Non-observable failure progression

Transcription

1 Non-observable failure progression 1

2 Age based maintenance policies We consider a situation where we are not able to observe failure progression, or where it is impractical to observe failure progression: Examples Wear of a light bulb filament Wear of balls in a ball-bearing Result an increasing hazard rate Hazard rate t z(t) t time, t 2

3 Weibull model Hazard rate z(t) = ()(t) -1 t -1 Re-parameterization introducing MTTF and aging parameter z(t) = ()(t) -1 =[ (1+1/)/MTTF ] t -1 Effective failure rate, E (), is the expected number of failures per unit time for a unit put into a god as new state each time units Assuming that only one failure could occur in [0, >, the average failure rate is E () = -1 0 [ (1+1/)/MTTF ] t -1 dt = [ (1+1/)/MTTF ] -1 3

4 Weibull standard PM model MTTF WO = Mean Time To Failure Without Maintenance = Aging parameter C PM = Cost per preventive maintenance action C CM = Cost per corrective maintenance action C EU = Expected total unavailability cost given a component failure C ES = Expected total safety cost given a component failure Total cost per unit time C() = C PM / + E () [C CM + C EU + C ES ] 4

5 Optimal maintenance interval C() = C PM / + E () [C CM + C EU + C ES ] = C PM / + [ (1+1/)/MTTF wo ] -1 [C CM + C EU + C ES ] C()/ = 0 MTTF C (11 / ) CCM CEU C ES ( 1) WO PM 1/ 5

6 Exercise Prepare an Excel sheet with the following input cells: MTTF WO = Mean Time To Failure Without Maintenance = Aging parameter C PM = Cost per preventive maintenance action C CM = Cost per corrective maintenance action C EU = Expected total unavailability cost given a component failure C ES = Expected total safety cost given a component failure Implement the formula for optimal maintenance interval 6

7 Exercise continued Timing belt Change of timing belt MTTF WO = km = 3 (medium aging) C PM = NOK C CM = NOK

8 Exercise continued Additional information Pr(Need to rent a car Breakdown) = 0.1 Cost of renting a car = NOK 5000 Pr(Overtaking Breakdown) = Pr(Collision Overtaking Breakdown)=0.2 C Collision = NOK 25 million Find optimal interval 8

9 Age replacement policy- ARP The age replacement policy (model) is one of the classical optimization models: The component is replaced periodically when it reaches a fixed age If the component fails within a maintenance interval, the component is replaced, and the maintenance clock is reset Usually replace the component after a service time of In some situations the component fails in the maintenance interval, indicated by the failure times T 1 and T 2 9

10 ARP, steps in optimization Assume all components are as good as new after a repair or a replacement Usually we assume Weibull distributed failure times Repair time could be ignored with respect to length of a maintenance cycle The length of a maintenance cycle (T MC ) is a random quantity E( T ) tf ( t)dt Pr( T ) 1 F ( t) dt MC Effective failure rate ( ) E 0 T 0 E(failure in the cycle) FT ( ) E(Cycle length) 1 F ( t) dt 0 10 T T

11 ARP, cont Rate of PM actions: 1/E(T MC )- E () Cost model C() = C PM [1/E(T MC )- E ()] + E () [C CM + C EU + C ES ] where ET ( ) tf ()d t t Pr( T ) 1 F () t dt MC ( ) E 0 0 T 0 F T ( ) 1 F ( t) dt T T 11

12 Exercise Use the ARP.xls file to solve the timing belt problem with the ARP Compare the expression for the effective failure rate with the standard Weibull model 12

13 Block replacement policy - BRP The block replacement policy (BRP) is similar to the ARP, but we do not reset the maintenance clock if a failure occurs in a maintenance period The BRP seems to be wasting some valuable component life time, since the component is replaced at an age lower than if a failure occurs in a maintenance period This could be defended due to administrative savings, or reduction of set-up cost if many components are maintained simultaneously Note that we have assumed that the component was replaced upon failure within one maintenance interval In some situations a minimal repair, or an imperfect repair is carried out for such failures 13

14 BRP Steps in optimization Effective failure rate ( ) E Expected number of failures in [0, ) W ( ) Where W(t) is the renewal function 14

15 How to find the renewal function Introduce F X (x) = the cumulative distribution function of the failure times f X (x) = the probability density function of the failure times From Rausand & Høyland (2004) we have: W ( t) F X t ( t) W ( t u) f 0 X ( u) du With an initial estimate W 0 (t) of the renewal function, the following iterative scheme applies: W ( t) i F X t ( t) W ( t u) f 0 i1 X ( u) du 15

16 3 levels of precision For small ( < 0.1MTTF WO ) apply: E () = [ (1+1/)/MTTF wo ] -1 For up to 0.5MTTF WO apply (Chang et al 2008) (11/ ) 1 E ( ) (,,MTTF) MTTF where the () is a correction term given by ( ) (,,MTTF) 1 MTTF 2 MTTF For > 0.5MTTF WO implement the Renewal function 16

17 BRP - Solution Numerical solution by the Excel Solver applies for all precision levels For small ( < 0.1MTTF WO ) we already know the analytical solution For up to 0.5MTTF WO an analytical solution could not be found, but an iterative scheme is required (or solver ) For > 0.5MTTF WO only numerical methods are available (i.e., E () =W()/ ) 17

18 BRP Iteration scheme Fix-point iteration scheme MTTF C (1 1/ ) [ ] ( 1) (,,MTTF ) '(,,MTTF ) WO PM i1 CPM CEU CES i WO i i WO Where () is the derivative of the correction term: '(,,MTTF) 2 MTTF MTTF 18

19 MRP = Minimal repair strategy Assume that time to first failure (TTFF) is Weibull distributed Upon a failure, the item is repaired to a bad as old level I.e., the repair (corrective task) will not influence the failure rate In such a model the failure rate is denoted ROCOF (Rate Of Occurrence Of Failures) ROCOF= w(t) = rate of failure for a system with (global) age t W(t) is the expected number of failures in a time interval [0, A common model is the power law model, where and This corresponds to TTFF is Weibull distributed with aging parameter 19

20 Cost model Cost elements C F = Cost of failure, i.e., each time we do a minimal repair C R = Cost of renewal (when our car is old, and we buy a new car) = Renewal interval (i.e., interval between buying a new car) Expected cost per unit time: Taking derivative of C( ) wrt gives

21 Shock model Consider a component that fails due to external shocks Thus, the failure times are assumed to be exponentially distributed with failure rate Further assume that the function is hidden With one component the probability of failure on demand, PFD is given by PFD = /2 The function is demanded by a demand rate f D 21

22 Cost model C I = cost of inspection C R =cost of repair/replacement upon revealing a failure during inspection C H = cost of hazard, i.e. if the hidden function is demanded, and, the component is in a fault state Average cost per unit time: C() C I / + C R (- 2 /2)+ C H /2 f D 22

23 Cost model for koon configuration Often, the safety function is implemented by means of redundant components in a koon voting, i.e.; we need k out of N of the components to report on a critical situation PFD for a koon structure is given by PFD 1 We may replace the /2 expression with this expression for PFD in the previous formula for the total cost In case of common cause failures, we add /2 to the expression for PFD to account for common cause failures, is the fraction of failures that are common to all components 23

24 How to calculate koon... For example 5 2 In MS Excel 10 Combinx,y PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2) + beta*lambda*tau/2 24

25 Exercise We are considering the maintenance of an emergency shutdown valve (ESDV) The ESDV has a hidden function, and it is considered appropriate to perform a functional test of the valve at regular intervals of length The cost of performing such a test is NOK If the ESDV is demanded in a critical situation, the total (accident) cost is NOK Cost of repair is NOK The rate of demands for the ESDV is one every 5 year. The failure rate of the ESDV is (hrs -1 ) Determine the optimum value of by Finding an analytical solution Plotting the total cost as a function of Minimising the cost function by means of numerical methods (Solver) 25

26 Exercise, continued In order to reduce testing it is proposed to install a redundant ESDV The extra yearly cost of such an ESDV is NOK Determine the optimum test interval if we assume that the second ESDV has the same failure rate, but that there is a common cause failure situation, with = 0.1 Will you recommend the installation of this redundant ESDV? 26

27 Exercise, continued part 2 The failure rate of the ESDV equal to (hrs -1 ) is the effective failure rate if the component is periodically overhauled every 3 years The aging parameter of the valve is = 3 The cost of an overhaul is NOK Find out whether it pays off to increase the overhaul interval Find the optimal strategy for functional tests and overhauls 27

28 Solution Analytical solution, one valve: 28