availability and maintainability...• the laplace transform of the restoration density is g (t)...
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11Piero Baraldi
Availability and Maintainability
22Piero Baraldi
Introduction: reliability and availability
• Non maintained systems: theycannot be repaired after a failure(a telecommunication satellite, aF1 engine, a vessel of a nuclearpower plant).
• Maintained systems: they can berepaired after the failure (pump ofan energy production plant, acomponent of the reactoremergency cooling systems)
System types
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Introduction: reliability and availability
• Non maintained systems: theycannot be repaired after a failure(a telecommunication satellite, aF1 engine, a vessel of a nuclearpower plant).
• Maintained systems: they can berepaired after the failure (pump ofan energy production plant, acomponent of the reactoremergency cooling systems)
Reliability (𝑹𝑹(𝑻𝑻𝑴𝑴))It quantifies the system capabilityof satisfying a specified missionwithin an assigned period of time(𝑇𝑇𝑀𝑀): 𝑅𝑅 𝑇𝑇𝑀𝑀 = 𝑃𝑃 𝑇𝑇 > 𝑇𝑇𝑀𝑀
Availability (𝑨𝑨(𝒕𝒕))It quantifies the system ability to fulfill the assigned mission at any specific moment of the lifetime
Performance parametersSystem types
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Definition of Availability
• X(t) = indicator variable such that:• X(t) = 1, system is operating at time t
• X(t) = 0, system is failed at time t
• Instantaneous availability p(t) and unavailability q(t)
X(t)
1
0
F F F
R R R t
F = FailedR = under Repair
[ ] [ ]( ) ( ) 1 ( )p t P X t E X t= = = [ ]( ) ( ) 0 1 ( )q t P X t p t= = = −
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Contributions to Unavailability
• RepairA component is unavailable because under repair after a failure
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Contributions to Unavailability
• RepairA component is unavailable because under repair
• Testing / preventive maintenanceA component is removed from the system because:
a. it must undergo preventive maintenance
TRANSMISSION BELT
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Contributions to Unavailability
• RepairA component is unavailable because under repair
• Testing / preventive maintenanceA component is removed from the system because:
a. it must undergo preventive maintenance
b. it has to be tested (safety system, standby components)
EMERGENCY CORE COOLING SYSTEM
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Contributions to Unavailability
• RepairA component is unavailable because under repair
• Testing / preventive maintenanceA component is removed from the system because:
a. it must undergo preventive maintenance
b. it has to be tested (safety system, standby components)
• Unrevealed failureA stand-by component fails unnoticed. The system goes on without noticing the component failure until a test on the component is made or the component is demanded to function
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Average availability descriptors
• How to compare different maintenance strategies?
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Average availability descriptors
• How to compare different maintenance strategies?
• We need to define quantities for an average description of the system probabilistic behavior:• If after some initial transient effects, the instantaneous availability assumes a time
independent value → Limiting or steady state availability:
)(lim tppt ∞→
=
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Average availability descriptors
• How to compare different maintenance strategies?
• We need to define quantities for an average description of the system probabilistic behavior:• If after some initial transient effects, the instantaneous availability assumes a time
independent value → Limiting or steady state availability:
• If the limit does not exist → Average availability over a period of time Tp:
∫
∫
===
===
p
p
p
p
T
ppT
T
ppT
TDOWNTIMEdttq
Tq
TUPTIMEdttp
Tp
0
0
...)(1
...)(1
)(lim tppt ∞→
=
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Average Availability12
p(t)
p(t)
p(t)
t
t
1
1
1
t
𝑇𝑇𝑝𝑝
𝑇𝑇𝑝𝑝
𝑇𝑇𝑝𝑝
∫pT
dttp0
)(
p
p
Tp
T
pT
dttp=∫0 )(
∫pT
dttp0
)(
UPTIME DOWNTIME
2𝑇𝑇𝑝𝑝
∫pT
dttp0
)(
∫pT
dttp0
)(
∫pT
dttp0
)(
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Maintenability
• How fast a system can be repaired after failure?
