Supplement for Repairable System Reliability PDF = probability_density_function = ft() Reference: Probability and Reliability for Engineers, Miller, TA340.M648 1985 ⌠t primarily chapter 15 CDF = cumulative_distribution_function = () = ⎮ () NIST e-book Engineering Statistics Handbook,Ft fx dx⌡0 sections labelled 8.1.2.2 Reliability or Survival function Reliability_function = probability_unit_survives_beyond_t Rt() = 1 − Ft() Ft() = 1 − R t()or ... 8.1.2.3 Failure (or Hazard) rate ht( ) = failure_rate = ft() = ft() conditional probability 1 − F t() Rt() therefore ... ft = R t ⋅ ()() ()htnow ... Rt() = 1 − Ft() dFt() = −d Rt() = ft() dt dt d Rt() ht() = ft() = − dt = −d ( ()ln R t ) Rt() Rt() dt ⌠t integrate from 0 to t ⎮ hx dx = − ( ()) () ln R t⌡0 ⌠t ⎮ hx dx− ()⌡ exponentiate .. Rt() = e0 ⌠t ⎮ hx dx− ()⌡ therefore ... ft = R t ⋅ () = ht()e 0() ()ht ⋅ now .. if assume (observe) failure rate h(t) = constant = λ ht() := λ λ⌠t ⎮ hx dx− ()⌡ ft() := ()e 0 hh t ⋅ ⋅ ) t ⋅t(− λ )⋅ Ft()t := ⌠⎮ t λ⋅e(− λ ) xdx → ⎡⎣−e(− λ ⋅⎤⎦+ 1 Ft() := 1 − e − λλft() → λ⋅e ⌡0 have exponential assumption of probability of failure times 1 11/27/2006exponential pdf and cdf example − λ⋅t1 − λ⋅tPDF_of_time_to_next_failure = f t() = λ⋅e λ := t := 0 .. 10000 ft() := λ⋅e 1000 probability density of time to next failure (lambda = 1/1000) 0.001 probability5 .10 4 0 0 2000 4000 6000 8000 1 .104 time to next failure − λ⋅tCDF = F t = − ⋅t() 1e = CDF_of_waiting_time_to_next_failure F t() 1 − e − λ:=() or .. CDF of "interarrival" time between failures cumulative probability density of time to next failure (lambda = 1/1000) 1 probability0.5 0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1 .104 time to next failure exponential pdf and cdf example reset variables interpret time to failure as a waiting time, it can be shown that this can be represented as a Poisson process, if a component which fails is immediately replaced with a new one having the same failure rate λ. Some results from this observation: 1mean_waiting_ttime_between_successive_failures = = MTBF λ Some results for exponential model 2 11/27/2006Reliability or Survival function Rt() 1− FF t():= (− λReliability_function = probability_unit_survives_beyond_t Rt()→ e) te.g. if component has a failure rate of 0.05/1000 hours, probability that − 0.05 ⋅10000it will survive at least 10,000 hrs is given by: 1000 ⋅ e = 0.607 n components in series if a system consists of n components in series, with respective failure rates λ1, λ2 ... λn n i ∑ i 1=∏= −λi⋅t n so it also is an exponential distribution ...− λ i⋅t Rs t() = e = e and the MTBF for the system is: 1 1 1 = =MTBFseries_system n n ∑ i 1= 1⎛⎜⎝ ⎞⎟⎠ ∑ i 1= for a parallel system ... with respective failure rates λ1, λ2 ... λn in this case we need to deal with "unreliabilities" λi MTBFi is probability component i will failFi 1 − Ri = n ∏=i 1 and probability of survival =Rp(t) Rp()t 1 probability_all_will_fail = unreliability Fp Fi = = ⎤⎦ ⎡⎣∏= n n Fi()t1 i ⎤⎦ ⎡⎣∏= 1 i 1 this will not show exponential distribution ... − Fp()t 1− 1 Ri()t= = = in this case; exponential probability of failure ) ∏= F ()⎤ti⎦ ⎡⎣∏= ⋅t i 1 i 1 d fp()tFp()t dthp()t = Rp()t= Rp()tdifficult to evaluate, but notice at least it is f(t). Rp(t) difficult to obtain in general, but when all components have same failure rate 3 11/27/2006 ( n n− λiFp()t 1 −= = e −−p ∏⎣ it ⎤⎦∏ − λ⋅t − λ⋅tR ()t= 1 − n ⎡1 − R () = 1 − n (1 − e )= 1 −(1 −e )n i = 1 i = 1 n!n_choose_k n k, ) binomial coefficient( := k!⋅(n − k)! − λ⋅t − λ⋅t −λ⋅2 t1 −(1 − e )n = 1 −(1 − ( , ⋅ + ( , ⋅ )n_choose_k n 1) e n_choose_k n 1) e − ........ − λ⋅t − λ⋅t − −λ⋅n1 n⋅ tRp t = n_choose_k n 1( , ⋅ − ( , )e + ......................... ()1 ⋅ ( ) ) e n_choose_k n 2 ⋅ − e binomial theorem from mathworld.wolfram.com/BinomialTheorem.html it can be shown see reference page 460 after differentiating to find fp(t) fp t() t Rp t()d d = and then calculating the mean (MBTFparallel) MBTFparallel = 1 α 1 +⎛ ⎜ ⎝ ⋅ 1 2 .............+ + 1 n ⎞ ⎟ ⎠ e.g. if use two identical components in parallel increase of 50% not doubleMBTFparallel 1 α 3 2 ⋅= 1 4 k ∑ = 1 k 2.083= four to double another example if time permits on board 4