1

 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 z ( t )

 t

 t time, t
2

 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

-1 dt = [

(1+1/

)/MTTF ]
  
-1
3

 MTTF
WO
= Mean Time To Failure Without Maintenance
 
= Aging parameter
 C
PM
= Cost per preventive maintenance action
 C
CM
= Cost per corrective maintenance action
 C
EU
= Expected total unavailability cost given a component failure
 C
ES
= Expected total safety cost given a component failure
 Total cost per unit time
 C (

) = C
PM
/

+

E
(

) [ C
CM
+ C
EU
+ C
ES
]
4

 C (

) = C
PM
/

+

E
(

) [ C
CM
+ C
EU
+ C
ES
]
= C
PM
/

+ [

(1+1/

)/MTTF wo
]
  
-1 [ C
CM
+ C
EU
+ C
ES
]
 
C (

)/
 
= 0

 
MTTF
WO




C
PM
C
CM

C
EU

C
ES

(
 
1)


 1/

5

 Prepare an Excel sheet with the following input cells:


MTTF
WO
= Mean Time To Failure Without Maintenance

= Aging parameter
 C
PM
= Cost per preventive maintenance action
 C
CM
= Cost per corrective maintenance action
 C
EU
= Expected total unavailability cost given a component failure
 C
ES
= Expected total safety cost given a component failure
 Implement the formula for optimal maintenance interval
6

 Change of timing belt




MTTF
WO
= 175 000 km

= 3 (medium aging)
C
PM
= NOK 7 000
C
CM
= NOK 35 000
7

 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
 C
Collision
= NOK 25 million
 Find optimal interval
8

 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 T
1 and T
2 t = 0
  
T
1
 
T
2
     
Tid
9

 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 ( T
MC
) is a random quantity
E T
MC
)
 
0

T
( )d
 
Pr( T
 
)
 
0
 
1

F t
T
 t
 Effective failure rate
  
E
E (failure in the cycle)
E (Cycle length)


0

F
T

1

F t
T
 t
10

 Rate of PM actions: 1/ E ( T
MC
)-

E
(

)
 Cost model
 C (

) = C
PM
[1/ E ( T
MC
)-

E
(

)] +

E
(

) [ C
CM
+ C
EU
 where
+ C
ES
]
E T
MC
)
 
0

T
( )d
 
Pr( T
 
)
 
0
 
1

F t
 t
  
E 
0

F
T

1

F t
T
 t
11

 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
12

 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
T
1
T
2
Time t = 0

2

3

4

5

6

7

8

9

10

11

12

13

 Effective failure rate
E


 Where
W ( t ) is the renewal function

W

14

 Introduce
 F
X
( x ) = the cumulative distribution function of the failure times
 f
X
( x ) = the probability density function of the failure times
 From Rausand & Høyland (2004) we have:
W ( t )

F
X
( t )
 t
0

W ( t
 u ) f
X
( u ) du
 With an initial estimate W
0
( t ) of the renewal function, the following iterative scheme applies:
W i
( t )

F
X
( t )
 t
0

W i

1
( t
 u ) f
X
( u ) du
15

 For small

E

(

< 0.1MTTF
(

) = [

(1+1/

)/MTTF wo
]

WO
 
-1
) apply:
 For
 up to 0.5MTTF
WO apply (Chang et al 2008)

 

E
MTTF
  
1    
 where the

() is a correction term given by
  


1
0.1

2
MTTF
2

(0.09
 
0.2)

MTTF


 For

> 0.5MTTF
WO implement the Renewal function
16

 Numerical solution by the Excel Solver applies for all precision levels
 For small

(

< 0.1MTTF
WO
) we already know the analytical solution
 For
 up to 0.5MTTF
WO an analytical solution could not be found, but an iterative scheme is required (or “solver”)
 For

> 0.5MTTF
(i.e.,

E
(
WO only numerical methods are available

) = W (

)/

)
17

 Fix-point iteration scheme
 i

1

MTTF
WO
  

(1 1 / ) [ C
PM

C
EU

C
ES
    
C
PM
] (
 
( , , MTTF i WO
)
    i
'( , , MTTF ) i WO

 Where
 ’ () is the derivative of the correction term:


0.2

MTTF
2

0.09
 
MTTF
0.2


18

 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, 𝑡
 A common model is the power law model , where 𝑤 𝑡 = 𝜆𝛽𝑡 𝛽−1 and 𝑊 𝑡 = 𝜆𝑡 𝛽
 This corresponds to TTFF is Weibull distributed with aging parameter

19

 Cost elements

 C
F
= Cost of failure, i.e., each time we do a minimal repair
 C
R
= 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:
 𝐶 𝜏 =
𝐶
R
+ 𝑊 𝜏 𝐶
F =
𝐶
R
+𝜆𝜏 𝛽
𝐶
F =
𝐶
R + 𝜆𝜏 𝜏 𝜏 𝜏
 Taking derivative of C (

) wrt
 gives 𝛽−1 𝐶
F
 𝐶 ′ 𝜏 = −
𝐶
R 𝜏 2
+ 𝜆 𝛽 − 1 𝜏 𝛽−2 𝐶
F
= 0
1 𝛽
 ⇒ 𝜏 =
𝐶
R
λ(𝛽−1 𝐶
F
20

 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 f
D
21

 C
I
= cost of inspection
 C
R
=cost of repair/replacement upon revealing a failure during inspection
 C
H
= 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 (

)

C
I
/

+ C
R
(

-

2

/2)+ C
H

/2
 f
D
22

 Often, the safety function is implemented by means of redundant components in a k oo N voting, i.e.; we need k out of N of the components to “report” on a critical situation
 PFD for a k oo N structure is given by
 PFD ≈
𝑁
𝑁 − 𝑘 + 1 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
23

 𝑥 𝑦 = 𝑥∙ 𝑥−1 ∙...∙(𝑥−𝑦+1
1∙2∙3∙⋯∙𝑦
 For example
 In MS Excel
5 𝑥
2
=
5∙4
1∙2
= 10 𝑦 = Combin(x, y
 PFD=COMBIN(N,N-k+1)*((lambda*tau)^(N-k+1))/(N-k+2)
+ beta*lambda*tau/2
24

 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)
25

 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?
26

 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
27

 Analytical solution, one valve: 𝜏 =
2𝐶
I 𝜆(𝑓
D
𝐶
H
−𝜆𝐶
R
28