ASTM E3291-21

Standard Guide for Reliability Demonstration Testing

SIGNIFICANCE AND USE
4.1 Reliability demonstration testing is a methodology for qualifying or validating a product’s performance capability. Demonstration methods are useful for components, devices, assemblies, materials, processes, and systems. Many industries require demonstration testing either for new product development and product introduction, in validating a change to an existing product or as part of an audit. Test plans generally try to answer the questions, “How long will a product last?” or “What is its reliability?”, under stated conditions at some specific time. When time is being used as a life variable, it must be cast in some kind of “time” units. Typical time units are hours (or minutes), cycles of usage, calendar time or some variation of these. In certain cases, “time” can be accelerated in order to reduce a plan’s completion time. In the automotive industry mileage may be used as the time variable. Certain means of accelerating tests involve the use of increased power, voltage, mechanical load, humidity, vibration, or temperature (often in the form of thermal cycling).  
4.2 Two fundamental objectives in reliability test planning are: (a) demonstrating that a product meets a specific life requirement, and (b) demonstrating what a product can do – its life capability. In the first case, a requirement is specified; in the second case a series of test results are used to state a result at the present time – its current capability. Both cases share similar inputs and outputs.  
4.3 Often a life distribution model is specified such as the Weibull, the exponential, the lognormal or the normal distribution. In addition, for the specific distribution assumed, a parameter is typically assumed (or a range of values for a parameter). For example, in the Weibull case, the shape parameter, β, is assumed; in the lognormal case the scale parameter, σ, is assumed and in the normal case the standard deviation, σ, is assumed. In other cases, a non-parametric analysis can be used. N...
SCOPE
1.1 This standard covers fundamental concepts, applications and mathematical relationships associated with the planning of reliability demonstration tests as applied to components and materials testing.  
1.2 The system of units for this guide is not specified. Quantities and examples are presented only as illustrations of a method or a calculation. Any examples used are not binding on any particular product or industry.  
1.3 This standard does not purport to address all of the safety concerns, if any, associated with its use. It is the responsibility of the user of this standard to establish appropriate safety, health, and environmental practices and determine the applicability of regulatory limitations prior to use.  
1.4 This international standard was developed in accordance with internationally recognized principles on standardization established in the Decision on Principles for the Development of International Standards, Guides and Recommendations issued by the World Trade Organization Technical Barriers to Trade (TBT) Committee.

General Information

• Status: Published
• Publication Date: 30-Apr-2021
• ICS: 21.020 - Characteristics and design of machines, apparatus, equipment
• Technical Committee: E11 - Quality and Statistics
• Drafting Committee: E11.40 - Reliability

Relations

• Refers: ASTM E456-13a(2022)e1 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 01-Apr-2022
• Refers: ASTM E2555-07(2018) - Standard Practice for Factors and Procedures for Applying the MIL-STD-105 Plans in Life and Reliability Inspection
• Effective Date: 01-Apr-2018
• Refers: ASTM E456-13A(2017)e1 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 01-Oct-2017
• Refers: ASTM E456-13A(2017)e3 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 01-Oct-2017
• Refers: ASTM E456-13ae2 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 15-Nov-2013
• Refers: ASTM E456-13ae3 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 15-Nov-2013
• Refers: ASTM E456-13ae1 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 15-Nov-2013
• Refers: ASTM E456-13a - Standard Terminology Relating to Quality and Statistics
• Effective Date: 15-Nov-2013
• Refers: ASTM E456-13 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 15-Aug-2013
• Refers: ASTM E2696-09(2013) - Standard Practice for Life and Reliability Testing Based on the Exponential Distribution
• Effective Date: 01-Apr-2013
• Refers: ASTM E2555-07(2012) - Standard Practice for Factors and Procedures for Applying the MIL-STD-105 Plans in Life and Reliability Inspection
• Effective Date: 01-May-2012
• Refers: ASTM E456-12 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 01-May-2012
• Refers: ASTM E456-12e1 - Standard Terminology Relating to Quality and Statistics
• Effective Date: 01-May-2012
• Refers: ASTM E2696-09e1 - Standard Practice for Life and Reliability Testing Based on the Exponential Distribution
• Effective Date: 01-May-2009
• Refers: ASTM E2696-09 - Standard Practice for Life and Reliability Testing Based on the Exponential Distribution
• Effective Date: 01-May-2009

Overview

ASTM E3291-21: Standard Guide for Reliability Demonstration Testing is a key international standard developed by ASTM to provide comprehensive guidance on qualifying or validating the reliability and performance capability of products. This standard is applicable to a broad range of industries and product forms, including components, devices, assemblies, materials, processes, and complete systems. Reliability demonstration testing is vital during new product development, validation of changes to existing products, or audits to ensure ongoing quality and performance.

ASTM E3291-21 outlines the foundational concepts, practical applications, and key mathematical relationships necessary for effective reliability demonstration test planning. The guide assists users in answering critical questions such as “How long will a product last?” and “What is its reliability under stated conditions?” By following this standard, organizations can confidently qualify their products, meet industry life requirements, and support robust product introduction initiatives.

Key Topics

• Purpose of Reliability Demonstration Testing

  • Qualification or validation of product reliability and life capability
  • Supports product development, change validation, and audit processes

• Life Distribution Models

  • Use of statistical models such as Weibull, exponential, lognormal, and normal distributions
  • Choice of appropriate model and parameters based on the nature of failure modes

• Test Plan Elements

  • Determining sample size, mission time, reliability targets, and confidence levels
  • Consideration of time units: hours, cycles, calendar time, or other relevant units
  • Accelerated life testing methods, including increased power, voltage, load, humidity, vibration, or temperature

• Objectives in Reliability Testing

  • Demonstrating compliance with specific product life requirements
  • Assessing current product life capability through test results

• Types of Failure Modes

  • Random failures, wear-out mechanisms, and early “infant mortality” failures
  • Understanding the “bathtub curve,” representing different failure rate periods during a product’s lifecycle

• Statistical Analysis Approaches

  • Parametric (specifying distribution and parameter) versus non-parametric (attribute or pass/fail) methods
  • Use of binomial and beta distributions for non-parametric cases

Applications

Reliability demonstration testing, as outlined in ASTM E3291-21, is widely used across industries that demand high product integrity and consistency, such as automotive, aerospace, electronics, defense, medical devices, and industrial manufacturing. The guide provides a structured approach for:

• New Product Qualification: Ensuring a product meets its design life and reliability standards prior to market introduction.
• Product Modification Validation: Verifying that changes or improvements do not adversely affect product reliability.
• Supplier or Process Audits: Demonstrating ongoing compliance and reliability of supplied components and processes.
• Accelerated Testing: Reducing time-to-market by simulating life usage under heightened stress or environmental conditions.
• Setting Warranty Policies: Using reliability data to inform warranty terms and after-sales support.

