Question

5- In a national park in Texas, vehicles begin to arrive at a parking lot at 7:30 A.M. with arrival rate function (in vehicles per minute) of (t) = 4 - 0.02t, where t is in minutes. 7:40 A.M. the parking lot opens and processes vehicles at a rate of 10 per minute. a) (7%) What is the average delay per vehicle? b) (8%) What is the maximum queue? c) (8%) What is the waiting time of 20th vehicle arrives at the parking lot?

Answer 1

Page- Departure rat (vehlmin Time period Aytinal Kode (veh min) 7:30 AM - 7:40AM A-0.02t Co to 10 min) .. 7:40 A.M onwards 10 4-0.02t (lo to Tc) let Te is time time in minutes at which queue is to dissipated, at this time cumulative arrivals equal cumulative departures, s (4-0.02t) dt s lodt > ATC-0.02 To 2 1OTc-100 M Tc=16,23 minutes 7846.23 AM queue is dissipated. 80, at time @ total delay Average delay per vehicle - No, cumulative arrivals or No. cumulative departure Tc Total delay = 1 arrival ratext at S departure rabext och 16.23 =164-0.0 2t)tat - 5 Coldt TC O o 10

Page-2 - 1498.3245-817.0645) 318.74 veh minutes 318.74 16.23 50, Average delay - (4-0.02t dt 5.12 minutes Avg. delay Maximum queue is at 7:40 AM, because after that only queue start to clear. = queue No. of arrivals oplo 7:40 AM 30 Maximum ES (4-0.02t) dt 0 39 vehicles haiting time of 20th vehicle = departure time of 20th asrival time of 20th vehicle. vehicle 10 minutes & minute 10 : Departure time of 20th vehicle a 12 minutes. tro 20= (4-0.02t) de 20 4120 -0.02 0.027,

Rage-3 to 11 5.06 minutes 30, waiting time of 20th vehicle = 12-5.06 = 6.94 minutes