# Game-Theoretic Power and Rate Control in IEEE 802.11\(p\) Wireless Networks
## 3 System Model
### Nash Equilibrium Properties  
In the game formulation, specific properties can be shown of the Nash equilibrium that exists by definition of the potential game. Initially, the existence of multiple Nash equilibria are shown. Thereafter, we try to identify the best one.  
**Lemma 2**.: _The best Nash equilibrium is when the nodes play with \(p_{i}^{1}=p_{i}^{2}=p_{max}\), resulting in \(R_{i}^{1}=R_{j}^{2}=R(SINR_{max})\) that satisfies the constraints of (5)._  
Proof.: We denote \(c_{1}\subset N\) and \(c_{2}\subset N\) as two subsets of the playing nodes \(N\) participating in our game formulation. Furthermore, we denote \(p_{1}=p^{*}\) as the transmission powers, which is the best response of the nodes in subset \(c_{1}\) and \(R_{1}=R^{*}\) denotes the best response rates. Moreover, we denote \(p_{2}=\tilde{p}\) and \(R=\tilde{R}\) as the best response transmission power and rates of subset \(c_{2}\), respectively. Finally, we denote \(\gamma(p)\geq\gamma_{thr}\), where \(\gamma_{thr}\) is the SINR threshold for the successful reception of a packet. As we have seen from [21], there exists two transmission powers \(p^{1}<p^{2}\) that two different sender/receiver pairs use, which results in both messages successfully received for a sparse configuration of the nodes. In addition, duty cycling of the nodes may result in equal or no interference; hence, we have for a link \((i,j)\) Equations (14) and (15) satisfied, where \(I_{t}=I_{q}\) if we duty cycle the nodes to keep interference to the same level.  
In the case that \(I_{t}=I_{q}=0\), the duty cycling mechanism ensures that the denominator of the SINR equation gets smaller; hence, the reception threshold is exceeded, and the \(PRR=1\). Even if we raise the transmission power, the PRR will be 1, as we can see below,  
\[\gamma_{i,j}=\frac{H_{ij}p_{i}}{N_{0}}\geq\gamma_{thr} \tag{16}\]  
and using transmission power \(p_{2}\), this ensures that (16) is satisfied as well.  
The transmission powers \(p^{*}\) and \(\tilde{p}\) set the \(PRR=1\), with a smaller transmission power than \(p_{max}\). Since (11) reaches its maximum value with maximum SINR, the transmission powers used by the two subsets of nodes is \(p^{*}=\tilde{p}=p_{max}\). Hence, the two subsets of nodes will transmit with \(p_{max}\), which is the best case equilibrium.  
As such, increasing the rate of such configuration will further increase the throughput of the network and maximise Equation (5). In conjunction with the transmission powers \(p_{1}\) and \(p_{2}\), the rates of the nodes in the subsets \(c_{1},c_{2}\) are \(R_{i}^{1}=R_{j}^{2}=R(SINR_{max})\), by using the maximum transmission power that satisfies the constraints. This is due to the maximum rate including the maximum \(PRR=1\), as we can see from the constraints of (5) when interference is mitigated or diminished using duty cycling, as we can see in Figure 4.  
**Lemma 3**.: _The price of stability when the best case equilibrium is played is 1._  
Proof.: We have seen that the best case equilibrium is when \(p_{i}^{1}=p_{i}^{2}=p_{max}\) and \(R_{i}^{1}=R_{j}^{2}=R(SINR_{max})\). If we suppose that there is a more efficient equilibrium, then we have that a node deviates to another transmission power \(p_{j}^{*}\),  
\[\sum_{i\in N,j\
eq j}^{N}u(p_{max}^{i})+u(p_{max}^{j})\] \[<\sum_{i\in N,j\
eq j}^{N}u(p_{max}^{i})+u(p_{j}^{*}),\]which contradicts that the \(p_{i}^{1}=p_{i}^{2}=p_{max}\) and \(R_{i}^{1}=R_{j}^{2}=R(SINR_{max})\) is the best case equilibrium. Hence, the price of stability is 1.