A robust distance-based relay selection for message dissemination in vehicular network

Abstract

The relay-node selection plays a decisive impact on the message dissemination in vehicular network. However, in some scenarios, due to lack of the reliable and stable relay-node selection, the message dissemination suffers from an intolerable delay, even a failure. In this paper, we focus on a design of the robust relay selection, which aims at (1) achieving a maximum message dissemination speed in general scenarios, and (2) assuring an acceptable dissemination speed in the adverse scenario. Two adverse scenarios are first introduced for the message dissemination when the distance-based relay selection is applied in multi-hop broadcast. To tackle the challenge, we propose a robust distance-based relay selection by optimizing the exponent-based partitioning broadcast protocol (our previous work) and incorporating a proposed mini-black-burst-assisted mechanism. Moreover, we develop analytic models for the robust approach performances in terms of contention latency and packet delivery ratio (PDR). Simulations are used to verify these analytic models, demonstrate the acceptable performances of the proposal in adverse scenarios, and compare it with the state-of-the-art approaches in general scenarios. Results show an increase of more than 11.01% in terms of message dissemination speed independent of vehicle density and a stable PDR of more than 99.99%.

Appendix: Optimization of EPBP

This section provides the optimization of EPBP with A. As mentioned in Sect. 4.1, we determine the optimal value ${A_{{\mathrm{opt}}}}$ of A to maximize the average message dissemination speed ${v_{{\mathrm{ave}}}}$ when $N_{{\mathrm{iter}}}=1$. ${A_{{\mathrm{opt}}}}$ can be derived in the following equation,

$$\begin{aligned} \sum \limits _{i = 1}^{{N_\lambda }} {\left( {\frac{1}{{{T_{{\mathrm{d,}}i}}}}\frac{{{\mathrm{d}} {\beta _i}}}{{{\mathrm{d}} {A_{{\mathrm{opt}}}}}} - \frac{{{\beta _i}}}{{T_{{\mathrm{d,}}i}^2}}\frac{{{\mathrm{d}} {T_{{\mathrm{d,}}i}}}}{{{\mathrm{d}} {A_{{\mathrm{opt}}}}}}} \right) } = 0. \end{aligned}$$

(26)

The index i of the i-th vehicle density ${\lambda _i}$ for all metrics is omitted for simple presentation.

1.1 Message progress $\beta$

$\beta$ can be attained as

$$\begin{aligned} \beta = \sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {{M_k}{P_{{\mathrm{seg\_sel}}}}\left( k \right) } \end{aligned}$$

(27)

where

$$\begin{aligned} {M_k} = \sum \limits _{n = k + 1}^{{N_{{\mathrm{part}}}}} {{W_{{\mathrm{seg}}}}(n)} + {W_{{\mathrm{seg}}}}(k)/2 \end{aligned}$$

(28)

is the average message progress if the k-th segment is the final segment, and

$$\begin{aligned} {P_{{\mathrm{seg\_sel}}}}\left( k \right) = {e^{ - {\mu _{{\mathrm{seg\_bro}}}}(k)}}(1 - {e^{ - {\mu _k}}}) \end{aligned}$$

(29)

is the probability of the selection of the k-th segment. ${\mu _{{\mathrm{seg\_bro}}}}(k)$ in (29) is the vehicle number in other segments in the message broadcasting direction of the k-th segment, which can be achieved as

$$\begin{aligned} {\mu _{{\mathrm{seg\_bro}}}}(k)= \lambda R\sum \limits _{n =1}^{k-1} {{W_{{\mathrm{seg}}}}({N_{{\mathrm{iter}}}},n)} \end{aligned}$$

(30)

1.2 One-hop delay ${T_{\mathrm{d}}}$

${T_{\mathrm{d}}}$ can be attained as

$$\begin{aligned} {T_{\mathrm{d}}} = {T_{{\mathrm{init}}}} + {T_{{\mathrm{part}}}} + {T_{{\mathrm{cont}}}} + {T_{{\mathrm{data}}}} \end{aligned}$$

