Anti-jamming Strategies: A Stochastic Game Approach

Abstract

Due to their shared and open-access design, wireless networks are very vulnerable to many malicious attacks, ranging from passive eavesdropping to active interfering. In this paper, using stochastic game modeling we study anti-jamming strategies and their effectiveness against two types of interference attacks: (i) a random jammer, where the malicious user combines jamming modes with sleep modes; and (ii) a sophisticated jammer, where the malicious user uses the network for a two-fold purpose: law-obedient communication with other users and non-obedient jamming against a specific (primary) user. We focus our research on constructing the optimal maxmin anti-jamming transmission strategy and an optimal strategy against a selfish malicious user. Further, employing the suggested models we demonstrate that incorporating silent modes into the anti-jamming transmission protocol, where the primary user does not transmit signals for the purpose of helping an intrusion detection system identify the source of a jamming attack, can improve communication reliability. Further, since the equilibrium strategies are obtained explicitly, we identify several interesting properties that can guide designing such anti-jamming transmission protocols.

Appendix: Proof of Theorem 1

First note that $(T_J,J)$ is an equilibrium if and only if

$$\begin{aligned} u^1=a^1_{T_JJ}+\delta \gamma u^1 \ge a^1_{SJ}+\delta \gamma _S u^1, \end{aligned}$$

(14)

$$\begin{aligned} u^2=a^2_{T_JJ}+\delta \gamma u^2 \ge \delta u^2. \end{aligned}$$

(15)

Thus, $u^1$ and $u^2$ have to be given by (6). (15) always hold, and (14) holds if and only if (5) holds.

It is clear that there is no other pure equilibrium. Now look for mixed equilibrium. Let (7) hold. By (1) and (2), a couple of probability vectors $(\varvec{x}^1,\varvec{x}^2)$ is an equilibrium with payoffs $(u^1,u^2)$ if and only if it is a solution of the equations

$$\begin{aligned} u^1=U^1(u^1) \text{ and } u^2=U^2(u^2), \end{aligned}$$

(16)

where $(U^1(u^1),\varvec{x}^2)$ and $(U^2(u^2),\varvec{x}^1)$ are solution of the following LP problems:

$$\begin{aligned} \begin{array}{clll} \min U^1(u^1)&{}&{}&{}\\ L^1_{T_J}(u^1,x^2_S)&{}:=(a^1_{T_JS}+\delta u^1)x^2_S&{}+(a^1_{T_JJ}+\delta \gamma u^1) (1-x^2_S)&{}\le U^1(u^1), \\ L^1_S(u^1,x^2_S)&{}:=\delta u^1x^2_S&{}+(a^1_{SJ}+\delta \gamma _S u^1)(1-x^2_S) &{}\le U^1(u^1), \end{array} \end{aligned}$$

(17)

$$\begin{aligned} \begin{array}{clll} \min U^2(u^2)&{}&{}&{}\\ L^2_S(u^2,x^1_{T_J})&{}:=\delta u^2 x^1_{T_J}&{}+\delta u^2 (1-x^1_{T_J})&{}\le U^2(u^2), \\ L^2_J(u^2,x^1_{T_J})&{}:=(a^2_{T_JJ}+\delta \gamma u^2)x^1_{T_J}&{}+\delta \gamma _S u^2(1-x^1_{T_J}) &{}\le U^2(u^2) \end{array} \end{aligned}$$

(18)

with the complementary slackness conditions (3) and (4).

First, consider LP problem (17). By (7), we have that

$$\begin{aligned} a^1_{SJ}\ge a^1_{T_JJ}. \end{aligned}$$

(19)

Then, by (19), (see, Fig. 6) for any $u^1\in [0,\bar{u}^1]$, where $\bar{u}^1 =(a^1_{SJ}-a^1_{T_JJ})/(\delta (\gamma -\gamma _S))$, $U^1(u^1)$ is a solution of the equations

$$ L^1_{T_J}(u^1,x^2_S) =U^1(u^1) \text{ and } L^1_S(u^1,x^2_S)=U^1(u^1). $$

Fig. 6.

LP problem (17) for $u^1<\bar{u}^1$ (left) and LP problem (17) for $u^1=\bar{u}^1$ (right).

Thus,

$$\begin{aligned} U^1(u^1)=\frac{\delta ^2(\gamma -\gamma _S)(u^1)^2+\delta (a^1_{T_JJ}-a^1_{SJ}-\gamma _Sa^1_{T_JS})u^1-a^1_{SJ}a^1_{T_JS}}{ \delta (\gamma -\gamma _S)u^1+a^1_{T_JJ}-a^1_{SJ}-a^1_{T_JS}}. \end{aligned}$$

(20)

It is clear that

$$\begin{aligned} U^1(0)>0. \end{aligned}$$

(21)

By Fig. 6(b),

$$\begin{aligned} U^1(\bar{u}^1)=a^1_{T_JJ}+\delta \gamma \frac{a^1_{SJ}-a^1_{T_JJ}}{\delta (\gamma -\gamma _S)}<(\text{ by } (7))<\frac{a^1_{SJ}-a^1_{T_JJ}}{\delta (\gamma -\gamma _S)}=\bar{u}^1. \end{aligned}$$

(22)

Thus, by (21) and (22), since $U^1$ is continuous, the Eq. (16) has at least one root in $[0,\bar{u}^1]$. By (20), this equation is equivalent to the following quadratic equation

$$\begin{aligned} c^1_2(u^1)^2+c^1_1u^1+ c^1_0=0 \end{aligned}$$

(23)

with $c^1_i$, $i=0,1,2$ given by (10). Since $c^1_0>0$ and $c^1_2>0$, by (21) and (22), the equation has the unique root in $[0,\bar{u}^1]$, while the second root of this quadratic equation is greater than $\bar{u}^1$. Thus, (8) and (9) is solution of the LP problem (17). It is clear that (8) and (9) gives solution of the LP problem (17). Since the complementary slackness conditions (3) and (4) obviously hold, and the result follows.   $\blacksquare$