- #1
Probability problem(minmal paths sets and cuts sets
- Thread starter Sadeq
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The discussion centers on the application of minimal path sets and minimal cut sets in calculating circuit reliability and failure probabilities. Participants highlight the importance of correctly evaluating joint probabilities, emphasizing that events must be treated as dependent rather than independent when they occur multiple times in expressions. The principle of inclusion/exclusion is discussed as a necessary method for accurately calculating probabilities, particularly when events are not mutually exclusive. Confusion arises regarding the application of these principles, with a focus on simplifying expressions to avoid miscalculating joint probabilities. Ultimately, the conversation underscores the need for careful consideration of event dependencies in probability calculations.
- #2
Sadeq
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thank you in advance
- #3
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The same failure (e.g. of A) occurs at more than one place in your expression. Looks to me that your calculation treats these as independent, which would be wrong.
Try breaking it up as:
P[A fails] * P[circuit fails given that A fails] + P[A does not fail] * P[circuit fails given that A does not fail]
Try breaking it up as:
P[A fails] * P[circuit fails given that A fails] + P[A does not fail] * P[circuit fails given that A does not fail]
- #4
Sadeq
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First of all many thanks
But there is something that i can't understand,which is my tutor told me we use minmal path sets for determine reliability and mimmal cut sets to determine failure
and he implement this in the class method and it works however here does not work,
So can i conclude that the method he gave me can't be use anywhere or there is something missing in my mind
I attached the class example to have alook on it and understand what I am saying
But there is something that i can't understand,which is my tutor told me we use minmal path sets for determine reliability and mimmal cut sets to determine failure
and he implement this in the class method and it works however here does not work,
So can i conclude that the method he gave me can't be use anywhere or there is something missing in my mind
I attached the class example to have alook on it and understand what I am saying
- #5
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The example in the PDF made it look easy because it was such a simple circuit.
There is nothing wrong with the principle of cutsets/minimal paths; the problem is with evaluating the probability expression.
In the OP, you have P[(Af & Bf) v (Af & Df) v (Cf & Df)]. This is awkward because you have the same event occurring in more than one place in the expression. In order to multiply probabilities, the events must be independent; in order to add them they must be disjoint (or you must correct for the double counting using the principle of inclusion/exclusion as is done in the PDF).
In this more complicated case you can write
P[(Af & Bf) v (Af & Df) v (Cf & Df)] = P[(Af & Bf) v (Af & Df) v (Cf & Df) | Af] * P[Af] + P[(Af & Bf) v (Af & Df) v (Cf & Df) | ~Af] *(1-P[Af])
= P[Bf v Df v (Cf & Df)] * P[Af] + P[Cf & Df] *(1-P[Af])
= P[Bf v Df] * P[Af] + P[Cf & Df] *(1-P[Af])
From there it's straightforward.
There is nothing wrong with the principle of cutsets/minimal paths; the problem is with evaluating the probability expression.
In the OP, you have P[(Af & Bf) v (Af & Df) v (Cf & Df)]. This is awkward because you have the same event occurring in more than one place in the expression. In order to multiply probabilities, the events must be independent; in order to add them they must be disjoint (or you must correct for the double counting using the principle of inclusion/exclusion as is done in the PDF).
In this more complicated case you can write
P[(Af & Bf) v (Af & Df) v (Cf & Df)] = P[(Af & Bf) v (Af & Df) v (Cf & Df) | Af] * P[Af] + P[(Af & Bf) v (Af & Df) v (Cf & Df) | ~Af] *(1-P[Af])
= P[Bf v Df v (Cf & Df)] * P[Af] + P[Cf & Df] *(1-P[Af])
= P[Bf v Df] * P[Af] + P[Cf & Df] *(1-P[Af])
From there it's straightforward.
- #6
Sadeq
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Please bare with me,if i understand you ,you said that i should use inclusion and exclusion principle,and i agree with this,the following attachment show that i use this method,but answers still different,
Please be patient,because i need to understand the problem in my understanding
Please be patient,because i need to understand the problem in my understanding
Attachments
- #7
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The principle says you add up the probs of X1 and X2, for example, but because they're not mutually exclusive you have to subtract the probability of X1 & X2 (because you've counted that twice). If X1 and X2 are independent, the joint probability is the product of the two probabilities. But here, some of the terms you're treating that way are not independent (e.g.because they both refer to A). So it isn't simply X1 prob * X2 prob. That's why you get the wrong answer.
- #8
Sadeq
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Great ,i UNDERSATND NOW,big thanks to you
So because they are independent i can't use this method , so what will be the solution for this problem?
