Predicting product life using reliability analysis methods

Predicting Product
Predicting Product
Life Using Reliability
Life Using Reliability
y
Analysis Methods
Steven Wachs
©2011 ASQ & Presentation Steven
Presented live on Nov 09th ~ 11th, 2012

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Predicting Product Life
Using Reliability Analysis
Methods
Steven Wachs
Principal Statistician
Integral Concepts, Inc.
www.integral-concepts.com
248-884-2276
www.integral-concepts.com 1
©2012 Copyright

Agenda

 Reliability Concepts & Reliability Data
 Probability, Statistics, and Distributions
 Assessing & Selecting Models
 Estimating Reliability Statistics
 Systems Reliability
 Reliability Test Planning
 Accelerated Life Testing
 Warranty Analysis

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Motivation

Intense Global Competition

Customer Expectations

Customer Loyalty

Product Liability

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Defining Reliability

Reliability is the probability that a
material, component, or system will
perform its intended function under
defined operating conditions for a
specified period of time.

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Ambiguity in Definition

 What is the intended function?
 What are the defined operating
conditions?
 How should time be defined?

We must clearly define these characteristics
when defining reliability for a specific
application
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Reliability Data

 Life Data
 Time-To-Failure (TTF) Data
 Time-Between-Failure (TBF) Data
 Survival Data
 Event-time Data
 Degradation Data

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Unique Aspects of Reliability Data

 Presence of Censoring
 Reliability Models based on positive
random variables (e.g. exponential,
lognormal, Weibull, gamma)
 Interpolation and extrapolation often
required

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Repairable vs. Non-repairable

 The focus of this course is non-
repairable components or systems
(characterized by time to failure)

 Repairable systems are characterized
by time between failure

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The Bathtub Curve

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The Reliability Function

R(to ) = P(T > to )

where T =“time”
to failure

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Censored Data

 When exact failure times are not known

 Provides useful information for estimation
of reliability (Do NOT drop from analysis)

 Types of Censoring
– Right Censoring
– Left Censoring
– Interval Censoring

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Types of Censoring
Left Censoring Right Censoring

Interval Censoring

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Other Censoring Ideas

 Competing Risks

• Impact on Reliability estimates
• Alternatives (if extreme censoring exists)
– Use Accelerated Testing Conditions
– Use Degradation Data

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Agenda


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Describing Time to Failure

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Integrating the PDF

B

c fxdx
d
PX  B  PX  c, d 
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Reliability
Distribution Plot
0.0008

0.0007

0.0006

0.0005
Density

0.0004

0.0003

0.0002

0.0001 R(1500)

0.0000
0 1500
X

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Failure Probability (Cumulative)
Distribution Plot
0.0008
F(500)
0.0007

0.0006

0.0005
Density

0.0004

0.0003

0.0002

0.0001

0.0000
0 500
X

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The CDF

 fxdx
t
Ft  PX  , t 

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19

CDF/Reliability Relationship

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Hazard Function (Rate)

The propensity to fail in the next instant
given that it hasn’t failed up to that
time (“instantaneous failure rate
function”)

ft ft
ht  1Ft
 Rt

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Mean Time to Failure (MTTF)

 The Expected Value or the Mean of the Time
to Failure Random Variable
 The average time to failure (often
significantly larger than the median time to
failure)
 The MTTF can be misleading as often as
much as 70% of the population will fail
before the MTTF
 
MTTF  ET  0 tftdt  0 Rtdt
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B-Life (or Quantile)

The time at which a
specified proportion
of the population is
expected to fail

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Advantages of Parametric Models
 May be described concisely with a few
parameters

 Allows extrapolation (in time)

 Provide “smooth” estimates of failure
time distributions

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Common Distributions in Reliability

 Weibull
 Exponential
 Lognormal
 Gamma
 Binomial
 Loglogistic
 Etc.

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Exponential Distribution

 The simplest model used in reliability
analysis (and sometimes misused)

 Described by a single parameter, l which
is the hazard rate (inverse of MTTF)

 Key property: the hazard rate is constant
(the only distribution with this property)

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Exponential Distribution

• pdf: f(t) = le-lt
• cdf: F(t) = 1 - e-lt
• Reliability: R(t) = e-lt
• Hazard rate: h(t) = l
• MTTF = 1/l = q
• Quantile: F-1(p) = (1/l)[-ln(1-p)]

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Exponential Distribution Example

Light bulb lifetime may be described by an
exponential distribution. The MTTF = 12,000
hrs.

Find:
A. Hazard Rate
B. Proportion failing by 12,000 hrs
C. Proportion failing by 24,000 hrs

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Exponential Distribution Example

Solution
A. l = 1/12,000 = 0.000083 = 83 failures per
million hrs
12,000
B. F12, 000  1  e 12,000  1 1
e  0. 632
24,000
C. F24, 000  1  e 12,000  1 1
 0. 865
e2

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Exponential Distribution Guidelines

 Constant hazard rate implies that the
probability that a unit will fail in the next
instant does not depend on the unit’s age

 Reasonable for many electronic
components that do not wear out

 Usually inappropriate for modeling TTF of
mechanical components that are subject
to fatigue, corrosion, or wear
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The Weibull Distribution

 The most common model in reliability analysis

 Described by 2 parameters:
h = “scale” parameter
b = “shape” parameter

 Flexible model that can effectively model a
wide variety of failure distributions

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The Weibull Distribution (some functions)

 1  t 
 pdf: ft     t
 e 


 t
 cdf: Ft  1  e 


 t
 Reliability: Rt  e 

 1
 Hazard rate: ht     t


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The Weibull Shape Parameter

Failure Rate

?

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The Weibull Scale Parameter

?

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Weibull Characteristics

 h is also referred to as the 63.2nd
percentile
 To see this: set t = h in F(t)
 

Ft  F  1  e 

1 
 1e
 1  e  0. 632
1

• The value of b is irrelevant when t =
h
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Conditional Reliability

 An application of conditional probability
 Needed to estimate reliability when “burn-
in” is used or to estimate reliability after a
warranty period.

Rtt 0 
Rt|t 0   Rt 0 
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Agenda


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Selecting Models

 Given time-to-failure data (failures and
censored data), which distribution best
describes the data?
 Graphical Methods (probability
plotting) and/or Statistical Methods
may be utilized

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Probability Plots

 Graphical method to assess “fit”
 Fit is determined by how well the plotted
points align along a straight line
 Plotted variables are “transformed” so that
y is a linear function of x
– X axis: Plot observed failure times
– Y axis: Plot estimated cumulative probabilities
(p)

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Probability Plotting

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Constructing Probability Plots
• X-Axis – Observed/Transformed
Failure Times

• Y-Axis – Estimated/Transformed
Cumulative Probabilities

• Transformed quantities for plot
depend on the distribution

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Constructing Probability Plots

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Linearizing the CDF - Example

 Consider the Weibull distribution. Recall
that the Weibull cdf is:

 t
Ft  1  e 

• We need to transform F(t) to achieve
a linear function

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Linearizing the CDF - Example

 t
1  Ft  e 

t 
1
1Ft
e 


ln 1Ft     
1 t

ln ln 1Ft 
1
  lnt   ln

By setting: y  ln ln 1Ft 
1

x  lnt
C   ln
we have: y  x  C

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Graphical Estimation

2.75
b = 2.0/2.75
= 0.73

2.0

h

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Selecting Models
 Multiple Distributions may adequately
describe the time-to-failure data
 Sensitivity Analysis is recommended
to assess how reliability predictions
vary with alternative viable models
 Confidence Intervals on reliability
estimates do not include model
uncertainty

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Selecting a Distribution

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Handling Multiple Failure Modes

 Multiple Failure Modes should be
modeled separately (if data exists)
 Failure rates of the various failure
modes are typically different
 Overall Reliability may be predicted
using system reliability concepts
(series model)

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ReliaSoft Weibull++ 7 - www.ReliaSoft.com
Probability - W eibull
99.000
Probability-Weibull

Data 1
Weibull-2P
90.000 ML E SRM MED FM
F=40/S=0
Data Points
Probability Line

50.000
U n re l i a b i l i ty , F (t)

10.000

5.000

Steven Wachs
integral Concepts, Inc.
10/28/2011
1:05:27 PM
1.000
0.010 0.100 1.000 10.000 100.000 1000.000 10000.000
Time, ( t)
b   h    

