Normally, the research objects of our paper are deployed on the highway. The existence of the Markov property in the scenario of highway has been validated in [
23‐
25]. Since the beaconing results are directly related to the presences of the passing-by vehicles, we can model the traditional beaconing procedure as a continuous-time Markov process. We define the state space as
X={
x
1,
x
2}, with
x
1=0 and
x
2=1 indicating that the beaconing results are 0 and 1. The state transition probability matrix is defined as
\(P(t)=\left (\begin {array}{cc} P_{00}(t) & ~P_{01}(t) \\ P_{10}(t) & ~P_{11}(t) \end {array}\right)\), where
P
ij
(
t)=
P(
R
s+t
=
j|
R
s
=
i),
i,
j∈{0,1}.
4.1 The state transition probabilities when the beaconing period is t0
As shown in Fig.
3, we define
M as the number of the beaconing periods in the modeling phase. We define
R={
R
1,
R
2,⋯,
R
M
} with
R
m
∈
X for ∀
m∈{1,⋯,
M} as the beaconing results. After the
M beaconing periods, the roadside infrastructure can get the beaconing result samples denoted as
r
1,
r
2,⋯,
r
M
. Based on these samples, we can estimate the state transition probabilities of the Markov model. The state transition probability matrix is denoted as
\(P({t_{0}})=\left (\begin {array}{cc} P_{00}({t_{0}}) & P_{01}({t_{0}}) \\ P_{10}({t_{0}}) & P_{11}({t_{0}}) \end {array}\right)\), where
\(P_{ij}({t_{0}})=P(R_{s+{t_{0}}}=j|R_{s}=i), i,j\in \{0,1\}\phantom {\dot {i}\!}\),
s={1
t
0,2
t
0,⋯,(
M−1)
t
0}. Since it can be understood that the discrete-time Markov chain is the discretization of the continuous-time Markov chain, when the time slot is very small, they are identical in essence. According to this property, we use the maximum likelihood estimation method that is used in the discrete-time Markov chain to estimate transition probability. The maximum likelihood estimation is consistent and asymptotically unbiased. The counts of the occurrence of four different transition types such as (
r
k
,
r
k+1)=(0,0),(0,1),(1,0),(1,1) are defined as
n
00,
n
01,
n
10, and
n
11,
k={1,2,⋯,
M−1}. The likelihood function is given by
$${} \begin{aligned} L(P({t_{0}}))&=P(\mathbf{R};P({t_{0}}))\\ &= P\left(\prod_{k=1}^{M-1}P(R_{(k+1)t_{0}}=r_{k+1}\mid R_{kt_{0}}=r_{k};A\right)\\ &= P\left(R_{1}=r_{1};P({t_{0}}))\cdot({P_{00}({t_{0}})}^{n_{00}}\cdot {P_{01}({t_{0}})}^{n_{01}}\right.\\ &\left.\qquad\cdot {P_{10}({t_{0}})}^{n_{10}}\cdot {P_{11}({t_{0}})}^{n_{11}}\right). \end{aligned} $$
(1)
By solving
$$\begin{array}{@{}rcl@{}} \left\{\begin{array}{ll} \partial L(P({t_{0}}))/\partial P_{00}({t_{0}}) & = 0 \\ \partial L(P({t_{0}}))/\partial P_{01}({t_{0}}) & = 0 \\ \partial L(P({t_{0}}))/\partial P_{10}({t_{0}}) & = 0 \\ \partial L(P({t_{0}}))/\partial P_{11}({t_{0}}) & = 0, \end{array} \right. \end{array} $$
(2)
the maximum likelihood estimators
\({\hat {P_{00}}({t_{0}})}\),
\({\hat {P_{01}}({t_{0}})}\),
\({\hat {P_{10}}({t_{0}})}\), and
\({\hat {P_{11}}({t_{0}})}\) can be obtained as follows:
$$\begin{array}{@{}rcl@{}} \left\{\begin{array}{lll} {\hat{P_{00}}({t_{0}})} & = n_{00}/(n_{00}+n_{01})\\ {\hat{P_{01}}({t_{0}})} & = n_{01}/(n_{00}+n_{01})\\ {\hat{P_{10}}({t_{0}})} & = n_{10}/(n_{10}+n_{11})\\ {\hat{P_{11}}({t_{0}})} & = n_{11}/(n_{10}+n_{11}). \end{array}\right. \end{array} $$
(3)
Then, the estimation process of the state transition probabilities has been done.
4.2 The duration of the modeling phase
As described above, if we have the samples, we can make the estimation. It is obvious that the more samples we get, the more accurate the estimation is. An analysis of how many samples are needed to satisfy the desired accuracy is presented in this subsection.
The central limit theorem is utilized in our analysis. Since
\(\hat {P_{00}({t_{0}})}=1-\hat {P_{01}({t_{0}})}\),
\(\hat {P_{11}({t_{0}})}=1-\hat {P_{10}({t_{0}})}\), only
\(\hat {P_{01}({t_{0}})}\) and
\(\hat {{{P_{10}}({t_{0}})}}\) will be discussed in this subsection. We discuss
\({\hat {P_{01}}({t_{0}})}\) first. Assuming among
M samples there are
M
0 ones whose values are 0, then we have
$$\begin{array}{@{}rcl@{}} M_{0}=M\cdot\left(1-\frac{{P_{01}({t_{0}})}}{{P_{01}({t_{0}})}+{P_{10}({t_{0}})}}\right)\ as\ M\rightarrow\infty, \end{array} $$
(4)
where
\(\frac {{P_{01}({t_{0}})}}{{P_{01}({t_{0}})}+{P_{10}({t_{0}})}}\) represents the probability that the beaconing result is 1. We define
$$\begin{array}{@{}rcl@{}} C_{k}= \left\{\begin{array}{lll} &0 \qquad r_{k}=0,r_{k+1}=0\\ &1 \qquad r_{k}=0,r_{k+1}=1 \end{array},\right. (k=1,\cdots,M_{0}). \end{array} $$
(5)
{
C
k
} are independent and identically distributed variables. The mean value of {
C
k
} is
P
01(
t
0), and the variance is
P
01(
t
0)−
P
01(
t
0)
2. From (
3),
\({\hat {P_{01}}({t_{0}})}\) can be expressed as
$$\begin{array}{@{}rcl@{}} {\hat{P_{01}}({t_{0}})} & = n_{01}/(n_{00}+n_{01})=\frac{1}{M_{0}}\sum\limits_{k=1}^{M_{0}} C_{k}. \end{array} $$
(6)
We demand that the relative error
\(|{\hat {P_{01}}({t_{0}})}-P_{01}({t_{0}})|\) of an estimator should be below a relative estimation error
θ with the corresponding confidence probability
α. In mathematical terms, we request
$$\begin{array}{@{}rcl@{}} P\left(\left|\frac{{\hat{P_{01}}({t_{0}})}-P_{01}({t_{0}})}{P_{01}({t_{0}})}\right|<\theta\right)\geq\alpha, \end{array} $$
(7)
where
θ and
α are inputs to describe the desired accuracy in the estimation. If (
7) is fulfilled, the estimation achieves the desired accuracy. Combining (
4) and (
7), since
P
01(
t
0)≥0, we have following equations:
$$ \begin{aligned} P&\left(\left|\frac{{\hat{P_{01}}({t_{0}})}-P_{01}({t_{0}})}{P_{01}({t_{0}})}\right|<\theta\right)\\ =&P\left(\left|\frac{1}{M_{0}}\sum\limits_{k=1}^{M_{0}}C_{k}-P_{01}({t_{0}})\right|<\theta{P_{01}({t_{0}})}\right)\\ =&P\left(\left|\frac{\sum\limits_{k=1}^{M_{0}}C_{k}-M_{0} P_{01}({t_{0}})}{\sqrt{M_{0}}\cdot\sqrt{P_{01}({t_{0}})-{P_{01}({t_{0}})}^{2}}}\cdot\frac{\sqrt{P_{01}({t_{0}})-{P_{01}({t_{0}})}^{2}}}{\sqrt{M_{0}}}\right|<\theta{P_{01}({t_{0}})}\!\!\right)\\ =&P\left(\left|\frac{\sum\limits_{k=1}^{M_{0}}C_{k}-M_{0} P_{01}({t_{0}})}{\sqrt{M_{0}}\cdot\sqrt{P_{01}({t_{0}})-{P_{01}({t_{0}})}^{2}}}\right|<\theta\frac{\sqrt{M_{0}}}{{\sqrt{\frac{1}{P_{01}({t_{0}})}-1}}}\right)\\ =&P\left(\left|\frac{\sum\limits_{k=1}^{M_{0}}C_{k}-M_{0} P_{01}({t_{0}})}{\sqrt{M_{0}}\cdot\sqrt{P_{01}({t_{0}})-{P_{01}({t_{0}})}^{2}}}\right|<\theta\frac{\sqrt{M\cdot\left(1-\frac{{P_{01}({t_{0}})}}{{P_{01}({t_{0}})}+{P_{10}({t_{0}})}}\right)}}{{\sqrt{\frac{1}{P_{01}({t_{0}}\!)}-1}}}\right). \end{aligned} $$
(8)
Based on the central limit theorem, we have
$$\begin{array}{@{}rcl@{}} {\frac{\sum\limits_{k=1}^{M_{0}}C_{k}-M_{0} P_{01}({t_{0}})}{\sqrt{M_{0}}\cdot\sqrt{P_{01}({t_{0}})-{P_{01}({t_{0}})}^{2}}}}\rightarrow N(0,1)\ as\ M_{0}\rightarrow\infty. \end{array} $$
(9)
We denote
Φ(∙) as the standard normal cumulative distribution function. Then, we have
$$\begin{array}{@{}rcl@{}} 2\Phi\left(\theta\frac{\sqrt{M\cdot\left(1-\frac{{P_{01}({t_{0}})}}{{P_{01}({t_{0}})}+{P_{10}({t_{0}})}}\right)}}{\sqrt{\frac{1}{P_{01}({t_{0}})}-1}}\right)-1\geq\alpha. \end{array} $$
(10)
In (
10),
M>0, 0≤
P
01(
t
0)≤1 and 0≤
P
10(
t
0)≤1. We denote
\(M_{\text {min}}^{a}\) as the minimum required number of the beaconing result samples for
\({\hat {P_{01}}({t_{0}})}\). For any state transition probability values,
M
a
can be expressed as follows:
$$\begin{array}{@{}rcl@{}}{} M_{\text{min}}^{a}=\frac{\left(\Phi^{-1}(\frac{1+\alpha}{2})\right)^{2}}{\theta^{2}}(1-P_{01}({t_{0}}))\left({\frac{1}{P_{01}({t_{0}})}}+{\frac{1}{P_{10}({t_{0}})}}\right). \end{array} $$
(11)
For \({\hat {P_{10}}({t_{0}})}\), the analysis process is the same. The minimum required number of the beaconing result samples for \({\hat {P_{01}}({t_{0}})}\) is denoted as \(M_{\text {min}}^{b}\). We have
$$\begin{array}{@{}rcl@{}}{} M_{\text{min}}^{b}=\frac{\left(\Phi^{-1}\left(\frac{1+\alpha}{2}\right)\right)^{2}}{\theta^{2}}(1-P_{10}({t_{0}}))\left({\frac{1}{P_{01}({t_{0}})}}+{\frac{1}{P_{10}({t_{0}})}}\right). \end{array} $$
(12)
Therefore, when the desired accuracy
α and
θ are given, we have the following equations:
$$\begin{array}{@{}rcl@{}} M_{\text{min}}=\text{max}\left\{M_{\text{min}}^{a},M_{\text{min}}^{b}\right\}, \end{array} $$
(13)
where M
min is denoted as the minimum required number of the beaconing result samples for the estimation.
Actually, since we do not know the values of
P
01(
t
0) and
P
10(
t
0) before we start the modeling phase, we cannot calculate the required number of the samples. To tackle this problem, we have the following processes. At the beginning of the modeling phase, we send
M
default beacon packets firstly. The value of
M
default is analyzed in the later section. According to the
M
default beaconing result samples, we can calculate the
P
01(
t
0) and
P
10(
t
0). Then, we can get the
M
min. If
M
min≤
M
default, it means the estimation of the state transition probabilities reaches the desired accuracy, and we can stop the modeling phase. If not, the periodical beaconing continues. We denote
M
actual as the actual number of the beacon packets. After each of the continuous beaconing processes, we recalculate the
M
min and compare it with the
M
actual. Until
M
actual≥
M
min, we stop the periodical beaconing and start the applying phase. The total beaconing times of the modeling phase
M
total is equal to the current
M
actual. The total duration of the modeling phase is expressed as
$$\begin{array}{@{}rcl@{}} T_{\text{modeling}}=t_{0}M_{\text{total}}. \end{array} $$
(14)
Through the above analyses, we can summarize the following proposition: