Use the data and results obtained in Problem 9-18 to determine the level of service of a two-lane section if a passing lane 1.5 mi long is added. The passing lane begins 0.75 mi from the starting point of the analysis segment.

Question

What will be an ideal response?

Answer 1

To determine the level of service, lengths of the regions in the segment, PTSFpl

and ATSpl, must first be determined.

From Problem 9-16, PTSFd = 76.83% and ATSd = 37.88 mi/h

Step 1: Determine region lengths.

Region I: Lu = 0.75 mi

Region II: Lpl = 1.5 mi

Region III:

For PTSF, from Table 9.27, Lde = 5.8 mi; from Table 9.29, fpl = 0.62

For ATS, from Table 9.27, Lde = 1.7 mi; from Table 9.30, fpl = 1.11

Region IV:Use Equation 9.21.

Ld = Lt - (Lu +Lpl + Lde )

For PTSF, Ld = 4.5 – 0.75 –1.5 – 5.8 = –3.55; use Ld = 0

and L’de = 4.5 – 0.75 – 1.5 = 2.25

For ATS, Ld = 4.5 – 0.75 – 1.5 – 1.7 = 0.55

Step 2: Compute PTSFpl using Equation 9.23



Step 4: Determine level of service from Table 9.11 (for Class I highway): LOS E.