# Bernoulli single-server queueing process

by lowball
Tags: bernoulli, process, queueing, singleserver
 P: 2 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?
 Homework Sci Advisor HW Helper Thanks P: 9,656 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 ?
 P: 2 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!
HW Helper
Thanks
P: 4,959
Bernoulli single-server queueing process

 Quote by 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!
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

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