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

National Examination — 98-Civ-A6 Transportation Planning & Engineering, December 2014. Closed book (one two-sided aid sheet), 3 hours. Seven questions of equal value (20 marks); any five constitute a complete paper. All seven are solved here as a study resource.

Reference texts (subject):

Papacostas, C.S. & Prevedouros, P.D., Transportation Engineering and Planning, 3rd ed. (Prentice Hall).
Roess, R.P., Prassas, E.S. & McShane, W.R., Traffic Engineering, 4th ed. (Pearson).
Mannering, F.L. & Washburn, S.S., Principles of Highway Engineering and Traffic Analysis (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).
Transportation Association of Canada (TAC), Geometric Design Guide for Canadian Roads (Canadian context).

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 stream: free-flow speed $u_f=50$ km/h, capacity (maximum flow) $q_{\max}=1500$ veh/h. Approaching state A travels at $u_A=40$ km/h. Blockage duration $=6$ min $=0.1$ h.

Find. (a) jam density $k_j$ and density at capacity $k_m$; (b) maximum platoon length; (c) dissipation time after the debris is removed.

Greenshields flow–density parabola. Chord slopes are shock-wave speeds: A→jam is the stopping wave ($-10$ km/h); jam→capacity is the faster starting wave ($-25$ km/h).

Approach. Fix the parabola from $u_f$ and $q_{\max}$; find the approaching state A from its speed; then the shock-wave speed between any two states is the chord slope $u_w=\Delta q/\Delta k$. The platoon grows behind the stopping wave for 6 min, then is eaten by the faster starting wave.

Jam and capacity densities. For Greenshields $q_{\max}=u_fk_j/4$, so $$k_j=\frac{4q_{\max}}{u_f}=\frac{4(1500)}{50}=\boxed{120\ \text{veh/km}},\qquad k_m=\frac{k_j}{2}=60\ \text{veh/km}.$$ (The speed at capacity is $u_f/2=25$ km/h.)
Approaching state A. From $u=u_f(1-k/k_j)$: $40=50(1-k_A/120)\Rightarrow k_A=24$ veh/km, and $q_A=u_Ak_A=40(24)=960$ veh/h.
Stopping shock wave (A → jam). The debris drives traffic to the jam state (B: $k_j=120$, $q=0$). The wave speed is $$u_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{0-960}{120-24}=-10\ \text{km/h}\quad(\text{moves upstream}).$$
Maximum platoon length. During the 6-min ($0.1$ h) blockage the front of the queue is held at the debris while the stopping wave runs back at 10 km/h, so the queue reaches $$L_{\max}=|u_{AB}|\,t_{\text{stall}}=10(0.1)=\boxed{1.0\ \text{km}},$$ containing $k_j\,L_{\max}=120(1.0)=120$ vehicles. Once the debris is removed the (faster) starting wave shortens the platoon, so 1.0 km is the maximum.
Starting shock wave (jam → capacity). When released, vehicles discharge at capacity (C: $k_m=60$, $q=1500$): $$u_{BC}=\frac{q_C-q_B}{k_C-k_B}=\frac{1500-0}{60-120}=-25\ \text{km/h}.$$
Dissipation time after removal. The starting wave (25 km/h) overtakes the stopping wave (10 km/h); the platoon disappears after $$\tau=\frac{L_{\max}}{|u_{BC}|-|u_{AB}|}=\frac{1.0}{25-10}=\frac{1}{15}\ \text{h}=\boxed{4\ \text{minutes}}.$$

Question 4 — results
Quantity	Value
Jam density $k_j$	120 veh/km
Density at capacity $k_m$	60 veh/km
Approaching state A	$k_A=24$ veh/km, $q_A=960$ veh/h
Stopping / starting wave speed	$-10$ / $-25$ km/h
Maximum platoon length	1.0 km (120 vehicles)
Dissipation time after removal	4 minutes

Topic: Greenshields' linear speed–density model and shock-wave (kinematic-wave) analysis.

Key relations: $u=u_f(1-k/k_j)$, $q=uk$, $q_{\max}=u_fk_j/4$ at $k_m=k_j/2$; shock speed $u_w=\dfrac{q_2-q_1}{k_2-k_1}$.

Why this works: A shock wave is the boundary between two traffic states; conservation of vehicles across it forces its speed to equal the chord slope on the $q$–$k$ diagram. Because the jam→capacity chord is steeper than the approach→jam chord, the recovery wave always outruns the stopping wave and eventually clears the queue.

Common pitfall: Using the approach speed (40 km/h) as the discharge speed — the queue discharges at capacity (25 km/h, $k=60$), not at the pre-incident state; and mistaking the maximum platoon as occurring at full dissipation rather than at the instant of release.

Source: Roess, Prassas & McShane, Traffic Engineering, Ch. 5 (traffic-stream models & shock waves); May, Traffic Flow Fundamentals.

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