Module 3 : Microscopic Traffic Flow Modeling

Lecture 13 : Vehicle Arrival Models: Count

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Numerical Example

The hourly flow rate in a road section is 120 vph. Use Poisson distribution to model this vehicle arrival.

Solution:

The flow rate is given as ($\mu$) = 120 vph =  $\frac{120}{60}$= 2 vehicle per minute. Hence, the probability of zero vehicles arriving in one minute $p(0)$ can be computed as follows:

$$p(0) = \frac{\mu^x e^{-\mu}}{x!} = \frac{2^0.e^{-2}}{0!} = 0.135.$$

Similarly, the probability of one vehicles arriving in one minute $p(1)$ is given by,

$$p(1) = \frac{\mu^x e^{-\mu}}{x!} = \frac{2.e^{-2}}{1!} = 0.271.$$

Now, the probability that number of vehicles arriving is less than or equal to zero is given as

$$p(x\leq 0)~=~p(0)~=~0.135.$$

Similarly, probability that the number of vehicles arriving is less than or equal to 1 is given as:

$$p(x\leq 1)~=~p(0)~+~p(1)~=~0.135~+~0.275~=~0.406.$$

Again, the probability that the number of vehicles arriving is between 2 to 4 is given as:

$$p(2\leq x\leq 4) = p(2)+p(3)+p(4),$$

$$= .271+.18+.09~=~0.54.$$

Now, if the $p(0)~=~0.135$, then the number of intervals in an hour where there is no vehicle arriving is

$$F(x)~=~p(0)~\times~60~=~0.135~\times~60~=~8.12.$$

The above calculations can be repeated for all the cases as tabulated in Table 1.

Table 1: Probability values of vehicle arrivals computed using Poisson distribution

$n$	$p(n)$	$p(x\leq n)$	$F(n)$
0	0.135	0.135	8.120
1	0.271	0.406	16.240
2	0.271	0.677	16.240
3	0.180	0.857	10.827
4	0.090	0.947	5.413
5	0.036	0.983	2.165
6	0.012	0.995	0.722
7	0.003	0.999	0.206
8	0.001	1.000	0.052
9	0.000	1.000	0.011
10	0.000	1.000	0.011

The shape of this distribution can be seen from Figure 1 and the corresponding cumulative distribution is shown in Figure 2.

Figure 1: Probability values of vehicle arrivals computed using Poisson distribution

Figure 2: Cumulative probability values of vehicle arrivals computed using Poisson distribution