Question 7 of 7: Question 2 repeated with reduced saturation and pedestrian flows

Question 7 — Question 2 repeated with reduced saturation and pedestrian flows 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. Everything from Question 2, with two changes.

Find. The revised phasing, cycle length and phase lengths, and an explanation of how the two decreases affect the cycle length.

Approach. The two changes act on different sides of the design and must be separated before they are combined. A uniform change to every saturation flow scales the flow ratios but provably leaves the green split untouched, so it moves only Webster's optimum — hyperbolically. A change to pedestrian volumes moves only the pedestrian bound — linearly. Computing four cases (base, saturation change alone, pedestrian change alone, both together) and naming which control binds in each is what actually answers the question asked.

1. Show that a uniform saturation-flow factor cannot change the lane split or the green split. Writing the lane-balance condition with every saturation flow multiplied by the same factor $f$,
   $$\frac{v_L+x}{f\,s_1}=\frac{\left(v_T-x\right)+v_R}{f\,s_2}$$
   the factor cancels, so $x$ is unchanged on every approach — north still splits at 308.3 veh/h, west at 460.3 veh/h. Consequently every flow ratio scales by exactly $1/f$, the same two approaches remain critical, and the ratio $y_i/Y$ that sets the green split is identical to Question 2. A changed green split in this question is an arithmetic error, not a result.
2. Scale the critical flow ratios and re-form $Y$. With $f=0.85$,
   $$y_{\text{NS}}=\frac{0.3467}{0.85}=0.4079,\qquad y_{\text{EW}}=\frac{0.3733}{0.85}=0.4391$$
   $$Y'=\frac{Y}{0.85}=\frac{0.7200}{0.85}=\boxed{0.8471}$$
   The split is confirmed unchanged: $0.4079/0.8471=0.4815$ and $0.4391/0.8471=0.5185$, exactly the Question 2 proportions.
3. Recompute Webster's optimum, and note how violently it responds. The lost time is unaffected at $L=10.0$ s, so
   $$C_o'=\frac{1.5\left(10.0\right)+5}{1-0.8471}=\frac{20.0}{0.1529}=130.8\ \text{s}$$
   A 15 per cent loss of saturation flow has taken the vehicle optimum from 71.4 s to 130.8 s — an 83 per cent increase. The cause is that $C_o=(1.5L+5)/(1-Y)$ is hyperbolic in $Y$: as $Y\to1$ the denominator collapses, so the closer the intersection already is to capacity, the more explosively the cycle responds to any further capacity loss.
4. Recompute the pedestrian bound, which moves only slightly and only because of the volume reduction. Since $y_i/Y$ is unchanged, the pedestrian inequality is affected solely through $v_{ped}$. For phase A, with 1187.5 ped/h across the 15 m crosswalk,
   $$0.4815\left(C-10\right)\ \ge\ 15.70+\frac{2.7\left(1187.5\right)}{3600\left(4.0\right)}\,C=15.70+0.22266\,C$$
   $$\left(0.4815-0.22266\right)C\ \ge\ 20.515\quad\Longrightarrow\quad C\ \ge\ 79.2\ \text{s}$$
   and phase B gives $C\ge58.6$ s. The governing pedestrian requirement has fallen only from 83.0 s to 79.2 s — a 4.5 per cent reduction for a 5 per cent volume reduction, confirming the near-linearity.
5. Identify the governing control, which has switched. The design cycle is the larger of the two requirements:
   $$C=\max\left(C_o',\ C_{ped}'\right)=\max\left(130.8,\ 79.2\right)=130.8\ \text{s}$$
   In Question 2 the pedestrians governed at 83.0 s against Webster's 71.4 s. Here the vehicle requirement has vaulted past the pedestrian requirement, so control flips from pedestrian-governed to vehicle-governed. Rounding up to the next 5 s increment,
   $$\boxed{C=135\ \text{s}}$$
6. Split the green using the unchanged proportions. The effective green available is $135-10=125$ s, so
   $$g_{\text{NS}}=0.4815\left(125\right)=60.2\ \text{s},\qquad g_{\text{EW}}=0.5185\left(125\right)=64.8\ \text{s}$$
   which sum to 125.0 s. Adding back the 3.5 s of acceleration-and-deceleration lost time gives displayed greens of 63.7 s and 68.3 s, and with the two 1.5 s all-red intervals the cycle closes at 135.0 s. The phasing plan itself is unchanged — two phases with permitted left turns, the same lane assignment — and the protected option is now even further out of reach, since its $Y$ would be $1.468/0.85=1.73$.
7. Confirm the pedestrian intervals are still delivered, and the degree of saturation. At the longer cycle the pedestrian demand per cycle rises, so the check must be repeated:
   $$G_{p,\text{A}}=3.2+12.50+\frac{2.7\left(1187.5\times135/3600\right)}{4.0}=45.8\ \text{s}\ \le\ 60.2\ \text{s}$$
   $$G_{p,\text{B}}=3.2+10.00+\frac{2.7\left(1092.5\times135/3600\right)}{4.0}=40.9\ \text{s}\ \le\ 64.8\ \text{s}$$
   Both pass with ample margin now, which is the visible signature of vehicle control. The cost is hidden in the degree of saturation:
   $$x_{crit}'=\frac{Y'C}{C-L}=\frac{0.8471\left(135\right)}{125}=0.915$$
   up from 0.816. Webster's delay formula gives about 56 s/veh on the critical north lane and 52 s/veh on the critical west lane, roughly double the Question 2 values, taking the intersection from level of service C to level of service D–E.
8. Answer the question asked by decomposing the two effects. Running each change alone isolates its contribution:

   Case	$Y$	Webster $C_o$ (s)	Pedestrian bound (s)	Governing control	Cycle adopted (s)
   Base (Question 2)	0.7200	71.4	83.0	pedestrian	85
   Saturation flow −15 per cent only	0.8471	130.8	83.0	vehicle	135
   Pedestrian flow −5 per cent only	0.7200	71.4	79.2	pedestrian	80
   Both (Question 7)	0.8471	130.8	79.2	vehicle	135

   The decomposition makes the answer unambiguous. The 15 per cent saturation-flow reduction is responsible for the entire increase, driving the cycle from 85 s to 135 s and switching the governing control from pedestrians to vehicles. The 5 per cent pedestrian reduction has no effect at all on the final answer: acting alone it would have shortened the cycle from 85 s to 80 s, but once the vehicle requirement governs, lowering a non-binding constraint changes nothing. The two changes therefore do not partially offset each other — one of them simply stops mattering.
9. State the engineering consequence, not only the arithmetic. A 135 s cycle is arithmetically correct but sits above the roughly 120 s practical maximum used in Canadian urban practice, and at $x_{crit}=0.915$ the intersection has almost no reserve for demand variation. Three points follow. First, the reduced saturation flows may themselves be a symptom — of parked vehicles, transit stops, narrow lanes or heavy pedestrian interference with permitted left turns — and removing that cause recovers the capacity far more cheaply than lengthening the cycle. Second, because the pedestrian term enters as $N_{ped}/W_E$, a wider crosswalk is the correct lever whenever pedestrians govern, and it is worth noting that at 135 s the pedestrian intervals of 45.8 s and 40.9 s are long absolute waits even though they are not binding. Third, if the 15 per cent loss cannot be recovered, the honest recommendation is not a longer cycle but additional capacity — widening an approach to three lanes, which would also reopen the protected-left option that the present geometry forecloses.

Results.