Reliability of a system

[Kakashi]

Mod note: Moved from technical math section, so no homework template in post.
Is it correct that the success of one link is independent of thte success of other links. For example is P(C->E|A->C)=P(C->E) true?

Node A can connect to node B through 3 possible paths:
ACEB
ACFB
ADB

If A is connected to B through one of these paths, does this imply that the other paths have failed, or can multiple paths be operational at the same time?

Meaning of a single path

The probability that the event ACEB works means that all links are up and since links are independent
P(A∩C∩E∩B)=0.9*0.8*0.9=0.648=P(E->B | C->E ∩ A->C)*P(C->E ∩ A->C)=P(E->B)*P(C->E|A->C)*P(A->C)=P(E->B)*P*(C->E)*P(A->C)
Is this probability the probability that only ACEB occurs? Since I am conditioning on predecessor links being up, does this make the probability of ACFB and ADB equal to 0?

Intuitively, since the link A→C is already known to be up, it seems that ACFB should still be possible.


P(A->B)=P(ACEB U ACFB U ADB)=P(ACEB)+P(ACFB)+P(ADB)-P(ACEB∩ACFB)-P(ACEB ∩ ADB)-P(ACFB ∩ ADB)-P(ACEB ∩ ACFB ∩ ADB)

Are the intersection terms P(ACEB ∩ ADB), P(ACFB ∩ ADB),and P(ACEB ∩ ACFB ∩ ADB) equal to 0 because the paths use disjoint links?

 

[Mark44]

If A is connected to B through one of these paths, does this imply that the other paths have failed, or can multiple paths be operational at the same time?

I don't believe that one path being operational implies that other paths have failed. The probability you calculate is just pertaining to the links in that path.

Is this probability the probability that only ACEB occurs?

I don't believe so. See what I wrote above.

Since I am conditioning on predecessor links being up, does this make the probability of ACFB and ADB equal to 0?

In my understanding, no.

 

[FactChecker]

If A is connected to B through one of these paths, does this imply that the other paths have failed, or can multiple paths be operational at the same time?

The problem statement specifies that all failures are independent.

The probability that the event ACEB works means that all links are up and since links are independent

All links on that path are up. It indicates nothing about the ADB path. That path could be up or down.

P(A∩C∩E∩B)=0.9*0.8*0.9=0.648

Correct.

=P(E->B | C->E ∩ A->C)*P(C->E ∩ A->C)=P(E->B)*P(C->E|A->C)*P(A->C)=P(E->B)*P*(C->E)*P(A->C)

This is irrelevant.

Is this probability the probability that only ACEB occurs?

No, The problem states that all probabilities are independent.

Since I am conditioning on predecessor links being up, does this make the probability of ACFB and ADB equal to 0?

No.

Intuitively, since the link A→C is already known to be up, it seems that ACFB should still be possible.


P(A->B)=P(ACEB U ACFB U ADB)=P(ACEB)+P(ACFB)+P(ADB)-P(ACEB∩ACFB)-P(ACEB ∩ ADB)-P(ACFB ∩ ADB)-P(ACEB ∩ ACFB ∩ ADB)

Are the intersection terms P(ACEB ∩ ADB), P(ACFB ∩ ADB),and P(ACEB ∩ ACFB ∩ ADB) equal to 0 because the paths use disjoint links?

No.
You are making this more complicated than it needs to be. The problem statement is that all failure probabilities are independent. With that, what is the probability that all links are working?