Question 7 of 7: Signalised-Intersection Capacity and Green Allocation

Question 7: Signalised-Intersection Capacity and Green Allocation (20 marks — (a) and (c) 4 each, (b) and (d) 2 each, (e) 8)

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.

Find. The critical lane and critical flow ratio for the phase; the permitted-left-turn adjustment factor; the approach saturation flow and saturation headway; the approach capacity; and the two-phase green split for each of four intersections.

Approach. Every part is an application of the flow ratio $y = q/s$ and the capacity relation $c = s(g/C)$: the critical lane maximises $y$, the left-turn adjustment factor converts a shared lane's mixed traffic into through-vehicle equivalents, and an equal-saturation-flow green split is proportional to the phase flow ratios.

1. Part (a) — Critical lane and critical flow ratio. The demand a lane places on the phase is measured not by its flow but by its flow ratio, the fraction of green it needs to serve that flow: $$y = \frac{q}{s}$$ Computing all four lanes in turn: NB L,S gives $600/1200 = 0.5000$; NB R,S gives $500/1700 = 0.2941$; SB L,S gives $450/1330 = 0.3383$; and SB R,S gives $720/1600 = 0.4500$. The largest of these governs, so $$y_{\text{crit}} = \max\left(0.5000,\ 0.2941,\ 0.3383,\ 0.4500\right) = \boxed{0.500}$$ and the critical lane is the northbound left-turn / through lane. The instructive point is that the busiest lane is not the critical lane: SB R,S carries 720 pcu/h, 20 per cent more traffic than NB L,S, but it also has a 33 per cent higher saturation flow because it is unencumbered by opposed left turns, so it needs less green. This phase therefore requires at least half of the effective green in the cycle before any other phase or lost time is considered.
2. Part (b) — Permitted left-turn adjustment factor. A permitted left turn made from a shared lane blocks the through traffic behind it while its driver waits for a gap, so each left-turning vehicle consumes $E_{L}$ through-vehicle equivalents of green rather than one. The lane group's capacity is reduced by the factor $$f_{LT} = \frac{1}{1 + P_{LT}\left(E_{L} - 1\right)}$$ Substituting $P_{LT} = 0.12$ and $E_{L} = 5.25$, $$P_{LT}\left(E_{L}-1\right) = 0.12(4.25) = 0.51$$ $$f_{LT} = \frac{1}{1 + 0.51} = \frac{1}{1.51} = \boxed{0.662}$$ So although left turns are only 12 per cent of the traffic, they consume 34 per cent of the approach's discharge capability — the direct consequence of one left turn costing 5.25 through-vehicle equivalents.
3. Part (c) — Approach saturation flow and saturation headway. Applying the adjustment factor to the base saturation flow of all three lanes, $$s = s_{0}\,N\,f_{LT} = 1830 \times 3 \times 0.6623 = \boxed{3636\ \text{veh/h of green}}$$ against the $1830 \times 3 = 5490$ veh/h the approach would discharge if there were no left turns — a loss of 1854 veh/h. The corresponding saturation headway for the approach as a whole is the reciprocal of that flow, $$h_{s} = \frac{3600}{s} = \frac{3600}{3636} = \boxed{0.990\ \text{s/veh}}$$ Expressed per lane, which is the form usually quoted in the field, $h_{s,\text{lane}} = 3600/(1830 \times 0.6623) = 2.97$ s/veh, up from the unadjusted 1.97 s/veh. A one-second increase in per-lane saturation headway is a large operational penalty and is the quantitative case for providing an exclusive left-turn lane on this approach.
4. Part (d) — Approach capacity. Capacity is the saturation flow prorated by the fraction of the cycle for which the approach has effective green: $$c = s\,\frac{g}{C} = 3636 \times \frac{50}{80} = 3636 \times 0.625 = \boxed{2272\ \text{veh/h}}$$ The approach can therefore serve about 2270 veh/h. Had the left turns been given their own lane so that $f_{LT}$ applied to one lane only, the same green would deliver roughly 2900 veh/h — a 28 per cent gain from a lane-marking change alone.
5. Part (e) — Two-phase green split. When the two phases share the same saturation flow rate the flow ratios reduce to the flows themselves, $y_{i} = q_{i}/s$, so Webster's proportional-split rule $$g_{i} = G_{\text{tot}}\,\frac{y_{i}}{\sum y} = G_{\text{tot}}\,\frac{q_{i}}{q_{1} + q_{2}}$$ can be applied directly to the pcu flows. For intersection 1, $q_{1} + q_{2} = 600$ pcu/h, so $g_{1} = 75(400/600) = 50.0$ s and $g_{2} = 75(200/600) = 25.0$ s. For intersection 2, the total is 400 pcu/h, giving $g_{1} = 75(300/400) = 56.25$ s and $g_{2} = 18.75$ s. For intersection 3, the total is 200 pcu/h, giving $g_{1} = 60.0$ s and $g_{2} = 15.0$ s. For intersection 4, the total is 150 pcu/h, giving $g_{1} = 40.0$ s and $g_{2} = 35.0$ s. Every pair sums to the 75 s available, as it must. Two of these splits deserve comment: intersection 3's 60/15 split leaves phase 2 with only 15 s, which is unlikely to satisfy a pedestrian walk-plus-clearance minimum and would in practice be raised to about 20 s with the balance taken from phase 1; and intersection 4's nearly even 40/35 split reflects genuinely balanced demand, so it is the one location where the proportional rule needs no adjustment.