©2012 Copyright

ReliaSoft Weibull++ 7 - www.ReliaSoft.com
Probability - W eibull
99.000
Probability-Weibull

Data 1
Weibull-CFM
90.000 ML E SRM MED FM
CFM 1 Points
CFM 2 Points
CFM 1 L ine
CFM 2 L ine
Probability Line

50.000
U n re l i a b i l i ty , F (t)

10.000

5.000

Steven Wachs
10/28/2011
1:03:08 PM
1.000
0.010 0.100 1.000 10.000 100.000 1000.000 10000.000
Time, ( t)
b    h      b   h  

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Agenda


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Reliability Estimation

 From the time-to-failure distribution we
can estimate quantities like:
• Reliability at various times
• Time at which x% are expected to fail
• Failure (hazard) rates
• Mean time to failure

 Confidence Intervals or Bounds should
be included to account for estimation
uncertainty
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 Estimation Methods
• Maximum Likelihood Estimation (MLE):
nice statistical properties, handles
censored data well, biased estimates for
small sample sizes

• Rank Regression: Unbiased estimates but
poorer precision and does not handle
censored data as well as MLE

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Properties of Estimators

 Bias – The extent to which the estimator
differs on average from the true value. (An
unbiased estimator equals the true value
on average)

 Precision – The amount of variability in the
estimates.

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Properties of Estimators

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Estimation Methods

 Maximum Likelihood Estimation
– Generally preferred by statisticians (minimum
variance) although the estimates tend to be biased

– ML method finds parameter values which maximize
the likelihood function (the joint probability of
observing all of the data).

– The maximization of the likelihood function usually
must be done numerically (rather than
analytically).

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MLE Example (Weibull)
• Given failure time data, we need to
estimate h, b.
i1 fx i  fx 1 fx 2 . . . fx n 
n
L

L      e
n  x 1  xi
i 
i1

• We maximize likelihood function by
taking derivatives with respect to
each parameter

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Effect of Censored Data on the Likelihood Function

• With no censoring, the likelihood
function is:

i1 fx i   fx 1 fx 2 . . . fx n 
n
L

• Censored observations cannot
use the pdf since the failure
time is unknown

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• Suppose we have a right-censored
observation at time = 1500?

• What function indicates the probability of
this occurring?

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• Suppose we have a right-censored

this occurring?

• R(1500) gives the probability that a unit fails
at time 1500 or later.

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Distribution Plot
0.0008

0.0007

0.0006

0.0005
Density

0.0004

0.0003

0.0002

0.0001 R(1500)

0.0000
0 1500
X

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• Suppose we have a left-censored

this occurring?

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• Suppose we have a left-censored observation
at time = 500?

this occurring?

• F(500) gives the probability that a unit fails at
time 500 or earlier.

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Distribution Plot
0.0008
F(500)
0.0007

0.0006

0.0005
Density

0.0004

0.0003

0.0002

0.0001

0.0000
0 500
X

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• Suppose we have an interval censored
condition where the failure occurred
between 1000 and 1300.

this occurring?

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• Suppose we have an interval censored
condition where the failure occurred
between 1000 and 1300.

this occurring?

• F(1300)-F(1000) gives the probability that a
unit fails between 1000 and 1300

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Distribution Plot
0.0008

0.0007

0.0006

0.0005 F(1300)-F(1000)
Density

0.0004

0.0003

0.0002

0.0001

0.0000
0 1000 1300
X

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Estimation Methods

• Rank Regression
– Determines best fit line on the probability
plot by using least squares regression

– Fitted line is used to estimate parameters

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Failure Probability Plot

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Estimating with Multiple Failure Modes
Failure Time Failure Model Failure Time Failure Model
63 linkage 791 motor
384 linkage 874 linkage
485 motor 999 motor
655 motor 1299 motor
662 linkage 1481 linkage

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Linkage Failure Mode
Distribution Analysis: Failure Time

Variable: Failure Time
Failure Mode: fm = linkage

Censoring Information Count
Uncensored value 20
Right censored value 20

Estimation Method: Maximum Likelihood

Distribution: Weibull

Parameter Estimates

Standard 95.0% Normal CI
Parameter Estimate Error Lower Upper
Shape 1.34641 0.264909 0.915592 1.97994
Scale 1325.81 240.466 929.169 1891.76

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Motor Failure Mode
Distribution Analysis: Failure Time

Variable: Failure Time
Failure Mode: fm = motor

Censoring Information Count
Uncensored value 20
Right censored value 20

Estimation Method: Maximum Likelihood

Distribution: Weibull

Parameter Estimates

Standard 95.0% Normal CI
Parameter Estimate Error Lower Upper
Shape 4.17342 0.634609 3.09784 5.62245
Scale 1154.46 62.7168 1037.86 1284.17
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Multiple Failure Modes
Probability Plot for Failure Time
Complete Data - ML Estimates
F ailure M ode = linkage
Failure Mode = linkage Failure Mode = motor S hape S cale
Weibull - 95% CI Weibull - 95% CI 1.34641 1325.81

F ailure M ode = motor
95 95 S hape S cale
4.17342 1154.46
80 80

50 50
Percent

20 Percent 20

5 5

2 2

1 1
10 100 1000 10000 500 1000 2000
Failure Time Failure Time

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Survival Plot for Failure Time
F ailure M ode = linkage
Failure Mode = linkage Failure Mode = motor S hape S cale
Weibull - 95% CI Weibull - 95% CI 1.34641 1325.81

F ailure M ode = motor
100 100 S hape S cale
4.17342 1154.46

80 80

60 60
Percent

Percent
40 40

20 20

0 0

0 1500 3000 4500 500 1000 1500
Failure Time Failure Time

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Survival Plot for Failure Time
Multiple Distributions - 95% CI

100

80

60
Percent

40

20

0

0 200 400 600 800 1000 1200 1400 1600
Failure Time

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Confidence Intervals
• An interval (l, u) around the point estimate that
contains the true value with high probability

• The interval is said to be a P% confidence interval
if P percent of the intervals we might calculate
from replicated studies contain the true
parameter value

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Improving Precision of Estimates

 More Data (Failures) = Better Precision
(tighter confidence intervals)
 Can make more assumptions (assume
distribution parameters)
 Reduce confidence level (not a real
solution)

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Agenda


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System Reliability

 A System may be thought of as a collection of
components or subsystems

 System Reliability Depends on:
a. Component reliability
b. Configuration (redundancy)
c. Time

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A Series System

i1 R it  R 1 tR 2 tR 3 tR 4 t
4
R s t 

Example: If the component reliabilities are 0.9, 0.9. 0.8, 0.8 at 1 year
Then the System reliability at 1 year is: 0.9*0.9*0.8*0.8 = 0.52

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Series Model for Multiple Failure Modes

n
Rt   R it  R A tR BtR C tR Dt
i1

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A Parallel System (Redundant Components)

R s t  1  F s t
 1  F 1 tF 2 tF 3 t
 1  1  R 1 t1  R 2 t1  R 3 t

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A Parallel System (Redundant Components)

Example: If the component reliabilities are 0.9, 0.9. 0.9 at 1 year
Then the System reliability at 1 year is: 1 - (.1)*(.1)*(.1) = 0.999

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k-out-of-n Parallel Systems
 System consists of n components in which k of
the n components must function in order for
the system to function

 For example, if 2 of 4 engines are required to
fly, then the system will not fail if:
– All 4 engines operate
– Any 3 operate
– Any 2 operate

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If all components have the same reliability, R(t):

The probabilities of all possible combinations leading to success are
summed

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k-out-of-n Parallel Systems Example

Suppose a system consists of 6 identical pumps. For the
system to function, at least 4 of the 6 pumps must operate. If
the reliability of each pump at 3 years in service is 0.90, what
is the system reliability at 3 years?

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Effect of k on System Reliability

 As k increases, system reliability decreases

 If k = 1 Pure Parallel System

 If k = n Series System

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Effect of k on System Reliability
System Reliability vs k (k-out-of-6, R = 0.90)

1.0

0.9

0.8 k Reliability
Reliability

1 1.0000
2 0.9999
0.7 3 0.9987
4 0.9842
5 0.8857
0.6 6 0.5314

0.5
1 2 3 4 5 6
k

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 When the components in the k-out-of-n
parallel configuration do not share the same
reliability function, all possible combinations
must be computed

 Example follows

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k-out-of-n System Example

Three generators are
configured in parallel. At 0.90
least two of the
generators must
function in order for the 0.8 2/3
system to function. At 5 7
years: R1 = 0.90, R2 =
0.87, R3 = 0.80. What is 0.80
the System Reliability at
5 years?

