Peak Hour Factor and Flow Rates

In traffic analysis, traffic flow is not uniform throughout an hour. A roadway segment may experience high congestion during a brief portion of the hour (e.g., a 15-minute surge) even if the total hourly volume is within the roadway’s theoretical capacity. To account for this variation, the Highway Capacity Manual (HCM) uses the Peak Hour Factor (PHF) to convert hourly volumes into equivalent peak-period flow rates.

The Peak Hour Factor (PHF)

The Peak Hour Factor represents the relationship between the hourly volume and the peak rate of flow within that hour (typically measured in 15-minute intervals).

Formula:

$$PHF = \frac{V}{4 \times V_{15}}$$

Where:

  • $V$ = total hourly volume during the peak hour ($\text{vehicles/hour}$)
  • $V_{15}$ = volume during the single busiest 15-minute interval of the peak hour ($\text{vehicles/15-min}$)
  • $4$ = multiplier to scale the 15-minute count to an equivalent hourly rate ($4 \text{ quarters/hour}$)

Physical Boundaries of PHF:

  • Maximum PHF = 1.00: Occurs when traffic is distributed perfectly evenly across all four 15-minute intervals ($V_{15} = V/4$).
  • Minimum PHF = 0.25: Occurs when all hourly traffic arrives in a single 15-minute interval (with zero vehicles in the other three intervals).
  • Typical Ranges:
    • Urban/congested areas: $0.85$ to $0.98$ (more stable, continuous traffic).
    • Rural/low-volume areas: $0.70$ to $0.85$ (more sporadic traffic).

Peak Flow Rate ($v$)

To evaluate the operational capacity and level of service (LOS) of a facility, we use the equivalent hourly flow rate ($v$) based on the peak 15-minute interval. This represents the rate of flow if the peak 15-minute surge were to persist for a full hour.

$$v = \frac{V}{PHF}$$

Where:

  • $v$ = equivalent peak hourly flow rate ($\text{vehicles/hour}$ or $\text{passenger cars/hour}$)
  • $V$ = total hourly volume ($\text{vehicles/hour}$)
  • $PHF$ = Peak Hour Factor

Note: If the problem asks for flow rate in passenger cars per hour per lane ($\text{pc/h/ln}$ or $\text{pcphpl}$), we must also adjust the volume for lane counts, heavy vehicles ($f_{HV}$), and driver population ($f_p$) using the standard HCM capacity adjustment formulas:

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

Worked Example: Calculating PHF and Peak Flow Rate from 15-Minute Counts

Problem:
A direction of a freeway has the following traffic counts recorded during the peak hour:

  • 5:00 PM to 5:15 PM: $450 \text{ vehicles}$
  • 5:15 PM to 5:30 PM: $520 \text{ vehicles}$
  • 5:30 PM to 5:45 PM: $610 \text{ vehicles}$
  • 5:45 PM to 6:00 PM: $480 \text{ vehicles}$

Calculate:

  1. The total hourly volume ($V$).
  2. The peak 15-minute volume ($V_{15}$).
  3. The Peak Hour Factor (PHF).
  4. The equivalent peak hourly flow rate ($v$).

Solution:

Step 1: Calculate Total Hourly Volume ($V$)
Sum the volumes of the four 15-minute intervals:

$$V = 450 + 520 + 610 + 480 = 2,060 \text{ vehicles}$$

Step 2: Identify Peak 15-Minute Volume ($V_{15}$)
The busiest 15-minute interval is between 5:30 PM and 5:45 PM:

$$V_{15} = 610 \text{ vehicles}$$

Step 3: Calculate the Peak Hour Factor (PHF)

$$PHF = \frac{V}{4 \times V_{15}}$$

$$PHF = \frac{2,060}{4 \times 610} = \frac{2,060}{2,440} \approx 0.844$$

Step 4: Calculate the Peak Flow Rate ($v$)

$$v = \frac{V}{PHF}$$

$$v = \frac{2,060}{0.844} \approx 2,440 \text{ vehicles/hour}$$

Alternative Check:
The equivalent flow rate is simply the peak 15-minute count scaled up to an hour:

$$v = V_{15} \times 4 = 610 \times 4 = 2,440 \text{ vehicles/hour}$$

Both methods yield the same results.

Worked Example: Estimating Peak 15-Minute Volume

Problem:
A highway is carrying an hourly volume of $1,600 \text{ vehicles/hour}$. If the Peak Hour Factor (PHF) is $0.80$, calculate the number of vehicles that pass through the segment during the busiest 15-minute interval.

Solution:

Step 1: Rearrange the PHF equation to solve for $V_{15}$

$$PHF = \frac{V}{4 \times V_{15}} \implies V_{15} = \frac{V}{4 \times PHF}$$

Step 2: Substitute values and solve

  • $V = 1,600 \text{ vehicles}$
  • $PHF = 0.80$ $$V_{15} = \frac{1,600}{4 \times 0.80} = \frac{1,600}{3.2} = 500 \text{ vehicles}$$

Interpretation: During the peak hour, the busiest 15 minutes will see $500$ vehicles. If this rate persisted for the entire hour, the flow rate would be $500 \times 4 = 2,000 \text{ vehicles/hour}$, which is significantly higher than the actual hourly volume of $1,600$ vehicles.

Crucial Pitfalls and Exam Traps

  • Mathematical Bounds Check: The Peak Hour Factor must always be between $0.25$ and $1.00$. If you calculate a PHF outside this range (e.g. $1.15$), you have made an error (often putting the peak 15-minute rate in the numerator).
  • Hourly Volume vs. Peak Flow Rate in Capacity Analysis: When using HCM methods (e.g., finding the level of service of a freeway segment or a signalized intersection), never plug the raw hourly volume ($V$) directly into capacity equations. You must divide it by the PHF to find the peak flow rate ($v$). Using the unadjusted hourly volume will underestimate congestion and overestimate level of service.
  • Data Identification: Ensure that the 15-minute count you use is the maximum value of the four intervals, not the first or the last.