Question 2 of 7: Webster signal design for a four-approach intersection

Question 2: Webster signal design for a four-approach intersection (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Peak-hour approach volumes, conflicting pedestrian volumes and the peak-hour factor are tabulated on page 2; the saturation flows by lane type are tabulated on page 3. Every approach is 15 m wide curb to curb.

Find. An appropriate phasing system with its justification, the intersection geometry it implies, the optimum and adopted cycle length, and the green time given to each phase.

Approach. Convert the hourly volumes to peak-flow rates, settle the geometry from the stated width, price the candidate phasing plans against the sum of flow ratios before choosing one, balance the two lanes on each approach so both carry the same flow ratio, form Webster’s optimum cycle, test it against the pedestrian requirement of each phase, adopt the larger, and split the available green in proportion to the critical flow ratios.

1. Peak-flow rates. The peak-hour factor converts an hourly volume to the flow rate that the signal must actually serve within the peak fifteen minutes, $$q = \frac{V}{\text{PHF}}$$ With PHF = 0.75 throughout, every movement is divided by 0.75. The north approach, for example, becomes 266.67 left, 1066.67 through and 160.00 right, a total of 1493.33 veh/h; south, east and west come to 1446.67, 1226.67 and 1126.67 veh/h respectively. All subsequent arithmetic uses these flow rates, never the raw volumes.
2. Geometry implied by the 15 m width. The "(Width)" entry in the volume table is the full curb-to-curb width shared by both directions. At the TAC standard 3.5 m lane, 15 m is four lanes plus a 1.0 m median, i.e. two lanes per approach. Reading 15 m as a one-direction width would imply four lanes each way carrying only about 300 veh/h per lane, which no signal design would ever produce. This is declared as an assumption under the paper’s Note 1.
3. Price the candidate phasing plans before choosing one. "Determine an appropriate phasing system" asks for the justification, not just the plan, so each option is tested against the sum of critical flow ratios \(Y = \sum y_i\), which must stay below 1 for any timing to exist. A four-phase plan with exclusive left-turn bays leaves one through-right lane per approach carrying the whole through-plus-right demand, giving $$Y_{4} = 0.1337 + 0.5452 + 0.1337 + 0.4267 = 1.2392 \gt 1$$ A two-phase plan with one shared left-through-right lane per approach (s = 2500) gives $$Y_{1\text{-lane}} = 0.5973 + 0.4907 = 1.0880 \gt 1$$ Both are infeasible: no cycle length whatever can serve them. The remaining option — two phases, two lanes per approach, permitted left turns — is therefore forced by the geometry and the demand together, not merely preferred.
4. Balance the two lanes on each approach. With an inner left-through lane (s = 2015) and an outer through-right lane (s = 2250), drivers distribute the through movement so that both lanes are equally loaded in flow-ratio terms. Writing \(x\) for the through flow that uses the inner lane, $$\frac{q_L + x}{s_{LT}} = \frac{q_T - x + q_R}{s_{TR}} \;\Longrightarrow\; x = \frac{s_{LT}(q_T+q_R) - s_{TR}\,q_L}{s_{LT}+s_{TR}}$$ On the north approach \(x = 438.86\) veh/h, so the inner lane carries 705.53 veh/h and the outer 787.81 veh/h, and both give the same flow ratio \(y_N = 0.3501\). Repeating for the other three approaches gives \(y_S = 0.3392\), \(y_E = 0.2876\) and \(y_W = 0.2642\).
5. Critical movements and the sum of flow ratios. Each phase is governed by its heavier approach — and it is the flow ratio, not the volume, that decides: $$Y = \max(y_N, y_S) + \max(y_E, y_W) = 0.3501 + 0.2876 = \boxed{0.6378}$$ so the north and east approaches are critical. The total lost time is \(L = n(\ell + R) = 2(3.5 + 1.5) = 10.0\) s.
6. Webster’s optimum cycle. Substituting into $$C_{o} = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0)+5}{1-0.6378} = 55.21\ \text{s}$$ This is the cycle that minimises total intersection delay for the vehicle demand alone.
7. Pedestrian minimum — solved as an inequality, not evaluated once. The table gives conflicting pedestrian volumes, so the design may be pedestrian-controlled. Each phase must supply $$G_{p} = 3.2 + \frac{L_{c}}{S_{p}} + 2.7\,\frac{N_{ped}}{W_{E}}, \qquad N_{ped} = \frac{v_{ped}\,C}{3600}$$ Because \(N_{ped}\) itself grows with the cycle, a one-shot check at \(C_o\) always passes and always under-designs. Writing each phase’s green as its share of the available green, \((y_i/Y)(C-L)\), and requiring that to cover \(G_p\) gives a linear inequality in C. Two assumptions are declared under Note 2: a walking speed \(S_p = 1.2\) m/s (TAC/MUTCDC practice) and an effective crosswalk width \(W_E = 4.0\) m, which places the calculation on the HCM wide-crosswalk branch. The crosswalk each phase must clear is the one whose pedestrians walk parallel to that phase’s vehicles, so the north–south phase clears the east and west legs across the 15 m east–west carriageway at 100 ped/h, and the east–west phase clears the north and south legs at 235 ped/h. Solving gives \(C \ge 39.96\) s for the north–south phase and \(C \ge 49.67\) s for the east–west phase, so the binding pedestrian cycle is 49.67 s.
8. Adopt the cycle. The design cycle is the larger of the two controls, rounded up to the nearest 5 s: $$C = 5\left\lceil \frac{\max(55.21,\ 49.67)}{5} \right\rceil = \boxed{60\ \text{s}}$$ The vehicle optimum governs here; the pedestrians are comfortably accommodated inside a cycle chosen for the traffic.
9. Split the green. The green available for movement is \(g_{tot} = C - L = 60 - 10 = 50.0\) s, divided in proportion to the critical flow ratios: $$g_i = \frac{y_i}{Y}\,(C-L) \;\Longrightarrow\; g_{NS} = \frac{0.3501}{0.6378}(50.0) = 27.45\ \text{s}, \quad g_{EW} = 22.55\ \text{s}$$ Adding back the 3.5 s of acceleration/deceleration lost time that is displayed as green gives displayed greens of 30.95 s and 26.05 s, each followed by a 1.5 s all-red. The timing closes exactly: \(27.45 + 22.55 + 2(3.5) + 2(1.5) = 60.0\) s.
10. Check the adopted timing against the pedestrians and against capacity. At \(C = 60\) s the pedestrian intervals actually required are 16.83 s (north–south phase) and 18.34 s (east–west phase), against 27.45 s and 22.55 s of effective green: both pass, with 10.63 s and 4.21 s of slack. The critical degree of saturation is $$x_{crit} = \frac{Y\,C}{C-L} = \frac{0.6378(60)}{50} = 0.7653$$ and this must reappear as q/c on both critical approaches — the critical north lane has capacity \(s\,g/C = 2015(27.45)/60 = 921.9\) veh/h against 705.5 veh/h of demand, and the critical east lane 757.3 against 579.5, giving \(q/c = 0.7653\) in both cases. That one line validates the lane assignment, the flow ratios and the split together. Webster’s delay formula gives 17.6 s/veh on the critical north lane and 21.1 s/veh on the critical east lane, comfortably within the range expected of a well-timed two-phase signal.