Module 3 : Microscopic Traffic Flow Modeling

Lecture 12 : Vehicle Arrival Models: Headway

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	Numerical Example If the mean and standard deviation of certain observed set of headways is 2.25 and 0.875 respectively, then compute the probability that the headway lies in an interval of 1.5 to 2.0 seconds. Solution: The probability that headway lies between 1.5 and 2.0 can be obtained using equation , given that $\mu=2.25$ and $\sigma=0.85$ as: $$p[1.5\le t\le 2.0] = p[t\le 2.0]-p[t\le 1.5]$$ $$= p\left[t\le \frac{2.0-2.25}{0.875}\right] - p\left[t\le \frac{1.5-2.25}{0.875}\right]$$ $$= p\left[t\le -0.29 \right] - p\left[t\le -0.86 \right]$$ $$= 0.3859-0.1949 \mathrm{~~(from~tables)}$$ $$= 0.191.$$ Note that the $p(t\le -0.29)$ and $p(t\le -0.80)$ are obtained from the standard normal distribution tables. Since the normal distribution is defined from $-\alpha$ to $+\alpha$ unlike an exponential distribution which is defined only for positive number, it is possible that normal distribution may generate negative headways. A practical way of avoiding this is to shift the distribution by some value so that it will mostly generate realistic headways. The concept is illustrated in figure 1. Figure 1: Normal Distribution Suppose $\alpha$ is the minimum possible headway and if we set $\alpha= \mu-\sigma$ than about 60% of headway will be greater than $\alpha$. Alternatively, if we set $\alpha=\mu-2\sigma$, than about 90% of the headway will be greater than $\alpha$. Further, if we set $\alpha=\mu-3\sigma$, than about 99% of the headway will be greater than $\alpha$.  To generalize, $$\alpha = \mu-n\sigma$$ where n is 1, 2, 3, etc and higher the value of $n$, then is better the precision. From this equation, we can compute the value of $\sigma$ to be used in normal distribution calculation when the random variable cannot be negative as: $$\sigma = \frac{\mu-\alpha}{n}$$ (1)