## 4 The Optimal Relay Nodes Placement with Respect to Minimum Overall Power Usage
### Solution to OPT-1  
To solve this optimization problem, we might use the method of the Lagrange multiplier. However, the mathematical correctness and completeness of this method can only be ensured when finding extreme points for problems with equality constraints. The optimization problem formulated in Section 4.1 obviously has inequality constraints and an integer optimization variable. Therefore, the method of the Lagrange multiplier may not be applied easily without any criticism. In order to solve this problem, alternatively, we firstly divide OPT-1 into two sub-optimization problems. Secondly, we form the Lagrange dual problem of the first one and prove the strong duality of the prime and dual problems. Eventually, we combine these two optimization problems together to get the optimal number and placement for relay nodes. By substituting the objective function with Equations (1) and (2), we have:  
min \[\int_{{}_{T}}[\sum\limits_{i=0}^{n}(\Phi_{1}+\Phi_{2}d_{i}^{k})r_ {{}_{T}}^{i}(t)+\sum\limits_{i=1}^{n+1}\rho r_{{}_{R}}^{i}(t)]dt\] s.t. \[(3)\] \[(0\leq d_{i}\leq R,i=0\ldots n)\]The objective function can be further reshaped as:  
\[\sum\limits_{i=0}^{n}(\[PERSON]_{1}+\[PERSON]_{2}d_{i}^{k})\int_{\tau}r_{{}_{T}}^{i}(t)dt+ \sum\limits_{i=1}^{n+1}\rho\int_{\tau}r_{{}_{R}}^{i}(t)dt=\sum\limits_{i=0}^{n} (\[PERSON]_{1}+\[PERSON]_{2}d_{i}^{k})f_{{}_{T}}^{i}+\sum\limits_{i=1}^{n+1}\rho f_{{}_{R }}^{i} \tag{4}\]  
where \(f_{{}_{T}}^{i}=\int_{\tau}r_{{}_{T}}^{i}(t)dt\) and \(f_{{}_{R}}^{i}=\int_{\tau}r_{{}_{R}}^{i}(t)dt\) are the data that the i-th node transmits and receives during the time period \(\tau\). For the data integrity, during the time period \(\tau\), we have the following chain equation,  
\[f_{{}_{T}}^{0}=f_{{}_{R}}^{1}=f_{{}_{T}}^{1}=\cdots=f_{{}_{R}}^{n}=f_{{}_{T}}^ {n}=f_{{}_{R}}^{n+1}=f \tag{5}\]  
This chain equation indicates that during time period \(\tau\), the data transmitted by the i-th node are equal to the data received by it (in this article, we do not take into consideration the data aggregation and data compression at each node). After expanding the objective function, we have:  
\[\min (n+1)(\Phi_{1}+\rho)f+\Phi_{2}f\sum\limits_{i=0}^{n}d_{i}^{k}\] s.t. \[(3)\] \[(0\leqslant d_{i}\leqslant R,i=0\ldots n)\]  
The Constraints (1) and (2) are omitted here in that they have been already plugged into the objective function. The optimization variables are the number of relay nodes \(n\) and the distance between two successive nodes \(d_{i}\). To solve this optimization problem, we firstly divide it into two sub-optimization problems. For a given number \(n\), we can firstly solve the following optimization problem OPT-2,  
\[\min \Phi_{2}f\sum\limits_{i=0}^{n}d_{i}^{k}\] s.t. \[\sum\limits_{i=0}^{n}d_{i}-L=0\] \[d_{i}-R\  
In general cases, the optimal value of the dual problem always provides a loose lower bound of the optimal value of the prime one. However, in this case, we can prove that the strong duality holds for the prime and dual problems, that is the optimal values of the prime and dual problem are identical. The strong duality of the prime optimization problem OPT-2 and its Lagrange dual problem holds, i.e., the optimal values of these two problem are identical. The proof of this theorem is not trivial, and we leave it to the Appendix section for a better intelligibility. The next theorem will show a necessary condition for achieving the optimality of OPT-1. The optimality of OPT-1 can be achieved only if relay nodes are placed at the equal partition points between two sensor nodes. Since the prime optimization problem OPT-2 is a convex problem and the strong duality holds, we can use the KKT conditions to figure out the optimal value of both the prime and dual optimization problems. The KKT conditions for OPT-2 for a given n are listed below:  
\[\left\{