16-Civ-B10 Traffic Engineering · May 2015

Question 3 of 7: Freeway Service Flow Rate at Level of Service C, and the Greenshields Flow–Density Curve

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 3: Freeway Service Flow Rate at Level of Service C, and the Greenshields Flow–Density Curve (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) — basic freeway segment	Value
Cross-section	four-lane freeway, so N = 2 lanes per direction
Lane width	3.5 m
Lateral obstruction	1.5 m from the travelled pavement, one side only
Grade	4 %, 1.0 km long
Traffic composition	10 % heavy trucks and buses, 5 % recreational vehicles
Design speed	100 km/h
Target level of service	C

Part (b): free-flow (mean-free) speed $u_f = 100$ km/h; jam density $k_j = 100$ veh/km; a linear speed–density relationship.

Find. (a) The maximum flow rate, in vehicles per hour in one direction, that this section can carry while still operating at level of service C. (b) The flow–density curve implied by the linear speed–density law, and the slope $dq/dk$ at its start, its midpoint and its end.

Approach. Part (a) applies the standard basic-freeway-segment relation: take the ideal maximum service flow rate for the target level of service and design speed, multiply by the number of lanes, then degrade it for the sub-ideal cross-section and for the heavy vehicles struggling on the grade. Part (b) substitutes the linear speed–density law into the identity $q = uk$ and differentiates.

Part (a) — write the service-flow relation and pick the ideal rate. The service flow rate of a basic freeway segment is the ideal per-lane rate for the level of service, scaled by the number of lanes and by adjustment factors for geometry and traffic composition: $$SF_i = MSF_i \times N \times f_w \times f_{HV}$$ For a 100 km/h design speed, level of service C corresponds to a volume-to-capacity ratio of 0.72 against an ideal capacity of 2000 pc/h/lane, so $$MSF_C = 0.72\,(2000) = 1450 \text{ pc/h/lane}$$ (rounded as tabulated). A four-lane freeway carries $N = 2$ lanes in each direction, and the service flow rate is quoted per direction.
Interpolate the lane-width and lateral-clearance factor. With a single-sided obstruction, the tabulated factors bracket the given geometry: 1.00 for 3.6 m lanes with clearance of 1.8 m or more, 0.99 for 3.6 m lanes at 1.2 m, 0.97 for 3.3 m lanes at 1.8 m, and 0.96 for 3.3 m lanes at 1.2 m. Interpolating on clearance first, at 1.5 m, $$f_w(3.6\text{ m}) = 1.00 + \frac{1.5-1.8}{1.2-1.8}(0.99-1.00) = 0.995, \qquad f_w(3.3\text{ m}) = 0.97 + \frac{1.5-1.8}{1.2-1.8}(0.96-0.97) = 0.965$$ and then on lane width, at 3.5 m, $$f_w = 0.965 + \frac{3.5-3.3}{3.6-3.3}\,(0.995-0.965) = 0.965 + 0.020 = 0.985$$ The penalty is only 1.5 % because the section is close to ideal: near-standard lanes and a clearance that is only 0.3 m short of the point at which obstructions cease to matter.
Build the heavy-vehicle factor for the specific grade. On a sustained upgrade a truck loses speed and occupies road space far out of proportion to its length, so it is counted as several passenger cars. For a 4 % grade of 1.0 km the tabulated equivalents are $E_T = 5$ for trucks and buses and $E_R = 3$ for recreational vehicles. With two classes present, $$f_{HV} = \frac{1}{1 + P_T(E_T-1) + P_R(E_R-1)}$$ Substituting $P_T = 0.10$, $P_R = 0.05$, $$f_{HV} = \frac{1}{1 + 0.10(4) + 0.05(2)} = \frac{1}{1 + 0.40 + 0.10} = \frac{1}{1.50} = 0.667$$ The grade is doing far more damage than the geometry: it removes a third of the section’s capacity where the cross-section removes only 1.5 %.
Assemble the service flow rate. Multiplying the four factors together, $$SF_C = 1450 \times 2 \times 0.985 \times 0.667$$ $$SF_C = 2900\,(0.985)\,(0.6667) = 2856.5\,(0.6667)$$ $$\boxed{\;SF_C \approx 1900 \text{ veh/h in one direction}\;}$$ More precisely 1904 veh/h. Because level of service C is a design target rather than a capacity, this is the flow at which the section still delivers stable operation with reasonable freedom to manoeuvre — roughly 72 % of the section’s adjusted capacity of about 2640 veh/h.

Check — table interpolation. Capacity-manual tables are conventionally read to the next lower tabulated cell rather than interpolated. Taking $f_w = 0.96$ (3.3 m lanes at 1.2 m clearance) instead of the interpolated 0.985 gives $SF_C = 1856$ veh/h, about 2.5 % lower. The interpolated value is used here because 3.5 m and 1.5 m both sit strictly between tabulated entries; the conservative reading is stated for completeness under the paper’s NOTE 1.

The second half of the question turns from capacity accounting to the underlying flow model that gives capacity its meaning.

