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Bernoulli single-server queueing process

by lowball
Tags: bernoulli, process, queueing, singleserver
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lowball
#1
Nov10-12, 09:23 PM
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
[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]

3. 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?
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haruspex
#2
Nov10-12, 09:58 PM
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
?
lowball
#3
Nov10-12, 10:04 PM
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!

Ray Vickson
#4
Nov11-12, 03:50 AM
Sci Advisor
HW Helper
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P: 4,959
Bernoulli single-server queueing process

Quote Quote by lowball View Post
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.
[tex] \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 [/tex]
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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