# Wireless Networks under a Backoff Attack: A Game Theoretical Perspective
## 4 Problem Modelling as a Static Game
### Two Player Case: \(N_{p}=2\)  
Now, we center in the case when \(n_{2}=1\), that is, there are only two players in the game: the server and one client. We proceed to describe the payoffs for each player. We denote by \(S_{1}^{ns}\) the throughput that the client can obtain by playing \(ns\). In that case, the \(n_{1}\) normal stations will obtain each a throughput \(S_{n_{1}}^{ns}=S_{1}^{ns}=S^{ns}\), that is, all stations obtain the same amount of throughput (cases (a,e) in Figure 1). If the client plays \(s\), it obtains a throughput \(S_{i}^{1}\) if the server plays \(nd\). This causes the normal stations to have a lower throughput, \(S_{n_{1}}^{s}<S_{1}^{s}\), as observed in Figure 1.  
We define \(-k_{d}\) (with \(k_{d}>0\)) as the cost of detecting malicious behavior for the server. We model the cost function for client and server as a linear function of the throughput, with \(k_{s}\) and \(k_{j}\) as a constant  
Figure 2: Network scheme for the case that there are \(n_{1}\) normal stations (NS) and \(n_{2}\) malicious stations (MS). NS respect 802.11 binary exponential backoff, whereas MS can choose to use it or to use a uniform backoff. for the server and the client respectively. Finally, we will assume that both players are maximizers, hence, they try to maximize the payoff function, that we define as follows:  
* If they play \((nd,s)\) (the first action corresponds to the server, the second to the client), the client is modifying its backoff and hence, the throughputs in the network. The server does not detect this modification, and hence, does not punish the client. Thus, the client obtains a throughput increment, which provides it a gain of \(k_{1}(S_{1}^{s}-S^{ns})\). The server has a cost proportional the throughput loss that the normal stations suffer: \(k_{s}n_{1}(S_{n_{1}}^{s}-S^{ns})\). * If they play \((d,s)\), the client modifies its backoff, but it is caught by the server, who drops its packet. This causes the client a loss of \(k_{1}(0-S^{ns})=-k_{1}S^{ns}\). The server has a gain proportional to the throughput that the normal stations would have lost minus the cost of the detection: \(k_{s}n_{1}(S^{ns}-S_{n_{1}}^{s})-k_{d}\). * If they play \((d,ns)\), the client does not modify its backoff and hence, does not affect the throughput. Hence, it has no gain nor lose. However, the server detects and hence, it incurs in the cost of detection, expressed as \(-k_{d}\). * If they play \((nd,ns)\), the client again has no gain nor lose. The server does not detect and hence, it incurs in no cost since the client is behaving properly: it also has no gain nor lose. All of these payoffs do not vary along the game, provided that there is no modification of the game conditions (e.g., the number of players). Since \(N_{p}=2\), we can pose the game as a bimatrix, non-zero sum, static game, whose game payoffs as a function of the player actions are in Table 2.  
In order to simplify, we will substitute the payoff values in Table 2 for the following constants, where \(R_{1}\) is the payoff matrix for player 1 and \(R_{2}\) for player 2:  
\[R_{1}=\left( \begin{array}{cc}-\alpha_{m}&0\\ \alpha_{c}&-\alpha_{f}\end{array} \right)\quad R_{2}=\left( \begin{array}{cc} \beta_{s}&0\\ -\beta_{c}&0\end{array} \right) \tag{13}\]  
Observe that all parameters in (13) are strictly positive, that is, \(\alpha_{c},\alpha_{m},\alpha_{f},\beta_{s},\beta_{c}\in(0,+\infty)\). This arises because:  
* \(k_{1},k_{c},k_{d},n_{1}\) and all the throughput values (\(S_{n_{1}}^{s},S_{1}^{s},S^{ns}\)) are all strictly positive parameters. * and from Figure 1, we see that this is not the case if there is only one client (\(n_{2}=1\)). This means that \(S_{1}^{s}>S^{ns}\). * The throughput of the normal stations must decrease when the client behaves selfishly with respect to their throughput when the client follows the binary exponential backoff. As we observe in Figure 1, that is indeed the case if there are malicious stations (i.e., \(n_{2}\geq 1\)). This means that \(S^{ns}>S_{n_{1}}^{s}\).