16-Civ-B10 Traffic Engineering · May 2015

Question 4 of 7: Deterministic Queueing at an Incident Bottleneck

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

Notes on this paper

Paper format. 98-Civ-B10 Traffic Engineering, National Examinations, May 2015. 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 grading scheme; the paper requires a total of five solutions. All seven are worked below, because this document is a study resource rather than an exam script.

Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering — Ch. 4 (traffic-engineering studies: volume counts, peak-hour factor, spot-speed studies), Ch. 6 (fundamental principles of traffic flow: the Greenshields model, deterministic and stochastic queueing, the moving-observer method) and Ch. 8 (intersection control: saturation flow, change and clearance intervals). Transportation Research Board, Highway Capacity Manual — basic-freeway-segment and two-lane-highway level-of-service criteria and the lane-width, lateral-clearance and heavy-vehicle adjustment factors. Transportation Association of Canada, Geometric Design Guide for Canadian Roads and the Manual of Uniform Traffic Control Devices for Canada — Canadian lane widths, crosswalk practice and clearance-interval policy. Institute of Transportation Engineers, Traffic Engineering Handbook — queueing applied to parking and toll facilities.

Check — declared assumptions, invoked under the paper’s own NOTE 1 and NOTE 2 (“any data required, but not given, can be assumed”, with a clear statement of the assumption). This paper is open-book and deliberately withholds several standard table values. The following are used consistently throughout and each is restated where it is first applied:

Bus (heavy-vehicle) passenger-car equivalent on the level intersection approaches of Question 1: E_B = 2.0.
Two-phase signal operation at the Question 1 intersection (one phase for the north–south pair, one for the east–west pair), with no start-up lost time given, so the whole intergreen is taken as lost time.
Maximum service flow rates and volume-to-capacity ratios for basic freeway segments at a 100 km/h design speed: 700/1000/1450/1800/2000 pc/h/lane is the 110 km/h family, so the 100 km/h family 600/1000/1450/1800/2000 pc/h/lane (v/c = 0.30/0.50/0.72/0.90/1.00) is used, with an ideal capacity of 2000 pc/h/lane.
Two-lane-highway ideal capacity 2800 pc/h both directions, with level-terrain v/c ratios 0.15/0.27/0.43/0.64/1.00 at 0 % no-passing zones, and a 50/50 directional split (f_d = 1.00).
Passenger-car equivalents on specific grades: E_T = 5 for trucks on both the 4 % / 1.0 km grade and the 3 % / 1.5 km grade, E_R = 3 for recreational vehicles on the 4 % grade, E_B = 1.6 for intercity buses on grades of 4 % or flatter, and E_T = 2 for trucks in general level terrain.

Different editions of the capacity manual tabulate these cells slightly differently. Where an alternative reading changes an answer materially, the alternative is computed and stated alongside; the method and the arithmetic are what carry the marks.

Question 4: Deterministic Queueing at an Incident Bottleneck (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. Arrival (demand) flow a constant $\lambda = 3500$ veh/h throughout. Service capacity: zero from 8:00 to 8:15 a.m. (freeway closed); $\mu_1 = 2500$ veh/h from 8:15 to 8:30 a.m. (partially open); $\mu_2 = 4500$ veh/h from 8:30 a.m. onwards (fully open). Normal capacity 4500 veh/h, so the freeway is uncongested before the incident.

Find. The cumulative arrival–departure diagram, the instant the queue clears, the greatest queue length, the total delay, the average delay per vehicle, and the longest wait experienced by any single vehicle.

Approach. Convert every rate to vehicles per minute, build the cumulative arrival curve (one straight line) and the cumulative departure curve (three straight segments), then read each answer off the diagram: the queue is a vertical gap, an individual delay a horizontal gap, the total delay the enclosed area, and dissipation the point where the curves meet.

Question 4 — cumulative arrival and departure curves for the incident. The queue peaks at 1125 vehicles when full capacity is restored at 8:30, not when the closure ends; the shaded area is the total delay and the horizontal gap is an individual wait.

Reduce every rate to vehicles per minute and set the origin at 8:00 a.m. Working in minutes keeps the arithmetic clean: $$\lambda = \frac{3500}{60} = 58.333 \text{ veh/min}, \quad \mu_1 = \frac{2500}{60} = 41.667 \text{ veh/min}, \quad \mu_2 = \frac{4500}{60} = 75.0 \text{ veh/min}$$ Before the incident the freeway ran at 3500 against a capacity of 4500, so there was no queue at $t = 0$ — the queue we are about to build starts from nothing.
Write the two cumulative curves. Arrivals accumulate at one constant rate, so the arrival curve is a single ray from the origin: $$A(t) = 58.333\,t$$ Departures are zero while the freeway is closed, then run at the restricted capacity, then at the full capacity: $$D(t) = \begin{cases} 0, & 0 \le t \le 15\\ 41.667\,(t-15), & 15 < t \le 30\\ 625 + 75.0\,(t-30), & t > 30\end{cases}$$ The constant 625 is the number discharged during the partially-open period, $41.667 \times 15 = 625$ vehicles, and carrying it forward correctly is what keeps the third segment on the curve.
Track the queue through the two bottleneck phases. The queue is the vertical separation $Q(t) = A(t) - D(t)$. During the closure it grows at the full arrival rate, $$Q(15) = 58.333(15) - 0 = 875 \text{ vehicles}$$ During the partially-open period demand still exceeds capacity, so it keeps growing, but more slowly, at $58.333 - 41.667 = 16.667$ veh/min: $$Q(30) = 875 + 16.667(15) = 875 + 250 = 1125 \text{ vehicles}$$ $$\boxed{\;Q_{\max} = 1125 \text{ vehicles at 8:30 a.m.}\;}$$ The maximum occurs at the instant full capacity is restored, not at the end of the closure — the partial re-opening slows the growth but does not reverse it, because 2500 veh/h is still below the 3500 veh/h demand.
Find when the queue dissipates. After 8:30 the discharge rate exceeds the arrival rate, and the queue drains at $$\mu_2 - \lambda = 75.0 - 58.333 = 16.667 \text{ veh/min}$$ Setting the queue expression for $t>30$ to zero, $$Q(t) = 58.333t - \left[625 + 75.0(t-30)\right] = 1625 - 16.667\,t = 0 \;\Rightarrow\; t = \frac{1625}{16.667} = 97.5 \text{ min}$$ $$\boxed{\;\text{the queue dissipates 97.5 min after 8:00, that is at 9:37:30 a.m.}\;}$$ Confirming with both curves: $A(97.5) = 58.333(97.5) = 5687.5$ and $D(97.5) = 625 + 75(67.5) = 625 + 5062.5 = 5687.5$. Note how long the recovery takes — a 30-minute disruption needs a further 67.5 minutes to clear, because the surplus capacity available for recovery (1000 veh/h) is much smaller than the capacity that was lost.
Integrate the queue to get the total delay. Total delay is the area between the curves, which the diagram divides into a triangle, a trapezoid and a triangle: $$\begin{aligned} \text{closure, 0 to 15 min:} &\quad \tfrac12(15)(875) = 6562.5\\ \text{partial, 15 to 30 min:} &\quad \tfrac12(875+1125)(15) = 15\,000\\ \text{recovery, 30 to 97.5 min:} &\quad \tfrac12(67.5)(1125) = 37\,968.75 \end{aligned}$$ Adding the three pieces, $$\boxed{\;\text{Total delay} = 59\,531.25 \text{ veh}\cdot\text{min} = 992.2 \text{ veh}\cdot\text{h}\;}$$ Nearly two-thirds of that total accrues during the recovery, long after the freeway is physically clear — the delay outlives the incident by a wide margin.
Divide by the number of vehicles affected to get the average delay. Every vehicle arriving before the queue clears is delayed, and that count is the arrival ordinate at dissipation: $$n = A(97.5) = 5687.5 \text{ vehicles}$$ $$\bar{w} = \frac{59\,531.25}{5687.5} = \boxed{\;10.47 \text{ min per vehicle}\;}$$ The average is modest because the great majority of those 5687 vehicles arrive late in the episode and are delayed only briefly.
Identify the vehicle with the longest wait. The queue discharges first-in-first-out, so a vehicle’s wait is the horizontal gap between the curves at its own position in the queue — not the maximum queue divided by anything. The worst-off vehicle is the last one released while the bottleneck is still restrictive, that is the vehicle discharged at the very instant full capacity returns at $t = 30$. Its position in the queue is the cumulative departure count at that moment, 625, and it arrived at $$t_{arr} = \frac{625}{\lambda} = \frac{625}{58.333} = 10.71 \text{ min after 8:00 (8:10:43 a.m.)}$$ so its wait was $$w_{\max} = 30 - 10.71 = \boxed{\;19.29 \text{ min, about 19.3 min}\;}$$ It is worth seeing why no vehicle does worse. For a vehicle numbered $n \le 625$ the wait is $15 + n/41.667 - n/58.333$, which increases with $n$; for $n > 625$ it is $30 + (n-625)/75 - n/58.333$, which decreases with $n$. The two expressions agree at $n = 625$, so the maximum sits exactly at the changeover. Vehicles arriving after that instant are joining a queue that is already shrinking, and every one of them waits less than its predecessor.

