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Non-observable failure progression. 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

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Non-observable failure progression


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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() =  -10 [(1+1/)/MTTF]t -1dt = [(1+1/)/MTTF] -1

    4. 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]

    5. Optimal maintenance interval • C() = CPM / + E() [CCM+ CEU + CES] = CPM / + [(1+1/)/MTTFwo] -1 [CCM+ CEU + CES] •  C()/   = 0

    6. 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

    7. Exercise continued – Timing belt • Change of timing belt • MTTFWO = 175 000 km •  = 3 (medium aging) • CPM = NOK 7 000 • CCM = 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 • CCollision = 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 T1 and T2

    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 (TMC) is a random quantity • Effectivefailure rate

    11. ARP, cont • Rate of PM actions: 1/E(TMC)-E() • Cost model • C() = CPM [1/E(TMC)-E()] + E() [CCM+ CEU + CES] • 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 • WhereW(t) is the renewal function

    15. 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: • With an initial estimate W0(t) of the renewal function, the following iterative scheme applies:

    16. 3 levels of precision • For small (< 0.1MTTFWO)apply: E() = [(1+1/)/MTTFwo] -1 • For up to 0.5MTTFWO apply (Chang et al 2008) • where the  () is a correction term given by • For > 0.5MTTFWO implement the Renewal function

    17. 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()/ )

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

    19. 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

    20. 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

    21. Cost model for kooNconfiguration • Often, the safety function is implemented by means of redundant components in a kooNvoting, i.e.; we need kout of N of the components to “report” on a critical situation • PFD for a kooNstructure is given by • We may replace the /2expression 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

    22. How to calculate kooN • For example • In MS Excel • PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2) + beta*lambda*tau/2

    23. 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)

    24. 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?

    25. 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

    26. Solution • Analytical solution, one valve: