---
title: "Webster two-phase signal with saturation headways"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:webster-two-phase-four-legged-table"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/webster-two-phase-four-legged-table
site: BCT Engineering
---

# Webster two-phase signal with saturation headways

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2079 Bhadra, 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-four-legged-table
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The design hour traffic and saturation headway at a four-legged intersection are:

| Approach | North (N) | South (S) | East (E) | West (W) |
|---|---|---|---|---|
| Design hour flow | 450 | 430 | 350 | 315 |
| Saturation headway (sec) | 2.25 | 2.33 | 2.5 | 2.75 |

Design a two phase signal for the intersection based on Webster method and draw the phase diagram. Use amber time of 2 sec on each phase for clearance and start-up loss time phase per phase of 1.5 sec.

## Answer

Phases: Phase 1 = North-South, Phase 2 = East-West. Saturation flow from saturation headway $s = 3600/h$.


#### 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 | North | 450 | 1600 | 0.281 (critical) |
| Phase 1 | South | 430 | 1545 | 0.278 |
| Phase 2 | East | 350 | 1440 | 0.243 (critical) |
| Phase 2 | West | 315 | 1309 | 0.241 |

$$
Y = 0.281 + 0.243 = 0.524
$$

#### Step 2: Lost time

Lost time per phase = start-up loss 1.5 s + amber 2 s = 3.5 s.

$$
L = (1.5+2.0) + (1.5+2.0) = 7.0\ \text{s}
$$

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

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(7.0) + 5}{1 - 0.524} = 32.6\ \text{s}
$$

Adopt **C = 35 s**.

#### Step 4: Green times

Total effective green $= C - L = 35 - 7.0 = 28.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.281}{0.524}(28.0) = 15.0\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.243}{0.524}(28.0) = 13.0\ \text{s}
$$

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 | 15.0 | 1.5 | 17 | 2 |
| Phase 2 | 13.0 | 1.5 | 14 | 2 |

Check: 17 + 2 + 14 + 2 = 35 s.

#### Phase diagram

```text
|      P1 green 17      |  A2 |    P2 green 14    |  A2 |
```

## Check yourself

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

1. How is saturation flow found from saturation headway h?
   A. s = 3600/h
   B. s = h/3600
   C. s = 60/h
   D. s = h x 3600
   Answer: A. For North, s = 3600/2.25 = 1600 veh/h. (Key point: Saturation flow)

2. What is Y for the intersection?
   A. 0.281 + 0.278 + 0.243 + 0.241 = 1.04
   B. 0.281 + 0.243 = 0.524
   C. 0.281 + 0.278 = 0.559
   D. 0.243 + 0.241 = 0.484
   Answer: B. Y is the sum of critical y of each phase: N-S 0.281 and E-W 0.243. (Key point: Critical ratios)

3. With start-up loss 1.5 s and amber 2 s per phase, L is:
   A. 3.5 s
   B. 7.0 s
   C. 4.0 s
   D. 14 s
   Answer: B. (1.5 + 2) per phase for two phases gives 7.0 s. (Key point: Lost time)

4. C0 = (1.5L + 5)/(1 - Y) = ?
   A. About 32.6 s, adopted as 35 s
   B. About 15 s
   C. About 70 s
   D. About 12 s
   Answer: A. (1.5 x 7 + 5)/(1 - 0.524) = 15.5/0.476 = 32.6 s. (Key point: Optimum cycle)

5. The displayed greens are:
   A. Phase 1 = 17 s, Phase 2 = 14 s
   B. Phase 1 = 15 s, Phase 2 = 13 s
   C. Phase 1 = 14 s, Phase 2 = 17 s
   D. Phase 1 = 28 s, Phase 2 = 28 s
   Answer: A. Effective g = 15 and 13 s are split by y/Y, then 1.5 s start-up loss added and rounded: 17 s and 14 s, total 17 + 2 + 14 + 2 = 35. (Key point: Green times)

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