Repair Chain (Markov Chain Sample Model)

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A machine has $3$ critical parts that are subject to failure, but can function as long as two of these parts are working. When two are broken, they are replaced and the machine is back to working order the next day. To formulate a Markov chain model we declare its state space to be parts that are broken $0,1,2,3,12,13,23$. If we assume that parts $1,2$ and $ 3 $ fail with probabilities $0.01, 0.02,$ and $ 0.04$ ,respectively but no two parts fail at the same day, then we arrive at the following transition matrix.

$$
beginmatrix
& 0 & 1 & 2 &3 &12 &13 &23 \
0 & 0.93 & 0.01 & 0.02 &0.04 &0 &0 &0 \
1 & 0 &0.94 & 0 & 0 &0.02 &0.04 & 0 \
2 & 0 & 0 & 0.95 & 0 &0.01 &0 &0.04 \
3 & 0 & 0 & 0 &0.97 &0 &0.01 &0.02 \
12 & 1 & 0 & 0 &0 &0 &0 &0\
13 & 1 & 0 & 0 &0 &0 &0 &0\
23 & 1 & 0 & 0 &0 &0 &0 &0\
end
matrix$$

I don't have any problem with the interpretation of $p(0,0), p(0,1), p(0,2)$ and $p(0,3)$. It says there that given everything is working fine, there is $93$% chance everything will still be fine. $1$% of part $1$ failing, $2$% of part $2$ failing, and $4$% of part $3$ failing. What I don't get is everything underneath.

asked Mar 12 '15 at 12:02

maskleon

858

Like for p(1,1), My interpretation is given part 1 already failed, 94% is the chance of what? Failing again?
â€“Â maskleon
Mar 12 '15 at 12:05

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix
â€“Â Did
Mar 12 '15 at 12:18

Thanks a lot, Did. Now, I get it.
â€“Â maskleon
Mar 12 '15 at 13:07

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix Could you explain this a bit more Did please?
â€“Â Kan
Aug 6 at 22:11

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asked Mar 12 '15 at 12:02

maskleon

858

Like for p(1,1), My interpretation is given part 1 already failed, 94% is the chance of what? Failing again?
â€“Â maskleon
Mar 12 '15 at 12:05

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix
â€“Â Did
Mar 12 '15 at 12:18

Thanks a lot, Did. Now, I get it.
â€“Â maskleon
Mar 12 '15 at 13:07

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix Could you explain this a bit more Did please?
â€“Â Kan
Aug 6 at 22:11

add a commentÂ |Â

up vote
-2
down vote

favorite

asked Mar 12 '15 at 12:02

maskleon

858

asked Mar 12 '15 at 12:02

maskleon

858

asked Mar 12 '15 at 12:02

maskleon

858

asked Mar 12 '15 at 12:02

maskleon

858

asked Mar 12 '15 at 12:02

maskleon

858

Like for p(1,1), My interpretation is given part 1 already failed, 94% is the chance of what? Failing again?
â€“Â maskleon
Mar 12 '15 at 12:05

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix
â€“Â Did
Mar 12 '15 at 12:18

Thanks a lot, Did. Now, I get it.
â€“Â maskleon
Mar 12 '15 at 13:07

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix Could you explain this a bit more Did please?
â€“Â Kan
Aug 6 at 22:11

add a commentÂ |Â

Like for p(1,1), My interpretation is given part 1 already failed, 94% is the chance of what? Failing again?
â€“Â maskleon
Mar 12 '15 at 12:05

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix
â€“Â Did
Mar 12 '15 at 12:18

Thanks a lot, Did. Now, I get it.
â€“Â maskleon
Mar 12 '15 at 13:07

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix Could you explain this a bit more Did please?
â€“Â Kan
Aug 6 at 22:11

Like for p(1,1), My interpretation is given part 1 already failed, 94% is the chance of what? Failing again?
â€“Â maskleon
Mar 12 '15 at 12:05

If in state 1, either part 2 fails and one goes to state 12, or part 3 fails and one goes to state 13, or neither part 2 nor part 3 fail and one stays in state 1. Likewise for every line of the transition matrix
â€“Â Did
Mar 12 '15 at 12:18

Thanks a lot, Did. Now, I get it.
â€“Â maskleon
Mar 12 '15 at 13:07

add a commentÂ |Â

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First, I am going to relabel your states to that the indexing can work a little more consistently (this isn't really necessary, but it helps me work through the problem):

Everything is in working order.

Only Part 1 is broken.

Only Part 2 is broken.

Only Part 3 is broken.

Parts 1 and 2 are broken.

Parts 1 and 3 are broken.

Parts 2 and 3 are broken.

The transition matrix $A$ is given by
$$A =
beginpmatrix
0.93 & 0.01 & 0.02 &0.04 &colorblue0 &colorblue0 &colorblue0 \
0 &0.94 & 0 & 0 &colorpurple0.02 &0.04 & 0 \
0 & 0 & 0.95 & 0 &0.01 &0 &0.04 \
0 & 0 & 0 &0.97 &0 &0.01 &0.02 \
colorred1 & 0 & 0 &0 &0 &0 &0\
colorred1 & 0 & 0 &0 &0 &0 &0\
colorred1 & 0 & 0 &0 &0 &0 &0\
endpmatrix.$$
Then $a_i,j$, the element of $A$ in the $i$-th row and $j$-th column, is the probability of transitioning from state $i$ into state $j$. For example, $colorpurplea_2,5 = 0.02$ (in purple) is the probability of transitioning from state 2 (only part 1 is broken) to state 5 (both parts 1 and 2 are broken). In other words, if part 1 is already broken on a given day, then $a_2,5$ is the probability that part 2 breaks down on the next day.

