Queues with random back-offs

Abstract

We consider a broad class of queueing models with random state-dependent vacation periods, which arise in the analysis of queue-based back-off algorithms in wireless random-access networks. In contrast to conventional models, the vacation periods may be initiated after each service completion, and can be randomly terminated with certain probabilities that depend on the queue length. We first present exact queue-length and delay results for some specific cases and we derive stochastic bounds for a much richer set of scenarios. Using these, together with stochastic relations between systems with different vacation disciplines, we examine the scaled queue length and delay in a heavy-traffic regime, and demonstrate a sharp trichotomy, depending on how the activation rate and vacation probability behave as function of the queue length. In particular, the effect of the vacation periods may either (i) completely vanish in heavy-traffic conditions, (ii) contribute an additional term to the queue lengths and delays of similar magnitude, or even (iii) give rise to an order-of-magnitude increase. The heavy-traffic trichotomy provides valuable insight into the impact of the back-off algorithms on the delay performance in wireless random-access networks.

Appendix: Preliminary results and proofs

This appendix contains a few technical lemmas and some proofs that have been relegated from the main text. To make this appendix self-contained we restate some results from the main text.

Lemma 7.1

(i):

If $X_0 {\,{d \over =}\,}X$, then,

$$\begin{aligned} \prod \limits _{i=0}^{\infty }Y(a^ir), \end{aligned}$$

with $0 \le a <1$, converges for all $r\in [0,1]$.

(ii):

If $X_0 >_{\mathrm{st}} X$, then,

$$\begin{aligned} \sum \limits _{j=0}^{\infty }K(a^jr)\prod \limits _{i=0}^{j-1}Y(a^ir), \end{aligned}$$

with $0 \le a <1$, converges for all $r\in [0,1]$.

Proof

To prove case $\mathrm{(i)}$ first note that this infinite product converges if and only if

$$\begin{aligned} \sum \limits _{i=0}^{\infty }(Y(a^ir)-1) \end{aligned}$$

converges. To prove convergence of this infinite series we will use the ratio test (d’Alembert’s criterion). We have, with $h(r)=\tilde{B}(\lambda (1-r))$ and $k(r)=\tilde{B}(\lambda (1-r))G_X(r)$,

$$\begin{aligned} \lim _{i\rightarrow \infty } \left| \frac{Y(a^{i+1}r)-1}{Y(a^ir)-1}\right|&= \lim _{i\rightarrow \infty } \frac{(-a^i r + h(a^i r ))(a^{i+1} r - k(a^{i+1}r ))}{(-a^{i+1} r + h(a^{i+1} r ))(a^{i} r - k(a^{i}r))}\\&= \lim _{i\rightarrow \infty } \frac{a^{i+1} r - k(a^{i+1}r)}{a^{i} r - k(a^{i}r)}. \end{aligned}$$

By l’Hôpital’s rule,

$$\begin{aligned}&\lim _{i\rightarrow \infty } \frac{a^{i+1} r - k(a^{i+1}r)}{a^{i} r - k(a^{i}r)}\\&\quad = \lim _{i\rightarrow \infty } a \frac{1+\lambda G_X(a^{i+1}r)\tilde{B}'(\lambda (1-a^{i+1}r))-\tilde{B}(\lambda (1-a^{i+1}r))G'_X(a^{i+1}r)}{1+\lambda G_X(a^{i}r)\tilde{B}'(\lambda (1-a^{i}r))-\tilde{B}(\lambda (1-a^{i}r))G'_X(a^{i+1}r)}. \end{aligned}$$

We thus find

$$\begin{aligned} \lim _{i\rightarrow \infty } \left| \frac{Y(a^{i+1}r)-1}{Y(a^ir)-1}\right| = a < 1, \end{aligned}$$

proving case $\mathrm{(i)}$.

