Greenshields Model and Shock Waves behind a Slow Truck: 16-Civ-B7 Transportation Planning and Engineering, December 2019, Q4

Question 4: Greenshields Model and Shock Waves behind a Slow Truck (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 single-lane highway obeying the Greenshields linear speed–density relationship, temporarily blocked by a slow truck.

Traffic states on the single-lane highway
Quantity	Symbol	Value
Free-flow speed	$u_f$	80 km/h
Jam density	$k_j$	100 veh/km
Normal (upstream) speed — state A	$u_A$	64 km/h
Truck speed and platoon speed — state B	$u_B$	8 km/h
Platoon density — state B	$k_B$	90 veh/km
Platoon flow — state B	$q_B$	720 veh/h
Distance travelled by the truck on the highway	$d$	0.5 km

Find. The capacity and the density at capacity; the length of the platoon at the instant the truck leaves; and the time from that instant until the platoon has completely dissipated.

Figure 4.1 — Time–space diagram of the moving bottleneck. The shaded region is the congested platoon (state B), bounded by the truck trajectory, the rear shock at −8 km/h and, after the truck exits at t = 3.75 min, the release wave at −32 km/h. The two boundaries close on each other 2.5 min later.

Approach. Fit the Greenshields parabola from $u_f$ and $k_j$ to get capacity and the critical density, characterise the upstream state A from its speed, compute the rear shock speed between states A and B from the chord slope of the flow–density curve, obtain the platoon length from the separation of the truck and the rear shock over the blockage duration, and then find the dissipation time from the closing speed of the rear shock and the release wave.

(a) Fit the Greenshields relations and obtain capacity. Greenshields assumes speed falls linearly with density, $u = u_f\left(1-\dfrac{k}{k_j}\right)$, so $q = uk = u_f\left(k - \dfrac{k^2}{k_j}\right)$ is a parabola whose maximum lies at $k_c = k_j/2$: $$k_c=\frac{k_j}{2}=\frac{100}{2}=50 \text{ veh/km}, \qquad u_c=\frac{u_f}{2}=40 \text{ km/h}$$ $$q_{\max}=\frac{u_f k_j}{4}=\frac{80 \times 100}{4}$$ $$\boxed{q_{\max}=2000 \text{ veh/h at } k_c=50 \text{ veh/km}}$$
Characterise the upstream state A. Inverting the speed–density relation at $u_A = 64$ km/h, $$k_A=k_j\left(1-\frac{u_A}{u_f}\right)=100\left(1-\frac{64}{80}\right)=100(0.20)=20 \text{ veh/km}$$ $$q_A=u_A k_A=64 \times 20 = 1280 \text{ veh/h}$$ Before checking anything further, confirm that the given platoon state is consistent with the same curve: $u_f(1-k_B/k_j)=80(1-90/100)=8$ km/h and $8 \times 90 = 720$ veh/h, which reproduce both printed values exactly. The data are therefore internally consistent and no alternative reading of the curve is needed.
Compute the rear (stopping) shock speed. The shock between two states is the chord slope of the flow–density curve joining them: $$w_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{720-1280}{90-20}=\frac{-560}{70}$$ $$\boxed{w_{AB}=-8 \text{ km/h (upstream, i.e. against the direction of travel)}}$$ The negative sign means the back of the platoon marches upstream at 8 km/h even while the platoon itself creeps forward at 8 km/h — the two boundaries separate at $u_B - w_{AB} = 16$ km/h.
Find how long the blockage lasted. The truck occupies the highway for the time it needs to cover 0.5 km at 8 km/h: $$t_b=\frac{d}{u_B}=\frac{0.5}{8}=0.0625 \text{ h}=3.75 \text{ min}$$
(b) Obtain the platoon length. The front of the platoon travels with the truck at $+8$ km/h and the rear boundary travels at $-8$ km/h, both starting from the truck's entry point at $t=0$, so the platoon grows at their relative speed: $$L=(u_B-w_{AB})\,t_b=\bigl(8-(-8)\bigr)(0.0625)=16 \times 0.0625$$ $$\boxed{L=1.0 \text{ km immediately after the truck exits}}$$ The vehicles stored in it are $N=k_B L = 90 \times 1.0 = 90$ vehicles. Verify that count independently from the arrival side: vehicles join the queue at the relative speed $u_A-w_{AB}$ at upstream density $k_A$, giving $(64+8)(20)(0.0625) = 90$ vehicles — the two counts agree, which confirms the shock speed. Note that the naive count $q_A t_b = 1280 \times 0.0625 = 80$ vehicles is wrong, because it counts only the vehicles passing a fixed point and ignores the shock advancing upstream to meet them.
Compute the release (starting) wave. Once the truck leaves, the vehicles at the head of the platoon accelerate away into an uncongested road, so they discharge at capacity — state C is $q_C=q_{\max}=2000$ veh/h at $k_C=k_c=50$ veh/km, moving at $u_c=40$ km/h. The wave separating the platoon from the discharging traffic is $$w_{BC}=\frac{q_C-q_B}{k_C-k_B}=\frac{2000-720}{50-90}=\frac{1280}{-40}=-32 \text{ km/h}$$ which runs upstream four times faster than the rear shock — the platoon is being eaten from the front faster than it is being fed from the back.
(c) Dissipation time. Both boundaries move upstream, so the platoon is consumed at their closing speed, not at $|w_{BC}|$ alone: $$t_d=\frac{L}{|w_{BC}|-|w_{AB}|}=\frac{1.0}{32-8}=\frac{1.0}{24}=0.04167 \text{ h}$$ $$\boxed{t_d=2.5 \text{ minutes after the truck exits}}$$ Checking the geometry: in 2.5 min the release wave travels $32(0.04167)=1.333$ km upstream and the rear shock travels $8(0.04167)=0.333$ km upstream, and the difference is exactly the 1.0 km platoon. The two waves meet 0.833 km upstream of the truck's exit point, 6.25 min after the truck entered. In total $(u_A-w_{AB})k_A(t_b+t_d)=72 \times 20 \times 0.10417 = 150$ vehicles were caught in the disturbance.

Check: state C is taken as capacity flow (2000 veh/h at 50 veh/km), which is the standard treatment of a queue discharging into an uncongested downstream road and is what the question's closing assumption directs. If instead the platoon were assumed to recover directly to the upstream state A, the release wave would be $(1280-720)/(20-90)=-8$ km/h — identical to the rear shock — and the platoon would never dissipate, which contradicts the question. Capacity discharge is therefore the only consistent reading.

Question 4 — final results
Quantity	Symbol	Value
Capacity	$q_{\max}$	2000 veh/h
Density at capacity	$k_c$	50 veh/km
Speed at capacity	$u_c$	40 km/h
Upstream density and flow (state A)	$k_A,\,q_A$	20 veh/km, 1280 veh/h
Blockage duration	$t_b$	0.0625 h (3.75 min)
Rear shock speed	$w_{AB}$	−8 km/h (upstream)
Platoon length at truck exit	$L$	1.0 km (90 vehicles)
Release wave speed	$w_{BC}$	−32 km/h (upstream)
Dissipation time after exit	$t_d$	0.04167 h (2.5 min)
Total vehicles affected	$N_{\text{tot}}$	150 vehicles

Topic. Greenshields' linear speed–density model and shock-wave analysis of a moving bottleneck.

Key relations. $u=u_f(1-k/k_j)$; $q=uk$; $k_c=k_j/2$, $u_c=u_f/2$, $q_{\max}=u_fk_j/4$; shock speed $w_{ij}=(q_j-q_i)/(k_j-k_i)$; platoon length $L=(u_B-w_{AB})t_b$; dissipation time $t_d=L/(|w_{BC}|-|w_{AB}|)$.

Why this works. Conservation of vehicles across a moving discontinuity in the flow–density plane gives the Rankine–Hugoniot condition: a boundary between two uniform traffic states propagates at the slope of the chord joining them, not at the slope of the tangent (which is the kinematic wave speed within a single state). Everything else follows from geometry on a time–space diagram. The platoon is the region bounded above by the truck trajectory and below by the rear shock, so its length is the separation of those two lines; after the truck leaves, the upper boundary is replaced by the much faster release wave and the region closes at the difference of the two upstream speeds. Because the two boundaries travel in the same direction, subtracting their magnitudes — not adding them — gives the closing rate.

Common pitfall. The two classic errors are dividing by $|w_{BC}|$ alone in part (c), which roughly halves the dissipation time (1.875 min instead of 2.5), and counting the trapped vehicles as $q_A t_b$ instead of $(u_A-w_{AB})k_A t_b$, which undercounts by 10 vehicles here. A third is assuming a queue boundary must always travel upstream: the sign comes from the chord slope, and if the platoon flow exceeded the approach flow the rear shock would run downstream. Always cross-check $k_B L$ against $(u_A-w_{AB})k_A t_b$ — if they disagree, the shock speed is wrong.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §6.3–6.4 (Greenshields model, shock waves); Papacostas & Prevedouros, Transportation Engineering and Planning, 3rd ed., §5.3–5.5.

Quantity	Symbol	Value
Free-flow speed	\(u_f\)	80 km/h
Jam density	\(k_j\)	100 veh/km
Normal (upstream) speed — state A	\(u_A\)	64 km/h
Truck speed and platoon speed — state B	\(u_B\)	8 km/h
Platoon density — state B	\(k_B\)	90 veh/km
Platoon flow — state B	\(q_B\)	720 veh/h
Distance travelled by the truck on the highway	\(d\)	0.5 km

Quantity	Symbol	Value
Capacity	\(q_{\max}\)	2000 veh/h
Density at capacity	\(k_c\)	50 veh/km
Speed at capacity	\(u_c\)	40 km/h
Upstream density and flow (state A)	\(k_A,\,q_A\)	20 veh/km, 1280 veh/h
Blockage duration	\(t_b\)	0.0625 h (3.75 min)
Rear shock speed	\(w_{AB}\)	−8 km/h (upstream)
Platoon length at truck exit	\(L\)	1.0 km (90 vehicles)
Release wave speed	\(w_{BC}\)	−32 km/h (upstream)
Dissipation time after exit	\(t_d\)	0.04167 h (2.5 min)
Total vehicles affected	\(N_{\text{tot}}\)	150 vehicles

Question 4 of 7: Greenshields Model and Shock Waves behind a Slow Truck

Question 4: Greenshields Model and Shock Waves behind a Slow Truck (20 marks)