Question

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

Answer 1

Let x be the interarrival time between i^th arriving car and (i - 1)^th and { x } be the independent variable.

if x_{2 $\geq$} c, the second arriving car need not do wait, then we have

P { x_{2 $\geq$} c } e^{- $\lambda$ c}

if x₂ $\propto$ c, it needs to wait (c - x₂) minutes,

Let x₂ = t, then we have

E (c - t) = $\int_{0}^{c}$ (c - t) $\lambda$ e^{- $\lambda$ t} dt + $\int_{c}^{+\infty }$ $\lambda$ e^{- $\lambda$ t} dt

Let y be the inspection time for the first arriving car.

using a condition, Y = y

if x_{2 $\geq$} 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 { x_{2 $\geq$} Y } = $\int_{c}^{+\infty }$ P { x_{2 $\geq$} Y | Y

P { x_{2 $\geq$} Y } = y $\int_{c}^{+\infty }$ f (y) dy

P { x_{2 $\geq$} Y } = $\int_{c}^{+\infty }$ e^{- $\lambda$ y} _$\mu$ e^{- _$\mu$ y} dy

P { x_{2 $\geq$} Y } = _$\mu$ / ( $\lambda$ + _$\mu$ )

And the mean waiting time which will be given as :

P { x_{2 $\geq$} Y } = $\int_{c}^{+\infty }$ $\int_{0}^{y}$ (y - t) $\lambda$ e^{- $\lambda$ t} dt . _$\mu$ e^{- _$\mu$ y} dy

P { x_{2 $\geq$} Y } = $\int_{c}^{+\infty }$ [y + (1 / $\lambda$ ) e^{- $\lambda$ y} - (1 / $\lambda$ )] _$\mu$ e^{- _$\mu$ y} dy

P { x_{2 $\geq$} Y } = $\lambda$ / _$\mu$ ( $\lambda$ + _$\mu$ )