16-Civ-A6 Highway Design, Construction, and Maintenance · December 2015

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: 98-Civ-A6 Transportation Planning & Engineering, National Examination, December 2015. Seven questions, each of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.

Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley); Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall); Roess, Prassas & McShane, Traffic Engineering (Pearson); Ortúzar & Willumsen, Modelling Transport (Wiley) for the demand-model chapters (trip generation, distribution, mode choice, assignment).

Question 4: Greenshields Model and Shock Waves (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. Greenshields traffic stream plus a moving bottleneck (the truck).

Given data
Quantity	Value
Free-flow speed $u_f$	60 km/h
Jam density $k_j$	144 veh/km
Normal (upstream) speed, state A	45 km/h
Truck speed / platoon speed, state B	10 km/h
Platoon density / flow, state B	120 veh/km / 1200 veh/h
Truck trip on highway	1.0 km at 10 km/h

Find. Capacity speed and density; platoon length when the truck exits; time for the platoon to clear.

Approach. Use Greenshields $u=u_f(1-k/k_j)$ and $q=ku$ for the capacity point and for the normal state A. Shock-wave speeds between states follow $u_w=\Delta q/\Delta k$. The platoon length is set by the stopping wave (behind) and the truck (ahead); dissipation is when the faster recovery wave overtakes the stopping wave.

Greenshields $q$–$k$ parabola. The slope of the chord between two states equals the shock-wave speed: A→B (stopping) is $-5$ km/h, B→C (recovery to capacity) is $-20$ km/h.

(a) Capacity point. For Greenshields the maximum flow is at half the jam density: $$k_m=\tfrac{k_j}{2}=72\ \text{veh/km},\qquad u_m=\tfrac{u_f}{2}=30\ \text{km/h},$$ $$\boxed{q_{\max}=u_m k_m = \tfrac{u_f k_j}{4}=\frac{60\times144}{4}=2160\ \text{veh/h}}$$
State A (normal flow). From $u=u_f(1-k/k_j)$ with $u=45$: $$k_A=k_j\!\left(1-\frac{u_A}{u_f}\right)=144\!\left(1-\tfrac{45}{60}\right)=36\ \text{veh/km},\quad q_A=45\times36=1620\ \text{veh/h}.$$ State B (platoon) is given: $k_B=120$, $q_B=1200$, $u_B=10$ km/h.
Stopping shock wave (back of the platoon). $$u_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{1200-1620}{120-36}=\frac{-420}{84}=-5\ \text{km/h}.$$ The back of the platoon moves upstream at 5 km/h while the truck is on the road.
(b) Platoon length when the truck exits. The truck is on the highway for $$t_{\text{truck}}=\frac{1.0\ \text{km}}{10\ \text{km/h}}=0.1\ \text{h}=6\ \text{min}.$$ Its front (at exit) is at $x=1.0$ km; the stopping wave has run back to $x=u_{AB}t_{\text{truck}}=-0.5$ km. Hence $$L=1.0-(-0.5)=\boxed{1.5\ \text{km}\ \ (=k_B L = 180\ \text{vehicles})}$$
Recovery shock wave (front of the platoon after release). Once the truck leaves, the head of the platoon discharges into the open road at capacity (state C: $k=72$, $q=2160$): $$u_{BC}=\frac{q_C-q_B}{k_C-k_B}=\frac{2160-1200}{72-120}=\frac{960}{-48}=-20\ \text{km/h}.$$ This recovery wave also travels upstream, but four times faster than the stopping wave.
(c) Dissipation time. The recovery wave ($-20$ km/h) chases the stopping wave ($-5$ km/h); the platoon of length $L=1.5$ km closes at the relative speed $|{-}20|-|{-}5|=15$ km/h: $$t_{\text{diss}}=\frac{L}{|u_{BC}|-|u_{AB}|}=\frac{1.5}{15}=0.1\ \text{h}=\boxed{6\ \text{min after the truck exits}}$$

Question 4 — results
Quantity	Value
Capacity speed $u_m$ / density $k_m$	30 km/h / 72 veh/km
Capacity flow $q_{\max}$	2160 veh/h
Stopping-wave speed $u_{AB}$	−5 km/h
Platoon length at exit	1.5 km (180 vehicles)
Recovery-wave speed $u_{BC}$	−20 km/h
Time to dissipate	6 min

Topic: Greenshields speed–density model and Lighthill–Whitham–Richards shock waves from a moving bottleneck.

Key relations: $u=u_f(1-k/k_j)$; $q=ku$; $q_{\max}=u_fk_j/4$ at $k_j/2$; shock speed $u_w=\Delta q/\Delta k$ (the chord slope on the $q$–$k$ curve).

Why this works: A shock is the interface between two traffic states; conservation of vehicles across it forces its speed to equal $\Delta q/\Delta k$. A slow truck creates a congested platoon (state B) bounded upstream by the stopping wave and downstream by the truck. When the truck leaves, the head discharges at capacity, launching a faster recovery wave; the platoon disappears the instant that wave overtakes the still-advancing stopping wave.

Common pitfall: Using the truck's 1 km travel distance as the platoon length — the stopping wave has run a further 0.5 km upstream, so the platoon is 1.5 km. Also, forgetting that both waves move upstream, so dissipation depends on their difference in speed, not their sum.

Source: Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (traffic flow, Greenshields model and shock waves).

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