Because several complaints have been received from local drivers, you have been assigned to determine the level of service for a one-lane approach at a local signalized intersection. The cycle length is 80 seconds, and 30 seconds of effective green time are enjoyed by the approach in question. The actual flow rate of traffic through the approach is 400 pcu/hr and the saturation flow rate for the approach is 1750 pcu/hr. What is the LOS for this approach?
[Solution Shown Below]
Before we can calculate the delay for the approach, we need to know the green ratio (g/C), the capacity (c), and the ratio V/c (X). The green ratio for this approach is 30/80 or 0.375. The capacity is (g/C)*s which equals 0.375*1750 or 656 pce/hr. The ratio V/c is 400/656 or 0.609. The average vehicle delay is given by the equation below.
d= [0.38C(1-g/C)2]/[1-(g/C)(X)] + 173X2[(X-1) + [(X-1)2 + (16X/C)]1/2]
By placing the values calculated above into the equation, we obtain an average vehicle delay of 23 seconds. This corresponds to the level of service grade "C".