Question

3. A multilane highway, with three lanes in each direction, operates at maximum level of services D conditions during the peak hour with an observed speed of 48 mph (with 3690 veh/h observed). The highway is on rolling terrain and the traffic stream consist of cars and some percentage of heavy vehicles. The peak hour factor is 0.85. If heavy vehicles were banned from the highway and car traffic remained the same, what would be the LOS

Answer 1

Solution :

Given :

The number of lanes in each of the direction is N = 3

The terrain is rolling. And the traffic stream is cars and heavy vehicles. And the peak hour factor is given as 0.85

Now the observed flow speed is S  = 48 mph

Flow rate of vehicles is  V = 3690 veh/h

Highway operates at the maximum LOS (level of service) D condition.

The density value that is allowed at this condition is D = 35 pc/mi/lane

Therefore, $D=\frac{v_p}{S}$

or $v_p = D \times S$

        = 35 x  48

        = 1680 pcphpl

$v_s= \frac{V}{PHF \times N \times f_{HV} \times f_p}$

$1680= \frac{3690}{0.85 \times 3 \times f_{HV} \times 1}$

$f_{HV}=0.8613$

$f_{HV}=\frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$0.8613=\frac{1}{1+P_T(2.5-1)}$

$P_T=0.1073$

So the percentage of trucks is 23.33% of the total volume and is equal to 754 vehicles per hour.

If the heavy vehicles are banned from the highway, the number of vehicles will reduce to 2936 vehicles per hour.

The value of $f_{HV} = 1$

$v_p=\frac{2936}{0.85 \times 3 \times 1 \times 1}$

   = 1151.3 pcphpl

Density value is , $D=\frac{v_p}{S}$

                               = $\frac{1151.3}{48}$

                               23.98 pc/mi/lane

Therefore, the level of service improves to LOS C, when the heavy vehicles are banned in the highways while the car traffic remains the same.