16-Civ-B10 Traffic Engineering · May 2014

Question 3 of 6: Overload probability, delay components, person delay and queue length

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

Notes on this paper

Paper format. National Examinations, May 2014 — 98-Civ-B10 Traffic Engineering. Three-hour, open-book examination; any non-communicating calculator is permitted. Six questions are printed and five complete solutions are required, all questions being of equal value (20 marks each). The printed grading scheme is Q1 (a) to (d) 5 marks each; Q2 (a) to (e) 4 marks each; Q3 (a) to (e) 4 marks each; Q4 (a) and (b) 10 marks each; Q5 (a) to (e) 4 marks each; Q6 (10 + 5 + 5) marks. The paper states that if doubt exists as to the interpretation of a question the candidate should submit with the answer paper a clear statement of any assumptions made, and that any data required but not given can be assumed. All six questions are worked below.

Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed. — Ch. 4 (traffic engineering studies: volume studies, peak-hour factor), Ch. 6 (fundamental principles of traffic flow: Poisson arrivals, deterministic and stochastic queueing, M/M/1 and M/M/N channels), Ch. 8 (intersection control: cycle length, phasing, change and clearance intervals, progression and time–space diagrams) and Ch. 10 (capacity and level of service at signalised intersections). This is the principal reference for the subject. Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory Technical Paper No. 56 — the optimum-cycle and three-term delay formulae used throughout Questions 1 to 3. Transportation Research Board, Highway Capacity Manual — saturation-flow adjustment factors and the pedestrian-green requirement. Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) and Geometric Design Guide for Canadian Roads — Canadian signal-timing, clearance-interval and crosswalk practice. Institute of Transportation Engineers, Traffic Engineering Handbook — controller types and coordinated timing plans.

Check: the four values assumed under the paper's own NOTE 2 (“any data required, but not given, can be assumed”). They are declared here once and used consistently in every question.

Bus passenger-car equivalent $E_B = 2.0$ pcu (HCM value for buses on level terrain). The paper gives bus occupancies but no pcu equivalent; note that the pcu-based and vehicle-based flow ratios in Question 1 agree exactly for any choice of $E_B$, so this assumption does not affect Questions 2 or 3.
Pedestrian walking speed $S_p = 1.2$ m/s (HCM 2000 / MUTCDC design value; the more recent 1.1 m/s would lengthen the pedestrian interval by about 1 s).
Lost time convention: the whole intergreen (amber + all-red) is taken as lost, so effective green equals displayed green. This is the conservative reading and needs no assumed start-up lost time. Webster's alternative, $L = n\ell + R$ with $\ell \approx 2$ s, would give $L = 10$ s instead of 12 s and lengthen each green by about 1 s.
Delay model: Webster's three-term formula, with the first two terms reported separately in Question 3(b) as the uniform and overflow components.

Question 3: Overload probability, delay components, person delay and queue length (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 timing plan settled in Question 2 — $C = 75$ s, $G_{NS} = 34$ s, $G_{EW} = 29$ s, $I = 6$ s — together with:

Approach	Lane demand $q$ (pcu/h)	Green $g$ (s)	$x$	Persons/h
North-bound	415	34	0.509	1975
South-bound	362	34	0.444	1580
East-bound	250	29	0.359	1000
West-bound	350	29	0.503	1400

Saturation flow $s_0 = 1800$ pcu/h/lane; pedestrian demand 140 per hour in each of the four crosswalks; arrivals assumed Poisson, which is the standard model for an isolated signal.

Find. For each approach, the probability that a cycle's arrivals exceed what the green can discharge; the uniform and overflow components of delay and their flow-weighted totals; the total person-hours of delay to vehicle occupants; the average pedestrian delay; and the average queue standing at the end of red.

Approach. Treat each cycle as an independent Poisson trial with mean $qC$ arrivals against a fixed discharge capacity $sg$; split Webster's delay into its deterministic and stochastic parts; convert vehicle delay to person delay through the occupancies of Question 1; and get the queue directly from the deterministic arrival–departure diagram.

Question 3(e) — deterministic cumulative arrival and departure curves for the north-bound lane. The queue at the end of red is the vertical gap between the two curves at t = r.

