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Reliability and MTTF, what's going on?

  1. Dec 7, 2013 #1
    I am reading a reference that says, with [itex]f(x)[/itex] being the failure probability density function, reliability of a system with that PDF can be expressed as [itex]1 - \int_0^t f(x) dx[/itex]. For a given t, this gives a number. Next, the same reference says that the mean time to failure of the system, [itex]MTTF = \int_0^\infty R(t) dt[/itex]. How is this possible? R(t) returns a number, integrating to infinity over a constant returns infinity...

    I guess I am missing some concept.

    Thanks
     
  2. jcsd
  3. Dec 7, 2013 #2

    AlephZero

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    ##R(t)## is not "a number", it is a function of ##t##. ##R(t)## will decrease to ##0## as ##t## goes to infinity, so ##\int_0^\infty R(t)\,dt## can be finite.

    For example if ##f(x) = ae^{-ax}## for some constant ##a##

    ##R(t) = 1 - (1/a)(ae^{-at} - a) = e^{-at}##

    and the MTTF = ##1/a##.
     
  4. Dec 7, 2013 #3
    Thanks
     
  5. Dec 7, 2013 #4
    Also I notice that a lot of sites online show examples with the mean=0. Since it's a PDF, we want only zero onward. So why are there so many examples with mean=0?
     
  6. Dec 9, 2013 #5
    MTTF = 0 implies that the component is already broken: you must be misinterpreting the data - perhaps 0 means that there is no data available for the component? Can you link to an example?
     
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