Modeling Traffic Characteristics

Lecture notes in Transportation Systems Engineering

4 August 2009

Modeling time-headway's

Modeling inter arrival time or time headway or simply headway is he time interval between the successive arrival of two vehicles at a given point. This is a continuous variable and can be treated as a random variable.

Classification of headway distribution

One can observe three types of flow in the field:

Low volume flow
1. Headway follow a random process as there is no interaction between the arrival of two vehicles.
2. The arrival of one vehicle is independent of the arrival of other vehicle.
3. The minimum headway is governed by the safety criteria.
4. A negative exponential distribution can be used to model such flow
High volume flow
1. This is characterized by 'near' constant headway
2. The flow is very high and is near to the capacity
3. The mean is very low and so is the variance
4. A normal distribution can used to model such flow
Intermediate flow
1. Some vehicle travel independently and some vehicle has interaction
2. More difficult to analyzed and has more application in the field.
3. Pearson Type III Distribution can be used which is a very general case of negative exponential distribution and normal distribution.

Negative exponential distribution

PDF of negative exponential function

$\displaystyle f(t)$ $\textstyle =$ $\displaystyle \frac{1}{\beta}e^{\frac{-t}{\beta}}$ (1)

$\textstyle =$ $\displaystyle e^{-t} \mathrm{if } \beta =1$
The pb of h/w greater than of equal to 0

$\displaystyle p(h\ge0)$ $\textstyle =$ $\displaystyle \int_{0}^{\infty} e^{-t} dt=\left. -e^{-t}\right\vert^\infty_0=1$ (2)
The pb of h/w greater than of equal to h
$\begin{eqnarray*} p(h\ge t)&=&1-p(h<t)\\ &=&1-p\left[\int_{0}^{h} e^{-t} dt\right]\\ &=&1-\left[\left.-e^{-t}\right\vert^h_0\right]\\ &=&e^{-h} \end{eqnarray*}$
From the PDF of possson distribution we have the probability of vehicle arriving in time interval as and is given by

$\displaystyle p(x)$ $\textstyle =$ $\displaystyle \frac{m^x e^{-m}}{x!}$ (3)

where is the mean arrival rate in the time interval . The mean arrival rate is given as $m=\frac{V}{3600}t$ where is the hourly flow rate.
If the probability of no vehicle in the interval is given as , then this probability is same as the probability that the headway greater than or equal to . Therefore,

$\displaystyle p(x=0)$ $\textstyle =$ $\displaystyle e^{-m}$ (4)

$\textstyle =$ $\displaystyle p(h\ge t)$ (5)

$\textstyle =$ $\displaystyle e^{\frac{-t}{\mu}}$ (6)

where $\mu=\frac{1}{{\frac{V}{3600}}}$ whcih is same as the mean headway.
$p[t\le h \le t+\delta t] = p[h\ge t]-p[h\ge t+\delta t]$

Normal distribution

for high volume traffic
probability density function for normal distribution is

$\displaystyle f(t)$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(t-\mu)^2}{2\sigma^2}}$ (7)
The probability for headway less than a given time

$\displaystyle p(h\le t)$ $\textstyle =$ $\displaystyle \int_{-\infty}^{t} \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(t-\mu)^2}{2\sigma^2}}$ (8)

$\displaystyle p(h\le t+\delta t)$ $\textstyle =$ $\displaystyle \int_{-\infty}^{t+\delta t} \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(t+\delta t-\mu)^2}{2\sigma^2}}$

$\displaystyle p(t\le h \le t+\delta t)$ $\textstyle =$ $\displaystyle p(h\le t+\delta t) - p(h\le t)$
The normal distribution has two parameters; $\mu$ and $\sigma$
The $\mu$ is simple mean headway or inverse of average flow rate given by

$\displaystyle \mu$ $\textstyle =$ $\displaystyle \frac{1}{(\frac{3600}{V})}$ (9)
Since headway's cannot be negative, $\sigma$ has to be calculated, not from observation, but empirically. First specify the theoretical minimum headway possible, say $\alpha$ and set it as mean minus twice standard deviation.

$\displaystyle \alpha$ $\textstyle =$ $\displaystyle \mu-2\times \sigma$ (10)

$\displaystyle \sigma$ $\textstyle =$ $\displaystyle \frac{\mu-\alpha}{2}$ (11)

For example, if the hourly flow rate is 1500 vehicles and the minimum expected headways is 0.5 second, then,

$\displaystyle \sigma$ $\textstyle =$ $\displaystyle \frac{\frac{3600}{1500}-0.5}{2}=0.95$ (12)
The integration of the pdf of normal distribution is not available in a closed form solution. Therefore, one has to numerically integrate or normalize the given distribution having mean $\mu$ and sd $\sigma$ to a standard normal distribution ( $\mu=0$ and sd $\sigma=1$ ) whose values are available as standard tables. The transformation is give as:

$\displaystyle p[t\le h\le (t+\delta t)]$ $\textstyle =$ $\displaystyle p\left[\frac{t-\mu}{\sigma}\le \frac{h-\mu}{\sigma}\le \frac{(t+\delta t)-\mu}{\sigma}\right]$ (13)

$\textstyle =$ $\displaystyle p\left[h\le \frac{(t+\delta t)-\mu}{\sigma}\right] - p\left[h\le \frac{t-\mu}{\sigma}\right]$ (14)
For example if q=1600 veh/hr, then $\mu=\frac{3600}{1600}=2.25$ . Therefore $\sigma=\frac{2.25-0.5}{2}=0.875$ (assuming $\alpha=0.5$ sec).

$\displaystyle p[1.5\le h\le 2.0]$ $\textstyle =$ $\displaystyle p[h\le 2.0]-p[h\le 1.5]$ (15)

$\textstyle =$ $\displaystyle p\left[h\le \frac{2.0-2.25}{0.875}\right] - p\left[h\le \frac{1.5-2.25}{0.875}\right]$ (16)

$\textstyle =$ $\displaystyle p\left[h\le -0.29 \right] - p\left[h\le -0.86 \right]$ (17)

$\textstyle =$ $\displaystyle 0.3859-0.1949 \mathrm{ (from tables)}$ (18)

$\textstyle =$ $\displaystyle 0.191.$ (19)

Pearson type III distribution

The pdf is given by

$\displaystyle \begin{array}{cclll} f(t)&=&\frac{\lambda}{\Gamma(K)} \left[ \lam... ...ambda e^{-\lambda t} &\mathrm{if }K=1 & \mathrm{Neg. Exp.}\nonumber \end{array}$

where $\lambda$ is parameter that is function of $\mu, K,$ and $\alpha$ ; is a user specified parameter between and $\infty$ ; $\alpha$ is a user selected parameter greater than 0 and called as the shift patmeter; and $\Gamma()$ is the gamma function ( $\Gamma(K)=(K-1)!$ ).
Then

$\displaystyle p(h\ge t)$ $\textstyle =$ $\displaystyle \int_{t}^{\infty} f(t) dt$ (20)
Prob of h/w lying b/w and $t+\delta t$ is

$\displaystyle p(t\le h < (t+\delta t) ) =\int_{t}^{\infty} f(t) dt - \int_{(t+\delta t)}^{\infty} f(t) dt$ (21)
Approximating

$\displaystyle p(t\le h < (t+\delta t) )$ $\textstyle =$ $\displaystyle \left[ \frac{f(t)+f(t+\delta t)}{2} \right] \delta t$ (22)
The solution steps
1. Input $\mu,\sigma$
2. $\alpha=0.5$ ensuring h/w is greater than 0.5
3. $K=\frac{\mu-\alpha}{\sigma}$
4. $\lambda=\frac{K}{\mu -\alpha}$
5. Determination of $\Gamma(K)$
  
  $\displaystyle \Gamma(K)$ $\textstyle =$ $\displaystyle \left \{ \begin{array}{ll} \int_{0}^{\infty}e^{-x}x^{K-1}dx, & \mathrm{if} 1\le K \le 2 \\ (K-1)\Gamma(K-1) & \mathrm{if} K>2 \end{array}\right .$ (23)
6. Solve for using Equation 20
7. Solve for $p(t\le h < (t+\delta t) )$ using Equation 22

Example

An obseravtion from 3424 samples is given table below. Mean headway observed was 3.5 seconds and the standard deviation 2.6 seconds. Fit a (i) negatie expoetial distribution, (ii) normal distrbution and (iii) Person Type III Distribution.