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 Here, k = 2, n = 3

 The following combinations of events lead to a
reliable system at 5 years in service:
– generator 1,2 operate and generator 3 fails
– All three generators operate

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R1 = 0.90
R2 = 0.87
R3 = 0.80
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Reliability Block Diagrams

Used to Model System and
Estimate System Reliability

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Reliability Allocation Problems

Given a reliability target for the system, how
should subsystem and/or component level
reliability requirements be established so that the
system objective is met?

Typical Goals
a. Maximize the System Reliability for a given cost
b. Minimize the Cost for a given System Reliability

Improve component reliability or add
redundancies?

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Agenda


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Reliability Test Planning

 Estimation Test Plans
 Determine sample size needed to
estimate reliability characteristics with a
specified precision

 Planning information such as assumed
distribution parameters, testing time, and
censoring scheme is required

 Failures during testing are required

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Sample Sizes for Desired
Precision
 We select sample size to achieve the desired
precision in our estimates

 Larger sample size  Greater precision

 Greater precision  Smaller confidence
intervals

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Sample Size Calculations

Calculation Depends On:
 Distribution used to model the failure data
 Level of precision desired
 Confidence level
 Presence of censoring
 Length of test (for Type I censoring)
 Failure proportion (for Type II censoring)

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Estimation Test Plan
Type I right-censored data (Single Censoring)

Estimated parameter: 50th percentile
Calculated planning estimate = 124.883
Target Confidence Level = 95%

Planning distribution: Weibull
Scale = 150, Shape = 2

Actual
Censoring Sample Confidence
Time Precision Size Level
100 62.435 8 96.2010

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Reliability Test Planning

 Demonstration Test Plans
 Determine sample size (or testing time)
needed to demonstrate reliability
characteristics (e.g. lower bound on
reliability)

 Planning information such as assumed
distribution and parameter is required

 Failures during testing are not required

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Reliability Demonstration

 Evaluates the following hypothesis
H0: The reliability is less than or equal to a
specified value

H1: The reliability is greater than a specified
value

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Types of Test Plans

 Zero-Failure Test Plans
– Test demonstrates reliability if zero failures
are observed during test
– Useful for highly reliable items

 M-Failure Test Plans
– Test demonstrates reliability if no more
than m failures occur
– Permit verification of test design
assumptions
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Planning Information

 Assumptions needed:
– Distribution
– Shape Parameter (for Weibull)
– Scale Parameter (for other distributions such
as lognormal, loglogistic, logistic, extreme
value)
– Assumptions based on expert opinions, prior
studies, similar products
– Sensitivity analysis is recommended

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Computing Test Time or Sample Size

 We specify either the sample size or the
testing time allocated for each unit (the
other quantity is computed)

 Demonstration Test Plan consists of:
– The maximum number of failures allowed
– The sample size
– The testing time for each unit

©2012 Copyright

Example: Demonstration Test Plan

 Reliability Goal: 1st percentile > 80,000 mi

 TTF estimated by Weibull w/ b = 2.5

 Can test for 120,000 miles

 How many units are needed for zero-failure
and 1-failure test plans?
©2012 Copyright


Demonstration Test Plans

Reliability Test Plan
Distribution: Weibull, Shape = 2.5
Percentile Goal = 80000,Target Confidence Level = 95%

Actual
Failure Testing Sample Confidence
Test Time Size Level
0 120000 108 94.9768

©2012 Copyright



Percentile Goal = 80000,Target Confidence Level = 95%

Actual
1 120000 172 95.0241

©2012 Copyright


 Suppose we can only test 50 units?
Percentile Goal = 80000,Actual Confidence Level = 95%

Failure Sample Testing
Test Size Time
0 50 163392

©2012 Copyright

Probability of Passing (POP)
Likelihood of Passing for Weibull Model
Maximum Failures = 0, Target Alpha = 0.05
Time = 120000, N = 108, Actual alpha = 0.0502316

100

80

60
Percent

40

20

0

2 4 6 8 10
Ratio of Improvement

©2012 Copyright

Probability of Passing (POP)
Likelihood of Passing for Weibull Model
Maximum Failures = 1, Target Alpha = 0.05
Time = 120000, N = 172, Actual alpha = 0.0497587

