Nearly Optimal Probabilistic Coverage for Roadside Data Dissemination in Urban VANETs

Abstract

Data disseminations based on Roadside Access Points (RAPs) in vehicular ad-hoc networks attract lots of attentions and have a promising prospect. In this paper, we focus on a roadside data dissemination, including three basic elements: RAP Service Provider (RSP), mobile vehicles and requesters. The RSP has deployed many RAPs at different locations in a city. A requester wants to rent some RAPs, which can disseminate their data to vehicles with some probabilities. Then, it tries to select the minimal number of RAPs to finish the data dissemination, in order to save the expenses. Meanwhile, the selected RAPs need to ensure that the probability of each vehicle receiving data successfully is no less than a threshold. We prove that this RAP selection problem is NP-hard, since it’s a meaningful extension of the classic Set Cover problem. To solve this problem, we propose a greedy algorithm and give its approximation ratio. Moreover, we conduct extensive simulations on real world data to prove its good performance.

Appendix: The Detailed Proof of Theorem 2

We first give two important properties of our utility function $f(\cdot )$.

Lemma 1

$f(\emptyset )\!=\!0$ and $f(\cdot )$ is an increasing function.

Proof

1. We define $f_i(S)\!\!=\!\!min\{Pr(v_{i}|S), \tau \}$. If $S\!=\!\emptyset$, $Pr(v_{i}|S)\!\!=\!\!0$ for $\forall v_i\!\in \!V$. Hence, $f_i(S)\!=\!0$ for $\forall v_i\!\in \!V$. Further, $f(S)\!=\!\sum _{v_i\in V}f_i(S)\!=\!0$.

2. Giving any two sets X and Y, and suppose $X\!\subseteq \!Y\!\subseteq \!A$, obviously we have $Pr(v_i|X)\!\le \!Pr(v_i|Y)$. Consequently, we have $f_i(X)\!\le \!f_i(Y)$ for $\forall v_i\!\in \!V$. Moreover, because of $\theta \!>\!0$, $f(X)\!=\!\theta *\sum _{v_i\in V}f_i(X)\!\le \!\theta *\sum _{v_i\in V}f_i(Y)\!=\!f(Y)$ when $X\!\subseteq \!Y\!\subseteq \!R$. Therefore, $f(\cdot )$ is an increasing function.

Then, we prove that our utility function $f(\cdot )$ is submodular by giving the below Lemma 2, since we all know that, submodular function relates closely to greedy algorithm.

Lemma 2

$f(\cdot )$ is a submodular function.

Proof

Given $X\!\subseteq \!Y\!\subseteq \!R$, $\forall r_k\!\in \!R\backslash \!Y$, if $\varDelta _{r_k}f(X)\!=\!f(X+\{r_k\})-f(X)\!\ge \!\varDelta _{r_k}f(Y)\!=\!f(Y+\{r_k\})-f(Y)$, $f(\cdot )$ is a submodular function. Before proving this conclusion, we first compare the size of $\varDelta _{r_k}f_i(X)\!=\!f_i(X+\{r_k\})-f_i(X)$ and $\varDelta _{r_k}f_i(Y)\!=\!f_i(Y+\{r_k\})-f_i(Y)$, by dividing all possibilities into the following six cases. Note that, $Pr(v_i|(X+\{r_k\}))\!\ge \!Pr(v_i|X)$, $Pr(v_i|(Y+\{r_k\}))\!\ge \!Pr(v_i|Y)$, $Pr(v_i|Y)\!\ge \!Pr(v_i|X)$ and $Pr(v_i|(Y+\{r_k\}))\!\ge \!Pr(v_i|(X+\{r_k\}))$.