The standard enables practitioners to develop statistically sound reliability test plans, calculate confidence intervals, and make data-driven decisions regarding product durability.

Related Standards

ASTM E3291-21 references and aligns with several other internationally recognized standards, including:

• ASTM E456 – Terminology Relating to Quality and Statistics
• ASTM E2555 – Practice for Factors and Procedures for Applying the MIL-STD-105 Plans in Life and Reliability Inspection
• ASTM E2696 – Practice for Life and Reliability Testing Based on the Exponential Distribution
• ASTM E3159 – Guide for General Reliability
• ISO 3534-1 – Statistics – Vocabulary and symbols, Part 1: Probability and general statistical terms
• ISO Guide 73 – Risk management vocabulary

These related standards provide additional definitions, statistical principles, and best practices for quality, reliability, and risk assessment to ensure consistency and compliance in reliability engineering.

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Keywords: ASTM E3291-21, reliability demonstration testing, reliability qualification, life testing, product validation, accelerated testing, failure modes, Weibull distribution, reliability standards, confidence interval, quality control, product reliability

Standards Content (Sample)

This international standard was developed in accordance with internationally recognized principles on standardization established in the Decision on Principles for the
Development of International Standards, Guides and Recommendations issued by the World Trade Organization Technical Barriers to Trade (TBT) Committee.
Designation: E3291 − 21 An American National Standard
Standard Guide for
Reliability Demonstration Testing
This standard is issued under the fixed designation E3291; the number immediately following the designation indicates the year of
original adoption or, in the case of revision, the year of last revision.Anumber in parentheses indicates the year of last reapproval.A
superscript epsilon (´) indicates an editorial change since the last revision or reapproval.
1. Scope 3. Terminology
1.1 Thisstandardcoversfundamentalconcepts,applications 3.1 Definitions:
and mathematical relationships associated with the planning of
3.1.1 Unless otherwise noted, terms relating to quality and
reliability demonstration tests as applied to components and
statistics are as defined in Terminology E456. Other general
materials testing.
statistical terms and terms related to risk are defined in ISO
3534-1 and ISO Guide 73.
1.2 The system of units for this guide is not specified.
3.1.2 B life, n—for continuous variables, the life at which
Quantities and examples are presented only as illustrations of p
there is a probability, p, (expressed as a percentage) of failure
a method or a calculation.Any examples used are not binding
at or less than this value. E3159
on any particular product or industry.
1.3 This standard does not purport to address all of the 3.1.3 failuremode,n—thewayinwhichadevice,processor
system has failed. E3159
safety concerns, if any, associated with its use. It is the
responsibility of the user of this standard to establish appro-
3.1.4 hazard rate, n—differential fraction of items failing at
priate safety, health, and environmental practices and deter-
time t among those surviving up to time t, symbolized by h(t).
mine the applicability of regulatory limitations prior to use.
E2555
1.4 This international standard was developed in accor-
3.1.5 mean time between failures (MTBF), n—the average
dance with internationally recognized principles on standard-
time to failure for a repairable item. E3159
ization established in the Decision on Principles for the
Development of International Standards, Guides and Recom- 3.1.6 mean time to failure (MTTF), θ, n—in life testing, the
mendations issued by the World Trade Organization Technical average length of life of items in a lot. E2696
Barriers to Trade (TBT) Committee.
3.1.7 reliability, n—the probability that a component,
device, product, process or system will function or fulfill a
2. Referenced Documents
function after a specified duration of time or usage under
2.1 ASTM Standards:
specified conditions. E3159
E456Terminology Relating to Quality and Statistics
3.2 Symbols:
E2555Practice for Factors and Procedures forApplying the
3.2.1 The following symbols are used extensively in the
MIL-STD-105 Plans in Life and Reliability Inspection
discussion.
E2696Practice for Life and ReliabilityTesting Based on the
3.2.2 C—confidence coefficient (decimal value between 0
Exponential Distribution
and 1).
E3159Guide for General Reliability
3.2.3 n—sample size, (positive integer at least 1).
2.2 ISO Standards:
ISO 3534-1Statistics – Vocabulary and symbols, Part 1:
3.2.4 p—failure probability (decimal between 0 and 1, or
Probability and general statistical terms
percentage between 0 and 100).
ISO Guide 73Risk management vocabulary
3.2.5 R—reliability. R=1– por(100– p)%for pexpressed
as a percentage.
This guide is under the jurisdiction of ASTM Committee E11 on Quality and 3.2.6 r—number of failures allowed (0 ≤ r < n).
Statistics and is the direct responsibility of Subcommittee E11.40 on Reliability.
3.2.7 β—Weibull shape parameter, also referred to as the
Current edition approved May 1, 2021. Published January 2022. DOI: 10.1520/
E3291-21. “Weibull slope.”
For referenced ASTM standards, visit the ASTM website, www.astm.org, or
3.2.8 θ—fortheexponentialmodel,themean(MTTF)ofthe
contact ASTM Customer Service at [e-mail]. For Annual Book of ASTM
Standards volume information, refer to the standard’s Document Summary page on distribution.
the ASTM website.
3.2.9 η—for the Weibull model, the characteristic life or
Available fromAmerican National Standards Institute (ANSI), 25 W. 43rd St.,
4th Floor, New York, NY 10036, http://www.ansi.org. scale parameter.
Copyright © ASTM International, 100 Barr Harbor Drive, PO Box C700, West Conshohocken, PA 19428-2959. United States
E3291 − 21
3.2.10 λ—for the exponential model, the failure rate; also time, t, and a maximum number of failures, r, allowed by the
equal to 1 / θ. plan. A test concludes and is successful if the n units tested
resultinnotmorethan rfailuresbytime t.Inanotherscenario,
3.2.11 f—scatter factor equal to the ratio B /B
50 0.1
the sample size, number of failures allowed and confidence
4. Significance and Use
value are first stated and the plan returns the test time
4.1 Reliability demonstration testing is a methodology for requirement.
