Busra_Keles

Maintenance Policies For A Deterioration System
Subject To Non-Self-Announcing Failures
Salih Tekin
N. Onur Bakır, Bü¸sra Kele¸s
TOBB University of Economics and Technology
Ankara, Turkey
June 6, 2014
Salih Tekin N. Onur Bakır, Bü¸sra Kele¸s (TOBB University of Economics and Technology Ankara, Turkey)Maintenance Policies For A Deterioration System Subject To Non-Self-Announcing FailuresJune 6, 2014 1 / 35

System Maintenance
Maintenance is crucial to improve system availability performance with
a minimum cost, especially when

System Maintenance
I the system fails stochastically

System Maintenance
I the degree of deterioration, in addition to
failure, are known only through inspections

System Maintenance
I optional preventive maintenance is
suggested

System Maintenance
I optional preventive maintenance is
suggested
I maintenance is not perfect (except
replacement)

Maintenance Categories in Literature
Table: Maintenance Categories in Literature
Maintenance policy System structure Maintenance degree
Age replacement
Block replacement
Periodic 3
Sequential
Control limit
Single-unit 3
Multi-unit
Perfect 3
Imperfect 3
Minimal 3
Worse
Maintenance cost Optimization criteria Modelling tools
Constant 3
Random
Complex
Minimize cost rate 3
Minimize availability
Minimize downtime
Renewal theory 3
Markov chain 3
Poisson process
Planning horizon Dependence System information
Inﬁnite 3
Finite
Discrete
Continuous
Economic
Failure
Probability
State 3
Perfect 3
Imperfect

Outline
1 Introduction
2 Mathematical Formulations
3 Optimal Inspection Time and Optimal Policy
4 Numerical Example
5 References

Introduction
Problem Deﬁnition
System structure
For a stochastically failing system in which the degree of deterioration
are known only through inspections,

Introduction
Problem Deﬁnition
System structure
I At each inspection epoch k⌧, k = {0, 1, 2, ...} the system occupies
one of three states: good (1), poor (2), failed (3) and the decision maker
chooses an available action: do nothing, repair and replace.

Introduction
Problem Deﬁnition
System structure
S = {1, 2, 3}; State space

Introduction
Problem Deﬁnition
System structure
T = {0, 1, 2, 3, ...}; Time horizon

Introduction
Problem Deﬁnition
System structure
T = {0, 1, 2, 3, ...}; Time horizon
A = {do nothing, repair, replace}; Action space

Introduction
Problem Deﬁnition
System structure
T = {0, 1, 2, 3, ...}; Time horizon
A = {do nothing, repair, replace}; Action space
It is called minor repair when the repair is performed at state 2 and major
repair when it is performed at state 3.

Introduction
Problem Deﬁnition
Deterioration process
I The deterioration is modelled by three-state continuous time Markov
chain

Introduction
Problem Deﬁnition
I The deterioration is modelled by three-state continuous time Markov
chain
If {Yt, t 0} represents the system state at time t, then we assume
that {Yt, t 0} is a continuous time Markov chain with transition
probabilities,
Pij(⌧) = P{Yt+⌧ = j | Yt = i}, 8 t, ⌧ 2 R (1)
when it is currently at state i = {1, 2, 3} at time t, will be at state j = {2, 3}
at time t + ⌧.
(Markovian assumption)

Introduction
Problem Deﬁnition
Good
(1)
Poor
(2)
Failed
(3)
𝐏𝟏𝟏(𝛕) 𝐏𝟐𝟐(𝛕) 𝐏𝟑𝟑(𝛕)
𝐏𝟏𝟑(𝛕)
𝐏𝟏𝟐(𝛕) 𝐏𝟐𝟑(𝛕)
Figure: Transition Probabilities

Introduction
Problem Deﬁnition
Good
(1)
Poor
(2)
Failed
(3)
𝐏𝟏𝟏(𝛕) 𝐏𝟐𝟐(𝛕) 𝐏𝟑𝟑(𝛕)
𝐏𝟏𝟑(𝛕)
𝐏𝟏𝟐(𝛕) 𝐏𝟐𝟑(𝛕)
Figure: Transition Probabilities
I The event that causes a transition within an inspection period from state i
to j, where i 6= j, occur after an exponential amount of time with rate ij.

Introduction
Problem Deﬁnition
Assumptions
I The time horizon considered is inﬁnite and the system with good state
is put into service in time 0.

Introduction
Problem Deﬁnition
Assumptions
I Transition from state i = 1, 2 to state j = 2, 3 occurs according to a Poisson
Process with rate ij.

Introduction
Problem Deﬁnition
Assumptions
I The system state is monitored only through inspections and inspection time
is negligible.

Introduction
Problem Deﬁnition
Assumptions
I The system state is monitored only through inspections and inspection time
is negligible.
I Time to inspect is negligible; however, inspection cost is assumed to be a
monotonically non-increasing function of time interval between two
consecutive inspections [4] to prevent frequent inspections.

Introduction
Problem Deﬁnition
Assumptions
I If the system is identiﬁed as good, "do nothing" action is required;
otherwise decision maker may choose other actions.

Introduction
Problem Definition
Assumptions
I Minor (Major) repair takes the system to state 1, with probability pmn (pmj),
and to state 2 with probability qmn = 1 pmn (qmj = 1 pmj) after a fixed time
tmn (tmj) with a fixed cost cmn (cmj).

Introduction
Problem Definition
Assumptions
I Replace always returns the system to state 1 with a fixed amount cost crpl
after a fixed maintenance time trpl.

Introduction
Problem Deﬁnition
Assumptions
I The system is not in working condition during repair or replace and hence,
the inspection cost does not occur.

Introduction
Problem Deﬁnition
Assumptions
I The system is not in working condition during repair or replace and hence,
the inspection cost does not occur.
I The system state is known instantaneously after each repair.

Introduction
Problem Deﬁnition
Maintenance Policies
Five available maintenance policies are proposed shown to emphasise how
the maintenance parameters affect the optimum inspection period and policy.
Table: Policy Description
Policy State 1 State 2 State 3
⇧1 Do nothing Do nothing Replacement
⇧2 Do nothing Do nothing Major repair
⇧3 Do nothing Minor repair Major repair
⇧4 Do nothing Minor repair Replacement
⇧5 Do nothing Replacement Replacement

Introduction
Problem Deﬁnition
Optimization criteria
Optimizing the maintenance policies is based on Renewal Theory.

Introduction
Problem Deﬁnition
I A cycle ends (a renewal) after a maintenance of the failed system,
while the maintenance brings the system to "good" state.

Introduction
Problem Deﬁnition
I Then cost rate is the ratio of the expected total cost occurred during
the renewal cycle over the expected cycle length.

Introduction
Problem Deﬁnition
I Then cost rate is the ratio of the expected total cost occurred during
the renewal cycle over the expected cycle length.
I The objective is when to inspect the system periodically and what
action to be made by minimizing the expected cost rate in the long run.

Mathematical Formulations
Mathematical Calculations
Transition probabilities
Let Xij be the holding time between state i = {1, 2, 3} and state j = {2, 3} .
By Poisson arrivals assumption leads to:

Transition probabilities
Let Xij be the holding time between state i = {1, 2, 3} and state j = {2, 3} .
By Poisson arrivals assumption leads to:
P11(⌧) = P{min{X12, X13} > ⌧} = e 1⌧
(2)
P12(⌧) = P{X12  ⌧, X12 < X13, X12 + X23 > ⌧}
=
8
>><
>>:
12 e 2⌧ ⌧, 1 2 = 0
12
1 2
e 2⌧ e 1⌧ , 1 2 6= 0
(3)
P13(⌧) = P{X12  ⌧, X12 < X13, X12 + X23  ⌧}
=
8
>><
>>:
1 e 1⌧
12 e 2⌧ ⌧, 1 = 2
1 e 1⌧ 12
1 2
e 2⌧ + e 1⌧ , 1 6= 2
(4)
where the total transition rate out of a state 1 is 1 = 12 + 13 and of state 2
is 2 = 23.

