cursus betonvereniging 25 oktober 2005 design-by-testing beslistheorie tijdsafhankelijk falen
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Cursus Betonvereniging 25 Oktober 2005 Design-by-Testing Beslistheorie Tijdsafhankelijk falen. Pieter van Gelder TU Delft. Sterkte - design by testing. NEN 6700, par. 7.2 Experimentele modellen Rekening houden met: Vereenvoudigingen experimenteel model - PowerPoint PPT PresentationTRANSCRIPT
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Cursus Betonvereniging25 Oktober 2005
Design-by-TestingBeslistheorie
Tijdsafhankelijk falen
Pieter van Gelder
TU Delft
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Sterkte - design by testing
• NEN 6700, par. 7.2 Experimentele modellen
• Rekening houden met:• Vereenvoudigingen experimenteel model• Onzekerheden m.b.t. lange-duur effecten• Representatieve steekproeven• Statistische onzekerheden • Wijze van bezwijken (bros/taai)• Eisen m.b.t. detaillering• Bezwijkmechanismen
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Voorbeeld• Nieuw anker voor bevestiging gevelelementen.
• Onder horizontale (wind-)belasting
Mogelijke bezwijkmechanismen:• spreidanker in beton bezwijkt• anker zelf bezwijkt• ankerdoorn breekt uit
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Voorbeeld
• Sterkte anker meten in proefopstelling.• Resultaten (in N):• 4897• 2922• 3700• 4856• 3221
Wat is de karakteristieke waarde (5%)?
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Statistische zekerheid
• Situatie:• Sterkte R normaal verdeeld
• Veel metingen
• Formule voor sterkte: u : standaard normaal verdeelde variabele
mR: steekproefgemiddelde
SR: standaarddeviatie uit steekproef
RR SumR
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Tabel normale verdeling
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Statistische onzekerheid• Situatie:• Sterkte R normaal verdeeld• Weinig metingen (n)• Gemiddelde onbekend• Standaarddeviatie onbekend
• Bayesiaanse statistiek:
n
11StmR R1nR
n : aantal metingentn-1 : standaard student verdeelde variabele met n-1 vrijheidsgraden
mR: steekproefgemiddelde
SR: standaarddeviatie uit steekproef
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Student t verdeling
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Statistische onzekerheid• Situatie:• Sterkte R normaal verdeeld• Weinig metingen (n)• Gemiddelde onbekend• Standaarddeviatie bekend
• Bayesiaanse statistiek:•
n : aantal metingenu : standaard normaal verdeelde variabele
mR: steekproefgemiddelde
R: bekende standaarddeviatie
n
11umR RR
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Voorbeeld
• Gegeven:
• 3 metingen: 88, 95 en 117 kN
• Bekende standaarddeviatie 15 kN
• Vraag:
• Bereken de karakteristieke waarde (5%)
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Voorbeeld
• Gegeven:
• 3 metingen: 88, 95 en 117 kN
• Onbekende standaarddeviatie
• Vraag:
• Bereken de karakteristieke waarde (5%)
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Voorbeeld
• Gegeven:
• 100 metingen
• steekproefgemiddelde 100 kN
• Onbekende standaarddeviatie, uit steekproef: 15 kN
• Vraag:
• Bereken de karakteristieke waarde (5%)
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Voorbeeld
0 20 40 60 80 100 120 140 160 180 2000
0.005
0.01
0.015
0.02
0.025
0.03
0.035
sterkte (kN)
kan
sdic
hth
eid
(1
/kN
)m=100 kN, s = 15 kN
veel metingen3 metingen m onbekend, s bekend3 metingen m onbekend, s onbekend
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Voorbeeld
0 20 40 60 80 100 120 140 1600
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
sterkte (kN)
kan
sm=100 kN, s = 15 kN
veel metingen3 metingen m onbekend, s bekend3 metingen m onbekend, s onbekend
P=0.05
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Beslistheorie
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Rationeel beslissen: ijscoman
ijs
regen
zon
patat
regen
zon
€ 0
€ 1000
€ 2000
€ -500
P{zon} = P{regen} = 0.5
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Rationeel beslissen: ijscoman
ijs
regen
zon
patat
regen
zon
€ 0
€ 1000
€ 2000
€ -500
P{zon} = P{regen} = 0.5
0 * 0.5 + 100 * 0.5 = 500
2000 * 0.5 - 500 * 0.5 = 750
Verwachte opbrengst:
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Irrationeel beslissen
• Risico-avers voorbeeld uitwerken op bord
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Definitie van risico
• Risico = kans x gevolg
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Matrix of risks
• Small prob, small event
• Small prob, large event
• Large prob, small event
• Large prob, large event
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Evaluating the risk
• Testing the risk to predetermined standards
• Testing if the risk is in balance with the investment costs
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Decision-making based on risk analysis
• Recording different variants, with associated risks, costs and benefits, in a matrix or decision tree, serves as an aid for making decisions. With this, the optimal selection can be made from a number of alternatives.
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Deciding under uncertainties
• Modern decision theory is based on the classic “Homo Economicus” model
• has complete information about the decision situation;
• knows all the alternatives;
• knows the existing situation;
• knows which advantages and disadvantages each alternative provides, be it in the form of random variables;
• strives to maximise that advantage.
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But in reality
• The decision maker:
does not know all the alternatives; does not know all the effects of the alternatives; does not know which effect each alternative
has.
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A decision model
A: the set of all possible actions (a), of which one must be chosen;
N: the set of all (natural) circumstances (θ);
Ω: the set of all possible results (ω), which are functions of the actions and circumstances: ω = f(a, θ).
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Example 4.1• Suppose a person has EUR 1000 at his disposal and is given the
choice to invest this money in bonds or in shares of a given company.
• The decision model consists of:
• a1 = investing in shares
• a2 = investing in bonds
•θ1 = company profit 5 %
•θ2 = 5 % < company profit 10 %
•θ3 = company profit > 10 %
•ω1 = return (0 % ‑ 2 %) = ‑2 % per annum
•ω2 = return (3 % ‑ 2 %) = 1 % per annum
•ω3 = return (6 % ‑ 2 %) = 4 % per annum
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Decision tree (example 4.1)
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Utility space
θ1 θ2 θ3
a1 -2 % 1 % 4 %a2 1 % 1 % 1 %
risk neutral risk averse risk seekingθ1 θ2 θ3 θ1 θ2 θ3 θ1 θ2
a1 0 0.5 1 0 0.75 1 0 0.25a2 0.5 0.5 0.5 0.75 0.75 0.75 0.25 0.25
Results space
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Likelihood of the circumstances
The expected value of the return of action a1: “buying shares” amounts to: 0.2 (-2 %) + 0.3 (1 %) + 0.5 (4 %) = 1.9 %.This is larger than the 1 % return of action a2: “buying bonds”.
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From discrete to continuous decision models
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Dijkhoogte bepaling
• Op bord uitwerken
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• Tijdsafhankelijke faalkansen
• Door veroudering is onderhoud noodzakelijk:– Onderhoudsmodellen
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Levensduur T:is een stochastische variabele
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J.K. Vrijling and P.H.A.J.M. van Gelder, The effect of inherent uncertainty in time and space on the reliability of flood protection, ESREL'98: European Safety and Reliability Conference 1998, pp.451-456, 16 - 19 June 1998, Trondheim, Norway.
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•Haringvliet outlet sluices
Lifetime distribution for one component
•
Modellering Modellering
Timestartt t t t
Replacement strategies of large numbers of similar components in hydraulic structures
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Voorbeeld “leeftijd van mensen”: stochastische variable Lmens
• Lmens ~ N(78,6) of EXP(76,8)• P(Lmens >90)=...?• P(Lmens >90| Lmens >89)= P(Lmens >90)/P(Lmens
>89)=...• Uitwerken op bord
• Vervolgens: Modelvorming voor algemene situatie
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Verwachte resterende levensduur als functie van reeds bereikte leeftijd
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Hazard rate population in S-Africa: f(t) / [1 - F(t) ]
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T = time to failure
• The Hazard Rate, or instantaneous failure rate is defined as:
• h(t) = f(t) / [1 - F(t) ] = f(t) / R(t)• f(t) probability density function of time to failure,
• F(t) is the Cumulative Distribution Function (CDF) of time to failure,
• R(t) is the Reliability function (CCDF of time to failure).• From: f(t) = d F(t)/dt , it follows that:
• h(t) dt = d F(t) / [1 - F(t) ] = - d R(t) / R(t) = - d ln R(t)
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Integrating this expression between 0 and T yields an
expression relating the Reliability function R(t) and the
Hazard Rate h(t):
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Bathtub Curve
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Constant Hazard Rate• The most simple Hazard Rate model is to assume
that: h(t) = λ , a constant. This implies that the Hazard or failure rate is not significantly increasing with component age. Such a model is perfectly suitable for modeling component hazard during its useful lifetime.
• Substituting the assumption of constant failure rate into the expression for the Reliability yields:
• R(t) = 1 - F(t) = exp (- λt)• This results in the simple exponential probability
law for the Reliability function.
