The arrival of aircraft to an airport can be modeled as a Poisson process with a

Question

The arrival of aircraft to an airport can be modeled as a Poisson process with a rate of 10 arrivals per hour. What is the probability that exactly 5 aircrafts arrive in a 1-h period? What is the probability that at least 5 aircrafts arrive in a 1-h period? What is the probability that at least 10 aircrafts arrive in a 1-h period? The airport has a policy of requiring aircraft to wait in a holding position in the air if the number of arriving aircraft in any 10-min period exceeds 2. What is the occurrence probability of an aircraft holding event? Recompute and plot the occurrence probability of an aircraft holding event in part (d) as a function of the rate of the Poisson process for rates ranging from 10 to 50.

Explanation / Answer

Poisson Distribution equation P(k,µ) = (µk * e-µ) / k!

where, k = number of success resulting from the experiment

µ = average number of success in a given period

In this case, µ = 10 arrivals per hour

a.) k = 5

P(5,10) = (105 * e-10 ) / 5!

   = 0.03

b.) k >= 5

P (at least 5 aircraft ) = 1 - P( less than 5 aircraft)

   = 1 - [ P(0,5) + P(1,5) + P(2,5) + P(3,5) +P(4,5) ]

Using the above formula and calculating similarly, we get

P( k >=5 ) = 0.97

c.) k >= 10

P(k >= 10) = 1 - P( k < 10)

We can calculate this exaclty similarly as previous one.

P(k >= 10) = 0.5421

d.) Using elementry method we can see the average number of aircraft arrive in 10 min period

Average number of aircraft per minute = 10 / 60

Average number of aircraft every 10 minute = (10 / 60) * 10

= 1.667

So the new µ = 1.667

Now we need to calculate P( k >= 2 ) = 1 - P(k<2)

We will solve the in exactly the same way we solved previous parts but we will use new µ, i.e. 1.667

P( k<2) = 0.503

P( k>=2) = 1 - 0.503 = 0.497

e.) We will do it similarly as we did previous question. We will just have to calculate different µ at different points. Lets consider unit cell of 5 aircraft. Now we will calculate the number of average aircraft in 10 minutes for different rate of poisson process, similarly as previous part.

µ10 = 1.667

µ15 = 2.5

µ20 = 3.33

We can calculate till µ50 and plot a graph between P( aircraft holding event) vs rates of Poisson process

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