**Table 1:** Obsered headway distribution
t	t+dt	p (obs)
0.0	0.5	0.012
0.5	1.0	0.064
1.0	1.5	0.114
1.5	2.0	0.159
2.0	2.5	0.157
2.5	3.0	0.130
3.0	3.5	0.088
3.5	4.0	0.065
4.0	4.5	0.043
4.5	5.0	0.033
5.0	5.5	0.022
5.5	6.0	0.019
6.0	6.5	0.014
6.5	7.0	0.010
7.0	7.5	0.012
7.5	8.0	0.008
8.0	8.5	0.005
8.5	9.0	0.007
9.0	9.5	0.005
9.5		0.033
Total		1.00

Solutions

**Table 2:** Solution using negative exponential distribution
t	t+dt	p (obs)	Pobs=p*N		N-exp (p)	P=p*N
0.0	0.5	0.012	41.1	1.000	0.133	455.8
0.5	1.0	0.064	219.1	0.867	0.115	395.1
1.0	1.5	0.114	390.3	0.751	0.100	342.5
1.5	2.0	0.159	544.4	0.651	0.087	296.9
2.0	2.5	0.157	537.6	0.565	0.075	257.4
2.5	3.0	0.130	445.1	0.490	0.065	223.1
3.0	3.5	0.088	301.3	0.424	0.056	193.4
3.5	4.0	0.065	222.6	0.368	0.049	167.7
4.0	4.5	0.043	147.2	0.319	0.042	145.4
4.5	5.0	0.033	113.0	0.276	0.037	126.0
5.0	5.5	0.022	75.3	0.240	0.032	109.2
5.5	6.0	0.019	65.1	0.208	0.028	94.7
6.0	6.5	0.014	47.9	0.180	0.024	82.1
6.5	7.0	0.010	34.2	0.156	0.021	71.2
7.0	7.5	0.012	41.1	0.135	0.018	61.7
7.5	8.0	0.008	27.4	0.117	0.016	53.5
8.0	8.5	0.005	17.1	0.102	0.014	46.4
8.5	9.0	0.007	24.0	0.088	0.012	40.2
9.0	9.5	0.005	17.1	0.076	0.010	34.8
9.5		0.033	113.0	0.066	0.066	226.8
Total		1.000	3424		1.000	3424

**Table 3:** Solution using normal distribution
h	t+dt	p (obs)	P=p*N			P=p*N
0.0	0.5	0.012	41.09	0.010	0.013	44.289
0.5	1.0	0.064	219.14	0.023	0.025	85.738
1.0	1.5	0.114	390.34	0.048	0.043	148.673
1.5	2.0	0.159	544.42	0.091	0.067	230.928
2.0	2.5	0.157	537.57	0.159	0.094	321.299
2.5	3.0	0.13	445.12	0.252	0.117	400.433
3.0	3.5	0.088	301.31	0.369	0.131	447.033
3.5	4.0	0.065	222.56	0.500	0.131	447.033
4.0	4.5	0.043	147.23	0.631	0.117	400.433
4.5	5.0	0.033	112.99	0.748	0.094	321.299
5.0	5.5	0.022	75.33	0.841	0.067	230.928
5.5	6.0	0.019	65.06	0.909	0.043	148.673
6.0	6.5	0.014	47.94	0.952	0.025	85.738
6.5	7.0	0.01	34.24	0.977	0.013	44.289
7.0	7.5	0.012	41.09	0.990	0.006	20.492
7.5	8.0	0.008	27.39	0.996	0.002	8.493
8.0	8.5	0.005	17.12	0.999	0.001	3.153
8.5	9.0	0.007	23.97	1.000	0.000	1.048
9.0	9.5	0.005	17.12	1.000	0.000	0.312
9.5		0.033	112.99	1.000	0.010	33.716
		3424