Question 4 — results
Quantity	Result
Queue at 8:15 a.m. (end of closure)	875 vehicles
Longest queue length	1125 vehicles, at 8:30 a.m.
Time of queue dissipation	9:37:30 a.m. (97.5 min after 8:00)
Total delay	59 531 veh·min = 992.2 veh·h
Vehicles affected	5687.5 (say 5688)
Average delay per vehicle	10.47 min
Longest wait of any vehicle	19.29 min (arrived 8:10:43, discharged 8:30)

Topic. Deterministic (cumulative-plot) queueing analysis of an incident bottleneck with a time-varying service rate.

Key relations. $Q(t) = A(t) - D(t)$ (vertical gap = queue); individual delay = horizontal gap at a given cumulative count; total delay $= \int_0^{t_d} Q\,dt$ = area between the curves; dissipation where $A(t_d) = D(t_d)$; average delay = total delay / $A(t_d)$; the worst individual wait belongs to the last vehicle released while the bottleneck is still restrictive, $w_{\max} = T_2 - D(T_2)/\lambda$.

Why this works. The cumulative diagram is exact rather than approximate, because both curves are simply running totals of known constant rates, and every quantity the question asks for is a geometric feature of the same picture. That is the real power of the method: one drawing answers five questions, and the answers are mutually consistent by construction. Two structural insights are worth internalising. First, the queue peaks when the service rate finally rises above the demand rate, not when the incident ends — any partial re-opening below demand merely slows the growth. Second, the recovery time is governed by the surplus capacity, $\mu_2 - \lambda$, and not by the size of the disruption; because that surplus (1000 veh/h) is far smaller than the capacity that was lost, a half-hour incident takes over an hour to shake off, and the total delay is dominated by the recovery limb. This asymmetry is the quantitative case for rapid incident clearance on high-volume facilities.

Common pitfall. Computing the longest individual wait as the maximum queue divided by the discharge rate, or as the queue divided by the arrival rate — both give the wrong answer because they mix vehicles that are in the queue at different times. The wait is a horizontal gap and must be traced through a specific vehicle’s position in the discharge order. The other classic slip is dropping the 625-vehicle constant when writing the third departure segment, which detaches the curve and moves the dissipation time; and dividing the total delay by the maximum queue length rather than by the 5687 vehicles actually delayed.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 6 (deterministic queueing analysis; cumulative arrival–departure diagrams applied to incidents and work zones); TRB Highway Capacity Manual, freeway-facility oversaturated analysis.

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