• Repair time TR depends from many factors such as:
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Maintenability
• How fast a system can be repaired after failure?
• Repair time TR depends from many factors
• Repair time TR varies statistically from one failure to another, depending on the conditions associated to the particular maintenance events
• Let TR denotes the downtime random variable, distributed according to the pdf 𝑔𝑔𝑇𝑇𝑅𝑅(𝑡𝑡)
• Maintenability:
14
( ) ( )∫=≤t
TR dgtTPR0
ττ
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Failure classification
• Failure can be:
• revealed
• unrevealed (stanby components, safety systems)
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REVEALED FAILURE
• Hypotheses:• Constant failure rate 𝜆𝜆
• Restoration starts immediately after the component failure
• Probability density function of the random time duration TR of the repair process = 𝑔𝑔𝑇𝑇𝑅𝑅(𝑡𝑡)
• Objective:• Computation of the availability p(t) of a component
continuously monitored
• Method• Conceptual Experiment with𝑁𝑁0 identical components that start
at time t = 0
Balance equation for the number N(t) of components working between time t and time t+∆t
Availability of a continuously monitored component (1)
working at t+Δt -
Failurebetween tand t+ Δt
working at t
∫ ∆⋅−⋅∆⋅⋅⋅+∆⋅⋅⋅−⋅=∆+⋅t
ttgpNttpNtpNttpN0
)()()()()( ττλτλ
Repair between t and t+ Δt
+=
(1) (2) (3) (4)
1) Number of items UP at time t+∆t
2) Number of items UP at time t
3) Number of items failing during the interval ∆t
4) Number of items that had failed in (τ, τ+∆τ) and whose restoration terminates in (t, t+∆t);
Availability of a continuously monitored component (3)
0 0 0 0
• The integral-differential form of the balance
• The solution can be obtained introducing the Laplace transforms
∫ ⋅−⋅⋅+⋅−=t
dtgptpdt
tdp
0
)()()()( τττλλ 1)0( =p
Availability of a continuously monitored component (3)
𝑝𝑝 𝑡𝑡
𝑑𝑑 𝑝𝑝 𝑡𝑡𝑑𝑑𝑡𝑡
𝑝𝑝 𝑡𝑡 ∗ 𝑔𝑔 𝑡𝑡 = �0
𝑡𝑡𝑝𝑝 𝜏𝜏 𝑔𝑔 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏
𝐿𝐿 𝑝𝑝 𝑡𝑡 = �0
+∞𝑒𝑒−𝑠𝑠𝑡𝑡𝑝𝑝 𝑡𝑡 𝑑𝑑𝑡𝑡 = �𝑝𝑝 (𝑠𝑠)
𝐿𝐿𝑑𝑑 𝑝𝑝 𝑡𝑡𝑑𝑑𝑡𝑡
= 𝑠𝑠 � �𝑝𝑝 𝑠𝑠 − 𝑝𝑝 0
𝐿𝐿 𝑝𝑝 𝑡𝑡 ∗ 𝑔𝑔 𝑡𝑡 = �𝑝𝑝 (𝑠𝑠) � �𝑔𝑔 (𝑠𝑠)
Time domain Laplace Transform
• Applying the Laplace transform we obtain:
which yields:
• Inverse Laplace transform → p(t)
• Limiting availability:
)(~)(~)(~1)(~ sgspspsps ⋅⋅+⋅−=−⋅ λλ
( ))(~11)(~
sgssp
−⋅+=
λ
[ ] ( )
−⋅+
=⋅==→→∞→∞ )(~1
lim)(~lim)(lim00 sgs
sspstppsst λ
Availability of a continuously monitored component (4)
• As s → 0, the first order approximation of �𝑔𝑔(𝑠𝑠) is:
Rs sdgsdgsdgesg ττττττττττ ⋅−=⋅−≅+⋅−== ∫∫∫
∞∞∞− 1)(1)()...1()()(~
000
[ ]R g RE Tτ =
cycle""pair failure/rea of timeaverageUP iscomponent the timeaverage
=+
=τ+λλ
=τ⋅λ+
=τ⋅⋅λ+
=→∞ MTTRMTTF
MTTF1
11
1ss
slimpRRR
0s
General result!
