## 3 Game Analysis
### Game II: Internet of Things-Service Providers (IoT-SPs) and Users
#### 3.2.2 IoT-SPs Pricing Stage  
In this stage, each IoT-SP want to maximize its own profit given by (7). Given the solution of the previous stage (28) and the solution for Game I in (21) the providers' profits in the Nash equilibrium are going to be analyzed. With the solution of OP-Sinks game and users subscription game we can re-write the profit for the IoT-SP\({}_{i}\) as:  
\[\Pi_{\text{IoT-SP}_{i}}(f_{1},f_{2})=\frac{f_{i}M\left(\frac{R_{i}}{f_{1}} \right)^{\alpha}}{\left(\frac{R_{1}}{f_{1}}\right)^{\alpha}+\left(\frac{R_{2}} {f_{2}}\right)^{\alpha}+1}-pR_{i},\qquad i=1,2. \tag{29}\]  
In order to find the Nash equilibrium we use the best response functions for both operators defined as follows:  
\[BR_{1}(f_{2}) =f_{1}^{*}(f_{2})=\arg\max\limits_{f_{1}>0}\Pi_{\text{IoT-SP}_{1}} (f_{1},f_{2}),\] \[BR_{2}(f_{1}) =f_{2}^{*}(f_{1})=\arg\max\limits_{f_{2}>0}\Pi_{\text{IoT-SP}_{2}} (f_{1},f_{2}).\]The Nash equilibrium is obtained from the equation system  
\[ \begin{split} f_{1}^{*}&=\operatorname*{argmax}_{f_{1}} \Pi_{\text{IoT-SP}_{1}}(f_{1},f_{2}^{*})\quad s.t.\quad f_{1}>0\\ f_{2}^{*}&=\operatorname*{argmax}_{f_{2}}\Pi_{\text{IoT- SP}_{1}}(f_{1}^{*},f_{2})\quad s.t.\quad f_{2}>0.\end{split}  \tag{30}\]  
In order to obtain the optimum prices we equal the partial derivatives to zero  
\[\frac{\partial\Pi_{\text{IoT-SP}_{1}}(f_{1},f_{2})}{\partial f_{1}} =\frac{M\left(\frac{R_{1}}{f_{1}}\right)^{\alpha}\left(-\alpha+ \left(\frac{R_{1}}{f_{1}}\right)^{\alpha}-(\alpha-1)\left(\frac{R_{2}}{f_{2}} \right)^{\alpha}+1\right)}{\left(\left(\frac{R_{1}}{f_{1}}\right)^{\alpha}+ \left(\frac{R_{2}}{f_{2}}\right)^{\alpha}+1\right)^{2}}=0,\] \[\frac{\partial\Pi_{\text{IoT-SP}_{2}}(f_{1},f_{2})}{\partial f_{2}} =\frac{M\left(\frac{R_{2}}{f_{2}}\right)^{\alpha}\left(\left( \frac{R_{2}}{f_{2}}\right)^{\alpha}-(\alpha-1)\left(\left(\frac{R_{1}}{f_{1}} \right)^{\alpha}+1\right)\right)}{\left(\left(\frac{R_{1}}{f_{1}}\right)^{ \alpha}+\left(\frac{R_{2}}{f_{2}}\right)^{\alpha}+1\right)^{2}}=0.\]  
With the change \(Ai=\left(\frac{R_{i}}{f_{i}}\right)^{\alpha}\) and simplifying the system we obtain  
\[A1 =(\alpha-1)(A2+1), \tag{31}\] \[A2 =(\alpha-1)(A1+1). \tag{32}\]  
Solving the previous equation system we obtain  
\[A1^{*}=A2^{*}=\frac{1-\alpha}{\alpha-2}. \tag{33}\]  
Given that \(R_{i}\) and \(f_{i}\) are positive, Ai has to be positive. From Equation (33) we can infer that \(1<\alpha\leq 2\), otherwise Ai would be negative, and there would be no real solutions for \(f_{i}\). In addition, we see that there is only one pricing equilibrium different than \((f_{1}^{*}=0,f_{2}^{*}=0)\) if and only if \(1<\alpha<2\). Figure 4(a) shows a particular solution when \(1<\alpha<2\), where we observe that there is a Nash equilibrium where the best response functions intersect. On the other hand, Figure 4(b) shows how the best response functions of both operators only intersect in \((0,0)\), and therefore the only possible solution when \(\alpha=2\) is \((f_{1}^{*}=0,f_{2}^{*}=0)\). Figure 5: Best responses with different values of \(\alpha\). Reverting the change \(f_{i}=R_{i}\text{Ai}^{-\frac{1}{\alpha}}\), we get the pricing strategies for both IoT-SPs in the equilibrium  
\[ \begin{split} f_{1}^{*}&=\left(\tfrac{1}{2-\alpha}-1 \right)^{-1/\alpha}R_{1}\quad s.t.\quad 1<\alpha<2,\\ f_{2}^{*}&=\left(\tfrac{1}{2-\alpha}-1\right)^{-1/ \alpha}R_{2}\quad s.t.\quad 1<\alpha<2.\end{split}  \tag{34}\]  
Replacing (34) in (28) we obtain the users distribution in the equilibrium, that depends only on \(\alpha\). \[y_{0}^{*}=\frac{2}{\alpha}-1,\qquad y_{1}^{*}=\frac{\alpha-1}{\alpha},\qquad y _{2}^{*}=\frac{\alpha-1}{\alpha}. \tag{35}\]  
Figure 6 shows how when the value of \(\alpha\) is close to 1 the percentage of users that subscribe is very small and the prices of the providers are very high. This could be counter-intuitive, but it explains cases where some users are willing to pay huge amounts of money even without clear evidence of a good quality of service.