The trial cycle method asks you to guess a cycle time and check it. Webster’s method does something better — it calculates the cycle time that produces the least delay, directly.
That is why it is described as the most rational method of design, and why it appears more often than any other signal design question in GATE, SSC JE and RRB JE.
The Idea Behind the Method
In Webster’s method, the optimum cycle time is calculated from the least total delay at a signalised intersection.
Why should there be an optimum at all? Because delay increases at both extremes.
- A very short cycle wastes time. Every phase change costs a fixed amount of lost time — vehicles taking a moment to start moving, and the amber period. Cycle rapidly and you pay that overhead again and again.
- A very long cycle also wastes time. Vehicles arriving just after their green ends must wait through the entire remaining cycle before getting another turn.
Between these two failures lies a cycle length at which total delay is minimum. Webster’s formula finds it.
The Optimum Cycle Time Formula
C0 = (1.5 L + 5) / (1 − Y)
where C0 is the optimum cycle time in seconds.
Two quantities feed into it — L, the total lost time, and Y, the sum of the flow ratios. Each needs its own explanation.
Lost Time, L
L = n tL + R
| Symbol | Meaning |
|---|---|
| n | Number of phases |
| tL | Start-up loss time, plus loss due to the falling discharge rate during the amber period |
| R | All red time |
For the average signal cycle, the lost time tL amounts to around 2 seconds. Substituting this gives the working form:
L = 2n + R
Where Lost Time Comes From
The definition of tL names two separate losses, and both are real.
Start-up loss. When the light turns green, the queue does not move off instantly. The first driver reacts, then the second, and the line takes a few seconds to reach full discharge rate. Those seconds of green are paid for but not used at full efficiency.
Falling discharge during amber. At the other end of the phase, drivers begin hesitating as amber appears. Flow tails off before the light actually turns red, so the last part of the phase is also not fully used.
All red time (R) is the additional period when every approach shows red simultaneously, provided so the junction can clear completely between phases.
Lost time is therefore the unavoidable overhead of switching between phases — and note that it is multiplied by n. Every extra phase adds about 2 more seconds of pure waste, which is a strong argument for keeping the number of phases small.
Flow Ratios, y and Y
The flow ratio for each phase compares the traffic actually arriving with the maximum the approach could discharge:
y1 = q1 / S1 y2 = q2 / S2
Y = y1 + y2 + …
| Symbol | Meaning |
|---|---|
| q1 | Critical lane volume for phase one — the maximum volume per lane |
| S1 | Saturation flow for phase one |
| Y | Sum of the flow ratios across all phases |
Note that q is the critical lane volume — the busiest lane, not the average. The signal must serve the worst-loaded lane, since clearing that one clears the rest.
Saturation Flow
S (veh/hr) = 3600 / time headway (sec)
Saturation flow is the rate at which vehicles discharge from a queue once it is moving freely. If vehicles cross at 2 second headways, then 3600/2 = 1,800 vehicles per hour could pass if the light stayed green permanently.
The Critical Condition
Y should not be greater than one.
This limit is worth understanding rather than memorising. Each y is the fraction of a cycle that phase needs to clear its traffic. If the fractions sum to more than 1, the phases collectively need more than the whole cycle — which is impossible. The intersection is over capacity, and no signal timing can fix it.
The formula reflects this directly. As Y approaches 1, the denominator (1 − Y) approaches zero and C0 shoots towards infinity. The mathematics is telling you the junction cannot cope.
Saturation Flow Values
Saturation flow may be taken from the following table:
| Road Width | Saturation Flow for Full Carriageway (PCU/hr) |
|---|---|
| 3 m | 1850 |
| 3.5 m | 1890 |
| 4 m | 1950 |
| 4.5 m | 2250 |
| 5 m | 2350 |
| 5.5 m | 2990 |
| Above 5.5 m | 525 per metre width |
Notice the pattern. Between 3 m and 4 m the values rise only slightly — a road that wide still carries a single line of vehicles, so extra width adds little. From 4.5 m onward the jumps grow, because the width begins to permit vehicles to move side by side. Beyond 5.5 m the relationship becomes a simple 525 PCU/hr per metre.
Distributing the Green Time
Once the optimum cycle is known, the usable green time — the cycle minus the lost time — is shared between phases in proportion to their flow ratios:
G1 = (C0 − L) × y1 / (y1 + y2 + …)
G2 = (C0 − L) × y2 / (y1 + y2 + …)
Read the structure and the fairness of it becomes obvious. (C0 − L) is the green time actually available after paying the lost-time overhead. Each phase then receives a share equal to its own flow ratio divided by the total — so a phase carrying twice the relative load gets twice the green.
