# Throughput Capacity of Selfish Wireless _Ad Hoc_ Networks with General Node Density
## 3 Achievable Rate
### The Rate for Transporting a Packet
According to Theorem 3, we can derive the achievable rate in each phase. Comparing the achievable rate of each phase, we get the rate bottleneck. (1) _Draining phase:_ Since the network is divided into squares with side length of \(l(n)=c_{0}\sqrt{\frac{1}{\zeta p(n)}}\), we partition the network area into slices with dimension \(\sqrt{A}\times l(n)\). Similar to Lemma 2 in [5], we bound the number of nodes \(N_{s}\) in each slice uniformly, which bounds the number of nodes accessing a crossing highway. Note that there are a total of \(\frac{\sqrt{A}}{\sqrt{2}l(n)}\) slices, each node in the \(i\)-th slice transmits directly to an entry point located on the \(i\)-th crossing highway, as shown in Figure 2.  
**Lemma 4**. : _Let \(N_{s}\) denote the number of nodes in each slice. Then,_  
\[\lim_{n\rightarrow\infty}P\left(N_{s}\leq 2\sqrt{2n/p(n)},\ \ \forall s\right)=1 \tag{11}\]  
Proof. : The proof follows from the Chernoff and Union bound. The next lemma illustrates the achievable rate in the draining phase. **Lemma 5**. : _The transmitter inside each square can achieve a rate to an entry node on the highway of_  
\[R^{Dr}= \begin{cases}\Omega\left(p(n)\left(\frac{\log^{2}np(n)}{\sqrt{\zeta p( n)}}\right)^{\alpha}\right)&ifl(n)=\Omega(\log np(n))\\ \Omega\left(\frac{p(n)}{\log^{2}np(n)}\right)&ifl(n)=O(\log np(n))\end{cases}  \tag{12}\]  
Proof. : Similar to [5], the area \(\mathbb{A}\) is divided into rectangular slabs of dimensions \(\sqrt{A}\times\sqrt{2}l(n)\left(\kappa\log m(n)+\epsilon_{n}\right)\), where \(m(n)\) is defined in (4) and \(\kappa\) is chosen such that there are at least \(\lceil\delta\log m(n)\rceil\) crossing paths in each slab. The crossing paths are denoted as \(1,\cdots,N_{n}\). In order to balance the load across the highways, we slice the each slab into \(\delta\log m(n)\) smaller strips, each of dimensions \(\sqrt{A}\times\sqrt{2}\omega l(n)\), where \(\omega\) is a constant and chosen appropriately. Note that each crossing highway may not be fully contained in its corresponding strip, but it may deviate from it. Once the source nodes are mapped to crossing paths, we choose the entry points for each source as follows: The entry point is chosen from only these open squares containing one AN. The transmitter drains the information to the entry point directly, and each transmitter finds its highway within the same slab. Hence, the distance between transmitter and entry point is never larger than \(\kappa l(n)\log m(n)+\sqrt{2}l(n)\). To compute the rate that node can transport to the entry point on the highway, let \(d=\kappa l(n)\log m(n)+\sqrt{2}l(n)\) and apply the Theorem 3. We can obtain that a node can communicate to its entry point at rate  
\[R^{Dr}=R\left(\kappa l(n)\log m(n)+\sqrt{2}l(n)\right) =R\left(\kappa l(n)\log\left(\sqrt{np(n)}\right)+\sqrt{2}l(n)\right) \tag{13}\] \[= \begin{cases}\Omega\left(\left(\frac{\log^{2}np(n)}{\sqrt{\zeta p (n)}}\right)^{\alpha}\right)&\text{if }l(n)=\Omega(\log np(n))\\ \Omega\left(1/\log^{2}np(n)\right)&\text{if }l(n)=O(\log np(n))\end{cases} \]  
Next, we derive the achievable rate on the highway phase. **Lemma 6**: _The information along the highways can achieve a per-node rate of \(R^{H}\) w.h.p., where \(R^{H}\) is_  
\[R^{H}= \begin{cases}\Omega\left(\frac{1}{\sqrt{n}}p^{\frac{1}{2}}(n)(\zeta p(n))^ {\alpha}\right)&\text{if }l(n)=\Omega(1)\\ \Omega\left(\frac{1}{\sqrt{n}}p^{\frac{1}{2}}(n)\right)&\text{if }l(n)=O(1)\end{cases}  \tag{14}\]  
**Proof. ** The information transported on the highway is forwarded by multi-hop routing. Under the pairwise coding and decoding, the transporting is performed along the horizontal highways first, until it reaches the crossing with the target vertical highway. Then, the same is performed along the vertical highways until it reaches the appropriate exit point for delivery. According to Lemma 2, a node on the horizontal highway must relay for at most \(2l(n)\sqrt{A}\) nodes, and the maximal distance between two hops is \(\sqrt{2}l(n)\).