Part (b) — write the linear speed–density law from the two stated end points. A linear relationship between speed and density is pinned down by its two extremes: at zero density traffic runs at the free-flow speed, and at jam density it is stopped. Hence $$u = u_f\left(1 - \frac{k}{k_j}\right) = 100\left(1 - \frac{k}{100}\right) = 100 - k$$ with $u$ in km/h and $k$ in veh/km. This is the Greenshields model.
Substitute into the flow identity to obtain the volume–density curve. Flow, density and speed are related by the definitional identity $q = u\,k$, so $$q = k\,u_f\left(1-\frac{k}{k_j}\right) = 100k\left(1-\frac{k}{100}\right) = 100k - k^2$$ This is a downward parabola through the origin and through $k = k_j$, plotted below. Its vertex gives the capacity of the stream: $$q_{\max} = \frac{u_f k_j}{4} = \frac{100(100)}{4} = 2500 \text{ veh/h at } k = 50 \text{ veh/km}, \; u = 50 \text{ km/h}$$

Question 3(b) — the Greenshields flow–density parabola q = 100k − k² for u_f = 100 km/h and k_j = 100 veh/km, with tangents at the beginning, middle and end. The slopes are the free-flow speed, zero, and the negative of the free-flow speed.

Differentiate to get the slope at a general density. The slope of the flow–density curve is $$\frac{dq}{dk} = \frac{d}{dk}\left(100k - k^2\right) = 100 - 2k \qquad \text{(km/h)}$$ This derivative is not merely a geometric slope: it is the speed at which a small disturbance in flow propagates through the traffic stream, the kinematic-wave speed.
Evaluate the slope at the beginning, middle and end of the curve. Substituting the three densities in turn, $$\begin{aligned} k = 0: &\quad \frac{dq}{dk} = 100 - 0 = +100 \text{ km/h} = +u_f\\ k = k_j/2 = 50: &\quad \frac{dq}{dk} = 100 - 100 = 0\\ k = k_j = 100: &\quad \frac{dq}{dk} = 100 - 200 = -100 \text{ km/h} = -u_f \end{aligned}$$ $$\boxed{\;\left.\frac{dq}{dk}\right|_{k=0} = +100 \text{ km/h}, \quad \left.\frac{dq}{dk}\right|_{k=50} = 0, \quad \left.\frac{dq}{dk}\right|_{k=100} = -100 \text{ km/h}\;}$$ The three results need no arithmetic once the structure is seen: for any linear speed–density model the end slopes are exactly the free-flow speed and its negative, and the slope vanishes at capacity. Physically, at very low density a disturbance travels forward with the traffic at the free-flow speed; at capacity it is stationary, which is why capacity operation is unstable and why a small perturbation there triggers a breakdown; and at jam density the disturbance travels backward at 100 km/h, which is the shock wave that runs upstream through a stopped queue.

Question 3 — results
Quantity	Result
(a) MSF at LOS C, 100 km/h design speed	1450 pc/h/lane (v/c = 0.72)
(a) f_w (3.5 m lanes, 1.5 m clearance, one side)	0.985
(a) f_HV (E_T = 5, E_R = 3)	0.667
(a) Service flow rate at LOS C	1904 ≈ 1900 veh/h, one direction
(b) Speed–density law	u = 100 − k (km/h)
(b) Flow–density law	q = 100k − k² (veh/h)
(b) Capacity (vertex)	2500 veh/h at k = 50 veh/km, u = 50 km/h
(b) Slope at k = 0 / k = 50 / k = 100	+100 / 0 / −100 km/h

Topic. Basic-freeway-segment level-of-service analysis, and the Greenshields linear speed–density model with its kinematic-wave interpretation.

Key relations. $SF_i = MSF_i \cdot N \cdot f_w \cdot f_{HV}$ with $MSF_i = (v/c)_i \times$ ideal capacity; $f_{HV} = 1/[1+\sum P_j(E_j-1)]$ over each heavy-vehicle class. Greenshields: $u = u_f(1-k/k_j)$, $q = uk = u_f k(1-k/k_j)$, $q_{\max} = u_f k_j/4$ at $k = k_j/2$ and $u = u_f/2$, wave speed $dq/dk = u_f(1-2k/k_j)$.

Why this works. The capacity calculation is a chain of independent multiplicative degradations from an ideal benchmark, and it works because each factor is defined as the ratio of the real throughput to the ideal throughput with only that one feature changed — so the factors compose. The passenger-car-equivalent structure of $f_{HV}$ deserves particular attention: the denominator $1 + \sum P_j(E_j-1)$ is the number of passenger-car units occupied by an average one-vehicle sample of the mixed stream, so its reciprocal converts a pcu capacity to a vehicle capacity exactly. In part (b), the flow–density parabola is not an empirical fit but a consequence of putting a linear speed law into the exact identity $q = uk$; and differentiating $q$ with respect to $k$ yields the kinematic-wave speed by the conservation equation of traffic flow. That is why the three requested slopes are so clean: they are $+u_f$, zero and $-u_f$ for any linear speed–density model, regardless of the numerical values of $u_f$ and $k_j$.

Common pitfall. In part (a), using the full four lanes instead of the two per direction, which doubles the answer; and applying general-terrain truck equivalents ($E_T \approx 2$) when a specific grade and length are given, which understates the penalty severely — $f_{HV}$ would come out near 0.87 instead of 0.667. In part (b), the frequent error is to report the slope at $k=0$ as the capacity flow, or to give the slope at jam density as $+100$; the sign matters, because the negative slope is the upstream-propagating shock wave. A related conceptual slip is reading the maximum of the parabola as the maximum speed rather than the maximum flow.

Source. TRB Highway Capacity Manual, basic freeway segments (level-of-service criteria, lane-width and lateral-clearance factors, passenger-car equivalents on specific grades); Garber & Hoel, Traffic and Highway Engineering, Ch. 6 (the Greenshields model, capacity and shock waves) and Ch. 9 (capacity and level of service).

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