# 1 Non-observable failure progression. 2 Age based maintenance policies We consider a situation where we are not able to observe failure progression, or.

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1 Non-observable failure progression

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

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

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 ]

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 

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

7 Exercise continued – Timing belt Change of timing belt MTTF WO = 175 000 km  = 3 (medium aging) C PM = NOK 7 000 C CM = NOK 35 000

8 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 C Collision = NOK 25 million Find optimal interval

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

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 Effective failure rate

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

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

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

14 BRP – Steps in optimization Effective failure rate Where W(t) is the renewal function

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: With an initial estimate W 0 (t) of the renewal function, the following iterative scheme applies:

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) where the  () is a correction term given by For  > 0.5MTTF WO implement the Renewal function

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(  )/  )

18 BRP – Iteration scheme Fix-point iteration scheme Where  ’() is the derivative of the correction term:

19 MRP = Minimal repair strategy

20 Cost model

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

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

23 Cost model for kooN configuration

24 How to calculate kooN

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 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 2  10 -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)

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 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?

27 Exercise, continued part 2 The failure rate of the ESDV equal to 2  10 -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

28 Solution

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