- #1

lowball

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## Homework Statement

Performance of a car wash center is modeled by the single-server Bernoulli queueing process with 2-minute frames. Cars arrive every 10 minutes, on the average. The average service time is 6 minutes. Capacity is unlimited. If there are no cars at the center at 10 am, compute the probability that one car is being washed and another car is waiting at 10:04 am.

## Homework Equations

[itex]Δ = 2[/itex] min

[itex]λ_A = .1[/itex] min[itex]^{-1}[/itex]

[itex]λ_S = .167[/itex] min[itex]^{-1}[/itex]

[itex]p_A = λ_AΔ = .2[/itex]

[itex]p_S = λ_SΔ = .333[/itex]

[itex]p_{00} = 1-p_A = .8[/itex]

[itex]p_{01} = p_A = .2[/itex]

[itex](1-p_A)p_S = .267[/itex]

[itex](1-p_S)p_A = .133[/itex]

[itex]1 - .267 - .133 = .6[/itex]

## The Attempt at a Solution

Using the above calculations I formed this transition probability matrix:

[itex]P = \begin{pmatrix}

.8 & .2 & 0 & 0 & \dots\\

.267 & .133 & .6 & 0 & \dots\\

0 & .267 & .133 & .6 & \dots\\

0 & 0 & .267 & .133 & \dots\\

\vdots & \vdots & \vdots & \vdots & \ddots

\end{pmatrix}[/itex]

With no cars in the system, the initial distribution is:

[itex]P_0 = \begin{pmatrix}1&0&0&0\end{pmatrix}[/itex]

With a frame size of 2 minutes, 10:04 is 2 frames away from 10:00, thus the distribution after 2 frames is:

[itex]P_2 = P_0P^6 = \begin{pmatrix}1&0&0&0\end{pmatrix}\cdot P^2[/itex]

[itex]= \begin{pmatrix}.6934&.1866&.12&0\end{pmatrix}[/itex]

And the probability for two cars to be in the system after 2 frames is [itex]P_2(2) = .12[/itex]

But that's not accepted as the right answer. The answer in the back of the book says [itex]\frac{2}{75}[/itex], but that's not even anywhere in the matrix of [itex]P^2[/itex]. Any idea what I'm doing wrong?