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1 Stracener_EMIS 7305/5305_Spr08_01.29.08 Reliability Models & Applications Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis Systems Engineering Program Department of Engineering Management, Information and Systems

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Page 1: Stracener_EMIS 7305/5305_Spr08_01.29.08 1 Reliability Models & Applications Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7305/5305

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Reliability Models & Applications

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7305/5305Systems Reliability, Supportability and Availability Analysis

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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Reliability Model

• Mathematical Model: a set of rules or equations which representthe behavior of a physical system or process

• Reliability Math Model: a mathematical relationship whichdescribes the reliability of an item in terms of itscomponents and their associated failure rates.

• Objective: to represent the real world by expressing physicalor functional relationships in mathematical form

• Approaches:AnalyticalSimulation

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Comparison of Reliability Modeling Approaches

TYPE MODEL

Analytical Monte Carlo

Advantages

A. Gives exact results (given the assumptions of the model).

B. Once the model is developed, output will generally be rapidly obtained.

C. It need not always be implemented on a computer; paper analyses may suffice.

A. Very flexible. There is virtually no limit to the analysis. Empirical distributions can be handled.

B. Can generally be easily extended and developed as required.

C. Easily understood by non-mathematicians.

Disadvantages

A. Generally requires restrictive assumptions to make the problem trackable.

B. Because of (A) it is less flexible than Monte Carlo. In particular, the scope for expending or developing a model may be limited.

C. Model might only be understood by mathematicians. This may cause credibility problems if output conflicts with preconceived ideas of designers or management.

A. Usually requires a computer; can require considerable computer time to achieve the required accuracy.

B. Calculations can take much longer than analytical models.

C. Solutions are not exact but depend on the number of repeated runs used to produce the output statistics. That is, all outputs are ‘estimates’.

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Reliability Modeling Development Stages

• System AnalysisModel output requirementsInput data requirementsStrike a balance between model complexity and accuracy

• Model DevelopmentFormulate the problem in mathematical termsSpecify assumptions and ground rulesDetermine modeling approach; analytical or simulation

• Model ValidationUse test cases to validate the mathematical formulationPerform sensitivity analysis to confirm realismIdentify modeling constraints

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Reliability Model Selection

• Model selection is basic to reliability modeling and analysis

• Selection criteria should be based onPhysical lawsExperienceStatistical goodness of fitSystem configuration and complexityType of analysis

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Reliability Models:

• Time to Failure Models:The random variable T is time to (or between) failure.

• Number of Successes Model:The random variable X is the number of successes thatoccur in N trials.

• Number of Failures Model:The random variable Y is the number of failures that occurin a period of time, t.

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Statistical Models

• Time to Failure Models:

The Exponential Model The Weibull ModelThe Normal (or Gaussian) ModelThe Lognormal Model

• Discrete Event Models:

The Binomial ModelThe Poisson Model

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The Exponential Model:

DefinitionA random variable T is said to have the ExponentialDistribution with parameters , where > 0, if the failure density of T is:

, for t 0

, elsewhere

t

e1

0

f (t) =

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The Exponential Model:

RemarksThe Exponential Model is most often used in Reliability applications, partly because of mathematical convenience due to a constant failure rate.

The Exponential Model is often referred to as the Constant Failure Rate Model.

The Exponential Model is used during the ‘Useful Life’ period of an item’s life, i.e., after the ‘Infant Mortality’period before Wearout begins.

The Exponential Model is most often associated withelectronic equipment.

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Properties of the Exponential Model:

• Probability Distribution Function

• Reliability Function

• MTBF (Mean Time Between Failure)

t

e-1 )t(F

t

e )t(R

MTBF

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Properties of the Exponential Model:

• Standard Deviation of Time to Failure:

• Failure Rate

• Cumulative Failure Rate

• P(T > t1 + t2 | T > t1) = P (T > t2), i.e.,

the Exponential Distribution is said to be without memory

1

)t(h

)t(H

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The Exponential Model - Example

1000

1000

1)(

t

etf

3

1

1000

1000

10x368.0

1000

1000)000,1(

e

ef

θ=1,000 Hours

Failure Density

0 1,000 2,000 3,000

f(t)

1000

1

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1000)(t

etR

368.0

)1000(1

1000

1000

e

eR

θ=1,000 Hours

Reliability Function

0 1,000 2,000 3,000

R(t)

1.0

0.8

0.6

0.4

0.2

0

0.3679

0.1353

0.0498

The Exponential Model – Example (continued)

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The Weibull Model:

• DefinitionA random variable T is said to have the Weibull Probability Distribution with parameters and , where > 0 and > 0, if the failure density of T is:

, for t 0

, elsewhere

• Remarks is the Shape Parameter is the Scale Parameter (Characteristic Life)

t

et 1

0

f (t) =

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The Weibull Model - Distributions:

Probability Density Functionf(t)

t

t is in multiples of

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

β=0.5

β=5.0

β=3.44

β=2.5β=1.0

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

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Properties of The Weibull Model:

• Probability Distribution Function

, for t 0

Where F(t) is the Fraction of Units Failing in Time t

• Reliability Function

t

e-1 )t(F

t

e )t(R

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The Weibull Model - Distributions:

Reliability Functions

R(t)

t

t is in multiples of

β=5.0

β=1.0

β=0.5

1.0

0.8

0.6

0.4

0.2

00 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

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The Weibull Model - Weibull Probability Paper (WPP):

• Derived from double logarithmic transformation of the Weibull Distribution Function.

