Certified Reliability Engineer
Exam Preparation
Presented at:
8th Annual Reliability & Maintainability Training Summit
3-4 Nov 2015
Dr Bill Wessels PE CRE
Introductions
Dr Bill Wessels PE CRE, Consulting Reliability Engineer
• Mechanical Engineer
• 40-years experience (1975 – present)
You
•
•
•
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Employer
Current Field:
Engineering Discipline
Experience in Years
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Test Question
A. I can vote for an answer
B. Not an answer
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A. Design/Prototype Test
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C. Sustainability for Fielded Systems
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Engineering Discipline (BS Degree)
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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
Probability and Statistics for Reliability
Reliability in Design and Development
Reliability Modeling and Predictions
Reliability Testing
Maintainability and Availability
Data Collection and Use
Sum
???
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{19 questions
{25 questions
{25 questions
{25 questions
{23 questions
{17 questions
{18 questions
172 questions
9
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-cycle
Describe 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:
A. Matching product design with customer’s application
B. Applying predictive and preventive maintenance
C. Optimize burn-in time
D. All of the above
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Examples of reliability program benefits include:
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A. Matching product design
with customer’s application
B. Applying predictive and
preventive maintenance
C. Optimize burn-in time
D. All of the above
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Examples of reliability program benefits include:
A.
B.
C.
D.
Matching product design with customer’s application
Applying predictive and preventive maintenance
Optimize burn-in time
All 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-Reliability
• FRACAS
Product life-cycle costs
•
•
•
•
Life-cycle stages with relationship to reliability
Maintenance costs
Life expectation
Duty cycle
Design evaluation
Requirements 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:
A. Existence
B. Quantity
C. Effectiveness
D. 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
Effectiveness
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:
A.
B.
C.
D.
Existence
Quantity
Effectiveness
Frequency
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 program
• Risk assessment tools
o FMECA
o PRAT
o FTA
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Product Safety and Liability
Reliability engineering applied failure consequences analysis to
address:
A. Product failure expectation
B. Personal injury concerns
C. Training requirements
D. Product 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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Product Safety and Liability
Reliability engineering applied failure consequences analysis to
address:
A. Product failure expectation
B. Personal injury concerns
C. Training requirements
D. 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 concepts
•
•
•
•
•
•
Independence
Mutually exclusive
Complementary
Conditional
Joint occurrence
Expected value
Discrete and continuous
distributions
• Binomial
• Poisson (homogeneous & nonhomogeneous)
• Exponential
• Lognormal
• Weibull
• Normal
• Bathtub curve
Statistical Process Control
• Terminology – SPC definitions
• Relationship to reliability
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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 mean
• Measure of central tendency
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The measure of central tendency is:
Mean
Mode
Median
All of the above
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The Central Limit Theory
A. Treats all sample data as
normally distributed
B. Is applicable to only normally
distributed data
C. States that the means of the
samples are normally
distributed
D. None of the above
on
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The measure of central tendency is:
The range
The Variance
The interquartile range
All of the above
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The measure of central tendency is:
A.
B.
C.
D.
The range
The variance
The interquartile range
All 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:
• SX
• SX2
•N
Then:
• Var = [SX2 – (SX)2/n]/(n-1)
• Std Dev = sqrt(VAR)
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Find Std Dev Given SX = 35.8, SX2 = 257.14, n = 5
0.40299
0.812
0.203
0.451
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Find Std Dev Given SX = 35.8, SX2 = 257.14, n = 5
VAR = [SX2 – (SX)2/n]/(n-1)
VAR = [257.14 – (35.8)2/5]/(5-1) = 0.812/4 = 0.203
Std Dev = sqrt(0.203) = 0.45056
A.
B.
C.
D.
0.40299 – 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 - Addition
Intersection of events –  – stated as events A and B occur –
Multiplication
Additive 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
nPr = n!/(n-r)! – NOTE: 0! = 1
• A B C can be arrange as ABC, ACB, BAC, BCA, CAB, CBA, AB, BA, AC, CA, BC, CB,
A, B, C
• Permutation of n = 3 at r = 3: 3P3 = 3!/(3-3)! = (3)(2)(1)/1 = 6
• Permutation of n = 3 at r = 2: 3P2 = 3!/(3-2)! = (3)(2)(1)/1 = 6
• Permutation of n = 3 at r = 1: 3P1 = 3!/(3-1)! = (3)(2)(1)/(2)(1) = 3
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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
nCr = n!/r!(n-r)! – NOTE: 0! = 1
•
•
•
•
A B C can be arrange as ABC, AB, AC, BC, A, B, C
Combination of n = 3 at r = 3: 3C3 = 3!/3!(3-3)! = (3)(2)(1)/(3)(2)(1) = 1
Combination of n = 3 at r = 2: 3C2 = 3!/2!(3-2)! = (3)(2)(1)/(2)(1) = 3
Combination of n = 3 at r = 1: 3C1 = 3!/1!(3-1)! = (3)(2)(1)/(2)(1) = 3
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Advanced Concepts
Statistical interval estimates
• Confidence intervals
• Tolerance intervals
Hypothesis testing
•
•
•
•
Means
Variances
Proportions
Type I & II error
Bayesian 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 = 0.9510
R = 0.95
R > 0.95
Not enough Information
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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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Six test articles were put on test for 1000-hrs each with 2 failures with
replacement. What is the point estimate of the MTBF.
A.
B.
C.
D.
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 ta,n = 1.4
166.04
197.55
202.45
233.96
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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 ta,n = 1.4
A.
B.
C.
D.
166.4
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.
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Find the 95% lower confidence limit for the MTBF for three test
article failures for 1200 hours total test time.
A.
B.
C.
D.
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/c2a,2r+2 = (2)(1200)/15.5 = 154.84 ~ 155.
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Reliability in Design and Development
Reliability Design Techniques
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Reliability Design Techniques
Use factors
• Environmental
o Temperature
o Humidity
o Vibration
o Corrosive
o Pollutants
• Stresses
o Static & dynamic loads
o Severity of service
o Electrostatic discharge
o Radio frequency interference
Stress-strength analysis
FMEA in design
Dr Bill Wessels PE CRE
FTA in design
Tolerance and worst-case
analyses
Robust design approaches (DOE)
• Independent and dependent
variable
• Factors and levels
• Error
• Replication
• Experimental design – full &
fractional factorials
Human factors reliability
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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 principles
• S-N diagram
• Stress-life relationship
Establishing 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.
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0.9963
0.0037
0.75
0.5
7
A.
B.
C.
D.
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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.
A.
B.
C.
D.
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 table
R = 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 techniques
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Serial Assembly – Reliability Calculation
Reliability Block Diagram
Assembly
LRU1
LRU2
LRUn
n
RASSY (t )   RLRUi  t    RLRU1  t   RLRU 2  t 
i 1
Let:
Then:
R1(t) = 0.90
R2(t) = 0.95
R3(t) = 0.99
If:
Then:
 RLRU n  t 
R1(t) = R2(t) = R3(t) = R(t) = 0.95
RASSY(t) = (R(t))3 = (0.95)3 = 0.8574
RASSY(t) = (0.90)(0.95)(0.99) = R1(t)R2(t)R3(t) = 0.8465
Serial Design Configuration
Reliability Rule-of-Thumb
RASSY(t) ≤ MIN{Ri(t)}
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Given the following RBD, the assembly reliability is…
0.97
0.93
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…
0.97
A.
B.
C.
D.
0.95
0.93
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
CRE Exam Prearation
92
5
0%
0.
85
7
0%
0.
0.
Dr Bill Wessels PE CRE
99
9
Answer Now
0%
9
0%
95
Response
Counter
0.
A.
B.
C.
D.
86
30
Given the following RBD, the assembly reliability is…
0.97
A.
B.
C.
D.
0.95
0.93
0.95
0.9999
0.857
0.925
R = (0.97)(0.95)(0.93) = 0.857
Dr Bill Wessels PE CRE
CRE Exam Prearation
87
Serial Assembly – Equal Reliability Allocation
Fault Tree Analysis
Reliability Block Diagram
Assembly in
Down State
Assembly
OR
LRU1
LRU1 Fails
LRU2 Fails
LRUn
LRU3 Fails
Given:
n
RASSY (t )   RLRU i (t )
i 1
RLRUi (t )  n RASSY (t )
Dr Bill Wessels PE CRE
LRU2
RASSY(t) = 0.99884
nLRU = 3
Then:
RLRU (t )  3 0.99884  0.999628
CRE Exam Prearation
88
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
99
6
99
9
0.
CRE Exam Prearation
0.
0.
Dr Bill Wessels PE CRE
0%
78
5
Answer Now
0%
0.
0%
99
9
0%
97
Response
Counter
65
5
A.
B.
C.
D.
89
30
Find the reliability allocation for B, C & D in a serial design
configuration given that the reliability requirement for A is 99%
A.
B.
C.
D.
0.97
0.996655
0.999999
0.785
R Allocation = RA(1/n) = 0.99(1/3) = 0.996655
Note: RA = RBRCRD = R Allocation3 = 0.9966553 = 0.99
Dr Bill Wessels PE CRE
CRE Exam Prearation
90
Serial System – Equal Reliability Allocation
Rsys  0.99
n subsys  3
Rsubsys 
nsubsys
Rsys  0.996655
n assy  3
Rassy 
nassy
Rsubsys  0.998884
n LRU  3
RLRU 
Dr Bill Wessels PE CRE
nLRU
Rassy  0.999628
CRE Exam Prearation
91
Given the following RBD, the assembly reliability is…
0.97
0.95
0.93
st
ha
n
0.
93
&
0.
95
0.
93
0%
Le
s
n
0.
95
&
0.
th
an
0%
0.
97
CRE Exam Prearation
Be
tw
ee
Dr Bill Wessels PE CRE
Answer Now
Gr
ea
te
r
Response
Counter
0%
97
0%
n
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
Be
tw
ee
A.
B.
C.
D.
92
30
Given the following RBD, the assembly reliability is…
0.97
0.95
A.
B.
C.
D.
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
Dr Bill Wessels PE CRE
0.93
CRE Exam Prearation
93
Given the following RBD, the assembly reliability is…
0.97
0.95
0.93
st
ha
n
0.
93
&
0.
95
0.
93
0%
Le
s
n
0.
95
&
0.
th
an
0%
0.
97
CRE Exam Prearation
Be
tw
ee
Dr Bill Wessels PE CRE
Answer Now
Gr
ea
te
r
Response
Counter
0%
97
0%
n
Greater than 0.97
Between 0.95 & 0.97
Between 0.93 & 0.95
Less than 0.93
Be
tw
ee
A.
B.
C.
D.
94
30
Given the following RBD, the assembly reliability is…
0.97
0.95
A.
B.
C.
D.
0.95
0.9999
0.857
0.925
0.93
R = 1 - (0.03)(0.05)(0.07) = 0.9999
Dr Bill Wessels PE CRE
CRE Exam Prearation
95
Parallel Assembly Reliability Calculation
Fault Tree Analysis
Reliability Block Diagram
Assembly in
Down State
Assembly
AND
LRU1
LRU1 Fails
LRU2 Fails
LRU2
LRU3 Fails
3
U ASSY (t )   U LRU i  t   U LRU1  t   U LRU 2  t   U LRU 3 (t ) 
LRU3
i 1
RASSY (t )  1  U ASSY (t )  1  U LRU1  t   U LRU 2  t   U LRU 3 (t ) 
U LRU (t )  1  RLRU (t )



