## 3 Content Acquisition Process and Caching Algorithm Optimization
### Analysis of Success Probability of Content Acquisition  
In our strategy, vehicles cache content in their caching space based on content's popularity, the caching space can accommodate content with maximum number of \(N\), and the probability that a vehicle caches the content ranked \(\tau-th\) is \(f^{\tau}(N)\). Assume that a vehicle \(V_{x}\) on lane \(i\) requests the content with a rank of \(\tau\) in popularity, and \(V_{x}\) is within or is going to enter the coverage area of \(R_{m}\). As described in Section 3.1, there are five cases for the request vehicle to obtain desired content: Obtaining content from vehicles that are a one-hop distance away from the request vehicle in the same lane, obtaining content from vehicles that are a two-hop distance away from request vehicle in the same lane, obtaining content from vehicles that are a one-hop or two-hop distance away from the request vehicle on the front opposite lane, obtaining content directly from an RSU, and obtaining content from the cluster with content cached by the RSU. The analysis of those five cases is shown as follows. * Obtaining content from vehicles that are a one-hop distance away from the request vehicle \(V_{x}\) drives after content source vehicle \(V_{y}\), the speed of each vehicle is, respectively, \(s_{i}^{x}\) and \(s_{i}^{y}\). The average transmission rate between vehicles is \(V^{a}\), and the distance between the request vehicle and content source vehicle is \(\frac{r^{v}}{2}\). When \(s_{i}^{x}>s_{i}^{y}\), the condition of content success transmission is \(\frac{r^{v}+\frac{r^{v}}{2}}{s_{i}^{x}-s_{i}^{y}}\cdot V^{a}\geq W\), and the feasible region is \(G1=\left\{(s_{1}^{x},s_{i}^{y})\Big{|}s_{i}^{x}-s_{i}^{y}\leq\frac{3r^{v}V^{a} }{2W}\right\}.\) When \(s_{i}^{x}<s_{i}^{y}\), the condition of content success transmission is \(\frac{r^{v}-\frac{r^{v}}{2}}{s_{i}^{y}-s_{i}^{x}}\cdot V^{a}\geq W\), and the feasible region is \(G2=\left\{(s_{i}^{x},s_{i}^{y})\Big{|}s_{i}^{y}-s_{i}^{x}\leq\frac{r^{v}V^{a} }{2W}\right\}.\)\(s_{i}^{x}\) and \(s_{i}^{y}\) are independent of each other and obey the same Truncated Normal Distribution [23], so \(p\Big{\{}s_{i}^{x}>s_{i}^{y}\Big{\}}=p\Big{\{}s_{i}^{x}<s_{i}^{y}\Big{\}}= \frac{1}{2}\). According to Formula (3), the probability that a vehicle on lane \(i\) can obtain content from another vehicle that is a one-hop distance away from itself is derived as follows:  
\[ \begin{array}{ll}P_{i}^{one}&=p\{s_{i}^{x}>s_{i}^{y}\}\cdot\iint\limits_{G1 }f\Big{(}s_{i}^{x},s_{i}^{y}\Big{)}\ d\sigma+p\{s_{i}^{x}<s_{i}^{y}\}\cdot \iint\limits_{G2}f\Big{(}s_{i}^{x},s_{i}^{y}\Big{)}\ d\sigma\\ &=\frac{1}{2}\cdot\iint\limits_{G1+G2}f\Big{(}s_{i}^{x},s_{i}^{y}\Big{)}\ d \sigma\end{array}  \tag{11}\]  
Let \(K_{1}\) denotes the number of vehicles that are a one-hop distance away from the request vehicle in the same lane. The probability that a request vehicle on lane \(i\) can obtain desired content from a vehicle that is a one-hop distance away from itself is derived as follows:  
\[ \begin{array}{ll}P_{i,\tau}^{self}&=\sum\limits_{k=1}^{+\infty}P_{i}\{K_{1 }=k\}\cdot\{f^{\tau}(N)\cdot 1+(1-f^{\tau}(N))\cdot[1-(1-f^{\tau}(N))^{k-1}] \cdot P_{i}^{one}\}\\ &=\sum\limits_{k=1}^{+\infty}P_{i}\{K_{1}=k\}\cdot\{f^{\tau}(N)+[1-f^{\tau}( N)-(1-f^{\tau}(N))^{k}]\cdot P_{i}^{one}\}\\ &=[P_{i}^{one}+f^{\tau}(N)-P_{i}^{one}\cdot f^{\tau}(N)]\cdot\sum\limits_{k=1 }^{+\infty}P_{i}\{K_{1}=k\}\cdot(1-f^{\tau}(N))^{k}\\ &=[P_{i}^{one}+f^{\tau}(N)-P_{i}^{one}\cdot f^{\tau}(N)]\cdot(1-e^{-2\lambda_{i }D^{v}})-P_{i}^{one}\cdot(e^{-2\lambda_{i}D^{v}f^{\tau}(N)}-e^{-2\lambda_{i}D ^{v}})\end{array}  \tag{12}\]  
Obtaining content from vehicles that are a two-hop distance away from the request vehicle  
Assumed that the distance from relay vehicle to request vehicle and content source vehicle is equal, it's \(\frac{3r^{v}}{4}\), and the probability of content success transmission from relay vehicle to request vehicle and content source vehicle is also equal.