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A two-runway (one runway for landing, one runway for taking off) airport is being designed for propeller-driven aircraft. The time to land an airplane is known to be exponentially distributed, with a mean of 1.5 minutes. If airplane arrivals are assumed to occur at random, what arrival rate can be tolerated if the average wait in the sky is not to exceed 3 minutes? (Hint: A single landing strip airport is modeled by an M/M/1 queue.)

Answer :

tatendagota

Answer:

[tex]\frac{4}{9}[/tex]

Step-by-step explanation:

First, we gather the data:

The model is a single [tex]M/M/1[/tex] landing strip with [tex]\mu = \frac{2}{3}[/tex]

This based on the assumption that service times are exponential.

Then, we must find the maximum rate such that [tex]w_{q} \leq 3[/tex]

Using the [tex]M/M/1[/tex] table, we find that:

[tex]w_{p} = \frac{\rho }{(\mu (1 -\rho) } \\ \frac{\lambda }{\mu (1 - \lambda } \leq 3[/tex]

if and only if

[tex]\lambda\leq \frac{3\mu ^{2} }{(1 + 3\mu) }[/tex]

=[tex]\frac{4}{9}[/tex]

lhmarianateixeira

In this exercise we have to use the exponential knowledge to calculate the time that an airplane can wait in the air, in this way we have to:

[tex]\lambda= 4/9[/tex]

So we have from the information given in the text we find that:

  • The model is a single [tex]M/M/1[/tex] landing strip with [tex]\mu=2/3[/tex]
  • The maximum rate such that [tex]w_q\leq 3[/tex]

Using the formula given below we find that:

[tex]w_p=\frac{\rho}{\mu(1-\rho)} \\\frac{\lambda}{\mu(1-\lambda)}\leq 3\\\lambda\leq \frac{3\mu^2}{(1+3\mu)} \\=4/9[/tex]

See more about exponential at brainly.com/question/2193820

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