• 1. $\tau \!<\!Pr(v_i|X)$, $\tau \!<\!Pr(v_i|Y)$. Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)$ $=[min\{Pr(v_i|(X\!+\!\{r_k\})),\tau \}\!-\!min\{Pr(v_i|X),\tau \}]-[min\{Pr(v_i|(Y\!+\!\{r_k\})),\tau \}\!-\!min\{Pr(v_i|Y),\tau \}]$ $=(\tau \!-\!\tau )\!-\!(\tau \!-\!\tau )\!=\!0$;

• 2. $Pr(v_i|X) \!\le \!\tau \!<\!Pr(v_i|(X+\{r_k\}))$, $\tau \!<\!Pr(v_i|Y)$. Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)$ $=(\tau \!-\!Pr(v_i|X))\!-\!(\tau \!-\!\tau )\ge 0$;

• 3. $Pr(v_i|X) \!\le \!\tau \!<\!Pr(v_i|(X+\{r_k\}))$, $Pr(v_i|Y)\!\le \!\tau \!<\!Pr(v_i|(X+\{r_k\}))$. Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)=(\tau \!-\!Pr(v_i|X))\!-\!(\tau \!-\!Pr(v_i|Y))$ $=Pr(v_i|Y)\!-\!Pr(v_i|X) \!\ge \!0$;

• 4. $Pr(v_i|(X+\{r_k\}))\!\le \!\tau$, $\tau \!<\!Pr(v_i|Y).$ Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)$ $=[Pr(v_i|(X+\{r_k\}))\!-\!Pr(v_i|X)]\!-\!(\tau \!-\!\tau )\!\ge \!0$;

• 5. $Pr(v_i|(X+\{r_k\}))\!\le \!\tau$, $Pr(v_i|(Y+\{r_k\}))\!\le \!\tau$. Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)$ $=[Pr(v_i|(X+\{r_k\}))\!-\!Pr(v_i|X) ]\!-\![Pr(v_i|(Y+\{r_k\}))\!-\!Pr(v_i|Y)]$;

• 6. $Pr(v_i|(X+\{r_k\}))\!\le \!\tau$, $Pr(v_i|Y)\!\le \!\tau \!<\!Pr(v_i|(Y+\{r_k\}))$. Hence, $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)$ $=[Pr(v_i|(X+\{r_k\}))\!-\!Pr(v_i|X)]\!-\!$ $[\tau \!-\!Pr(v_i|Y)]\ge [Pr(v_i|(X+\{r_k\}))\!-\!Pr(v_i|X)]\!-\![Pr(v_i|(Y+\{r_k\}))\!-\!Pr(v_i|Y)]$.

For the cases 5 and 6 above, we need to continue analyzing the size of $Pr(v_i|(S+\{r_k\}))\!-\!Pr(v_i|S)$ for $\forall v_i\!\in \!V$ and $\forall r_k\!\in \!R\backslash \!S$. We define $\varDelta _{r_k}Pr(v_i|S)\!=\!Pr(v_i|(S+\{r_k\}))\!-\!Pr(v_i|S)$. In fact, $\varDelta _{r_k}Pr(v_i|S)$ for $\forall v_i$ is the increment of its probability of receiving data successfully when a new RAP $r_k$ is added to S. It is obvious that $r_k$ only influences the vehicles in $V(r_k)$. In order to compute $\varDelta _{r_k}Pr(v_i|S)$, there exists three possible case below:

• 1. $v_i$ is covered by $r_k$ and (part or all of) the RAPs in S. Hence, $\varDelta _{r_k}Pr(v_i|S)$ $=[1-\prod _{r_j \in (S\!+\!\{r_k\})(v_i)}(1-p_{j}^{i})]\!-\!$ $[1-\prod _{r_j \in S(v_i)}(1-p_{j}^{i})]$ $=\prod _{r_j \in S(v_i)}(1-p_{j}^{i})\!-\!(1\!-\!p^{i}_{k})\prod _{r_j \in S(v_i)}(1-p_{j}^{i})$ $=p_{k}^{i}*\prod _{r_j \in S(v_i)}(1-p_{j}^{i})$;

• 2. $v_i$ is covered by $r_k$ but isn’t covered by any RAPs in S. Hence, $\varDelta _{r_k}Pr(v_i|S)$ $=[1-(1-p_{k}^{i})]-0\!=\!p_{k}^{i}$;

• 3. $v_i$ isn’t covered by $r_k$, namely, wether or not $r_k$ is selected has no influence on $v_i$. Obviously, $\varDelta _{r_k}Pr(v_i|S)\!=\!0$.

