A gas station is providing the state auto inspection service for the general public. As soon as the gas station opens at 7AM, cars arrive for inspection following a Poisson process with rate λ (arrivals/minute). (a) Assume that each inspection takes a constant amount time, namely c (minutes). Let W2 denote the random time that the second vehicle waits between arriving and the time at which its inspection starts. What is the probability that W2 = 0? What is the expected value of W2? (b) Answer the same questions above assuming that the service time for the first customer is an exponentially distributed random variable with mean 1/k
Let x be the interarrival time between ith arriving car and (i - 1)th and { x } be the independent variable.
if x2 c, the second arriving car need not do wait, then we have
P { x2 c } e-c
if x2 c, it needs to wait (c - x2) minutes,
Let x2 = t, then we have
E (c - t) = (c - t) e-t dt + e-t dt
Let y be the inspection time for the first arriving car.
using a condition, Y = y
if x2 y, the second will not need to wait. otherwise, it will wait (y - t) minutes.
So, the probability that it will not wait is given by :
P { x2 Y } = P { x2 Y | Y
P { x2 Y } = y f (y) dy
P { x2 Y } = e-y e-y dy
P { x2 Y } = / ( + )
And the mean waiting time which will be given as :
P { x2 Y } = (y - t) e-t dt . e-y dy
P { x2 Y } = [y + (1 / ) e-y - (1 / )] e-y dy
P { x2 Y } = / ( + )
A gas station is providing the state auto inspection service for the general public. As soon...