1 Non-observable failure progression
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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, t
Haz
ard
rate
t
z(t) t
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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 -1dt = [ (1+1/)/MTTF ] -1
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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]
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Optimal maintenance interval
C() = CPM / + E() [CCM + CEU + CES]
= CPM / + [ (1+1/)/MTTFwo] -1 [CCM + CEU + CES]
C()/ = 0
1/
WO PM
CM EU ES
MTTF(1 1/ ) C ( 1)
CC C
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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
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Exercise continued – Timing belt
Change of timing belt MTTFWO = 175 000 km = 3 (medium aging) CPM = NOK 7 000 CCM = NOK 35 000
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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
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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
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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
0
( )(failure in the cycle)( )(Cycle length) 1 ( ) d
TE
T
FEE F t t
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ARP, cont
Rate of PM actions: 1/E(TMC)-E() Cost model
C() = CPM [1/E(TMC)-E()] + E() [CCM + CEU + CES]
where
MC 0 0( ) ( )d Pr( ) 1 ( ) dT TE T tf t t T F t t
0
( )( )1 ( ) d
TE
T
F
F t t
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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
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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
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BRP – Steps in optimization
Effective failure rate
WhereW(t) is the renewal function
Expected number of failures in [0, ) ( )( )EW
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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:
t
XX duufutWtFtW0
)()()()(
t
XiXi duufutWtFtW0
1 )()()()(
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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
1(1 1/ )( ) ( , , MTTF)MTTFE
2
2
0.1 (0.09 0.2)( , ,MTTF) 1MTTF MTTF
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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()/ )
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BRP – Iteration scheme
Fix-point iteration scheme
Where ’() is the derivative of the correction term:
WO PM
1PM EU ES WO WO
MTTF(1 1/ ) [ ] ( 1) ( , ,MTTF ) '( , , MTTF )i
i i i
CC C C
2
0.2 0.09 0.2'( , ,MTTF)MTTF MTTF
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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
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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 1 0
⇒⁄
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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
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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
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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 /2to the expression for PFD to account for common cause failures, is the fraction of failures that are common to all components
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How to calculate kooN
∙ ∙...∙
∙ ∙ ∙⋯∙
For example 52∙∙
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In MS Excel Combin x, y
PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2) + beta*lambda*tau/2
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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)
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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?
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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
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Solution
Analytical solution, one valve: