16-Civ-B10 Traffic Engineering · May 2015

Question 2 of 7: Parking-Space Availability and a Toll-Booth Queue

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 2: Parking-Space Availability and a Toll-Booth Queue (2 × 10 = 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. Part (a): $N = 4$ parking spaces; mean occupancy time 5 min per space, exponentially distributed, so a service rate of $\mu = 60/5 = 12$ vehicles per hour per space; Poisson arrivals at $\lambda = 18$ customers per hour. Part (b): booth opens at 4:00 a.m.; arrivals begin at 3:45 a.m. at 6 veh/min until 4:15 a.m., then 2 veh/min; service rate a constant 6 veh/min once the booth opens.

Find. (a) The probability that an arriving driver finds all four spaces occupied. (b) The instant the queue clears, the total delay accumulated, and the greatest number of vehicles in the queue at any one time.

Approach. Part (a) is a multi-channel stochastic queue: build the traffic intensity $\rho = \lambda/\mu$, form the steady-state probability that all four servers are busy, and state which of the two standard conventions — wait-for-a-space or drive-away — is being answered. Part (b) is deterministic: draw the cumulative arrival and departure curves, and read the three answers off as an intersection, an area, and a vertical gap.

Part (a) — establish the traffic intensity and check that the system is stable. Each parking space is a server, so the store is a four-channel queue. With arrivals and service both exponential, $$\rho = \frac{\lambda}{\mu} = \frac{18}{12} = 1.5$$ Note carefully that $\rho$ here is $\lambda/\mu$ and not $\lambda/(N\mu)$; the latter is the utilisation per server, $\rho/N = 0.375$, and substituting it into the state equations silently halves the answer. Since $\rho/N = 0.375 < 1$ the system is stable: on average 1.5 of the four spaces are occupied.
Write the probability that the system is empty. For an M/M/N queue the normalising condition on the state probabilities gives $$P_0 = \left[\sum_{n=0}^{N-1}\frac{\rho^{\,n}}{n!} \; + \; \frac{\rho^{\,N}}{N!\left(1-\rho/N\right)}\right]^{-1}$$ Evaluating the truncated series with $\rho = 1.5$, $$\sum_{n=0}^{3}\frac{\rho^{\,n}}{n!} = 1 + 1.5 + \frac{1.5^2}{2} + \frac{1.5^3}{6} = 1 + 1.5 + 1.125 + 0.5625 = 4.1875$$ and the tail term, which is the piece representing a full system, $$\frac{\rho^{4}}{4!\,(1-1.5/4)} = \frac{5.0625}{24(0.625)} = \frac{0.21094}{0.625} = 0.3375$$ so that $$P_0 = \frac{1}{4.1875 + 0.3375} = \frac{1}{4.525} = 0.2210$$
Evaluate the probability that all four spaces are occupied. An arriving driver fails to find an open space exactly when the system is in state $n \ge N$, whose probability is the tail term scaled by $P_0$: $$P(n \ge 4) = \frac{\rho^{\,N}}{N!\left(1-\rho/N\right)}\,P_0 = 0.3375\,(0.2210)$$ $$\boxed{\;P(\text{no open space}) = 0.0746 \approx 7.5\ \%\;}$$ So roughly one arrival in thirteen during the busiest hour finds the lot full. Four spaces are adequate but not generous; a fifth space would push the figure to about 2 %.
State the competing convention explicitly. The wording “will not find an open parking space” describes the state of the lot, which is what the M/M/N result above gives, and it presumes that a driver who finds the lot full waits briefly rather than leaving. If instead the store is modelled as a pure loss system — the driver sees a full lot and drives away, so no queue can form — the correct model is M/M/N/N and the answer is the Erlang-B blocking probability $$P_B = \frac{\rho^{N}/N!}{\sum_{n=0}^{N}\rho^{\,n}/n!} = \frac{0.21094}{4.1875+0.21094} = \frac{0.21094}{4.39844} = 0.0480$$ That is 4.8 %, appreciably lower, because forbidding a queue removes the states in which arrivals pile up behind a full lot.

