Module 3 : Microscopic Traffic Flow Modeling
Lecture 12 : Vehicle Arrival Models: Headway
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Negative exponential distribution

The low flow traffic can be modeled using the negative exponential distribution. First, some basics of negative exponential distribution is presented. The probability density function $ f(t)$ of any distribution has the following two important properties: First,
$\displaystyle p[-\infty < t < +\infty]$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{+\infty}f(t)~dt=1$ (1)

where $ t$ is the random variable. This means that the total probability defined by the probability density function is one. Second:

$\displaystyle p[a \le t\le b]=\int_{a}^{b}f(t)~dt$ (2)

This gives an expression for the probability that the random variable $ t$ takes a value with in an interval, which is essentially the area under the probability density function curve. The probability density function of negative exponential distribution is given as:

$\displaystyle f(t)=\lambda e^{-\lambda t},~~~t\ge 0$ (3)

where $ \lambda $ is a parameter that determines the shape of the distribution often called as the shape parameter. The shape of the negative exponential distribution for various values of $ \lambda $ (0.5, 1, 1.5) is shown in figure 1.
Figure 1: Shape of the Negative exponential distribution for various values of $ \lambda $
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The probability that the random variable $ t$ is greater than or equal to zero can be derived as follow,
$\displaystyle p(t\ge0)$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\lambda~e^{-\lambda t}~dt$ (4)
  $\displaystyle =$ $\displaystyle \lambda~\int_{0}^{\infty} e^{-\lambda t}~dt$  
  $\displaystyle =$ $\displaystyle \lambda \left\vert\frac{e^{-\lambda t}}{-\lambda}\right\vert^{\infty}_{0}$  
  $\displaystyle =$ $\displaystyle \left\vert-e^{-\lambda t}\right\vert^{\infty}_{0}$  
  $\displaystyle =$ $\displaystyle -e^{-\lambda \infty}+e^{-\lambda 0}$  
  $\displaystyle =$ $\displaystyle 0+1 = 1$  

The probability that the random variable $ t$ is greater than a specific value $ h$ is given as
$\displaystyle p(t\ge h)$ $\displaystyle =$ $\displaystyle 1-p(t<h)$ (5)
  $\displaystyle =$ $\displaystyle 1-\int_{0}^{h}\lambda.e^{-\lambda t}~dt$  
  $\displaystyle =$ $\displaystyle 1-\lambda \left[\frac{e^{-\lambda t}}{-\lambda}\right]^h_0$  
  $\displaystyle =$ $\displaystyle 1+\left\vert e^{-\lambda t}\right\vert^h_0$  
  $\displaystyle =$ $\displaystyle 1+\left[e^{-\lambda h}-e^{-\lambda 0}\right]$  
  $\displaystyle =$ $\displaystyle 1+e^{-\lambda h}-1$  
  $\displaystyle =$ $\displaystyle e^{-\lambda h}$  

Unlike many other distributions, one of the key advantages of the negative exponential distribution is the existence of a closed form solution to the probability density function as seen above. The probability that the random variable $ t$ lies between an interval is given as:
$\displaystyle p[h\le t \le~(h+\delta h)]$ $\displaystyle =$ $\displaystyle p[t\ge h]-p[t\ge (h+\delta h)]$ (6)
  $\displaystyle =$ $\displaystyle e^{-\lambda~h}-e^{\lambda~(h+\delta h)}$  

This is illustrated in figure 2.
Figure 2: Evaluation of negative exponential distribution for an interval
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The negative exponential distribution is closely related to the Poisson distribution which is a discrete distribution. The probability density function of Poisson distribution is given as:

$\displaystyle p(x)=\frac{\lambda^x~e^{-\lambda}}{x!}$ (7)

where, $ p(x)$ is the probability of $ x$ events (vehicle arrivals) in some time interval ($ t$), and $ \lambda $ is the expected (mean) arrival rate in that interval. If the mean flow rate is $ q$ vehicles per hour, then $ \lambda=\frac{q}{3600}$ vehicles per second. Now, the probability that zero vehicle arrive in an interval $ t$, denoted as $ p(0)$, will be same as the probability that the headway (inter arrival time) greater than or equal to $ t$. Therefore,
$\displaystyle p(x=0)$ $\displaystyle =$ $\displaystyle \frac{\lambda^0~e^{-\lambda}}{0!}$  
  $\displaystyle =$ $\displaystyle e^{-\lambda}$  
  $\displaystyle =$ $\displaystyle p(h\ge t)$  
  $\displaystyle =$ $\displaystyle e^{-\lambda~t}$  

Here, $ \lambda $ is defined as average number of vehicles arriving in time $ t$. If the flow rate is $ q$ vehicles per hour, then,

$\displaystyle \lambda=\frac{q\times~t}{3600}=\frac{t}{\mu}$ (8)

Since mean flow rate is inverse of mean headway, an alternate way of representing the probability density function of negative exponential distribution is given as

$\displaystyle f(t)=\frac{1}{\mu}~e^\frac{-t}{\mu}$ (9)

where $ \mu=\frac{1}{\lambda}$ or $ \lambda=\frac{1}{\mu}$. Here, $ \mu$ is the mean headway in seconds which is again the inverse of flow rate. Using equation 6 and equation 5 the probability that headway between any interval and flow rate can be computed. The next example illustrates how a negative exponential distribution can be fitted to an observed headway frequency distribution.