qualifying or validating a product’s performance capability.
4.5 The “RC” nomenclature for specifying a test require-
Demonstration methods are useful for components, devices,
ment is often used, where R stands for reliability and C for
assemblies, materials, processes, and systems. Many industries
confidence. For example, to state a requirement of 2000 hours
require demonstration testing either for new product develop-
at R99C90 means that the requirement is to demonstrate 99 %
ment and product introduction, in validating a change to an
reliability at 2000 hours with 90 % confidence. Alternatively,
existing product or as part of an audit. Test plans generally try
this also means to demonstrate a B life of 2000 hours with 90
to answer the questions, “How long will a product last?” or
% confidence.
“What is its reliability?”, under stated conditions at some
4.6 This guide considers, the Weibull, lognormal and nor-
specifictime.Whentimeisbeingusedasalifevariable,itmust
mal parametric cases as well as the basic non-parametric case
be cast in some kind of “time” units. Typical time units are
for attribute reliability. The common exponential case is a
hours (or minutes), cycles of usage, calendar time or some
Weibull distribution with assumed shape parameter β = 1, but
variationofthese.Incertaincases,“time”canbeacceleratedin
is considered as a separate case, distinct from the Weibull.
order to reduce a plan’s completion time. In the automotive
industry mileage may be used as the time variable. Certain
5. General Introduction
means of accelerating tests involve the use of increased power,
voltage, mechanical load, humidity, vibration, or temperature
5.1 Before reliability can be assessed, and a test plan
(often in the form of thermal cycling).
specified,itisusefultoknowsomethingaboutthenatureofthe
failure modes that might occur. The type of failure mode is
4.2 Two fundamental objectives in reliability test planning
important in selecting a life distribution and any assumed
are: (a) demonstrating that a product meets a specific life
parameter where variable (continuous distribution) data is
requirement,and (b)demonstratingwhataproductcando–its
used.
life capability. In the first case, a requirement is specified; in
5.1.1 Ingeneral,therearethreeclassesoffailuremodesthat
the second case a series of test results are used to state a result
operate in electro-mechanical systems: (a) random, (b) wear
at the present time – its current capability. Both cases share
out, and (c) “infant mortality”. These are related to the failure
similar inputs and outputs.
rates operative in field applications.
4.3 Often a life distribution model is specified such as the
5.2 In a random type failure mode, the failure mechanism is
Weibull, the exponential, the lognormal or the normal distri-
not related to the age of the object tested or fielded, and the
bution. In addition, for the specific distribution assumed, a
failure rate is constant throughout this portion of life (constant
parameter is typically assumed (or a range of values for a
failure rate or CFR). A new object and any unfailed object
parameter). For example, in the Weibull case, the shape
having seen usage each have the same propensity to fail in the
parameter, β, is assumed; in the lognormal case the scale
future. Causes for this type of failure mode are often related to
parameter, σ, is assumed and in the normal case the standard
some external stimulus, operability robustness or to a complex
deviation, σ, is assumed. In other cases, a non-parametric
ofrarefactorsthatmightcometogether,rarely,injusttheright
analysis can be used. Non-parametric cases typically require a
mix to cause a failure.
larger sample size than parametric cases. This standard will
5.2.1 Random stress spikes under field conditions, such as
discuss conditions under which distributions and associated
those driven by electrical, mechanical, chemical or thermal
parameters can be assumed.
shock may cause random failures at any point in a product life
4.4 Generally, a life requirement is cast as a mission time
cycle. Operating a product outside of specification limits at
and associated reliability, for example, to demonstrate a
randomtimesmayrenderaproductpronetofailureatanytime
reliability of 99% at time t=1000 hours of usage. In another
initslife.Otheroutsidestimulisuchasthosecausedbyforeign
case the requirement might be cast as a B life requirement,
p
object damage or biological interference are possible and may
such as the B life. For example, if B = 10 000 cycles are
5 5
lead to random type failures.
specified, this means to demonstrate a reliability of 95 % at t =
10 000 cycles. Other life requirements might be a mean life, a 5.3 Wearoutmechanismsarerelatedtotheageoftheobject
median life (B ) or a failure rate not to be exceeded at a andhaveincreasingfailurerateswithtime/usage.Thisiscalled
specified time t. In other cases, the requirement might mean anincreasingfailurerate(IFR).Themoreusagetheobjecthas,
withstanding a load for some duration. Demonstration neces- the greater the likelihood of failures for the surviving popula-
sarily means to demonstrate with some statistical confidence. tion. Wear-out or performance degradation is generally a
Thus, a confidence value is a standard input in any plan. gradual process as usage increases. Ultimately, this results in
Commonly used confidence values are 99 %, 95 %, 90 %, and loss of robustness and eventual failure. In electro-mechanical
63.2 %. applications, causes of this type of failure may be driven by
4.4.1 Whenarequirementandaconfidencevaluehavebeen gradual chemical, thermal, mechanical, electrical or radiation
stated, a derived plan will determine a sample size, n, a test stresses.
E3291 − 21
5.4 An “infant mortality” type failure mode relates to be used to describe this phase. The number of such random
certain severe conditions possessed by some objects in a failuresoccurringwithinthisperiodisPoissondistributedwith
population rendering these select units prone to early failure, somemeannumberofevents.Afailurerateparameter, λ,inthe
while the remaining population may last many times longer. unitsoffailureperunittimeisusedtocharacterizethisportion
The older a unit is, the less likely such failures would appear of life.
becausetheaffectedunitwouldmostlikelyhavefailedearlyin 5.5.3 Wear Out Period—As components begin to fatigue or
the life cycle if it had the severe condition. Essentially, there is
wear out, failures occur at increasing rates for a specified