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Non-Constant Hazard Rate
• One of the more common non-constant Hazard Rate models used for evaluation of component aging phenomenon, is to assume a Weibull distribution for the time to failure:
• Using the definition of the Hazard function and substituting in appropriate Weibull distribution terms yields:
• h(t) = f(t) / [1 - F(t) ] = β t β -1 / β
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• For the specific case of: β = 1.0 , the Hazard Rate h(t) reverts back to the constant failure rate model described above, with: λ . The specific value of the β parameter determines whether the hazard is increasing or decreasing.
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β values, 0.5, 1.0, and 1.5.
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β values, 0.5, 1.0, and 1.5.
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Maintenance in Civil Engineering
– Many design and build projects in the past– Nowadays many maintenance projects
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Good Detoriation
Model?
Statedependent
ContainsEffect of Loading?
Consequenceof
failure
CorrectiveMaint.
Usedependent
Timedependent
Loaddependent
yes
yessmall
large no
no
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Hydraulic Engineering
• corrective maintenanceis not advised in view of the risks involved
• preventive maintenance
• time based
• failure based
• load based
• resistance based
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resistance
loadtimefailure
Ro
resistance
loadtime
Ro
resistance
load
cum. loadtime
Ro
load
time
repairRo
Resistance
based
Load
based
Time
based
Failure
based
Rmin
repair
repair
repair
Δt
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Dike Settlement
S.L.S h0 – A ln t = h(t)
U.L.S. h(t) – HW
timetopt
R,Sh0
hmin
S
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Condition based maintenance
bad
good
Inspection
Repair
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Maintenance
• A case study
• Some concepts
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Maintenance strategies Maintenance strategies
of large numbersof large numbers
of similar componentsof similar components
in hydraulic structuresin hydraulic structures
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IntroductionIntroduction
• Maintenance replacement
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IntroductionIntroduction
– Maintenance replacement
• Large numbers of similar components
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IntroductionIntroduction
– Maintenance replacement
• Large numbers of similar components
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IntroductionIntroduction
– Maintenance replacement
• Large numbers of similar components
• Same lifetime-distribution
• Same age
• Same function
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ModellingModellingCase study
Conclusions
• Variables of a replacement scenario
– Start date of the (start) replacements
– Replacement interval (t)
– Number of preventive ( ) replacements
Modellering Modellering
Timestartt t t t
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ModellingModellingCase study
Conclusions
• Finding the optimal strategy
– Balance between risk costs and costs of preventive replacements
– Replacement capacity
– Capacity of the supplier
Modellering Modellering
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Modelling
Case studyCase studyConclusions
• Probability of failure for different scenarios
Casestudie Casestudie
0,00
0,02
0,040,06
0,08
0,10
0,12
0,140,16
0,18
0,20
2014
2016
2018
2020
2022
2024
2026
2028
2030
2032
2034
2036
2038
2040
2042
Start date of scenario (years)
Pro
babi
lity
of s
yste
m fa
ilure
(ent
ire s
cena
rio)
4 prev. repl.
8 prev. repl.
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ReliabilityReliability MaintainabilityMaintainability
AvailabilityAvailability
The Concept of Availability
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Maintainability
Maintainability is the probability that a Maintainability is the probability that a process or a system that has failed will process or a system that has failed will be restored to operation effectiveness be restored to operation effectiveness
within a given time.within a given time.
M(M(tt) = 1 - ) = 1 - ee--tt
where where is repair (restoration) rate is repair (restoration) rate
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Availability
Availability is the proportion of the Availability is the proportion of the process or system “Up-Time” to the process or system “Up-Time” to the total time (Up + Down) over a long total time (Up + Down) over a long
period.period.
Availability =Availability =Up-TimeUp-Time
Up-Time + Down-TimeUp-Time + Down-Time
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UpUp
DownDownA1A1 A3A3A2A2
tt
B1B1 B2B2 B3B3
Up:Up: System up and runningSystem up and runningDown:Down: System under repairSystem under repair
System Operational States
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MTTF is defined as the mean time of the occurrence of the MTTF is defined as the mean time of the occurrence of the first failure after entering service.first failure after entering service.
MTTF = MTTF = B1 + B2 + B3B1 + B2 + B333
UpUp
DownDownA1A1 A3A3A2A2
tt
B1B1 B2B2 B3B3
Mean Time To Fail (MTTF)
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MTBF is defined as the mean time between successive MTBF is defined as the mean time between successive failures.failures.
