Chapter 5 Reliability of Systems Hey! Can you tell us how to analyze complex systems? Serial Configuration Parallel Configuration Combined Series-Parallel C. Ebeling, Intro to Reliability & Maintainability Engineering, Chapter 5 2 nd ed. Waveland Press, Inc. Copyright 2010 1
Serial Configuration Reliability Block Diagram E 1 = the event, component 1 does not fail, and E 2 = the event, component 2 does not fail, then P{E 1 } = R 1 and P{E 2 } = R 2 where R 1 = the reliability of component 1, and R 2 = the reliability of component 2. Therefore assuming independence: R s = P{E 1 E 2 } = P{E 1 } P{E 2 } = R 1 R 2 Chapter 5 2
Multiple Components in Series Generalizing to n mutually independent components in series; R s (t) = R 1 (t) x R 2 (t) x... x R n (t) and R s (t) min {R 1 (t), R 2 (t),..., R n (t)} Chapter 5 3
Component Count vs. System Reliability Component Number of Components Reliability 10 100 1000.900.3487. 266x10-4. 1748x10-45.950.5987.00592. 5292x10-22.990.9044.3660. 443x10-4.999.9900.9048.3677 System Reliability Chapter 5 4
Constant Failure Rate Components n λi t n n - t - t R s(t) = R() t = i e = e i=1 = λi λs e i=1 i=1 where λ s = λ i n i=1 Chapter 5 5
Weibull Components R s(t) = n e F H G i=1 n F H t Iβi - = e i=1 θi K J t θi I K βi λ (t) = e - n β t I i HG θ i K J i=1 F e - L NM n i=1 β θ n β t I i HG θ i K J i=1 F i i F HG t θ i I K J β i -1 O QP = n F i t i H G β I -1 i i K J θ θ i=1 β Chapter 5 6
Components in Series - Example A communications satellite consists of the following components: Probability Shape Characteristic Component Distribution Parameter life Power unit Weibull 2.7 43,800 hr. Receiver Weibull 1.4 75,000 hr. Transmitter Weibull 1.8 68,000 hr. Antennae Exponential MTTF = 100,000 hr. Chapter 5 7
λ S () t Components in Series - Example 17.. 4. 8 27. t 14. t 18. t 1 43800, 43800, 75000, 75000, 68000, 68000, 100000, = F H G I K J + F H G I K J + F H G I K J + R () t e e e e s 27. 1.4 1.8 t t t t 43, 800 75, 000 68, 000 100, 000 = F H G I K J F H G I K J F H G I K J 17, 520 27. 1.4 1.8 17, 520 17, 520 F H G I K J F H G I K J F H G I K J 17, 520 R (, ) 43, 800 75, 000 68, 000 100, 000 s 17 520 = e e e e =. 62 Chapter 5 8
Class Exercise The failure distribution of the main landing gear of a commercial airliner is Weibull with a shape parameter of 1.6 and a characteristic life of 10,000 landings. The nose gear also has a Weibull distribution with a shape parameter of.90 and a characteristic life of 15000 landings. What is the reliability of the landing gear system if the system is to be overhauled after 1000 landings? Chapter 5 9
Class Exercise - solution 1000 1000, 15, 000 10 000 R( 1000) e e 1.6. 9 (. 975) (. 916). 893 = F H G I K J F H G I K J = = Animated Chapter 5 10
Parallel Configuration R s =P{E1 E2} = 1 - P{(E1 E2) c } R 1 R 2 =1-P{E1 c E2 c } =1-P{E1 c }P{E2 c } = 1 - (1-R 1 ) (1-R 2 ) R n where E1 = event, component 1 does not fail E2 = event, component 2 does not fail, and Pr{E1} = R 1 and Pr{E2} = R 2 The probability at least one component does not fail! Chapter 5 11
