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16-Civ-B10 Traffic Engineering · December 2016

Question 4 of 7: Sensitivity of the Cycle to Saturation Flow and Pedestrian Demand

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2016 — 98-Civ-B10 Traffic Engineering. Three-hour duration; OPEN BOOK, any non-communicating calculator permitted. Seven questions, all of equal value (20 marks each), with the mark split printed in the paper’s own grading scheme: Q1 and Q2 and Q5 are (a) to (e) at 4 marks each, Q3 and Q4 are single 20-mark questions, Q6 is (a) 6 marks with (b) and (c) 7 marks each, and Q7 is (a) to (h) at 2.5 marks each. The paper states that a total of five solutions is required and that only the first five in the answer book will be marked. All seven questions are solved here, because this set is a study resource rather than a sitting. The paper also permits assumptions — “Any data required, but not given, can be assumed” — and every assumption made below is stated explicitly where it is used.

Reference texts. Garber, N. J. & Hoel, L. A., Traffic and Highway Engineering, 5th ed. (Cengage) — the core reference for this exam code; Transportation Association of Canada, Geometric Design Guide for Canadian Roads (TAC GDG); Institute of Transportation Engineers (Canadian District), Canadian Capacity Guide for Signalized Intersections, 3rd ed.; Transportation Research Board, Highway Capacity Manual (HCM) — pedestrian crossing-time model; AASHTO, A Policy on Geometric Design of Highways and Streets (the “Green Book”, 2001 edition — the source of the stopping-sight-distance table printed on page 4 of this paper); Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC). Canadian practice governs wherever the paper does not name a specific standard.

Question 4: Sensitivity of the Cycle to Saturation Flow and Pedestrian Demand (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. The Question 3 intersection unchanged in geometry and demand, but with every saturation flow rate raised by 20 per cent and every pedestrian volume raised by 25 per cent.

Revised inputs
QuantityQuestion 3Question 4
North–south approach saturation flow $s_{NS}$3,900 veh/h green4,680 veh/h green
East–west approach saturation flow $s_{EW}$6,300 veh/h green7,560 veh/h green
Governing pedestrian volume per leg1,350 ped/h1,687.5 ped/h
Lost time $L$ (unchanged)11.0 s11.0 s

Find. The revised sum of critical flow ratios, the revised Webster cycle, the revised pedestrian-governed cycle, and an explanation of which of the two changes actually moves the answer.

Approach. Re-run the Question 3 chain with the two scaled inputs, keeping the geometry, the demand and the lost time fixed, and compare the vehicular and the pedestrian requirement separately so the cause of the change is visible.

  1. Rescale the flow ratios. The demand is unchanged and every saturation flow rises by the same factor, so every $y = v/s$ simply falls by that factor: $$Y' = \frac{Y}{1.20} = \frac{0.5066}{1.20} = \boxed{0.4222}$$ Critically, the ratio between the two critical flow ratios is untouched, so the green split proportions 0.5541 and 0.4459 are exactly the same as before.
  2. Recompute the Webster vehicle optimum. With the lost time unchanged at 11.0 s, $$C_o' = \frac{1.5(11.0) + 5}{1 - 0.4222} = \frac{21.5}{0.5778} = 37.2\ \text{s}$$ so the extra saturation flow shortens the delay-optimal cycle from 43.6 s to 37.2 s, a reduction of about 15 per cent. This is the expected direction: a wider throat needs less green to pass the same demand, and a shorter cycle then wastes less time in queues.
  3. Recompute the pedestrian requirement. Each crosswalk now receives $N_{ped} = 1{,}687.5\,C/3{,}600 = 0.4688\,C$ pedestrians per cycle, so the east–west phase condition becomes $$0.4459\,(C - 11.0) \;\ge\; 3.2 + \frac{17.0}{1.2} + \frac{2.7(0.4688\,C)}{4.0}$$ which reduces to $0.1295\,C \ge 22.27$ and therefore $C \ge 172.0$ s; the north–south condition returns $C \ge 112.7$ s. Rounding up, $$\boxed{C = 175\ \text{s}}$$
  4. Verify the 175 s design. The effective green available is $175 - 11 = 164$ s, split $g_{NS} = 0.5541(164) = 90.9$ s and $g_{EW} = 0.4459(164) = 73.1$ s. At this cycle each crosswalk collects $N_{ped} = 1{,}687.5(175)/3{,}600 = 82.0$ pedestrians, so $G_{p,NS} = 3.2 + 17.5 + 55.4 = 76.1$ s against 90.9 s available, and $G_{p,EW} = 3.2 + 14.2 + 55.4 = 72.7$ s against 73.1 s available. Both are satisfied, the east–west phase again by the narrowest of margins.
  5. Answer the question asked. The two changes push in opposite directions and one of them wins outright. The 20 per cent saturation-flow increase lowers the vehicular optimum from 43.6 s to 37.2 s, but that optimum was never the binding constraint, so the reduction has no effect on the design whatsoever — it merely buys additional spare capacity, dropping every degree of saturation to about 0.35. The 25 per cent pedestrian increase raises the binding constraint from 115.5 s to 172.0 s. The net result is that the cycle lengthens from 120 s to 175 s, an increase of about 46 per cent, driven entirely by pedestrians.
  6. Judge the result and recommend. A 175 s cycle is beyond acceptable practice: Canadian guidance keeps urban cycles at or below about 120 s, because longer cycles produce pedestrian waits that provoke non-compliance and vehicle queues that spill into upstream intersections. The remedy is to attack the term that is growing rather than to keep adding green. Widening the effective crosswalk from 4.0 m to 5.0 m reduces the platoon-discharge term by one fifth and brings the requirement back to $$C \ge 115.5\ \text{s} \rightarrow C = 120\ \text{s}$$ — the original design cycle. Staging the 21 m crossing through a median refuge, which halves $L_c$ for the east–west phase, would relieve it further, and an exclusive pedestrian (scramble) phase becomes worth evaluating once pedestrian demand at this level is confirmed.
Webster vehicle optimum (base)43.6 sWebster vehicle optimum (+20% s)37.2 sPedestrian-governed (base)120 sPedestrian-governed (+25% peds)175 sCycle length is set by the pedestrian requirement, not by the vehicle optimum
Which constraint governs. The vehicle optimum falls when saturation flow rises; the design cycle follows the pedestrian curve, which rises.
Question 4 — final results
QuantityQuestion 3Question 4
Sum of critical flow ratios $Y$0.50660.4222
Webster optimum cycle (vehicles only)43.6 s37.2 s
Pedestrian-governed cycle requirement115.5 s172.0 s
Design cycle length $C$120 s175 s
North–south green60 s90.9 s
East–west green49 s73.1 s
Cycle if crosswalks widened to 5.0 m—120 s