Module 3 : Microscopic Traffic Flow Modeling

Lecture 12 : Vehicle Arrival Models: Headway

1

2

3

4

5

6

7

8

9

10

11

12

Numerical Example

An observation from 2434 samples is given table below. Mean headway observed was 3.5 seconds and the standard deviation 2.6 seconds. Fit a Person Type III Distribution.

Table 1: Observed headway distribution

$h$	$h+dh$	$p_i^o$
0.0	1.0	0.012
1.0	2.0	0.178
2.0	3.0	0.316
3.0	4.0	0.218
4.0	5.0	0.108
5.0	6.0	0.055
6.0	7.0	0.033
7.0	8.0	0.022
8.0	9.0	0.013
9.0	$>$	0.045
Total		1.00

Solutions

Given that mean headway ($\mu$) is 3.5 and the standard deviation ($\sigma$) is 2.6. Assuming the expected minimum headway ($\alpha$) is 0.5, $K$ can be computed as

$$K = \frac{\mu-\alpha}{\sigma} = \frac{3.5-0.5}{2.6}=1.15$$

and flow rate term $\lambda$ as

$$\lambda = \frac{K}{\mu -\alpha}=\frac{1.15}{3.5-0.5}=0.3896$$

Now, since $K=1.15$ which is between 1 and 2, $\Gamma(K)$ can be obtained directly from the gamma table as $\Gamma(K)=0.93304$. Here, the probability density function for this example can be expressed as

$$f(t) = \frac{0.3846}{0.93304}\times\left[~0.3846\times(t-0.5)~\right]^{1.15-1}\times e^{-0.3846~(t-0.5)}$$

The given headway range and the observed probability is given in column (2), (3) and (4). The observed frequency ($f_i^o$) for the first interval (0 to 1) can be computed as the product of observed proportion $p_i^o$ and the number of observations ($N$). That is, $f_i^o = p_i^o\times N = 0.012\times2434 = 29.21$ as shown in column (5). The probability density function value for the lower limit of the first interval (h=0) is shown in column (6) and computed as:

$$f(0) = \frac{0.3846}{0.93304}\left[~0.3846\times(0-0.5)\right]^{1.15-1}\times e^{-0.3846~(0-0.5)}\approx 0.$$

Note that since $t-\alpha~(0-0.5)$ is negative and $K-1~(1.15-1)$ is a fraction, the above expression cannot be evaluated and hence approximated to zero (corresponding to t=0.5). Similarly, the probability density function value for the lower limit of the second interval (h=1) is shown in column 6 and computed as:

$$f(1) = \frac{0.3846}{0.93304}\left[0.3846 (1-0.5)\right]^{1.15-1}\times e^{-0.3846(1-0.5)}=0.264$$

Now, for the first interval, the probability for headway between 0 and 1 is computed by equation  as $p_i^c(0\le t\le 1)=\left(\frac{f(0)+(f(1)}{2}\right)\times(1-0)=(0+0.0264)/2\times~1=0.132$ and is given in column (7). Now the computed frequency $f_i^c$ is $p_i^c\times N = 0.132\times2434=321.1$ and is given in column (8). This procedure is repeated for all the subsequent items. It may be noted that probability of headway $>$ 9 is computed by 1-probability of headway less than 9 $= 1-(0.132+0.238+\dots)=0.044.$ The comparison of the three distribution for the above data is plotted in Figure 1.

Table 2: Solution using Pearson Type III distribution

$No$	$h$	$h+\delta~h$	$p_i^o$	$f_i^o$	$f(t)$	$p_i^c$	$f_i^c$
(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)
1	0	1	0.012	29.2	0.000	0.132	321.2
2	1	2	0.178	433.3	0.264	0.238	580.1
3	2	3	0.316	769.1	0.213	0.185	449.5
4	3	4	0.218	530.6	0.157	0.134	327.3
5	4	5	0.108	262.9	0.112	0.096	233.4
6	5	6	0.055	133.9	0.079	0.068	164.6
7	6	7	0.033	80.3	0.056	0.047	115.3
8	7	8	0.022	53.6	0.039	0.033	80.4
9	8	9	0.013	31.6	0.027	0.023	55.9
10	$>$9		0.045	109.5	0.019	0.044	106.4
	Total		1.0	2434		1.0	2434

Figure 1: Comparison of distributions