Problem 1: Queueing Theory (20 Points) At 7:30 pm, vehicles are travelling on a
ID: 376862 • Letter: P
Question
Problem 1: Queueing Theory (20 Points)
At 7:30 pm, vehicles are travelling on a two-lane segment of US99 at a rate of 2400 vph. The capacity of the freeway is 1800 vph per lane. At 8:00 pm, an accident happens and completely blocks one of the two lanes. It takes Caltrans 20 minutes to clear the accident and open the blocked lane back to traffic. Answer the following.
Draw the queueing diagram
Calculate the following statistics:
A. Time queue dissipates.
B. The delayed vehicle suffering the longest delay.
C. Delay of the 1000th delayed vehicle.
D. Average delay.
Explanation / Answer
A.
Queue length from 7:30 to 8pm = (arrival rate - capacity of freeway) * time (=0.5 hour) = (2400-1800)*0.5 = 300
vehicles added to queue from 8 pm to 8:20 (when accident is cleared and blocked lane is opened) = (2400-900)*20/60 = 500
Total queue length at 8:20 pm = 300+500 = 800
Capacity of the freeway is 1800 vph, and vehicles are travelling at a higher rate of 2400 vph. If the vehicles keep travelling at this rate, then the queue will never dissipate, but keep on increasing perpetually, leading to queue explosion.
However, assuming that the vehicles stop coming on the freeway after 8:20 pm. The time it takes for queue to dissipate completely = 800/1800 = 0.44 hours = 26.67 minutes
B. The longest delay = 26.67 minutes
C. The queue has only 800 vehicles. So, there is no 1000th delayed vehicle.
D. Average delay = (26.67+0)/2 = 13.33 minutes