Availability of a continuously monitored component (5)
MTTR
MTTR=
Exercise 1
• Find instantaneous and limiting availability for an exponential component whose restoration probability density is
tetg ⋅−⋅= µµ)(
Exercise 1: Solution
• Find instantaneous and limiting availability for an exponential component whose restoration probability density is
• The Laplace transform of the restoration density is
tetg ⋅−⋅= µµ)(
[ ]µ
µ+
==s
tgLsg )()(~
)(1)(~
λµµ
µλ ++⋅
+=
+⋅+
=sss
sss
sp 𝑝𝑝 𝑡𝑡 = 𝐿𝐿−1 �𝑝𝑝(𝑠𝑠) = 𝐿𝐿−1𝑠𝑠 + 𝜇𝜇
𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
Exercise 1: Solution
• Find instantaneous and limiting availability for an exponential component whose restoration probability density is
tetg ⋅−⋅= µµ)(
�𝑝𝑝 𝑠𝑠 =𝐴𝐴𝑠𝑠
+𝐵𝐵
(𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆)=𝐿𝐿−1
𝑠𝑠 + 𝜇𝜇
𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
𝐿𝐿 1 =1𝑠𝑠
𝐿𝐿1
𝑠𝑠 + 𝑎𝑎= 𝑒𝑒−𝑎𝑎𝑡𝑡
𝑝𝑝 𝑡𝑡 = 𝐴𝐴𝐿𝐿−11𝑠𝑠
+ 𝐵𝐵𝐿𝐿−11
𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇= 𝐴𝐴 + 𝐵𝐵𝑒𝑒− 𝜆𝜆+𝜇𝜇 𝑡𝑡
Exercise 1: Solution
• Find instantaneous and limiting availability for an exponential component whose restoration probability density is
tetg ⋅−⋅= µµ)(
�𝑝𝑝 𝑠𝑠 =𝐴𝐴𝑠𝑠
+𝐵𝐵
(𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆)=𝐴𝐴 𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆 + 𝐵𝐵𝑠𝑠𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
=
𝑠𝑠(𝐴𝐴 + 𝐵𝐵) + 𝐴𝐴𝜇𝜇 + 𝐴𝐴𝜆𝜆𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
=𝑠𝑠 + 𝜇𝜇
𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
� 1 = 𝐴𝐴 + 𝐵𝐵𝜇𝜇 = 𝐴𝐴𝜆𝜆 + 𝐴𝐴𝜇𝜇
𝐴𝐴 =𝜇𝜇
𝜇𝜇 + 𝜆𝜆
𝐵𝐵 =𝜆𝜆
𝜇𝜇 + 𝜆𝜆𝑝𝑝 𝑡𝑡 = 𝐴𝐴 + 𝐵𝐵𝑒𝑒− 𝜇𝜇+𝜆𝜆 𝑡𝑡 =
𝜇𝜇𝜇𝜇 + 𝜆𝜆
+𝜆𝜆
𝜇𝜇 + 𝜆𝜆𝑒𝑒− 𝜇𝜇+𝜆𝜆 𝑡𝑡
𝐿𝐿−1𝑠𝑠 + 𝜇𝜇
𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)
𝐿𝐿 1 =1𝑠𝑠
𝐿𝐿1
𝑠𝑠 + 𝑎𝑎= 𝑒𝑒−𝑎𝑎𝑡𝑡
0 20 40 60 80 100 120 140 160 180 2000.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1
t [day]
p(t)
constant failure rate =1/50 [day-1]constant repair rate = 1/5 [day-1]
Exercise 1: Solution
UNREVEALED FAILURE
Safety Systems: Examples
Fire extinguisher
Risk: Ignition of the gases in the oil tank, lightning strike or generation of sparks due to electrostatic charges.
Fire protection
Exercise 2: Instantaneous Availability of an Unattended Component
• Find the instantaneous unavailability of an unattended component (no repair is allowed) whose cumulative failure time distribution is F(t)
• Find the instantaneous availability of an unattended component (no repair is allowed) whose cumulative failure time distribution is F(t)
SOLUTION
The istantaneous unavailability, i.e. the probability q(t) that at time t the component is not functioning is equal to the probability that it fails before t
)()( tFtq ≡
Exercise 2: Solution
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Availability of a safety system under periodic maintenance (1)
• Safety systems are generally in standby until accident and their components must be periodically tested (interval between two consecutive test = Tp)
• The instantaneous availability is a periodic function of time
p(t)
R(t)1
τ 2τ 3τ tTp 2Tp 3Tp
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Availability of a safety system under periodic maintenance (1)
• Safety systems are generally in standby until accident and their components must be periodically tested
• The instantaneous unavailability is a periodic function of time
• The performance indicator used is the average unavailability over a period of time Tp
complete maintenance cycle
Average time the system is not working
pp
T
T TDOWNtime
T
dttqq
p
p== ∫0
)(
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EX. 3: Availability of a component under periodic maintenance
• Unavailability is due to unrevealed random failures, e.g. with constantfailure rate λ
• Assume also instantaneous and perfect testing and maintenanceprocedures performed every Tp hours
• Draw the qualitative time evolution of the instantaneous availability
• Find the average unavailability of the component
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Ex 3. Solution (1)
• Unavailability is due to unrevealed random failures, e.g. with constantrate λ
• Assume also instantaneous and perfect testing and maintenanceprocedures
• The instantaneous availability within a period Tp coincides with thereliability
p(t)
R(t)1
τ 2τ 3τ tTp 2Tp 3Tp
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• The average unavailability and availability are then:
For different systems, we can compute 𝑞𝑞𝑇𝑇𝑝𝑝 and 𝑝𝑝𝑇𝑇𝑝𝑝 by first computing their failure probability distribution FT(t) and reliability R(t) and then applying the above expressions.
Ex. 3: Solution (2)
p
T
T
p
T
T T
dttF
T
dttqq
pp
p
∫∫ === 00)()(
p
T
p
T
T T
dttR
T
dttpp
pp
p
∫∫ == 00)()(
pp TT qp −=1
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• For a system with constant faiure rate:
Ex. 3: Solution (3)
pTT Tqppp
λ2111 −=−=
Good approximationwhen 𝜆𝜆𝑡𝑡 < 0.1
pp
p
p
T
p
T t
p
T
T
p
T
T TTT
T
dtt
T
dte
T
dttF
T
dttqq
pppp
pλ
λλλ
21
2
)1()()( 20000 ==≈
−==== ∫∫∫∫ −
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Ex.4: A more realistic case 41
