5.14 Vehicles begin to arrive at a park entrance at 7:45 A. M. at a constant rate of six per minute and at a
constant rate of four vehicles per minute from 8:00 A. M.
on. The park opens at 8:00 A. M. and the manager wants
to set the departure rate so that the average delay per
vehicle is no greater than 9 minutes (measured from the
time of the first arrival until the total queue clears).
Assuming D/D/1 queuing, what is the minimum
departure rate needed to achieve this?

step 1: you first have to ensure that you are working with the same units throughout the solution, and in this case, i choose to work with minutes, so i will convert the service rate of 30 per hour into minutes per hour. if 60mins = 30 serviced request what of 1 min = ? ((30*1)/60) = a service rate of 0.5 per min. step 2: now that we do have the average service rate at0.5 per minute and our arrival rate is at 6 per minute, we can go ahead to find the traffic intensity, which is arrived at by dividing the average arrival rate by the average service rate. therefore: 5/0.5 = 10 the traffic intensity(ratio of the average arrival rate to the average service rate) is 10.

Explanation:

false

a hangover caused by drinking a lethal dosage of alcohol

answer:

the answer is d

explanation: