## 4 Load-Similar Node Distribution
### Analysis of Load in Uniform Distribution Scheme  
Assume static sensor nodes are uniformly distributed with node density \(\rho\) within a \(2R\)\(\times\)\(2R\) square area, and the sink is located at the center of the area. Each source node generates and sends \(\lambda\) bits of data per unit time. The ratio of the number of source nodes to the total number of sensor nodes is \(\mu\). Figure 4 illustrates the geometric relationship of layer \(i\) (with radius \(r\) which is equal to the transmission range) within a \(2R\)\(\times\)\(2R\) square area. Assume the distance from the sink to the outer boundary of layer \(i\) is \(d\). First, we deduce the average load per node at layers within the range of \(r<d\leq R\), as shown in Figure 4(a). According to the energy model, the total energy consumed for data reception and transmission in a unit time by all the sensor nodes at layer \(i\) (_i_\(>\) 1), \(E_{ri}\) and \(E_{ti}\), are given by:  
\[E_{ri}=E_{elec}\lambda\rho\mu(4R^{2}-\pi l^{2}),\ \ \ \ i>1,\ \ \ \ r<d\leq R, \tag{5}\]\[E_{ti}=(E_{elec}+\sigma^{2})\lambda\rho\mu(4R^{2}-\pi(d-r)^{2})\,,\ \ \ \ \ i>1,\ \ \ \ \ r<d\leq R. \tag{6}\]  
Assume \(N_{i}\) is the number of sensor nodes at layer \(i\), which can be calculated by \(\rho\pi(d^{2}-(d-r)^{2})\). Then the average load per node at layer \(i\) (for \(i\)\(>\)1 and \(r\)\(<\)\(d\)\(\leq\)\(R\)) in a unit time is given by:  
\[ \begin{split}& L_{i}=\frac{E_{ri}+E_{ti}}{N_{i}}+\frac{E_{ci}}{T_{r}} \\ &=\frac{E_{elec}\lambda\rho\mu(4R^{2}-\pi d^{2})}{\rho\pi(d^{2}- (d-r)^{2})}+\frac{(E_{elec}+\sigma^{2})\lambda\rho\mu(4R^{2}-\pi(d-r)^{2})}{ \rho\pi(d^{2}-(d-r)^{2})}+\frac{(l_{cp}*n_{cp}+l_{bp}*n_{bp}+l_{hp}*n_{hp}+l_{ ccp}*n_{ccp})E_{elec}}{T_{r}},\end{split}  \tag{7}\]  
where \(T_{r}\) is the average lifetime per round, and \(1/T_{r}\) gives the number of network formations performed in a unit time. For \(R<d\leq R+r\), shown as the shaded area in Figure 4(b), the average load per node at layer \(i\) in a unit time can be calculated as:  
\[L_{i}=\frac{E_{ri}+E_{ti}}{N_{i}}+\frac{E_{ci}}{T_{r}}=\frac{4\lambda\mu E_{ elec}A}{S_{1}}+\frac{\lambda\mu(E_{elec}+\sigma^{2})(4A+S_{1})}{s_{1}}+\frac{(l_{cp}*n_{cp}+ l_{bp}*n_{bp}+l_{hp}*n_{hp}+l_{ccp}*n_{ccp})E_{elec}}{T_{r}}, \tag{8}\]  
where \(A\) gives the area of each corner outside the shaded area (shape CDE in Figure 4(b)), which can be calculated by \(\left({{R-\sqrt{d^{2}-R^{2}}}}\right)\)\({{R-\left({\frac{\pi}{4}-a\cos\left({\frac{R}{d}}\right)}\right)}}\)\({{d^{2}}}\), and \(S_{I}\) is the area of the shaded layer which can be calculated by \(\frac{}{}_{4(R^{2}-\frac{\pi}{4}(d-r)^{2}-A)}\). For \(R+r<d<\sqrt{2}R\), shown as the shaded area in Figure 4(c), the average load per node at layer \(i\) in a unit time can be calculated as:  
\[L_{i}=\frac{E_{ri}+E_{ti}}{N_{i}}+\frac{E_{ci}}{T_{r}}=\frac{\lambda\mu E_{ elec}A}{A-A}+\frac{\lambda\mu(E_{elec}+\sigma^{2})A^{i}}{A-A}+\frac{\lambda\mu(E_{ elec}+\sigma^{2})A^{i}}{A-A}+\frac{\lambda\mu_{cp}*n_{cp}+l_{bp}*n_{bp}+l_{bp}*n_{ hp}+l_{ccp}*n_{ccp})E_{elec}}{T_{r}}, \tag{9}\]  
where \(A^{\prime}\) gives the area of shape BCDEF, which can be calculated as \(\left({{R-\sqrt{(d-r)^{2}-R^{2}}}}\right)\)\({{R-\left({\frac{\pi}{4}-d\left({\frac{R}{d-r}}\right)}\right)}}\)\({{d-r}}\)\({{d^{2}}}\), and the calculation of the area of each corner \(A\) follows that in (5-2). For \(d=\sqrt{2}R\), _i.e._, at the outermost layer, the average load per node at layer \(i\) in a unit time can be calculated as:  
\[L_{i}=\frac{E_{ti}}{N_{i}}+\frac{E_{ci}}{T_{r}}=\mu\lambda(E_{elec}+\sigma^{2} )+\frac{(l_{cp}*n_{cp}+l_{bp}*n_{bp}+l_{ccp}*n_{ccp})E_{elec}}{T_{r}}. \tag{10}\]  
For sensor nodes at layer 1, they will receive and forward the data coming from outside layer 1 and also send the data generated from layer 1.