• The test is performed after a time 𝜏𝜏 from the end of the previous test
• Assume a finite test time τR
• Unavailability is due to unrevealed random failures, e.g. with constant rate λ
• Draw the qualitative time evolution of the instantaneous availability
• Estimate the average unavailability and availability over the complete maintenance cycle period Tp =τ + τR
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42Safety System under periodic maintenance: a more realistic case
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Ex.4: Solution (1) 43
• Assume a finite test time τR:
• Draw the qualitative time evolution of the instantaneous availability
t
p(t)
1
τR τR τR
τ
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• Assume a finite test time τR
• The average unavailability and availability over the complete maintenance cycle period Tp = τ +τR are:
[ ]0 0
( ) ( )R T R T
RR
F t dt F t dtq
τ τ
τ ττ τ
τ τ τ
+ += ≈ << ≈
+
∫ ∫
[ ]0 0
( ) ( )
RR
R t dt R t dtp
τ τ
τ ττ τ τ
= ≈ << ≈+
∫ ∫
Ex. 4: Solution (2)
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Safety System under periodic maintenance: a more realistic case
• Objective: computation of the average unavailability over the lifetime [0, TM]
• Hypoteses:• The safety system is initially working: q(0) = 0; p(0) = 1
• 𝜏𝜏 time interval between two consecutive maintenance interventions
• 𝜏𝜏𝑟𝑟 duration of the maintenance intervention
• Failure causes:
• random failure at any time T ∼ FT(t)• on-line switching failure on demand ∼ Q0
• maintenance disables the component ∼ γ0 (human error during inspection, testing or repair)
MT T
DOWNtimeqM=],0[
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𝑞𝑞[0,𝑇𝑇𝑀𝑀] = ∫0𝑇𝑇𝑀𝑀 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡
𝑇𝑇𝑀𝑀= ∫0
𝐴𝐴 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+∫𝐴𝐴𝑇𝑇𝑀𝑀 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡
𝑇𝑇𝑀𝑀
=∫0𝐴𝐴 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+𝐾𝐾 ∫𝐴𝐴
𝐵𝐵 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+∫𝐵𝐵𝐶𝐶 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡
𝑇𝑇𝑀𝑀=
= 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸0𝐴𝐴+𝐾𝐾 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸𝐴𝐴𝐵𝐵+𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸𝐵𝐵𝐶𝐶𝑇𝑇𝑀𝑀
=𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸0𝑇𝑇𝑀𝑀
𝑇𝑇𝑀𝑀
A B τ C Ft
T0 τ τ+τR 2τ+2τR2τ+τR
maintenance maintenance
𝑘𝑘 =𝑇𝑇𝑀𝑀 − 𝜏𝜏𝜏𝜏 + 𝜏𝜏𝑟𝑟
Safety System under periodic maintenance: a more realistic case
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A B τ C Ft
T0 τ τ+τR 2τ+2τR2τ+τR
maintenance maintenance
(1) (2) (3) (4)
Safety System under periodic maintenance: a more realistic case
Period from O to A
Possible failure causes: random failure (𝐸𝐸1) on-line switching failure on demand (𝐸𝐸2)
∀𝑡𝑡 ∈ O, A → 𝑞𝑞0𝐴𝐴 𝑡𝑡 = 𝑃𝑃 𝐸𝐸1 + 𝑃𝑃 𝐸𝐸2 − 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 = 𝐹𝐹 𝑡𝑡 + 𝑄𝑄0 − 𝐹𝐹 𝑡𝑡 𝑄𝑄0
[ ](0 ) 0 0 0 0 00 0 0
