---
title: "Peak hour volume and PHF from counts"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:peak-hour-volume-phf-urban-approach"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/peak-hour-volume-phf-urban-approach
site: BCT Engineering
---

# Peak hour volume and PHF from counts

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2080 Baisakh, 4 marks
- Video: [Peak Hour Factor (PHF) on a road and its calculation and application. Why do we need PHF? watch it.](https://www.youtube.com/watch?v=nQkAJikoL4w) — Transportation Engg Talks | Satish IITR
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/peak-hour-volume-phf-urban-approach
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Table below shows volume counts on an approach of an urban intersection from 8.30 AM to 10.30 AM.

| Time | Count (PCU) |
|---|---|
| 8:30 - 8:45 | 500 |
| 8:45 - 9:00 | 390 |
| 9:00 - 9:15 | 300 |
| 9:15 - 9:30 | 300 |
| 9:30 - 9:45 | 260 |
| 9:45 - 10:00 | 350 |
| 10:00 - 10:15 | 400 |
| 10:15 - 10:30 | 490 |

Determine (i) peak hour (ii) peak hourly volume (iii) peak hour factor, and (iv) the design flow rate for the approach.

## Short answer

Peak hour is 9:30-10:30 AM with peak hourly volume 1500 PCU. PHF = 1500/(4x490) = 0.765, and design flow rate = 1500/0.765 = 1960 PCU/h.

## Answer

The peak hour is found by adding the 15-minute counts over every four consecutive intervals (a sliding one-hour total) and choosing the largest.

| One-hour period | Volume (PCU) |
|---|---|
| 8:30 - 9:30 | 1490 |
| 8:45 - 9:45 | 1250 |
| 9:00 - 10:00 | 1210 |
| 9:15 - 10:15 | 1310 |
| 9:30 - 10:30 | 1500 |

#### (i) Peak hour
The largest one-hour total is 1500 PCU, so the peak hour is **9:30 - 10:30 AM**.

#### (ii) Peak hourly volume
**Peak hour volume = 1500 PCU/h**

#### (iii) Peak hour factor
The highest 15-minute count within the peak hour is 490 PCU (10:15 - 10:30).

$$
PHF = \frac{V_{peak\ hour}}{4 \times V_{15,max}} = \frac{1500}{4 \times 490} = 0.765
$$

#### (iv) Design flow rate
$$
v = \frac{V}{PHF} = \frac{1500}{0.765} = 1960\ \text{PCU/h}\quad (= 4 \times 490)
$$

**Answer: peak hour 9:30-10:30 AM, peak hour volume 1500 PCU, PHF = 0.765, design flow rate = 1960 PCU/h.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the peak hour found from 15-minute counts?
   A. Pick the hour starting at the first count
   B. Sum every four consecutive intervals and choose the largest total
   C. Take the largest single 15-minute count times 4
   D. Average all counts
   Answer: B. A sliding one-hour total is computed and the largest is the peak hour. (Key point: Finding the peak hour)

2. Which hour has the peak volume and what is it?
   A. 8:30-9:30, 1490 PCU
   B. 9:00-10:00, 1210 PCU
   C. 9:30-10:30, 1500 PCU
   D. 9:15-10:15, 1310 PCU
   Answer: C. The sliding totals show 9:30-10:30 as the largest at 1500 PCU. (Key point: Peak hour result)

3. PHF is calculated as:
   A. V / (4 x V15,max) = 1500/(4 x 490) = 0.765
   B. V15,max / V = 0.33
   C. V / 4 = 375
   D. 4 x V / V15,max = 12.2
   Answer: A. PHF = peak hour volume divided by four times the highest 15-minute count within that hour. (Key point: Peak hour factor)

4. The design flow rate is obtained as:
   A. Peak volume times PHF
   B. Peak volume divided by PHF, about 1960 PCU/h
   C. Peak volume minus 15-min maximum
   D. The highest 15-min count only
   Answer: B. v = V/PHF = 1500/0.765 = 1960 PCU/h, equal to 4 x 490. (Key point: Design flow rate)

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