Freeway Incident Shockwave — Greenshields Queue and Delay Worked Example
A crash blocks two of three lanes for 30 minutes during moderate traffic. Using the Greenshields model we find the traffic state in the queue, how fast the back of the queue moves upstream, how long the queue gets, how fast it dissolves once the lanes reopen, and the total delay. The shockwave picture and the simpler cumulative-count picture are compared at the end.
The incident
| Freeway | 3 lanes in the direction of travel |
| Greenshields | Free-flow speed vf = 60 mi/h, jam density kj = 180 veh/mi/ln |
| Demand | 4,200 veh/h (1,400 veh/h/ln), steady |
| Incident | Two lanes blocked for 30 min; the open lane flows at its full capacity |
Traffic flow tool: vf 60, kj 180; evaluate q = 1,400 and q = 900 to read both densities.
Open the traffic flow calculator →Per lane and at the bottleneck
Bottleneck (1 lane open) = 2,700 veh/h < demand 4,200 veh/h → a queue forms
Arrivals (state A) and the queue (state Q)
State A, q = 1,400: k = 90[1 − √(1 − 0.519)] = 27.6 veh/mi/ln, v = 60(1 − 27.6/180) = 50.8 mi/h
State Q: the queue passes 2,700 veh/h over 3 lanes = 900 veh/h/ln, congested root: k = 90[1 + √(1 − 0.333)] = 163.5 veh/mi/ln, v = 5.5 mi/h
Shockwave between A and Q
Queue length when the lanes reopen (t = 0.5 h): 3.68 × 0.5 = 1.84 mi
Vehicles stored: (163.5 − 27.6) × 1.84 × 3 lanes = 750 veh (check: (4,200 − 2,700) × 0.5 = 750 ✓)
Front of the queue after the lanes reopen
uQD = (2,700 − 900)/(90 − 163.5) = −24.5 mi/h (upstream, much faster than the back of the queue)
Front catches back: 3.68t = 24.5(t − 0.5) → t = 0.588 h = 35.3 min, at 2.16 mi upstream (the maximum queue length)
Afterwards the D/A boundary moves downstream at (2,700 − 1,400)/(90 − 27.6) = +20.8 mi/h
Queueing diagram at the bottleneck
Then shrinks at 8,100 − 4,200 = 3,900 veh/h → 750/3,900 = 0.192 h → cleared at 41.5 min
Total delay = triangle area = ½ × 750 × 0.692 = 260 vehicle-hours (2,908 vehicles delayed, 5.4 min average)
Tools used in this example
Traffic flow (Greenshields) for the states and capacity; Stopbar to sketch the network.
Frequently asked questions
How do you calculate shockwave speed?
u = (q2 − q1)/(k2 − k1), the change in flow divided by the change in density across the boundary between two traffic states. A negative value means the wave moves upstream. Here the boundary between free-flowing arrivals (1,400 veh/h/ln, 27.6 veh/mi/ln) and the queue (900 veh/h/ln, 163.5 veh/mi/ln) moves upstream at 3.7 mi/h.
Why is the queue state congested and not free-flowing?
The queue discharges through the bottleneck at the bottleneck capacity. That flow, spread over all lanes, can be carried at either a low or a high density; upstream of a bottleneck the stored vehicles are in the high-density (congested) root of the flow–density curve.
Why do the shockwave and queueing-diagram clearing times differ?
The queueing (point-queue) diagram counts vehicles at the bottleneck and says the backlog is gone at 41.5 minutes. The shockwave picture tracks the queue in space; the congested region vanishes at 35.3 minutes, but vehicles behind it are still traveling at the 30 mi/h capacity speed, below their free speed, so they are still delayed. Both are consistent; they answer different questions.
How realistic is Greenshields for incident analysis?
It is a teaching and first-estimate model. It puts capacity at half the free-flow speed and assumes a lane next to a crash flows at full capacity; in reality the lane next to a crash carries well below normal capacity, which makes queues much longer. Use HCM capacity-reduction factors or a simulation for operational decisions.