Light-traffic analysis of random access systems without collisions

Abstract

We consider a retrial queueing model for random access protocols arising in local area networks such as carrier sense multiple access networks. In our model, one channel is shared among multiple nodes. Each node accesses the channel according to a Poison process and the holding time of the channel is exponentially distributed. Blocked users join the orbit and retry to access again after an exponentially distributed time depending on the number of retrials so far. As the model under study is not amenable for exact analysis, we focus on its performance in the light traffic regime. In particular, we describe a fast algorithm for calculating the terms in the Maclaurin series expansion of various performance measures, the arrival rate being the independent parameter of the expansion. We illustrate our approach by various numerical examples and verify the accuracy of the light traffic approximation by means of simulation.

Stability of the infinite population model

We now formally prove the existence of the stationary distribution for the infinite population model, using the following Foster–Lyapunov criterion from Chen (1991).

Theorem 1

Given a Markov process with countable state space E and irreducible generator matrix $Q=[q_{ij}]_{i,j \in E}$. Suppose that there exist a compact function $g: E \rightarrow \mathbb {R}^+$ and constants $\kappa \ge 0$ and $\eta > 0$ such that

$$\begin{aligned} f(i) \doteq \sum _{j \in E} q_{ij}(g(j) - g(i)) \le \kappa - \eta g(i) \,, \quad {for }\, i \in E \, . \end{aligned}$$

(5)

then the Markov chain is positive recurrent and hence has a unique stationary distribution.

Compactness of a function $E \rightarrow \mathbb {R}^+$ means that for each $d \in \mathbb {R}^+$, the set $\{i \in E; g(i) \le d\}$ is finite.

Let $[{\mathbf {x}}, y] = [x_0,\ldots ,x_{K-1},y] \in \mathcal S \times \{0,1\}$ denote an element of the state space of the infinite population model. Here, $x_k$ is the number of customers in orbit that have retried k times so far and y is the number of customers being served (y is either 1 or 0). To prove stability, we consider the following function,

$$\begin{aligned} g([{\mathbf {x}}, y]) \doteq \sum _{k=0}^{K-1} (K-k) x_k \, . \end{aligned}$$

(6)

It is easily verified that this function is indeed compact. We now calculate an upper bound for the left-hand side in (5), differentiating between states where a customer is being served and states where this is not the case. When no customer is being served, the possible transitions are retrials and new arrivals, such that the left-hand side of (5) can be written as

$$\begin{aligned} f([{\mathbf {x}},0]) = \widetilde{\lambda }(g([{\mathbf {x}}, 1])-g([{\mathbf {x}}, 0])) + \sum _{k=0}^{K-1} \gamma _k x_k (g([{\mathbf {x}} - {\mathbf {e}}_k, 1])-g([{\mathbf {x}}, 0])) \, . \end{aligned}$$

Plugging (6) into the former expression further yields,

$$\begin{aligned} f([{\mathbf {x}},0]) = -\sum _{k=0}^{K-1} (K-k) \gamma _k x_k \le -\widetilde{\gamma }\sum _{k=0}^{K-1} (K-k) x_k = -\widetilde{\gamma }\, g([{\mathbf {x}},0]) \end{aligned}$$

where $\widetilde{\gamma }= \inf _{k} \gamma _k > 0$ is the smallest retrial rate. When a customer is being served, the possible transitions are new arrivals, retrials and service completions, leading to,

$$\begin{aligned} f([{\mathbf {x}},1]) = \mu (g([{\mathbf {x}}, 0])-g([{\mathbf {x}}, 1])) + \widetilde{\lambda }(g([{\mathbf {x}} + {\mathbf {e}}_0, 1])-g([{\mathbf {x}}, 1])) \\ \begin{aligned}&+ \sum _{k=0}^{K-2} \gamma _k x_k h_k (g([{\mathbf {x}} - {\mathbf {e}}_k, 1])-g([{\mathbf {x}}, 1]))\\&+ \sum _{k=0}^{K-2} \gamma _k x_k (1-h_k) (g([{\mathbf {x}} - {\mathbf {e}}_k +{\mathbf {e}}_{k+1}, 1])-g([{\mathbf {x}}, 1])) \end{aligned}\\ + \gamma _{K-1} x_{K-1} (g([{\mathbf {x}} - {\mathbf {e}}_{K-1}, 1])-g([{\mathbf {x}}, 1])) \, . \end{aligned}$$

Accounting for (6), we get the following upper bound,

$$\begin{aligned} f([{\mathbf {x}},1])&= \widetilde{\lambda }K - \sum _{k=0}^{K-2} \gamma _k x_k h_k (K-k) - \sum _{k=0}^{K-2} \gamma _k x_k (1-h_k) - \gamma _{K-1} x_{K-1} \\&= \widetilde{\lambda }K - \sum _{k=0}^{K-1} \gamma _k x_k h_k (K-k) - \sum _{k=0}^{K-1} \gamma _k x_k (1-h_k) \\&\le \widetilde{\lambda }K - \widetilde{\gamma }\sum _{k=0}^{K-1} x_k h_k (K-k) - \widehat{\gamma }\sum _{k=0}^{K-1} x_k (1-h_k) (K-k) \\&\le \widetilde{\lambda }K - \widehat{\gamma }\sum _{k=0}^{K-1} x_k (K-k) = \widetilde{\lambda }K - \widehat{\gamma }g([{\mathbf {x}},1]) \end{aligned}$$

with $\widehat{\gamma }= \inf _k \frac{\gamma _k}{K-k} > 0$. Combining both cases then show that (5) holds for $\kappa = \widetilde{\lambda }K$ and $\eta = \widehat{\gamma }$ such that the Markov process has a unique stationary distribution.