Peak-Hour Factor

PHF = \frac{V}{4 V _{15}}

PHF accounts for peaking within the analysis hour. V is the hourly volume (veh/h) and V15 is the highest 15-minute count in that hour. PHF ≤ 1; urban peaks often 0.80–0.95. HCM service-flow rate uses v = V / PHF (or the peak 15-min rate expanded ×4). A PHF of 1.00 means a perfectly uniform hour.

$PHF$: Peak-hour factor
$V$ (veh/h): Hourly volume
$V_{15}$ (veh): Peak 15-minute volume

Step-by-step solved example

A peak hour counts 1,800 veh. The highest 15 min is 500 veh. Find PHF and the 15-min flow rate.

The peak 15 min is denser than the hourly average along the lane.

1. PHF
PHF = 1800 / (4×500) = 1800/2000 = 0.90.
$PHF = 0.90$
2. Flow rate
v = 4×500 = 2000 veh/h, or V/PHF = 1800/0.90 = 2000 veh/h.
$v = 2000 veh/h$

Answer: PHF = 0.90, v = 2000 veh/h

10 practice questions

0/10 correct

1.V = 1600 veh/h, V15 = 500. PHF =

2.PHF cannot exceed

3.A lower PHF means the peak 15 min is

4.HCM uses the peak 15-min rate because

5.V15 = 420 veh, PHF = 0.95. Hourly volume V =

6.Typical urban peak PHF is nearest

7.If all four quarters equal 400 veh, PHF =

8.Design flow v = V/PHF. V = 1800, PHF = 0.75 → v =

9.The 4 in PHF = V/(4 V15) converts 15 min to

10.PHF is dimensionless because