Poisson Distribution. Poisson Distribution Example

Transcription

1 Poisson Distribution Can b usd to valuat th probability of an isolatd vnt occurring a spcific numbr of tims in a givn tim intrval,.g. # of faults, # of lightning stroks tim intrval Rquirmnts: -Evnts must b random -Hazard rat must b constant Exprssion for Poisson Distribution: P x Only applis to th usful lif priod of a systm componnt x t P x (t): probability of vnt occurring x tims in tim t : constant hazard rat (known as failur rat) Expctd valu of Poisson distribution, E(x) = μ = t t Probability of zro failurs in tim t, P R =! t 9 Poisson Distribution Exampl If th avrag numbr of cabl faults pr yar pr 1 km of cabl is.5, valuat th probabilitis of, 1,,.. faults occurring in (a) yar priod (b) 4 yar priod P x Failur Rat, =.5 f/yr (a) For a yar priod, t = yr Expctd # of failurs, E(x) = t =.5 x = 1. P x # of failurs Probability probability Failur Dnsity Function # of failurs probability x t x 1 (1) Failur Distribution Function # of failurs

2 Standby Systms - Poisson Distribution On or mor componnts ar in standby mod rady to oprat whn th main oprating componnt fails. A snsing and changovr dvic is rquird to bring th standby unit into opration whn th main componnt fails. A S A: main componnt : standby componnt S: snsing and changovr dvic Failur of standby componnt is dpndnt on th failur or succss of th main componnt A. If th componnts ar in thir usful lif priod, thir tims to failur ar xponntially distributd. Poisson Distribution can, thrfor, b applid. 3 Prfct Switching Assumptions: - snsing & changovr dvic is 1% rliabl - standby componnts do not fail in standby mod - componnts ar idntical with failur rat A S Th -componnt standby systm is quivalnt to a singl unit which is allowd to fail onc. If thr is a nd failur, th quivalnt unit also fails. Applying Poisson s Distribution: Probability of x componnts failing in tim t, P x x t P ( fail) = P t P (1 fail) = P 1 t. t Systm Rliability, R s P (t) + P 1 t (1 + t) 33

3 Prfct Switching : -Standby Componnts A S This systm is quivalnt to a singl unit which is allowd to fail twic. t C R s P (t) + P 1 (t) + P (1 + t + ) 3 For a systm with n standby componnts, n x t R s x= n + 1 i.. th sum of first n trms of th Poisson Distribution. Th systm MTTF is th sum of individual componnt MTTF valus. (sinc ach componnt gts its turn aftr complt opration of th prvious componnt) 34 Exampl Compar th rliability and MTTF of a -componnt systm ach having a failur rat of. f/hr aftr a tim of 1 hr if thy ar (a) paralll rdundant (b) standby rdundant with 1% rliabl snsing and changovr dvic. (a) Paralll systm Using product rul of unrliability, Q s Q 1 (t). Q (t) t = (1- )(1- ).x1 = (1- ) Q s (1h) =.3859 R s (1h) = = MTTF = + + = 75 h t (b) Standby systm R s t (1 + t) R s (1h) = MTTF = = 1 h 35

4 Effct of Spar Componnts Spar componnt systm bcoms similar to standby systm (prfct switching) if - rplacmnt tim of a faild componnt by a spar is insignificant - faild componnts ar not rpaird With ths assumptions, Poisson Distribution mthod can also b applid to valuat th ffct of spar componnts. Applying Poisson s Distribution: Probability of x componnts failing in tim t, P x x t 36 Standby Systms: Imprfct Switching Snsing & changovr dvic (S) has a probability of succssful opration, P s # of succssful oprations whr P s = # of rqustd oprations A S For a -componnt standby systm: P ( comp. fails) = P t P (1 fails AND S oprats succssfully) = P 1 (t). P s = P s.t. t Systm Rliability, R s P (t) + P s.p 1 t (1 + t. P s ) R s MTTF = ( t) 1 + P s 37

5 Exampl Compar th rliability and MTTF of a -componnt systm ach having a failur rat of. f/hr aftr a tim of 1 hr if thy ar a) paralll rdundant b) standby rdundant with th snsing and changovr dvic rliability btwn 1. and.9 (a) Paralll systm Using product rul of unrliability, Q s Q 1 (t). Q (t) R s (1h) = MTTF = 75 h (b) Standby systm t R s (1 + t. P s ) R s (1) =. (1 +. P s ) Ps Rs (1 hr) P s MTTF = = MTTF (hr) Ps. 38