Question 1 of 6: Intersection flows, saturation flow, clearance intervals and flow ratio
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 98-Civ-B10
Traffic Engineering. Three-hour, open-book examination; any non-communicating
calculator is permitted. Six questions are printed and five complete solutions are
required, all questions being of equal value (20 marks each). The printed grading scheme
is Q1 (a) to (d) 5 marks each; Q2 (a) to (e) 4 marks each; Q3 (a) to (e) 4 marks each;
Q4 (a) and (b) 10 marks each; Q5 (a) to (e) 4 marks each; Q6 (10 + 5 + 5) marks. The paper
states that if doubt exists as to the interpretation of a question the candidate should
submit with the answer paper a clear statement of any assumptions made, and that any data
required but not given can be assumed. All six questions are worked below.
Reference texts. Garber, N. J. and Hoel, L. A.,
Traffic and Highway Engineering, 5th ed. — Ch. 4 (traffic engineering
studies: volume studies, peak-hour factor), Ch. 6 (fundamental principles of traffic flow:
Poisson arrivals, deterministic and stochastic queueing, M/M/1 and M/M/N channels),
Ch. 8 (intersection control: cycle length, phasing, change and clearance intervals,
progression and time–space diagrams) and Ch. 10 (capacity and level of service at
signalised intersections). This is the principal reference for the subject.
Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory
Technical Paper No. 56 — the optimum-cycle and three-term delay formulae used
throughout Questions 1 to 3. Transportation Research Board, Highway Capacity Manual
— saturation-flow adjustment factors and the pedestrian-green requirement.
Transportation Association of Canada, Manual of Uniform Traffic Control Devices for
Canada (MUTCDC) and Geometric Design Guide for Canadian Roads — Canadian
signal-timing, clearance-interval and crosswalk practice. Institute of Transportation
Engineers, Traffic Engineering Handbook — controller types and coordinated
timing plans.
Check: the four values assumed under the paper's own
NOTE 2 (“any data required, but not given, can be assumed”). They are
declared here once and used consistently in every question.
Bus passenger-car equivalent $E_B = 2.0$ pcu (HCM value for buses on
level terrain). The paper gives bus occupancies but no pcu equivalent; note that
the pcu-based and vehicle-based flow ratios in Question 1 agree exactly for any choice of
$E_B$, so this assumption does not affect Questions 2 or 3.
Pedestrian walking speed $S_p = 1.2$ m/s (HCM 2000 / MUTCDC design
value; the more recent 1.1 m/s would lengthen the pedestrian interval by about 1 s).
Lost time convention: the whole intergreen (amber + all-red) is taken
as lost, so effective green equals displayed green. This is the conservative reading and
needs no assumed start-up lost time. Webster's alternative, $L = n\ell + R$ with
$\ell \approx 2$ s, would give $L = 10$ s instead of 12 s and lengthen each green by about
1 s.
Delay model: Webster's three-term formula, with the first two terms
reported separately in Question 3(b) as the uniform and overflow components.
Question 1: Intersection flows, saturation flow, clearance intervals and flow ratio (20 marks)
Find. The four approach demands expressed both in passenger-car
units and in persons per hour; the heavy-vehicle-adjusted saturation flows on the two
bus-carrying approaches; the all-red, intergreen and total lost time implied by the
intersection geometry; and the intersection flow ratio $Y$ that will drive the cycle-length
calculation in Question 2.
Question 1 — intersection layout. Two 3.75 m lanes in each direction on both streets, 3 m crosswalks immediately outside the junction box, and stop lines set back 1 m from each crosswalk.
Approach. Convert every approach to passenger-car units and to
persons; obtain the saturation flow from the basic rate through the heavy-vehicle factor;
size the all-red from the distance a stopped car must cover to be clear of the far
crosswalk; and form the flow ratio from the critical lane on each of the two phases.
Part (a) — convert each approach to passenger-car units.
A bus is taken as $E_B = 2.0$ passenger-car units, so the demand on an approach is
$$q_{\text{pcu}} = N_{\text{car}} + E_B\,N_{\text{bus}}.$$
North-bound: $800 + 2.0(15) = 830$ pcu/h. South-bound: $700 + 2.0(12) = 724$ pcu/h. The
east- and west-bound approaches carry no buses, so they are already 500 and 700 pcu/h.
Adding the four approaches,
$$\boxed{\;\sum q_{\text{pcu}} = 830 + 724 + 500 + 700 = 2754\ \text{pcu/h}\;}$$
Part (a) — convert the same demands to persons per hour.