A couple of key points to think about:

Note that $colorbluea_1,j = 0$ for $j=4,5,6$ (these are the blue entries). This makes sense, as these three entries give the probability of transitioning from state 1 (everything is working fine) to a state where two parts are broken. In the statement of the problem, it is asserted that two parts cannot fail on the same day, thus the probability of going from "everything's fine!" to "two parts are broken!" is zero.

Next, note that $colorreda_i,1 = 1$ for $i=4,5,6$ (these are the red entries). Again, this makes sense. These three entries denote the probability of going from "two parts are broken" to "everything is fine". Since the machine is fixed whenever two parts break, the only possible transition out of "two parts are broken" is into a working state. Thus if two parts are broken, there is a 100% probability of transitioning to state 1 (everything's fine!)

Can you reason about the remaining entries?

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

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up vote
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First, I am going to relabel your states to that the indexing can work a little more consistently (this isn't really necessary, but it helps me work through the problem):

Everything is in working order.

Only Part 1 is broken.

Only Part 2 is broken.

Only Part 3 is broken.

Parts 1 and 2 are broken.

Parts 1 and 3 are broken.

Parts 2 and 3 are broken.

A couple of key points to think about:

Note that $colorbluea_1,j = 0$ for $j=4,5,6$ (these are the blue entries). This makes sense, as these three entries give the probability of transitioning from state 1 (everything is working fine) to a state where two parts are broken. In the statement of the problem, it is asserted that two parts cannot fail on the same day, thus the probability of going from "everything's fine!" to "two parts are broken!" is zero.

Next, note that $colorreda_i,1 = 1$ for $i=4,5,6$ (these are the red entries). Again, this makes sense. These three entries denote the probability of going from "two parts are broken" to "everything is fine". Since the machine is fixed whenever two parts break, the only possible transition out of "two parts are broken" is into a working state. Thus if two parts are broken, there is a 100% probability of transitioning to state 1 (everything's fine!)

Can you reason about the remaining entries?

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

add a commentÂ |Â

up vote
0
down vote

First, I am going to relabel your states to that the indexing can work a little more consistently (this isn't really necessary, but it helps me work through the problem):

Everything is in working order.

Only Part 1 is broken.

Only Part 2 is broken.

Only Part 3 is broken.

Parts 1 and 2 are broken.

Parts 1 and 3 are broken.

Parts 2 and 3 are broken.

A couple of key points to think about:

Note that $colorbluea_1,j = 0$ for $j=4,5,6$ (these are the blue entries). This makes sense, as these three entries give the probability of transitioning from state 1 (everything is working fine) to a state where two parts are broken. In the statement of the problem, it is asserted that two parts cannot fail on the same day, thus the probability of going from "everything's fine!" to "two parts are broken!" is zero.

Next, note that $colorreda_i,1 = 1$ for $i=4,5,6$ (these are the red entries). Again, this makes sense. These three entries denote the probability of going from "two parts are broken" to "everything is fine". Since the machine is fixed whenever two parts break, the only possible transition out of "two parts are broken" is into a working state. Thus if two parts are broken, there is a 100% probability of transitioning to state 1 (everything's fine!)

Can you reason about the remaining entries?

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

add a commentÂ |Â

up vote
0
down vote

First, I am going to relabel your states to that the indexing can work a little more consistently (this isn't really necessary, but it helps me work through the problem):

Everything is in working order.

Only Part 1 is broken.

Only Part 2 is broken.

Only Part 3 is broken.

Parts 1 and 2 are broken.

Parts 1 and 3 are broken.

Parts 2 and 3 are broken.

A couple of key points to think about:

Note that $colorbluea_1,j = 0$ for $j=4,5,6$ (these are the blue entries). This makes sense, as these three entries give the probability of transitioning from state 1 (everything is working fine) to a state where two parts are broken. In the statement of the problem, it is asserted that two parts cannot fail on the same day, thus the probability of going from "everything's fine!" to "two parts are broken!" is zero.

Next, note that $colorreda_i,1 = 1$ for $i=4,5,6$ (these are the red entries). Again, this makes sense. These three entries denote the probability of going from "two parts are broken" to "everything is fine". Since the machine is fixed whenever two parts break, the only possible transition out of "two parts are broken" is into a working state. Thus if two parts are broken, there is a 100% probability of transitioning to state 1 (everything's fine!)

Can you reason about the remaining entries?

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

First, I am going to relabel your states to that the indexing can work a little more consistently (this isn't really necessary, but it helps me work through the problem):

Everything is in working order.

Only Part 1 is broken.

Only Part 2 is broken.

Only Part 3 is broken.

Parts 1 and 2 are broken.

Parts 1 and 3 are broken.

Parts 2 and 3 are broken.

A couple of key points to think about:

Note that $colorbluea_1,j = 0$ for $j=4,5,6$ (these are the blue entries). This makes sense, as these three entries give the probability of transitioning from state 1 (everything is working fine) to a state where two parts are broken. In the statement of the problem, it is asserted that two parts cannot fail on the same day, thus the probability of going from "everything's fine!" to "two parts are broken!" is zero.

Next, note that $colorreda_i,1 = 1$ for $i=4,5,6$ (these are the red entries). Again, this makes sense. These three entries denote the probability of going from "two parts are broken" to "everything is fine". Since the machine is fixed whenever two parts break, the only possible transition out of "two parts are broken" is into a working state. Thus if two parts are broken, there is a 100% probability of transitioning to state 1 (everything's fine!)

Can you reason about the remaining entries?

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

edited Aug 8 at 1:32

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

answered Aug 8 at 1:24

Xander Henderson

13.1k83150

add a commentÂ |Â

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