MTBF =MTBF = (A1 + B1) + (A2 + B2) + (A3 + B3)(A1 + B1) + (A2 + B2) + (A3 + B3)33
UpUp
DownDownA1A1 A3A3A2A2
tt
B1B1 B2B2 B3B3
Mean Time Between Failure (MTBF)
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Mean Time To Repair (MTTR)
MTTR =MTTR = A1 + A2 + A3A1 + A2 + A333
MTTR is defined as the mean time of restoring a process or MTTR is defined as the mean time of restoring a process or system to operation condition.system to operation condition.
UpUp
DownDownA1A1 A3A3A2A2
tt
B1B1 B2B2 B3B3
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Availability
Availability is defined as:Availability is defined as:
A =A = Up-TimeUp-TimeUp-Time + Down-TimeUp-Time + Down-Time
Availability is normally expressed in terms of MTBF and Availability is normally expressed in terms of MTBF and MTTR as:MTTR as:
A =A = MTBFMTBFMTBF + MTTRMTBF + MTTR
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Reliability/Maintainability Measures
(Failure Rate)(Failure Rate) = 1 / MTBF = 1 / MTBF
R(t) = R(t) = ee--tt
Reliability R(t)Reliability R(t)
Maintainability M(t)Maintainability M(t)
(Maintenance Rate)(Maintenance Rate) = 1 / MTTR = 1 / MTTR
M(t) = 1 - M(t) = 1 - ee--tt
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Types of Redundancy
• Active Redundancy
• Standby Redundancy
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Active Redundancy
A
B
DividerDivider
OutputOutputInputInput
Both A and B subsystems are operative at all timesBoth A and B subsystems are operative at all times
Div
Note: the dividing device is a Series ElementNote: the dividing device is a Series Element
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Standby Redundancy
A
B
SW
SwitchSwitch
OutputOutputInputInput
StandbyStandby
The standby unit is not operative until a failure-sensing device The standby unit is not operative until a failure-sensing device senses a failure in subsystem A and switches operation to senses a failure in subsystem A and switches operation to subsystem B, either automatically or through manual selection.subsystem B, either automatically or through manual selection.
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Series System
A1 A2 An
ppss = p = p11 + p + p22 +……. + p +……. + pn n - (-1)- (-1)nn joint probabilities joint probabilities
ppss : : Probability of system failureProbability of system failure
ppii : : Probability of component failureProbability of component failure
For identical and independent elements:For identical and independent elements:
ppss ~ 1 - (1-p) ~ 1 - (1-p)nn < np (>p) < np (>p)
InputInput OutputOutput
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Parallel System
B
A
ppss = p = p11.p.p22 … … ppnn
Multiplicative RuleMultiplicative Rule
OutputOutputInputInput
ppss : : Probability of system failureProbability of system failure
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Series / Parallel System
B1
A1
B2
A2
CInputInput OutputOutput
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System with RepairsLet MTBF = Let MTBF = and system MTBF = and system MTBF = ss
s = ( 3s = ( 3 + + )/ ( 2)/ ( 22 2 ))
For Active Redundancy (Parallel or duplicated system)For Active Redundancy (Parallel or duplicated system)
s = s = / 2/ 222 = MTBF = MTBF22 / 2 MTTR / 2 MTTR
B
AOutputOutputInputInput
<< <<
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For Standby RedundancyFor Standby Redundancy
s = ( 2s = ( 2 + + )/ ()/ (2 2 ))
s = s = / / 22 = MTBF= MTBF22 / MTTR / MTTR
A
B
SW
SwitchSwitch
OutputOutputInputInput
StandbyStandbyNote: The standby Note: The standby system is normally system is normally inactive.inactive.
Note: The switch is a Note: The switch is a series element, neglect series element, neglect for now.for now.
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System without Repairs
s = ( 3s = ( 3 + + )/ ( 2)/ ( 22 2 ))
For Active RedundancyFor Active Redundancy
s = (3/2) s = (3/2) where where = 1/ = 1/s s MTBFMTBF
For Standby RedundancyFor Standby Redundancy
s = ( 2s = ( 2 + + )/ ()/ (2 2 ))
s = 2s = 2where where = 1/ = 1/s s MTBFMTBF
For systems without repairs, For systems without repairs, = 0 = 0
s = 3s = 3/ ( 2/ ( 22 2 ) = 3) = 3/ ( 2/ ( 2 ) )
s = 2s = 2/ / 2 2 = 2/ = 2/
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MTBFMTBF22 / 2 MTTR / 2 MTTR
MTBFMTBF22 / MTTR / MTTR
MTBFMTBF
MTBFMTBF
ActiveActive
StandbyStandby
With RepairsWith Repairs Without RepairsWithout RepairsTypeType
Summary
Redundancy techniques are used to increase the system MTBF