Bernoulli single-server queueing process

[lowball]

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

Δ = 2 min
λ_A = .1 min^{-1}
λ_S = .167 min^{-1}
p_A = λ_AΔ = .2
p_S = λ_SΔ = .333
p_{00} = 1-p_A = .8
p_{01} = p_A = .2
(1-p_A)p_S = .267
(1-p_S)p_A = .133
1 - .267 - .133 = .6

The Attempt at a Solution

Using the above calculations I formed this transition probability matrix:

P = \begin{pmatrix}<br /> .8 &amp; .2 &amp; 0 &amp; 0 &amp; \dots\\<br /> .267 &amp; .133 &amp; .6 &amp; 0 &amp; \dots\\<br /> 0 &amp; .267 &amp; .133 &amp; .6 &amp; \dots\\<br /> 0 &amp; 0 &amp; .267 &amp; .133 &amp; \dots\\<br /> \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots<br /> \end{pmatrix}

With no cars in the system, the initial distribution is:
P_0 = \begin{pmatrix}1&amp;0&amp;0&amp;0\end{pmatrix}

With a frame size of 2 minutes, 10:04 is 2 frames away from 10:00, thus the distribution after 2 frames is:
P_2 = P_0P^6 = \begin{pmatrix}1&amp;0&amp;0&amp;0\end{pmatrix}\cdot P^2
= \begin{pmatrix}.6934&amp;.1866&amp;.12&amp;0\end{pmatrix}

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

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

 

[haruspex]

Shouldn't the second row of the matrix read:
10: 0.8*1/3 = .267
11: 0.8*2/3 + .2*1/3 = 0.6
12: 0.2*2/3 = .133
?

 

[lowball]

Ah... now I see. The example in my book had it ordered somewhat oddly. After swapping those around, I do indeed get the correct answer of .0266. Thanks!

 

[Ray Vickson]

Ah... now I see. The example in my book had it ordered somewhat oddly. After swapping those around, I do indeed get the correct answer of .0266. Thanks!

You should not use such inaccurate numbers, especially not at the beginning.
\lambda_A = 1/10,\; \lambda_S = 1/6\\<br /> p_A = 1/5,\; p_S = 1/3\\<br /> p_{00} = 4/5, \; p_{01} = p_A = 1/5\\<br /> (1-p_A)p_S = (4/5)(1/3) = 4/15, \; (1-p_S)p_A = (2/3)(1/5) = 2/15\\<br /> 1 - 4/15 - 2/15 = 9/15 = 3/5
Later, you can round off, and to get *accurate* multi-step probabilities (via Pⁿ) you should keep a lot more digits in P---full machine floating-point accuracy would be best. (Of course, maybe you did keep all those figures and just rounded off for presentation purposes, in which case you should say so.)

RGV