Check — which convention is boxed. The M/M/N value of 7.5 % is boxed, on two grounds: it is the multi-channel formula set that the standard traffic-engineering text develops for exactly this parking problem, and a customer at a convenience store who finds four spaces taken realistically idles for a moment rather than abandoning the trip. If the marker intends a blocked-calls-cleared loss system, the Erlang-B answer is 4.8 %. Both are stated under the paper’s NOTE 1.

The second half of the question replaces randomness with a known, piecewise-uniform demand, so the queue can be tracked exactly rather than in expectation.

Question 2(b) — cumulative arrival and departure curves at the toll booth. The queue is the vertical gap; it is frozen at 90 vehicles between 4:00 and 4:15 because the arrival and service rates are both 6 veh/min, then drains at 4 veh/min.

Part (b) — set a time origin and write the cumulative arrival function. Measure $t$ in minutes from 3:45 a.m., when the first vehicle arrives. The arrival rate is 6 veh/min for the first 30 minutes (to 4:15 a.m.) and 2 veh/min thereafter, so the cumulative count of arrivals is $$A(t) = \begin{cases} 6t, & 0 \le t \le 30\\ 180 + 2(t-30), & t > 30\end{cases}$$ By 4:00 a.m. ($t = 15$) some $A(15) = 90$ vehicles have arrived, and by 4:15 a.m. ($t = 30$) the total is 180.
Write the cumulative departure function. Nothing is served until the booth opens at 4:00 a.m., after which vehicles are processed at a constant 6 veh/min: $$D(t) = \begin{cases} 0, & 0 \le t \le 15\\ 6(t-15), & t > 15 \text{ (while a queue exists)}\end{cases}$$
(iii) Identify the maximum queue length. The queue at any instant is the vertical gap $Q(t) = A(t) - D(t)$. Over $0 \le t \le 15$ nothing departs, so the queue grows linearly to $$Q(15) = 90 - 0 = 90 \text{ vehicles}$$ Between 4:00 and 4:15 the arrival rate and the service rate are both 6 veh/min, so the gap neither grows nor shrinks: $$Q(t) = 6t - 6(t-15) = 90 \quad \text{for } 15 \le t \le 30$$ $$\boxed{\;Q_{\max} = 90 \text{ vehicles, held constant from 4:00 to 4:15 a.m.}\;}$$ This flat top is the signature of a bottleneck whose demand exactly equals its capacity: the queue that formed before opening is frozen, neither served down nor added to.
(i) Locate the instant the queue dissipates. After 4:15 the arrival rate drops to 2 veh/min while service continues at 6 veh/min, so the queue drains at 4 veh/min. Writing the gap for $t > 30$, $$Q(t) = \left[180 + 2(t-30)\right] - 6(t-15) = 210 - 4t$$ Setting $Q(t) = 0$, $$t = \frac{210}{4} = 52.5 \text{ min after 3:45 a.m.}$$ $$\boxed{\;\text{the queue dissipates at 4:37:30 a.m.}\;}$$ As a check, $A(52.5) = 180 + 2(22.5) = 225$ and $D(52.5) = 6(37.5) = 225$ — the two cumulative curves meet, as they must.
(ii) Integrate the queue to obtain the total delay. Total delay is the area between the two cumulative curves, which the diagram splits into three simple pieces. From 3:45 to 4:00 the queue rises linearly from 0 to 90; from 4:00 to 4:15 it is a rectangle at 90; from 4:15 to 4:37:30 it falls linearly from 90 to 0 over 22.5 min: $$\begin{aligned} \text{Total delay} &= \tfrac12(15)(90) \;+\; (90)(15) \;+\; \tfrac12(22.5)(90)\\ &= 675 + 1350 + 1012.5 \end{aligned}$$ $$\boxed{\;\text{Total delay} = 3037.5 \text{ veh}\cdot\text{min} = 50.6 \text{ veh}\cdot\text{h}\;}$$
Reduce the total to a per-vehicle figure and to the worst individual wait. All 225 vehicles that arrive before the queue clears are delayed, so $$\bar{w} = \frac{3037.5 \text{ veh}\cdot\text{min}}{225 \text{ veh}} = 13.5 \text{ min per vehicle}$$ For the individual worst case, first-in-first-out means the vehicle arriving at time $t$ (the $6t$-th vehicle, for $t \le 30$) is served at $15 + 6t/6 = 15 + t$, a wait of exactly 15 min. Every vehicle arriving before 4:15 therefore waits 15 minutes — the delay caused by the booth opening 15 minutes late is passed on undiminished — and vehicles arriving afterwards wait progressively less. The longest individual wait is 15 min, and the average of 13.5 min is close to it because most of the traffic arrives in the heavy first half-hour.

Question 2 — results
Quantity	Result
(a) Traffic intensity, ρ = λ/μ	1.5 (utilisation ρ/N = 0.375)
(a) P₀	0.2210
(a) P(no open space), M/M/N	0.0746 (7.5 %)
(a) Alternative: Erlang-B loss system	0.0480 (4.8 %)
(b)(i) Queue dissipates	4:37:30 a.m. (52.5 min after 3:45)
(b)(ii) Total delay	3037.5 veh·min = 50.6 veh·h
(b)(ii) Average delay per vehicle (225 vehicles)	13.5 min
(b)(iii) Maximum queue length	90 vehicles (4:00–4:15 a.m.)
(b) Longest individual wait	15.0 min

Topic. Multi-channel stochastic queueing applied to parking supply, and deterministic queueing analysis at a temporary bottleneck.

Key relations. M/M/N: $\rho = \lambda/\mu$, $P_0 = \left[\sum_{n=0}^{N-1}\rho^n/n! + \rho^N/(N!(1-\rho/N))\right]^{-1}$, $P(n\ge N) = \rho^N P_0/[N!(1-\rho/N)]$. Erlang-B loss system: $P_B = (\rho^N/N!)/\sum_{n=0}^{N}\rho^n/n!$. Deterministic queueing: $Q(t) = A(t)-D(t)$, total delay $= \int Q\,dt$ = area between the cumulative curves, average delay $=$ total delay divided by the number of vehicles delayed.

Why this works. The two halves of this question are the two complementary ways of handling a queue, and it is worth seeing why each is legitimate. The parking problem has genuinely random arrivals and departures, so no single trajectory can be traced; instead the birth–death balance equations are solved for the long-run fraction of time the system spends in each state, and the answer is a probability. The toll booth has a demand that is known minute by minute, so the cumulative arrival and departure curves are exact functions and every quantity of interest is a geometric feature of them: the queue is a vertical gap, the delay of one vehicle is a horizontal gap, the total delay is the enclosed area, and the queue clears where the curves intersect. The flat-topped queue between 4:00 and 4:15 is a particularly instructive feature: with the arrival and service rates equal, the pre-existing queue is preserved intact, which is the graphical statement that a bottleneck at exactly capacity can hold a backlog forever without either clearing it or making it worse.

Common pitfall. In part (a), substituting $\rho = \lambda/(N\mu)$ into the state formulas instead of $\rho = \lambda/\mu$; this is the single most common error in multi-channel work and it roughly halves the blocking probability. In part (b), integrating only to 4:15 and forgetting the draining triangle, which loses a third of the total delay; and dividing the total delay by the maximum queue length rather than by the 225 vehicles actually delayed. A further trap is assuming the queue peaks when arrivals slow at 4:15 — it peaks then only because the rates happened to be equal beforehand, not as a general rule.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 6 (single- and multi-channel queueing; deterministic queueing analysis with cumulative plots); ITE Traffic Engineering Handbook, queueing applied to parking and toll facilities.

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