a failure mode and some items in the population have the interval. An increasing or sometimes sharp rise in failure rate
failure conditions more severely than others, resulting in early
can be the result of fatigue and other physical actions.As time
failure. The failure rate in these cases is decreasing with time. goes on, failures occur more and more frequently to a point
This is called a decreasing failure rate (DFR). There are
where it may no longer be practical to continue operating the
numerous causes for this case ranging from assembly issues to product. Several distributions may be appropriate to model the
material problems.
wear-out period. The Weibull and lognormal distributions are
often used. For the Weibull distribution, an increasing failure
5.5 The three general classes of failure are depicted in the
rate occurs where the shape parameter β>1.
so-called bathtub curve shown in Fig. 1. Fig. 1 is a highly
idealizedportraitofaproduct’slifecycle.Itsuggeststhatthere
5.6 Forpurposesofvalidationofanewproductintroduction
are three phases to life: (a) the burn-in period is the trouble
or for a change to an existing product it is common practice to
shooting or corrective action phase where the failure rate is
use the exponential distribution which covers the useful
decreasing,(b)therandomperiodistheusefullifephasewhere
(random) portion of product life. This assumption is conserva-
the failure rate is constant, and (c) the wear out period with an
tive in that it takes more test time and/or a larger sample size
increasing failure rate leads to end of life or retirement. Not
to validate a specific requirement for the random life case than
everyproductwillexperienceallthreephasesandthedegreeof
forthewearoutcase.Ingeneral,theinfantmortalitycaseisnot
each phase depicted may be of very dissimilar proportion.
typical in validating procedures (although in theory, this case
5.5.1 Infant Mortality—Failures of this type are generally
would be valid). The infant mortality case is more often used
the result of some components that do not meet specifications
in working with a fielded population that shows signs of early
orworkmanshipstandards.Thesetypesoffailuresaretypically
failures. It may also be used to develop a burn-in or screening
not design-related issues, but quality-related issues. As a
periodthataproductisexposedtopriortoitsfieldapplication,
product is put into service, early failures are observed as a
in an attempt to find early type failures before customer use.
result of these or similar conditions. Corrections are gradually
5.7 Insomecases,thefailureratemayexhibitincreasesand
made, and the failure rate decreases for a given time until it
decreases as a function of age. This is typically distribution-
may reach a steady state or constant rate. The infant mortality
specificandwouldbethecaseincertainparametercaseswhen
period, then, is characterized by a decreasing failure rate. The
the lognormal distribution is used.
Weibull distribution with shape parameter 0<β<1 is commonly
used to characterize infant mortality conditions.
6. The Nonparametric Case
5.5.2 Constant Failure Rate—Once the failures due to
componentsandworkmanshipareforthemostparteliminated, 6.1 In the nonparametric case (NP) there is no life distribu-
the constant failure rate period is entered, and called the tion being used to model variable data. Testing is of the
random failure rate period. “Random” means that failure pass-failtypeandgovernedbythebinomialdistribution.Inthis
eventsareproportionaltotimeinuse,butthisisnotdependent attribute reliability case, n is the sample size, and p is the
onageatthestartofanytimeinterval.Theconstantfailurerate failureprobabilityatthespecifiedtestcondition.Thereliability
period is the most common time frame for making reliability is cast as R=1– p. The parameters for this test case are C the
predictions, where the exponential distribution is used. The specified confidence, R the reliability, n, the sample size, and r
exponential–whichisequivalenttoaWeibullwith β=1–can the number of failures in n allowed by the plan.
FIG. 1 The “Bathtub” Curve
E3291 − 21
6.1.1 A random variable X is said to have a binomial 6.2.1 Eq4maybesolvednumericallyforanyvariablewhen
distributionif xistakesonintegervaluesbetween0and n,and the remaining two are known or assumed.
is a count of the number of occurrences of a defined condition,
6.2.2 Example 2—Suppose 1 failure in n = 22 tests are
whereeachofthensampleunitseitherhasordoesnothavethe observed. What is the reliability demonstrated using C=90%
condition. Each sample unit has a probability p of having the
confidence?UseEq4with n=22and C=0.9,anditeratingon
conditionandthensamplevaluesbehaveindependentlyofone R until the inequality in Eq 4 is just met, gives R = 0.8344 or
another. We say that X is binomially distributed with param-
about 83.4 % reliability. If the desire is to achieve 99 %
eters n and p. reliability using 90 % confidence, what sample size is required
6.1.2 A “failure event” is defined as the failure of a
ifweallownotmorethan1failurein n?UseEq4with C=0.9
component, a material specimen or other entity to meet a and R=0.99.Theniteratingon nuntiltheinequalityisjustmet
well-defined requirement. A requirement may be a variable
find that n = 388.
quantity such as a number of cycles or a minimum load
6.3 The variables in the general case include a sample size
condition in a fatigue test, a variable dimensional requirement,
n, confidence C, reliability R and r the number of allowable
or a basic attribute tested as go or no-go type of characteristic.
failures in n. The solution is a generalization of the r = 1 case
Eachtesteithermeetsorfailstomeettherequirement.Forpure
discussedinsection6.2.Thebetadistributionequivalenttothe
attribute pass or fail testing, a “zero failure” test plan is a
binomial probability expression P(X ≤ r) ≥1– C is used. Let
common theme. For the zero-failure test plan the following
pbethefailureprobability.Then R=1– p.Givenaconfidence
equation, based on the binomial distribution, relates sample
levelC,asamplesizen,andanumberofallowedfailuresr,the
size, n, confidence, C, and reliability, R (1, 2).
relationship is (3):
n
12R
R$ =1 2 C (1)
C$ * β~r 11, n 2 r!dy (5)
6.1.3 In Eq 1, C is the chosen confidence coefficient (0 < C
6.3.1 Eq 5 is a cumulative beta distribution with parameters
< 1), n is the sample size (n > 0), and R is the demonstrated
r + 1 andn–r evaluated at the point 1 – R. Refer to the cdf
reliability (0 < R < 1) when n tests result in zero failures. Eq 1
integral as G(1 – R). Then C = G(1 – R) and the parameter R
