MTBF or probability of failure

[carlosarce]

I'm engineer electrician and electronic, I'm from Argentina, maybe you would assist me in the following question:

I have 3 FM transmitters, A, B, C. A has 3 failures / 10000 hours (MTBF = 3333 hours. B has 2 failures / 10000 hours (MTBF = 5000 hours) and C has 1 failures / 10000 hours (MTBF = 10000). All transmitters are on air in differents frecuencies all year. Which is the probability that A and B, B and C, A and C and A, B, C has a failure in the same time ?, for eg. if I make (for A and B), (3/10000).(2/10000) = 6.10^(-8), MTBF (A,B) = 16,666...10.10^6 hours, Can I say that A and B could fail in the same time at least 1 time in 16 millions of hours ?. I apologize for my English.

Thanks so much
Carlos

 

[IRstuff]

You are confused, because MTBF is NOT a probability. I suggest you read the Wiki articles:

http://en.wikipedia.org/wiki/MTBF

http://en.wikipedia.org/wiki/Failure_rate

and

http://www.vicorpower.com/documents/quality/Rel_MTBF.pdf

Your analysis must include how long the transmitters are unavailable if the fail, i.e., what is the mean time to repair (MTTR), assuming that you attempt to repair.

TTFN
faq731-376

 

[carlosarce]

Hi,
I know is different, but I'm trying to pass MTBF to probability of failure, I have to justify that isn't necessary to buy 3 backup transmitters based in low probability of 2 or all transmitters break down in the same time. The model that I imagine is: I have a box with 10000 little balls, all are white exept 3 are red, 2 are blue and 1 is yellow, these colored ball represent failures in transmitters A,B and C, white balls represent no-failure. 10000are represent hours. Now, I go each hour and take three balls, how many hours I need to wait until I get one ball red, one ball blue and one white (failure in A and B) ?
The idea is to buy only one transmitter with fast change of frecuency instead three transmitter, based, how I said before, in the low probability of 2 or 3 transmitter have a failure in the same time.
I appreciate your response and links, thanks so much.
Carlos
P.S. After I took the three balls, I back the balls in the box.

 

[BigInch]

Maybe not quite the same, since MTBF could imply that P(Failure) increases as MTBF is approached, however if any transmitter can fail at any given moment within the 10,000 hour time span, then
P(1xFails) = max(P(1),P(2), P(3))
P(2xFail) = max(P(1) x P(2), P(1) x P(3), P(2) x P(3))
P(3xFail) = P(1) * P(2) * P(3)

MTBF must be converted to P(x), perhaps as you have done by using the average number of failures over the operating time, however it is not the true story, but it is a story you could tell ... somebody.

A failure would also have to be repaired, say within the hour, or the problem should include the higher probability of failure of 1 of the two remaining operating units during any MTTR period.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[IRstuff]

Sorry, not.
[λ] = 1/MTBF

Pfail(t,[λ]) = INT([λ]*exp(-[λ]*t))[t,t+MTTR] --> probability of failure during time period from t to t+MTTR is the integral of the weighted probability distribution function, which is assumed to be exponential in the conventional MTBF analysis.

The probability of any pair of transmitters failing is, therefore, I think, the convolution of their respective failure probabilities. Note that this formulation is not strictly correct, since the time duration is MTTR plus the time it took to detect the failure, which could be a while, if the failure is very subtle. Here, I would define failure to be anything that causes a repeatable inability to transmit correctly. This is to include obscure failures that only affect specific data or data patterns.

You might get by with just multiplying the failure probabilities together.

TTFN
faq731-376

 

[BigInch]

If the average is assumed to be constant over the MTBF period.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[IRstuff]

?? average of what?

TTFN
faq731-376

 

[BigInch]

If the probability of failure is constant over the time of interest;, ie does not increase as some end of lifetime is approached.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[IRstuff]

Right, neither infant mortality nor wearout is modeled in the standard exponential failure distribution; MTBF is defined only for the "constant failure rate" regime.

TTFN
faq731-376