(31)

where ${T_{{\mathrm{init}}}}$, ${T_{{\mathrm{part}}}}$, ${T_{{\mathrm{cont}}}}$ and ${T_{{\mathrm{data}}}}$ are the initial latency, the partitioning latency, the contention latency and the data transmission latency, respectively. Among the four latencies, the selection of A can affect the value of ${T_{{\mathrm{part}}}}$ and ${T_{{\mathrm{cont}}}}$. They can be attained as

$$\begin{aligned}&{T_{{\mathrm{part}}}} = {T_{{\mathrm{slot}}}}\left( {1 + \sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {{N_{{\mathrm{P}\_\mathrm{slot}}}}\left( k \right) {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) } } \right) \end{aligned}$$

(32)

$$\begin{aligned}&{T_{{\mathrm{cont}}}} = \sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {{T_{{\mathrm{cont}\_\mathrm{seg}}}}\left( k \right) {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) } \end{aligned}$$

(33)

where

$$ N_{{{\text{P}}\_{\text{slot}}}} \left( k \right) = \left\{ {\begin{array}{*{20}l} {k\,\bmod \,N_{{{\text{part}}}} \left( {k\,\bmod \,N_{{{\text{part}}}} } \right) < N_{{{\text{part}}}} } \hfill \\ {N_{{{\text{part}}}} - 1\left( {k\,\bmod \,N_{{{\text{part}}}} } \right) = N_{{{\text{part}}}} } \hfill \\ \end{array} } \right. $$

(34)

is the number of time slots spent when the k-th segment is selected. And

$$\begin{aligned} {T_{{\mathrm{cont}\_\mathrm{seg}}}}\left( k \right) = \left( {\frac{{1 - {e^{ - {\mu _k}p}}}}{{{\mu _k}p{e^{ - {\mu _k}p}}}} - 1} \right) {T_{{\mathrm{col}}}} + {T_{{\mathrm{suc}}}} \end{aligned}$$

(35)

is the average contention latency when the k-th segment is the final segment. It is noted that ${T_{{\mathrm{cont}\_\mathrm{seg}}}}$ in (35) differs from that in (16) since the exponential back-off timer isn’t adopted in the optimization here.

1.3 Expression of $\frac{{{\mathrm{d}} \beta }}{{{\mathrm{d}} A}}$

From (27), $\frac{{{\mathrm{d}} \beta }}{{{\mathrm{d}} A}}$ in (26) can be achieved as

$$\begin{aligned} \frac{{{\mathrm{d}} \beta }}{{{\mathrm{d}} A}} = \sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {\left[ {{M_k}\frac{{{\mathrm{d}} {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) }}{{{\mathrm{d}} A}} + {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) \frac{{{\mathrm{d}} {M_k}}}{{{\mathrm{d}} A}}} \right] } \end{aligned}$$

(36)

where

$$\begin{aligned} \frac{{{\mathrm{d}} {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) }}{{{\mathrm{d}} A}}&= {} {e^{ - {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k)}}{e^{ - {\mu _k}}}\frac{{{\mathrm{d}} {\mu _k}}}{{{\mathrm{d}} A}} \nonumber \\&- {e^{ - {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k)}}(1 - {e^{ - {\mu _k}}})\frac{{{\mathrm{d}} {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k)}}{{{\mathrm{d}} A}} \end{aligned}$$

(37)

$$\begin{aligned} \frac{{{\mathrm{d}} {M_k}}}{{{\mathrm{d}} A}}= & {} \frac{1}{{2{A^2}}}{\left( \left( {{{(1 + A)}^{\frac{{k - 1}}{{{N_{{\mathrm{part}}}}}}}} + {{(1 + A)}^{\frac{k}{{{N_{{\mathrm{part}}}}}}}}}\right) - \frac{A}{{(1 + A)}}\right. } \nonumber \\&\left( {\left. {\frac{{k - 1}}{{{N_{{\mathrm{part}}}}}}{{(1 + A)}^{\frac{{k - 1}}{{{N_{{\mathrm{part}}}}}}}} + \frac{k}{{{N_{{\mathrm{part}}}}}}{{(1 + A)}^{\frac{k}{{{N_{{\mathrm{part}}}}}}}}} \right) } \right) - \frac{1}{{{A^2}}} \end{aligned}$$

