Vehicular delay tolerant network routing algorithm based on trajectory clustering and dynamic Bayesian network

Abstract

Typically, delay tolerant network (DTN) suffers from frequent disruption, high latency, and lack of stable connections between nodes. As a special case of DTN, vehicular delay tolerant network (VDTN) has particular spatial-temporal characteristics. Different kinds of vehicles may have different movement ranges and movement patterns and the movements of vehicles exhibit significant dynamics from the temporal view. The movement patterns and dynamic characteristics of vehicles are difficult to be described accurately. To this end, a novel framework of VDTN routing algorithm based on trajectory clustering and dynamic Bayesian network (DBN) is proposed, which can capture the spatial-temporal characteristics and the movement patterns of vehicles in the real VDTN scenarios. Firstly, a K-means trajectory clustering (KTC) algorithm is adopted to cluster the trajectories of vehicles according to their spatial characteristics. Then, a KTC-based DBN structure learning algorithm is proposed to construct the prior network and transition network of DBN by an extended K2$^+$ algorithm to capture the temporal characteristics of VDTN, and multiple DBN models are established for different trajectory clusters to further improve the prediction accuracy. Finally, a VDTN routing algorithm is presented to forward message by the inference of DBN models. Simulation results show that the proposed VDTN routing algorithm has a higher delivery ratio as well as a lower overhead compared with other related routing algorithms, and the effectiveness of the trajectory clustering method and DBN models are verified.

Appendix A Proof

According to DBN Model, when $t = 0$, its conditional probability $f_0 (X_{1:6}^0 )$ is a function of the attribute variables $X_{1:6}^0$ described by $B_0$. Let

$$\begin{aligned} f_0 (X_{1:6}^0) = \prod \limits _{i=1}^6 P(X_i^0 |Pa(X_i^0)) \end{aligned}$$

(21)

where $X_{1:6}^0$ denotes the set of variables $\{X_1^0,\ldots ,X_5^0\}$, $Pa(X_i^0)$ represents the parent attribute variable of $X_i^0$, and $Pa(X_i^0) \in \{X_{1:i-1}^0\}$.

When $t > 0$, its conditional probability $f_\rightarrow (X_{1:6}^{t-1:t})$ related to time slice t is a function of the attribute variables $X_{1:6}^{t-1}$ and $X_{1:6}^t$ described by $B_\rightarrow$. Let

$$\begin{aligned} f_\rightarrow (X_{1:6}^{t-1:t}) = \prod \limits _{i=1}^6 P(X_i^t |Pa(X_i^t)) \end{aligned}$$

(22)

where $X_{1:6}^{t-1:t}$ denotes the set of variables $\{X_1^{t-1},\ldots ,X_6^{t-1},X_1^t,\ldots ,X_6^t\}$, $Pa(X_i^t)$ represents the parents attribute variable of $X_i^t$ and $Pa(X_i^t) \in \{X_{1:6}^{t-1},X_{1:i-1}^t\}$.

So, the joint probability $P(X_{1:6}^{0:t})$ of DBN model can be described as

$$\begin{aligned} P(X_{1:6}^{0:t}) = f_0 (X_{1:6}^0) f_\rightarrow (X_{1:6}^{0:1}) \ldots f_\rightarrow (X_{1:6}^{t-1:t}) \end{aligned}$$

(23)

and $P(X_6^t |X_{1:5}^{0:t})$ can be described as

$$\begin{aligned} \begin{aligned}&P(X_6^t |X_{1:5}^{0:t}) \propto P(X_6^t,X_{1:5}^{0:t}) = \sum _{X_6^{0:t - 1}}P(X_{1:6}^{0:t}) \\&\qquad = \sum _{X_6^{t - 1} \ldots X_6^0}f_0(X_{1:6}^0)f_{\rightarrow }(X_{1:6}^{0:1}) \ldots f_{\rightarrow }(X_{1:6}^{t - 1:t}) \end{aligned} \end{aligned}$$

(24)

According to Eq. (24), We can find that only two adjacent terms are related to the sum of $X_6^i,i \in [0,t-1]$. Therefore, we can gradually sum and calculate $X_6^i$ from $i=0$ to $t-1$. Let

$$\begin{aligned} g_0(X_6^1,X_{1:5}^{0:1}) = \sum _{X_6^0}f_0(X_{1:6}^0) f_\rightarrow (X_{1:6}^{0:1}) \end{aligned}$$

(25)

Because $X_{1:5}^{0:t}$ are observable variables, and $X_6^1$ cannot be the parent node of variables in time slice 0 or 1. Therefore, after summing all possible values of $X_6^0$ in Eq. (25), $g_0 (X_6^1,X_{1:5}^{0:1})$ will be the function of $X_6^1$,$X_{1:5}^{0:1}$. It contains up to $k_6$ (number of delivery levels) pending terms about $X_6^1$. The time complexity of $g_0(X_6^1,X_{1:5}^{0:1})$ is $O(q^2 k_6)$, q is the total number of random variables and $q = 6$ in this paper. Similarly, we can get

$$\begin{aligned} g_1(X_6^2,X_{1:5}^{0:2})= & {} \sum _{X_6^1}g_0 (X_6^1,X_{1:5}^{0:1})f_\rightarrow (X_{1:6}^{1:2}) \end{aligned}$$

(26)

$$\begin{aligned}{} & {} \ldots \nonumber \\ g_{t-1} (X_6^t,X_{1:5}^{0:t})= & {} \sum _{X_6^{t-1}} g_{t-2} (X_6^{t-1},X_{1:5}^{0:t-1})f_\rightarrow (X_{1:6}^{t-1:t}) \end{aligned}$$

(27)

Each of the above equation contains up to $k_6$ pending terms about $X_6^i$ and the time complexity is $O(q^2k_6)$ too.

According to Eqs. (24) and (27), we can obtain

$$\begin{aligned} P(X_6^t |X_{1:5}^{0:t}) \propto g_{t-1} (X_6^t,X_{1:5}^{0:t}) \end{aligned}$$

(28)

The calculation of $P(X_6^t |X_{1:5}^{0:t})$ needs to start from $g_0$ until $g_{t-1}$. Thus, the time complexity of calculating $P(X_6^t |X_{1:5}^{0:t})$ is $O(q^2k_6t)$. And if the intermediate calculation results are saved in each step, the actual time complexity of inference in this paper is only $O(q^2k_6)$.