---
title: "Fixed-time two-phase signal: optimum cycle and greens"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:fixed-time-two-phase-signal-optimum"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/fixed-time-two-phase-signal-optimum
site: BCT Engineering
---

# Fixed-time two-phase signal: optimum cycle and greens

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2075 Asoj, 8 marks
- Video: [Numerical of Webster method - Traffic Engineering - Transportation Engineering - GATE](https://www.youtube.com/watch?v=Wm5UYbOf3Dg) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/fixed-time-two-phase-signal-optimum
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

A four-legged right angled intersection is to be signalized with a fixed time 2-phase signal. The design hour flow and saturation flow are as under:

| | North (N) | South (S) | East (E) | West (W) |
|---|---|---|---|---|
| Design hour flow | 900 | 500 | 800 | 700 |
| Saturation flow | 2500 | 2000 | 3200 | 3000 |

The lost time is 2 seconds per phase due to starting delays and amber time for north-south and east-west are 3 seconds and 4 seconds respectively. Determine the optimum cycle time. Allocate the green times to the two phases.

## Short answer

Y=0.610, L=11 s, optimum Webster cycle 55.1 s, adopted 60 s. Phase 1 (N-S) green 31 s, amber 3 s; Phase 2 (E-W) green 22 s, amber 4 s.

## Answer

Phase 1 = North-South, Phase 2 = East-West.


#### 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 | 900 | 2500 | 0.360 (critical) |
| Phase 1 | South | 500 | 2000 | 0.250 |
| Phase 2 | East | 800 | 3200 | 0.250 (critical) |
| Phase 2 | West | 700 | 3000 | 0.233 |

$$
Y = 0.360 + 0.250 = 0.610
$$

#### Step 2: Lost time

Lost time per phase = start-up loss 2 s + amber (3 s for N-S, 4 s for E-W).

$$
L = (2.0+3.0) + (2.0+4.0) = 11.0\ \text{s}
$$

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

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(11.0) + 5}{1 - 0.610} = 55.1\ \text{s}
$$

Adopt **C = 60 s**.

#### Step 4: Green times

Total effective green $= C - L = 60 - 11.0 = 49.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.360}{0.610}(49.0) = 28.9\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.250}{0.610}(49.0) = 20.1\ \text{s}
$$

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 | 28.9 | 2.0 | 31 | 3 |
| Phase 2 | 20.1 | 2.0 | 22 | 4 |

Check: 31 + 3 + 22 + 4 = 60 s.

#### Phase diagram

```text
|       P1 green 31       | A3 |   P2 green 22   |  A4  |
```

## Check yourself

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

1. How is the critical y chosen in each phase?
   A. The smaller y
   B. The larger y of the phase
   C. The sum of both
   D. The average
   Answer: B. The approach with the larger q/s controls the phase: North 0.360 and East 0.250. (Key point: Critical flow ratio)

2. What is Y?
   A. 0.610
   B. 0.483
   C. 1.093
   D. 0.360
   Answer: A. Y = 0.360 + 0.250 = 0.610. (Key point: Total Y)

3. What is L here?
   A. 4 s
   B. 7 s
   C. 11 s
   D. 15 s
   Answer: C. (2+3) + (2+4) = 11 s. (Key point: Lost time)

4. What is the optimum cycle and its adopted value?
   A. 55.1 s, adopt 60 s
   B. 27 s, adopt 30 s
   C. 110 s, adopt 120 s
   D. 16.5 s, adopt 20 s
   Answer: A. C0 = (1.5x11 + 5)/(1 - 0.61) = 55.1 s, adopt 60 s. (Key point: Cycle time)

## Related in Transportation Engineering II

- [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
- [Two-phase signal cycle length and green times](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-cycle-length-two-phase-table) — Transportation Engineering II, chapter 1
- [Skid numerical: A from North, B from West](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-north-west-47-56) — Transportation Engineering II, chapter 1, asked 3×
- [Skid numerical: A from West, B from South (45, 50)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-45-50) — Transportation Engineering II, chapter 1, asked 2×
- [Skid numerical: A from West, B from South (51, 61)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-51-61) — Transportation Engineering II, chapter 1, asked 2×
- [Skid numerical: A from West, B from South (49, 27)](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-west-south-49-27) — Transportation Engineering II, chapter 1

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