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

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 Greenshields (linear speed–density) stream with a stated free-flow speed and capacity, an operating point on the approach, and a fixed red interval.

Find. The jam density and the density at capacity, the maximum queue length reached during red, and the time the queue takes to dissipate once green begins.

Approach. Fit the Greenshields curve from $u_f$ and $q_{\max}$, verify that the printed operating point lies on that curve, then apply the shock-wave (Rankine–Hugoniot) relation between the approach state and the jam state to get the stopping wave, and between the jam state and the capacity discharge state to get the starting wave.

1. Part (a) — fit the Greenshields curve from the free-flow speed and the capacity. Under Greenshields the speed falls linearly with density, $u = u_f\,(1-k/k_j)$, so flow is the parabola $q = u_f\,(k - k^2/k_j)$ whose maximum is $q_{\max}=u_f k_j/4$ at $k=k_j/2$. Inverting for the jam density, $$k_j=\frac{4\,q_{\max}}{u_f}=\frac{4\times 1200}{60}$$ $$\boxed{k_j = 80 \text{ veh/km}, \qquad k_c = \tfrac{1}{2}k_j = 40 \text{ veh/km}}$$ The corresponding speed at capacity is $u_c = u_f/2 = 30$ km/h, which will be needed in part (c).
2. Check the printed operating point against the fitted curve before going further. This one-line test decides whether the four supplied data are mutually consistent or whether the question is over-determined: $$u = u_f\left(1-\frac{k_A}{k_j}\right)=60\left(1-\frac{20}{80}\right)=60\times 0.75=45 \text{ km/h}$$ which reproduces the printed approach speed exactly. The data are consistent, so no competing interpretation has to be carried through the rest of the solution. The approach flow follows from the fundamental relation: $$q_A = u_A\,k_A = 45\times 20 = \boxed{900 \text{ veh/h}}$$ that is, 75 per cent of capacity — a busy but under-saturated approach.
3. Part (b) — identify the two states either side of the stopping shock. Upstream of the stop line during red the traffic is in state A, with $q_A = 900$ veh/h and $k_A = 20$ veh/km. Downstream of the shock, vehicles are stopped, so the jam state has $u=0$, $k=k_j=80$ veh/km and $q=0$. A shock wave is the boundary between two such states, and its speed is the slope of the chord joining them on the flow–density diagram: $$u_w=\frac{q_2-q_1}{k_2-k_1}$$
4. Compute the speed of the stopping wave. Substituting the jam state and the approach state, $$w_{AB}=\frac{0-900}{80-20}=\frac{-900}{60}$$ $$\boxed{w_{AB}=-15 \text{ km/h}}$$ The negative sign means the boundary travels upstream, away from the stop line, at 15 km/h — the queue "grows backwards" as arriving vehicles meet its tail.
5. Convert the red interval and obtain the queue length. With $t_r = 30\text{ s} = 30/3600 = 1/120$ h, the boundary has travelled $$L = |w_{AB}|\,t_r = 15\times\frac{1}{120}=0.125 \text{ km}$$ $$\boxed{L = 0.125 \text{ km} = 125 \text{ m at the end of red}}$$
6. Convert the length to a vehicle count and check it two independent ways. The queued vehicles are packed at jam density, so $$n = k_j\,L = 80\times 0.125 = 10 \text{ vehicles}$$ The same count must also equal the number of vehicles overtaken by the shock, which is the closing speed between the traffic and the boundary times the approach density and the elapsed time: $$n = (u_A-w_{AB})\,k_A\,t_r=(45+15)\times 20\times\frac{1}{120}=10 \text{ vehicles}$$ The two agree, which is the cheapest available confirmation that the wave speed is right. Note that the naive count $q_A t_r = 900/120 = 7.5$ vehicles is wrong, because it counts only the vehicles passing a fixed point and ignores that the shock advances upstream to meet the arriving traffic.
7. Part (c) — identify the discharge state and compute the starting wave. The question states there is no downstream congestion, so once the signal turns green the queue discharges at capacity: state C has $q = q_{\max} = 1{,}200$ veh/h and $k = k_c = 40$ veh/km, moving at $u_c = 30$ km/h. The boundary between the jam state and the discharge state is the starting (release) wave: $$w_{BC}=\frac{q_{\max}-0}{k_c-k_j}=\frac{1200-0}{40-80}=\frac{1200}{-40}$$ $$\boxed{w_{BC}=-30 \text{ km/h}}$$ It too runs upstream, but twice as fast as the stopping wave, so it must eventually catch it.
8. Close the two boundaries against each other, not against a fixed point. The stopping wave does not stop when the light turns green: traffic in state A is still arriving and still joining the back of the queue, so the tail keeps retreating at 15 km/h while the front is released at 30 km/h. The queue therefore shrinks at the closing speed of the two boundaries: $$\tau=\frac{L}{|w_{BC}|-|w_{AB}|}=\frac{0.125}{30-15}=\frac{0.125}{15}=0.008\overline{3} \text{ h}$$ $$\boxed{\tau = 30 \text{ s after the start of green}}$$ By that instant the stopping wave has reached $15\times(60/3600)=0.25$ km, i.e. 250 m upstream of the stop line, and 20 vehicles in total have been stopped in the cycle. Dividing by $|w_{BC}|$ alone would give 15 s and is the commonest error on this question; it treats the tail of the queue as though it were stationary.
9. Interpret the result against the signal timing. A 30-second dissipation time for a 30-second red means the green interval must be at least 30 s long simply to clear the queue formed on the previous red, before any further arrivals are served. That is the operational reason a cycle of this kind cannot be shortened without either reducing the red or accepting a residual queue that carries over from cycle to cycle — the condition that separates an under-saturated signal from an over-saturated one.