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.