# Adaptive Multi-Path Routing Protocol in Autonomous Vehicular Networks
## 5 Mathematical Analysis
### Probability of Finding a High-Throughput Path  
Figure 6 shows a scenario where node A is transmitting to node B. \(R_{1}\), \(R_{2}\), and \(R_{5.5}\) denote the maximum transmission range for data rates 1, 2, and 5.5 Mbps, respectively, and the values as shown in Table 3.  
Let us assume node A is using min-hop and is transmitting to node B, which is at the edge of the maximum transmission range \(R_{1}\). In order to leverage a higher throughput path, there need to exist a node in the middle which can relay the data to node B, and let that node be named as C. As explained in Section 4.3, the links AC and CB should enable at least the data rates of 2 Mbps and 5.5 Mbps to be effective. To enable this, relay node C should be placed in the overlapping area of the two circles, i.e., the shaded area as shown in Figure 6.  
We next compute the shaded area. First, we obtain the angles \(\theta_{A}\) and \(\theta_{B}\) via the cosine law as follows:  
\[R_{1}^{2}=R_{2}^{2}+R_{5.5}^{2}-2R_{2}R_{5.5}\cos\theta_{A},\ R_{5.5}^{2}=R_{1}^{ 2}+R_{2}^{2}-2R_{1}R_{2}\cos\theta_{B} \tag{5}\]  
Thus,  
\[\theta_{A}=\cos^{-1}(\frac{R_{2}^{2}+R_{5.5}^{2}-R_{1}^{2}}{2R_{2}R_{5.5}}),\ \theta_{B}=\cos^{-1}(\frac{R_{1}^{2}+R_{2}^{2}-R_{5.5}^{2}}{2R_{1}R_{2}}) \tag{6}\]  
Let us denote the left part of the shaded area as \(A_{1}\), and the right part as \(A_{2}\), and the total area as \(A_{T}\), thus \(A_{T}=A_{1}+A_{2}\). Using the results in Equation (6),  
\[A_{1}=\frac{R_{5.5}^{2}(2\theta_{A})}{2}-\frac{R_{5.5}^{2}\sin(2\theta_{A})}{2 },\ A_{2}=\frac{R_{2}^{2}(2\theta_{B})}{2}-\frac{R_{2}^{2}\sin(2\theta_{B})}{2} \tag{7}\]  
Here we assume that the nodes are randomly distributed following the Poisson distribution [33], with density \(\delta\). The average number of nodes in area \(A_{T}\) can be denoted as \(\lambda=\delta\cdot A_{T}\). Finally, the probability for a node to exist in the area \(A\) can be expressed as follows:  
\[p=P(X\geq 1)=1-P(X=0)=1-e^{-\lambda}=1-e^{-\delta\cdot A_{T}} \tag{8}\]  
Random variable \(X\) denotes the number of nodes in the area \(A\). Equation (8) presents the probability of finding a high-throughput path, thus the effectiveness of the MARV protocol.  
Next, let us assume another case where the distance between node D and E is \(R_{2}\) instead of \(R_{1}\), as shown in Figure 7. In order to exploit a higher throughput path, we again assume that a node F in the middle has to exist, and links DF and FE links should allow at least data rates of 5.5 Mbps and 11 Mbps, due to the MTM values shown in Table 3. We repeat the computation process above for the newly defined ranges.  
\[\theta_{D}=\cos^{-1}(\frac{R_{2}^{5.5}+R_{11}^{2}-R_{2}^{2}}{2R_{5.5}R_{11}}), \ \theta_{E}=\cos^{-1}(\frac{R_{2}^{2}+R_{5.5}^{2}-R_{11}^{2}}{2R_{2}R_{5.5}}) \tag{9}\]  
Again, let us denote the left part of the shaded area as \(A_{3}\), and the right part as \(A_{4}\), and the total area as \(A_{T}\), thus \(A_{T}=A_{3}+A_{4}\). Using the results in Equation (9),  
\[A_{1}=\frac{R_{11}^{2}(2\theta_{D})}{2}-\frac{R_{11}^{2}\sin(2\theta_{E})}{2}, \ A_{2}=\frac{R_{5.5}^{2}(2\theta_{E})}{2}-\frac{R_{5.5}^{2}\sin(2\theta_{E})}{2} \tag{10}\]  
Finally, we use the results of Equation (10), and again use Equation (8) to acquire the probability of finding a high-throughput path, \(p\). Note that, if the distance between the two nodes is equal or less than \(R_{5.5}\), then there are no relay nodes that can improve the throughput. In the next subsection, we present the analytical results of the model presented above.  
Figure 6: Node A is transmitting to node B, and the nodes are \(R_{1}\) distance apart. If a node exists in the shaded area, then MARV can take advantage of the high-throughput path.