Based on the analyses above, we can compute $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)$ by dividing it into following four possible cases. Note that, $X\!\subseteq \!Y$ and $X(v_i)\!\subseteq \!Y(v_i)$.

• 1. $v_i$ isn’t covered by any RAPs in Y, but is covered by $r_k$. Hence, $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)=$ $p^{i}_{k}\!-\!p^{i}_{k}\!=\!0$;

• 2. $v_i$ isn’t covered by any RAPs in X, but $r_k$ and RAPs in $Y\backslash \!X$ cover $v_i$. Hence, $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)=p^{i}_{k}\!-\!p^{i}_{k}*\prod _{r_j \in Y(v_i)}(1-p_{j}^{i})\!\ge \!0$;

• 3. $r_k$ and RAPs in X and Y cover $v_i$. Hence, $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)=p^{i}_{k}* [ \prod _{r_j \in X(v_i)}(1-p_{j}^{i})-\prod _{r_j \in Y(v_i)}(1-p_{j}^{i}) ]$ $\ge 0$ since $X(v_i)\!\subseteq \!Y(v_i)$;

• 4. $v_i$ isn’t covered by $r_k$. Hence, $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)=0\!-\!0\!=\!0$.

In summary, we have $\varDelta _{r_k}Pr(v_i|X) \!-\!\varDelta _{r_k}Pr(v_i|Y)\!\ge \!0$. Finally, we also can conclude that $\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)\!\ge \!0$ in all six cases. Moreover, as $\theta \!>\!0$, $\varDelta _{r_k}f(X)\!-\!\varDelta _{r_k}f(Y)\!=\!\theta *\sum _{v_i\in V}[\varDelta _{r_k}f_i(X)\!-\!\varDelta _{r_k}f_i(Y)]\!\ge \!0$. Therefore, we have the conclusion that $f(\cdot )$ is a submodular function.

According to Lemmas 1 and 2, we know that, $f(\cdot )$ is polymatroid since it’s an increasing submodular function with $f(\emptyset )\!=\!0$. Similarly, we can easily prove that the cardinality function $c(X)\!=\!|X|$ is polymatroid. Given two polymatroid functions $g(\cdot )$ and $h(\cdot )$ on $2^{E}$, the problem of Minimum Submodular Cover with Submodular Cost (MSC/SC) is defined, which is the minimization problem $min\{h(X)| g(X)\!=\!g(E), X\!\subseteq \!E\}$[15].

In this paper, for the utility function $f(\cdot )$, given a selected RAP set S, if $f(S)\!=\!\theta \!*\!\tau \!*\!|V|$, S is a feasible solution of the RAP selection problem and $f(S)\!\!=\!\!f(V)\!=\!\theta \!*\!\tau \!*\!|V|$. Therefore, the RAP selection problem can be described as: $min\{c(S)| f(S)\!=\!f(V), S\!\subseteq \!V\}$, where $c(\cdot )$ is the cardinality function. That is to say, our selection problem is a MSC/SC problem. Based on this fact, we give the following Theorem 3.

Theorem 3

[15] Suppose $g(\cdot )$ is a polymatroid function on $2^{E}$, and $g(E)\!\ge \!opt$ where opt is the cost of a minimum submodular cover. For a greedy algorithm, if the selected x in each round always satisfies that $\frac{g(X\!+\!\{x\})\!-\!g(X)}{c(\{x\})}\!\ge \!1$, then the greedy solution is a $1\!+\!\rho ln(\frac{g(E)}{opt})$-approximation, where $\rho \!=\!1$ if $c(\cdot )$ is modular (i.e., linear).