Parallel Configuration - Generalization R s n i= 1 [ R ] i (t) = 1-1- (t) R s (t) >= max {R 1 (t), R 2 (t),, R n (t)} Animated Chapter 5 12
Parallel Configuration - CFR Model R s n (t) = 1-1- - it e λ i= 1 Chapter 5 13
For n = 2: s Parallel Configuration 2-component CFR Model n - λ t s 1 i= 1 1- e i R (t) = - -λ t -λ t R (t) = 1 - (1 - e 1 )(1 - e 2 ) -λ t -λ t -( λ + λ )t = e + e - e 1 2 1 2 z z z z -λ1t -λ 2t -( λ MTTF = R (t)dt = e dt + e dt - 1+ λ 2 )t e dt 0 s 0 0 0 = 1 + 1 1 - λ 1 λ + 2 λ1 λ2 Chapter 5 14
Parallel Configuration - Weibull Model L i n t 1 1 s i= 1 N M F O P H G I Β K J θ Q R (t) = - - e i If n identical Weibull components are in parallel: R (t) = - s L NM t 1 1 F H G I Β K J θ - e O QP n Chapter 5 15
Parallel Configuration - Example A circuit breaker has a Weibull failure distribution (against a power surge) with beta equal to.75 and a characteristic life of 12 years. 1 12 F H G I K J. 75 R() 1 = e =. 856 Determine the one year reliability if two identical circuit breakers are redundant. Chapter 5 16
Parallel Configuration Example Solution Two redundant breakers have a reliability function of: Rt () L = 1 NM 1 e F t H G I K J 12 O. 75 QP 2 L = NM F 1 H G I K J O. 75 P Q 12 2 R() 1 1 1 e P = 1 ( 1. 856) =. 979 2 Question: how are two redundant circuit breakers physically configured? Chapter 5 17
Combined Series - Parallel Systems R 1 R 2 R 3 R 6 R 4 R 5 Chapter 5 18
Combined Series - Parallel Systems A B R 1 R 2 R 3 R 6 C R 4 R 5 Chapter 5 19
Combined Series - Parallel Systems A R 1 R 2 R A = [ 1 - (1 - R 1 ) (1 - R 2 )] Chapter 5 20
Combined Series - Parallel Systems A R A = [ 1 - (1 - R 1 ) (1 - R 2 )] R 1 R 2 A B R 1 R B = R A R 3 R 3 R 2 Chapter 5 21
Combined Series - Parallel Systems C R C = R 4 R 5 R 4 R 5 R B R 6 R C R s = [1 - (1 - R B ) (1 - R c ) ] R 6 Chapter 5 22
High Level Redundancy R high = 1 - (1-R 2 ) 2 = 1 - [1-2 R 2 +R 4 ] = 2R 2 -R 4 Animated Chapter 5 23
Low Level Redundancy R LOW = [1-(1-R) 2 ] 2 = [1-(1-2R+R 2 )] 2 = (2R-R 2 ) 2 Chapter 5 24
High vs Low Level Redundancy R low -R high =(2R-R 2 ) 2 -(2R 2 -R 4 ) =R 2 (2-R) 2 -R 2 (2 - R 2 ) =R 2 [4-4R+R 2-2+R 2 ] =2R 2 [R 2-2R+1]=2R 2 (R - 1) 2 0 Chapter 5 25
k-out-of-n Redundancy Let n = the number of redundant, identical and independent components each having a reliability of R. Let X = a random variable, the number of components (out of n components) operating. Then F n Pr{ X = x} = P(x) = H G I K J R x(1- R ) x n-x If k (<= n) components must operate for the system to operate: R = s N x = K P(x) Chapter 5 26
k-out-of-n Redundancy Exponential Distribution R (t) = s N N - xt x=k F H G I K J λ 1 -λt x e -e N-x z MTTF = R (t)dt = 1 N 1 0 s x=k λ x Chapter 5 27
A Very Good Example Out of the 12 identical AC generators on the C-5 aircraft, at least 9 of them must be operating in order for the aircraft to complete its mission. Failures are known to follow an exponential distribution with a mean of 100 operating hours. What is the reliability of the generator system over a 10 hour mission? Find the MTTF. Chapter 5 28
A Very Good Solution Let T i = time to failure of the i th generator 10/100 Pr{ Ti 10} = e =.9048 R 12.9048 x [ 1.9048 ] 12 -x (t) = - =.9782 x 12 s x= 9 M TTF 12 1 = 100 = 38.53 hours x x = 9 Chapter 5 29
Complex Configurations a. linked network: A B E C D Chapter 5 30
Decomposition Approach (b) Component E does not fail, R E : A C B D R (b) = [1-(1-R A )(1-R B )][1-(1-R C )(1-R D )] Chapter 5 31
Decomposition Approach (c ) Component E fails (1-R E ): A C B D R (c) = 1-(1-R A R C ) (1-R B R D ) Chapter 5 32
Decomposition Approach R S = R E R (b) + (1 - R E ) R (c ) if R A = R B =.9, R C = R D =.95, and R E =.80, then R (b) = [1-(1-.9) 2 ] [1-(1-.95) 2 ] =.99 x.9975 =.9875 R (c ) = 1 - [1-(.9) (.95)] 2 =.978975 R s =.8 (.9875) + (1-.8) (.978975) =.9858 Chapter 5 33
Enumeration Method S = success; F = failure A B C D E System Probability S S S S S S.58482 F S S S S S.06498 S F S S S S.06498 S S F S S S.03078 S S S F S S.03078 S S S S F S.146205 F F S S S F S F F S S S.00342 S S F F S F S S S F F S.007695 TOTAL.9858 Chapter 5 34
Example Automotive Braking System WC1 BP1 M C WC2 WC3 WC4 BP2 BP3 BP4 Chapter 5 35
WC1 BP1 M C WC2 WC3 BP2 BP3 Case I. BP3 fails and BP4 is operational. 1. P I = [1 - R(BP)] R(BP) 2. R f = R(M){1 - [1 - R(WC) R(BP)] 2 [1 - R(WC)]}. 3. R{mechanical} = R(C) 4. These two subsystems operate in parallel, therefore, R I =1-[1-R f ] [1 - R(C)]. Case II. BP4 fails and BP3 is operational. Then P II = P I and R II = R I (by symmetry of the network and assuming identical reliabilities) WC4 Chapter 5 36
WC1 BP1 WC2 BP2 M WC3 BP3 Case III. BP3 and BP4 have failed. C WC4 BP4 1. P III = [1 - R(BP)] 2 2. Cable system has failed. 3. R III = R(M) {1 - [1 - R(WC) R(BP)] 2 } Chapter 5 37
WC1 BP1 WC2 BP2 M WC3 C WC4 Case IV. Both BP3 and BP4 are operational. 1. P IV = [R(BP)] 2 2. R f = R(M){1 - [1 - R(WC) R(BP)]2 [1 - R(WC)]2}. 3. R{cable system} = R(C) 4. R IV = 1 - [1 - Rf] [1 - R(C)] The overall system reliability can now be found from: R S =P I R I +P II R II +P III R III +P IV R IV Chapter 5 38
A Common Everyday Reliability Block Diagram to Solve R A =.8 R D =.95 R E =.7 R B =.9 R C =.9 R F =.8 Help! Where can I find a common everyday reliability block diagram? Chapter 5 39
A Common Everyday Reliability Block Diagram solved R A =.8 R D =.95 R E =.7 R B =.9 R C =.9.81 R F =.8 1-(1-.7)(1-.8) =.94 1-(1-.8)(1-.81) =.962 R s = (.962) (.95) (.94) =.859066
Summary Series Configuration Parallel Configuration Combined Series-Parallel Configuration High / Low Level Redundancy K out-of-n Redundancy Complex Configurations linked networks Chapter 5 41