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...