( ) (1 ) ( ) (1 ) ( )D A A T TT q t dt Q Q F t dt Q Q F t dtτ τ τ
τ= = + − ⋅ = ⋅ + − ⋅∫ ∫ ∫],0[ ADOWNtime
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A B τ C Ft
T0 τ τ+τR 2τ+2τR2τ+τR
maintenance maintenance
(1) (2) (3) (4)
Safety System under periodic maintenance: a more realistic case
Period (2) From A to B
System under maintenance:
∀𝑡𝑡 ∈ A, B → 𝑞𝑞𝐴𝐴𝐴𝐴 𝑡𝑡 = 1
𝐷𝐷𝐷𝐷𝐷𝐷𝑁𝑁𝑇𝑇𝐷𝐷𝐷𝐷𝐸𝐸[A,B] = ∫𝐴𝐴𝐴𝐴 𝑞𝑞𝐴𝐴𝐴𝐴 𝑡𝑡 𝑑𝑑𝑡𝑡 = ∫𝜏𝜏
𝜏𝜏+𝜏𝜏𝑅𝑅 1 𝑑𝑑𝑡𝑡 = 𝜏𝜏𝑟𝑟
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A B τ C Ft
T0 τ τ+τR 2τ+2τR2τ+τR
maintenance maintenance
(1) (2) (3) (4)
Safety System under periodic maintenance: a more realistic case
Period (3) From B to C Possible failure causes: random failure (𝐸𝐸1) on-line switching failure on demand (𝐸𝐸2) maintenance disables the component (𝐸𝐸3)
∀𝑡𝑡 ∈ B, C → 𝑞𝑞𝐴𝐴𝐵𝐵 𝑡𝑡 = 𝑃𝑃 𝐸𝐸1 + 𝑃𝑃 𝐸𝐸2 + 𝑃𝑃 𝐸𝐸3 − 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 − 𝑃𝑃 𝐸𝐸2 ∩ 𝐸𝐸3 −−𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸3 + 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 ∩ 𝐸𝐸3 =
=𝛾𝛾0 + (1 − 𝛾𝛾0)(𝑄𝑄0 + 1 − 𝑄𝑄0 𝐹𝐹(𝑡𝑡))
( ) 0 0 0 00 0
( ) (1 ) (1 ) ( )D BC BC TT q t dt Q Q F t dtτ τ
γ τ γ τ
= = ⋅ + − ⋅ ⋅ + − ⋅
∫ ∫],[ CBDOWNtime
5050Piero Baraldi
(0 ) 0 0 0 0 0 00 0
(1 ) ( ) (1 ) (1 ) ( )MM
D T T R TR
TT Q Q F t dt Q Q F t dtτ τ
τ τ γ τ γ ττ τ
= + − ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫
(0 ) 0 00 0 0 0 0
0 0
1 1( ) (1 ) (1 ) ( )M
M
D TT T R T
M M M R
T Q Qq F t dt Q Q F t dtT T T
τ ττ τ γ τ γ ττ τ
− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫
DT
MTq
DOWNtime
MT T
DOWNtimeqM=
−𝜏𝜏
Safety System under periodic maintenance: a more realistic case
5151Piero Baraldi
(0 ) 0 00 0 0 0 0
0 0
1 1( ) (1 ) (1 ) ( )M
M
D TT T R T
M M M R
T Q Qq F t dt Q Q F t dtT T T
τ ττ τ γ τ γ ττ τ
− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫MTq
• 𝜏𝜏 ≪ 𝑇𝑇𝑀𝑀
• ∫0𝜏𝜏 𝐹𝐹𝑇𝑇 𝑡𝑡 𝑑𝑑𝑡𝑡 ≤ 𝜏𝜏
• 𝜏𝜏𝑅𝑅 ≪ 𝜏𝜏
(0 ) 0 00 0 0 0 0
0 0
1 1( ) (1 ) (1 ) ( )M
M
D TT T R T
M M M R
T Q Qq F t dt Q Q F t dtT T T
τ ττ τ γ τ γ ττ τ
− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫
MT T
DOWNtimeqM=
MT T
DOWNtimeqM=
Safety System under periodic maintenance: a more realistic case
5252Piero Baraldi
• Consider an exponential system with small, constant failure rate
λ ⇒• Since typically γ0<<1, Q0<<1, the average unavailability reads:
00 0 0 0
0
1(1 ) ( )M
RT T
Qq Q F t dtττ γ γ
τ τ −
≅ + + − ⋅ + ⋅
∫
0 0 012M
RTq Qτ γ λ τ
τ≅ + + + ⋅ ⋅
Error after test Switching failure on demand
Random, unrevealed failures
between testsMaintenance
( ) 1 tTF t e tλ λ− ⋅= − ≅ ⋅
MT T
DOWNtimeqM=
MT T
DOWNtimeqM=
Safety System under periodic maintenance: a more realistic case
5353Piero Baraldi
Where to study?
• Slides• Lecture 5: Reliability of simple systems
• Chapter 5 (Red Book): Theory• Chapter 5 (Green Book): Exercises
• Lecture 6: Availability and Maintenability• Chapter 6 (Red Book): Theory• Chapter 6 (Green Book): Exercises
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