Exam Preparation Presented at 8 th Annual Reliability amp Maintainability Training Summit 34 Nov 2015 Introductions Dr Bill Wessels PE CRE Consulting Reliability Engineer Mechanical Engineer ID: 646055
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Certified Reliability EngineerExam Preparation
Presented at:
8
th
Annual Reliability & Maintainability Training Summit
3-4 Nov 2015Slide2
Introductions
Dr Bill Wessels PE CRE, Consulting Reliability Engineer
Mechanical Engineer
40-years experience (1975 – present)You EmployerCurrent Field:Engineering DisciplineExperience in Years
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Test Question
I can vote for an answer
Not an answer
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30
Response Counter
Answer NowSlide4
Employer
Government
Government Contractor
Private Sector
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30
0 of 1
Answer NowSlide5
Current Field
Design/Prototype Test
Systems Engineering & Integration
/ Development-Operational TestSustainability for Fielded Systems
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Response Counter
Answer NowSlide6
Engineering Discipline (BS Degree)
Mechanical
Electrical
Civil
Chemical
Systems
Industrial
Computer
Software
Physics
Chemistry
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Response Counter
Answer NowSlide7
Experience (in years since BS)
0 – 4
4 – 8
8 – 12Over 12
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Response Counter
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Goal
Pass the exam
The Certified Reliability Engineer Exam
Duration: 4-hrs
Number of Questions: 150 {that is 96-sec / question
Multiple choice with 4 alternatives
Coverage: The body of knowledge as described in the Certified Reliability Engineer Primer
Open Book/Open Notes – Includes the ‘Primer’ without the question sheets and these notes
Bring pencils, scratch paper, calculator
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Reliability Engineering Body of Knowledge
Reliability Management {19 questions
Probability and Statistics for Reliability {25 questions
Reliability in Design and Development {25 questionsReliability Modeling and Predictions {25 questionsReliability Testing {23 questionsMaintainability and Availability {17 questionsData Collection and Use {18 questions
Sum ??? 172 questions
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Reliability Management
Strategic Management
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Strategic Management
Demonstrate how reliability engineering improves programs, processes, products and services
Define and describe quality and reliability and how they relate to each other
Demonstrate how reliability engineers interact with marketing, safety, product liability, engineering, manufacturing and logistics (analysis methods)Explain how reliability integrates with other development activities (program interactions)Explain reliability engineering role in determination of failure consequences and liability management (warrantee)
Determine impact of failures on service and cost throughout product’s life-cycleDescribe how reliability engineering provides feedback to determine customer needs and specify product and service requirements
Interpret basic project management tools and tools:
Gannt
, PERT, Critical Path, QFD
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Strategic Management
Examples of reliability program benefits include:
Matching product design with customer’s application
Applying predictive and preventive maintenanceOptimize burn-in time
All of the above
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Examples of reliability program benefits include:
Matching product design with customer’s application
Applying predictive and preventive maintenance
Optimize burn-in time
All of the above
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Response Counter
Answer NowSlide14
Examples of reliability program benefits include:
Matching product design with customer’s application
Applying predictive and preventive maintenance
Optimize burn-in timeAll of the above
Key word(s); reliability program benefits
Index lists page II-3
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Reliability Management
Reliability Program Management
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Reliability Program Management
Terminology – reliability definitions
Element of a reliability program
Design-for-ReliabilityFRACASProduct life-cycle costsLife-cycle stages with relationship to reliabilityMaintenance costsLife expectation
Duty cycleDesign evaluationRequirements management
Reliability training programs
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Reliability Program Management
The usefulness of FMEA as a design tool is largely dependent on the
_______________ of the communication within the design process:
ExistenceQuantityEffectiveness
Frequency
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The usefulness of FMEA as a design tool is largely dependent on the
_______________ of the communication within the design process:
Existence
Quantity
EffectivenessFrequency
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Response Counter
Answer NowSlide20
The usefulness of FMEA as a design tool is largely dependent on the _______________ of the communication within the design process:
Existence
Quantity
EffectivenessFrequency
Key word(s); FMEA, communication, design process
Index lists page
II-11
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Reliability Management
Product Safety and Liability
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Product Safety and Liability
Role of reliability engineering
Ethical issues
System safety programRisk assessment toolsFMECAPRATFTA
