)} f^{-1}(y)\mathrm{d}y$. Let $x=f^{-1}(y)$, so that $y=f(x)$ and $\mathrm{d}y=\mathrm{d}f(x)$. Thus, using integration by parts,

$$\begin{aligned} \int _{f(a)}^{f(b)} f^{-1}(y)\mathrm{d}y= & {} \int _a^b x\mathrm{d}f(x)\\= & {} \left. xf(x)\right| _a^b - \int _a^b f(x)\mathrm{d}x\\= & {} bf(b)-af(a)-\int _a^b f(x)\mathrm{d}x. \end{aligned}$$

$\square$

Suppose that $a=f(a)=0$. Then this lemma says the integral of f plus the integral of its inverse give the area of the rectangle with sides b and f(b).