Sade: competitive MAC under adversarial SINR

Abstract

This paper considers the problem of how to efficiently share a wireless medium which is subject to harsh external interference or even jamming. So far, this problem is understood only in simplistic single-hop or unit disk graph models. We in this paper initiate the study of MAC protocols for the SINR interference model (a.k.a. physical model). This paper makes two contributions. First, we introduce a new adversarial SINR model which captures a wide range of interference phenomena. Concretely, we consider a powerful, adaptive adversary which can jam nodes at arbitrary times and which is only limited by some energy budget. Our second contribution is a distributed MAC protocol called Sade which provably achieves a constant competitive throughput in this environment: we show that, with high probability, the protocol ensures that a constant fraction of the non-blocked time periods is used for successful transmissions.

A Proof of Lemma 3

We prove the lemma for the case that initially $p_S \le \rho _{green}$; the other case is analogous. Consider some fixed round t in $I'$. Let $p_S$ be the cumulative probability at the beginning of t and $p'_S$ be the cumulative probability at the end of t. Moreover, let $p_{S}^{(0)}$ denote the cumulative probability of the nodes $w \in S$ with a total interference of less than $\vartheta$ in round t when ignoring the nodes in S. Similarly, let $p_{S}^{(1)}$ denote the cumulative probability of the nodes $w \in S$ with a single transmitting node in $Z_1(w) {\setminus } S$ and additionally an interference of less than $\vartheta$ in round t, and let $p_{S}^{(2)}$ be the cumulative probability of the nodes $w \in S$ that do not satisfy the first two cases (which implies that they will not experience an idle channel, no matter what the nodes in S will do). Certainly, $p_S = p_S^{(0)}+p_S^{(1)}+p_S^{(2)}$. Our goal is to determine $p'_S$ in this case. Let $q_0(S)$ be the probability that all nodes in S stay silent, let $q_1(S)$ be the probability that exactly one node in S is transmitting, and let $q_2(S) = 1-q_0(S)-q_1(S)$ be the probability that at least two nodes in S are transmitting.

First, let us simplify our setting slightly and ignore the case that $c_v > T_v$ for a node $v \in S$ at round t. By examining the 9 different cases, we obtain the following result:

$$\begin{aligned} \mathbb {E}[p'_S]\le & {} q_0(S) \cdot [(1+\gamma ) p_S^{(0)} + (1+\gamma )^{-1} p_S^{(1)} + p_S^{(2)}]\\&+\,q_1(S) \cdot [(1+\gamma )^{-1} p_S^{(0)} + p_S^{(1)}+p_S^{(2)}]\\&+\,q_2(S) \cdot [p_S^{(0)} + p_S^{(1)} + p_S^{(2)}] \end{aligned}$$

To give an example (the other cases are similar), we consider $q_0(S)$ and $p_S^{(1)}$, i.e., all nodes in S are silent and for all nodes in $w \in S$ accounted for in $p_S^{(1)}$ there is exactly one transmitting node in $Z_1(w) {\setminus } S$ and the remaining interference is less than $\vartheta$. In this case, w is guaranteed to receive a message, so according to the $Sade$ protocol, it lowers $p_w$ by $(1+\gamma )$.

The upper bound on $\mathbb {E}[p'_S]$ certainly also holds if $c_v > T_v$ for a node $v \in S$, because $p_v$ will never be increased (but possibly decreased) in this case.

Now, consider the event $E_2$ that at least two nodes in S transmit a message. If $E_2$ holds, then $\mathbb {E}[p'_S] = p'_S = p_S$, so there is no change in the system. On the other hand, assume that $E_2$ does not hold. Let $q'_0(S)=q_0(S)/(1-q_2(S))$ and $q'_1(S)=q_1(S)/(1-q_2(S))$ be the probabilities $q_0(S)$ and $q_1(S)$ under the condition of $\lnot E_2$. We distinguish between three cases.

Case 1 $p_S^{(0)}=p_S$. Then

$$\begin{aligned} \mathbb {E}[p'_S]\le & {} q'_0(S) \cdot (1+\gamma ) p_S + q'_1(S) \cdot (1+\gamma )^{-1} p_S \\= & {} ((1+\gamma )q'_0(S) + (1+\gamma )^{-1}q'_1(S)) p_S. \end{aligned}$$

We know that $q_0(S) \le q_1(S) / p_S$, so $q'_0(S) \le q'_1(S) / p_S$. If $p_S \ge \rho _{green}$, then $q'_0(S) \le q'_1(S)/5$. Hence,

$$\begin{aligned} \mathbb {E}[p'_S]\le & {} ((1+\gamma )/6 + (1+\gamma )^{-1}5/6) p_S \le (1+\gamma )^{-1/2} p_S \end{aligned}$$

since $\gamma =o(1)$. On the other hand, $p'_S \le (1+\gamma ) p_S$ in any case.

