Reliability of Independent BatteriesDate: 02/08/2012 at 16:20:14 From: Marcelo Subject: Probability problem with exponentially distributed variables Consider a satellite that runs on a certain battery, A. It has an independent backup, B. The satellite performs its task until both A and B fail. The lifetimes of A and B are exponentially distributed with the mean lifetime of 10 years. a) What is the probability that the satellite will work for more than 10 years? b) Compute the expected working life of the satellite. I think I got a), but I want to confirm; and I'm not sure about how to compute b. a) P{A <= 10} = 1 - e^(lambda * x) = 1 - e^((1/10)*10) = (e - 1)/e With lambda = 1/10, I compute the probability of failure within 10 years for battery A; and since battery B has the same characteristics, A*B (chance of failure on both) equals (e - 1)^2/e^2 = 0.3995 So 1 - 0.3995 = 0.60. b) I don't know how to do this part mathematically. But if we consider a deviation of, for instance, 2 years, it would give rise to these four combinations: (A = 12, B = 12), (A = 12, B = 8), (A = 8, B = 12), (A = 8, B = 8) On the first three, the satellite would keep on going. It seems that lifetime expectation for the satellite must be greater than the expectations for independent batteries, but how is it computed? And what is the graphical interpretation of the expectations' combination? If I'm wrong and expectation remains 10 years, what if each battery had a different life expectancy? Excuse me for this lengthy question, but I would like to really understand its logical basis. Date: 02/14/2012 at 17:41:33 From: Doctor Anthony Subject: Re: Probability problem with exponentially distributed variables With 2 systems in parallel having the same mean time to failure of 10 years, the probability of lasting more than t years is given by P(T > t) = e^(-t/10) + e^(-t/10) - e^(-2t/10) So, P(T > 10) = e^(-1) + e^(-1) - e^(-2) = 0.60042 In part b), the formula for the expected lifetime of the two in parallel is 1 1 1 E(T) = ----- + ------ - --------------- (1/10) (1/10) (1/10) + (1/10) = 10 + 10 - 5 = 15 years - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 02/24/2012 at 19:52:39 From: Marcelo Subject: Probability problem with exponentially distributed variables First, I would like to thank you for the help. I would also like to ask if you can show me the derivation of the formula for calculating parallel expectations, i.e., item b) of my original question. Date: 02/25/2012 at 08:02:55 From: Doctor Anthony Subject: Re: Probability problem with exponentially distributed variables Let T = time to failure of the system. Reliability = R(t) = Prob(T > t) Assuming components function independently of each other, the reliability of the system, R(t), can be expressed in terms of the reliabilities of the components, say R1(t) and R2(t), as follows: R(t) = P(T > t) = 1 - P(T < t) = 1 - [(P(T1 < t) and (P(T2 < t)] = 1 - [1 - P(T1 > t)].[1 - P(T2 > t)] = 1 - [1 - R1(t)].[1 - R2(t)] = R1(t) + R2(t) - R1(t).R2(t) (1) If n components functioning independently are in parallel, then R(t) is given by R(t) = 1 - [1 - R1(t)].[1 - R2(t).[1 - R3(t)].....[1 - Rn)t)] If all the components have the same reliability, say Ri(t) = r(t), the above expression becomes R(t) = 1 - [1 - r(t)]^n If n = 2, we get expression (1) above: R(t) = R1(t) + R2(t)- R1(t).R2(t) = e^(-at) + e^(-bt) - e^[-(a + b)t] Here, a and b are the parameters for the two components, which equals 1/(mean failure times). The probability density function (PDF) of the failure time for the parallel system is f(t) = a.e^(-at) + b.e^(-bt) - (a + b).e^[-(a + b)t] Then the expected value of T is E(T) = 1/a + 1/b - 1/(a + b) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 03/11/2012 at 15:18:11 From: Marcelo Subject: Thank you (Probability problem with exponentially distributed variables) Thanks for the answer. It doesn't get clearer than that. |
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