**Table 4:** Solution using Pearson type III distribution
h	t+dt	p (obs)	P=p*N	f(t)		P=p*N
0.0	0.5	0.012	41.09		0.000	0.00
0.5	1.0	0.064	219.14	0.000	0.066	225.91
1.0	1.5	0.114	390.34	0.264	0.127	433.26
1.5	2.0	0.159	544.42	0.242	0.114	389.45
2.0	2.5	0.157	537.57	0.213	0.099	339.13
2.5	3.0	0.13	445.12	0.183	0.085	291.12
3.0	3.5	0.088	301.31	0.157	0.072	247.86
3.5	4.0	0.065	222.56	0.133	0.061	209.90
4.0	4.5	0.043	147.23	0.112	0.052	177.08
4.5	5.0	0.033	112.99	0.095	0.044	148.97
5.0	5.5	0.022	75.33	0.079	0.037	125.04
5.5	6.0	0.019	65.06	0.067	0.031	104.78
6.0	6.5	0.014	47.94	0.056	0.026	87.67
6.5	7.0	0.01	34.24	0.047	0.021	73.28
7.0	7.5	0.012	41.09	0.039	0.018	61.18
7.5	8.0	0.008	27.39	0.033	0.015	51.04
8.0	8.5	0.005	17.12	0.027	0.012	42.54
8.5	9.0	0.007	23.97	0.023	0.010	35.44
9.0	9.5	0.005	17.12	0.019	0.009	29.51
9.5		0.033	112.99	0.016	0.102	350.83
	x	3424

Comparison of distributions

Examples compared

Evaluating the selected distribution

Modeling vehicle arrival

Modeling speed

Estimation of population and sample means

Bibliography

1 Adolf D. May. Fundamentals of Traffic Flow. Prentice - Hall, Inc. Englewood Cliff New Jersey 07632, second edition, 1990.

Prof. Tom V. Mathew 2009-08-04

$\displaystyle f(t)$	$\textstyle =$	$\displaystyle \frac{1}{\beta}e^{\frac{-t}{\beta}}$	(1)
	$\textstyle =$	$\displaystyle e^{-t} \mathrm{if } \beta =1$

$\displaystyle p(x=0)$	$\textstyle =$	$\displaystyle e^{-m}$	(4)
	$\textstyle =$	$\displaystyle p(h\ge t)$	(5)
	$\textstyle =$	$\displaystyle e^{\frac{-t}{\mu}}$	(6)

$\displaystyle p(h\le t)$	$\textstyle =$	$\displaystyle \int_{-\infty}^{t} \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(t-\mu)^2}{2\sigma^2}}$	(8)
$\displaystyle p(h\le t+\delta t)$	$\textstyle =$	$\displaystyle \int_{-\infty}^{t+\delta t} \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(t+\delta t-\mu)^2}{2\sigma^2}}$
$\displaystyle p(t\le h \le t+\delta t)$	$\textstyle =$	$\displaystyle p(h\le t+\delta t) - p(h\le t)$

$\displaystyle \alpha$	$\textstyle =$	$\displaystyle \mu-2\times \sigma$	(10)
$\displaystyle \sigma$	$\textstyle =$	$\displaystyle \frac{\mu-\alpha}{2}$	(11)

$\displaystyle p[t\le h\le (t+\delta t)]$	$\textstyle =$	$\displaystyle p\left[\frac{t-\mu}{\sigma}\le \frac{h-\mu}{\sigma}\le \frac{(t+\delta t)-\mu}{\sigma}\right]$	(13)
	$\textstyle =$	$\displaystyle p\left[h\le \frac{(t+\delta t)-\mu}{\sigma}\right] - p\left[h\le \frac{t-\mu}{\sigma}\right]$	(14)

$\displaystyle p[1.5\le h\le 2.0]$	$\textstyle =$	$\displaystyle p[h\le 2.0]-p[h\le 1.5]$	(15)
	$\textstyle =$	$\displaystyle p\left[h\le \frac{2.0-2.25}{0.875}\right] - p\left[h\le \frac{1.5-2.25}{0.875}\right]$	(16)
	$\textstyle =$	$\displaystyle p\left[h\le -0.29 \right] - p\left[h\le -0.86 \right]$	(17)
	$\textstyle =$	$\displaystyle 0.3859-0.1949 \mathrm{ (from tables)}$	(18)
	$\textstyle =$	$\displaystyle 0.191.$	(19)