# 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: 9,839 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: 5,064
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

 Related Discussions Computers 18 Set Theory, Logic, Probability, Statistics 4 Calculus & Beyond Homework 0 Set Theory, Logic, Probability, Statistics 0 Introductory Physics Homework 0