Case 2 $p_S^{(1)}=p_S$. Then

$$\begin{aligned} \mathbb {E}[p'_S]\le & {} q'_0(S) \cdot (1+\gamma )^{-1} p_S + q'_1(S) p_S \\= & {} (q'_0(S)/(1+\gamma ) + (1-q'_0(S))) p_S\\= & {} (1 - q'_0(S) \gamma /(1+\gamma ))p_S. \end{aligned}$$

Now, it holds that $1-x \gamma /(1+\gamma ) \le (1+\gamma )^{-x/2}$ for all $x \in [0,1]$ because from the Taylor series of $e^x$ and $\ln (1+x)$ it follows that

$$\begin{aligned} (1+\gamma )^{-x/2} \ge 1- (x \ln (1+\gamma ))/2 \ge 1-(x (1-\gamma /2)\gamma )/2 \end{aligned}$$

and

$$\begin{aligned} 1 - x \gamma /(1+\gamma ) \le 1-(x (1-\gamma /2)\gamma )/2 \end{aligned}$$

for all $x, \gamma \in [0,1]$ as is easy to check. Therefore, when defining $\varphi = q'_0(S)$, we get $\mathbb {E}[p'_S] \le (1+\gamma )^{-\varphi /2} p_S$. On the other hand, $p'_S \le p_S \le (1+\gamma )^{\varphi } p_S$.

Case 3 $p_S^{(2)}=p_S$. Then for $\varphi =0$, $\mathbb {E}[p'_S] \le p_S = (1+\gamma )^{-\varphi /2} p_S$ and $p'_S \le p_S = (1+\gamma )^{\varphi } p_S$.

Combining the three cases and taking into account that $p_S^{(0)}+p_S^{(1)}+p_S^{(2)}=p_S$, we obtain the following result.

Claim

There is a $\phi \in [0,1]$ (depending on $p_S^{(0)}$, $p_S^{(1)}$ and $p_S^{(2)}$) so that

$$\begin{aligned} \mathbb {E}[p'_S] \le (1+\gamma )^{-\phi } p_S \quad \text{ and } \quad p'_S \le (1+\gamma )^{2\phi } p_S. \end{aligned}$$

(2)

Proof

Let $a=(1+\gamma )^{1/2}$, $b=(1+\gamma )^{\varphi /2}$ for the $\varphi$ defined in Case 2, and $c=1$. Furthermore, let $x_0 = p_S^{(0)}/p_S$, $x_1 = p_S^{(1)}/p_S$ and $x_2 =p_S^{(2)}/p_S$. Define $\phi = - \log _{1+\gamma } ((1/a)x_0 + (1/b)x_1 + (1/c)x_2)$. Then we have

$$\begin{aligned} \mathbb {E}[p'_S]\le & {} (1+\gamma )^{-1/2} p_S^{(0)} + (1+\gamma )^{-\varphi /2} p_S^{(1)} + p_S^{(2)}\\= & {} (1+\gamma )^{-\phi } p_S. \end{aligned}$$

We need to show that for this $\phi$, also $p'_S \le (1+\gamma )^{2\phi } p_S$. As $p'_S \le (1+\gamma ) p_S^{(0)} + (1+\gamma )^{\varphi } p_S^{(1)} + p_S^{(2)}$, this is true if

$$\begin{aligned} a^2 x_0 + b^2 x_1 + c^2 x_2 \le \frac{1}{((1/a)x_0 + (1/b)x_1 + (1/c)x_2)^2} \end{aligned}$$

or

$$\begin{aligned} ((1/a)x_0 + (1/b)x_1 + (1/c)x_2)^2(a^2 x_0 + b^2 x_1 + c^2 x_2) \le 1\nonumber \\ \end{aligned}$$

(3)

To prove this, we need two claims whose proofs are tedious but follow from standard math.

Claim

For any $a,b,c>0$ and any $x_0,x_1,x_2>0$ with $x_0+x_1+x_2=1$,

$$\begin{aligned} (a x_0 + b x_1 + c x_2)^2 \le (a^2 x_0 + b^2 x_1 + c^2 x_2) \end{aligned}$$

Claim

For any $a,b,c>0$ and any $x_0,x_1,x_2>0$ with $x_0+x_1+x_2=1$,

$$\begin{aligned} ((1/a)x_0 + (1/b)x_1 + (1/c)x_2)(a x_0 + b x_1 + c x_2) \le 1 \end{aligned}$$

Combining the claims, Eq. (3) follows, which completes the proof. $\square$

Hence, for any outcome of $E_2$, $\mathbb {E}[p'_S] \le (1+\gamma )^{-\varphi } p_S$ and $p'_S \le (1+\gamma )^{2\varphi } p_S$ for some $\varphi \in [0,1]$. If we define $q_S = \log _{1+\gamma } p_S$, then it holds that $\mathbb {E}[q'_S] \le q_S - \varphi$. For any time t in I, let $q_t$ be equal to $q_S$ at time t and $\varphi _t$ be defined as $\varphi$ at time t. Our calculations above imply that as long as $p_S \in [\rho _{green}, \rho _{yellow}]$, $\mathbb {E}[q_{t+1}] \le q_t - \varphi _t$ and $q_{t+1} \le q_t + 2\varphi _t$.

Now, suppose that within subframe I we reach a point t when $p_S > \rho _{yellow}$. Since we start with $p_S \le \rho _{green}$, there must be a time interval $I' \subseteq I$ so that right before $I'$, $p_S \le \rho _{green}$, during $I'$ we always have $\rho _{green} < p_S \le \rho _{yellow}$, and at the end of $I'$, $p_S > \rho _{yellow}$. We want to bound the probability for this to happen.

Consider some fixed interval $I'$ with the properties above, i.e., with $p_S \le \rho _{green}$ right before $I'$ and $p_S \ge \rho _{green}$ at the first round of $I'$, so initially, $p_S \in [\rho _{green}, (1+\gamma )\rho _{green}]$. We use martingale theory to bound the probability that in this case, the properties defined above for $I'$ hold. Consider the rounds in $I'$ to be numbered from 1 to $|I'|$, let $q_t$ and $\varphi _t$ be defined as above, and let $q'_t = q_t + \sum _{i=1}^{t-1} \varphi _i$. It holds that

$$\begin{aligned} \mathbb {E}[q'_{t+1}]= & {} \mathbb {E}[q_{t+1} + \sum _{i=1}^{t} \varphi _i]\\= & {} \mathbb {E}[q_{t+1}] + \sum _{i=1}^{t}\varphi _i \le q_t - \varphi _t + \sum _{i=1}^{t}\varphi _i\\= & {} q_t + \sum _{i=1}^{t-1} \varphi _i\\= & {} q'_t. \end{aligned}$$

Moreover, it follows from Inequality (2) that for any round t, $p'_S \le (1+\gamma )^{2\varphi _t} p_S$. Therefore, $q_{t+1} \le q_t + 2\varphi _t$, which implies that $q'_{t+1} \le q'_t + \varphi _t$. Hence, we can define a martingale $(X_t)_{t \in I'}$ with $\mathbb {E}[X_{t+1}] = X_t$ and $X_{t+1} \le X_t + \varphi _t$ that stochastically dominates $q'_t$. Recall that a random variable $Y_t$ stochastically dominates a random variable $Z_t$ if for any z, $\mathbb {P}[Y_t \ge z] \ge \mathbb {P}[Z_t \ge z]$. In that case, it is also straightforward to show that $\sum _i Y_i$ stochastically dominates $\sum _i Z_i$, which we will need in the following. Let $T = |I'|$. We will make use of Azuma’s inequality to bound $X_T$.

Fact 5

(Azuma Inequality) Let $X_0, X_1, \ldots$ be a martingale satisfying the property that $X_i \le X_{i-1} + c_i$ for all $i\ge 1$. Then for any $\delta \ge 0$,

$$\begin{aligned} \mathbb {P}[X_T > X_0 + \delta ] \le e^{-\delta ^2/(2 \sum _{i=1}^T c_i^2)}. \end{aligned}$$

Thus, for $\delta = 1/\gamma + \sum _{i=1}^T \varphi _i$ it holds in our case that

$$\begin{aligned} \mathbb {P}[X_T > X_0 + \delta ] \le e^{-\delta ^2/(2 \sum _{i=1}^T \varphi _i^2)}. \end{aligned}$$

This implies that

$$\begin{aligned} \mathbb {P}[q'_T > q'_0 + \delta ] \le e^{-\delta ^2/(2 \sum _{i=1}^T \varphi _i^2)}, \end{aligned}$$

for several reasons. First of all, stochastic dominance holds as long as $p_S \in [\rho _{green},\rho _{yellow}]$, and whenever this is violated, we can stop the process as the requirements on $I'$ would be violated, so we would not have to count that probability towards $I'$. Therefore,

$$\begin{aligned} \mathbb {P}[q_T > q_0 + 1/\gamma ] \le e^{-\delta ^2/(2 \sum _{i=1}^T \varphi _i^2)}. \end{aligned}$$

Notice that $q_T > q_0 + 1/\gamma$ is required so that $p_S > \rho _{yellow}$ at the end of $I'$, so the probability bound above is exactly what we need. Let $\varphi = \sum _{i=1}^T \varphi _i$. Since $\varphi _i \le 1$ for all i, $\varphi \ge \sum _{i=1}^T \varphi _i^2$. Hence,

$$\begin{aligned} \frac{\delta ^2}{2 \sum _{i=1}^T \varphi _i^2}\ge & {} \frac{(1/\gamma + \varphi )^2}{2 \varphi } \ge \left( \frac{1}{2\varphi \gamma ^2} + \frac{\varphi }{2} \right) . \end{aligned}$$

This is minimized for $1/(2\varphi \gamma ^2) = \varphi /2$ or equivalently, $\varphi =1/\gamma$. Thus,

$$\begin{aligned} \mathbb {P}[q_T > q_0 + 1/\gamma ] \le e^{-1/\gamma } \end{aligned}$$

Since there are at most ${f \atopwithdelims ()2}$ ways of selecting $I' \subseteq I$, the probability that there exists an interval $I'$ with the properties above is at most

$$\begin{aligned} {f \atopwithdelims ()2} e^{-1/\gamma } \le f^2 e^{-1/\gamma } \le \frac{1}{\log ^c n} \end{aligned}$$

for any constant c if $\gamma = O(1/(\log T + \log \log n))$ is small enough.