For case $\mathrm{(ii)}$ note that

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{K(a^{n+1}r) \prod _{i=0}^{n}Y(a^ir)}{K(a^{n}r) \prod _{i=0}^{n-1}Y(a^ir)}=Y(0)<1, \end{aligned}$$

for all $r$ as $0\le a < 1$. Thus, by the ratio test, the series in case $\mathrm{(ii)}$ converges. $\square$

Lemma 7.2

If $\alpha y + (1 - \alpha ) z = \alpha ' y' + (1 - \alpha ') z'$, with $0 \le \alpha , \alpha ' \le 1$ and $y' \le y \le z \le z'$, then

(i):

If $g(\cdot )$ is a concave function,

$$\begin{aligned} \alpha g(y) + (1 - \alpha ) g(z) \ge \alpha ' g(y') + (1 - \alpha ') g(z'). \end{aligned}$$

(54)

(ii):

If $g(\cdot )$ is a convex function,

$$\begin{aligned} \alpha g(y) + (1 - \alpha ) g(z) \le \alpha ' g(y') + (1 - \alpha ') g(z'). \end{aligned}$$

(55)

Proof

Since $y' \le y \le z \le z'$, there exist $0 \le \alpha _y, \alpha _z \le 1$, such that $y = \alpha _y y' + (1 - \alpha _y) z'$, and $z = \alpha _z y' + (1 - \alpha _z) z'$. It follows from the equality $\alpha y + (1 - \alpha ) z = \alpha ' y' + (1 - \alpha ') z'$ that $\alpha ' = \alpha \alpha _y + (1 - \alpha ) \alpha _z$, and $1 - \alpha ' = \alpha (1 - \alpha _y) + (1 - \alpha ) (1 - \alpha _z)$. Further, if $g(\cdot )$ is concave,

$$\begin{aligned} \alpha _y g(y') + (1 - \alpha _y) g(z') \le g(y), \end{aligned}$$

and

$$\begin{aligned} \alpha _z g(y') + (1 - \alpha _z) g(z') \le g(z). \end{aligned}$$

We may then write

$$\begin{aligned} \alpha g(y) + (1 - \alpha ) g(z)&\ge \alpha [\alpha _y g(y') + (1 - \alpha _y) g(z')] + (1 - \alpha ) [\alpha _z g(y') + (1 - \alpha _z) g(z')] \\&= [\alpha \alpha _y + (1 - \alpha ) \alpha _z] g(y') + [\alpha (1 - \alpha _y) + (1 - \alpha ) (1 - \alpha _z)] g(z') \\&= \alpha ' g(y') + (1 - \alpha ') g(z'), \end{aligned}$$

which completes the proof for case $\mathrm{(i)}$. The inequality in (55) follows by symmetry. $\square$

Corollary 7.3

For all $x$, if $a' \le a < 1$, $b' \ge b > 1$, then

(i):

If $g(\cdot )$ is a concave function, $\gamma _{a', b'}(x) \le \gamma _{a, b}(x) \le 1$ and thus $\kappa _{a', b'} \ge \kappa _{a, b} \ge 0$.

(ii):

If $g(\cdot )$ is a convex function, $\gamma _{a', b'}(x) \ge \gamma _{a, b}(x) \ge 1$ and thus $\chi _{a', b'} \le \chi _{a, b} \le 0$.

Proof

Taking $y = a x$, $y' = a' x$, $z = b x$, $z' = b' x$, $\alpha = (b - 1) / (b - a)$, and $\alpha ' = (b' - 1) / (b' - a')$ in (54), we obtain for $g(\cdot )$ concave,

$$\begin{aligned} \frac{(b - 1) g(a x) + (1 - a) g(b x)}{b - a}&= \alpha g(y) + (1 - \alpha ) g(z)\\&\ge \alpha ' g(y') + (1 - \alpha ') g(z')\\&= \frac{(b' - 1) g(a' x) + (1 - a') g(b' x)}{b' - a'}, \end{aligned}$$

which yields the statement for concave $g(\cdot )$.

The assertion for convex $g(\cdot )$ follows by symmetry. $\square$

Let $W$ henceforth be a nonnegative integer-valued random variable with probability distribution $p(x) = {\mathbb {P}}\{{ W = x }\}$. For any $y \ge 0$, define $F(y) = {\mathbb {P}}\{{ W \le y }\} = {\mathbb {P}}\{{ W \le \lfloor y \rfloor }\}$, with pseudo inverse

$$\begin{aligned} F^{- 1}(u) = \inf \{y: F(y) \ge u\} \end{aligned}$$

for any $u \in [0, 1]$, so that we may write

$$\begin{aligned} {\mathbb {E}}\{{ g(W) }\} = \mathop \int \limits _{u = 0}^{1} g(F^{- 1}(u)) du, \end{aligned}$$

and in particular

$$\begin{aligned} {\mathbb {E}}\{{ W }\} = \mathop \int \limits _{u = 0}^{1} F^{- 1}(u) du. \end{aligned}$$

For compactness, denote $\hat{F}^{- 1}(u) = F^{- 1}(u) / {\mathbb {E}}\{{ W }\}$,

$$\begin{aligned} x_1(\epsilon _1) = \frac{1}{\epsilon _1} \mathop \int \limits _{u = 0}^{\epsilon _1} \hat{F}^{- 1}(u) du, \end{aligned}$$

and

$$\begin{aligned} x_2(\epsilon _2) = \frac{1}{\epsilon _2} \mathop \int \limits _{u = 1 - \epsilon _2}^{1} \hat{F}^{- 1}(u) du. \end{aligned}$$

Lemma 7.4

Let $0 < \epsilon _1 \le F({\mathbb {E}}\{{ W }\})$, $0 < \epsilon _2 \le 1 - F({\mathbb {E}}\{{ W }\})$, so that $x_1(\epsilon _1) \le \hat{F}^{- 1}(\epsilon _1) \le 1$ and $x_2(\epsilon _2) \ge \hat{F}^{- 1}(1 - \epsilon _2) \ge 1$, with

$$\begin{aligned} \epsilon _1 x_1(\epsilon _1) + \epsilon _2 x_2(\epsilon _2) = \epsilon _1 + \epsilon _2, \end{aligned}$$

or equivalently,

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} \hat{F}^{- 1}(u) du = 1 - \epsilon _1 - \epsilon _2. \end{aligned}$$

(i):

If $g(\cdot )$ is a concave function,

$$\begin{aligned} (\epsilon _1 + \epsilon _2) \kappa _{x_1(\epsilon _1), x_2(\epsilon _2)} \le 1 - \frac{{\mathbb {E}}\{{ g(W) }\}}{g({\mathbb {E}}\{{ W }\})}. \end{aligned}$$

(56)

(ii):

If $g(\cdot )$ is a convex function,

$$\begin{aligned} (\epsilon _1 + \epsilon _2) \chi _{x_1(\epsilon _1), x_2(\epsilon _2)} \ge 1 - \frac{{\mathbb {E}}\{{ g(W) }\}}{g({\mathbb {E}}\{{ W }\})}. \end{aligned}$$

(57)

Proof

Write

$$\begin{aligned} {\mathbb {E}}\{{ g(W) }\} \!=\! \mathop \int \limits _{u = 0}^{\epsilon _1} g(F^{- 1}(u)) du + \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} g(F^{- 1}(u)) du + \mathop \int \limits _{u = 1 - \epsilon _2}^{1} g(F^{- 1}(u)) du.\qquad \end{aligned}$$

(58)

Because of Jensen’s inequality we find for concave $g(\cdot )$

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} g(F^{- 1}(u)) du&\le (1 - \epsilon _1 - \epsilon _2) g\left( \frac{1}{1 - \epsilon _1 - \epsilon _2} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} F^{- 1}(u) du\right) \\&= (1 - \epsilon _1 - \epsilon _2) g({\mathbb {E}}\{{ W }\}). \end{aligned}$$