Part (a) — set up the overload trial. “Discharge overload” means more vehicles arrive during a cycle than the green can release, so the residue carries over. Per lane the mean arrivals per cycle and the whole number of vehicles that a green can discharge are $$m = \frac{q\,C}{3600}, \qquad n = \left\lfloor \frac{s_0\,g}{3600} \right\rfloor,$$ and with Poisson arrivals the overload probability is the upper tail $$P(\text{overload}) = P(X > n) = 1 - \sum_{k=0}^{n} \frac{e^{-m} m^{k}}{k!}.$$ North-bound, $m = 415(75)/3600 = 8.65$ vehicles per cycle against $n = 1800(34)/3600 = 17$ discharged, so the tail beyond 17 is required.
Part (a) — evaluate the tail on each approach. Carrying out the summation gives the following.

Approach $m$ (arrivals/cycle) $n$ (discharged) $P(X > n)$

North-bound 8.65 17 0.0036

South-bound 7.54 17 0.00084

East-bound 5.21 14 0.00034

West-bound 7.29 14 0.0081

The largest value is west-bound, and it is instructive: that approach does not have the heaviest flow, but it has the shortest green, so its ratio $m/n$ is the worst. Overall, $$\boxed{\;P(\text{overload}) \le 0.8\ \% \text{ on every approach — roughly one cycle in 120 on the worst}\;}$$ which is an acceptable level of cycle failure for an urban signal.
Part (b) — separate the uniform and overflow components. The first term of Webster's formula is the delay a perfectly regular arrival stream would suffer, and the second is the extra delay caused by the randomness of real arrivals: $$d_1 = \frac{C\,(1 - \lambda)^2}{2\,(1 - \lambda x)}, \qquad d_2 = \frac{x^2}{2\,q\,(1 - x)}, \qquad d_3 = 0.65\left(\frac{C}{q^2}\right)^{1/3} x^{\,2 + 5\lambda}.$$ For the north-bound lane, $d_1 = 14.57$, $d_2 = 2.28$ and the empirical correction $d_3 = 0.65$ s/pcu. The four approaches give:

Approach $d_1$ uniform (s/pcu) $d_2$ overflow (s/pcu) $d_3$ correction $d$ total

North-bound 14.57 2.28 0.65 16.20

South-bound 14.03 1.76 0.39 15.39

East-bound 16.38 1.45 0.29 17.54

West-bound 17.51 2.62 0.87 19.26

Weighting each column by the approach demand in pcu/h, $$\boxed{\;\bar d_1 = 15.50,\quad \bar d_2 = 2.08,\quad \bar d_3 = 0.57,\quad \bar d = 15.50 + 2.08 - 0.57 = 17.0\ \text{s/pcu}\;}$$ The uniform term supplies about 88 % of the delay: at these degrees of saturation the intersection is dominated by the deterministic effect of stopping at red, not by randomness.
Part (c) — convert vehicle delay into person delay. Every occupant of a delayed vehicle is delayed with it, so the person-hours lost per hour on an approach are $$H_i = \frac{d_i \times (\text{persons per hour})_i}{3600}.$$ North-bound: $16.20(1975)/3600 = 8.89$ person-hours per hour. The remaining approaches give 6.76 (SB), 4.87 (EB) and 7.49 (WB), so $$\boxed{\;H_{\text{total}} = 8.89 + 6.76 + 4.87 + 7.49 = 28.0\ \text{person-hours per hour}\;}$$ Note the ranking: the north-bound approach has neither the longest delay nor the worst saturation, yet it dominates the person delay because of its 15 buses.
Part (d) — average delay to pedestrians. A pedestrian arriving at random in the cycle waits nothing if the crossing is already open and waits the remainder of the non-crossing time otherwise. Averaging over a uniform arrival instant gives the standard result $$d_p = \frac{(C - g_p)^2}{2C}.$$ Pedestrians on the north and south legs cross on the 29 s E–W green: $d_p = (75 - 29)^2/150 = 14.11$ s. Pedestrians on the east and west legs cross on the 34 s N–S green: $d_p = (75 - 34)^2/150 = 11.21$ s. With 140 pedestrians per hour in each of the four crosswalks the demands are equal, so $$\boxed{\;\bar d_p = \tfrac{1}{2}(14.11 + 11.21) = 12.7\ \text{s per pedestrian}\;}$$ equivalent to $4(140)(12.66)/3600 = 1.97$ person-hours of pedestrian delay per hour, which is worth quoting beside the 28.0 vehicle person-hours when the plan is defended.
Part (e) — queue standing at the end of red. On the deterministic diagram above, vehicles accumulate at the arrival rate throughout the red and none departs, so the queue at the instant green begins is simply $$Q_r = q\,r, \qquad r = C - g.$$ The N–S approaches see $r = 75 - 34 = 41$ s and the E–W approaches $r = 75 - 29 = 46$ s, giving $$\boxed{\;Q_{NB} = 4.7,\quad Q_{SB} = 4.1,\quad Q_{EB} = 3.2,\quad Q_{WB} = 4.5 \ \text{veh per lane}\;}$$ or about 9.5 vehicles across the two north-bound lanes. At roughly 8 m of kerb per queued car that is a standing queue of about 38 m in the worst lane — well within a normal block length, so no spill-back check is needed.

