## 3 Analysis and Numerical Results  
### SINR and Power Analysis  
The volume of data originating in the individual nodes is assumed to be the same. Because the \(i\)th node cumulatively receives data from the first to \((i-1)\)th nodes, the amount received by the \(i\)th node is proportional to \((i-1)\). The amount of data transmitted by the \(i\)th node is proportional to \(i\) including the node's data. Assume that the data rate for the transmission of each node is the transmission rate \(C\). If \(C_{i}\) is assumed to be the rate of data received from the \(i\)th node, \(C_{i}\)=\((i-1)C\) (\(i=2\),3,4,...,\(M+1\)). The SINR (signal-to-interference-plus-noise ratio) received by the \(i\)th node is assumed to be \(SINR_{i}\). The commonly used bandwidth is assumed to be \(W\) and the following Shannon capacity equation is approximately used:  
\[C_{i}\approx W\mid\log_{2}\left(1+SINR_{i}\right) \tag{1}\]  
where \(C_{i}\) is the best achievable data rate. A higher \(SINR_{i}\) or \(C_{i}\) requires a higher modulation or a lower code rate in practical design due to the same transmission time for all nodes. Because the required \(SINR_{i}\) that satisfies the data rate \(C_{i}\) is calculated from Equation (1) and the actual \(SINR_{i}\) should exceed or be equal to the calculated one, if the calculation is simplified with the bandwidth efficiency, \(C_{W}=C/W\), the following condition should be satisfied for the \(SINR_{i}\):\[ \begin{split}\text{1}\text{0}\text{9}_{2}\left(\text{1}+\text{SINR}_{i} \right)\geq\left(i-1\right)C_{W}\end{split}  \tag{2}\]  
Note that because each node operates as a half-duplex, \(C_{W,avg}\), which is the average \(C_{W}\), becomes \(\dfrac{1}{2}\)\(C_{W}\) in the case of Schemes 1 and 2 and \(\dfrac{1}{3}\)\(C_{W}\) in the case of Scheme 3. The SINRs at the individual nodes are calculated. Only the dominant interference and noise are considered approximately. Assume \(P_{i}\) is the power received by the \((i+1)\)th node, which is transmitted from the \(i\)th node. Because the distance between two nodes and the attenuation in distance are always the same, the ratio of power \(P_{i}\) to the transmitted power is always the same regardless of \(i\). First, for Scheme 1 in Figure 4a, the SINR received by the \(i\)th node is obtained as follows:  
\[ \begin{split}\text{SINR}_{i}=\dfrac{P_{i-1}}{P_{i+1}+P_{N}}, \hskip 14.226378 pti=2,3,...,M-1\end{split}  \tag{3}\]  
where \(P_{N}\) is the noise power. In most cases, the dominant interference comes from the box next to the right, as shown in Equation (3). \[ \begin{split}\text{SINR}_{M}&=\dfrac{P_{M-1}}{P_{M} \left(\frac{1}{2}\right)^{1.5}+P_{N}}\\ \text{SINR}_{M+1}&=\dfrac{P_{M}}{P_{M-1}\left(\frac {1}{2}\right)^{1.5}+P_{N}}\end{split}  \tag{4}\]  
As a boundary condition, the dominant interference comes from the second box to the right for the node immediately before the sink node in Equation (4) and the sink node in Equation (5) (\(P_{M+2}=P_{M}\), \(P_{M+3}=P_{M-1}\) ). The path loss exponent is assumed to be 1.5 [14]. Next, the \(\text{SINR}_{i}\) for Scheme 2 in Figure 4b is as follows:  
\[ \begin{split}\text{SINR}_{i}=\dfrac{P_{i-1}}{P_{N}},\hskip 14.226378 pti=23 \end{split}  \tag{5}\]  
The first two SINRs received are shown by Equation (5) because they have only noises without any interference:  
\[ \begin{split}\text{SINR}_{i}=\dfrac{P_{i-1}}{P_{i-3}\left(\dfrac {1}{3}\right)^{1.5}+P_{N}},\hskip 14.226378 pti=45,\ldots M-1\end{split}  \tag{6}\]  
In most cases, the dominant interference comes from the third box to the right as shown by Equation (7):  
\[ \begin{split}\text{SINR}_{M}&=\dfrac{P_{M-1}}{P_{M- 3}\left(\dfrac{1}{3}\right)^{1.5}+P_{M-1}\left(\dfrac{1}{3}\right)^{1.5}+P_{ N}}\\ \text{SINR}_{M+1}&=\dfrac{P_{M}}{P_{M-2}\left(\dfrac {1}{3}\right)^{1.5}+P_{M-1}\left(\dfrac{1}{2}\right)^{1.5}+P_{N}}\end{split}