Problem 2 Green Valley aisport has been in operation for several years and is be

Question

Problem 2 Green Valley aisport has been in operation for several years and is beginning to espuence light congestion. A study of aiport operations revealed that planes artive at an average rate of 12 per hour, with a poisso distribution. On the single runway, a plane can land and be cleared every 4 minutes on average, and service times are exponentially distributed. Planes are pr landings. Planes waiting to land are asked to circle the airport by the air traffic controller What is the expected number of aixplanes circling the airport, waiting in queue for cleatance to land? (5 points) a. b. What is the average time a plane spends circling, waiting for clearance to land? (5 points) c. What fraction of the time is the air traffic controller available to handle takeoffs? (5 points)

Explanation / Answer

Solution

Problems (a), (b) and (c) are solved as M/M/1 Queuing problems.

Back-up Theory

An M/M/1 queue system is characterized by arrivals following Poisson pattern with average rate , service time following Exponential Distribution with average service rate of µ and single service channel.

Let n = number of customers in the system and m = number of customers in the queue.

[Trivially, n = m + number of customers under service.]

Let (/µ) =

The steady-state probability of n customers in the system is given by Pn = n(1 - ) ……(1)

The steady-state probability of no customers in the system is given by P0 = (1 - ) ….…(2)

Average queue length = E(m) = (2)/{ µ(µ - )} ………………………………………..(3)

verage waiting time = E(w) = ()/{ µ(µ - )} ……………………………………………..(4)

Given, = 12/hr µ = 15/hr [given average service time as 4 minutes, µ = 60/4] and hence

= 12/15 = 0.8

Part (a)

Expected number of planes circling waiting in queue for clearance = E(m)

= (122)/{15(15 - 12) [vide (3) above]

= 144/45 = 16/5 = 3.2 ANSWER

Part (b)

Average time a plane spends circling waiting in queue for clearance = E(w)

= 12/{15(15 - 12)} [vide (4) above]

= 12/45 = 4/15 hour = 60(4/15) = 16 minutes ANSWER

Part (c)

Fraction of time air traffic controller is available to handle take-offs

= Fraction of time there is no plane landing or waiting for landing

= P0

= 1 – (12/15) [vide (2) above]

= 0.2 ANSWER

Part (d)

This problems is solved using M/M/2 Queuing system, i.e., Poisson-arrival-Exponential-service-2 service channels. Back-up Theory

Formulae for M/M/2

Let (/µ) =

The steady-state probability of n customers in the system is given by

Pn = P0 x n/n! for n < c ………………………………………………………………(1a)

Pn = P0 x n/{c!(cn-c) for n c ..………………………………………………………(1b)

The steady-state probability of no customers in the system is given by

P0 = 1/(S1 + S2), where S1 = [0,c - 1](n/n!) and S2 = (c)/[c!{1 - (/c)}]…………(2)

Given = 12, µ = 15, and hence = 0.8,

Vide (2) above. S1 = {1 + 0.8} = 1.8 and S2 = (0.8)2)/[2!{1 - (0.4)}] = 0.533.

So, P0 = 1/(1.8 + 0.533) = 0.429 = P0

Now, average number of planes circling waiting in queue for clearance = E(m)

= P0{(µ)(/µ)c}/{(c - 1)!(cµ - )2} = P0{(3/µ)}/{(2µ - )2} [for c =2]

= 0.429 x (115.2/81)

= 0.61 ANSWER

[Additional input: comapring Parts (a) and (d), efficacy of adding one more service channel can be seen,]

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