# Capacity and Delay Analysis for Large Social-Aware Mobile Ad Hoc Wireless Networks
## 5 Scheduling Scheme and Achieved Throughput Capacity
Let \(h_{i}\) denote the number of hops for each S-D pair, we get  
\[h_{i}=\Theta\left(\frac{E\{\mathcal{L}(S,D)\}}{\sqrt{m}}\right).\]  
where \(E\{\mathcal{L}(S,D)\}\) is the distance of S-D pairs. The total number of hops required for all S-D pairs to send data to corresponding destination node can be represented as \(N_{h}=\sum_{i=1}^{n+1}h_{i}\). If the \(k\)th hop originated from S-D pair \(i\) intersects the super-cell, the random variable \(I_{k}^{i}=1\), where \(1\leq i\leq n+1\) and \(1\leq k\leq h_{i}\); otherwise \(I_{k}^{i}=0\). Therefore, the total number of super-cells \(\mathbb{SD}\) lines passing through is \(I=\sum_{1}^{n}\sum_{i=1}^{h_{i}}I_{i}^{k}\). A given super-cell can be intersected by \(\mathbb{SD}\) lines at most once. \(I_{k}^{i}\) and \(I_{l}^{j}\) are independent for any \(1\leq i\
eq j\leq n\). Variable \(I\) reflects the relaying load of a super-cell, its mathematical expectation is  
\[E[I] =E_{N_{h}}[E[I|N_{h}]]\] \[=E_{N_{h}}[\sum_{i=1}^{n+1}\sum_{k=1}^{h_{i}}E[I_{1}^{1}]]\] \[=E_{N_{h}}[N_{h}E[I_{1}^{1}]]=E_{N_{h}}[N_{h}m]\] \[=mE\left[\sum_{i=1}^{n+1}\frac{\mathcal{L}(S,D)}{\sqrt{m}}\right]\] \[=nE\{\mathcal{L}(S,D)\}\sqrt{m}, \tag{18}\]  
where \(E[I_{1}^{1}]=m\), since any hop is equally likely to originate from any of the super-cells. Therefore, the number of super-cell that S-D lines passing through is \(O(nE\{\mathcal{L}(S,D)\}\sqrt{m})\). Next, we present that the interior zone of each super-cell achieves constant per-node throughput. In large social-aware mobile ad hoc wireless networks, the per-node throughput for the interior zone of each super-cell is \(\Theta(1)\), with multi-hop relay and slots allocation scheme. Let us assume that the total number of data transmitted for each S-D pairs in sub-slot A is \(D(t)\). Let \(C(t)\) denote the number of cells which occur a source node transmit data to relay node transmission. Then, we have \(D(t)=\sum_{1}^{nm}C(i,t)\). Since the transmission of source nodes occupied \(\frac{1}{3}\) time slot, we can obtain  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]0Next, we determine the value of \(E[C(t)]\) and \(E[D(t)]\). The probability that \(k\) nodes are in a general cell is \(P_{k}\), then  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]1  
Obviously,  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]2 \[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]3  
Thus,  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]4  
The event that an source node in interior cell transmit data to relay node transmission denote by indicator - \(\mathcal{I}_{i}\). \(\mathcal{I}_{i}\) contains two parts: One is the probability that cell contains exactly one source node, not a destination node represented by \(P_{1}(1-P_{0})\), The other is the probability that cell contains two or more nodes represented by \(1-P_{0}-P_{1}\). Thus, we can calculate the mathematical expectation of \(\mathcal{I}_{i}\),  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]5 \[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]6 \[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]7  
Therefore,  
\[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]8 \[\mathcal{L}(X_{k},X_{i}) \geq (1+\Delta)r(n);\]9 \[\mathcal{L}(X_{j},X_{i}) \leq r(n). \tag{17}\]0 \[\mathcal{L}(X_{j},X_{i}) \leq r(n). \tag{17}\]1  
where \(c_{3}=c_{2}/c_{1}\). Each node randomly chooses direction and randomly chooses speed in super-cell. Then, we know the state can model as an irreducible finite-state Markov Chain. Therefore, over a period of time slot \(N\), by the ergodicity of such a Markov Chain, we get  
\[\mathcal{L}(X_{j},X_{i}) \leq r(n). \tag{17}\]2 \[\mathcal{L}(X_{j},X_{i}) \leq r(n). \tag{17}\]3  
Thus, the total number of data transmitted from sources node is \(\Theta(nm)\), which equals to the number of nodes in super-cell. We obtain that each source node can transmit at rate of \(\Theta(1)\).