---
title: "Webster signal: roads H and F, 3 s and 4 s headways"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:webster-two-phase-h-f-400"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/webster-two-phase-h-f-400
site: BCT Engineering
---

# Webster signal: roads H and F, 3 s and 4 s headways

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2073 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/webster-two-phase-h-f-400
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The average normal flow of traffic on cross roads H and F during design period are 400 and 250 PCU per hour, the saturated headway on these roads are estimated as 3 secs and 4 secs respectively. The all red time required for pedestrian crossing is 15 secs. Design two phase traffic signal by Webster's method.

## Short answer

Saturation flows are 1200 and 900 PCU/h, giving Y=0.611. With assumed 2 s amber/lost time per phase and 15 s all-red, L=19 s, so optimum cycle 86.1 s, adopt 90 s. Effective green 71 s split as 38.7 and 32.3 s; displayed greens 39 and 32 s plus 2 s ambers.

## Answer

Saturation flow from the saturation headway: road H $s = 3600/3 = 1200$ PCU/h; road F $s = 3600/4 = 900$ PCU/h.

#### Step 1: Flow ratios
Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one.

| Phase | Approach | q | s | y = q/s |
|---|---|---|---|---|
| Phase 1 | Road H | 400 | 1200 | 0.333 |
| Phase 2 | Road F | 250 | 900 | 0.278 |

$$
Y = 0.333 + 0.278 = 0.611
$$

#### Step 2: Lost time

The lost time and amber are not given. An amber of 2 s per phase is assumed and the lost time per phase is taken as 2 s (equal to the amber), plus the all-red period for pedestrians $R$ = 15 s per cycle.

$$
L = (0.0+2.0) + (0.0+2.0) + 15 = 19.0\ \text{s}
$$

#### Step 3: Optimum cycle time (Webster)

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(19.0) + 5}{1 - 0.611} = 86.1\ \text{s}
$$

Adopt **C = 90 s**.

#### Step 4: Green times

Total effective green $= C - L = 90 - 19.0 = 71.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.333}{0.611}(71.0) = 38.7\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.278}{0.611}(71.0) = 32.3\ \text{s}
$$

Displayed green $G = g + $ start-up loss (so that $\sum G + \sum I + R = C$):

| Phase | Effective g (s) | Start-up loss (s) | Green G (s) | Amber (s) |
|---|---|---|---|---|
| Phase 1 | 38.7 | 0.0 | 39 | 2 |
| Phase 2 | 32.3 | 0.0 | 32 | 2 |

Check: 39 + 2 + 32 + 2 + 15 = 90 s.

#### Phase diagram

```text
Cycle = 90 s
Road H:      [ GREEN 39 ][A 2][          RED 49           ]
Road F:      [     RED 41     ][ GREEN 32 ][A 2][ ALL-RED 15 ]
Pedestrians: [          DON'T WALK 75           ][ WALK 15 ]
```

## Check yourself

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

1. How is saturation flow found from a saturated headway of 3 s?
   A. 3 x 3600 = 10800 PCU/h
   B. 3600/3 = 1200 PCU/h
   C. 60/3 = 20 PCU/h
   D. 3/3600 PCU/h
   Answer: B. Saturation flow is the number of vehicles per hour at saturated headway: 3600/h. (Key point: Saturation flow from headway)

2. What is Y for road H (400/1200) and road F (250/900)?
   A. 0.333 + 0.278 = 0.611
   B. 0.400 + 0.250 = 0.650
   C. 0.333 x 0.278 = 0.093
   D. 0.611/2 = 0.306
   Answer: A. y = q/s gives 0.333 and 0.278; Y is their sum, 0.611. (Key point: Flow ratio and Y)

3. How is total lost time L = 19 s obtained?
   A. 15 s all-red only
   B. 2 s per phase for two phases plus 15 s pedestrian all-red
   C. Amber 2 s plus 15 s
   D. 19 s is given in the question
   Answer: B. L = 2 + 2 (per-phase lost time including amber) + 15 (all red) = 19 s. (Key point: Lost time)

4. What is Webster's optimum cycle time here?
   A. C = (1.5L + 5)/(1 - Y) = 86.1 s, adopted as 90 s
   B. C = 1.5L + 5 = 33.5 s
   C. C = L/(1 - Y) = 48.8 s
   D. C = (1.5L + 5)(1 - Y) = 12.9 s
   Answer: A. C0 = (1.5 x 19 + 5)/(1 - 0.611) = 33.5/0.389 = 86.1 s, rounded to 90 s. (Key point: Webster cycle time)

## Close variants set in other exams

- [Intersection control methods; Webster signal 400/300 PCU](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aintersection-control-methods-webster-400-300) — Transportation Engineering II, chapter 1, asked 2×

## Related in Transportation Engineering II

- [Two-phase signal: 400 and 250 PCU/hr, 7 m roads](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Awebster-two-phase-400-250-width) — Transportation Engineering II, chapter 1
- [Webster signal: 450 and 350 PCU/hr, all-red 15 s](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Awebster-two-phase-450-350-pcu) — Transportation Engineering II, chapter 1
- [Webster two-phase signal with saturation headways](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Awebster-two-phase-four-legged-table) — Transportation Engineering II, chapter 1
- [Signal timing: roads A 15 m and B 7 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-pedestrian-road-a-15m-7m) — Transportation Engineering II, chapter 1