CIVL2700 · Transport Systems
Signal Timing II: Webster Cycle & Green-Time Allocation
Week 11 of CIVL2700 Transport Systems at the University of Sydney is the design calculation at the heart of traffic-signal engineering: turn the critical flow ratios of a phased intersection into a cycle length (Webster’s optimum for least delay, or the minimum cycle that holds a target degree of saturation), then split the available green so every critical movement runs at the same v/c and finally check the pedestrian minimum green. It builds directly on the clearance-time and saturation-flow ideas of the previous week and is examined as a long, method-marked numerical question in the comprehensive final exam.
What this chapter covers
- 01Effective vs displayed green: g = G + I − l, with intergreen I = Y + AR (yellow + all-red)
- 02Start-up and clearance lost time; total cycle lost time L = Σ lₓ is a fixed overhead
- 03Saturation flow s, flow ratio v/s, and the critical lane group (highest v/s) that sizes each phase
- 04Sum of critical flow ratios Y_c = Σ(v/s)_ci — must be < 1 for the intersection to be signalisable
- 05Webster optimum cycle C_opt = (1.5L + 5)/(1 − Y_c), rounded up to the nearest 5 s
- 06Minimum cycle C_min = L·X_c/(X_c − Σ(v/s)_ci) for a target degree of saturation
- 07Degree of saturation X_c = Y_c·C/(C − L); < 1 within capacity, ≥ 1 oversaturated
- 08Green-time allocation gₓ = (v/s)_ci·(C − L)/Y_c so every critical movement shares one X_c
- 09Converting effective green back to displayed green G = g + l − Y − AR
- 10Pedestrian minimum green (ITE): compare G_p with the parallel vehicle green and take the larger
Webster cycle and green split for a two-phase signal
- +1Sum of critical flow ratios. Y_c = 0.36 + 0.24 = 0.60 (< 1, so the intersection is signalisable).
- +1Total lost time. Two phases at 4 s each: L = Σ lₓ = 2 × 4 = 8 s.
- +2Webster optimum cycle. C_opt = (1.5L + 5)/(1 − Y_c) = (1.5·8 + 5)/(1 − 0.60) = (12 + 5)/0.40 = 17/0.40 = 42.5 s → round UP to the nearest 5 s ⇒ C = 45 s.
- +1Degree of saturation. X_c = Y_c·C/(C − L) = 0.60·45/(45 − 8) = 27/37 = 0.73 (< 1, within capacity).
- +2Effective green. Green available = C − L = 37 s. Split in proportion to the critical ratios, gₓ = (v/s)_ci·(C − L)/Y_c: g_A = 0.36·37/0.60 = 22.2 → 22 s, g_B = 0.24·37/0.60 = 14.8 → 15 s (Σg = 37 = C − L ✓).
- +1Displayed green. G = g + l − Y − AR = g + 4 − 3 − 1 = g (here the lost time equals the intergreen), so G_A = 22 s, G_B = 15 s. Cycle check: ΣG + ΣI = 37 + 2·4 = 45 s = C ✓.
Key terms
- Effective vs displayed green (g, G)
- Displayed green G is the green actually shown on the lantern; effective green g is the equivalent green that would discharge the same vehicles at the full saturation flow, g = G + I − l. Capacity is computed from effective, not displayed, green; the greens sum to C − L, and G = g + l − Y − AR converts back.
- Intergreen and lost time
- The intergreen I = Y + AR is the yellow plus all-red between phases. The lost time l per phase is the start-up plus clearance time that vehicles do not use productively; summed over phases it gives the fixed total cycle lost time L = Σ lₓ.
- Flow ratio (v/s)
- The demand flow of a lane group divided by its saturation flow. The critical lane group in a phase is the one with the highest v/s, and it sizes that phase's green.
- Sum of critical flow ratios (Y_c)
- Y_c = Σ(v/s)_ci, adding the single highest flow ratio in each phase. It measures the intersection's total demand pressure and must be less than 1 for a workable cycle to exist.
- Webster optimum cycle (C_opt)
- The cycle length that minimises average delay, C_opt = (1.5L + 5)/(1 − Y_c), rounded up to the nearest 5 s. Delay is a flat-bottomed U in cycle length, rising steeply below the optimum and only gently above it.