Now, we give the following crucial Lemma 3.

Lemma 3

Give the set $S\!\subseteq \!R$, which is the set of RAPs selected by Algorithm 1 after r-th round, and the RAP $r_k$ selected during $(r\!+\!1)$-th round, we have $\frac{f(S\!+\!\{r_k\})\!-\!f(S)}{c(\{r_k\})}\!\ge \!1$.

Proof

  1. It is obvious that the cardinality function $c(\cdot )$ is linear and $c(\{r_k\})\!=\!1$.

2. In each round of Algorithm 1, if $f(S)\!<\!\theta \!*\!\tau \!*\!|V|$, a new RAP will be selected and added into the set S. In fact, if Algorithm 1 doesn’t terminate after r-th round, there must exist a vehicle $v_i$ with $Pr(v_i|S)\!<\!\tau$. Otherwise, if $Pr(v_i|S)\!\ge \!\tau$ for $\forall v_i\!\in \!V$, then $f(S)\!=\!\theta \!*\!\tau \!*\!|V|$, consequently, Algorithm 1 will terminate. Moreover, for this vehicle $v_i$ and the selected RAP $r_k$, since $f(S\!+\!\{r_k\})$ is maximized during $(r\!+\!1)$-th round, we can suppose $r_k$ must cover this $v_i$, i.e., $p_{k}^{i}\!>\!0$, without loss of generality. That is to say, if the RAP $r_k$ is selected during $(r\!+\!1)$-th round, $r_k$ must cover at least one vehicle $v_i$ with $Pr(v_i|S)\!<\!\tau$. Otherwise, we have $f(S\!+\!\{r_k\})\!-\!f(S)\!=\!0$ for $\forall r_k\!\in \!R\!\backslash \!S$, which conflicts with the fact that there must exist a vehicle $v_i$ with $Pr(v_i|S)\!<\!\tau$ if Algorithm 1 doesn’t terminate. Based on these analyses, we have

$f(S\!+\!\{r_k\})\!-\!f(S)$ $= \theta \!*\!\sum _{v_i\!\in \!V}[min\{Pr(v_i|(S\!+\!\{r_k\})), \tau \}\!-\!min\{Pr(v_i|S), \tau \}]$

$\ge \!\theta *(min\{Pr(v_i|(S\!+\!\{r_k\})), \tau \}-min\{Pr(v_i|S), \tau \})$ $=\!\theta *(min\{Pr(v_i|(S\!+\!\{r_k\})), \tau \}-Pr(v_i|S))$ $=\!\theta *min\{Pr(v_i|(S\!+\!\{r_k\}))-Pr(v_i|S),\tau -Pr(v_i|S)\}$ $=\!\theta *min\{p_{k}^{i}*\prod _{r_j \in S(v_i)}(1-p_{j}^{i}),\tau -Pr(v_i|S)\}$ $\ge \!min\{\theta \!*\!\theta _{1}, \theta \!*\!(\tau -\theta _{2})\}\!\ge \!1$

To sum up, $\frac{f(S\!+\!\{r_k\})\!-\!f(S)}{c(\{r_k\})}\!\ge \!f(S\!+\!\{r_k\})\!-\!f(S)\!\ge \!1$

Finally, suppose that ${S}_{opt}$ is an optimal solution of the RAP selection problem and $c({S}_{opt})\!=\!|{S}_{opt}|\!=\!opt$ which is number of RAPs in ${S}_{opt}$. Consequently, we have $f(R)\!=\!\theta \!*\!\tau \!*\!|V|\!\ge \!\frac{\tau \!*\!|V|}{\theta _3}\!\ge \!|R|\!\ge \!opt$ since $\theta \!\ge \!\frac{1}{\theta _3}$. Therefore, we can conclude that our proposed Algorithm 1 is a $1+$ln$(\frac{\theta \!*\!\tau \!*\!|V|}{opt})$-approximation, by applying Theorem 3.