RASSY (t )  1  1  RLRU1 (t ) 1  RLRU 2 (t ) 1  RLRU 3 (t )
Let:
Then:
R1(t) = 0.90 U1(t) = 0.10
R2(t) = 0.95 U2(t) = 0.05
R3(t) = 0.99 U3(t) = 0.01
UASSY(t) = (0.10)(0.05)(0.01) = 0.00005
RASSY(t) = 1 - UASSY(t) = 0.99995

If:
Then:
UASSY(t) = (U(t))3 = (0.05)3 = 0.000125
RASSY(t) = 1 - UASSY(t) = 0.999875
Parallel Design Configuration Reliability Rule-of-Thumb:
Dr Bill Wessels PE CRE
R1(t) = R2(t) = R(t) = 0.95
RASSY(t) ≥ MAX{Ri(t)}
CRE Exam Prearation
96
Parallel Assembly Reliability Allocation
Let:
Rassy(t) = 0.99
Reliability Block Diagram
Uassy(t) = 0.01
Then: Rassy  0.99
Assembly
Uassy  1  Rassy  0.01
n LRU  3
LRU1
RLRU  1 
LRU2
nLRU
Uassy  0.785
OR
LRU3
ULRU 
nLRU
Uassy  0.215
RLRU  1  ULRU  0.785
Dr Bill Wessels PE CRE
CRE Exam Prearation
97
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
99
9
99
6
99
9
0.
CRE Exam Prearation
0%
78
5
0%
Answer Now
0.
Dr Bill Wessels PE CRE
0%
0.
0%
97
Response
Counter
0.
A.
B.
C.
D.
98
30
Find the reliability allocation for B, C & D in a parallel design
configuration given that the reliability requirement for A is 99%
A.
B.
C.
D.
0.97
0.996
0.999999
0.785
R Allocation = 1-UA(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
Dr Bill Wessels PE CRE
CRE Exam Prearation
99
Parallel Design – Equal Reliability Allocation
R (t )  1   1  R  t  
n
ith Higher Level
Design
Configuration
i
j
j
Assume : R1 (t )  R2 (t ) 
jth Lower Level
Design
Configuration
jth Lower Level
Design
Configuration
jth Lower Level
Design
Configuration
Rn (t )
Then : Ri (t )  1  1  R j  t  
n
U j (t )  1  R j  t  
Ri (t )  1  U j (t ) 
n
U (t )   1  R (t )
U (t )   1  R (t ) 
n
j
i
n
n
j
n
i
n
U i (t )
U j (t )  n 1  Ri (t )  n U i (t )
R j (t )  1  U j (t )
Dr Bill Wessels PE CRE
CRE Exam Prearation
100
Stand-by Design Logic
Perfect Switch
Imperfect Switch
Primary
Primary
Stand-by
Switch
Normally Open
Stand-by
Primary
Primary
Stand-by
Switch
Normally Closed
Stand-by
Primary
Primary
Stand-by
Switch
Switch Engaged
Dr Bill Wessels PE CRE
Stand-by
 Fail Open
 Fail Closed
 Fail in Operation
 Mechanical – typically Weibull
 Digital – typically exponential
 Mechanical failure mode - evident to
operator/maintainer
 Digital failure mode- hidden to operator/
maintainer
 Mechanical FD/FI - less potential for BIT/
BITE
 Digital FD/FI- high potential for BIT/
BITE
 Mechanical & digital subject to significant
pre-Log downtime
 Mechanical subject to significant post-Log
downtime
CRE Exam Prearation
101
Equal Failure Rates – Perfect Switch
Equal Failure Rates
Perfect Switch
Exponential Failure Model
Primary
Part

Rassy ( )  R p ( ) 1   p
SW
Standby
Part
R p(|t) = 0.990050
 p = 0.000125

 p = 0.010000
 = 80
1] Equal Failure Rate, Perfect Switch ( P =  S)
R(|t) = R p(|t)(1 +  p)
0.999950
Dr Bill Wessels PE CRE
CRE Exam Prearation
102
Unequal Failure Rates – Perfect Switch
Unequal Failure Rates
Exponential Failure Model
Perfect Switch
Primary
Part
 p
Rassy ( )  R p ( )  
 s   p

SW
Standby
Part
R p(|t) = 0.990050
 p = 0.000125
R s(|t) = 0.956571
 s = 0.000555

 R p ( )  Rs ( )




 p/( s- p) = 0.290698
 = 80
2] Unequal Failure Rate, Perfect Switch ( P <  S)
R(|t) = R p(|t) + [( p/( s -  p)][R p(|t) - R s(|t)]
0.999782
Dr Bill Wessels PE CRE
CRE Exam Prearation
103
Equal Failure Rates – Imperfect Switch
Exponential Failure Model
Primary
Part
Equal Failure Rates
Imperfect Switch

Rassy ( )  R p ( ) 1  Rsw ( ) p t
SW
Standby
Part
R p(|t) = 0.990050
 p = 0.000125

 p = 0.010000
 = 80
R sw(|t) = 0.941765
3] Equal Failure Rate, Imperfect Switch ( P =  S)
R(|t) = R p(|t)[1 + R sw(|t) p]
0.999374
Dr Bill Wessels PE CRE
CRE Exam Prearation
104
Unequal Failure Rates – Imperfect Switch
Unequal Failure Rates
Exponential Failure Model
Imperfect Switch
Primary
Part
 p
Rassy ( )  R p ( )  Rsw ( ) 
 s   p

SW
Standby
Part
R p(|t) = 0.990050
 p = 0.000125
R s(|t) = 0.956571
 s = 0.000555

 R p ( )  Rs ( )




 p/( s- p) = 0.290698
 = 80
R sw(|t) = 0.941765
4] Unequal Failure Rate, Imperfect Switch ( P <  S)
R(|t) = R p(|t) + R sw(|t)[ p/( s -  p)][R p(|t) - R s(|t)]
0.999215
Dr Bill Wessels PE CRE
CRE Exam Prearation
105
Complex RBD – Cut Set Method
A
B
E
C
Dr Bill Wessels PE CRE
D
Cut Set
Terms
AB
CD
AED
CEB
-ABCD
-ABDE
-ABCE
-ACDE
-BCDE
-(2)ABCDE
+(4)ABCDE
A
B
C
0.96 0.95 0.94
0.96 0.95
0.94
0.96
0.95 0.94
0.96 0.95 0.94
0.96 0.95
0.96 0.95 0.94
0.96
0.94
0.95 0.94
0.96 0.95 0.94
0.96 0.95 0.94
CRE Exam Prearation
D
E
Path
0.93 0.97 R/U
0.91
0.93
0.87
0.93 0.97 0.87
0.97 0.87
0.93
-0.80
0.93 0.97 -0.82
0.97 -0.83
0.93 0.97 -0.81
0.93 0.97 -0.81
0.93 0.97 -1.55
0.93 0.97 3.09
Reliability = 0.994
106
Reliability Modeling and Predictions
Reliability Predictions
Dr Bill Wessels PE CRE
CRE Exam Prearation
107
Reliability Predictions
Part count and part stress analysis
Advantages and limitations
Reliability prediction methods
Reliability apportionment and allocation
Dr Bill Wessels PE CRE
CRE Exam Prearation
108
Parts Count Method for MTBF & MTTR
A
B
C
D
Assembly
MTBF by Parts Count Method
Part MTBF FR
n
nFR
A
1000 0.0010
1 0.0010
B
300 0.0033
2 0.0067
C
500 0.0020
4 0.0080
D
200 0.0050
3 0.0150
SFR = 0.0113 SnFR = 0.0307
MTBF = 32.61
Dr Bill Wessels PE CRE
CRE Exam Prearation
MTTR
MTTR
FR*MTTR
2.5
0.0025
3
0.0100
5
0.0100
1
0.0050
SFR*MTTR =
0.0275
MTTR =
2.4265
109
Reliability Testing
Reliability Test Planning
Dr Bill Wessels PE CRE
CRE Exam Prearation
110
Reliability Test Planning
Elements of a reliability test plan
Types and applications of reliability testing
Test environment considerations
Dr Bill Wessels PE CRE
CRE Exam Prearation
111
Dr Bill Wessels PE CRE
CRE Exam Prearation
112
Dr Bill Wessels PE CRE
CRE Exam Prearation
113
Reliability Testing
Developmental Testing
Dr Bill Wessels PE CRE
CRE Exam Prearation
114
Development Testing
Accelerated life test – ALT
• Single stress
• Multiple stress
• Sequential stress
Step stress testing – HALT
Reliability growth testing
• AMSAA
• Duane
• TAAF
Software testing
• White box
• Fault injection
Dr Bill Wessels PE CRE
CRE Exam Prearation
115
Reliability Testing
Product Testing
Dr Bill Wessels PE CRE
CRE Exam Prearation
116
Product Testing
Assess purpose, advantage and limitations
• Qualification/demonstration testing
o Sequential
o Fixed length
• Product reliability acceptance test – PRAT
• Stress screening
o ESS
o HASS
o Burn-in
• Attribute testing
o Binomial
o Hypergeometric
• Degradation testing
• Software testing
Dr Bill Wessels PE CRE
CRE Exam Prearation
117
Dr Bill Wessels PE CRE
CRE Exam Prearation
118
Maintainability and Availability
Management Strategies
Dr Bill Wessels PE CRE
CRE Exam Prearation
119
Management Strategies
Maintainability and availability planning
Maintenance strategies
• Corrective maintenance – reactive
• Preventive maintenance – proactive
o RCM
Maintainability apportionment/allocation
Availability tradeoffs
Dr Bill Wessels PE CRE
CRE Exam Prearation
120
Availability
A = Uptime/(Uptime + Downtime)
Dr Bill Wessels PE CRE
CRE Exam Prearation
121
Maintainability and Availability
Analyses
Dr Bill Wessels PE CRE
CRE Exam Prearation
122
Analyses
Maintenance time distributions
• Lognormal
• Weibull
Preventive maintenance (PM) analysis
• Tasks
• Intervals
• Non-applicability
Corrective maintenance (CM) analysis
• Fault isolation time – AKA FD/FI
• Repair/replace time
• Skill level
Testability
Spare parts strategy
Dr Bill Wessels PE CRE
CRE Exam Prearation
123
Data Collection and Use
Data Collection
Dr Bill Wessels PE CRE
CRE Exam Prearation
124
Data Collection
Types of data
• Attribute (discrete) v continuous
• Compete v censored
Data sources
Collection methods
Data management
• Acquisition
• Media
• Database
Dr Bill Wessels PE CRE
CRE Exam Prearation
125
Data Collection and Use
Data Use
Dr Bill Wessels PE CRE
CRE Exam Prearation
126
Data Use
Analysis
• Algorithmic
• Graphic
FOR
• Math Modeling
• Trend Analysis
• ANOVA
Measures of effectiveness
Dr Bill Wessels PE CRE
CRE Exam Prearation
127
Data Collection and Use
Data and Failure Analysis Tools
Dr Bill Wessels PE CRE
CRE Exam Prearation
128
Data and Failure Analysis Tools
FMEA
Criticality analysis
• Hazards analysis
• Severity analysis
• Consequences analysis
FTA
FRACAS
Dr Bill Wessels PE CRE
CRE Exam Prearation
129