The arrival rate at a parking lot is 6 veh/ min. Vehicles start arriving at 6: 00 P. M., and when the queue reaches 36 vehicles, service begins. If company policy is that total vehicle delay should be equal to 500 veh-min, what is the departure rate? (Assume D/ D/ 1 queuing and a constant service rate.)

departure rate k = 7.653 vehicle per minute

Explanation:

given data

parking = 6 veh/ min

arriving  = 6: 00 P. M

queue reaches = 36 vehicles

total vehicle delay = 500 veh-min

solution

we know here that total vehicle in t min will be

total vehicle in t min = 6t vehicle

and

we consider here service rate is equal to departure rate

so service rate = k vehicle per min

and service is starting when reach vehicle 36

so that no of vehicle is

no of vehicle = 6t ,   ( 0 ≤ t ≤ 6 )

no of vehicle  = 6t - k (t -6 )   for t ≥ 6  

and

Queue dissipate when here 6t - k( t-6) will be zero

so

k =  ........................a

we know here

total delay is equal to 500 veh min

0.5 t  × 6t - 0.5 (t-6) × k × (t-6)  = 500

so solve it we get

t = 27.78 min

so k will be from equation a

k =

k = 7.653

so departure rate is 7.653 veh/min

the correct answer is c.stops after infancy and childhood.

myelin is the substance of the central nervous system (cns) that insulates nerve cell axons to increase the speed at which nerve impulse travels. myelination is a process of myelin formation. this process occurs postnatally, begins in the embryo during the midst of early development and finishes after birth.

answer:

i cant give me a pic

explanation: