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

Question 4 of 7: Greenshields Model and Shock Waves at a Signal

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination — 98-Civ-A6 Transportation Planning & Engineering, May 2015. 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, 6th ed. (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).

Question 4: Greenshields Model and Shock Waves at a Signal (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. A signalised approach obeying the Greenshields linear speed–density relation, with a known free-flow speed, capacity, approach state and red duration.

Given data
Quantity	Value
Free-flow speed, $u_f$	60 km/h
Capacity, $q_{\max}$	1200 veh/h
Approach speed, $u_A$	45 km/h
Approach density, $k_A$	20 veh/km
Red interval	30 s = 1/120 h

Find. (a) $k_j$ and $k_m$; (b) maximum queue length during red; (c) time to dissipate the queue after red.

Approach. Use the Greenshields parabola to get jam and optimum densities, then track the stopping and starting shock waves whose speeds are $u_w=\Delta q/\Delta k$.

Jam density and density at capacity. For Greenshields, $q_{\max}=u_f k_j/4$, so $$k_j=\frac{4\,q_{\max}}{u_f}=\frac{4(1200)}{60}=80\ \text{veh/km},\qquad k_m=\frac{k_j}{2}=40\ \text{veh/km}.$$ $$\boxed{k_j = 80\ \text{veh/km},\quad k_m = 40\ \text{veh/km.}}$$ Consistency check: $u_A=u_f(1-k_A/k_j)=60(1-20/80)=45$ km/h ✓, so the approach flow is $q_A=u_A k_A=45(20)=900$ veh/h.
Stopping shock wave (approach state A → stopped/jam state J). The stopped state has $k_J=80$, $q_J=0$. The wave speed is $$u_{w,\text{stop}}=\frac{q_J-q_A}{k_J-k_A}=\frac{0-900}{80-20}=-15\ \text{km/h},$$ i.e. the back of the queue propagates upstream at 15 km/h.
Maximum queue length during red. The stopping wave acts for the whole 30 s red: $$L_{\max}=|u_{w,\text{stop}}|\times t_{\text{red}}=15\times\frac{1}{120}=0.125\ \text{km}.$$ $$\boxed{L_{\max}=125\ \text{m}\ \ (=0.125\times 80 = 10\ \text{vehicles at jam density).}}$$
Starting shock wave (jam J → capacity discharge C). When the light turns green the queue discharges at capacity, $k_C=k_m=40$, $q_C=1200$. The release wave speed is $$u_{w,\text{start}}=\frac{q_C-q_J}{k_C-k_J}=\frac{1200-0}{40-80}=-30\ \text{km/h},$$ travelling upstream at 30 km/h — twice as fast as the stopping wave, so it overtakes the back of the queue.
Time to dissipate. Measuring position upstream from the stop line, the stopping wave front is at $x_{\text{stop}}=-15\,(t_{\text{red}}+\tau)$ and the starting wave (released at the end of red) is at $x_{\text{start}}=-30\,\tau$, where $\tau$ is time after the end of red. Setting them equal: $$-30\,\tau=-15\,(t_{\text{red}}+\tau)\;\Rightarrow\;-15\,\tau=-15\,t_{\text{red}}\;\Rightarrow\;\tau=t_{\text{red}}.$$ $$\boxed{\tau = 30\ \text{s after the end of red}\ (\text{the queue clears }60\ \text{s after red began}).}$$ The two waves meet 250 m upstream, which is where the last stopped vehicle is released.

Question 4 — Final Results
Quantity	Value
Jam density $k_j$	80 veh/km
Density at capacity $k_m$	40 veh/km
Maximum queue length	125 m (≈ 10 vehicles)
Stopping / starting wave speeds	−15 / −30 km/h (upstream)
Time to dissipate after red	30 s

Topic: Greenshields speed–density model and shock-wave analysis of a signal queue.

Key relations: $u=u_f(1-k/k_j)$; $q=uk$; $q_{\max}=u_fk_j/4$ at $k_m=k_j/2$; shock-wave 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; its speed is the slope of the chord joining the two states on the flow–density (fundamental) diagram. The stopping wave (approach→jam) sets how far back the queue reaches during red; the faster starting wave (jam→capacity) chases and consumes it. Because the release wave is faster than the formation wave, the queue always eventually clears, and here the geometry makes the clearance time equal the red time.

Common pitfall: Using speeds instead of the chord slope for wave speed, or forgetting that the stopping wave keeps moving upstream during the first part of green until the starting wave catches it.

Source: Mannering & Washburn, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (traffic flow, shock waves); Roess, Prassas & McShane, Traffic Engineering, Ch. 5–12.

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