Assume that at Denver National Airport there are 5 security

Assume that at Denver National Airport, there are 5 security lines, with each line capable of scanning 300 people per hour. People arrive in front of the security lines “waiting plaza” at the rate of 750 per hour. Make the standard assumptions of a Poisson distribution for arrivals, and an Exponential distribution for service times, and calculate the following: a) The probability of zero people in the waiting plaza. b) The average length (number of people) in queue.

Solution

The mean number of arrivals per time period = 750 per hour

The mean number of people or items served per time period = 300 = 1500 per hour.

( since we have 5 security lines)

a) Then the probability of zero people in the waiting plaza is

P₀ = probability of 0 units in the system (that is, the service unit is idle) = 1- ( / )

                                                                                                        = 1 - (750/1500)

                                                                                                        = 1-(1/2)

                                                                                                        = 1/2 = 0.5

Therefore, the probability of zero people in the waiting plaza is 0.5.

b) The average length (number of people) in queue is L_s = ² / ( - )

                                                                                = 750² / 1500(1500 - 750)

                                                                                = ( 750 x 750 ) /( 1500 x 750 )

                                                                                = 1 /2 = 0.5

So, on an average 0.5 people wait in the queue.