100

80

60
Percent

40

20

0

2 4 6 8 10
Ratio of Improvement

©2012 Copyright

Demonstration Test Plan (1st Percentile)
Percentile Goal = 80000, Target Confidence Level = 95%

Actual
0 120000 108 94.9768
1 120000 172 95.0241
2 120000 228 94.9669
3 120000 281 94.9567

©2012 Copyright

Test Units vs Test Time
772.775
0 Failures
1 Failures
2 Failures
3 Failures

639.853

506.932
N u m b e r o f T e s t U n i ts

374.010

241.088

Steven Wachs
10/28/2011
2:25:16 PM
108.167
80000.000 88000.000 96000.000 104000.000 112000.000 120000.000
Test Time

©2012 Copyright

Introduction to ALT
 Purpose:To estimate reliability on a
timely basis

 Inducefailures sooner by testing at
accelerated stress conditions

 Extrapolateresults obtained at
accelerated conditions to use conditions
(using acceleration models)

 Focus
on one or a small number of failure
modes
©2012 Copyright

Life-Stress Relationship
ReliaSoft AL TA 7 - www.ReliaSoft.com
Lif e vs Stress
100000.000
Life

Data 1
Ey ring
Weibull
323
F=30 | S=0
Eta L ine
1%
99%
393
Stress Lev el Points
Eta Point
Imposed Pdf
408
Eta Point
Imposed Pdf
423
Eta Point
Imposed Pdf
L i fe

10000.000

Steven Wachs
7/7/2011
3:07:17 PM
1000.000
300.000 328.000 356.000 384.000 412.000 440.000
Temperature
Beta=4.2918; A=-11.0878; B=1454.0864

©2012 Copyright

Accelerated Stress Testing

 Combination of Statistical Modeling and
understanding of Physics of Failure
 Care must be taken in designing tests to
yield useful information
 ALT models should be refined based on
correlation to actual results obtained at
normal use conditions

©2012 Copyright

Accelerated Life Testing - Topics
 Purpose and Key Concepts

 Accelerated Life Test Models

 One, Two, and Multiple Stress Models

 ALT Test Planning

 Accelerated Degradation Models

 Pitfalls, Guidelines, and Examples
©2012 Copyright

Introduction to ALT
 Purpose: To estimate reliability on a timely
basis

 Induce failures sooner by testing at accelerated
conditions

 Extrapolate results obtained at accelerated
conditions to use conditions (using acceleration
models)

 Focus on one or a small number of failure
modes
©2012 Copyright

Types of Accelerated Testing
– Units tested until failure
– Accelerating factor(s) are used to shorten the
time to failure

 Accelerated Degradation Testing
– Accelerating factor(s) are used to promote
degradation
– Amount of degradation observed during test
– Degradation data used to predict actual time
to failure at stressed conditions
©2012 Copyright

Accelerating Methods

1. Increase Usage Rate
– Increase usage rate from normal usage rate
– ex. Car door hinges have median lifetime of
44,000 cycles (15 years at 8 cycles per day)
– Increasing rate to 5000 cycles per day will
reduce median lifetime to 9 days.
– Assumes TTF is independent of usage rate
– Need to avoid unintended “stress” (e.g.
temp) caused by higher usage rate

©2012 Copyright

Accelerating Methods
2. Test Under Stress Conditions
• Test at higher levels of one or multiple stress
factors
• Common stress factors
– temperature
– thermal cycling
– voltage
– pressure
– mechanical load
– humidity

©2012 Copyright

Accelerated Life Test Models
ALT Models have 2 parts
1. Stochastic Part
• failure time distribution at each level of
stress
• Use distribution fitting to fit appropriate
models (Weibull, lognormal, etc.) at each
level of stress

2. Structural Part
• Life-stress relationship
• Use regression models to relate the stress
variable to the Time To Failure Distribution
©2012 Copyright

Acceleration Models

 Acceleration models relate accelerating
factors (e.g. temp, voltage) to the TTF
distribution.

 Model depends on acceleration method
(usage or stress) and the type of stress

 Physical models are based on physical or
chemical theory that describes the failure
causing process
©2012 Copyright

Life-Stress Models
 Increased stress promotes earlier failures and
life is predicted as a function of time and stress

 Common stress factors include:
– Temperature, Load, Pressure, Voltage, Current,
Thermal cycling, etc.

 The models assume stress levels are positive.
For temperature, use absolute temperature
(Kelvin) instead of Celsius or Farenheit

©2012 Copyright

Acceleration Factor

 Quantifies the degree to which a given
stress accelerates failure times

AF = Life at Use Condition / Life at Stress Condition

 Acceleration factor increases with stress

©2012 Copyright

Acceleration Factor
ReliaSoft ALTA 7 - www.ReliaSoft.com
Acc el erati on Factor vs Stres s
10.000
Acceleration Factor

Data 1
Arrhenius
Weibull
323
F= | S=
30 0
AF Line
8.000
A c c e le r a t io n F a c t o r

6.000

4.000

2.000

Steven Wachs
8/17/2011
9:47:29 PM
0.000
300.000 340.000 380.000 420.000 460.000 500.000
Temp erat u re
Beta=4.2916; B=1861.6187; C=58.9848

©2012 Copyright

Arrhenius Model (Temp Acceleration)

 Commonly used for products which fail as
a result of material degradation at
elevated temperatures

 Based on a kinetic model that describes
the effect of temperature on the rate of a
simple chemical reaction.

©2012 Copyright

Arrhenius Relationship

Rate = rate of a chemical reaction (rate is inversely
proportional to life)
tempK = absolute temperature in the Kelvin scale
= temp in deg C + 273.15
kB = Boltzmann’s constant = 8.6171x10-5= 1/11605
electron volts per deg C
Ea = activation energy in electron volts
g = a constant
(Ea and g are product or material characteristics)
©2012 Copyright

Arrhenius Model (ALTA Formulation)

Rate = rate of a chemical reaction (rate is inversely
proportional to life)
T = absolute temperature in Kelvin
kB = Boltzmann’s constant = 8.6171x10-5= 1/11605
electron volts per deg C
Ea = activation energy in electron volts
C = a constant

©2012 Copyright

Inverse Power Law Model

 Supports a variety of stress variables such as
voltage, temperature, load, etc.

 Assumes that the product life is proportional
to the inverse power of the stress induced

©2012 Copyright

Inverse Power Law Relationship

where:
T(V) = TTF at a given voltage
V = Voltage
A = constant (product characteristic)
a = constant (product characteristic)
(Voltage is the acceleration variable here)

©2012 Copyright

Other Models

 Some 2-stress and multiple stress models
will be mentioned later

 Many specific models have been developed
(for certain materials, failure modes, and
applications) although most may be
modeled with general formulations.

©2012 Copyright

Guidelines for ALT Models
 Acceleration Factor(s) should be chosen to
accelerate failure modes
 The amount of extrapolation between test stresses
and use condition should be minimized
 Different failure modes may be accelerated at
different rates (best to focus on one or two modes)
 The available data will generally provide little
power to detect model lack of fit. An
understanding of the physics is important.

©2012 Copyright

Guidelines for ALT Models
 Sensitivity analysis should be performed to assess
the impact of changing model assumptions
 ALT should be planned and conducted by teams
including personnel knowledgeable about the
product, its use environment, the
physical/chemical/mechanical aspects of the
failure mode, and the statistical aspects of the
design and analysis of reliability tests
 ALT results should be correlated with longer term
tests or field data

©2012 Copyright

Strategy for Analyzing ALT Data
1. Examine the data graphically
2. Generate multiple probability plots
3. Fit an overall model
4. Perform residual analysis
5. Assess reasonableness of the model
6. Utilize model for predictions (with
uncertainly quantified)

©2012 Copyright

Example – Analyzing ALT Data
 ALT of mylar-polyurethane insulation used
in high performance electromagnets*
 Insulation has a characteristic dielectric
strength which may degrade over time
 When applied voltage exceeds dielectric
strength a short circuit will occur
 Accelerating variable is voltage

*From Meeker & Escobar (1998)

©2012 Copyright

Time to Failure (Minutes) of Mylar-Polyurethane Insulation

Voltage Stress (kV/mm)
219.0 157.1 122.4 100.3

15.0 49.0 188.0 606.0
16.0 99.0 297.0 1012.0
36.0 154.5 405.0 2520.0
50.0 180.0 744.0 2610.0
55.0 291.0 1218.0 3988.0
95.0 447.0 1340.0 4100.0
122.0 510.0 1715.0 5025.0
129.0 600.0 3382.0 6842.0
625.0 1656.0
700.0 1721.0

©2012 Copyright

Fitting the Model

Model: Inverse Power Law
Std. = scale parameter for Distribution: Lognormal
Lognormal distribution Analysis: MLE
Std: 1.049793128
The location parameter
is a function of Voltage K: 1.149419255E-012
per the IPL model n: 4.289109625
LK Value: -271.4247009
Fail Susp: 36 0
©2012 Copyright

Probabi l i ty - Lognormal
99.000
Probability

Data 1
Inverse Power Law
Lognormal
100.3
F= | S=
8 0
Stress Level Points
Stress Level Line
122.4
F= | S=
8 0
Stress Level Points
Stress Level Line
157.1
F= | S=
10 0
Stress Level Points
Stress Level Line
219
F= | S=
10 0
U n r e lia b ilit y

Stress Level Points
Stress Level Line
50
50.000 Use Level Line

10.000

5.000

Steven Wachs
8/19/2011
3:15:43 PM
1.000
10.000 100.000 1000.000 10000.000 100000.000
Time
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Us e Level Probabi l i ty Lognormal
99.000
Use Level
CB@90% 2-Sided

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
Data Points
Use Level Line
Top CB-II
Bottom CB-II

50.000

10.000

5.000

Steven Wachs
8/19/2011
3:22:01 PM
1.000
1000.000 10000.000 100000.000 1000000.000 1.000E+7
Time
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

R el i abi l i ty vs Ti me
1.000
Reliability
CB@90% 2-Sided [R]

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
0.800 Data Points
Reliability Line
Top CB-II
Bottom CB-II

0.600
R e lia b ilit y

0.400

0.200

Steven Wachs
8/19/2011
3:25:11 PM
0.000
0.000 60000.000 120000.000 180000.000 240000.000 300000.000
Time
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Unrel i abi l i ty vs Ti me
1.000
Unreliability

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
Data Points
0.800 Unreliability Line

0.600

0.400

0.200

Steven Wachs
8/19/2011
3:19:32 PM
0.000
0.000 60000.000 120000.000 180000.000 240000.000 300000.000
Time
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Fai l ure R ate vs Ti me
5.000E-5
Failure Rate

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
Failure Rate Line
4.000E-5

3.000E-5
F a ilu r e R a t e

2.000E-5

1.000E-5

Steven Wachs
8/19/2011
3:52:01 PM
0.000
0.000 100000.000 200000.000 300000.000 400000.000 500000.000
Time
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Li fe vs Stres s
100000.000
Life

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
Median Line
100.3
Stress Level Points
10000.000 Median Point
Imposed Pdf
122.4
Stress Level Points
Median Point
Imposed Pdf
157.1
Stress Level Points
Median Point
Imposed Pdf
219
Stress Level Points
L ife

1000.000 Median Point
Imposed Pdf

100.000

Steven Wachs
8/19/2011
4:15:19 PM
10.000
10.000 100.000 1000.000
V olt ag e
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Acc el erati on Factor vs Stres s
600.000
Acceleration Factor

Data 1
Inverse Power Law
Lognormal
50
F= | S=
36 0
AF Line
480.000
A c c e le r a t io n F a c t o r

360.000

240.000

120.000

Steven Wachs
8/19/2011
4:17:07 PM
0.000
10.000 68.000 126.000 184.000 242.000 300.000
V olt ag e
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Standardi z ed R es i dual s
99.000
Standard Residuals

Data 1
Inverse Power Law
Lognormal
Residual Line
100.3
F= | S=
8 0
Residuals
122.4
F= | S=
8 0
Residuals
157.1
F= | S=
10 0
Residuals
219
F= | S=
10 0
Residuals
P r o b a b ilit y

50.000

10.000

5.000

Steven Wachs
8/19/2011
4:18:09 PM
1.000
-10.000 -6.000 -2.000 2.000 6.000 10.000
Resid u al
Std=1.0498; K=1.1494E-12; n=4.2891

©2012 Copyright

Predicting product life using reliability analysis methods

Predicting product life using reliability analysis methods

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