## 5 Performance Analysis
### Packet Delivery Probability  
It is assumed that the distance between two adjacent vehicles on the road follows an exponential distribution [39; 40]. Let \(\rho\) denote the density of vehicles. Then, the number \(n\) of vehicles over a length \(d\) of road follows a Poisson distribution, and the probability is calculated as:  
\[p(n,d)=\frac{{{{(pd)}^{n}}{e^{-pd}}}}{{n!}},n=0,1,2 \tag{7}\]  
Let \(N\) denote the number of nodes in OHS of the source node. Then, the probability distribution function (pdf) of \(N\) is given by:  
\[\Pr\{N=n\}=p(n,2R_{T})=\frac{{{{(2\rho R_{T})}^{n}}{e^{-2\rho R_{T}}}}}{{n!}},n= 0,1,2,\cdots. \tag{8}\]  
According to (2), a node in OHS of the source node receives a packet with the probability \(p_{s}(1)\) during a slot. Let \(V\) denote the number of nodes that receive the packet successfully. Given \(N=n\), \(V\) follows a binomial distribution and its conditional pdf is given by:  
\[\Pr\{V=v|N=n\}=C_{n-1}^{v}[p_{s}(1)]^{v}[1-p_{s}(1)]^{n-v-1} \tag{9}\]  
In (9), if \(n\leq 1\), no nodes receive the packet; if \(2\leq n\leq F\), up to \(n-1\) nodes can receive the packet successfully; if \(n>F\), only \(F-1\) nodes that have reserved their own time slots can receive the packet. Thus, the pdf of \(V\) can be calculated as:  
\[\Pr\{V=v|N=n\}=\left\{ \begin{array}{cc}1,&v=0,n\leq 1\\ C_{n-1}^{v}[p_{s}(1)]^{v}[1-p_{s}(1)]^{n-v-1},&0\leq v\leq n-1,2\leq n\leq F\\ C_{F-1}^{v}[p_{s}(1)]^{v}[1-p_{s}(1)]^{F-v-1},&0\leq v\leq F-1,n>F\end{array} \right. \tag{10}\]  
Let \(q\) denote the probability that a node that successfully receives the packet becomes a potential helper. That means the node does not have any two-hop neighbors that reserve the same slot as the source node. Let \(H\) denote the number of potential helpers. Given \(V=v\), \(H\) follows a binomial distribution and its conditional pdf is given by:  
\[\Pr\{H=h|V=v\}=C_{q}^{h}{h}^{h}(1-q)^{v-h} \tag{11}\]  
\(E_{K}\) is used to denote the event of existence of \(K\) helpers. From (9)-(11), the probability of the existence of \(K\) helpers is given by:  
\[ \begin{array}{rcl}\Pr(E_{K})&=&\Pr\{\text{H}\geq\text{K}\}\\ &=&\sum_{n=K+1}^{F}\sum_{v=K}^{n-1}\sum_{h=K}^{v}C_{p}^{h}{h}^{h}(1-q)^{v-h}C_ {n-1}^{v}[p_{s}(1)]^{v}[1-p_{s}(1)]^{n-v-1}\frac{{{{(2\rho R_{T})}^{n}}{e^{-2 \rho R_{T}}}}}{{n!}}\\ &+&\sum_{v=K}^{n-1}\sum_{h=K}^{v}C_{p}^{h}{h}^{h}(1-q)^{v-h}C_{F-1}^{v}[p_{s}( 1)]^{v}[1-p_{s}(1)]^{F-v-1}(1-\sum_{n=0}^{F}\frac{{{{(2\rho R_{T})}^{n}}{e^{-2 \rho R_{T}}}}}{{n!}})\end{array}  \tag{12}\]  
In (12), the value of \(q\) for each node depends on its position. As shown in Figure 7, the \(2R_{T}\) range of the helper can be divided into two parts: \(R_{1}\) and \(R_{2}\). The nodes in range \(R_{1}\) are also in \(2R_{T}\) range of the source node. Thus, they do not have the same slot with the source node according to the channel access rules (Section 3). Meanwhile, the nodes in range \(R_{2}\) may occupy the same slot with the source node. Let \(N_{R_{1}}\) and \(N_{R_{2}}\) denote the number of nodes in range \(R_{1}\) and \(R_{2}\), respectively. Let \(A_{R_{2}}\) denote the available slots for the nodes in range \(R_{2}\). \(E_{R_{2}}\) is used to denote the event that the nodes in range \(R_{2}\) do not have the same slot with the source node. Given \(N_{R_{1}}=n\), \(N_{R_{2}}=m\), the probability of \(E_{R_{2}}\) is:Let \(d\) denote the length of range \(R_{2}\), then the length of range \(R_{1}\) is \(2R_{T}-d\).