Module 3 : Microscopic Traffic Flow Modeling

Lecture 15 : Lane Changing Models

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	Gap acceptance A gap is defined as the gap in between the lead and lag vehicles in the target lane (see Figure 1). For merging into an adjacent lane, a gap is acceptable only when both lead and lag gap are acceptable. Drivers are assumed to have minimum acceptable lead and lag gap lengths which are termed as the lead and lag critical gaps respectively. These critical gaps vary not only among different individuals, but also for a given individual under different traffic conditions. Most models also make a distinction between the lead gap and the lag gap and require that both are acceptable. The lead gap is the gap between the subject vehicle and the vehicle ahead of it in the lane it is changing to. The lag gap is defined in the same way relative to the vehicle behind in that lane. The critical gap lengths are assumed to be log normally distributed The critical gap for driver n at time t is assumed to have the following relation. $$G_{n}^{g,cr} = e^{\left[X_n^{g}\beta + \alpha V_n + \varepsilon_n\right]}$$ where, $G_n^{g,cr}(t)$ is the critical gap measure for gap G perceived by driver n at time step t, $X_n^{g}(t)$ is the explanatory variable used to characterize mean $G_n^{g,cr}(t)$, $\varepsilon n$ is the random term follows log normal distribution, and $\alpha_{g}$ is the parameter of driver specific random term $v_n$. Figure 1: Definition of gaps Assuming $\varepsilon_n^g(t) \approx N(0,\sigma_{\varepsilon^{g}}^2)$ the conditional probability of acceptance of a gap p(g) considering that the probability of lead gap acceptance (p(lead)) and lag gap (p(lag)) acceptance as two independent events is probability that the lead and lag gaps are accepted. That is: $$P(g) = p(lead)\times p(lag)$$ $$= p(G_{tn}^{lead} \ge \log(G_{cr,tn}^{lead}) ~\mathrm{and}~ G_{tn}^{tag} \ge \log(G_{cr,tn}^{tag}))$$ $$= p(\log(G_{tn}^{lead}) \ge \log(G_{cr,tn}^{lead})) \times P(\log(G_{tn}^{tag}) \ge \log(G_{cr,tn}^{tag}))$$ $$= \Phi\left[\frac{\log(G_{tn}^{lead})-\beta_{lead}X_{tn}^{lead}-\alpha^{lead}V_n}{\sigma_{\varepsilon,lead}}\right] \times$$ $$\Phi\left[\frac{\log(G_{tn}^{lag})-\beta_{lag}X_{tn}^{lag}-\alpha^{lead}V_n}{\sigma_{\varepsilon,lag}}\right]$$ (1) where (X) means X follows standard normal distribution ,N(0,1) Numerical Example For the given state of traffic predict if the subject vehicle in the figure 5 would initiate a lane change.if yes what is the feasibility and probability of lane change. Given is the mid-block section of 2 lane highway with no other blocks in either of the lane. Neglect lateral acceleration. Consider update time 1 sec. Maximum deceleration driver ready to apply is -2 $m/s^2$ and maximum acceleration feasible is -2.2 $m/s^2$ Assume that lane change take 1 second. Given: $\sigma^{lead}$ =2, $\sigma^{lead}$ = 3 ,$G^{lead}$ =40m, $G^{lag}$ =50m , $\beta^{lead}$ = $\beta^{lag}$ = 1 ,$Xn^{lead}$ = $Xn^{lag}$ = 0.8, $V_{n}^{lead}$ = $V_{n}^{lag}$ = 0.7 , $\alpha^{lead}$ = $\alpha{lag}$ = 1.2 Solution Step 1. Decision to change the lane:  In the case of discretionary lane change, the decision to change the lane is taken by the driver when he finds higher utility in any other lane. Here, we consider higher speed or desired speed as higher utility. Let the desired speed be $25~m/s^{2}$. Considering the subject vehicle as vehicle n and the vehicle preceding it in the current lane as vehicle n-1, we calculate the minimum distance required by the subject vehicle to attain the desired speed in a time T $$D_x = x_{n-1}(t) - S_{n-1} - x_n(t)$$ $$V_n(t+T) = b_n~T+\sqrt{b_n^2T^2-b_n~(2D_x~-V_nT-V_n-1^2/b)}$$ $$25 = -2\times1 +\sqrt{-2^2+{(2~D_x~-19.4 +18^2/2.5)}}$$ The $D_x$ in this problem is 155 m, which means that the subject vehicle requires at least 155 m to reach his desired speed. But the gap available is 30 m. So decision is to change the lane or trigger DLC. Step 2. Check for the feasibility of lane change:  A lane change is said to be feasible if the subject vehicle is able to maintain maximum safe speed with respect to the preceding vehicle in the target line. In order to find the maximum safe speed possible for the subject vehicle to avoid collision we consider the subject vehicle as N and preceding vehicle in the target lane as N-1. Then we substitute the values in the second equation. $V_n(t+T) = -2\times 1 +[ -2^2+2{2(40)-19.4 +18^2/2.5}]^{1/2} =17.6 m/s$ And the deceleration required = (17.6-19.4)/1 = -1.79 $m/s^2$ Since -1.79 $m/s^2$ less than -2.2 $m/s^2$ the lane change feasible to avoid collision with the lead vehicle in the target lane. Now we have to check if the lag vehicle in the target line would be able to avoid the collision with the subject vehicle after the lane change. For this we take lag vehicle as $N^{th}$ vehicle and subject vehicle as $N-1^{th}$ vehicle Here, $D_x~ = 50 +19.4 -20.83 = 48.5 m$ as the lag vehicle and subject vehicle would have moved some distance during the lane change duration of 1 second. These distances are 20.83 m and 19.4 m respectively. $V_n(t+T) = -2x1 +[ -2^2+2{2(48.5)-20.83 +19.4^2/2.5}]^{1/2} = 19.38 m/s$ The deceleration required to be applied by the lag vehicle in the target lane to avoid collision with the subject vehicle = $(19.38 -20.83 )/1 = -1.44 m/s^2.$ Since -1.44 $m/s^2$ < -2.2 $m/s^2$ the lane change feasible to avoid collision of the lag vehicle in the target lane. Step 3. Check for the gap acceptance of lane change in the given state of traffic:  Here we find that the lag gap that was available is 50 m and the lead gap is 40 m.using the equation 2 we get, $$p(g) = \Phi\left[\frac{\log(G_{tn}^{lead})-\beta_{lead}X_{tn}^{lead}-\alpha^{lead}V_n}{\sigma_{\varepsilon,lead}}\right] \times$$ $$\Phi\left[\frac{\log(G_{tn}^{lag})-\beta_{lag}X_{tn}^{lag}-\alpha^{lead}V_n}{\sigma_{\varepsilon,lag}}\right]$$ $$= \Phi{(\log(40)-0.8-0.84)/2} \times \Phi{(\log(50)-0.8-0.84)/2}$$ $$= \Phi(1.02) \times \Phi(0.75)= 0.8212 \times 0.7734= 0.635.$$ This means that a given driver would opt for a lane change in the the given condition with a probability of 0.635.