Expected downtime
I How long ago could the failure have ﬁrst been noticed
by an inspection?
1
Inspection time
kτ
At the time system fails
τ' Downtime
3
Inspection time
(k+1)τ
Figure: Downtime

Expected downtime
D1(⌧) : the expected downtime within an inspection period, ⌧ , of the system
initially found at state 1

Expected downtime
D1(⌧) = DQ
(⌧) Q(⌧) + DH
(⌧) H(⌧) (5)
where
1
Inspection period, τ
32
Q(τ)
H(τ)
3
𝐏𝟑𝟑(τ)
Figure: Downtime

Expected downtime
DQ
(⌧) = ⌧ E[X13 | X13  ⌧, X13 < X12 ] =
⌧
1 e 1⌧
1
1
(6)
DH
(⌧) = ⌧ E[X12 | X12  ⌧, X12 < X13, X12 + X23  ⌧]
E[X23 | X12  ⌧, X12 < X13, X12 + X23  ⌧]
=
8
>>>><
>>>>:
⌧ 2
1
+e 1⌧
⇣
2
1
+⌧
⌘
1 e 1⌧
1e 1⌧ ⌧
, 1 2 = 0
( 1 2)
⇣
⌧ 1
1
1
2
⌘
1 2 1e 2⌧ + 2e 1⌧ +
e 2⌧
⇣
1
2
⌘
e 1⌧
⇣
2
1
⌘
1 2 1e 2⌧ + 2e 1⌧ , 1 2 6= 0
(7)
D2(⌧) = ⌧
1 e 2⌧
2
(8)

Average cost rate
Let i(⌧) be the cost rate for ⇡i, i = 0, .., 5,
The ﬁrst aim is to calculate each i(⌧) via optimizing the inspection
interval time ⌧.
The second is to determinate the optimal policy with the correspon-
ding minimum cost rate.

Average cost rate
For 8 ⇡i, i = 1, .., 5, we deﬁne
Li
j(⌧) as the expected time length the system spends to state 1 from
state j = {1, 2, 3}, and
Ci
j(⌧) as the expected maintenance cost the system spends to state
1 from state j = {1, 2, 3}.

Average cost rate
The cost rate of the system is then
i(⌧) =
Ci
1(⌧)
Li
1(⌧)
+ (⌧) (9)
where (⌧) is the inspection cost per time which is a non increasing
function of time interval between two consecutive inspections such that
lim
⌧!0
(⌧) = K < 1 ve lim
⌧!1
(⌧) = 0 (10)

Average cost rate
Cycle Length
Li
1(⌧) =
1
1 P11(⌧)
⇥
⌧ + P12(⌧)Li
2(⌧) + P13(⌧)Li
3(⌧)
⇤
(11)

Average cost rate
Cycle Length
Li
1(⌧) =
1
1 P11(⌧)
⇥
⌧ + P12(⌧)Li
2(⌧) + P13(⌧)Li
3(⌧)
⇤
(11)
where
Li
2(⌧) =
8
>>>><
>>>>:
1
1 P22(⌧)
⇥
⌧ + P23(⌧)Li
3(⌧)
⇤
i = 1, 2
tmn
pmn
i = 3, 4
trpl i = 5
(12)
Li
3(⌧) =
8
<
:
tmj + 1 pmj Li
2(⌧) i = 3, 4
trpl i = 5
(13)

Average cost rate
Cycle Cost
Ci
1(⌧) =
1
1 P11(⌧)
⇥
P12(⌧)Ci
2(⌧) + P13(⌧) cdD1(⌧) + Ci
3(⌧)
⇤
(14)

Average cost rate
Cycle Cost
Ci
1(⌧) =
1
1 P11(⌧)
⇥
P12(⌧)Ci
2(⌧) + P13(⌧) cdD1(⌧) + Ci
3(⌧)
⇤
(14)
where
Ci
2(⌧) =
8
>>><
>>>:
cdD2(⌧) + Ci
3(⌧), i = 1, 2
cmn+tmn[cd (⌧)]
pMn
, i = 3, 4
crpl + trpl [cd (⌧)], i = 5
(15)
Ci
3(⌧) =
8
>>><
>>>:
cmj + tmj [cd (⌧)]
+(1 pmj)Ci
2(⌧), i = 3, 4
crpl + trpl [cd (⌧)], i = 5
(16)

Optimal Inspection Time and Optimal Policy
Optimization Criteria
For any ⌧ 2 (0, 1), the parameters in (11-16) are all positive and ﬁnite,
then the cost rate is both positive and ﬁnite.

For any ⌧ 2 (0, 1), the parameters in (11-16) are all positive and ﬁnite,
then the cost rate is both positive and ﬁnite.
We need to examine the behaviour of the cost rate as
⌧ ! 0 (continuous inspection) and ⌧ ! 1 (never inspect)

Continuous inspection
I Continuously inspection ensures that the system state is always
known.

known.
I Failure is detected as soon as the system fails.

known.
I Failure is detected as soon as the system fails.
I The cost rate becomes
i(0) =
Ci
1(0)
Li
1(0)
+ K (17)

The cycle length leads to:
Li
1(0) =
12
1

1
12
+ Li
2(0) +
13
1

1
13
+ Li
3(0) (18)
where

The cycle length leads to:
Li
1(0) =
12
1

1
12
+ Li
2(0) +
13
1

1
13
+ Li
3(0) (18)
where
Li
2(0) =
8
>>><
>>>:
1
2
+ Li
3(0), i = 1, 2
tmn
pmn
, i = 3, 4
trpl, i = 5
(19)
Li
3(0) =
8
<
:
tmj + (1 pmj)Li
2(0), i = 3, 4
trpl, i = 5
(20)

The cycle maintenance cost leads to:
Ci
1(0) =
12
1
Ci
2(0) +
13
1
Ci
3(0) (21)
where

The cycle maintenance cost leads to:
Ci
1(0) =
12
1
Ci
2(0) +
13
1
Ci
3(0) (21)
where
Ci
2(0) =
8
>>><
>>>:
Ci
3(0), i = 1, 2
cmn+tmn(cd K)
pmn
, i = 3, 4
crpl + trpl(cd K), i = 5
(22)
Ci
3(0) =
8
<
:
cmj + tmj(cd K) + (1 pmj) Ci
2(0), i = 3, 4
crpl + trpl(cd K), i = 5
(23)

Never inspect
When ⌧ ! 1, the failure of the system occurs with probability 1 which
leads that the cycle cost consists of only downtime cost and inspection
cost.

Never inspect
cost.
The cost rate becomes

Never inspect
cost.
lim
⌧!1
i(⌧) = lim
⌧!1
hcdD1(⌧)
⌧
+ (⌧)
i
= cd + lim
t!1
(⌧)
= cd (24)

Never inspect
cost.
lim
⌧!1
i(⌧) = lim
⌧!1
hcdD1(⌧)
⌧
+ (⌧)
i
= cd + lim
t!1
(⌧)
= cd (24)
I i(⌧) is a continuous function and bounded on [0, min{ (0), i(1)}]

Corollary 1

Corollary 1
If there exists 9 ⌧ 2 [0, 1) such that i(⌧) < cd, then an optimal
inspection interval of each policy exists such that
⌧⇤
i 2 1
i ([0, min{ i(0), i(1)}])
otherwise , all policies converge each other (cd) in the long run.

Corollary 1
⌧⇤
i 2 1
i ([0, min{ i(0), i(1)}])
Corollary 2

Corollary 1
⌧⇤
i 2 1
i ([0, min{ i(0), i(1)}])
Corollary 2
If cd = 1 and other cost parameters set to zero, the average cost rate
gives the system unavailability.

Numerical Example
Numerical Example
We present several numerical examples which illustrate how system
parameters (transition rates, maintenance cots and times) could affect the
optimum inspection period and the optimum maintenance policy as well.

Numerical Example
Numerical Example
We present several numerical examples which illustrate how system
parameters (transition rates, maintenance cots and times) could affect the
optimum inspection period and the optimum maintenance policy as well.
Consider the deterioration model with inspection cost function as
(⌧) = Ke 0.1⌧
. Other input parameters are shown below of illustrated ﬁgures.

Numerical Example
Numerical Example
Transition rates
Replacement
Minor
Repair
Major
Repair
Fixed cost 750 300 500
Time to
maintenance
1 0.5 2
Probability of repair-
success
1 0.9 0.6
Inspection cost
parameter
1000
Figure: Relationships between transition rates and inspection interval

Numerical Example
Numerical Example
Replacement cost
5100 2 2 2 2 2 3 3 3 3 3
4900 2 2 2 2 2 3 3 3 3 3
4700 3 3 2 2 2 3 3 3 3 3
4500 3 3 3 3 3 3 3 3 3 3
4300 3 3 3 3 3 3 3 3 4 4
4100 3 3 3 3 3 3 3 3 4 4
3900 3 3 3 3 3 4 4 4 4 4
3700 3 3 3 3 3 4 4 4 4 4
3500 4 4 3 3 3 4 4 4 4 4
3300 4 4 4 4 4 4 4 4 5 4
3100 5 4 4 4 4 4 4 5 5 5
2900 5 4 4 4 4 5 5 5 5 5
2700 5 5 4 4 4 5 6 5 5 5
2500 6 5 5 5 5 5 6 6 6 6
2300 6 6 6 6 6 6 6 6 6 6
2100 7 7 7 7 7 𝛑 𝟎 6 7 7 7 7
1900 7 7 7 7 7 𝛑 𝟏 7 7 7 7 7
1700 8 8 8 8 8 𝛑 𝟐 8 8 8 8 8
1500 9 9 9 9 9 𝛑 𝟑 9 9 9 9 9
1300 10 11 11 11 11 𝛑 𝟒 11 11 11 11 11
1100 12 12 13 13 13 𝛑 𝟓 12 13 13 13 13
900 15 15 15 15 15 15 16 16 16 16
700 19 20 20 20 20 20 20 21 21 21
500 26 27 27 27 27 27 28 28 28 28
300 38 39 39 39 39 38 39 40 40 40
100 60 65 66 66 66 60 66 66 66 66
75 66 75 75 75 75 66 75 76 76 76
50 76 103 108 108 108 76 102 108 108 108
250 500 1000 2500 5000 250 500 1000 2500 5000
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 𝐜 𝐫𝐩𝐥 300 500
Time to
maintenance
0.001 0.001 0.001
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
1000
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 𝐜 𝐫𝐩𝐥 300 500
Time to
maintenance
1 0.5 2
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
1000
Cost of replacement, 𝐜 𝐫𝐩𝐥
Cost of
downtime, 𝐜 𝐝
Figure: Relationships between downtime cost and replacement cost

Numerical Example
Numerical Example
Inspection cost
Cost of
downtime, 𝐜 𝐝
Cost of Inspection Parameter, K
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 1000 300 500
Time to
maintenance
0.001 0.001 0.001
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
K
5100 0 1 2 13 25 0 0 3 13 26
4900 0 1 2 13 26 0 0 3 14 26
4700 0 1 2 14 27 0 0 3 15 27
4500 0 1 3 14 28 0 0 3 15 28
4300 0 1 3 15 29 0 0 3 16 29
4100 0 1 3 16 30 0 0 4 17 30
3900 0 2 3 17 31 0 0 4 17 31
3700 0 2 3 18 32 0 0 4 18 32
3500 0 2 3 19 33 0 0 4 19 34
3300 0 2 4 20 34 0 0 4 20 35
3100 0 2 4 21 36 0 0 5 22 36
2900 0 2 4 22 37 0 3 5 23 38
2700 0 2 4 24 39 0 3 5 24 39
2500 0 2 5 25 40 0 3 6 26 41
2300 0 3 6 27 42 0 3 6 28 43
2100 0 3 7 29 44 𝛑 𝟎 0 4 7 30 44
1900 0 3 7 32 46 𝛑 𝟏 0 4 7 32 46
1700 0 4 8 34 49 𝛑 𝟐 0 4 8 35 48
1500 0 4 9 37 50 𝛑 𝟑 0 5 10 37 51
1300 0 5 11 40 53 𝛑 𝟒 0 6 11 40 53
1100 0 6 13 43 56 𝛑 𝟓 0 7 13 44 56
900 2 7 15 47 59 0 8 16 48 59
700 2 10 20 52 64 0 10 21 53 64
500 3 14 27 58 69 4 15 28 60 69
300 5 24 39 67 77 6 25 40 67 77
100 20 55 66 89 98 21 55 66 89 98
75 31 65 75 97 106 32 65 76 97 106
50 72 98 108 128 135 73 99 108 126 135
100 500 1000 5000 10000 100 500 1000 5000 10000
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 1000 300 500
Time to
maintenance
1 0.5 2
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
K

Numerical Example
Numerical Example
Inspection cost
Cost of major repair, 𝐜 𝐦𝐣 Maintenance time for major repair, 𝐭 𝐦𝐣
5100 3 3 3 3 3 2 2 3 3 3
4900 3 3 3 3 3 2 2 3 3 3
4700 3 3 3 3 3 2 3 3 3 3
4500 3 3 3 3 3 3 3 3 3 3
4300 4 4 4 4 4 3 3 4 4 4
4100 4 4 4 4 4 3 3 4 4 4
3900 4 4 4 4 4 3 3 4 4 4
3700 4 4 4 4 4 3 3 4 4 4
3500 4 4 4 4 4 3 3 4 4 4
3300 5 5 5 5 5 4 4 5 5 5
3100 5 5 5 5 5 4 4 5 5 5
2900 5 5 5 5 5 4 4 5 5 5
2700 5 5 5 5 5 4 5 5 5 5
2500 6 6 6 6 6 5 5 6 6 6
2300 6 6 6 6 6 5 5 6 6 6
2100 7 7 7 7 7 6 6 7 7 7
1900 7 7 8 8 8 𝛑 𝟏 7 7 7 8 8
1700 8 8 9 9 9 𝛑 𝟐 7 7 8 9 9
1500 9 9 10 10 10 𝛑 𝟑 9 9 9 10 10
1300 11 11 11 12 12 𝛑 𝟒 10 10 11 12 12
1100 13 13 13 14 14 𝛑 𝟓 12 12 13 14 13
900 16 16 17 18 18 15 15 16 17 18
700 20 21 22 23 23 20 20 21 22 24
500 27 28 30 32 33 27 28 28 29 31
300 38 40 43 48 50 39 40 40 41 42
100 62 66 99 >1000 >1000 66 66 66 67 68
75 68 76 >1000 >1000 >1000 75 75 76 76 77
50 79 108 >1000 >1000 >1000 108 108 108 109 109
300 500 1000 1500 1750 0.001 0.1 2 5 10
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 2000 300 𝐜 𝐦𝐫
Time to
maintenance
1 0.5 2
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
1000
Repla-
cement
Minor
Repair
Major
Repair
Fixed cost 2000 300 𝟓𝟎𝟎
Time to
maintenance
1 0.5 𝐭 𝐦𝐫
Probability of
repair-success
1 0.9 0.6
Inspection cost
parameter
1000
Cost of
downtime, 𝐜 𝐝

Numerical Example
Numerical Example
Maintenance probabilities
Probability of Major
repair-success, 𝐩 𝐦𝐣
1
0,8
0,6
0,4
0,2
0,2 0,4 0,6 0,8 1 0,2 0,4 0,6 0,8 1
1
0,8
0,6
0,4
0,2
0,2 0,4 0,6 0,8 1 0,2 0,4 0,6 0,8 1
𝛑 𝟎
𝛑 𝟏
𝛑 𝟐
𝛑 𝟑
𝛑 𝟒
𝛑 𝟓
Probability of Minor
repair-success, 𝐩 𝐦𝐧
I. 𝐜 𝐫𝐩𝐥 = 𝟓𝟎𝟎 II. 𝐜 𝐫𝐩𝐥 = 𝟔𝟓𝟎
III. 𝐜 𝐫𝐩𝐥 = 𝟕𝟓𝟎 IV. 𝐜 𝐫𝐩𝐥 = 𝟏𝟎𝟎𝟎
Replace
ment
Minor
Repair
Major
Repair
Fixed
cost
𝐜 𝐫𝐩𝐥 150 500
Time to
maintenance
1 0.5 2
Probability
of repair-
success
1 𝐩 𝐦𝐧 𝐩 𝐦𝐣
Inspection cost
parameter
1000
Down Time Cost 400

Numerical Example
Numerical Example
Maintenance times
Maintenance time
for major repair, 𝐭 𝐦𝐣
Replace
ment
Minor
Repair
Major
Repair
Fixed
cost
750 150 500
Time to
maintenance
𝐭 𝐫𝐩𝐥 𝐭 𝐦𝐣 𝐭 𝐦𝐧
Probability
of repair-
success
1 0.9 0.6
Maintenance time
for minor repair,
𝐭 𝐦𝐧
Inspection cost
parameter
1000
Downtime Cost 400
𝛑 𝟎
𝛑 𝟏
𝛑 𝟐
𝛑 𝟑
𝛑 𝟒
𝛑 𝟓
I. 𝐭 𝐫𝐩𝐥 = 𝟎. 𝟎𝟏
II. 𝐭 𝐫𝐩𝐥 = 𝟎. 𝟏
III. 𝐭 𝐫𝐩𝐥 = 𝟎. 𝟓
IV. 𝐭 𝐫𝐩𝐥 = 𝟏
2,5
2
1,5
1
0,5
0,5 1 1,5 2 2,5 0,5 1 1,5 2 2,5
2,5
2
1,5
1
0,5
0,5 1 1,5 2 2,5 0,5 1 1,5 2 2,5

Numerical Example
Thank you!

References
References
Barlow, R., Hunter, L., Optimum preventive maintenance policies. Operations Research , 8 (1): 90-100, 1960
Barlow, R.E., Proschan, F., Mathematical theory of reliability. New York: Wiley 1965
Luss, H., Maintenance policies when deterioration can be observed by inspections. Operation Research, 24 (2): 359–366,
1976
Zuckerman, D., Inspection and replacement policies. Applied Probability, 17 (1): 168–177, 1980
Christer, A.H., Waller, W.M, Delay time models of industrial inspection maintenance models. Operational Research
Society, 35 (5): 401–406, 1984
Chelbi, A., Ait-Kadi, D., Replacement strategy for non self announcing failure equipment. Emerging Technologies and
Factory Automation 1, pp.423-430, 1994
Chelbi, A., Ait-Kadi, D., An optimal inspection strategy for randomly failing equipment. Reliability Engineering and System
Safety 63 (2), pp.127-131, 1999
Klutke, G.A., Yang, Y., The availability of inspected systems subject to shocks and graceful degradation. IEEE Transition
Reliability, 51 (3), pp.371-374, 2002
Wang, W., An inspection model for a process with two types of inspections and repairs. Reliability Engineering and
System Safety, 94 : 526–533, 2008.
Wang, W., An inspection model based on a three-stage failure process. Reliability Engineering and System Safety, 96
:838–848, 2011.

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