Traffic Signals

How the model works

Each road gets an effective green time g in a cycle of length C. Every phase change also loses L = 4 s while the junction clears, so C = g₁ + g₂ + 2L. While a road has green and a queue, cars leave at the saturation flow s = 1800 cars/h (one every 2 s). So a road can serve at most

capacity = s × g / C degree of saturation x = q / capacity

where q is the arrival rate. If x ≥ 1 the queue grows every cycle and never clears. Below that, Webster's formula estimates the average delay per car:

d ≈ 0.9 × [ C(1 − g/C)² / (2(1 − x g/C)) + x² / (2q(1 − x)) ]

The first term is the wait caused by red; the second is the extra delay from random arrivals. Short cycles waste a larger share of time on changes; long cycles make cars wait through long reds. Webster's optimum cycle is roughly C₀ = (1.5 × total lost time + 5) / (1 − Σ q/s).

One lane per road, no turning, Poisson arrivals, and cars leave at a constant headway. Webster's formula is an approximation and usually reads a little below simulation. Averages need several minutes of simulated time to settle.

Guided experiments

1. Balanced junction. 500 cars/h on each road, 25 s green each (cycle 58 s). Prediction: x ≈ 0.64 on both roads and average delay around 15–16 s per car. Queues reach 5–10 cars at the end of red and clear every green. Balanced
2. Unequal demand, equal greens. 900 cars/h north–south, 300 cars/h east–west, still 25 s each. Prediction: north–south capacity is only about 776 cars/h, so x ≈ 1.16 and its queue grows by about 2 cars every minute without limit, while east–west sits mostly empty. Equal greens
3. Re-time the signal. Same traffic, 35 s north–south and 15 s east–west. Prediction: both roads are now under capacity (x ≈ 0.83 and 0.64). North–south delay drops to roughly 15–20 s (it is near capacity, so the average wanders); east–west rises to about 23–25 s, a fair trade for a stable junction. Re-timed
4. Long cycle. Balanced traffic with 50 s green each (cycle 108 s). Prediction: delay rises to about 22–24 s per car even though capacity barely changes. Webster's best cycle for this traffic is about 38 s. Long cycle

Challenge

A junction has 700 cars/h on the main road and 350 cars/h on the side road, with s = 1800 cars/h and 4 s lost per phase. With a 60 s cycle, roughly how should the green time be split so that both roads have the same degree of saturation?

References

• Webster, F. V. (1958), Traffic Signal Settings, Road Research Technical Paper 39, HMSO.
• Transportation Research Board, Highway Capacity Manual, 7th ed. (2022), signalized intersections chapter.
• Roess, R. P., Prassas, E. S. and McShane, W. R., Traffic Engineering, 5th ed. (2019), ch. 18–19.