may be solved for either C or n giving.
is calculated using the inverse beta cdf evaluated at C. This is
n
C$ 1 2 R (2)
–1
G (C) ≥ 1– R. Then the demonstrated reliability R is
ln~1 2 C!
calculated as:
n 5 (3)
ln R
~ !
R$ 1 2 G C (6)
~ !
6.1.4 In Eq 2, the confidence in meeting the reliability
6.3.2 This calculation is carried out numerically using for
requirement R using sample size n is at least C.In Eq 3, the
example a spreadsheet-type program having beta distribution
sample size needed to claim reliability R with confidence C is
capability.When R, n,and rarespecified,Eq5isuseddirectly
n. In each case r = 0 failures are observed in a sample of size
to calculate the confidence demonstrated. If R, C and r are
n.
specified or if R, C and n are specified, then Eq 5 is iterated on
6.1.5 Example 1—Suppose n = 22 units have been run to a
the variable being solved for until the inequality in Eq 6 is just
specific requirement and 0 failures were observed. Then at
met. It is important, that r must be strictly less than n in all
C=90% confidence the minimum reliability demonstrated is
cases for these types of plans. In addition, for specified R, C
=120.9=0.90or90%,usingEq1.Thisisoftenreferredtoas
and n, there may not exist a suitable r that meets the RC
an “R90C90” value (reliability at least 90 % with 90 %
requirement. In that case, the sample size needs to increase.
confidence). Other “RC” combinations are in use. Suppose
6.3.3 Example 3—Atestplanwasconductedusingasample
further that R = 90 % is not good enough and the practitioner
of n = 250 units with r = 3 failures observed. What reliability
requires 95 % reliability, what sample size is necessary? Use
is demonstrated at C = 95 % confidence? Use Eq 6 with beta
Eq3with R=0.95and C=0.9giving n=ln(1–0.9)/ln(0.95)
distribution parameters r+1=4andn–r = 247. This results
= 44.89 or n = 45 units required without failure. What is the
in R ≥ 0.969 or 96.9 %. If an R95C95 plan is desired that
confidenceinmeetingareliabilityofatleast95%using n=22
allows 3 failures in n, what sample size is required? Use Eq 6
tests where 0 failure have occurred? Use Eq 2 with n=22and
with C = 0.95, R = 0.95 and r = 3. Iterate n in the beta
R = 0.95 giving C ≥ 1 – 0.95 = 0.676 or about 67.6 %
parameters until the inequality in Eq 6 is just met. This results
confidence.
in n = 153 as the required sample size.
6.2 Another commonly occurring case is to allow 1 failure
6.3.4 Example 4—If n = 450 units are available for test and
in n(notmorethan1failureamong nunitstested).Inthatcase,
R95C90 is the requirement, what number of failures is al-
the relation among n, C and R, based on a binomial model and
lowed? Use Eq 6 with R = 0.95, C = 0.9 and n = 450, iterating
simplified from the relation between a binomial and beta
on r until the inequality in Eq 6 is just met. Find that r=16is
distribution is (1):
the maximum number of failures allowed. Note that this
n21 n
combination of n and r is one of several possible combinations
nR 2 n 2 1 R $ 1 2 C (4)
~ !
that would satisfy the R95C90 requirement. For example,
n=282 and r=9or n = 209 and r = 6 would also work. A
smaller sample size would be possible and more economical.
Theboldfacenumbersinparenthesesrefertothelistofreferencesattheendof
this standard. Process capability should be taken into account, where
E3291 − 21
feasible, when deciding on any plan. Table 1 is a short time tforthepopulationofsurvivingunitsatthattime.Forthe
summary comparing sample sizes for the r=0, r = 1 and r = exponential, h(t)= λ, a constant throughout time.
2 cases and several common “RC” combinations.
7.2.6 The exponential distribution has the following key
properties:
7. The Exponential Case
7.2.7 The mean and standard deviation are the same value.
7.1 The exponential distribution case is one the most com-
7.2.8 Approximately 63.2 % of the area under the curve
monly required testing scenarios found in reliability. In this
falls below the mean (θ). This further means that the failure
section, theory and methods are outlined for the exponential
probability at the mean life is F(θ) = 63.2 % leaving the
lifecasewherethefailuremodeisassumedtobeoftherandom
reliability at the mean life of 100 – 63.2 = 36.8 %. This
type. Further related information is given in the Appendices.
property is not greatly appreciated by practitioners and can be
The exponential distribution is shown in Fig. 2. It is used for
the source of numerous types of errors and misunderstandings
predicting the reliability of items in the constant failure period
concerning reliability with this model.
(see Fig. 1). This is typically the starting point in design
7.2.9 The hazard function (failure rate) is constant through-
reliability requirements determination.
out life.
7.2 The exponential probability density function (pdf) is:
7.2.10 The quantity λt is called the cumulative hazard. It is
1 the expected number of failures at time t. For given time
2t⁄θ
f t 5 e , t.0 (7)
~ !
interval t, the number of failures in the interval [0,t] is Poisson
θ
distributed with mean λt provided the random behavior re-
7.2.1 The parameter θ is often called the mean time to
mainshomogeneousthroughtime t.Fortheexponential,thisis
failure (or MTTF) and is the mean of the distribution of t for
independent of the starting time of the interval.
single use items or first occurrence failure times. In repairable
7.2.11 Memoryless Property—For an exponentially distrib-
systems, units are repaired upon failure and reinstalled until
uted life, the probability that an item of age t, still surviving,
failure a second or a third time, the average time between
lasts an additional time s without failing is identical to a
repaircyclesiscalledthemeantimebetweenfailures(MTBF).
brand-new item lasting through a time s without failure.
Ingeneral,thisshouldnotbeconfusedwiththeMTTFbecause
Essentially,anitem’spropensitytofailisnotagedependent.A
repair does not always render a repaired item in like new
surviving item at any age has the same probability of failing
condition;however,fortheexponentialcase,itdoes,atleastin
withinorsurvivingafuturetimedurationsasanewitemdoes.
theory, and we have that MTTF = MTBF.
Thus, there is no wear out mechanism.
7.2.2 Eq7mayberecastusing λ=1/ θwhere λiscalledthe
7.2.12 A key property that holds only for the exponential
failure rate (failures per unit time).
distribution is:
2λt
f t 5 λe (8)
~ !
R t 1 s 5 R t R s (11)
~ ! ~ ! ~ !
7.2.3 Integrating Eq 8 gives the cumulative distribution
7.2.12.1 This can be manipulated further as:
function (cdf), F(t).
t n
R~ns! 5 $R ~s!% (12)
2λt
* f~y!dy 5 F~t! 5 1 2 e (9)
7.2.12.2 For example, if we know that the reliability at 1
7.2.4 Where, again, λ is substituted for 1 / θ. F(t) is the
cycle is 0.9999 (s = 1), then the reliability at 1000 cycles (n =
failureprobabilityattimetforobjectshavingaconstantfailure
1000) is 0.9999 or about 0.905.
rate λ. The quantity 1 – F(t)= R(t) is the reliability function at
7.2.13 Reproductive or Closure Property—The smallest or-
time t.Anothernameforreliabilityissurvivalprobability,used
der statistic in a sample of size n from an exponential
particularly in the life sciences.
distribution with mean θ is also exponentially distributed with
2λt
R t 5 1 2 F t 5 e (10)
~ ! ~ !
mean θ / n. The smallest in n inherits the exponential property
with mean proportional to 1 / n.
7.2.5 The hazard function, h(t), of any failure time distribu-
tion is the ratio f(t)/R(t). It is the instantaneous failure rate at 7.2.14 B-life Formula—The B life of a failure distribution
p
is that time by which p% of the population is expected to fail
by. For the exponential distribution this is found by inverting
A
TABLE 1 Sample Sizes for Nonparametric RC Test Plan Cases
the cdf, Eq 7:
with r Failures Allowed
p
RC case r=0 r=1 r=2
B 52θln 1 2 (13)
S D
p
R90C90 22 38 52
R90C95 29 46 61
7.2.15 Simulation—A random observation from an expo-
R90C99 44 64 81
R95C90 45 77 105
nential distribution with mean θ is determined using:
R95C95 59 93 124
R95C99 90 130 165 t52θln u (14)
~ !
R99C90 230 388 531
R99C95 299 473 628 where:
R99C99 459 662 838
u = a random observation from a uniform distribution on
A
“R” = reliability, “C” = confidence
[0,1].
E3291 − 21
FIG. 2 The Exponential Distribution
7.2.16 Both the time units and the failure rate may be 7.2.21.1 First, convert to the failure metric to failures per
–5
expressed in several ways. Hours or cycles are the typical time hour, λ. Multiply %/K by 10 to get λ = 4E-7.The probability
units used. Several commonly used metrics are: of failure by 15 000 hours is F(15 000) or:
(1)Failures/hr,
2~4 E 2 7!~15000!
F~15 , 000! 5 1 2 e 5 0.006
(2)Failures per 100k hours,
7.2.21.2 The 1 % failure time is the B life. Use Eq 13 with
(3)Failures per million hours, and 1
θ=1/ λ = 1 / (4E-7) = 2 500 000 hours.
(4)%/K hours.
7.2.17 The last expression, %/K hours, represents % failing
B 522,500,000 ln 1 2 5 25,126
S D
in 1000 hours of service. For example, if the rate is 1.5 %/K
and 10 000 units each operate for 500 hours we would expect
9 9
7.2.21.3 The FIT measure of failure is 10 λ=10 (4E-7) =
75failuresbecause10000unitsat500hourseachisequivalent
400 failures per billion operating hours.
to 5 000 units at 1000 hours each. 1.5 % of 5000 is 75 failures.
7.2.22 Example 6—Acertain type of aerospace system uses
7.2.18 The standard rate in failures per hour is then 75 / (10
five components operating simultaneously in series. The com-
000 × 500) = 1.5E-5. The relationship between %/K and the
5 ponent MTTF is known to be 250,000 cycles.The system fails
failure rate λ in failures per hour is: %/K=10 λ.
ifanyofthe5componentsfail.Itsmissiontimeis50cyclesof
7.2.19 Another variation is “failures in time” abbreviated as
usage. Calculate the system reliability at the mission time.
FIT and equivalent to parts per million per 1000 hours of
Calculate the B life of the system.
0.1
service or PPM/K. 1 PPM/K means 1 failure in 1 million units
7.2.22.1 The system will fail when the first failure in n=5
each operating for 1000 hours or 1 failure per billion operating
occurs. The first of n = 5 distribution is also exponential with
hours. This is equivalent to FIT = 10 000(%/K). In the above
MTTF = θ/n=250000/5=50000 cycles (see 7.2.13). The
example, the FIT number is 10 000X(1.5%/K) = 15 000 FIT.
probability of failure by the mission time of 50 cycles is:
Table 2 shows a conversion among λ,%/K and FITin the time
250⁄50000
units of hours. F 50 5 1 2 e 5 0.001
~ !
7.2.20 Eq 9 can be re-parameterized in the following way
7.2.22.2 Thus, the t = 50 cycle value is the B system life.
0.1
substitutingany B lifeinfor θ.Thisequationisusefulinsome
p
7.3 Test Planning—Atest plan for the exponential distribu-
applications.
tion requires several key design parameters.
t
F t 5 1 2 1 2 p ⁄ 100 B (15)
~ ! ~ !
p
(1)Sample size (n) and number of failures allowed (r),
(2)Objective MTTF (θ) or failure rate (λ), and
7.2.21 Example 5—A certain type of transistor is known to
(3)Confidence level (C).
have a constant failure rate with rate 0.04 %/K. What is the
7.3.1 As an alternative objective, a mission time and asso-
probability that one of these transistors fails before 15 000
ciated reliability may be specified, and this is equivalent to
hoursofuse?Howlongdowehavetowaittoexpect1%ofthe
specifying a B life. For example, if a reliability of 0.99 is
population to have failed? What is the FIT measure of the
p
required for a mission time of 1200 hours, this is equivalent to
failure rate?
stating a B life requirement of 1200 hours.
7.3.2 We may also ask what MTTF, B life or reliability at
A p
TABLE 2 Equivalent Failure Rates in Different Metrics
a stated mission time T has been demonstrated given a set of
rate, λ %/K FIT
failure time data at the present time during a test. In addition,
1E-04 10 100000
we can solve for the demonstrated confidence in meeting a
1E-05 1 10000
1E-06 0.1 1000
requirement.
1E-07 0.01 100
7.3.3 The most commonly occurring questions about test
1E-08 0.001 10
planning are: (1) what sample size (n) should I use? And (2)
1E-09 0.0001 1
A what test time (t) is appropriate?Associated with both of these
λ = 1E-5X(%/K) = 1E-9X(FIT); FIT = 1E4X(%/K)
is the number of failures allowed (r). This analysis starts with
E3291 − 21
a relationship involving, n, r, the confidence value C and the themaximumnumberoffailuresforthesameconfidencelevel,
beta distribution on the interval [0,1]. Let p be beta distributed C = 0.90, and objective MTTF of 50 000.
with parameters r+1 and n-r. Its cumulative distribution is
7.3.9.1 Again, solve Eq 18 for p , using θ = 50 000 and t =
c
denoted G(p). Set G(p)= C and solve for p. This uses the
600.This result is p = 0.011928. Next, solve the beta function
c
inverse beta integral: G(p)= Ctargeting C=0.90,using n=1200heldconstantand
c
21 adjusting r in the formula until G(p ) = 0.9 results.
c
p 5 G ~C! (16)
c
Solution Exhibit for Example 8
7.3.4 Alternatively, we have the beta integral itself:
rG(p )=C
c
6 0.9886
G p 5 C (17)
~ !
c
7 0.9741
8 0.9479
7.3.5 Eq 16 and Eq 17 can be solved using many software
9 0.9060
programs including spreadsheet-type programs. For clarity of
10 0.8458
notation, a subscript “c” is attached to p indicating that it
7.3.9.2 Here we see that r = 9 is the solution.
satisfiesEq16andEq17.Thequantity p istheupper100C%
c
7.3.10 To solve for the test time per unit required first use
non-parametric confidence bound on a failure probability p
Eq 16 solving for p with an assumed confidence, C, sample
c
when n units are tested and r failures occur where 0 ≤ r < n.
size n,andnumberoffailuresallowed, r.Thenusing p andthe
c
7.3.6 Let F(t) be the cumulative exponential failure prob-
objective MTBF, θ, solve Eq 18 for t. This results in:
ability where t is the test time and θ is an objective MTTF (see
t52θln~1 2 p ! (19)
c
7.2). Note that we could use λ, θ,ora B life, converting one
p
from the other. Use the relation p = F(t).
c
7.3.11 Example 9—Test time required; n, r, C and θ speci-
t
2 fied. Suppose we want to demonstrate a B life of 2500 cycles
p 5 1 2 e θ (18)
c
using 95 % confidence. We will use a sample size of n = 120
7.3.7 Eq 18 then involves the key variables of the test plan
units and allow r = 2 failures. What test time should be used?
and may be solved or manipulated for any variable. It is noted
7.3.11.1 First, use Eq 13 with p = 1 and B = 2500, solving
that Eq 18 depends on the ratio of test time to the objective
for θ,andfinding θ=248748.Wesimplifythisbyroundingup
MTTF (t / θ). So long as that remains the same, the test plan
to 249 000. Next, use Eq 16 with n = 120, r = 2 and C = 0.95
will be the same. This further means that for any test plan, the
finding p = 0.051534. Then use Eq 18, solving for t finding t
c
test time will be proportional to the objective MTTF desired.
= 13 174.4 cycles.
For example, if t / θ = 0.1 then we could use a test time of t =
7.3.12 Example 10—Confidence demonstrated required; n,
100 to demonstrate θ = 1000 or t = 10 to demonstrate θ = 100.
r, t and θ specified. If 150 units on test have resulted in 3
7.3.8 Example 7—Sample size required; r, C, t, θ specified. failures at t = 100 cycles, what confidence can we claim in
AnengineerwantstoclaimanMTTFof θ=50000hourswith
stating an MTTF requirement of 2500 cycles has been met?
90 % confidence (C = 0.90). He is willing to allow up to r=5
7.3.12.1 Use Eq 17 directly with p = 0.63212. This is the
c
failuresduringthetestandcanruneachofthesamplesfor500
failure probability in theory at t=θ. Use the beta integral with
hours. What sample size, n, is required?
parameters r + 1 andn–r or 4 and 147 respectively. The beta
7.3.8.1 First solve Eq 18 for p , using θ = 50 000 and t = integral evaluates to 0.8433 making the confidence demon-
c
strated 84.3 %. Iterating on sample size, n, shows that n = 169
500.This results in p = 0.00995. Next, solve the beta function
c
would be the requirement to state at least 90% confidence with
G(p)= C targeting C = 0.90, using r = 5 held constant and
c
r = 3 failures.
adjusting n in the formula until G(p ) = 0.9 results. It is useful
c
to use a spreadsheet type of calculator in executing the 7.3.13 The Zero-Failure Case—Many organizations require
calculation. The exhibit below shows several steps toward the zero failure test plans. In such a plan, a sample of n units are
solution.We want to solve G(p ) = 0.9 for n using C = 0.9 and tested, where the test times may be different for different units
c
p = 0.00995. tested, and zero failures is the requirement. To determine a
c
plan, is to determine the total test time, T, required to
Solution Exhibit for Example 7
demonstrate an MTTF of θ using confidence C.Atotal time T
nG(p )= C
c
canbecalculatedusingnt=Twheretisthetimeperunittested
900 0.88278
910 0.88879
in for each of n units. Sample size and test time can be
915 0.89169
interchanged as long as nt = T is preserved. Further we can
920 0.89453
have variable test times as long as T = Σt is preserved.
925 0.89730
926 0.89785
7.3.13.1 The equation governing this case is:
929 0.89948
930 0.90002 T52θln~1 2 C! (20)
7.3.8.2 Hereweseethatn=930isthesolution.Thismethod
7.3.13.2 In using Eq 20, T is the total time on test for all
may also be used if r is the unknown. In this case, n remains
units tested. Once we have T we can decide on the number of
fixed and r is varied until Eq 16 is true.
test units and distribute the test time in any way that is
convenient.
7.3.9 Example 8—Number of failures (r) required; n, C, t, θ
specified. Suppose in Example 7 that n = 1200 units can be 7.3.14 Example 11—Determine the total test time, T, re-
tested for the increased time of t = 600. We want to determine quiredforaplantodemonstratethatthefailurerateisnotmore
E3291 − 21
than λ=2.5E-5failurespercycleatconfidence95%withzero below. For the Weibull, demonstration test planning is gener-
failures required. Here, C = 0.95, θ=1/ λ = 40 000 cycles. ally used where the Weibull takes on wear out type failure
Using Eq 20, find that T = 119 829.3 cycles. If we wanted to modes. Infant mortality type failure modes are not generally
use equal test times for n = 10 units we should use a test time assumed for product test. They are more likely manifested as
of T / 10 = 11 983 cycles per unit. field data cases during an initial product release. The two
7.3.14.1 The C = 50 % zero failures estimate is sometimes parameter Weibull model cdf is Eq 21.
used as a simple point estimate for λ. It is a value of λ that
t β
S D
F t 5 1 2 e η (21)
~ !
makes the likelihood of obtaining zero failures in the given
experiment 50 %. Using this is as if one has observed 0.693
8.1.1 The reliability function is:
failures in the sample. This 50 % confidence estimate of λ
t β
S D
R t 5 1 2 F t 5 e η (22)
would then be 0.693 / T. Alternatively, when zero failures are ~ ! ~ !
observed in a total test time of T, the assumption of x=1
8.1.2 ThetwoparametersshowninEq21andEq22arethe
failure is sometimes assumed. In that case the resulting
Weibull shape parameter, β, and Weibull scale parameter, η.
confidenceintheestimate1/ Tisabout63.2%.Whenthetotal
The parameter β is also referred to as the “Weibull slope”.The
testtimeis Tandzerofailureshavebeenobserved,Table3can
parameter ηisalsoreferredtoasthe“characteristiclife”andis
be used to develop the estimate of λ.
formally, the 63.2th percentile of the Weibull distribution. The
7.3.15 Whenusingazero-failureplan,itisimportanttonote
parameter β is related to the type of failure mode being
thatthesetestplanshavealowpassprobabilitywhentheactual
modeled and the variation in failure times for this model. The
true parameter value is at or even substantially larger than the
