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        riddles >> medium >> Circuit Failure Analysis
        
(Message started by: Yournamehere on Sep 20^th, 2002, 2:38pm)

Title: Circuit Failure Analysis
Post by Yournamehere on Sep 20^th, 2002, 2:38pm

Let w_k = 1-p_k ; w_k is the probability of a component working at time T.

(a) P_{failure} = 1-P_{working}
= 1-(P_{upper_path_works} + P_{lower_path_works} - P_{both_paths_work})
= 1-(w_4 w_1 + w_4 w_3 w_2 - w_4 w_3 w_2 w_1)

You can trivially expand this out to an expression in p_k, if you wish.

(b) If component C_k fails, substitute 0 for w_k in the above expression; don't forget that you're now asked for P_{working}, so use 1-P_{failure}:

C_1 fails: P_{working} = w_4 w_3
C_2 fails: P_{working} = w_4 w_1
C_3 fails: P_{working} = w_4 w_1
C_4 fails: P_{working} = 0

I'm not sure what the "subtlety" is in the problem is, unless you're referring to the use of the principle of inclusion-exclusion.

Title: Re: Circuit Failure Analysis
Post by william wu on Sep 22^nd, 2002, 6:08pm

Your answer to (a) is correct -- that's pretty straightforward -- but the answer to (b) is slightly incorrect. Can't just substitute zero ...

Problem statement reprinted here for convenience:

Consider the circuit shown below. If all components along a path from In to Out are working, then the whole system is considered to be working.

Through appropriate experimentation and modelling, it has been determined that at time T, component C_k fails with probability p_k. Also assume that component failure are independent.

http://www.ocf.berkeley.edu/~wwu/images/riddles/ee126_hw3_circuit.gif

(a) What is the probability the system has failed at time T?

(b) Your maintenance technician reports that at time T, component C_k has failed, but she hasn't been able to check any other components yet. What is the probability the system is working? Repeat for k = 1, 2, 3, 4.

Title: Re: Circuit Failure Analysis
Post by Yournamehere on Sep 23^rd, 2002, 2:17pm

on 09/22/02 at 18:08:35, william wu wrote:

but the answer to (b) is slightly incorrect. Can't just substitute zero ...

I see I made a typo: For C1 failure, P_{working} = w_4 w_3 w_2. This is what one would get by substituting 0 for w_1 in the original equation.

I don't see why substituting 0 for w_k doesn't work.

Title: Re: Circuit Failure Analysis
Post by william wu on Sep 23^rd, 2002, 10:59pm

You're right. Sorry.

Well, regarding this subtlety I referred to, my solution used the following definition of conditional probability:

P(A|C) = P(A intersection C) / P(C)

So in the context of the problem, let

A = event that the system works
C2 = event that C₂ failed

For P(A|C2), I first computed P(A intersection C2) as follows:

P(A intersection C2) = (1-p₁)(1-p₄)

Then since P(C2) = p₂, my final answer was:

P(A|C2) = (1-p₁)(1-p₄) / p₂

which is different from your answer. So I thought most people would not remember to divide by p₂.

However, there is another subtlety in this problem which cancels out the other subtlety, and makes my answer wrong and everyone else's right. The formula for P(A intersection C2) actually should be:

P(A intersection C2) = (1-p₁)(1-p₄)p₂

because A intersection C2 is the probability that C1 and C4 are working, AND C2 has failed. So after multiplying the numerator of P(A|C2) by p₂, we cancel out the p₂ in the denominator, and you just get (1-p₁)(1-p₄), an expression you could arrive at by both forgetting about dividing by p₂, and incorrectly computing (A intersection C).

Most disturbingly, my GSI was trying to convince me that my answers were right. One student who didn't like my answers pointed out that it didn't make sense that I had different expressions for P(A|C2) and P(A|C3):

(wrong answers)

P(A intersection C2) = (1-p₁)(1-p₄)p₂
P(A intersection C3) = (1-p₁)(1-p₄)p₃

Intuitively, you'd think that the probability the system still works given that C2 failed should be the same as if C3 failed. In both cases, the only potential path from In to Out is the top branch, which does not depend on C2 nor C3. So it turns out that this intuition is right.

However my GSI then offered the following argument. Consider the components to be light bulbs. Let's say C2 is known to be an extremely reliable bulb -- more reliable than C1, C3, and C4 -- so the probability of its failure is very small. And let's say C3 is a very unreliable bulb. Then the probability that the system works, given that C2 failed, is very low since if your most reliable bulb has gone out, it is likely that the less reliable bulbs C1 and C4 have gone out too. Likewise, if C3 has gone out, the probability the system has failed is lower than it would be had C2 gone out, since it was expected that C3 would go out eventually anyways.

I didn't like this argument because it implied that the bulbs were dependent, even though the problem explicitly says that the components are independent. But then he said that the bulbs become dependent when they are linked as a system.

Whatever. Guess I made things too complicated.