16-Civ-B10 Traffic Engineering · December 2013

Question 3 of 7: Mean Speeds and the Linear Speed–Density Relationship

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

Notes on this paper

Paper format. National Examinations, December 2013 — 98-Civ-B10 Traffic Engineering. Three-hour, open-book examination; any non-communicating calculator is permitted. Seven questions are printed and five complete solutions are required, all questions being of equal value. The printed grading scheme is Q1 (a) 6, (b) 6, (c) 8; Q2 (a) 10, (b) 10; Q3 (a) 10, (b) 10; Q4 20; Q5 20; Q6 (a)–(d) 5 each; Q7 (a)–(d) 5 each. The paper also states that if doubt exists as to the interpretation of a question the candidate should submit a clear statement of any assumptions made, and that any data required but not given can be assumed. All seven questions are worked below.

Reference texts.

Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed. — Ch. 4 (traffic engineering studies: spot-speed statistics, volume and travel-time studies, the moving-vehicle method), Ch. 6 (fundamental principles of traffic flow: time-mean and space-mean speed, speed–density models, Poisson arrivals, deterministic queueing), Ch. 8 (intersection control and signalisation: progression and time–space diagrams), Ch. 9 (capacity and level of service for highway segments), Ch. 10 (capacity and level of service at signalised intersections). This is the principal reference for the subject.
Transportation Research Board, Highway Capacity Manual (HCM) — basic freeway segments and signalised-intersection saturation flow.
Transportation Association of Canada (TAC), Geometric Design Guide for Canadian Roads — Canadian lane-width, shoulder and clearance practice.
TAC, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — Canadian signal warrants, timing and coordination practice.
Institute of Transportation Engineers, Traffic Engineering Handbook — signal-system progression and alternate-system design.

Check: assumed reference-table values. Questions 2(a) and 2(b) are capacity problems whose input list — lane width, lateral obstruction, per-cent heavy vehicles, a specific grade, design speed and a target level of service — is exactly the argument list of the classical Highway Capacity Manual equations, but the paper does not reproduce the lookup tables. Consistent with the paper's own instruction that any data required but not given may be assumed, every table value used is stated explicitly at the point of use, drawn from one coherent edition family (HCM 1985/1994, ideal capacity 2,000 pc/h/ln for the freeway segment). Substituting another edition's tables rescales the final flow rate but changes neither the method nor the arithmetic chain; the sensitivity is discussed in the Question 2(a) concept note.

Question 3: Mean Speeds and the Linear Speed–Density Relationship (20 marks: (a) 10, (b) 10)

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. (a) Three vehicles crossing a 100 m study section at constant speeds of 30, 32 and 40 m/s. (b) A linear (Greenshields) speed–density model with free-flow speed $u_{f}=100$ km/h and jam density $k_{j}=100$ veh/km.

Find. (a) the time-mean speed, the space-mean speed and the variance of speeds about the space-mean speed; (b) the volume–density curve, and its slope at zero density, at the density that maximises volume, and at jam density.

Approach. Part (a) turns on the distinction between averaging speeds over vehicles passing a point and averaging over the time each vehicle spends in a section: the first is an arithmetic mean, the second a harmonic mean, and the gap between them is fixed by the speed variance. Part (b) substitutes the linear speed law into the identity $q=u\,k$ and differentiates.

