16-Civ-A6 Highway Design, Construction, and Maintenance · May 2014
Question 4 of 7: Greenshields Model and Shock Waves
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination, 98-Civ-A6 Transportation Planning & Engineering, May 2014 — 3 hours, closed book (one two-sided aid sheet). Seven questions; any five constitute a complete examination and all are of equal value (20 marks). All seven are solved below as a study resource.
Given. One‑lane road, Greenshields model. Free‑flow speed $u_f=60$ km/h, capacity $q_{\max}=2160$ veh/h. Normal state A: $u_A=45$ km/h, $k_A=36$ veh/km. A slow vehicle at $10$ km/h travels $1.0$ km then exits; following vehicles slow to $10$ km/h (state B).
Traffic states (Greenshields, $u_f=60$, $k_j=144$)
State
Speed u (km/h)
Density k (veh/km)
Flow q (veh/h)
A — normal
45
36
1620
B — behind slow veh
10
120
1200
C — capacity discharge
30
72
2160
Find. (a) jam density and density at capacity; (b) platoon length when the slow vehicle exits; (c) time for the platoon to dissipate.
Greenshields flow–density parabola. Chord slopes are shock‑wave speeds: A–B (stopping wave, −5 km/h) and B–C (recovery wave, −20 km/h). Negative slopes mean both waves travel upstream.
Approach. Fix the Greenshields parameters from $u_f$ and $q_{\max}$, obtain each traffic state, then track the stopping and starting shock waves whose speeds are chord slopes $u_w=\Delta q/\Delta k$ on the $q$–$k$ curve.
Greenshields parameters (a). At capacity $q_{\max}=u_f k_j/4$, so the jam density is $k_j=4q_{\max}/u_f=4(2160)/60=\boxed{144\ \text{veh/km}}$ and the density at capacity is $k_m=k_j/2=\boxed{72\ \text{veh/km}}$. Check: the normal state gives $u=u_f(1-k/k_j)=60(1-36/144)=45$ km/h, matching the stated 45 km/h.
Congested state B. Behind the slow vehicle $u_B=10$ km/h, so $k_B=k_j(1-u_B/u_f)=144(1-10/60)=120\ \text{veh/km}$ and $q_B=u_Bk_B=1200\ \text{veh/h}$. Normal state A has $q_A=45\times36=1620\ \text{veh/h}$.
Stopping shock A–B and platoon length (b). The back of the platoon is the shock between approaching state A and queued state B: $u_{AB}=\dfrac{q_B-q_A}{k_B-k_A}=\dfrac{1200-1620}{120-36}=-5\ \text{km/h}$ (upstream). The slow vehicle (front of the platoon) covers $1.0$ km at $10$ km/h, i.e. it is on the road for $t_s=1.0/10=0.1\ \text{h}$. In that time the front advances at $+10$ and the back at $-5$ km/h, so the platoon grows to $L=(10-(-5))\times0.1=\boxed{1.5\ \text{km}}$, containing $L\,k_B=1.5\times120=180$ vehicles.
Recovery shock B–C. When the slow vehicle exits, the front vehicles accelerate and discharge at capacity (state C: $k_C=72$, $q_C=2160$). The starting wave is $u_{BC}=\dfrac{q_C-q_B}{k_C-k_B}=\dfrac{2160-1200}{72-120}=-20\ \text{km/h}$ (upstream, faster than the back shock).
Dissipation time (c). At the instant of exit the platoon is $1.5$ km long. The recovery wave chases the back shock, closing the gap at $|u_{BC}|-|u_{AB}|=20-5=15\ \text{km/h}$. The platoon clears after $\tau=L/(|u_{BC}|-|u_{AB}|)=1.5/15=0.1\ \text{h}=\boxed{6\ \text{min}}$.
