Question 4 of 7: Sensitivity of the design to saturation flow and pedestrian volume

Question 4 — Sensitivity of the design to saturation flow and pedestrian volume 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.

Approach. The two changes push the cycle in opposite directions, so the honest answer is not one number but an identification of which control governs. Recompute the flow ratios (they scale exactly, because both lane saturation flows fall by the same factor), recompute Webster's optimum, recompute the pedestrian lower bound with the reduced pedestrian flows, and adopt whichever is larger — then decompose the change to attribute it.

1. Recompute the flow ratios — they scale exactly by $1/0.8$. Because both shared-lane saturation flows are multiplied by the same factor 0.80, the lane-balance equation
   $$\frac{v_L + x}{0.8 \times 1900} = \frac{(v_T - x) + v_R}{0.8 \times 2080}$$
   has the identical solution $x$ as before — the 0.80 cancels — so the lane assignment is unchanged and every flow ratio simply rises by $1/0.80 = 1.25$:

   Approach	$y$ (Question 3)	$y$ (Question 4)
   North	0.3142	0.3928
   South	0.3094	0.3868
   East	0.3536	0.4420
   West	0.3758	0.4698

   The same two approaches remain critical, so the phasing decision of Question 3 stands: two phases with permitted left turns. (The four-phase plan, already infeasible at $Y = 1.10$, is now worse still at $Y = 1.37$.)
2. Sum the critical ratios and recompute Webster's optimum.
   $$Y = 0.3928 + 0.4698 = 0.8626 \quad (\text{still} \lt 1, \text{ so a two-phase plan remains feasible})$$
   $$C_o = \frac{1.5(10.0) + 5}{1 - 0.8626} = \frac{20.0}{0.1374} = \boxed{145.5\ \text{s}}$$
   The 20 % loss of saturation flow has driven the vehicular optimum from 64.5 s to 145.5 s. This is the crux of the question: $C_o$ depends on $1/(1-Y)$, which is violently non-linear as $Y$ approaches unity, so a 25 % rise in $Y$ has produced a 126 % rise in $C_o$.
3. Recompute the pedestrian lower bound with the reduced pedestrian flows. The green split is unchanged, because all four flow ratios scaled by the same factor:
   $$\frac{y_A}{Y} = 0.4554, \qquad \frac{y_B}{Y} = 0.5446 \quad (\text{identical to Question 3})$$
   With $v_{ped}$ reduced by 10 % — Phase A now clears 1210.5 ped/h across the 20 m carriageway, Phase B 900 ped/h across 18 m — the two requirements become
   $$0.4554\,(C-10) \ge 19.87 + 0.2270\,C \;\Longrightarrow\; C \ge 106.9\ \text{s}$$
   $$0.5446\,(C-10) \ge 18.20 + 0.1688\,C \;\Longrightarrow\; C \ge 62.9\ \text{s}$$
   The pedestrian bound has fallen, from 120.2 s to 106.9 s.
4. Identify the governing control and adopt the cycle. Comparing the two controls:
   $$C_o = 145.5\ \text{s} \;\gt\; C_{ped} = 106.9\ \text{s}$$
   so control has shifted from the pedestrians to the vehicles — the opposite of Question 3. Rounding up to the nearest 5 s,
   $$C = \boxed{150\ \text{s}}$$
5. Distribute the green and confirm both checks. With $C - L = 140$ s of effective green,
   $$g_A = 0.4554 \times 140 = \boxed{63.7\ \text{s}}, \qquad g_B = 0.5446 \times 140 = \boxed{76.3\ \text{s}}$$
   $$G_A = 67.2\ \text{s}, \qquad G_B = 79.8\ \text{s}, \qquad 67.2 + 1.5 + 79.8 + 1.5 = 150.0\ \text{s} \;\checkmark$$
   The pedestrian requirements at $C = 150$ s are $G_{p,A} = 19.87 + 0.2270(150) = 53.9$ s and $G_{p,B} = 18.20 + 0.1688(150) = 43.5$ s, both comfortably satisfied. The critical degree of saturation is now
   $$x_{crit} = \frac{0.8626 \times 150}{140} = 0.924$$
   which is the operational cost of the change: the intersection has moved from a relaxed $x_{crit} = 0.750$ to a value close to capacity, where small demand fluctuations produce large delays and cycle failures.
6. Attribute the change — decompose the two effects. Applying each change on its own isolates its contribution:

   Scenario	Webster $C_o$	Pedestrian bound	Governing $C$
   Question 3 (base)	64.5 s	120.2 s	125 s (pedestrian)
   Saturation flow $-20$ % only	145.5 s	120.2 s	150 s (vehicle)
   Pedestrian flow $-10$ % only	64.5 s	106.9 s	110 s (pedestrian)
   Both changes (this question)	145.5 s	106.9 s	150 s (vehicle)

   Answer to "how do these decreases effect the cycle length": the cycle length increases, from 125 s to 150 s, and the two decreases act in opposition. The 20 % reduction in saturation flow is entirely responsible: it raises every flow ratio by 25 %, drives $Y$ from 0.690 to 0.863, and because $C_o \propto 1/(1-Y)$ it more than doubles the vehicular optimum to 145.5 s. The 10 % reduction in pedestrian volume works the other way, lowering the pedestrian bound from 120.2 s to 106.9 s, and on its own it would have shortened the cycle to 110 s. Because the saturation-flow effect is much the larger, the net result is a longer cycle — and, more importantly for the designer, the governing control has changed: the intersection is no longer pedestrian-controlled but vehicle-controlled, and the engineering response changes with it. In Question 3 the remedy for a long cycle was a wider crosswalk; here it is more capacity — additional lanes, protected phasing where geometry permits, or recovery of the lost saturation flow (parking removal, transit-stop relocation, bus bays, better lane discipline).