Markov Processes

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February 24th 2010, 08:00 AM
Dogod11

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Markov Processes

A seamstress works exclusively in one phase of the process of producing a garment. Exactly ½ hour delay to implement its work on garments. Each ½ hour passes a messenger to collect the clothes are ready and leave, by the way, new clothes them to the seamstress do its work. New sewing garments for the messenger is random at each visit is as follows: the
30% do not carry any, 50% of the time takes 1 and 20% of the time takes 2 garments. The messenger are instructed to never leave more than three pieces by the seamstress sew.

Calculate the percentage of time that the seamstress is idle. R// 14%.

The exercise provides the transition matrix:
http://www.mathhelpforum.com/math-he...1&d=1267030748

I wonder if it is possible that calculated under the following process:

$<br /> P = \begin{bmatrix}{p_{00}}&{p_{01}}&{p_{02}}&{p_{03}} \\{p_{10}}&{p_{11}}&{p_{12}}&{p_{13}}\\{p_{20}}&{p _{21}}&{p_{22}}& {p_{23}}\\{p_{30}}&{p_{31}}&{p_{32}}& {p_{33}}\end{bmatrix}$ ,

where, for example, $p_{00}$ is the probability that the messenger does not let clothes

given that the seamstress had no clothes ... and so on.

But I have problems to get these conditional probabilities can not remember, especially in the third line, which would be understood as:

$<br /> p_ {20}$ = Probability of leaving nothing since there are two garments.

$p_ {21}$ probability of leaving a given that there are two.

$p_ {22}$ probability of leaving two since there are two.

$p_ {23}$ likely leave three because there are two.

thank you buddy,

A greeting
February 24th 2010, 05:46 PM
Aileys.

The seamstress completes work on 1 garment with every transition.

Therefore,

new number of garments = old number of garments - 1 + number of garments brought by messenger.

^This with the constraint that the number of garments must remain between zero and three.

So for example, $p_{20}$ =probability of having two garments, then no garments. How can this happen? It cannot. This probability is zero.

$p_{21}$ =probability of having two garments, then one garments. How can this happen? Only if the messenger brings no new garments - the seamstress will complete one of the two garments during the transition, leaving one remaining. This is probability 0.3

etc
February 24th 2010, 06:16 PM
Dogod11

Hi, thanks, I had not stopped to think about this because of lack of time,

I thought that the answer to the question of leisure time

question is a "long term", so I calculate $P_{00}$ in $P^n$ . Do not you think?

A greeting and thanks again.