Question 2 of 7: Freeway Service Flow Rate and Signalised-Intersection Capacity
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 2: Freeway Service Flow Rate and Signalised-Intersection Capacity
(20 marks: (a) 10, (b) 10)
Given. Two independent capacity calculations, one for a
freeway mainline segment and one for a signalised approach.
Quantity
(a) Freeway segment
(b) Signalised approach
Lanes in the direction analysed
2 (four-lane freeway)
2 approach lanes
Lane width
3.5 m
3.5 m
Lateral obstruction
1.0 m, one side only
—
Grade
5 % up, 1.5 km long
3 % down
Heavy vehicles
10 % trucks and buses, 3 % RVs
8 % heavy trucks
Design speed
100 km/h
—
Area type
—
Central Business District
Parking / bus stop / turns
—
none / none / through only
Signal timing
—
C = 75 s, g/C = 0.50
Target
Level of service C
Capacity
Find. (a) the service flow rate the segment can carry
while still operating at level of service C, in vehicles per hour in the
direction analysed; and (b) the capacity of the signalised approach in vehicles
per hour.
Approach. Both parts follow the same logic: start from an
ideal flow rate per lane, multiply by the number of lanes, then apply
multiplicative adjustment factors for every way in which the real facility falls
short of ideal — geometry, heavy vehicles, grade, and for the signal the
fraction of each cycle that is actually green.
Part (a) — state the governing service-flow equation.
For a basic freeway segment the service flow rate at level of service \(i\) is
$$SF_{i}=MSF_{i}\times N\times f_{w}\times f_{HV}\times f_{p}$$
where \(MSF_{i}\) is the maximum service flow rate per lane at that level of
service, \(N\) the number of lanes in one direction, \(f_{w}\) the lane-width and
lateral-clearance factor, \(f_{HV}\) the heavy-vehicle factor and \(f_{p}\) the
driver-population factor.
Read the maximum service flow rate for level of service C.
For a 100 km/h design speed on a four-lane freeway the assumed table values are an
ideal capacity of 2,000 pc/h/ln and a volume-to-capacity ratio of 0.70 at level of
service C, so
$$MSF_{C}=\left(\frac{v}{c}\right)_{C}\times c_{j}=0.70\times 2000
=1400\ \text{pc/h/ln}$$
This is a rate in passenger cars per hour per lane under ideal geometry;
the factors that follow convert it into mixed vehicles on the real cross-section.
Evaluate the geometric factor. The lanes are 3.5 m rather
than the ideal 3.6 m, and a lateral obstruction stands 1.0 m from the travelled
pavement on one side rather than the ideal 1.8 m. Interpolating the lane-width and
lateral-clearance table for a four-lane freeway with an obstruction on one side
gives
$$f_{w}=0.97$$
a 3 % penalty. Physically, narrow lanes and a close obstruction make drivers shy
away from the edge and increase their following distances.
Evaluate the heavy-vehicle factor on the specific grade.
A sustained 5 % grade 1.5 km long is long enough that trucks reach crawl speed, so
specific-grade passenger-car equivalents apply rather than the much smaller
general-terrain values. Taking \(E_{T}=10\) for trucks and buses and \(E_{R}=4\) for
recreational vehicles on this grade and length,
$$f_{HV}=\frac{1}{1+P_{T}(E_{T}-1)+P_{R}(E_{R}-1)}
=\frac{1}{1+0.10(9)+0.03(3)}=\frac{1}{1.99}$$
$$f_{HV}=0.5025$$
The denominator of 1.99 says that the 13 % of the traffic stream that is heavy
occupies almost as much road space as the whole passenger-car stream — the
single dominant effect in this problem.
Assemble the service flow rate. Taking \(f_{p}=1.00\) for a
familiar, largely commuting driver population and substituting all four factors,
$$SF_{C}=1400\times 2\times 0.97\times 0.5025\times 1.00$$
$$\boxed{SF_{C}\approx 1{,}365\ \text{veh/h in the direction analysed}}$$
which is about 682 veh/h per lane. Set against an ideal capacity of
2,000 pc/h/ln, the grade and the heavy vehicles between them have cost this
segment roughly two thirds of its nominal ability: a two-lane direction that would
carry 2,800 pc/h at level of service C on level terrain manages fewer than 1,400
real vehicles per hour on this climb. That is the engineering message, and it is
the standard justification for a climbing lane.
Part (b) — state the capacity equation for a signalised
approach. Capacity is the saturation flow rate reduced by the fraction of
the cycle during which the approach actually has green:
$$c=s\times\frac{g}{C},\qquad
s=s_{o}\,N\,f_{w}\,f_{HV}\,f_{g}\,f_{p}\,f_{bb}\,f_{a}\,f_{RT}\,f_{LT}$$
where \(s_{o}\) is the ideal saturation flow per lane and the remaining factors
adjust for width, heavy vehicles, grade, parking, bus blockage, area type and
turning movements.
Evaluate the adjustment factors one by one. Taking
\(s_{o}=1900\) pc/h/ln of green: the width factor for a 3.5 m lane is
$$f_{w}=1+\frac{W-3.6}{9}=1+\frac{3.5-3.6}{9}=0.9889$$
the heavy-vehicle factor with 8 % trucks at \(E_{T}=2.0\) is
$$f_{HV}=\frac{100}{100+\%HV\,(E_{T}-1)}=\frac{100}{100+8(1)}=0.9259$$
and the grade factor for a 3 % down grade is
$$f_{g}=1-\frac{\%G}{200}=1-\frac{-3}{200}=1.015$$
The down grade is a small bonus, because vehicles accelerate away from the
stop line more briskly. There is no parking and no bus stop, and the movement is
through only, so \(f_{p}=f_{bb}=f_{RT}=f_{LT}=1.00\); the Central Business District
location carries the area-type penalty \(f_{a}=0.90\) for narrow lanes, pedestrian
friction and general activity.
Compute the saturation flow rate. Multiplying the ideal
rate by the two lanes and by every factor in turn,
$$s=1900\times 2\times 0.9889\times 0.9259\times 1.015\times 0.90
=3{,}178\ \text{veh/h of green}$$
i.e. about 1,589 veh/h of green per lane, or a saturation headway close to
2.3 seconds per vehicle per lane.
Reduce to capacity over the whole cycle. With a green-to-cycle
ratio of 0.50 the approach is discharging for only half of each 75-second cycle,
so
$$c=s\times\frac{g}{C}=3{,}178\times 0.50$$
$$\boxed{c\approx 1{,}589\ \text{veh/h}}$$
Note that the cycle length itself does not enter the capacity — only the
ratio \(g/C\) does. The 75-second cycle matters instead for the
quantum of discharge: there are 48 cycles per hour, each with 37.5 s of
effective green, so capacity corresponds to about 33 vehicles per cycle across the
two lanes, or roughly 17 per lane per cycle. A queue longer than about 17 vehicles
per lane will therefore not clear in one green, which is the practical check an
engineer applies next.