## 4 Network Interference Probability Analysis
### Heterogeneous Node Deployment Probability Analysis
#### 4.2.2 The Node Distance is \(0\leq d\leq r/2\)  
As shown in Figure 10b, when the transmission range of node \(v\) is smaller than \(r/2\), the whole coverage area of node \(v\) will be included by node \(u\); so the overlapping coverage area is \(\pi d^{2}\), which is also the quadratic function. According to Figure 4, in this scenario, when the transmission ranges are reduced, the probability of interference-optimality is 1. Similar to the analysis in the first part of this section, the feasible regions are \(IFY\) and \(HDX\) in Figure 14.  
As shown in Figure 14 and Table 1, the coordinate of point \(B\) is \((0.99r,0.495r)\), the coordinates of point \(D\) are \((r,0.5r)\), and the coordinates of point \(C\) are \((r,0.343r)\). Thus, the areas of \(HBX\) and \(HDX\) are similar, which means that we can use \(HBX\) to represent \(HDX\) to simplify the probability calculation in heterogeneous networks. The relationship between \(IFY\) and \(IEY\) is similar to that between \(HBX\) and \(HDX\). **Theorem 5**. : _In heterogeneous networks, the probability of interference-optimality by adjusting the transmission range is:_  
\[p_{\rm het}=\int_{0.343r}^{0.84r}(e^{-\pi p(r-r_{1})^{2}}-e^{-\pi p(0.445\sqrt {45^{2}-4r_{1}}+0.31r_{1})^{2}})dr_{1}+\int_{0.84r}^{r}\left(e^{-\pi p(r-r_{1} )^{2}}-e^{-\pi p(0.343r_{1}+\sqrt{4^{2}-r_{1}})^{2}}\right)dr_{1}\]  
Proof. : In Section 4.2, we investigate the constraints of interference-optimality by adjusting the transmission ranges in heterogeneous networks. The whole feasible region is shown in Figure 14 (the shade area). As shown in Figure 14, when \(0<r_{1}<0.343r\), the value range of \(d_{1}\) is \((r-r_{1},\,r)\); when \(0.343r<r_{1}<0.84r\), the value range is \((r-r_{1},\,0.445\cdot\sqrt{4.5r^{2}-4r_{1}}^{2}+0.31r_{1})\); when \(0.84r<r_{1}<r\), this will be \((r-r_{1},\,0.343r_{1}+\sqrt{r^{2}-r_{1}}^{2})\). Therefore, similarly with the analysis in Section 3 and according to the PDF (19), the probability of interference-optimality by adjusting the transmission range can be calculated as:  
\[p_{\rm het}=\int_{0.343r}^{0.84r}(e^{-\pi p(r-r_{1})^{2}}-e^{-\pi p(0.445\sqrt {45^{2}-4r_{1}}+0.31r_{1})^{2}})dr_{1}+\int_{0.84r}^{r}\left(e^{-\pi p(r-r_{1} )^{2}}-e^{-\pi p(0.343r_{1}+\sqrt{4^{2}-r_{1}})^{2}}\right)dr_{1}\, \tag{36}\] \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline
**Points** & \(A\) & \(B\) & \(C\) & \(D\) & \(E\) & \(F\) & \(G\) \\ \hline Coordinate & \((0.84r,0.84r)\) & \((0.99r,0.495r)\) & \((r,0.343r)\) & \((r,0.5r)\) & \((0.495r,0.99r)\) & \((0.5r,r)\) & \((0.343r,r)\) \\ \hline \hline \end{tabular}
\end{table} 
Table 2: The coordinates of the points in Figure 14.  
Figure 14: The feasible region in Heterogeneous networks. In Equation (36), the probability relates to both the original transmission range and the node density. The properties of this probability under different \(r\) and \(\rho\) will be investigated in the simulation part in detail. Similar to the probability in Equation (20), the probability that is shown in Equation (36) is just a general probability under different original transmission ranges. In terms of a different original transmission range, this probability means the possibility that there exist optimal solutions which can guarantee low network interference if the transmission range is changed. However, the probability of interference-optimality under specific transmission range will be different from that shown in Equation (36). **Theorem 6**. : _In heterogeneous networks, the probability that the network is interference-optimal under specific transmission range is:_  
\[P_{\rm hets}=\left\{ \begin{array}{l}\int_{r-r_{1}}^{r}2\pi\rho r_{1}\cdot e^ {-\pi\rho r_{1}}{}^{2}dr_{1},\left(0<r_{1}<0.343r\right)\\ \int_{r-r_{1}}^{0.445\cdot\sqrt{4.5r^{2}-4r_{1}}{}^{2}+0.31r_{1}}2\pi\rho r_{1} \cdot e^{-\pi\rho r_{1}}{}^{2}dr_{1},\left(0.343r<r_{1}<0.84r\right)\\ \int_{r-r_{1}}^{0.343r+\sqrt{r^{2}-x^{2}}}2\pi\rho r_{1}\cdot e^{-\pi\rho r_{1} }{}^{2}dr_{1},\left(0.84r<r_{1}<r\right)\end{array} \right..\]  
Proof.