Question

Traffic is moving on a 6-lane dual carriageway road (3 lanes per direction). Traffic volume per direction during the peak hour (08:00 am to 09:00 am) is 6000 veh/h, distributed uniformly across the lanes in each direction. Just at 08:00 am, a truck goes out of order on the middle lane of one side, disrupting traffic on that lane. The lane capacity under normal conditions is 2000 veh/h/ln, and under queue formation it is 1600 veh/h/ln. The traffic resumes at 08:30 am on removing the truck from the middle lane. Hourly traffic volume after 09:00 am reduces to 5000 veh/h/dir. The number of vehicles in the queue at 10:00 am is ______ (in integer).

Hint:

Track cumulative arrivals vs departures in three phases; once a queue forms, the discharge rate stays capped at the reduced (queue) capacity even after the blockage clears.

Answer 1

Correct Answer: 2200

Step 1: Set up the three time phases.
The road has 3 lanes per direction, each rated at 2000 veh/h/ln under free flow, so the normal capacity \(=3\times2000=6000\) veh/h, exactly matching the 08:00-09:00 demand of 6000 veh/h. So no queue exists until the incident starts. Three phases follow the truck breakdown: Phase 1 (08:00-08:30), truck blocks the middle lane, only 2 lanes work; Phase 2 (08:30-09:00), truck removed, all 3 lanes physically open, but a queue is already present; Phase 3 (09:00-10:00), demand falls to 5000 veh/h/dir.

Step 2: Key idea - the capacity drop persists once a queue exists.
Once a bottleneck causes a queue to form, the actual rate at which vehicles pass through stays capped at the lower "queue discharge" value of 1600 veh/h/ln for every lane discharging queued traffic, even after the truck is removed, until the queue fully empties. This is why the problem gives two separate capacities (2000 free-flow vs 1600 under queue conditions).

Step 3: Phase 1 (08:00-08:30 am), duration 0.5 h.
Only 2 lanes work, and a queue is forming, so capacity \(=2\times1600=3200\) veh/h.
Arrivals \(=6000\times0.5=3000\) vehicles. Departures \(=3200\times0.5=1600\) vehicles.
Queue at 08:30 am \(=3000-1600=1400\) vehicles.

Step 4: Phase 2 (08:30-09:00 am), duration 0.5 h.
The truck is removed and all 3 lanes are physically open, but since 1400 vehicles are already queued, the discharge rate stays reduced: \(3\times1600=4800\) veh/h.
Arrivals \(=6000\times0.5=3000\) vehicles. Departures \(=4800\times0.5=2400\) vehicles.
Queue at 09:00 am \(=1400+3000-2400=2000\) vehicles.

Step 5: Phase 3 (09:00-10:00 am), duration 1 h.
Demand is now 5000 veh/h/dir; capacity is still the reduced discharge rate of 4800 veh/h since the queue (2000 vehicles) has not cleared. Because demand (5000) still exceeds this rate (4800), the queue keeps growing through the whole hour, so departures equal the full 4800 veh/h for all 60 minutes.
Arrivals \(=5000\times1=5000\) vehicles. Departures \(=4800\times1=4800\) vehicles.
Queue at 10:00 am \(=2000+5000-4800=2200\) vehicles.

Final Answer:
\[ \boxed{2200 \text{ vehicles}} \]