# Bernoulli single-server queueing process

1. Nov 10, 2012

### lowball

1. The problem statement, all variables and given/known data
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.

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

3. The attempt at a solution
Using the above calculations I formed this transition probability matrix:

$P = \begin{pmatrix} .8 & .2 & 0 & 0 & \dots\\ .267 & .133 & .6 & 0 & \dots\\ 0 & .267 & .133 & .6 & \dots\\ 0 & 0 & .267 & .133 & \dots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$

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

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

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?

2. Nov 10, 2012

### 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
?

3. Nov 10, 2012

### 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!

4. Nov 11, 2012

### Ray Vickson

You should not use such inaccurate numbers, especially not at the beginning.
$$\lambda_A = 1/10,\; \lambda_S = 1/6\\ p_A = 1/5,\; p_S = 1/3\\ p_{00} = 4/5, \; p_{01} = p_A = 1/5\\ (1-p_A)p_S = (4/5)(1/3) = 4/15, \; (1-p_S)p_A = (2/3)(1/5) = 2/15\\ 1 - 4/15 - 2/15 = 9/15 = 3/5$$
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