Performance of CSMA in multi-channel wireless networks

Abstract

We analyze the performance of CSMA in multi-channel wireless networks, accounting for the random nature of traffic. Specifically, we assess the ability of CSMA to fully utilize the radio resources and in turn to stabilize the network in a dynamic setting with flow arrivals and departures. We prove that CSMA is optimal in the ad-hoc mode, when each flow goes through a unique dedicated wireless link from a transmitter to a receiver. It is generally suboptimal in infrastructure mode, when all data flows originate from or are destined to the same set of access points, due to the inherent bias of CSMA against downlink traffic. We propose a slight modification of CSMA that we refer to as flow-aware CSMA, which corrects this bias and makes the algorithm optimal in all cases. The analysis is based on some time-scale separation assumption which is proved valid in the limit of large flow sizes.

Appendix

Proof of Lemma 1

For any class k, let

$$\beta_k=\min_{j:k\in V_j}\beta_{kj}. $$

Note that β _k >0. We have for all $y\in\mathcal{Y}(x)$:

$$w(x,y)\ge\prod_{k:x_k>0} {x_k(x_k-1) \cdots(x_k-y_k+1)\over x_k^{y_k}} {\beta_k^J} u(x,y). $$

If x _k ≤2J, we have

$${x_k(x_k-1)\cdots(x_k-y_k+1) \over x_k^{y_k}}\ge{1\over x_k^{y_k}} \ge{1\over(2J)^J}. $$

Otherwise, we have using the fact that y _k ≤J for all k=1,…,K

$${x_k(x_k-1)\cdots(x_k-y_k+1) \over x_k^{y_k}}\ge \biggl({x_k-y_k+1 \over x_k} \biggr)^{y_k} \ge{1\over2^J}. $$

Combining these results, we obtain the existence of some constant m>0 such that

$$\forall y\in\mathcal{Y}(x),\quad w(x,y)\ge m u(x,y). $$

Now let

$$\mathcal{Z}(x)= \biggl\{y\in\mathcal{Y}(x): \log\bigl(u(x,y)\bigr)\ge\biggl(1-{ \epsilon \over 2}\biggr) \log\bigl(u(x)\bigr) \biggr\}. $$

We have

$$\sum_{y\in\mathcal{Z}(x)} \pi(x,y)\log\bigl(u(x,y)\bigr)\ge \biggl(1-{\epsilon \over2}\biggr) \log\bigl(u(x)\bigr)\sum _{y\in\mathcal{Z}(x)} \pi(x,y). $$

Using the fact that w(x,y)≤u(x,y) for all $y\in\mathcal{Y}(x)$, we get

where M denotes the total number of schedules (that is, the cardinality of $\mathcal{Y}$). Since u(x) tends to +∞ when |x|≡∑ _k x _k tends to +∞, this quantity is less than ϵ/2 for all states x but some finite number. In those states, we have

$$\sum_{y\in\mathcal{Z}(x)} \pi(x,y)\ge1-{\epsilon\over2}. $$

We deduce that in all states x but some finite number:

 □

Proof of Lemma 2

Let

$$v(x,y)= \prod_{k:x_k\ge J}(x_k{ \alpha_k})^{y_k}. $$

There are some positive constants m,M such that

$$\forall x\in{\mathbb{N}}^K,\ \forall y\in\mathcal{Y},\quad m\le {u(x,y)\over v(x,y)}\le M. $$

The proof then follows from the fact that:

 □

Proof of Theorem 1

If the vector of traffic intensities lies in the interior of the capacity region, there exist some ϵ>0 and some probability measure π on $\mathcal{Y}$ such that

$$ \forall k=1,\ldots,K,\quad\rho_k= \varphi_k(1-2\epsilon)\sum_{y\in \mathcal{Y}} \pi(y)y_k. $$

(11)

Note that we can choose π(y)>0 for all $y\in\mathcal{Y}$.

Define the Lyapunov function:

$$F(x)=\sum_{k:x_k>0} {x_k \sigma_k\over\varphi_k} \log(x_k \alpha_k). $$

The corresponding drift is given by

In particular, we have ΔF(x)=G(x)+H(x) with

where we use the convention 0log(0)≡0. Since ϕ _k (x)≤Jφ _k , the function H(x) is bounded. Regarding G(x), it follows from (5) and (11) that

By Lemma 1, we have for all states x but some finite number:

Since π(y)>0 for all $y\in\mathcal{Y}$, the first term tends to −∞ when |x|≡∑ _k x _k tends to +∞. By Lemma 2, the second term is bounded. We deduce the existence of some δ>0 such that ΔF(x)≤−δ for all states x but some finite number. The proof then follows from Foster’s criterion. □

Proof of Theorem 2

In the following, we consider (X ^N(t)) _N≥1 as a sequence of stochastic processes in the space $\mathcal{D}_{{\mathbb{N}}^{K}}([0,\infty[)$ of càd-làg functions with values in ℕ^K with the Skorohod topology.

First, we have to prove the tightness of the sequence (X ^N(t)). It is enough to remark that, for all N≥1, $X^{N}_{k}(t)$ is stochastically dominated by a Poisson process of intensity λ _k and stochastically dominates an M/M/1 queue with arrival rate λ _k and service rate φ _k /σ _k . Thus, the conditions of the Arzelà–Ascoli theorem are fulfilled and the sequence (X ^N(t)) is tight (see [4, Theorem 12.3]).

We now consider a bounded function f on ℕ^K. Denote by Ω ^N the infinitesimal generator of the Markov process (X ^N(t),Y ^N(t)). For all x∈ℕ^K and $y\in\mathcal{Y}$, we have

$$\varOmega^N(f) (x,y) = \sum_{k=1}^K \lambda_k \bigl(f(x+e_k)-f(x)\bigr) - \sum _{k=1}^K\varphi_k/\sigma_k \sum_{j=1}^Jy_{kj} \bigl(f(x-e_k)-f(x)\bigr). $$

Note that Ω ^N(f)(x,y) does not depend on N. We then define Ω ^∞(f)(x,y)=Ω ^N(f)(x,y). According to the Martingale characterization of Markov jump processes (see [25]), the process:

$$M_f^N(t) = f\bigl(X^N(t)\bigr) - f \bigl(X^N(0)\bigr) - \int_0^t \varOmega^N(f) \bigl(X^N(s),Y^N(s)\bigr) \,\mathrm{d}s $$

is a locale martingale and, since the process X ^N(t) is not exploding on [0,t] (it is stochastically dominated by a Poisson process), it is a martingale.

For each N≥1, define the random measure:

Γ ^N is a random variable with value in the set $\mathcal {L}(\mathcal{Y} )$ of the random measures on $[0,\infty[\times\mathcal{Y}$ such that if $\mu\in\mathcal {L}(\mathcal{Y})$ then $\mu([0,t]\times\mathcal{Y}) = t$ for all t≥0. Since $\mathcal{Y}$ is finite, the set $\mathcal {L}(\mathcal{Y})$ is compact and then the sequence (Γ ^N) _N≥1 is relatively compact.

Assume that the sequence (X ^N(t),Γ ^N) _N≥1 tends to some limit (X(t),Γ). Since

$$\int_0^t \varOmega^N(f) \bigl(X^N(s),Y^N(s)\bigr)\,\mathrm {d}s = \int_0^t \sum _{y\in\mathcal{Y}}\varOmega^N(f) \bigl(X^N(s),y \bigr) \varGamma^N(\mathrm{d}s\times \mathrm{d}y) $$

and f is bounded, this random variable tends in distribution to

$$\int_0^t \sum_{y\in\mathcal{Y}} \varOmega^\infty(f) \bigl(X(s),y\bigr) \varGamma ( \mathrm{d}s\times\mathrm{d}y). $$

It remains to characterize Γ. According to Lemma 1.3 of [16], there exists a set of random probability measures ϑ(t,.) on $\mathcal{Y}$ such that

$$\varGamma\bigl([0,t]\times B\bigr)= \int_0^t \vartheta(s,B)\,\mathrm{d}s ,\quad\text {for } B\subset \mathcal{Y}. $$

