## 6 Results and Discussion
### Theoretical Analysis of Constant \(d_{2 hops}\)  
Here, we present a theoretical analysis when \(d_{2 hops}\) is fixed because this scenario appears to show the optimal point in Figures 14-16. The reason for the optimized performances, except for Case 2, when \(d_{1 st}=d_{2 nd}=d_{2 hop}/2\) is described in this section. The probability of transmission failure in the two-hop case \(P_{fail,2 hop}\) is expressed using the probability of transmission failure at each hop \(P_{fail,1 st}\), \(P_{fail,2 nd}\) as follows:  
\[P_{fail,2 hop}=P_{fail,1 st}+\ \left(1-P_{fail,1 st}\right)P_{fail,2 nd} \tag{22}\]  
\[P_{fail,1 st}(d_{1 st})=\prod_{i=1}^{[\overline{tr_{N1}\to N2}-0.5]}P_{fail,i}(d_ {1 st}) \tag{23}\]  
\[P_{fail,2 nd}(d_{1 st})=\prod_{i=1}^{[\overline{tr_{N2}\to H}-0.5]}P_{fail,i} \Big{(}d_{2 hop}-d_{1 st}\Big{)} \tag{24}\]  
\[0<P_{fail,1 st}(d_{1 st}),\ P_{fail,2 nd}(d_{1 st})<1 \tag{25}\]  
\[0<P_{fail,1 st}(d_{1 st})^{\prime} \tag{26}\]  
\[P_{fail,2 nd}(d_{1 st})^{\prime}<0 \tag{27}\]  
Here, \(P_{fail,1 st}(d_{1 st})^{\prime}=\frac{\mathrm{d}P_{fail,1 st}(d_{1 st})}{\mathrm{ d}(d_{1 st})}\) and \(P_{fail,2 nd}(d_{1 st})^{\prime}=\frac{\mathrm{d}P_{fail,2 nd}(d_{1 st})}{\mathrm{ d}(d_{1 st})}\). The communication distance in two hops \(d_{2 hops}\) is defined as follows:  
\[d_{2 hops}=d_{1 st}+d_{2 nd} \tag{28}\]  
\[0<d_{1 st},\ d_{2 nd}<d_{2 hop} \tag{29}\]  
\(P_{fail,2 hop}\) is differentiated by \(d_{1 st}\) as follows:  
\[ \begin{split} P_{fail,2 hop}{}^{\prime}=P_{fail,1 st}{}^{\prime}+ P_{fail,1 st}{}^{\prime}P_{fail,2 nd}+\left(1-P_{fail,1 st}\right)P_{fail,2 nd} {}^{\prime}\\ =\left(1-P_{fail,2 nd}\right)P_{fail,1 st}{}^{\prime}+\left(1-P_{ fail,1 st}\right)P_{fail,2 nd}{}^{\prime}.\end{split}  \tag{30}\]  
Figure 16: Average number of transmissions with constant \(d_{2 hops}\). Here, the case that (30) = 0 is considered. Equation (30) is modified by the following equation:  
\[\frac{1}{1-P_{fail,1 st}}P_{fail,1 st}{}^{\prime}=-\frac{1}{1-P_{fail,2 nd}}P_{fail,2 nd}{}^{\prime} \tag{31}\]  
Both sides of (31) are integrated by \(d_{1 st}\) as follows:  
\[ \begin{split}-\ln\Bigl{(}1-P_{fail,1 st}\Bigr{)}=\frac{1}{1-P_{ fail,2 nd}}\Bigl{(}-\frac{\mathrm{d}P_{fail,2 nd}}{\mathrm{d}d_{2 nd}}\frac{\mathrm{d}d_{2 nd}}{ \mathrm{d}d_{1 st}}\Bigr{)}\\ =-\ln\Bigl{(}1-P_{fail,2 nd}\Bigr{)}+C.\end{split}  \tag{32}\]  
Here, \(d_{c}\) where \(P_{fail,1 st}(d_{c})=P_{fail,2 nd}(d_{c})\) is considered. Under this condition, \(C=0\). Thus, (32) is rewritten as follows:  
\[ \begin{split}\prod\limits_{i=1}^{[\overline{tr_{N1 \to N2}}-0.5]}P_{fail,i}(d_{1 st})-\prod\limits_{i=1}^{[\overline{tr_{N2\to H}}-0.5]}P_{fail,i}\Bigl{(}d_{2 hop}-d_{1 st}\Bigr{)}=0\end{split}  \tag{33}\]  
\(d_{1 st}\), \(\overline{tr_{N1\to N2}}\), and \(\overline{tr_{N2\to H}}\) that satisfies (33) are considered. Here, it is assumed that \(d_{1 st}=d_{2 nd}=\frac{d_{2 hop}}{2}\). Under this condition, (33) is satisfied when \(\overline{tr_{N1\to N2}}=\overline{tr_{N2\to H}}\) in the computer simulations (except for Case 2) because, when \(\overline{tr_{N1\to N2}}\
eq\overline{tr_{N2\to H}}\), (27) is modified as follows:  
\[\prod\limits_{i=1}^{[\overline{tr_{N2\to H}}-0.5]}P_{fail,i}(\frac{d_{2 hop}}{ 2})\left\{1-\prod\limits_{i=[\overline{tr_{N2\to H}}-0.5]}^{[\overline{tr_{N1 \to N2}}-0.5]}p_{fail,i}\Bigl{(}\frac{d_{2 hop}}{2})\right\}=0 \tag{34}\]  
When (26) is satisfied,  
\[\prod\limits_{i=1}^{[\overline{tr_{N2\to H}}-0.5]}P_{fail,i}(\frac{d_{2 hop}}{ 2})=0 \tag{35}\]  
or  
\[\prod\limits_{i=[\overline{tr_{N1\to N2}}-0.5]}^{[\overline{tr_{N1\to N2}}-0.5]}p_{fail,i}(\frac{d_{2 hop}}{2})=1 \tag{36}\]  
However, due to (27)-(29), (35) and (36) are not satisfied. Thus, it can be said that \(\overline{tr_{N1\to N2}}=\overline{tr_{N2\to H}}\). On the other hand, for Case 2, (33) is not satisfied when \(\overline{tr_{N1\to N2}}=\overline{tr_{N2\to H}}\) because \(P_{fail,1 st,i}(\frac{d_{2 hop}}{2})\
eq P_{fail,2 nd,i}(\frac{d_{2 hop}}{2})\).