Since the denominator is Y itself, the expressions can also be written as G1 = (C0 − L) y1/Y.
The Complete Procedure
- Find the critical lane volume q for each phase.
- Find the saturation flow S for each phase, from the table or from 3600/headway.
- Compute each y = q/S, and sum them to get Y. Confirm Y < 1.
- Compute L = 2n + R.
- Compute C0 = (1.5L + 5)/(1 − Y).
- Distribute (C0 − L) between phases in proportion to their y values.
Formula Summary
| Quantity | Expression |
|---|---|
| Optimum cycle time | C0 = (1.5L + 5)/(1 − Y) |
| Lost time | L = n tL + R = 2n + R |
| Flow ratio | y = q / S |
| Sum of flow ratios | Y = y1 + y2 + …, must be less than 1 |
| Saturation flow | S = 3600 / time headway |
| Green time for a phase | G = (C0 − L) × y / Y |
| Lost time per phase | About 2 sec |
Quick Revision Notes
- Webster’s method finds the cycle time giving least total delay.
- It is described as the most rational method of design.
- C0 = (1.5L + 5)/(1 − Y).
- L = n tL + R, and with tL about 2 sec, L = 2n + R.
- tL covers start-up loss and falling discharge rate during amber.
- R is the all red time.
- y = q/S, with q the critical lane volume and S the saturation flow.
- Y = sum of all y values, and must not exceed 1.
- S = 3600 / time headway.
- Green time is shared in proportion to y: G = (C0 − L) y / Y.
- Saturation flow above 5.5 m width = 525 PCU/hr per metre.
Mistakes Students Commonly Make
- Writing the numerator as 1.5L − 5 or 1.5(L + 5). It is 1.5L + 5.
- Forgetting to multiply tL by the number of phases. Lost time is 2n + R, not 2 + R.
- Leaving out the all red time R from L.
- Using the average lane volume instead of the critical (maximum) lane volume for q.
- Distributing the whole cycle C0 between phases instead of (C0 − L). Lost time must be deducted first.
- Proceeding when Y comes out greater than 1. That signals an over-capacity intersection, and the formula breaks down.
- Inverting the saturation flow relation. It is 3600 divided by headway, not headway divided by 3600.
Conclusion
Webster’s method turns signal design from guesswork into calculation. Total delay rises if the cycle is too short, because lost time is paid too often, and rises again if it is too long, because waiting vehicles sit through longer reds. The formula C0 = (1.5L + 5)/(1 − Y) locates the minimum between those two, using just two inputs — the lost time overhead and the sum of flow ratios. Get L and Y right, remember to deduct L before distributing green, and check that Y stays below 1, and every Webster numerical follows.
Frequently Asked Questions
What is the Webster method?
A signal design method in which the optimum cycle time is calculated from the condition of least total delay at a signalised intersection. It is regarded as the most rational method of signal design.
What is the Webster formula for optimum cycle time?
C0 = (1.5L + 5)/(1 − Y), where L is the total lost time in seconds and Y is the sum of the flow ratios.
How is lost time calculated?
L = n tL + R, where n is the number of phases, tL is the lost time per phase and R is the all red time. Taking tL as about 2 seconds gives L = 2n + R.
What causes lost time?
Start-up loss when a queue takes time to reach full discharge rate after the green appears, and the falling discharge rate during the amber period as drivers begin to hesitate.
What is the flow ratio y?
The critical lane volume divided by the saturation flow for that phase, y = q/S.
Why must Y be less than one?
Because each y represents the fraction of the cycle that phase needs. If the fractions sum to more than one, the phases together need more than a whole cycle, which is impossible — the intersection is beyond capacity.
How is saturation flow calculated?
S = 3600 divided by the time headway in seconds, giving vehicles per hour.
What is the saturation flow for a road wider than 5.5 m?
525 PCU per hour per metre of width.
How is green time distributed among phases?
The available green, (C0 − L), is shared in proportion to the flow ratios, so G = (C0 − L) × y/Y for each phase.
Why is lost time subtracted before distributing green?
Because lost time is an unavoidable overhead of switching between phases and is not available as effective green. Only the remainder can be shared between the traffic movements.