Occupancy is applied to the vehicles, not to the passenger-car units:
$$P = N_{\text{car}}\,O_{\text{car}} + N_{\text{bus}}\,O_{\text{bus}}.$$
North-bound: $800(2.0) + 15(25) = 1600 + 375 = 1975$ persons/h. South-bound:
$700(2.0) + 12(15) = 1400 + 180 = 1580$ persons/h. East-bound $500(2.0) = 1000$ and
west-bound $700(2.0) = 1400$ persons/h. The intersection therefore serves
$$\boxed{\;\sum P = 1975 + 1580 + 1000 + 1400 = 5955\ \text{persons/h}\;}$$
Note that the north-bound approach carries only 30 % more passenger-car units than the
east-bound approach but almost twice the people — the person-based view is what
justifies transit priority, and it reappears in Question 3(c).
Part (b) — reduce the basic saturation flow for heavy vehicles.
The 1800 pc/h figure is a passenger-car rate; the equivalent rate in mixed
vehicles per hour follows from the heavy-vehicle adjustment factor
$$f_{HV} = \frac{1}{1 + P_{HV}\,(E_B - 1)}, \qquad s = s_0\,f_{HV},$$
where $P_{HV}$ is the proportion of buses in the traffic stream. North-bound there are 15
buses in $800 + 15 = 815$ vehicles, so $P_{HV} = 0.01840$ and
$f_{HV} = 1/(1 + 0.01840) = 0.9819$. South-bound, $P_{HV} = 12/712 = 0.01685$ and
$f_{HV} = 0.9834$. Hence
$$\boxed{\;s_{NB} = 1800(0.9819) = 1767\ \text{veh/h/lane},\qquad
s_{SB} = 1800(0.9834) = 1770\ \text{veh/h/lane}\;}$$
or 3535 and 3540 veh/h over the two lanes of each approach. A useful internal check: the
flow ratio is the same whether it is formed in pcu against 1800 or in vehicles against the
adjusted rate, because $f_{HV}$ is exactly the conversion between the two
($415/1800 = 407.5/1767.5 = 0.2306$).
Part (c) — measure the distance a departing car must clear.
The last car to enter on amber starts at the stop line and is not clear of the conflict
area until its rear bumper has passed the far crosswalk. Reading the chain across the
intersection,
$$W + L_v = \underbrace{1.0}_{\text{set-back}} + \underbrace{3.0}_{\text{near crosswalk}}
+ \underbrace{4(3.75)}_{\text{cross-street carriageway}} + \underbrace{3.0}_{\text{far crosswalk}}
+ \underbrace{6.0}_{\text{car length}} = 28.0\ \text{m}.$$
Question 1(c) — the clearing distance is measured from the stop line to the far side of the far crosswalk, plus one car length.
Part (c) — convert the clearing distance to an all-red interval.
At the stated clearing speed of 30 km/h = 8.333 m/s,
$$R = \frac{W + L_v}{v} = \frac{28.0}{8.333} = 3.36\ \text{s}
\;\Rightarrow\; R = 3\ \text{s (nearest second)}.$$
The intergreen is the amber plus the all-red, and with two phases the intersection lost
time is the sum of the two intergreens:
$$I = A + R = 3.0 + 3 = 6\ \text{s},\qquad
\boxed{\;L = n\,I = 2(6) = 12\ \text{s per cycle}\;}$$
Twelve seconds of every cycle therefore buys no capacity at all, which is exactly why
Question 2 shows the cycle cannot be shortened indefinitely.
Part (d) — form the flow ratio from the critical lane on each phase.
With two equal lanes per approach and no turning bans stated, the demand splits evenly,
and the flow ratio of an approach is
$$y = \frac{q_{\text{lane}}}{s_0}
= \frac{q_{\text{pcu}}/2}{1800}.$$
This gives $y_{NB} = 415/1800 = 0.2306$, $y_{SB} = 362/1800 = 0.2011$,
$y_{EB} = 250/1800 = 0.1389$ and $y_{WB} = 350/1800 = 0.1944$. Each phase is governed by
its heaviest approach — north-bound on the N–S phase and west-bound on the
E–W phase — so
$$\boxed{\;Y = y_{NS} + y_{EW} = 0.2306 + 0.1944 = 0.425\;}$$
Because $Y < 1$ the intersection is workable; the margin $1 - Y = 0.575$ is what the lost
time and the delay formulae will consume.