Invoking Jensen’s inequality once again,

$$\begin{aligned}&\mathop \int \limits _{u = 0}^{\epsilon _1} g(F^{- 1}(u)) du + \mathop \int \limits _{u = 1 - \epsilon _2}^{1} g(F^{- 1}(u)) du \\&\quad \le \epsilon _1 g\left( \frac{1}{\epsilon _1} \mathop \int \limits _{u = 0}^{\epsilon _1} F^{- 1}(u) du\right) + \epsilon _2 g\left( \frac{1}{\epsilon _2} \mathop \int \limits _{u = 1 - \epsilon _2}^{1} F^{- 1}(u) du\right) \\&\quad = \epsilon _1 g(x_1(\epsilon _1) {\mathbb {E}}\{{ W }\}) + \epsilon _2 g(x_2(\epsilon _2) {\mathbb {E}}\{{ W }\}) \\&\quad = \gamma _{x_1(\epsilon _1), x_2(\epsilon _2)}({\mathbb {E}}\{{ W }\}) (\epsilon _1 + \epsilon _2) g({\mathbb {E}}\{{ W }\}) \\&\quad \le (1 - \kappa _{x_1(\epsilon _1), x_2(\epsilon _2)}) (\epsilon _1 + \epsilon _2) g({\mathbb {E}}\{{ W }\}). \end{aligned}$$

Substituting the above two inequalities in (58) we obtain the statement of the lemma for concave $g(\cdot )$. The assertion for convex $g(\cdot )$ follows from symmetry. $\square$

Lemma 7.5

Let $0 < \epsilon _1 \le F({\mathbb {E}}\{{ W }\})$, $0 < \epsilon _2 \le 1 - F({\mathbb {E}}\{{ W }\})$, so that $x_1(\epsilon _1) \le \hat{F}^{- 1}(\epsilon _1) \le 1$ and $x_2(\epsilon _2) \ge \hat{F}^{- 1}(1 - \epsilon _2) \ge 1$, with

$$\begin{aligned} \epsilon _1 x_1(\epsilon _1) + \epsilon _2 x_2(\epsilon _2) = \epsilon _1 + \epsilon _2, \end{aligned}$$

or equivalently,

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} \hat{F}^{- 1}(u) du = 1 - \epsilon _1 - \epsilon _2. \end{aligned}$$

(i):

If $g(\cdot )$ is a concave function,

$$\begin{aligned} \kappa _{x_1(\epsilon _1), x_2(\epsilon _2)} \ge \max \{\kappa _{\hat{F}^{- 1}(\epsilon _1), 1 + \frac{\epsilon _1}{\epsilon _2} (1 - \hat{F}^{- 1}(\epsilon _1))}, \kappa _{1 - \frac{\epsilon _2}{\epsilon _1} (\hat{F}^{- 1}(1 - \epsilon _2) - 1), \hat{F}^{- 1}(1 - \epsilon _2)}\}. \end{aligned}$$

(ii):

If $g(\cdot )$ is a convex function,

$$\begin{aligned} \chi _{x_1(\epsilon _1), x_2(\epsilon _2)} \le \min \{\chi _{\hat{F}^{- 1}(\epsilon _1), 1 + \frac{\epsilon _1}{\epsilon _2} (1 - \hat{F}^{- 1}(\epsilon _1))}, \chi _{1 - \frac{\epsilon _2}{\epsilon _1} (\hat{F}^{- 1}(1 - \epsilon _2) - 1), \hat{F}^{- 1}(1 - \epsilon _2)}\}. \end{aligned}$$

Proof

Observing that

$$\begin{aligned} x_2(\epsilon _2) \ge \hat{F}^{- 1}(1 - \epsilon _2), \end{aligned}$$

we obtain

$$\begin{aligned} \epsilon _1 x_1(\epsilon _1) \le \epsilon _1 + \epsilon _2 - \epsilon _2 \hat{F}^{- 1}(1 - \epsilon _2) = \epsilon _1 + \epsilon _2 (1 - \hat{F}^{- 1}(1 - \epsilon _2)). \end{aligned}$$

