Chapter 5 Reliability of Systems

Transcription

1 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

2 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

3 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

4 Component Count vs. System Reliability Component Number of Components Reliability x x x x System Reliability Chapter 5 4

5 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

6 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

7 Components in Series - Example A communications satellite consists of the following components: Probability Shape Characteristic Component Distribution Parameter life Power unit Weibull ,800 hr. Receiver Weibull ,000 hr. Transmitter Weibull ,000 hr. Antennae Exponential MTTF = 100,000 hr. Chapter 5 7

8 λ S () t Components in Series - Example t 14. t 18. t , 43800, 75000, 75000, 68000, 68000, , = 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 t t t t 43, , , , 000 = F H G I K J F H G I K J F H G I K J 17, , , 520 F H G I K J F H G I K J F H G I K J 17, 520 R (, ) 43, , , , 000 s = e e e e =. 62 Chapter 5 8

9 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 landings. What is the reliability of the landing gear system if the system is to be overhauled after 1000 landings? Chapter 5 9

10 Class Exercise - solution , 15, R( 1000) e e (. 975) (. 916). 893 = F H G I K J F H G I K J = = Animated Chapter 5 10

11 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

12 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

13 Parallel Configuration - CFR Model R s n (t) = it e λ i= 1 Chapter 5 13

14 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 z z z z -λ1t -λ 2t -( λ MTTF = R (t)dt = e dt + e dt - 1+ λ 2 )t e dt 0 s = λ 1 λ + 2 λ1 λ2 Chapter 5 14

15 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

16 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 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

17 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() e P = 1 ( ) = Question: how are two redundant circuit breakers physically configured? Chapter 5 17

18 Combined Series - Parallel Systems R 1 R 2 R 3 R 6 R 4 R 5 Chapter 5 18

19 Combined Series - Parallel Systems A B R 1 R 2 R 3 R 6 C R 4 R 5 Chapter 5 19

20 Combined Series - Parallel Systems A R 1 R 2 R A = [ 1 - (1 - R 1 ) (1 - R 2 )] Chapter 5 20

21 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

22 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

23 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

24 Low Level Redundancy R LOW = [1-(1-R) 2 ] 2 = [1-(1-2R+R 2 )] 2 = (2R-R 2 ) 2 Chapter 5 24

25 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

26 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

27 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

28 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

29 A Very Good Solution Let T i = time to failure of the i th generator 10/100 Pr{ Ti 10} = e =.9048 R x [ ] 12 -x (t) = - =.9782 x 12 s x= 9 M TTF 12 1 = 100 = hours x x = 9 Chapter 5 29

30 Complex Configurations a. linked network: A B E C D Chapter 5 30

31 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

32 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

33 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 = R s =.8 (.9875) + (1-.8) ( ) =.9858 Chapter 5 33

34 Enumeration Method S = success; F = failure A B C D E System Probability S S S S S S F S S S S S S F S S S S S S F S S S S S S F S S S S S S F S F F S S S F S F F S S S S S F F S F S S S F F S TOTAL.9858 Chapter 5 34

35 Example Automotive Braking System WC1 BP1 M C WC2 WC3 WC4 BP2 BP3 BP4 Chapter 5 35

36 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

37 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

38 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

39 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

40 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) =

41 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