Question

A six- lane freeway (three lanes in each direction) in a scenic area has a measured...

A six- lane freeway (three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hour factor is 0.80, and there are 7% SUTs and 3% TTs in the traffic stream. One upgrade is 4.5% and 1.5 mi long. Free-flow speed is 70 mph. If the peak- hour traffic volume is 3900 vehicles, determine level of service?

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Answer #1

Free Flow Speed (FFS) = 70 mph

Measured free flow speed = 55 mph

Number of lanes in each direction = 3

Peak Flow, V = 3900 veh/hr

Peak-hour factor = 0.80

SUTs = 7%

TTs = 3 %

4.5% Upgrade 1.5 mi long

fHV = 1/ (1 + 0.07 (3.5-1) + 0.03 (4-1)) = 1/1.265 = 0.79

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 3900/ (0.80*3*0.79*1.0) = 2056.96 ~ 2057 veh/hr/ln

S = FFS

S = 55 mph

Density = Vp/S = (2057) / (55) = 37.4 veh/mi/ln

LOS E

Density of LOS E should lie between 36 – 45 veh/mi/ln

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