In addition,

$$\begin{aligned} x_1(\epsilon _1) \le \hat{F}^{- 1}(\epsilon _1), \end{aligned}$$

yielding

$$\begin{aligned} x_1(\epsilon _1) \le \min \{\hat{F}^{- 1}(\epsilon _1), 1 - \frac{\epsilon _2}{\epsilon _1} (\hat{F}^{- 1}(1 - \epsilon _2) - 1)\}. \end{aligned}$$

Likewise,

$$\begin{aligned} x_2(\epsilon _2) \ge \max \{\hat{F}^{- 1}(1 - \epsilon _2), 1 + \frac{\epsilon _1}{\epsilon _2} (1 - \hat{F}^{- 1}(\epsilon _1))\}. \end{aligned}$$

Combining the above two inequalities and using Corollary 7.3 completes the proof. $\square$

Proposition 7.5

Assume $g(\cdot )$ is concave and $\kappa _{a, b} >0$ for any $a < 1$ and $b > 1$, or $g(\cdot )$ is convex and $\chi _{a, b} < 0$ for any $a < 1$ and $b > 1$. If

$$\begin{aligned} \lim _{\rho \uparrow 1} \frac{{\mathbb {E}}\{{ g(W) }\}}{g({\mathbb {E}}\{{ W }\})} = 1, \end{aligned}$$

then

$$\begin{aligned} \frac{W}{{\mathbb {E}}\{{ W }\}} \xrightarrow {\;d\;}1 \hbox { as } \rho \uparrow 1. \end{aligned}$$

Proof

Take $\delta > 0$ and $\epsilon _1 = F((1 - \delta ) {\mathbb {E}}\{{ W }\})$. Then either $\epsilon _1 = 0$, or $0 < \epsilon _1 \le F({\mathbb {E}}\{{ W }\})$ and $x_1(\epsilon ) \le \hat{F}^{- 1}(\epsilon _1) \le 1 - \delta$. In the latter case, define $\epsilon _2^* = 1-\hat{F}^{-1}({\mathbb {E}}\{{ W }\})$, and observe that

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2^*} \hat{F}^{- 1}(u) du \le 1 - \epsilon _1 - \epsilon _2^*, \end{aligned}$$

while

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1} \hat{F}^{- 1}(u) du \ge 1 - \epsilon _1. \end{aligned}$$

Hence, by continuity, there must exist an $\epsilon _2 \in (0, \epsilon _2^*)$ with $x_2(\epsilon _2) > 1$ and

$$\begin{aligned} \mathop \int \limits _{u = \epsilon _1}^{1 - \epsilon _2} \hat{F}^{- 1}(u) du = 1 - \epsilon _1 - \epsilon _2, \end{aligned}$$

so that the assumptions of Lemmas 7.4 and 7.5 are satisfied. Applying these two lemmas then yields for concave $g(\cdot )$

$$\begin{aligned} \kappa _{\hat{F}^{- 1}(\epsilon _1), 1 + \epsilon _1 (1 - \hat{F}^{- 1}(\epsilon _1))} \le \kappa _{\hat{F}^{- 1}(\epsilon _1), 1 + \frac{\epsilon _1}{\epsilon _2} (1 - \hat{F}^{- 1}(\epsilon _1))} \rightarrow 0 \hbox { as } \rho \uparrow 1. \end{aligned}$$

This means that $\epsilon _1 = {\mathbb {P}}\{{ W \le (1 - \delta ) {\mathbb {E}}\{{ W }\} }\} \rightarrow 0$ as $\rho \uparrow 1$. A similar argument shows that ${\mathbb {P}}\{{ W \ge (1 + \delta ) {\mathbb {E}}\{{ W }\} }\} \rightarrow 0$ as $\rho \uparrow 1$. It now follows from the definition of convergence in probability that $\frac{W}{{\mathbb {E}}\{{ W }\}}$ converges to $1$ in probability. Hence we conclude that $\frac{W}{{\mathbb {E}}\{{ W }\}} \xrightarrow {\;d\;}1$ as $\rho \uparrow 1$ if $g(\cdot )$ is concave.

The proof for convex $g(\cdot )$ follows by symmetry. $\square$