Document

```Non-observable failure
progression
1
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
Hazard rate
 Result  an increasing hazard rate
z(t)t
t time, 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

  -1dt = [ (1+1/)/MTTF ]  -1
0 [ (1+1/)/MTTF ] t
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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/

MTTFWO 
CPM



(1  1/  )   CCM  CEU  CES  (  1) 
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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
 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



T1


T2 





Tid
t=0
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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


0
0
E (TMC )   tfT (t )dt   Pr(T   )   1  FT (t )  dt
 Effective failure rate
E ( ) 
E (failure in the cycle)

E (Cycle length)

FT ( )
 1  F (t )  dt
0
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T
ARP, cont
 Rate of PM actions: 1/E(TMC)-E()
 Cost model
 C() = CPM [1/E(TMC)-E()] + E() [CCM + CEU + CES]
 where


0
0
E (TMC )   tfT (t )dt   Pr(T   )   1  FT (t )  dt
E ( ) 

FT ( )
 1  F (t )  dt
0
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
T2
T1
t=0

Time
2
3
4
5
6
7
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8
9
10
11
12
BRP – Steps in optimization
 Effective failure rate
E ( ) 
Expected number of failures in [0, )

 Where
W(t) is the renewal function
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
W ( )

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:
t
W (t )  FX (t )   W (t  u ) f X (u )du
0
 With an initial estimate W0(t) of the renewal function, the
following iterative scheme applies:
t
Wi (t )  FX (t )  Wi 1 (t  u ) f X (u )du
0
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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)

 (1  1/  )   1
E ( )  
    ( ,  , MTTF)
 MTTF 
 where the  () is a correction term given by
 0.1 2 (0.09  0.2) 
 ( ,  , MTTF)  1 


2
MTTF
MTTF


 For  > 0.5MTTFWO implement the Renewal function
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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
 i 1 
MTTFWO
CPM

(1  1/  ) [CPM  CEU  CES ]  (  1)   ( i ,  , MTTFWO )   i   '( i ,  , MTTFWO ) 
 Where ’() is the derivative of the correction term:
0.09  0.2 
 0.2
 '( , , MTTF)   

2
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   =
−1 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:
   =
R +   F

=
R +  F

=
R

+  −1 F
 Taking derivative of C( ) wrt  gives
 ′  = −
 ⇒=
R
2
+   − 1  −2 F = 0
1
R
λ(−1 F
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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
−+1

 PFD ≈
−+2
−+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 /2 to 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

∙
  =
−1 ∙...∙(−+1
1∙2∙3∙⋯∙
5∙4
5
=
= 10
1∙2
2

 In MS Excel  = Combin(x, y
 PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2)
+ beta*lambda*tau/2
 For example
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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 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)
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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 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
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Solution
 Analytical solution, one valve:  =
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2I
(D H −R
```