Part (a) — compute the time-mean speed. The time-mean speed is the arithmetic mean of the individual spot speeds, as would be measured by a radar at a point: $$\bar{u}_{t}=\frac{1}{n}\sum_{i=1}^{n}u_{i}=\frac{30+32+40}{3}=\frac{102}{3}$$ $$\boxed{\bar{u}_{t}=34.0\ \text{m/s}\ (122.4\ \text{km/h})}$$
Compute the travel times over the section. The space-mean speed is defined through time spent in the section, so first obtain each travel time from $t_{i}=L/u_{i}$ with $L=100$ m: $$t_{1}=\frac{100}{30}=3.3333\ \text{s},\quad t_{2}=\frac{100}{32}=3.1250\ \text{s},\quad t_{3}=\frac{100}{40}=2.5000\ \text{s}$$ The mean travel time is $\bar{t}=8.9583/3=2.9861$ s.
Compute the space-mean speed. Space-mean speed is the section length divided by the mean travel time, which is algebraically the harmonic mean of the spot speeds: $$\bar{u}_{s}=\frac{L}{\bar{t}}=\frac{n}{\sum(1/u_{i})} =\frac{3}{\tfrac{1}{30}+\tfrac{1}{32}+\tfrac{1}{40}}=\frac{3}{0.0895833}$$ $$\boxed{\bar{u}_{s}=33.49\ \text{m/s}\ (120.6\ \text{km/h})}$$ Both routes give the same number, which is the arithmetic check: $100/2.9861$ also equals 33.49 m/s. As it must, the space-mean speed is the smaller of the two, because the slow vehicles occupy the section for longer and so carry more weight in a space average.
Estimate the variance about the space-mean speed. The two means are linked by the standard relationship $$\bar{u}_{t}=\bar{u}_{s}+\frac{\sigma_{s}^{2}}{\bar{u}_{s}}$$ which rearranges to give the variance directly from the pair of means: $$\sigma_{s}^{2}=\bar{u}_{s}\left(\bar{u}_{t}-\bar{u}_{s}\right) =33.49\,(34.00-33.49)=33.49\times 0.5116$$ $$\boxed{\sigma_{s}^{2}\approx 17.1\ \text{m}^{2}/\text{s}^{2} \quad(\sigma_{s}\approx 4.1\ \text{m/s})}$$ As a cross-check, forming the deviations directly about the space-mean speed gives $\sum(u_{i}-\bar{u}_{s})^{2}/n=18.9$ m²/s². The two agree to about 10 %, which is as close as three observations allow: the relationship above is asymptotic in the sample size, and the question asks for an estimate. The estimate from the means is quoted as the answer because it is the one the relationship supplies.
Part (b) — write the linear speed–density law. Greenshields assumed speed falls linearly from free flow to zero at jam density: $$u=u_{f}\left(1-\frac{k}{k_{j}}\right)=100\left(1-\frac{k}{100}\right)=100-k$$ with $u$ in km/h and $k$ in veh/km. The numerical coincidence $u=100-k$ arises only because $u_{f}$ and $k_{j}$ happen to share the value 100.
Form the volume–density relationship. Volume, speed and density are tied by the fundamental identity $q=u\,k$, so substituting the speed law gives a parabola through the origin: $$q=u\,k=k\,(100-k)=100k-k^{2}$$ This is the curve requested, plotted below. It vanishes at both ends — no vehicles at $k=0$, and no movement at jam density — and peaks in between.
Locate the maximum. Differentiating and setting the derivative to zero, $$\frac{dq}{dk}=100-2k=0\quad\Rightarrow\quad k=50\ \text{veh/km}$$ at which the speed is $u=100-50=50$ km/h and the volume is $$q_{max}=50\times 50=\boxed{2{,}500\ \text{veh/h}}$$ Capacity therefore occurs at exactly half the jam density and half the free-flow speed, the signature result of the linear model.
Evaluate the three requested slopes. The slope of the volume–density curve is $$\frac{dq}{dk}=100-2k\ \ \text{(km/h)}$$ so at the beginning, middle and end of the curve: $$\left.\frac{dq}{dk}\right|_{k=0}=+100\ \text{km/h},\qquad \left.\frac{dq}{dk}\right|_{k=50}=0,\qquad \left.\frac{dq}{dk}\right|_{k=100}=-100\ \text{km/h}$$ Each slope has physical meaning: it is the speed of a kinematic (shock) wave in the traffic stream. At the beginning the wave travels forward at the free-flow speed — a disturbance moves downstream with the vehicles. At capacity the wave is stationary, which is why capacity operation is unstable and why a bottleneck queue forms at a fixed location. At jam density the wave travels backward at 100 km/h, i.e. the tail of a stopped queue propagates upstream as fast as free-flowing traffic moves forward.

Volume–density curve q = 100k − k² for the linear speed–density model (u_f = 100 km/h, k_j = 100 veh/km). Tangents are drawn at the three requested points; their slopes are the kinematic-wave speeds — +100 km/h at free flow, zero at capacity, and −100 km/h at jam density.

Part	Quantity	Result
(a)	Time-mean speed ũ_t	34.0 m/s (122.4 km/h)
(a)	Space-mean speed ũ_s	33.49 m/s (120.6 km/h)
(a)	Variance about ũ_s	≈ 17.1 m²/s² (σ_s ≈ 4.1 m/s)
(b)	Speed–density law	u = 100 − k (km/h)
(b)	Volume–density law	q = 100k − k² (veh/h)
(b)	Capacity	q_max = 2,500 veh/h at k = 50 veh/km, u = 50 km/h
(b)	Slope at k = 0 (beginning)	+100 km/h
(b)	Slope at k = 50 (middle)	0
(b)	Slope at k = 100 (end)	−100 km/h

Topic. The two mean speeds of a traffic stream, and the Greenshields linear speed–density model with its parabolic volume–density curve.

Key relations. $\bar u_{t}=\frac{1}{n}\sum u_{i}$; $\bar u_{s}=n/\sum(1/u_{i})=L/\bar t$; $\bar u_{t}=\bar u_{s}+\sigma_{s}^{2}/\bar u_{s}$; the fundamental identity $q=u\,k$; Greenshields $u=u_{f}(1-k/k_{j})$, hence $q=u_{f}k-(u_{f}/k_{j})k^{2}$, $k_{cap}=k_{j}/2$, $u_{cap}=u_{f}/2$, $q_{max}=u_{f}k_{j}/4$, and wave speed $dq/dk$.

Why this works. Space-mean speed is the physically meaningful average for flow work because it is the one consistent with $q=u\,k$: only the harmonic mean, weighted by time spent in the section, reproduces the correct volume when multiplied by density. Time-mean speed always exceeds it (or equals it when every vehicle travels at the same speed), and the excess is exactly the variance divided by the space-mean speed — so the two means coincide only for a perfectly uniform stream, and diverge as the speed distribution spreads. For part (b), the parabola is not an empirical curve fitted to data but an algebraic consequence of assuming speed falls linearly with density; every one of its landmark results ($k_{j}/2$, $u_{f}/2$, $u_{f}k_{j}/4$) follows from that single assumption. The slope $dq/dk$ deserves its own emphasis: it is the celerity of a kinematic wave, and its change of sign at capacity is the mathematical statement of why queues form and propagate upstream on congested facilities.

Common pitfall. The dominant error is computing the space-mean speed as an arithmetic mean, or worse, computing the time-mean speed as the harmonic mean — if your space-mean speed comes out larger than your time-mean speed you have swapped them. In the variance part, mixing the two candidate formulas is common: $\sigma_{s}^{2}=\bar u_{s}(\bar u_{t}-\bar u_{s})$ is an estimate from the two means, while $\sum(u_{i}-\bar u_{s})^{2}/n$ is the direct computation, and with only three observations they will not agree to more than a significant figure or two — quote one, name it, and note the other. In part (b) the frequent slip is reporting the slope at jam density as $+100$ rather than $-100$, thereby losing the entire physical point about upstream wave propagation, or confusing the slope of the $q$–$k$ curve (a wave speed) with the space-mean speed (the chord $q/k$ from the origin).

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 (Fundamental Principles of Traffic Flow) — time-mean and space-mean speed, the fundamental diagram, and Greenshields' model.

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