- Minimum cycle (C_min)
- The shortest cycle that keeps the critical movement at or below a target degree of saturation X_c: C_min = L·X_c/(X_c − Σ(v/s)_ci), rounded up to the nearest 5 s. Used when a capacity ceiling, not least delay, is the goal.
- Degree of saturation (X_c)
- The critical volume-to-capacity ratio of the intersection, X_c = Y_c·C/(C − L). Below 1 it operates within capacity (0.85–0.95 is a comfortable target); at or above 1 it is oversaturated and queues grow without bound.
- Pedestrian minimum green (G_p)
- The ITE minimum walk-plus-clearance green for a crossing: G_p = 3.2 + L_cw/S_p + 0.27·N_ped for a narrow crosswalk (W_E ≤ 3.05 m) or 3.2 + L_cw/S_p + 0.81·N_ped/W_E for a wider one. It is a floor: take the larger of G_p and the parallel vehicle green.
Signal Timing II: Webster Cycle & Green-Time Allocation FAQ
When do I use Webster's optimum cycle and when do I use the minimum cycle?
Use Webster's C_opt = (1.5L + 5)/(1 − Y_c) when the question asks you to minimise delay — it sits at the bottom of the delay-versus-cycle U. Use the minimum cycle C_min = L·X_c/(X_c − Σ(v/s)_ci) when you are given a target degree of saturation (a v/c ceiling such as 0.90) and want the shortest cycle that still holds it. Usually C_opt is the longer of the two, and both round up to the nearest 5 s. Watch the denominators: Y_c must be below 1 for C_opt, and the target X_c must exceed Σ(v/s)_ci for C_min to stay positive.
Why does a cycle that is too short make delay worse, not better?
Because the total lost time L is a fixed overhead paid once per cycle no matter how short the cycle is. Shorten the cycle and L becomes a bigger fraction of it, so less of each cycle is usable green; the greens shrink until the critical movements cannot clear their queues and delay climbs steeply. The Webster curve is a flat-bottomed U for exactly this reason — err a little long rather than short. A cycle that is much too long only adds delay gently, through extra red time.
Can AI help me with Webster cycle and green-time allocation in CIVL2700?
Yes — Sia can explain it step by step. Paste a signal-timing question and Sia will walk through summing the critical flow ratios to Y_c, adding the lost time L, sizing the cycle with Webster's formula (rounding up to 5 s), computing the degree of saturation X_c, splitting the green in proportion to the critical ratios, and converting effective green to displayed green — and it will remind you to check Σg = C − L and the pedestrian minimum. Sia teaches the method step by step so you can reproduce it yourself; it never hands over answers to assessed work, completes your assignment for you, or guarantees a grade, and you should confirm the model parameters against your Canvas materials.
Exam move
Drill the design routine until it is automatic and always in the same order: (1) Y_c = Σ(v/s)_ci using one critical ratio per phase, checking Y_c < 1; (2) total lost time L = Σ lₓ; (3) cycle length — Webster's C_opt = (1.5L + 5)/(1 − Y_c) for least delay, or C_min = L·X_c/(X_c − Σ(v/s)_ci) for a target v/c, rounding up to the nearest 5 s; (4) degree of saturation X_c = Y_c·C/(C − L); (5) split the green gₓ = (v/s)_ci·(C − L)/Y_c so Σg = C − L; (6) convert to displayed green G = g + l − Y − AR; (7) check the pedestrian minimum G_p and take the larger of it and the vehicle green. Carry the two built-in checks (Σg = C − L and ΣG + ΣI = C) so an arithmetic slip costs one line, not the whole question, and label every number with its unit and which green it is. Because these questions are method-marked, write each formula before substituting. The final exam runs 2.5 hours and is a hurdle (you must score at least 40% on it to pass the unit); budget time in proportion to marks — about 1.5 minutes per mark on a 100-mark paper — and confirm the exact date, room and open- or closed-book status on Canvas and the University exam timetable.
Working through Signal Timing II: Webster Cycle & Green-Time Allocation in CIVL2700? Sia is AskSia’s AI Engineering tutor — ask any CIVL2700 Signal Timing II: Webster Cycle & Green-Time Allocation question and get a clear, step-by-step explanation grounded in how CIVL2700 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