For any function g on $\mathcal{Y}$, we define the martingale:

$$\bar{M}_g^N(t) = \frac{1}{N} \biggl(g \bigl(Y^N(t)\bigr) - g\bigl(Y^N(0)\bigr) - \int _0^t \varOmega^N(g) \bigl(X^N(s),Y^N(s)\bigr)\,\mathrm{d}s \biggr). $$

For x∈ℕ^K and $y\in\mathcal{Y}$, we have

The increasing process of this martingale is

It tends to 0 on all compact sets so that the martingale tends in distribution to 0. Since $\mathcal{Y}$ is finite, g is bounded and (g(Y ^N(t))−g(Y ^N(0)))/N also tends to 0. Finally, we get that

$${1\over N}\int_0^t \varOmega^N(g) \bigl(X^N(s),Y^N(s)\bigr)\,\mathrm{d}s $$

converges in distribution to 0. This implies

and for almost every s in [0,t], we have

The probability distribution ϑ(s,.) is then the stationary distribution given by (3).

It follows that

$$\int_0^t \varOmega^N(f) \bigl(X^N(s),Y^N(s)\bigr)\,\mathrm{d}s $$

converges in distribution to:

$$\int_0^t \varOmega(f) \bigl(X(s)\bigr)\, \mathrm{d}s $$

where Ω is the infinitesimal generator of the Markov process described in Sect. 5. For x∈ℕ^K, we have

$$\varOmega(f) (x) = \sum_{k=1}^K \lambda_k \bigl(f(x+e_k)-f(x) \bigr) + \phi_k(x)/\sigma_k \bigl(f(x-e_k)-f(x) \bigr), $$

where ϕ _k (x) is the mean throughput of class k in state x, given by (4).

By dominated convergence, $M_{f}^{N}(t)$ tends in distribution to

$$M_f(t) = f\bigl(X(t)\bigr) - f\bigl(X(0)\bigr) - \int _0^t \varOmega(f) \bigl(X(s)\bigr)\,\mathrm{d}s, $$

and M _f (t) is a martingale. Using the characterization of the Markov jump processes, we get that the process X(t) is a Markov process with infinitesimal generator Ω.

This concludes the proof. □

Proof of Proposition 2

For this proof, we will need the notion of fluid limits. For any $x\in {\mathbb{R}}_{+}^{K}$, we define $|x|\equiv\sum_{k=1}^{K} x_{k}$. For all n≥1, let (X ^N,n(t),Y ^N,n(t)) be the evolution of the Markov process when starting from some initial state such that |X ^N,n(0)|=n. A fluid limit is a limiting point $\bar{X}^{N}(t)$ of the laws of the processes {X ^N,n(nt)/n} _n≥1 in the set of probability measures on the space $\mathcal{D}$ of càd-làg functions on ℝ₊ with values in ${\mathbb{R}}_{+}^{K}$ with the Skorohod topology [4]. It is not difficult to show that the set of processes {X ^N,n(nt)/n,n≥1} is tight in the set of probability distributions on the space $\mathcal{D}$ endowed with the metric associated to the uniform norm on compact sets. Therefore, there exists at least one fluid limit and any fluid limit is continuous. Since the process Y ^N,n(nt) has its values in a finite space for all n≥1, it can be proved as in [9, 24] that, if there exists a deterministic time T>0 such that $\bar{X}^{N}(t)=0$ for all t≥T, then the Markov process (X ^N(t),Y ^N(t)) is ergodic.

