Module 1 : Traffic Stream Characteristics

Lecture 03 : Traffic Stream Models

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Calibration of Greenshield's model

In order to use this model for any traffic stream, one should get the boundary values, especially free flow speed ($v_f$) and jam density ($k_j$). This has to be obtained by field survey and this is called calibration process. Although it is difficult to determine exact free flow speed and jam density directly from the field, approximate values can be obtained from a number of speed and density observations and then fitting a linear equation between them. Let the linear equation be $y = a + bx$ such that $y$ is density $k$ and $x$ denotes the speed $v$. Using linear regression method, coefficients $a$ and $b$ can be solved as,

$$b = \frac{n\sum_{i=1}^{n}{x_iy_i}-\sum_{i=1}^{n}{x_i}.\sum_{i=1}^{n}{y_i}}{n.\sum_{i=1}^{n}{x_i}^2-({\sum_{i=1}^{n}{x_i}})^2}$$ (1)

$$a = \bar{y}-b\bar{x}$$ (2)

Alternate method of solving for b is,

$$b = \frac{\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^{n}{(x_i - \bar{x})}^2}$$ (3)

where $x_i$ and $y_i$ are the samples, $n$ is the number of samples, and $\bar x$ and $\bar y$ are the mean of $x_i$ and $y_i$ respectively.

Numerical example

For the following data on speed and density, determine the parameters of the Greenshield's model. Also find the maximum flow and density corresponding to a speed of 30 km/hr.

k	v
171	5
129	15
20	40
70	25

Solution

Denoting y = v and x = k, solve for a and b using equation 2 and equation 3. The solution is tabulated as shown below.

x(k)	y(v)	( $x_i-\bar{x}$)	( $y_i-\bar{y}$)	( $x_i-\bar{x}$)( $y_i-\bar{y}$)	( $x_i-\bar{x}^{2}$)
171	5	73.5	-16.3	-1198.1	5402.3
129	15	31.5	-6.3	-198.5	992.3
20	40	-77.5	18.7	-1449.3	6006.3
70	25	-27.5	3.7	-101.8	756.3
390	85			-2947.7	13157.2

$\bar{x} =\frac{\Sigma{x}}{n} =\frac{390}{4}$ = 97.5, $\bar{y} = \frac{\Sigma{y}}{n} = \frac{85}{4}$ = 21.3. From equation 3, b = $\frac{-2947.7}{13157.2}$ = -0.2 $a = y - b\bar{x}$ = 21.3 + 0.2$\times$97.5 = 40.8 So the linear regression equation will be,

$$v = 40.8 - 0.2 k$$ (4)

Here $v_f$ = 40.8 and $\frac{v_f}{k_j}$ = 0.2. This implies, $k_j = \frac{40.8}{0.2}$ = 204 veh/km. The basic parameters of Greenshield's model are free flow speed and jam density and they are obtained as 40.8 kmph and 204 veh/km respectively. To find maximum flow, use equation , i.e., $q_{max} = \frac{40.8 \times 204}{4}$ = 2080.8 veh/hr Density corresponding to the speed 30 km/hr can be found out by substituting $v = 30$ in equation 4. i.e, 30 = 40.8 - 0.2 $\times$ k Therefore, k = $\frac{40.8 - 30}{0.2}$ = 54 veh/km.