## 4 Network Interference Probability Analysis
### Homogenous Node Deployment Probability Analysis
Therefore, the coverage area of node \(n\), node \(u\), and node \(v\) can be calculated as:  
\[s_{t}=\pi r^{2}-(r+1)\cdot d+o[(\frac{d}{2r})^{3}] \tag{6}\]4 \[s_{t}=\pi r^{2}-(r+1)\cdot d+o[(\frac{d}{2r})^{3}] \tag{6}\]5  
In (14), there are three parts: (1) \(\pi{r_{1}}^{2}\) is the area of circle \(u\); (2) \(\pi{r_{1}}^{2}-\left(\pi{r_{1}}^{2}-1.93r_{1}d_{0}\right)\) is the area of circle \(n\) after removing the overlapping area between circle \(u\) and circle \(n\); (3) \(\pi{r_{1}}^{2}-\left(\pi{r_{1}}^{2}-1.93r_{1}d_{1}\right)\)  
\[s_{t}=\pi r^{2}-(r+1)\cdot d+o[(\frac{d}{2r})^{3}] \tag{6}\]6 The fitting results of Equations (5) and (11). is the area of circle \(v\) after removing the overlapping area between circle \(n\) and circle \(v\). Moreover, in Equation (14), since \(d_{0}\) and \(d_{1}\) are all variables, for simplifying the calculation, we assume that the distance \(d_{0}\) is equal to the transmission range of node \(u\), i.e., \(r_{1}=d_{0}\). This assumption is reasonable. Since the distance equal to the transmission range means the node which is farthest from the source node will be chosen as the relay node. For instance, as shown in Figure 4, when the node \(u\) sends data packet to node \(v\), the source node \(u\) will choose node \(n\) which is in the boundary of the coverage area as the relay node; therefore, the transmission range is equal to the distance \((u,n)\), i.e., the distance \((u,n)\) is the maximum value among all the neighbors' distances. Thus, when the transmission ranges are reduced, the coverage area can be rewritten as:  
\[ \begin{array}{l}s_{r_{1}}=\pi{r_{1}}^{2}+\pi{r_{1}}^{2}-(\pi{r_{1}}^{2}-1.93 {r_{1}}^{2})+\pi{r_{1}}^{2}-(\pi{r_{1}}^{2}-1.93{r_{1}}d_{1})\\ =5.07\tau_{1}^{2}+1.93{r_{1}}d_{1}\end{array}  \tag{15}\]  
Additionally, in Equation (15), \(r_{1}\) is the transmission range after being reduced and equal to the maximum value between distance \((u,n)\) and distance \((u,v)\). For instance, as shown in Figure 4, distance \((n,v)\) is smaller than distance \((u,n)\) and equal to \(d_{1}\); so \(r_{1}\) is equal to distance \((u,n)\) and larger than \(d_{1}\) (the situation that \(r_{1}\) is smaller than \(d_{1}\) will be investigated following). In this scenario, the transmission ranges of node \(u\), node \(v\), and node \(n\) are all the same and equal to \(r_{1}\). Based on the principle of linear programming, the boundary condition of this issue is \(\rho s_{r}=\rho s_{r_{1}}\) (as shown in Section 3, the interference can be calculated by Equation (4), the boundary condition means the interference before transmission range adjustment is equal to the interference that after the transmission range adjustment), which can be expressed as:  
\[d_{1}=2.63\cdot(\frac{r^{2}}{r_{1}}-r_{1}), \tag{16}\]  
In addition, when the transmission ranges are reduced, for guaranteeing the network connection, \(r_{1}\) and \(d_{1}\) should satisfy the constraints as follows:  
\[\left\{ \begin{array}{l}0<r_{1}<r\\ 0<d_{1}<r\\ r_{1}>d_{1}\\ r_{1}+d_{1}>r\end{array} \right., \tag{17}\]  
In the calculation of the coverage area, we assume that \(r_{1}\) is larger than \(d_{1}\), i.e., \(0<d_{1}<r_{1}\), so based on Equations (16) and (17), the feasible region when \(0<d_{1}<r_{1}\) is the triangle \(DBC\) in Figure 9.  
Figure 9: The feasible region in homogeneous networks. In the coverage area calculation, \(r_{1}=d_{0}\) and the role of \(d_{0}\) and \(d_{1}\) are equal in Equation (14), i.e., \(d_{0}\) and \(d_{1}\) can be replaced by each other in Equation (14). Therefore, when distance (\(u\), \(n\)) is smaller than the distance (\(n\), \(v\)), i.e., \(r_{1}<d_{1}\), the feasible region should be symmetrical to \(DBC\) based on line \(d_{1}=r_{1}\). So, according to Equation (16), when \(r_{1}\) is smaller than \(d_{1}\), the boundary will be \(d_{1}=0.19(\sqrt{{r_{1}}^{2}+27.67r^{2}}-r)\) (this boundary condition can be obtained by variable swapping between \(r_{1}\) and \(d_{1}\) in Equation (16)).