Many thanks again
So because they are independent i can't use this method , so what will be the solution for this problem?
Many thanks again
- #9
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Sadeq said:
You can still use the inclusion/exclusion principle, but you have to calculate the joint probabilities correctly. The problem is that they are NOT independent in this circuit.
To sum up:
Principle of inclusion/exclusion, quite general (two event case for simplicity):
P[X v Y] = P[X] + P[Y] - P[X & Y]
There are three special cases which make this simpler:
If X and Y are mutually exclusive, P[X & Y] = 0
If X and Y are independent, P[X & Y] = P[X]*P[Y]
If X implies Y, P[X & Y] = P[X]
For the circuit you have, none of these apply. Can you understand the method I used a couple of posts ago?
- #10
Sadeq
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Exactly it works man,your are brilliant and thank you
May i ask you why you choose A and consider it as bridge(onetime works and other fails)
Why spesfically A
I think because A make the whole problem and it makes events non independent
I'm i right!
May i ask you why you choose A and consider it as bridge(onetime works and other fails)
Why spesfically A
I think because A make the whole problem and it makes events non independent
I'm i right!
- #11
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I chose A because it appeared frequently in the expression. Fixing its value was sure to simplify things quickly. Glad to have helped.
- #12
- #13
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He's doing the same thing a slightly different way.
First, he adds up the four probabilities for the OR conditions. But then he has to subtract out the probs of the pairwise combinations of these. This is the inclusion/exclusion principle at work. Look at one example of these pairs: (A&B) & (A&C&E). He reduces this (behind the scenes) to (A&B&C&E) before computing the probability: qA*qB*qC*qE.
In my method, I simplified the logic combination before applying inclusion/exclusion. It's a matter of whichever you're more comfortable with.
First, he adds up the four probabilities for the OR conditions. But then he has to subtract out the probs of the pairwise combinations of these. This is the inclusion/exclusion principle at work. Look at one example of these pairs: (A&B) & (A&C&E). He reduces this (behind the scenes) to (A&B&C&E) before computing the probability: qA*qB*qC*qE.
In my method, I simplified the logic combination before applying inclusion/exclusion. It's a matter of whichever you're more comfortable with.
- #14
Sadeq
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Thank you very much,
Unfortuantily now i am confused so please bar with me, i use exactly the book method to solve the problem i rise in the upper posts and the answer were differnent if u try to solve them usiing minmal bath sets and minmal cut sets as he did, i also inclusion and exclusion principle in my solution
Seems strange
PLEASE HELP ME i have a lot of coursework to be done and i have to understand this in order to be able to proceed
Unfortuantily now i am confused so please bar with me, i use exactly the book method to solve the problem i rise in the upper posts and the answer were differnent if u try to solve them usiing minmal bath sets and minmal cut sets as he did, i also inclusion and exclusion principle in my solution
Seems strange
PLEASE HELP ME i have a lot of coursework to be done and i have to understand this in order to be able to proceed
- #15
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Please post your working for the two methods (or send it to me through the Forum's email service) and I'll see if I can spot what's going wrong.
- #16
- #17
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OK, you're still making the mistake of calculating joint probability as though events are independent when they're not.
P((A&D)v(A&C)v(B&D)) = P(X1vX2vX3) = P(X1) + P(X2) + P(X3) - P(X1&X2) - P(X2&X3) - P(X3&X1) + P(X1&X2&X3)
So far so good. But then you turn P(X1&X2) into P(X1)*P(X2). You can't do that because X1 and X2 are not independent.
P(X1&X2) = P((A&D)&(A&C)), which simplifies to P(A&D&C). Do you see why?
So P(X1&X2) = P(A)*P(D)*P(C) = 0.729 (because A, C and D are independent).
So with p = 0.9, P((A&D)v(A&C)v(B&D)) = 3*p^2 - p^3 - p^3 - p^4 + p^4 = 0.972
P((A&D)v(A&C)v(B&D)) = P(X1vX2vX3) = P(X1) + P(X2) + P(X3) - P(X1&X2) - P(X2&X3) - P(X3&X1) + P(X1&X2&X3)
So far so good. But then you turn P(X1&X2) into P(X1)*P(X2). You can't do that because X1 and X2 are not independent.
P(X1&X2) = P((A&D)&(A&C)), which simplifies to P(A&D&C). Do you see why?
So P(X1&X2) = P(A)*P(D)*P(C) = 0.729 (because A, C and D are independent).
So with p = 0.9, P((A&D)v(A&C)v(B&D)) = 3*p^2 - p^3 - p^3 - p^4 + p^4 = 0.972
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