---
title: "Two-phase signal cycle length and green times"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-cycle-length-two-phase-table"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-cycle-length-two-phase-table
site: BCT Engineering
---

# Two-phase signal cycle length and green times

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2080 Bhadra, 8 marks
- Video: [Two-phase traffic signals by webster method (Math)](https://www.youtube.com/watch?v=W3Taijdxp1Q) — Zahid hasan Prince
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-cycle-length-two-phase-table
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Using the data in the table below, come up with cycle length and actual green for all 2 phase signal. Assume lost time per phase = 3 seconds, yellow time of 3 seconds and all red time of 4 sec.

| Veh/hr | East | North | West | South |
|---|---|---|---|---|
| Flow | 488 | 338 | 115 | 217 |
| Saturation flow | 1725 | 1725 | 1725 | 1725 |

## Short answer

Critical flow ratios are 0.283 for EW and 0.196 for NS, Y=0.479. Lost time L=20 s; Webster cycle C=70 s, effective greens 29.5 and 20.5 s; displayed greens 33 s and 23 s with 7 s clearance each.

## Answer

Two phases: Phase 1 = East-West, Phase 2 = North-South. The question asks for the cycle length and the actual (displayed) green of each phase.


#### 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 | East | 488 | 1725 | 0.283 (critical) |
| Phase 1 | West | 115 | 1725 | 0.067 |
| Phase 2 | North | 338 | 1725 | 0.196 (critical) |
| Phase 2 | South | 217 | 1725 | 0.126 |

$$
Y = 0.283 + 0.196 = 0.479
$$

#### Step 2: Lost time

For each phase: start-up lost time 3 s, plus yellow 3 s and all-red 4 s (clearance interval $I$ = 7 s), all treated as lost to traffic movement.

$$
L = (3.0+7.0) + (3.0+7.0) = 20.0\ \text{s}
$$

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

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(20.0) + 5}{1 - 0.479} = 67.2\ \text{s}
$$

Adopt **C = 70 s**.

#### Step 4: Green times

Total effective green $= C - L = 70 - 20.0 = 50.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.283}{0.479}(50.0) = 29.5\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.196}{0.479}(50.0) = 20.5\ \text{s}
$$

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 + all-red (s) |
|---|---|---|---|---|
| Phase 1 | 29.5 | 3.0 | 33 | 7 |
| Phase 2 | 20.5 | 3.0 | 23 | 7 |

Check: 33 + 7 + 23 + 7 = 70 s.

#### Phase diagram

```text
|     P1 green 33     |   A7   | P2 green 23  |   A7   |
```

## Check yourself

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

1. Which flow ratio of each phase is used to find Y?
   A. The average of both approaches
   B. The larger (critical) y of the phase, i.e. East 0.283 and North 0.196
   C. The sum of all four approaches
   D. The smaller y of each phase
   Answer: B. Y is the sum of critical flow ratios: 0.283 + 0.196 = 0.479. (Key point: Critical flow ratio)

2. With start-up lost time 3 s, yellow 3 s and all-red 4 s per phase, total lost time L for two phases is:
   A. 10 s
   B. 14 s
   C. 20 s
   D. 26 s
   Answer: C. Each phase loses 3 + 3 + 4 = 10 s, so L = 20 s. (Key point: Total lost time)

3. Which result follows from C0 = (1.5L + 5)/(1 - Y) with L = 20 s and Y = 0.479?
   A. About 67 s, adopted as 70 s
   B. About 35 s, adopted as 35 s
   C. About 48 s, adopted as 50 s
   D. About 100 s, adopted as 100 s
   Answer: A. C0 = 35/0.521 = 67.2 s, rounded to 70 s. (Key point: Webster cycle length)

4. How is the total effective green of 50 s divided between phases?
   A. Equally, 25 s each
   B. In proportion to the critical flow ratios y/Y
   C. In proportion to the amber times
   D. In proportion to the saturation flows only
   Answer: B. g = (y/Y)(C - L), giving 29.5 s for Phase 1 and 20.5 s for Phase 2. (Key point: Green split)

5. How are the displayed greens of 33 s and 23 s obtained?
   A. Effective green plus start-up loss, rounded (29.5 + 3, 20.5 + 3)
   B. Effective green minus start-up loss
   C. Effective green plus amber and all-red
   D. Cycle length divided by 2
   Answer: A. G = g + start-up loss so that sum G + sum I = C: 33 + 7 + 23 + 7 = 70 s. (Key point: Displayed green)

## Related in Transportation Engineering II

- [Fixed-time two-phase signal: optimum cycle and greens](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afixed-time-two-phase-signal-optimum) — 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
- [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 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
- [Time and space mean speed from 14 travel times](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Atime-space-mean-speed-14-vehicles) — Transportation Engineering II, chapter 1

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