The proof is then very similar to that given in [10] for random capture algorithms. Consider a fluid limit $\bar{X}^{N}(t)$. We say that a class k is non-empty at time t if $\bar{X}^{N}_{k}(t)>0$. As long as there is a non-empty class, the J channels are used. Moreover, if some class in C _l takes channel j, all other non-empty classes in C _l use this channel. Let α _jl (t) be the fraction of time channel j is used by classes in C _l at time t. For any non-empty class k∈C _l , we have

$$ {\mathrm{d}\bar{X}_k^N \over\mathrm{d}t}(t)=\lambda_k-{ \varphi_k\over\sigma_k}\sum _{j=1}^J\alpha_{jl}(t). $$

(12)

Now define

$$W^N(t) = \sum_{l=1}^L \max_{k \in C_l} \biggl(\bar{X}^N_k(t){ \sigma_k\over\varphi_k} \biggr). $$

Note that W ^N(t)=0 if and only if $\bar{X}^{N}(t)=0$. Moreover,

$$W^N(0)\le M\equiv\sum_{l=1}^L \max_{k \in C_l} \biggl({\sigma_k\over\varphi_k} \biggr). $$

Using (12) and the fact that:

$$\sum_{j=1}^J\sum _{l=1}^L\alpha_{jl}(t)=J, $$

at any time t such that W ^N(t)>0, we deduce

$$ W^N(t) \leq\max \Biggl(0, M+ \Biggl(\sum _{l=1}^L \max_{k\in C_l} \frac{\rho_k}{\varphi_k} \Biggr)t - Jt \Biggr). $$

(13)

In L-partite networks, the capacity region is given by the throughput vectors ϕ for which there exist a set {π ₁,…,π _J } of probability distributions over {C ₁,…,C _L } such that

$$\forall k\in C_l,\quad\phi_k\le\varphi_k \sum_{j=1}^J \pi_j(C_l). $$

In particular,

$$\sum_{l=1}^L\max_{k \in C_l} \frac{\phi_k}{\varphi_k} \leq\sum_{l=1}^L\sum _{j=1}^J \pi_j(C_l)=J. $$

Since ρ lies inside the capacity region, it follows from (13) that W ^N(t)=0 for all t≥T, with

$$T= {1\over J-\sum_{l=1}^L \max_{k \in C_l} \rho_k}, $$

which implies the ergodicity of the Markov process (X ^N(t),Y ^N(t)). □

Proof of Proposition 4

Define the throughput vector $\tilde{\phi}$ such that $\tilde{\phi }_{3}(x)=\phi_{3}(x)$ and for all k≠3. The two following inequalities can be verified:

(14)

(15)

Now consider the coupling of the stochastic processes X(t) and $\tilde{X}(t)$ describing the evolution of the queues for the throughput vectors ϕ and $\tilde{\phi}$, respectively, starting from the same initial state $X(0)=\tilde{X}(0)$. The inequality (14) and the monotonicity property (15) imply that $\tilde{X}(t) \le X(t)$ a.s. at any time t≥0. In particular, the transience or the null recurrence of $\tilde{X}(t)$ implies that of X(t).

For the modified system, queues 1, 2, 4, 5 are independent M/M/1 queues with load ρ ₁. If ρ ₁≥1, the Markov process $\tilde{X}(t)$ is null recurrent or transient. Note that (10) then reduces to ρ ₃≥0.

Assume now that ρ ₁<1. To prove the transience of $\tilde{X}(t)$, we use fluid limits. Since ρ ₁<1 and queues 1, 2, 4, 5 are independent M/M/1 queues with load ρ ₁ in the modified system, there exists some finite time after which, for any initial conditions, the corresponding components of the fluid limit are null. We then consider the fluid limits with the initial condition $\bar{X}_{3}(0)=1$ and $\bar{X}_{k}(0)=0$ for all k≠3. In this case, Proposition 9.14 of [24, p. 241] applies and the fluid limit satisfies

$$\bar{X}_3(t)= 1+ \biggl( \lambda_3 - \frac{\bar{\phi}_3}{\sigma_3} \biggr)t, $$

as long as this function is positive, where $\bar{\phi}_{3}$ is the throughput of link 3 averaged over the states of other links. Since each other link is active with probability ρ ₁, it follows from (9) that:

$$\bar{\phi}_3=\frac{1}{3}\rho_1^4 - \frac{2}{3}\rho_1^3 - \frac {2}{3} \rho_1^2+1. $$

In particular, $\bar{X}_{3}(t)$ increases linearly to infinity whenever inequality (10) is satisfied and, according to [18], the Markov process $\tilde{X}(t)$ is transient. □