(38)

In (37),

$$\begin{aligned} \frac{{{\mathrm{d}} {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k)}}{{{\mathrm{d}} A}} = \frac{{\lambda R}}{{{A^2}}}\left( {\frac{{k - 1}}{{{N_{{\mathrm{part}}}}}}A{{(1 + A)}^{\frac{{k - 1 - {N_{{\mathrm{part}}}}}}{{{N_{{\mathrm{part}}}}}}}} - {{(1 + A)}^{\frac{{k - 1}}{{{N_{{\mathrm{part}}}}}}}} + 1} \right) , \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{{{\mathrm{d}} {\mu _k}}}{{{\mathrm{d}} A}} = \frac{{{\mathrm{d}} {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k + 1)}}{{{\mathrm{d}} A}} - \frac{{{\mathrm{d}} {\mu _{{\mathrm{seg}\_\mathrm{bro}}}}(k)}}{{{\mathrm{d}} A}}. \end{aligned}$$

(40)

1.4 Expression of $\frac{{{\mathrm{d}} {T_{\mathrm{d}}}}}{{{\mathrm{d}} A}}$

From (31), $\frac{{{\mathrm{d}} {T_{\mathrm{d}}}}}{{{\mathrm{d}} A}}$ in (26) can be achieved as

$$\begin{aligned} \frac{{{\mathrm{d}} {T_{\mathrm{d}}}}}{{{\mathrm{d}} A}} = \frac{{{\mathrm{d}} {T_{{\mathrm{part}}}}}}{{{\mathrm{d}} A}} + \frac{{{\mathrm{d}} {T_{{\mathrm{cont}}}}}}{{{\mathrm{d}} A}} \end{aligned}$$

(41)

where

$$\begin{aligned} \frac{{{\mathrm{d}} {T_{{\mathrm{part}}}}}}{{{\mathrm{d}} A}}= & {} {T_{{\mathrm{slot}}}}\sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {{N_{{\mathrm{P}\_\mathrm{slot}}}}\left( k \right) \frac{{{\mathrm{d}} {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) }}{{{\mathrm{d}} A}}}\end{aligned}$$

(42)

$$\begin{aligned} \frac{{{\mathrm{d}} {T_{{\mathrm{cont}}}}}}{{{\mathrm{d}} A}}= & {} \sum \limits _{k = 1}^{{N_{{\mathrm{part}}}}} {\left( {{T_{{\mathrm{cont}\_\mathrm{seg}}}}\left( k \right) \frac{{{\mathrm{d}} {P_{{\mathrm{seg\_sel}}}}\left( k \right) }}{{{\mathrm{d}} A}} }\right. }\nonumber \\&+{\left. { {P_{{\mathrm{seg}\_\mathrm{sel}}}}\left( k \right) {T_{{\mathrm{col}}}}\frac{{\mathrm{d}} }{{{\mathrm{d}} A}}\left( {\frac{{1 - {e^{ - {\mu _k}p}}}}{{{\mu _k}p{e^{ - {\mu _k}p}}}}} \right) } \right) }. \end{aligned}$$

(43)

In (43),

$$\begin{aligned} \frac{{\mathrm{d}} }{{{\mathrm{d}} A}}\left( {\frac{{1 - {e^{ - {\mu _k}p}}}}{{{\mu _k}p{e^{ - {\mu _k}p}}}}} \right) = \frac{{{\mu _k}p + {e^{ - {\mu _k}p}} - 1}}{{\mu _k^2p{e^{ - {\mu _k}p}}}}\frac{{{\mathrm{d}} {\mu _k}}}{{{\mathrm{d}} A}}. \end{aligned}$$

(44)