If the arrival of major road vehicles can be described by the poisson distribution, and the peak hour volume is 1000 veh/hr. the value of the critical gap is 4 seconds. the expected number of available gaps during the peak hour is equal to:
expected no. of vehicles per hour = 1000 expected no. of vehicles per 4 minute interval = 1000/(3600/4) = 10/9 = λ
Using the poisson distribution, probability that no vehicles within the 4 second interval P(0)= λ ^0 e^(- λ )/ 0! = (10/9)^0 e^(-10/9) / 1 = e^(-10/9) = 0.3292 (approx.)
In an hour, there are n=3600/4=900 such intervals, each with probability of p=0.3292 occurring.
Applying binomial distribution, expected number of gaps =np =900*0.3292 = 296.3
Answer: the expected number of gaps is 296 (to the nearest unit).
Note: in fact, the actual expected number will be higher because gaps do not line up at 4 second intervals. If a vehicle passes at the 1 second of the interval, a gap can commence at the next second. Therefore the expected value could/should be slightly higher.
