---
title: "Two-phase signal: 400 and 250 PCU/hr, 7 m roads"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:webster-two-phase-400-250-width"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/webster-two-phase-400-250-width
site: BCT Engineering
---

# Two-phase signal: 400 and 250 PCU/hr, 7 m roads

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2075 Chaitra, 8 marks
- Video: [Numerical of Webster Method - Traffic Engineering - Transportation Engineering - I](https://www.youtube.com/watch?v=ygZwnOyKUMg) — Ekeeda
- 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-400-250-width
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The average normal flow of traffic on cross roads A and B, of width 7m both, during design period are 400 and 250 PCU/hr; the saturation flow values on these roads are estimated as 1250 and 1000 PCU per hour respectively. The all-red time is provided for pedestrian crossing with speed 1 m/sec and initial walk time 6 sec. Design two phase traffic signal.

## Short answer

All-red time 13 s (6+7/1). Y=0.570, L=17 s, Webster cycle 70.9 s adopted 75 s. Phase 1 green 33 s, amber 2 s; Phase 2 green 25 s, amber 2 s; all-red 13 s.

## Answer

Both roads are 7 m wide. Pedestrian crossing time (all-red period $R$) = initial walk 6 s + width/speed = $6 + 7/1.0 = 13$ s.

#### 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 A | 400 | 1250 | 0.320 |
| Phase 2 | Road B | 250 | 1000 | 0.250 |

$$
Y = 0.320 + 0.250 = 0.570
$$

#### Step 2: Lost time

The lost time and amber are not given. As in the usual Webster design, 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 pedestrian period $R$ = 13 s. Because the lost time equals the amber, the displayed green equals the effective green.

$$
L = (0.0+2.0) + (0.0+2.0) + 13 = 17.0\ \text{s}
$$

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

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(17.0) + 5}{1 - 0.570} = 70.9\ \text{s}
$$

Adopt **C = 75 s**.

#### Step 4: Green times

Total effective green $= C - L = 75 - 17.0 = 58.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.320}{0.570}(58.0) = 32.6\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.250}{0.570}(58.0) = 25.4\ \text{s}
$$

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 | 32.6 | 0.0 | 33 | 2 |
| Phase 2 | 25.4 | 0.0 | 25 | 2 |

Check: 33 + 2 + 25 + 2 + 13 = 75 s.

#### Phase diagram

```text
Cycle = 75 s
Road A:      [ GREEN 33 ][A 2][          RED 40           ]
Road B:      [     RED 35     ][ GREEN 25 ][A 2][ ALL-RED 13 ]
Pedestrians: [          DON'T WALK 62           ][ WALK 13 ]
```

## Check yourself

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

1. How is the all-red time R computed?
   A. Initial walk + width/speed = 6 + 7/1 = 13 s
   B. Width x speed = 7 s
   C. Initial walk only = 6 s
   D. Width/speed only = 7 s
   Answer: A. R = 6 s walk + 7 m / 1 m/s = 13 s. (Key point: Pedestrian all-red)

2. What are y for roads A and B and Y?
   A. 0.25, 0.32, 0.57
   B. 0.32, 0.25, 0.57
   C. 0.32, 0.25, 0.32
   D. 0.40, 0.25, 0.65
   Answer: B. yA = 400/1250 = 0.32, yB = 250/1000 = 0.25, Y = 0.57. (Key point: Flow ratios)

3. What is total lost time L used?
   A. 4 s
   B. 13 s
   C. 17 s
   D. 30 s
   Answer: C. 2 s per phase (4 s) plus the 13 s all-red gives 17 s. (Key point: Lost time)

4. What is C0 and how is green divided?
   A. C0 = 70.9 s, adopt 75 s; effective green (C - L) divided in proportion to y
   B. C0 = 17 s; green equal
   C. C0 = 75 s; green in proportion to saturation flow
   D. C0 = 58 s; green by road width
   Answer: A. C0 = (1.5x17 + 5)/0.43 = 70.9 s; with C = 75 s, C - L = 58 s shared in proportion to y. (Key point: Cycle and green)

## Related in Transportation Engineering II

- [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×
- [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 signal: roads H and F, 3 s and 4 s headways](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Awebster-two-phase-h-f-400) — Transportation Engineering II, chapter 1, asked 2×
- [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: Road 1 14 m, Road 2 7 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-webster-road1-14m-road2-7m) — Transportation Engineering II, chapter 1
- [Signal design: roads X 15 m and Y 10.5 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-design-roads-x-15m-y) — Transportation Engineering II, chapter 1