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Product Safety and Liability
Reliability engineering applied failure consequences analysis to address:
Product failure expectation
Personal injury concernsTraining requirementsProduct testing requirements
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Reliability engineering applied failure consequences analysis to address:
Product failure expectation
Personal injury concerns
Training requirements
Product testing requirements
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Response Counter
Answer NowSlide26
Product Safety and Liability
Reliability engineering applied failure consequences analysis to address:
Product failure expectation
Personal injury concernsTraining requirements
Product testing requirements
Key word(s);
Product safety, consequences
Index lists page
II-17
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Probability and Statistics for Reliability
Basic Concepts
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Basic Concepts
Terminology –
statistical definitions
Basic probability conceptsIndependenceMutually exclusiveComplementaryConditionalJoint occurrenceExpected value
Discrete and continuous distributions
Binomial
Poisson (homogeneous & non-homogeneous)
Exponential
Lognormal
Weibull
NormalBathtub curveStatistical Process ControlTerminology – SPC definitionsRelationship to reliabilityCRE Exam Prearation
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Reliability Terminology
Table of contents and index have entry for “Reliability Terminology”
Primer provides a multi-page table that includes all of the terms and definitions
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Basic Concepts
Distinction between:
Mean
Point estimate of the meanMeasure of central tendencyCRE Exam Prearation
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The measure of central tendency is:
Mean
Mode
Median
All of the above
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Response Counter
Answer NowSlide34
The Central Limit Theory
Treats all sample data as normally distributed
Is applicable to only normally distributed data
States that the means of the samples are normally distributed
None of the above
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Response Counter
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The measure of central tendency is:
The range
The Variance
The interquartile range
All of the aboveCRE Exam Prearation
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30
Response Counter
Answer NowSlide37
The measure of central tendency is:
The range
The variance
The interquartile rangeAll of the above
Primer only mentions range and variance by finding page in index for “measure of central tendency”. But right answer is either one of the first three or D, yet A and B are correct, therefore the answer must be D.
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Shortcut Formula for Standard Deviation
Given:
S
XSX2NThen:
Var = [SX
2
– (
S
X)
2
/n]/(n-1)Std Dev = sqrt(VAR)CRE Exam Prearation
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Find Std Dev Given
S
X = 35.8,
SX2 = 257.14, n = 5
0.40299
0.812
0.203
0.451
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Response Counter
Answer NowSlide40
Find Std Dev Given
S
X = 35.8,
SX2 = 257.14, n = 5VAR = [
SX2
– (
S
X)
2/n]/(n-1)
VAR = [257.14 – (35.8)2/5]/(5-1) = 0.812/4 = 0.203Std Dev = sqrt(0.203) = 0.450560.40299 – divided VAR by n = 5, not n-1 = 4
0.812 – did not divide numerator by n-1
0.203 – did not take sqrt of VAR
0.451
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Probability Concepts
Union of events –
– stated as events A or B occur - AdditionIntersection of events – – stated as events A and B occur – MultiplicationAdditive Rule P(A
B) = P(A and B) = P(A) + P(B) – P(A and B)Multiplication Rule P(A
B) = P(A or B) = P(A)P(B)
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Counting
Permutation – number of ways that n items can be arranged taking them r at time where order matters
n
Pr = n!/(n-r)! – NOTE: 0! = 1A B C can be arrange as ABC, ACB, BAC, BCA, CAB, CBA, AB, BA, AC, CA, BC, CB
, A, B, CPermutation of n = 3 at r = 3:
3
P
3
= 3!/(3-3)! = (3)(2)(1)/1 = 6
Permutation of n = 3 at r = 2:
3P2 = 3!/(3-2)! = (3)(2)(1)/1 = 6Permutation of n = 3 at r = 1: 3P1
= 3!/(3-1)! = (3)(2)(1)/(2)(1) =
3CRE Exam Prearation
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Counting
Combination– number of ways that n items can be arranged taking them r at time where order does not matter
n
Cr = n!/r!(n-r)! – NOTE: 0! = 1A B C can be arrange as ABC, AB, AC, BC, A, B, CCombination of n = 3 at r = 3: 3
C3 = 3!/3!(3-3)! = (3)(2)(1)/(3)(2)(1) = 1Combination
of
n = 3 at r =
2:
3
C2
= 3!/2!(3-2)! = (3)(2)(1)/(2)(1) = 3Combination of n = 3 at r = 1: 3C1
= 3!/1!(3-1)! = (3)(2)(1)/(2)(1) =
3CRE Exam Prearation
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Advanced Concepts
Statistical interval estimates
Confidence intervals
Tolerance intervalsHypothesis testingMeansVariancesProportionsType I & II errorBayesian technique
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Given: Reliability of a part with an exponential reliability calculated to be 0.95 for a 10-hr mission duration. Find the reliability of the part for another mission given that the part has accumulated 100 hours (10 missions)? R(10|t = 100)
R = 0.95
10
R = 0.95
R > 0.95
Not enough Information
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Response CounterSlide51
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Probability and Statistics for Reliability
Statistical Inference
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Six test articles were put on test for 1000-hrs each with 2 failures with replacement. What is the point estimate of the MTBF.
500
1500
3000
6000
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Response Counter
Answer NowSlide60
Six test articles were put on test for 1000-hrs each with 2 failures with replacement. What is the point estimate of the MTBF.
500
1500
3000
6000
MTBF =
nT
/r = (6)(1000)/2 = 3000
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The sample MTBF is 200 hrs
with a standard deviation of 24 hrs. Sample size is 8 test articles. Find the 90% lower confidence limit for the MTBF given
t
a,n = 1.4
166.04
197.55
202.45
233.96
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Response Counter
Answer NowSlide64
The sample MTBF is 200 hrs with a standard deviation of 24 hrs. Sample size is 8 test articles. Find the 90% lower confidence limit for the MTBF given
t
a,n
= 1.4166.4
197.575
202.425
233.6
LCL = 200 – 1.4SQRT(24/8) = 197.575
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Find the lower confidence limit for the MTBF for three test article failures for 1200 hours total test time.
400
155
77
1200
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Response Counter
Answer NowSlide68
Find the 95% lower confidence limit for the MTBF for three test article failures for 1200 hours total test time.
400
155
77
1200
Chi-square from the table for a = 0.05 and 8 degrees of freedom (2r+2) is 15.5. MTBF = 2T/
c
2
a
,2r+2
= (2)(1200)/15.5 = 154.84 ~ 155.CRE Exam Prearation
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Reliability in Design and Development
Reliability Design Techniques
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Reliability Design Techniques
Use factors
Environmental
TemperatureHumidityVibrationCorrosivePollutantsStressesStatic & dynamic loads
Severity of serviceElectrostatic discharge
Radio frequency interference
Stress-strength analysis
FMEA in design
FTA in design
Tolerance and worst-case analyses
Robust design approaches (DOE)Independent and dependent variableFactors and levelsErrorReplicationExperimental design – full & fractional factorials
Human factors reliabilityCRE Exam Prearation
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Reliability in Design and Development
Parts and Systems Management
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Parts and Systems Management
Parts Selection
Materials selection and control
Derating methods and principlesS-N diagramStress-life relationshipEstablishing specifications
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Analysis finds part strength has a mean of 1200 with a
std
dev of 50. Mean stress is 900 with a
std dev of 100. Find the reliability.
0.9963
0.0037
0.75
0.5
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Response Counter
Answer NowSlide80
Analysis finds part strength has a mean of 1200 with a std
dev of 50. Mean stress is 900 with a
std
dev of 100. Find the reliability.0.9963
0.0037 - unreliability
0.75 =900/1200
0.5 =50/100
z
= (1200 – 900)/
sqrt
(502 + 1002) = 2.68
U = 0.0037 from the standard normal tableR = 1 – U = 0.9963
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Reliability Modeling and Predictions
Reliability Modeling
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Reliability Modeling
Sources of reliability data
Reliability block diagrams and models
Simulation techniquesCRE Exam Prearation
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Serial Assembly – Reliability Calculation
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Given the following RBD, the assembly reliability is…
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
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Response Counter
Answer NowSlide85
Given the following RBD, the assembly reliability is…
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than
0.93
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Given the following RBD, the assembly reliability is…
0.95
0.9999
0.857
0.925
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Response Counter
Answer NowSlide87
Given the following RBD, the assembly reliability is…
0.95
0.9999
0.857
0.925
R = (0.97)(0.95)(0.93) = 0.857
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Serial Assembly – Equal Reliability Allocation
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Find the reliability allocation for B, C & D in a serial design configuration given that the reliability requirement for A is 99%
0.97
0.996655
0.999999
0.785
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Response Counter
Answer NowSlide90
Find the reliability allocation for B, C & D in a serial design configuration given that the reliability requirement for A is 99%
0.97
0.996655
0.999999
0.785R Allocation = R
A
(1/n)
= 0.99
(1/3)
= 0.996655
Note: RA = RBRCRD = R Allocation3 = 0.9966553
= 0.99
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Serial System – Equal Reliability Allocation
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Given the following RBD, the assembly reliability is…
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
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30
Response Counter
Answer NowSlide93
Given the following RBD, the assembly reliability is…
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than
0.93
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Given the following RBD, the assembly reliability is…
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
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30
Response Counter
Answer NowSlide95
Given the following RBD, the assembly reliability is…
0.95
0.9999
0.857
0.925
R = 1 - (0.03)(0.05)(0.07) = 0.9999
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Parallel Assembly Reliability Calculation
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Parallel Assembly Reliability Allocation
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Find the reliability allocation for B, C & D in a parallel design configuration given that the reliability requirement for A is 99%
0.97
0.996
0.999999
0.785
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30
Response Counter
Answer NowSlide99
Find the reliability allocation for B, C & D in a parallel design configuration given that the reliability requirement for A is 99%
0.97
0.996
0.999999
0.785R Allocation = 1-U
A
(1/n)
= 1 – (1-0.99)
(1/3)
= 0.785
NOTE: RA = 1 – (1 – R Allocation)3 = 1 – (1 – 0.785)3 = 1 – (0.215)3 = 0.99
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Parallel Design – Equal Reliability Allocation
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Stand-by Design Logic
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Equal Failure Rates – Perfect Switch
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Unequal Failure Rates – Perfect Switch
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Equal Failure Rates – Imperfect Switch
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Unequal Failure Rates – Imperfect Switch
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Complex RBD – Cut Set Method
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Reliability Modeling and Predictions
Reliability Predictions
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Reliability Predictions
Part count and part stress analysis
Advantages and limitations
Reliability prediction methodsReliability apportionment and allocationCRE Exam Prearation
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Parts Count Method for MTBF & MTTR
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Reliability Testing
Reliability Test Planning
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Reliability Test Planning
Elements of a reliability test plan
Types and applications of reliability testing
Test environment considerationsCRE Exam Prearation
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Reliability Testing
Developmental Testing
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Development Testing
Accelerated life test – ALT
Single stress
Multiple stressSequential stressStep stress testing – HALTReliability growth testingAMSAADuane
TAAFSoftware testingWhite boxFault injection
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Reliability Testing
Product Testing
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Product Testing
Assess purpose, advantage and limitations
Qualification/demonstration testing
SequentialFixed lengthProduct reliability acceptance test – PRATStress screeningESS
HASSBurn-inAttribute testingBinomial
Hypergeometric
Degradation testing
Software testing
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Maintainability and Availability
Management Strategies
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Management Strategies
Maintainability and availability planning
Maintenance strategies
Corrective maintenance – reactivePreventive maintenance – proactiveRCMMaintainability apportionment/allocationAvailability tradeoffs
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Availability
A = Uptime/(Uptime + Downtime)
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Maintainability and Availability
Analyses
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Analyses
Maintenance time distributions
Lognormal
WeibullPreventive maintenance (PM) analysisTasksIntervalsNon-applicabilityCorrective maintenance (CM) analysis
Fault isolation time – AKA FD/FIRepair/replace timeSkill level
Testability
Spare parts strategy
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Data Collection and Use
Data Collection
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Data Collection
Types of data
Attribute (discrete) v continuous
Compete v censoredData sourcesCollection methodsData managementAcquisitionMediaDatabase
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Data Collection and Use
Data Use
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Data Use
Analysis
Algorithmic
GraphicFORMath ModelingTrend AnalysisANOVAMeasures of effectiveness
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Data Collection and Use
Data and Failure Analysis Tools
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Data and Failure Analysis Tools
FMEA
Criticality analysis
Hazards analysisSeverity analysisConsequences analysisFTAFRACAS
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