# The Effects of Topology on Throughput Capacity of Large Scale Wireless Networks
## 3 The Intra-Cluster Transmission
### Per-Node Rate of Intra-Cluster Traffic
\begin{split} I(d)&\leq\int_{j=1}^{\infty}\left(\left(2 j+1\right)^{3}-\left(2j-1\right)^{3}\right)\cdot P_{i+4 jd}\cdot\left(\frac{1}{ \sqrt{3}\left(r_{i+4 jd}-r_{i}\right)}\right)^{\alpha}\\ &\leq\sum_{j=1}^{\infty}\left(\left(2j+1\right)^{3}-\left(2j-1 \right)^{3}\right)\cdot P_{i+4 jd}\cdot\left(\frac{1}{4\sqrt{3}jdr_{i}\cdot \frac{2\pi}{y}}\right)^{\alpha}\\ &=\sum_{j=1}^{\infty}\left(\left(2j+1\right)^{3}-\left(2j-1 \right)^{3}\right)\cdot P_{0}\cdot\left(\frac{1}{4\sqrt{3}jd}\cdot\left(1+ \frac{2\pi}{y}\right)^{4 jd}\right)^{\alpha}\\ &=P_{0}(4\sqrt{3}d)^{-\alpha}\cdot\sum_{j=1}^{\infty}\frac{\left( 2j+1\right)^{3}-\left(2j-1\right)^{3}}{j^{\alpha}}\cdot\left(1+\frac{2\pi}{y} \right)^{4 jd\alpha}.\end{split}  \tag{17}\]  
Since \(y=\Theta\left(\left(\frac{n^{1-\alpha}}{\left(1-\alpha\right)\log n}\right) ^{\frac{1}{3}}\right)\), \(\alpha>3\) and \(d=o\left(\left(\frac{n^{1-\alpha}}{\left(1-\alpha\right)\log n}\right)^{\frac {1}{3}}\right)\), the summation will be converged to some constant. According to the tessellation of a cluster, the distance (Euclidean distance) between the transmitter and the receiver located at \(d\) quasi-cubelets away is at most \(\frac{2\pi r_{i}}{y}d\). Thus, the received power at the receiver is bounded by:  
\[S\left(d\right)\geq P_{i}\bigg{(}\frac{2\pi r_{i}}{y}d\bigg{)}^{-\alpha}=P_{0} \cdot d^{-\alpha}. \tag{18}\]  
Substituting Equations (17) and (18) into Equation (4), we obtain an asymptotic lower bound of the rate \(R(d)\) of  
\[R\left(d\right)=\log\left(1+\frac{S\left(d\right)}{N_{0}+I\left(d\right)} \right)=\Omega(d^{-\alpha}). \tag{19}\]  
Since the time is divided into \(k^{3}=8(d+1)^{3}\) time slots, the final rate for each subcube is  
\[R\left(d\right)=\Omega(d^{-\alpha-3}). \tag{20}\]  
\(\square\)  
In Equation (20), we have derived a general equation to calculate the data rate. Using Equation (20) and substituting \(d\) with different transmission distances, we can get the rate of each routing phase. In addition, [PERSON] et al. [2] had proved that the data rate bottleneck is located at the highway phase; thus, we will only compare the data rate of the highway phase. Lemma 1: _Every node on the RH can achieve a rate of \(\Omega\left(1/(n^{1-\alpha}((1-v)\log n\ )^{2})^{\frac{1}{3}}\right)\) w.h.p._  
Proof: According to the routing scheme, packets delivered on the highway are hop-by-hop, i.e., the packet is transmitted from one quasi-cubelet to its adjacent quasi-cubelet, which means that the distance \(d\) (number of quasi-cubelet) is one. Using Equation (20), let \(d=1\), and we can get that the rate over each hop is \(\Omega(1)\). However, a highway may not sever one node; all the nodes on the highway need to share it. To derive the number of nodes sharing a highway, we employ Chernoff's bound to derive it. Chernoff Bound: Let \(X\) be a Poisson random variable of parameter \(\lambda\). We have  
\[Pr\left[X\geq\xi\right]\leq\frac{e^{-\lambda}(e\lambda)^{\xi}}{\delta^{\xi}}, \ \ \text{for}\ \ \xi>\lambda, \tag{21}\]\[Pr\left[X\leq\xi\right]\leq\frac{e^{-\lambda}\left(e\lambda\right)^{\xi}}{\xi^{ \xi}},\ \ \text{for}\ \ \xi<\lambda. \tag{22}\]  
For \(0<\delta<1\), Chernoff bounds given in Equations (21) and (22) can be combined and simplified to  
\[Pr\left[|X-\lambda|\geq\delta\lambda\right]\leq 2e^{\frac{-\beta^{2}\lambda}{2}}. \tag{23}\]  
Chernoff's bound indicates that there are at most \(\frac{2n}{my^{2}}\) nodes in each cone w.h.p. Therefore, we can get that the data rate on the RH is  
\[R_{1} =\Omega\left(1/(n^{1-v}((1-v)\log n\ )^{2})^{\frac{1}{3}}\right). \tag{24}\]  
\(\square\)  
Similar to Lemma 1, we derive the rate on SH. After comparing the rate on RH and SH, we obtain the rate bottleneck, which is also the throughput capacity of intra-cluster transmission. **Lemma 2**.