Question

Please someone I really could use some help with this problem!

5.43 Vehicles arrive at a toll both starting at 7:00 A.M. at a rate of 1(t) = 5.1 – 0.05t [with (t) in veh/min and t in minutes after 7:00 A.M.]. The first operator processes cars at a rate of 3 veh/minute 7:00 A.M. until 7:15 A.M when the person leaves because of illness. From 7:15 A.M to 7:25 A.M, no one is at the toll booth but a new operator arrives at 7:25 A.M and processes at a rate of u(t) = 8 + 0.3t [with y(t) in veh/min and t in minutes after 7:25 A.M.]. Assuming D/D/1 queuing, what is the maximum queue length (in vehicles) and the average delay per vehicle?

Answer 1

Solecton, Time arival rate eparstuxe yate 0-15 5.1-0.05t 3 veh/min 15-25 51-005 t 95t 8tost S.1-0105t Maxinum queue length wexage delay pes vehicle - Let is tme in minutes at which gueue s dini paer aHte :00AM; so, al this time comulafi ve axoival Cumulotve depathue t +51-0st) dt dt +sot)dt s1-0105t) dt-us+(o3t) dt 15 7 25 5t05t 1950372s 4548tt .3 Jaliiny Ft 30 32 minltes Gaph to cumalative asivaly and Depasturer wth ine a03ival tuave Bl.844 dipahuve cusve Mainaum ueve ength in minu) 25 30. 32 Ceemulafive Count

F30mm gaphy mauimum queue long h is at 25 minuter ates 00 AM Maximum queee length 25 (5-1-0.05t) dt - l8 166.8 45 vehiley

Total vehicle dela tavtvalate t dt- departuse at xt dt 3013) 51-0stdt Satdt tot) de 30.32 18y.66-335-8605 4.46 vehice-minule hverage vehicle deloyTetal Vehicle delay umulatve anival Cas) Cumulative depastee 145544 30 32 y5546 131 6494 1LL 05 minule