## 3 Analysis of Energy Efficiency and Network Connectivity
### Analysis of the Energy Efficiency
 \tag{8}\]  
where at the third condition \((\mathbb{U}>r)\cap\bar{\mathcal{G}}\), there is no transmission since no route is established to the sink, and thus the progress is equal to zero. As a result, the one-hop energy-distance ratio or the average energy consumption (denoted by \(\epsilon(r)\) in J/m) is given by  
\[\epsilon(r) =\frac{\eta(r)}{E[\mathbb{X}]} \tag{9}\] \[=\frac{\eta(r)}{E[\mathbb{X}](\mathbb{U}\leq r)\cup((\mathbb{U}>r )\cap\mathcal{G})]}\]  
Note that  
\[E[\mathbb{X}|(\mathbb{U}\leq r)\cup((\mathbb{U}>r)\cap\mathcal{G })]\] (10) \[= \int_{0}^{r}\text{Pr}\{\mathbb{X}>x|(\mathbb{U}\leq r)\cup(( \mathbb{U}>r)\cap\mathcal{G})\}dx\] \[= \int_{0}^{r}\frac{\text{Pr}\{(\mathbb{X}>x)\cap(\mathbb{U}\leq r )\}+\text{Pr}  
where there are four unknown terms: \(\text{Pr}\{(\mathbb{X}>x)\cap(\mathbb{U}\leq r)\}\), \(\text{Pr}\{(\mathbb{X}>x)\cap((\mathbb{U}>r)\cap\mathcal{G})\}\), \(\text{Pr}\{\mathbb{U}\leq r\}\) and \(\text{Pr}\{(\mathbb{U}>r)\cap\mathcal{G}\}\). For the first unknown term, using Equation (8) gives \(\text{Pr}\{(\mathbb{X}>x)\cap(\mathbb{U}\leq r)\}=\text{Pr}\{x<\mathbb{U}\leq r\}\). Then, from Figure 2, we see that  
\[\text{Pr}\{x<\mathbb{U}\leq r\}=\left\{ \begin{aligned} &\frac{r^{2}-x^{2}}{a^{2}},& \text{if}\;\;x<r\\ & 0,&\text{otherwise}\end{aligned} \right. \tag{11}\]  
where \(a\) is the largest possible distance between the sink \(D\) and any senor. Similarly, based on Figure 2, the third unknown term can be obtained by  
\[\text{Pr}\{\mathbb{U}\leq r\}=\frac{r^{2}}{a^{2}} \tag{12}\]  
To compute the fourth unknown term, we denote by \(A_{s}\) the area of the overlapping region between the circle centered at \(S\) with radius \(r\) and the circle centered at \(D\) with radius \(u\), _i.e._, the sum of the shaded regions \(A_{\alpha}\) and \(A_{\beta}\) in Figure 2. Then  
\[\text{Pr}\{(\mathbb{U}>r)\cap\mathcal{G}\} =\int_{r}^{a}\text{Pr}\{\text{at least one neighbor exists in }A_{s}\}f(u)du \tag{13}\] \[=\int_{r}^{a}(1-e^{-\rho A_{s}(u,r)})\frac{2u}{a^{2}}du\] \[=1-\frac{r^{2}}{a^{2}}-\frac{2}{a^{2}}\int_{r}^{a}ue^{-\rho A_{s} (u,r)}du\]  
where \(f(u)\) is the probability density function (PDF) of \(\mathbb{U}\), and differentiating Equation (12) gives \(f(u)=\frac{2u}{a^{2}}\). Note that the second identity in Equation (13) comes from the assumption that the probability of having \(n\) nodes in an area of size \(A\) complies with a Poisson distribution, _i.e._, \((\rho A)^{n}e^{-\rho A}/n!\), where \(\rho\) is the density parameter indicating the number of sensors per unit area [6]. The geometry of Figure 2 gives \(A_{s}(u,r)=r^{2}\text{cos}^{-1}(\frac{r}{2u})+u^{2}\text{cos}^{-1}(1-\frac{r ^{2}}{2u^{2}})-\frac{1}{2}r\sqrt{(2u+r)(2u-r)}\). According to Equation (8) and Figure 2, the second unknown term can be determined by  
\[\text{Pr}\{(\mathbb{X}>x)\cap((\mathbb{U}>r)\cap\mathcal{G}  
It is to note that the event {_there is no neighbor in \(A_{\beta}\)_} in Equation (14) will be {_at least one neighbor in \(A_{\alpha}\)_} if the MFR strategy is adopted as in [17]. Substituting Equations (11), (12), (13) and (14) into Equation (10), then we obtain the result of Equation (10). Finally, we complete the calculation of Equation (9) by combing Equations (7) and (10).