## 3 Game Analysis
### Game I: OP and Sinks
#### 3.1.2 OP Pricing Stage  
In this stage, the OP wants to maximize its profit given by Equation (4). Given the three cases obtained from (11) we analyze the case where the maximum profit is reached. \[\Pi_{\text{OP}}= \begin{cases}pNr&\text{if}\quad 0<p\leq c\left(1-\tau Nr \right)\\ p\frac{c-p}{c\tau}&\text{if}\quad c\left(1-\tau Nr\right)\leq p\leq c\\ 0&\text{if}\quad p\geq c\end{cases}  \tag{12}\]  
* Case \(0<p\leq c\left(1-\tau Nr\right)\): In this case, the maximum profit is obtained solving the optimization problem \[\underset{p}{\text{max}} \Pi_{\text{OP}_{1}}^{*}=pNr\] (13) subject to \[0<p\leq c\left(1-\tau Nr\right).\] The solution for the previous problem is \[\Pi_{\text{OP}_{1}}^{*}=c\left(1-\tau Nr\right)Nr\quad\text{if}\quad 0\leq \tau<\tfrac{1}{Nr}\quad\text{with}\quad p^{*}=c\left(1-\tau Nr\right).\] (14) Note that if \(\tau Nr>1\) the upper limit \(c\left(1-\tau Nr\right)\) is negative and, therefore, there is not possible solution for \(p^{*}\) in this case. * Case \(c\left(1-\tau Nr\right)\leq p\leq c\): In this case, the maximum profit is obtained solving the optimization problem \[ \begin{array}{ll}\max_{p}&\Pi_{OP_{2}}^{*}=\left(p^{\frac{c-p}{c\tau}}\right) \\ \text{subject to}&c\left(1-\tau Nr\right)\leq p\leq c\end{array} \] (15) The problem in (15) is solved using Karush-Kuhn-Tucker (KKT) conditions and its solution is: \[\Pi_{OP_{2}}^{*}= \begin{cases}c\left(1-\tau Nr\right)Nr&\text{if}&0\leq\tau< \frac{1}{2 Nr}\\ \frac{c}{4\tau}&\text{if}&\tau\geq\frac{1}{2 Nr}\end{cases} \quad.\] (16)
* Case \(p\geq c\): In this case, for any value of \(p\) the maximum profit is \[\Pi_{OP_{3}}=0.\] (17) Combining (14) and (16) the OP optimal profit can be summarized as: \[\Pi_{OP}^{*}= \begin{cases}c\left(1-\tau Nr\right)Nr&\text{if}&\tau<\frac{1}{2 Nr}&\text{with}&p^{*}=c\left(1-\tau Nr\right)\\ \max\left(\left\{cNr\left(1-\tau Nr\right),\frac{c}{4\tau}\right\}\right)& \text{if}&\frac{1}{2 Nr}\leq\tau\leq\frac{1}{Nr}&\text{with}&p^{*}=\left[c \left(1-\tau Nr\right),\frac{c}{2}\right]\\ \frac{c}{4\tau}&\text{if}&\frac{1}{Nr}<\tau&\text{with}&p^{*}=\frac{c}{2} \end{cases} \quad.\] (18)  
The expression for the profit in (18) can be simplified given that \(\frac{c}{4\tau}\geq cNr\left(1-\tau Nr\right)\) for any value of \(c\), \(N\), \(r\) and \(\tau\). To prove this we analyze the expressions for any value of \(\tau\)  
\[\frac{c}{4\tau}\geq cNr\left(1-\tau Nr\right), \tag{19}\]  
re-writing with \(A=\tau Nr\), the previous expression is simplified to:  
\[1\geq 4A-4A^{2}.\]  
We can demonstrate that  
\[\max_{A}\left(4A-4A^{2}\right)\leq 1 \tag{20}\]  
\[\frac{\partial\left(4A-4A^{2}\right)}{\partial A}=0\] \[\frac{\partial\left(4A-4A^{2}\right)}{\partial A}=4-8A\] \[4-8A^{*}=0\] \[A^{*}=\frac{1}{2}\] \[4A^{*}-4A^{*2}=1,\]which proves (19). Figure 4 shows a particular case of the demonstration, where we can see how \(\frac{c}{4\tau}\geq cNr\left(1-\tau Nr\right)\) for the range of interest \(\tau\geq\frac{1}{2 Nr}\). With the previous demonstration, OP optimal profit can be simplified to:  
\[\Pi_{OP}^{*}= \begin{cases}c\left(1-\tau Nr\right)Nr&\text{if}\quad\tau<\frac{1 }{2 Nr}\\ \frac{c}{4\tau}&\text{if}\quad\frac{1}{2 Nr}\leq\tau\end{cases} \quad. \tag{21}\]  
\[p^{*}= \begin{cases}c\left(1-\tau Nr\right)&\text{if}\quad\tau<\frac{1}{2 Nr}\\ \frac{c}{2}&\text{if}\quad\frac{1}{2 Nr}\leq\tau\end{cases} \quad. \tag{22}\]  
\[x_{1}^{*}= \begin{cases}1&\text{if}\quad\tau<\frac{1}{2 Nr}\\ \frac{1}{2\tau Nr}&\text{if}\quad\frac{1}{2 Nr}\leq\tau\end{cases} \quad. \tag{23}\]  
In order to understand better the behavior of the first game we can re-write the equations in terms of the maximum amount of data generated by sensors normalized by the system capacity, which we define as maximum system load \(L\):  
\[L=\tau Nr=\frac{Nr}{\mu}\]  
obtaining  
\[\Pi_{OP}^{*}= \begin{cases}c\left(1-L\right)L\mu&\text{if}\quad L<\frac{1}{2} \\ \frac{c\mu}{4}&\text{if}\quad L\geq\frac{1}{2}\end{cases} \quad. \tag{24}\]  
\[p^{*}=