Approach	$m$ (arrivals/cycle)	$n$ (discharged)	$P(X > n)$
North-bound	8.65	17	0.0036
South-bound	7.54	17	0.00084
East-bound	5.21	14	0.00034
West-bound	7.29	14	0.0081

Approach	$d_1$ uniform (s/pcu)	$d_2$ overflow (s/pcu)	$d_3$ correction	$d$ total
North-bound	14.57	2.28	0.65	16.20
South-bound	14.03	1.76	0.39	15.39
East-bound	16.38	1.45	0.29	17.54
West-bound	17.51	2.62	0.87	19.26

Final Results

Part	Quantity	Result
(a)	$P$(discharge overload) NB / SB / EB / WB	0.0036 / 0.00084 / 0.00034 / 0.0081
(b)	Average overall uniform delay	15.50 s/pcu
(b)	Average overall overflow delay	2.08 s/pcu
(b)	Average overall delay	17.0 s/pcu
(c)	Person delay NB / SB / EB / WB	8.89 / 6.76 / 4.87 / 7.49 person-h per hour
(c)	Total person delay (excluding pedestrians)	28.0 person-hours per hour
(d)	Pedestrian delay, N/S legs and E/W legs	14.11 s and 11.21 s
(d)	Average pedestrian delay	12.7 s per pedestrian (1.97 person-h per hour)
(e)	Queue at end of red, NB / SB / EB / WB	4.7 / 4.1 / 3.2 / 4.5 veh per lane

Topic. Stochastic cycle failure, the decomposition of signal delay, and the deterministic queue at a fixed-time signal.

Key relations. Poisson overload $P(X > n)$ with $m = qC/3600$ and $n = sg/3600$; uniform delay $d_1 = C(1-\lambda)^2/[2(1 - \lambda x)]$; overflow delay $d_2 = x^2/[2q(1-x)]$; pedestrian delay $d_p = (C - g_p)^2/2C$; queue at end of red $Q_r = q\,r$.

Why this works. Arrivals at an isolated signal are close to Poisson because the upstream stream is unmetered, so each cycle is an independent trial and the number of arrivals in one cycle has mean and variance both equal to $qC/3600$. Overload occurs when that count exceeds the deterministic discharge, and because $n$ is roughly twice $m$ here the tail is small. The two delay terms separate cleanly for the same reason: $d_1$ is what you would compute from the triangular arrival–departure diagram with no randomness at all, while $d_2$ is the queueing-theory penalty that blows up as $x \to 1$. The pedestrian delay formula is the same triangular argument applied to a walker who arrives uniformly in the cycle.

Common pitfall. Using $x$ (the degree of saturation) directly as the overload probability — they are unrelated: $x = 0.51$ here but overload happens in under 1 % of cycles. A second pitfall in part (e) is quoting the queue per approach when the arrival rate used was per lane, or vice versa; state which one you mean. In part (c), remember to weight by persons and not by vehicles, or the buses vanish from the answer.

Source. Garber & Hoel, 5th ed., §6.5 (queueing and Poisson arrivals) and §10.5 (delay components at signalised intersections); Webster & Cobbe, RRL TP 56, delay derivation; HCM, uniform and incremental delay terms.

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