three general failure mode classes are all possible using the
parameter minimum requirement. They are very conservative.
Weibulldistribution.Thethreecasesaredefinedbyvaluesof β.
For example, a zero-failure plan used to validate an MTTF of
8.1.3 β < 1 is the Infant mortality condition where there is a
1000 hours with C = 95 % confidence requires a total test time
failure mode that is manifested more severely in some units
of about 3000 hours without failure. The test can be allocated
that in others and can cause early failure in those units having
in any configuration of sample size and individual test time as
the severe condition. The failure rate is decreasing under this
long as the total test time equals 3000 hours. If the actual
condition.
MTTF is truly just at 1000 hours, the test plan will only
8.1.4 β = 1 is the random failure mode condition where the
achievea5% pass probability. In fact, the pass probability in
failure rate is constant throughout life. This is the exponential
this scenario will only reach 50 % when the true MTTF is
model of Section 7.
about 4330 – more than 4 times the parameter value minimum
8.1.5 β>1isthewearoutconditionwhereunitsarewearing
requirement.
out in time with an increasing failure rate.
7.3.16 Compare the zero-failure plan to a plan that allows 1
8.2 TheWeibull has a closed form formula for its mean and
failure in 5. In that case the test time will be 1071 hours per
variance but these quantities are not typically used for test
unittestedmakingthetotaltimebetween4284and5355hours.
planning purposes. The Weibull hazard function, h(t), and
The r = 1 plan can result in as much as a 77.8 % increase in
cumulative hazard function, H(t), are also available in closed
required test time; however, the r = 1 plan achieves a pass
form (see 7.2) as:
probability of 50 % at an MTTF = 2840. Fig. 3 shows
β21
comparison between the two plans with the 50 % pass
β t
h~t! 5 (23)
S DS D
probability indicated.
η η
7.3.17 The general rule is that the true MTTF has to be
β
t t
much better than the minimum requirement to have a reason- H t 5 h y dy 5 (24)
~ ! * ~ ! S D
0 η
able pass probability. This same trend holds for other sample
8.2.1 The Weibull B life is:
sizes and requirements as well as other distributional assump-
p
tions. 1⁄β
p
B 5 η 2 ln 1 2 (25)
H S DJ
p
8. The Weibull Case
8.2.2 Theinterpretationofthe B lifeisthatthereliabilityat
p
8.1 The Weibull distribution is widely used to model field
t = B is (100 – p) %. For the cumulative distribution function,
p
data exhibiting the full range of wear out, infant mortality and
the interpretation is F(B)= p %.TheWeibull distribution also
p
random type failure modes. Section 7 discussed failure modes
sharesthereproductivepropertyoftheexponentialdistribution
of the random type – the exponential distribution. That model
(see 7.2.13). In a sample of size n, from a Weibull distribution
is a special case of the Weibull distribution as will be seen
withparameters βand η,thefirstorderstatistic(smallestvalue)
1/β
has a Weibull distribution with parameters β and η / n .
8.2.3 An alternative parameterization for F(t) that substi-
TABLE 3 Rate Estimate When Observing Zero Failures in Time T
tutes a B life for η is useful in some applications:
p
Assumed Failures Rate Estimate λ Confidence
t β
0.693 0.693/T 50%
S D
F~t! 5 1 2 ~1 2 p ⁄ 100! B (26)
p
1 1/T 63.2%
2.3 2.3/T 90%
8.3 In a field data analysis application or a test plan, the
3 3/T 95%
Weibull shape parameter may sometimes be assumed. This is
4.6 4.6/T 99%
done because there is historical evidence for or industrywide
E3291 − 21
FIG. 3 Comparison of Pass Probability, β =1, r = 0 and r =1in n = 5 tests; C = 95 % for MTTF$ 1000 hours
agreementfortheassumedvalue.Forexample,bearingspallin 8.3.4 Thus,iftheengineercanspecifyascatterfactorofsay
both bench testing and in the field has been observed over and f=3,then βforthatcasewillequalabout8.8.Alternatively,Eq
over again in many companies to exhibit a Weibull shape 27 can be solved directly for β giving its exact value as a
parameter between 1.5 and 2. In aerospace components, high function of f.
cycle fatigue commonly exhibits a Weibull shape between 4.5 8.3.5 The Weibull coefficient of variation, cv, is the ratio of
and 6. When β is assumed, the Weibull distribution becomes a the standard deviation to the mean. For the Weibull, this ratio
one parameter model. That single parameter is traditionally is related to β and is:
taken as the characteristic life, η, but any other B life may be
p
~
=Γ 1 12⁄ β 2 Γ 1 11⁄ β
~ ! ~ !
substituted using Eq 25 and Eq 26.
cv 5 (28)
Γ~1 11⁄ β!
8.3.1 In the Weibull test planning application, β must be
assumed. Either a specific value or a range of plausible values
8.3.6 Where Γ(z) denotes the gamma function (see Appen-
for β is assumed. There are two alternative quantities that can
dices). The functional relationship between cv and β can be
be used in place of the assumed β. Each of these is a function
solved using a variety of software programs, for example, a
of β, and by their selection we are essentially assuming β. The
spreadsheet-type program. Table 5 gives several values of β as
two alternative quantities are the Weibull scatter factor, f, and
related to cv using Eq 28.
coefficient of variation cv.
8.3.7 It is noted that for the exponential distribution
8.3.2 The Weibull scatter factor, f, used in some industries,
(Weibull β = 1), cv = 1 meaning that the mean and standard
istheratioB /B .Thisissometimesknownincertaintypes
50 0.1 deviation are equal in that case. Using Eq 27 and Eq 28 a
of materials testing, for specific materials. The functional
relationship can also be established between the scatter factor
relationship between f and β is:
and the coefficient of variation.
1⁄β
f 5 692.8 (27)
8.4 Test Planning—To specify a test plan for the Weibull
model one needs to first select β. In some cases, as shown
8.3.3 Table 4 gives several values of f as related to β using
below,thetestplanwillbeindependentof β.Inthoseinvariant
Eq 27.
cases the same sample size would be adequate for any value of
β. These plans are essentially identical to the non-parametric
case. Most test plans are designed to demonstrate a mission
timeatanassociatedreliabilityusingsomeconfidencevalueC.
TABLE 4 Scatter Factor, f=B /B , as Related to the Shape
50 0.1
In this standard the mission time is synonymous with the B
Parameter, β, for the Weibull Distribution
p
β f β f β f
1.00 692.80 4.50 4.28 11.00 1.81 TABLE 5 Weibull Coefficient of Variati
...