---
title: "Webster signal: 450 and 350 PCU/hr, all-red 15 s"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:webster-two-phase-450-350-pcu"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/webster-two-phase-450-350-pcu
site: BCT Engineering
---

# Webster signal: 450 and 350 PCU/hr, all-red 15 s

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2076 Chaitra, 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-450-350-pcu
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The average normal flow of traffic on the cross roads 1 and 2 during design period are 450 and 350 PCU/hr. The saturation headway on these roads are estimates as 2.5sec and 3.75 sec respectively. The all red time required for pedestrian crossing is 15sec. Design two phase signal by Webster's method with neat phase diagram. Take amber time of 2 sec on each phase for clearance and start-up loss time of 2 sec and 3 sec for roads 1 & 2 respectively.

## Short answer

Saturation flows: 1440 pcu/h (Road 1), 960 pcu/h (Road 2). Y = 0.677, lost time 24 s. Webster optimum cycle = 130 s. Displayed greens: Road 1 = 51 s, Road 2 = 60 s; amber 2 s each, all‑red 15 s.

## Answer

Saturation flow from headway: $s = 3600/h$. Road 1: 3600/2.5 = 1440 PCU/h; Road 2: 3600/3.75 = 960 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 1 | 450 | 1440 | 0.312 |
| Phase 2 | Road 2 | 350 | 960 | 0.365 |

$$
Y = 0.312 + 0.365 = 0.677
$$

#### Step 2: Lost time

Lost time per phase = start-up loss (2 s for road 1, 3 s for road 2) + amber 2 s. The all-red period $R$ = 15 s for pedestrian crossing occurs once in the cycle and is added.

$$
L = (2.0+2.0) + (3.0+2.0) + 15 = 24.0\ \text{s}
$$

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

$$
C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(24.0) + 5}{1 - 0.677} = 127.0\ \text{s}
$$

Adopt **C = 130 s**.

#### Step 4: Green times

Total effective green $= C - L = 130 - 24.0 = 106.0$ s, divided in proportion to $y$:

$$
g_{1} = \frac{y}{Y}(C-L) = \frac{0.312}{0.677}(106.0) = 48.9\ \text{s}
$$

$$
g_{2} = \frac{y}{Y}(C-L) = \frac{0.365}{0.677}(106.0) = 57.1\ \text{s}
$$

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 | 48.9 | 2.0 | 51 | 2 |
| Phase 2 | 57.1 | 3.0 | 60 | 2 |

Check: 51 + 2 + 60 + 2 + 15 = 130 s.

#### Phase diagram

```text
Cycle = 130 s
Road 1:      [ GREEN 51 ][A 2][            RED 77             ]
Road 2:      [     RED 53     ][ GREEN 60 ][A 2][ ALL-RED 15 ]
Pedestrians: [            DON'T WALK 115           ][ 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 saturation headway?
   A. s = h/3600
   B. s = 3600/h
   C. s = 3600 x h
   D. s = h x 60
   Answer: B. s = 3600/h, giving 1440 PCU/h for 2.5 s and 960 PCU/h for 3.75 s. (Key point: Saturation flow from headway)

2. What are the flow ratios and Y?
   A. y1 = 0.312, y2 = 0.365, Y = 0.677
   B. y1 = 0.365, y2 = 0.312, Y = 0.677
   C. y1 = 0.312, y2 = 0.365, Y = 0.365
   D. y1 = 0.45, y2 = 0.35, Y = 0.8
   Answer: A. y1 = 450/1440 = 0.312 and y2 = 350/960 = 0.365, so Y = 0.677. (Key point: Total flow ratio)

3. How is the total lost time L = 24 s obtained?
   A. Only 2 + 3 start-up losses
   B. (2+2) + (3+2) + 15 all-red
   C. 2 + 2 + 15
   D. Amber times only
   Answer: B. Start-up loss plus amber for each phase, plus the 15 s all-red pedestrian period. (Key point: Lost time)

4. What does Webster's formula give?
   A. C0 = (1.5L + 5)/(1 - Y) = about 127 s, adopt 130 s
   B. C0 = (1.5L + 5)/Y = about 60 s
   C. C0 = 24/(1 - Y) = about 74 s
   D. C0 = 1.5L = 36 s
   Answer: A. (1.5x24 + 5)/(1 - 0.677) = 127 s, rounded to 130 s. (Key point: Optimum cycle)

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