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 -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] = PMCost / + [ (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 = km
 = 3 (medium aging)
CPM = NOK 7 000
CCM = NOK

8. Exercise continued Additional information Find optimal interval
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
Effective failure rate

11. ARP, cont Rate of PM actions: 1/E(TMC)-E() Cost model where
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 (PRP) 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
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 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 ≈ 𝑛 𝑛−𝑘 𝜆𝜏 𝑛−𝑘+1 𝑛−𝑘+2
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

22. How to calculate KooN 𝑥 𝑦 = 𝑥∙ 𝑥−1 ∙...∙(𝑥−𝑦+1) 1∙2∙3∙…∙𝑦
𝑥 𝑦 = 𝑥∙ 𝑥−1 ∙...∙(𝑥−𝑦+1) 1∙2∙3∙…∙𝑦
For example = 5∙4 1∙2 =10
In MS Excel 𝑥 𝑦 =Combin(x,y)
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
If the ESDV is demanded in a critical situation, the total (accident) cost is NOK
Cost of repair is NOK
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
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 NOK
Find out whether it pays off to increase the overhaul interval
Find the optimal strategy for functional tests and overhauls