Chapter 9 Part II Maintainability

Transcription

1 Chapter 9 Part II Maintainability 9.4 System Repair Time 9.5 Reliability Under Preventive Maintenance 9.6 State-Dependent Systems with Repair C. Ebeling, Intro to Reliability & Maintainability Chapter 9 Engineering, nd ed. Waveland Press, Inc. Copyright 1 010

2 9.4 System Repair Time MTTR i = the mean time to repair the ith unique subsystem, f i = the expected number of failures of the ith unique subsystem over the system design life, q i = the number of identical subsystems of type i, MTTR s = n i=1 qf i n i=1 i q i f MTTR i i f i = R S T toi MTT zt oi 0 i ρ( tdt ) for renewal process for minimal repair Chapter 9

3 Redundant System Repair Time A out of 3 redundant system with each component having a constant repair rate equal to 1/MTTR. Restore when one is repaired Repair one at a time MTTR Repair simultaneously MTTR/ Restore when both are MTTR 1.5 MTTR repaired system mean repair time Chapter 9 3

4 System MTTR Derivation Under Simultaneous Repair Case: Restore when one is repaired and repair simultaneously Assume: r = 1/MTTR and independent repair times T 1 = the time to repair component 1 and T = the time to repair component T = a random variable, the time of the first repair; T = minimum {T 1, T } Pr{T t } = Pr{ T 1 t or T t} = Pr{T 1 t} + Pr{ T t} - Pr{T 1 t} Pr{ T t} Under CR, then Pr{T i t } = 1 Exp(-rt) and Pr{T t }= [1 Exp(-rt)] - [1 Exp(-rt)] = exp(-rt) [ 1 Exp(-rt) + Exp(-rt) ] = 1 Exp(-rt) which is exponential with rate r and MTTR s = 1/(r) = MTTR / Chapter 9 4

5 The Rate Diagram λ λ ailure State State 1 Both Operating State one Operating State 3 None Operating r = 1/MTTR r = 1/MTTR one-at-a-time r = 1/MTTR r = /MTTR simultaneously Chapter 9 5

6 9. 5 Reliability under Preventive Maintenance R(t) = system reliability without maintenance T = interval of time between preventive maintenance R m (t) = reliability of the system with preventive maintenance R (t) = R(t) for 0 t < T m n R m(t) = R(T ) R(t - nt) for nt t < (n+ 1)T Prob of surviving n PM intervals of length T Repair to as good as new Chapter 9 6

7 Reliability under Preventive Maintenance (n+1)t = z z 0 m nt n=0 MTT R (t) dt = R (t) dt = z n R(T ) R(t - nt)dt nt n=0 (n+1)t = R(T ) z n n=0 (n+1)t nt R(t - nt)dt = z n R(T ) R(t )d t where t = t - nt 0 n=0 n=0 n R(T ) T Therefore MTT = z T 0 R(t)dt 1-R(T) is an infinite geometric series having as its sum m 1 1-R(T) Chapter 9 7

8 CR Model m -λt R(t) = e R (t) = ( e -λt n - λ(t-nt) ) e = e -λnt e -λt e λnt -λt = e = R(t) Another example of the memoryless property of the Exponential Distribution. Chapter 9 8

9 Weibull Example β -n T - R t-nt m(t) = e HG I θ K J e H G I θ K J, nt t (n+1)t numerical example -n 0 I - R t-0n m(t) = e HG 100 K J e H G I 100 K J, 0n t 0(n+1) β To find R m (90), observe that n = 4. Then m HG I K J H G I 100 K J R (90) = e e =. Chapter 9 9

10 Weibull Example ility b0.6 lia e R0.4 PM CUM PM NO PM Time Chapter 9 10

11 Weibull Example -n 0 e HG I K J 0 ind the.90 design life: n = (- ln.9 0) HG I K J = I - R t- 0 m(t) = e HG 100 K J 4 e H G I 100 K J, 40 t < 60 HG - t- 4 0 =.931 e 100 =.90 I K J L NM HG I 1 K JO QP =.90 t = 100 -ln days.931 Chapter 9 11

12 Maintenance-induced ailures m n n R (t) = R(T ) (1- p ) R(t -nt), nt t <(n+1)t Chapter 9 1

13 Maintenance-induced ailures - lognormal example L NM n R(T ) = 1-Φ HG 1 s ln t T MED IO K J n QP R(t - nt) = 1-Φ HG 1 s ln t-nt t MED I K J With t med = 5,000 hr and s = 1.0: R(5000) = 1- HG Φ ln I K J = 1-.5 =.50 Assume p =.005 and T = 500 hr. L NM R m(5000) = 1- HG Φ ln I K J O QP ( ) =.854 Chapter 9 13

14 DR and PM Weibull with beta = 0.5 and theta = 100 days ility 0.5 b lia e 0.4 R TIME PM CUM PM NO PM Chapter 9 14

15 9.6 State Dependent Systems with Repair dp1 () t dt dp () t dt dp3 () t dt λ λ 3 r = λp() t + rp () t = λp() t ( r+ λ) P () t = λp () t Chapter 9 15

16 State Dependent Systems with Repair - solution where P 1(t) = λ +r+x x x 1 e xt λ +r+x x x e xt λ λ x e xt P (t) = e x 1 x x1 P 3(t) = 1 x x-x e x t + xt x1 x-x e 1 x1, x = ( 3λ + r) ± λ + 6λr+ r x t Chapter 9 16

17 z 0 MTT = HG State Dependent Systems with Repair - solution x1 Rt P t x-x e xt () = 1 3() = x1 x-x e x x-x e x t 1 x x-x e x x x x r x t KJ dt = QP x-x x x = + λ + = 1 xx λ xt I L NM O MTT = HG MTT MTTR c c I KJ MTT c Chapter 9 17

18 State Dependent Systems with Repair - example A computer system consists of two active parallel processors each having a constant failure rate of.5 failures per day. Repair of a failed processor requires an average of one half a day (exponential distribution). MTT = [3(.5) + ] / [() (.5)] = 7 days R() 1 = e e = where x 1 = and x = without repair, MTT = 3 days and R(1) =.845 Chapter 9 18

19 Standby System with Repair d P 1(t) dt d P (t) dt d P 3(t) dt = -( λ1+ λ - )P 1(t)+ r P (t) = λ1 P 1(t)-( λ+ r) P (t) - = λ P 1(t)- λ1 P 3(t) Chapter 9 19

20 Standby System with Repair No failures in standby d P 1(t) dt d P (t) dt = - P (t)+ r P (t) λ1 = λ1 P 1(t)-( λ+r)p (t) Chapter 9 0

21 Standby System with Repair No failures in standby - solution P 1(t) = λ +r+x x1- x 1 λ λ e + +r+ x x- x1 x t e x t λ + x - x e 1 x t 1 P (t) = x 1- x e 1 x t Rt () = P() t + P() t = xt ( k1+ x1) e ( k1+ x) e x x xt Chapter 9 x 1, x = - k + _ k1-4k k 1 = λ 1+ λ +r k = λ λ 1

22 Standby System with Repair No failures in standby-example An on-board computer system has, through the use of built-in-test equipment (BITE), the capability of being restored when a failure occurs. A standby computer is available for use whenever the primary fails. λ 1 =.0005, r =.1, and λ =.00 failures per hour. K 1 = = K = (.0005)(.00) = 10 _ -6 x 1,x = (.105 ) -4x 10 6 = x10, Chapter 9

23 Standby System with Repair No failures in standby-example R(t) = x10 t t e -(9.757x10 )e R(1000) = R(000) = R(3000) =.9715 R(4000) =.9618 R(5000) =.9548 Chapter 9 3

24 The End I fix things in accordance with the lognormal probability distribution Yes, but your MTTR is unacceptable! Chapter 9 4