Non-observable failure progression

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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

time, tH

Weibull model

Hazard rate z(t) = ()(t) -1 t -1

Re-parameterization introducing MTTF and aging parameter z(t) = ()(t) -1 =[ (1+1/)/MTTF ]

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 -1dt = [ (1+1/)/MTTF ] -1

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]

Optimal maintenance interval

C() = CPM / + E() [CCM + CEU + CES]

= CPM / + [ (1+1/)/MTTFwo] -1 [CCM + CEU + CES]

C()/ = 0

CM EU ES

MTTF(1 1/ ) C ( 1)

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

Exercise continued – Timing belt

Change of timing belt MTTFWO = 175 000 km = 3 (medium aging) CPM = NOK 7 000 CCM = NOK 35 000

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

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

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

Effective failure rate

MC 0 0( ) ( )d Pr( ) 1 ( ) dT TE T tf t t T F t t

( )(failure in the cycle)( )(Cycle length) 1 ( ) d

FEE F t t

ARP, cont

Rate of PM actions: 1/E(TMC)-E() Cost model

C() = CPM [1/E(TMC)-E()] + E() [CCM + CEU + CES]

MC 0 0( ) ( )d Pr( ) 1 ( ) dT TE T tf t t T F t t

( )( )1 ( ) d

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

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

2 3 4 5 6 7 8 9 10 11 12

BRP – Steps in optimization

Effective failure rate

WhereW(t) is the renewal function

Expected number of failures in [0, ) ( )( )EW

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:

XX duufutWtFtW0

)()()()(

XiXi duufutWtFtW0

1 )()()()(

3 levels of precision

For small ( < 0.1MTTFWO) apply:E() = [ (1+1/)/MTTFwo]

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

1(1 1/ )( ) ( , , MTTF)MTTFE

0.1 (0.09 0.2)( , , MTTF) 1MTTF MTTF

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()/ )

BRP – Iteration scheme

Fix-point iteration scheme

Where ’() is the derivative of the correction term:

1PM EU ES WO WO

MTTF(1 1/ ) [ ] ( 1) ( , , MTTF ) '( , , MTTF )i

CC C C

0.2 0.09 0.2'( , ,MTTF)MTTF MTTF

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 [

A common model is the power law model, where and This corresponds to TTFF is Weibull distributed with aging

parameter

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:

Taking derivative of C( ) wrt gives

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

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

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

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

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

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)

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?

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

Solution

Analytical solution, one valve:

Non-observable failure progression

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