## 4 The Energy Balance Problem
### The General Solution  
We next provide a lemma useful in the estimation of the total energy dissipation in a sector. **Lemma 9**: The expected total energy dissipation in sector \(i\) is:  
\[E[\mathcal{E}_{i}]=E[h_{i}]\cdot E[\epsilon_{ik}]\]  
**Proof:**  
\[ \begin{array}{rcl}E[\mathcal{E}_{i}]&=&E\left[\sum_{k=1}^{h_{i}}\epsilon_{ik }\right]\\ &=&\sum_{n=0}^{N}E\left[\sum_{k=0}^{h_{i}}\left(\epsilon_{ik}\cap h_{i}=n \right)\right]\\ &=&\sum_{n=0}^{N}E\left[\sum_{k=1}^{h_{i}}\epsilon_{ik}\Big{|}h_{i}=n\right] \cdot\mathbf{P}\{h_{i}=n\}\end{array} \]Furthermore,  
\[E\left[\sum_{k=1}^{h_{i}}\epsilon_{ik}\Big{|}h_{i}=n\right] = E\left[\sum_{k=1}^{n}\epsilon_{ik}\right]=\sum_{k=1}^{n}E[\epsilon_ {ik}]\] \[= n\cdot E[\epsilon_{ik}]\]  
Thus, we have from the above that:  
\[E[\mathcal{E}_{i}] = \sum_{n=0}^{N}E\left[\sum_{k=1}^{h_{i}}\epsilon_{ik}\Big{|}h_{i}= n\right]\cdot\mathbf{P}\{h_{i}=n\}\] \[= E[\epsilon_{ik}]\cdot E[h_{i}]\]  
\(\diamond\)  
**Definition 10**: Let \(g_{i}\) be the number of the messages that are _generated_ in the area \(T_{i}\). Note that messages are generated in an area only when events occur in this area. **Definition 11**: Let \(f_{i}\) be the number of the messages that are _forwarded to_ the area \(T_{i}\). We note that messages are forwarded to a ring sector (say \(i\)) only because of an event generation at a sector \(j>i\) and successive one-hop propagations from sector \(j\) to sector \(i\). We notice the following important relation:  
\[h_{i}=g_{i}+f_{i} \tag{4}\]  
which means that the number of messages that area \(T_{i}\) handles equals the number of the messages that are generated in \(T_{i}\), plus the number of messages that are forwarded to it. Because of event generation according to a probability distribution and also because of the probabilistic nature of message propagation in, all three quantities above are random variables. By linearity of expectation, we get:  
**Lemma 10**: \(E[h_{i}]=E[g_{i}]+E[f_{i}]\)__  
We establish a relationship between \(E[f_{i}]\) and \(E[h_{i+1}]\). **Lemma 11**: \(E[f_{i}]=p_{i+1}\cdot E[h_{i+1}]\)__  
**Proof:** Let \(\delta_{i,j}\) an indicator random variable that is equal to 1 if area \(T_{i}\) forwards the message j to the area \(T_{i-1}\) and 0 otherwise. Thus:  
\[\delta_{i,j}=\left\{ \begin{array}{ll}1&\quad\mbox{with probability $p_{i}$}\\ 0&\quad\mbox{with probability $1-p_{i}$}\end{array} \right.\]  
Clearly, \(\delta_{i,j}\) depends only on \(i\), but we add \(j\) for counting purposes. Obviously, \(E[\delta_{i,j}]=p_{i}\). It is:  
\[f_{i}=\sum_{j=0}^{h_{i+1}}\delta_{i+1,j}\]  
Similarly to the proof of Lemma 9, we get:  
\[E[f_{i}]=\sum_{n=0}^{N}E\left[\sum_{j=0}^{h_{i+1}}\delta_{i+1,j}\Big{|}h_{i+ 1}=n\right]\cdot\mathbf{P}\{h_{i+1}=n\}\]and the proof is completed. \(\diamond\)  
Recall that, according to Definition 9, to achieve the same on the average energy dissipation _per area unit_ (and thus per sensor) in the network area, the following equality should hold:  
\[E\left[\frac{\sum_{k=1}^{h_{i}}\epsilon_{ik}}{S_{i}}\right]=E\left[\frac{\sum_{k =1}^{h_{j}}\epsilon_{jk}}{S_{j}}\right]\quad\forall i,j\in\{1,\ldots n\} \tag{5}\]  
i.e. the average energy consumption per sensor should be equal in any two ring sectors. By induction, it suffices to guarantee this for any two adjacent sectors. In what follows, we guarantee the above balance property, requiring a certain recurrence relation to hold. This recurrence basically relates 3 successive terms of the \(E[f_{i}]\) sequence (the \(E[g_{i}]\) terms depend only on \(i\) and on input parameters).