Module 3 : Microscopic Traffic Flow Modeling

Lecture 12 : Vehicle Arrival Models: Headway

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	Pearson Type III distribution As noted earlier, the intermediate flow is more complex since certain vehicles will have interaction with the other vehicles and certain may not. Here, Pearson Type III distribution can be used for modelling intermediate flow. The probability density function for the Pearson Type III distribution is given as $$\begin{array}{cclll} f(t)&=&\frac{\lambda}{\Gamma(K)} \left[ ... ...^{K-1} e^{-\lambda(t-\alpha)}, &K,\alpha \in R &\\ \end{array}$$ (1) where $\lambda$ is a parameter which is a function of $\mu$, $K$ and $\alpha$, and determine the shape of the distribution. The term $\mu$ is the mean of the observed headways, K is a user specified parameter greater than 0 and is called as a shift parameter. The $\Gamma()$ is the gamma function and given as $$\Gamma(K)=(K-1)!$$ (2) It may also be noted that Pearson Type III is a general case of Gamma, Erlang and Negative Exponential distribution as shown in below: $$\begin{array}{cclll} f(t)&=&\frac{\lambda}{\Gamma(K)} \left[ ... ...a t} &\mathrm{if~}K=1 & \mathrm{Neg.~Exp.}\nonumber \end{array}$$ The expression for the probability that the random headway (t) is greater than a given headway (h), $p(t\ge h)$, is given as: $$p(t\ge h) = \int_{h}^{\infty} f(t)~dt$$ (3) and similarly $p(t > h+\delta h)$ is given as: $$p(t > h+\delta h) = \int_{(h+\delta h)}^{\infty} f(t)~dt$$ (4) and hence, the probability that the headway between $h$ and $h+\delta h$ is given as $$p(h\le t\le (h+\delta h) ) = \int_{h}^{\infty} f(t) dt - \int_{(h+\delta h)}^{\infty} f(t)~dt$$ (5) It may be noted that closed form solution to equation 3 and equation 4 is not available. Numerical integration is also difficult due to computational requirement. Using table as in the case of Normal Distribution is difficult, since the table will be different for each $K$. A common way of solving this is by using the numerical approximation to equation 5. The solution to equation 5 is essentially the area under the curve defined by the probability density function between $h$ and $h+\delta h$. If we assume that line joining $f(h)$ and $f(h+\delta h)$ is linear, which is a reasonable assumption if $\delta h$ is small, than the are under the curve can be found out by the following approximate expression: $$p(h\le t\le (h+\delta h) ) \approx \left[ \frac{f(h)+f(h+\delta h)}{2} \right]\times \delta h$$ (6) This concept is illustrated in figure 1 Figure 1: Illustration of the expression for probability that the random variable lies in an interval for Person Type III distribution Stepwise procedure to fit a Pearson Type III distribution Input required: the mean ($\mu$) and the standard deviation ($\sigma$) of the headways. Set the minimum expected headway ($\alpha$). Say, for example, 0.5. It means that the $p(t < 0.5)\approx 0$. Compute the shape factor using the mean ($\mu$) the standard deviation ($\sigma$) and the minimum expected headway ($\alpha$) $$K=\frac{\mu-\alpha}{\sigma}$$ Compute the term flow rate ($\lambda$) as $$\lambda=\frac{K}{\mu -\alpha}$$ Note that if $K=1$ and $\alpha=0$, then $\lambda=\frac{1}{\mu}$ which is the flow rate. Compute gamma function value for $K$ as: $$\begin{array}{llll} \Gamma(K) &= (K-1)!&\mathrm{if}~K\in I&(... ...1)~\Gamma (K-1)&\mathrm{if}~K\in R&(\mathrm{Real}) \end{array}$$ (7) Although the closed form solution of $\Gamma(K)$ is available, it is difficult to compute. Hence, it can be obtained from gamma table. For, example: $$\Gamma(4.785) = 3.785\times\Gamma(3.785)$$ $$= 3.785\times2.785\times\Gamma(2.785)$$ $$= 3.785\times2.785\times1.785\times\Gamma(1.785)$$ $$= 3.785\times2.785\times1.785\times0.92750$$ $$= 17.45$$ Note that the value of $\Gamma(1.785)$ is obtained from gamma table for $\Gamma(x)$ which is given for $1\le x\le 2$. Using equation 1 solve for $f(h)$ by setting $t=h$ where h is the lower value of the range and $f(h+\delta h)$ by setting $t=h+\delta h$ where $(h+\delta h)$ is the upper value of the headway range. Compute the probability that headway lies between the interval of $h$ and $h+\delta h$ using equation 6. The Gamma function can be evaluated by the following approximate expression also: $$Gamma(x)=x^x~e^{-x}~\sqrt \frac{2~\pi}{x} \left(1+\frac{1}{12~x}+\frac{1}{288x^2}+\dots\right)$$ (8)