Report

Non-observable failure progression 1 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 Hazard rate Result an increasing hazard rate z(t)t t time, t 2 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 -1dt = [ (1+1/)/MTTF ] -1 0 [ (1+1/)/MTTF ] t 3 Weibull standard PM model MTTFWO = Mean Time To Failure Without Maintenance = Aging parameter CPM = Cost per preventive maintenance action CCM = Cost per corrective maintenance action CEU = Expected total unavailability cost given a component failure CES = Expected total safety cost given a component failure Total cost per unit time C() = CPM / + E() [CCM + CEU + CES] 4 Optimal maintenance interval C() = CPM / + E() [CCM + CEU + CES] = CPM / + [ (1+1/)/MTTFwo] -1 [CCM + CEU + CES] C()/ = 0 1/ MTTFWO CPM (1 1/ ) CCM CEU CES ( 1) 5 Exercise Prepare an Excel sheet with the following input cells: MTTFWO = Mean Time To Failure Without Maintenance = Aging parameter CPM = Cost per preventive maintenance action CCM = Cost per corrective maintenance action CEU = Expected total unavailability cost given a component failure CES = Expected total safety cost given a component failure Implement the formula for optimal maintenance interval 6 Exercise continued – Timing belt Change of timing belt MTTFWO = 175 000 km = 3 (medium aging) CPM = NOK 7 000 CCM = NOK 35 000 7 Exercise continued Additional information Pr(Need to rent a car|Breakdown) = 0.1 Cost of renting a car = NOK 5000 Pr(Overtaking |Breakdown) = 0.005 Pr(Collision|Overtaking |Breakdown)=0.2 CCollision = NOK 25 million Find optimal interval 8 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 T1 and T2 T1 T2 Tid t=0 9 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 (TMC) is a random quantity 0 0 E (TMC ) tfT (t )dt Pr(T ) 1 FT (t ) dt Effective failure rate E ( ) E (failure in the cycle) E (Cycle length) FT ( ) 1 F (t ) dt 0 10 T ARP, cont Rate of PM actions: 1/E(TMC)-E() Cost model C() = CPM [1/E(TMC)-E()] + E() [CCM + CEU + CES] where 0 0 E (TMC ) tfT (t )dt Pr(T ) 1 FT (t ) dt E ( ) FT ( ) 1 F (t ) dt 0 T 11 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 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 T2 T1 t=0 Time 2 3 4 5 6 7 13 8 9 10 11 12 BRP – Steps in optimization Effective failure rate E ( ) Expected number of failures in [0, ) Where W(t) is the renewal function 14 W ( ) How to find the renewal function Introduce FX(x) = the cumulative distribution function of the failure times fX(x) = the probability density function of the failure times From Rausand & Høyland (2004) we have: t W (t ) FX (t ) W (t u ) f X (u )du 0 With an initial estimate W0(t) of the renewal function, the following iterative scheme applies: t Wi (t ) FX (t ) Wi 1 (t u ) f X (u )du 0 15 3 levels of precision For small ( < 0.1MTTFWO) apply: E() = [ (1+1/)/MTTFwo] -1 For up to 0.5MTTFWO apply (Chang et al 2008) (1 1/ ) 1 E ( ) ( , , MTTF) MTTF where the () is a correction term given by 0.1 2 (0.09 0.2) ( , , MTTF) 1 2 MTTF MTTF For > 0.5MTTFWO implement the Renewal function 16 BRP - Solution Numerical solution by the Excel Solver applies for all precision levels For small ( < 0.1MTTFWO) we already know the analytical solution For up to 0.5MTTFWO an analytical solution could not be found, but an iterative scheme is required (or “solver”) For > 0.5MTTFWO only numerical methods are available (i.e., E() =W()/ ) 17 BRP – Iteration scheme Fix-point iteration scheme i 1 MTTFWO CPM (1 1/ ) [CPM CEU CES ] ( 1) ( i , , MTTFWO ) i '( i , , MTTFWO ) Where ’() is the derivative of the correction term: 0.09 0.2 0.2 '( , , MTTF) 2 MTTF MTTF 18 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 = −1 and = This corresponds to TTFF is Weibull distributed with aging parameter 19 Cost model Cost elements CF = Cost of failure, i.e., each time we do a minimal repair CR = 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: = R + F = R + F = R + −1 F Taking derivative of C( ) wrt gives ′ = − ⇒= R 2 + − 1 −2 F = 0 1 R λ(−1 F 20 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 fD 21 Cost model CI = cost of inspection CR =cost of repair/replacement upon revealing a failure during inspection CH = 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() CI/ + CR(- 2/2)+ CH /2 fD 22 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 −+1 PFD ≈ −+2 −+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 How to calculate kooN ∙ = −1 ∙...∙(−+1 1∙2∙3∙⋯∙ 5∙4 5 = = 10 1∙2 2 In MS Excel = Combin(x, y PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2) + beta*lambda*tau/2 For example 24 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 10 000 If the ESDV is demanded in a critical situation, the total (accident) cost is NOK 10 000 000 Cost of repair is NOK 50 000 The rate of demands for the ESDV is one every 5 year. The failure rate of the ESDV is 210-6 (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 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 15 000 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 Exercise, continued part 2 The failure rate of the ESDV equal to 210-6 (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 40 000 NOK Find out whether it pays off to increase the overhaul interval Find the optimal strategy for functional tests and overhauls 27 Solution Analytical solution, one valve: = 28 2I (D H −R