• Of the form

where

• Any straight line on WPP represents a Weibull Distribution with

Slope = & Intercept = - ln

)/t(e1)t(Fbaxy

)t(F11lnlny

a tlnx lnb

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The Weibull Model - Weibull Probability Paper (WPP):

Weibull Probability Paper links

http://engr.smu.edu/~jerrells/courses/help/resources.html

http://perso.easynet.fr/~philimar/graphpapeng.htm

http://www.weibull.com/GPaper/index.htm

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Use of Weibull Probability Paper:

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Properties of the Weibull Model:

• p-th Percentile

and, in particular

1

P p)-ln(1- t

t 632.0

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1

1

MTBF

• MTBF (Mean Time Between Failure)

•Standard Deviation

2

1

2 11

12

Properties of the Weibull Model:

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The Gamma Function

0

1ax dxxe)a(

)a(a)1a(

y=a (a) a (a) a (a) a (a)1 1 1.25 0.9064 1.5 0.8862 1.75 0.9191

1.01 0.9943 1.26 0.9044 1.51 0.8866 1.76 0.92141.02 0.9888 1.27 0.9025 1.52 0.887 1.77 0.92381.03 0.9836 1.28 0.9007 1.53 0.8876 1.78 0.92621.04 0.9784 1.29 0.899 1.54 0.8882 1.79 0.92881.05 0.9735 1.3 0.8975 1.55 0.8889 1.8 0.93141.06 0.9687 1.31 0.896 1.56 0.8896 1.81 0.93411.07 0.9642 1.32 0.8946 1.57 0.8905 1.82 0.93691.08 0.9597 1.33 0.8934 1.58 0.8914 1.83 0.93971.09 0.9555 1.34 0.8922 1.59 0.8924 1.84 0.94261.1 0.9514 1.35 0.8912 1.6 0.8935 1.85 0.9456

1.11 0.9474 1.36 0.8902 1.61 0.8947 1.86 0.94871.12 0.9436 1.37 0.8893 1.62 0.8959 1.87 0.95181.13 0.9399 1.38 0.8885 1.63 0.8972 1.88 0.95511.14 0.9364 1.39 0.8879 1.64 0.8986 1.89 0.95841.15 0.933 1.4 0.8873 1.65 0.9001 1.9 0.96181.16 0.9298 1.41 0.8868 1.66 0.9017 1.91 0.96521.17 0.9267 1.42 0.8864 1.67 0.9033 1.92 0.96881.18 0.9237 1.43 0.886 1.68 0.905 1.93 0.97241.19 0.9209 1.44 0.8858 1.69 0.9068 1.94 0.97611.2 0.9182 1.45 0.8857 1.7 0.9086 1.95 0.9799

1.21 0.9156 1.46 0.8856 1.71 0.9106 1.96 0.98371.22 0.9131 1.47 0.8856 1.72 0.9126 1.97 0.98771.23 0.9108 1.48 0.8858 1.73 0.9147 1.98 0.99171.24 0.9085 1.49 0.886 1.74 0.9168 1.99 0.9958

2 1

Values of theGamma Function

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Properties of the Weibull Model:

• Failure Rate

Notice that h(t) is a decreasing function of t if < 1a constant if = 1an increasing function of t if > 1

1-t )t(h

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Properties of the Weibull Model:

• Cumulative Failure Rate

• The Instantaneous and Cumulative Failure Rates, h(t) and H(t), are straight lines on log-log paper.

• The Weibull Model with = 1 reduces to the Exponential Model.

)t(ht )t(H

1-

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The Weibull Model - Distributions:

Failure Rates

h(t)

t

t is in multiples of h(t) is in multiples of 1/

3

2

1

0

0 1.0 2.0

β=5

β=1

β=0.5

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Properties of the Weibull Model:

• Conditional Probability of Surviving Time t2, given survival to time t1, where t1 < t2,

, if > 1

• Mode - The value of time (age) that maximizes the failure density function.

12112

ttettR

1emod 11t

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The Weibull Model - Expected Time to First Failure:

• n identical items are put on test or into service under identical conditions and at age of zero time. The failure distribution of each item is Weibull with parameter and .

• The expected time to first failure is:

• The expected time to second failure is:

E(T)n

1/11/)(

β1

1 nTE

1

1

n

1n

1n

n)T(E

/112

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The Weibull Model

WEIBULL FIT - A

Based on 2failures (Excludes128 hr failure)

Based on 3failuresWEIBULL FIT - B

New SpacersNew Engine Build - Undisturbed

New SpacersNew Engine Build - UndisturbedPercent

Survivingat 1Flight Hours

Age (Flight Hours)0 100 200 300 400 500 600 700 800 900

100

99

98

97

96

95

Turbine Spacer Life Expectancy

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The Weibull Model – Example 1

Time to failure of an item has a Weibull distribution with characteristic life = 1000 hours. Formulate the reliability function, R(t), and the Failure Rate, h(t), as a function of time (age) and plot, for:

(a) = 0.5

(b) = 1.0

(c) = 1.5

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The Weibull Model – Example 1 Solution

(a)

(b)

5.0)1000/()( tetR

t

tth

/01581.0

)1000/5.0()( 5.0

1000/te)t(R

001.0

1000/1)(

th

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5.1)1000/t(e)t(R

t

tth

000047434.0

)1000/5.1()( 5.05.1

(c)

The Weibull Model – Example 1 Solution continued

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The Weibull Model – Example 2

Time to failure of an item follows a Weibull distribution with = 2 and = 1000 hours.

(a) What is the reliability, R(t), for t = 200 hours?

(b) What is the hazard rate, h(t), (instantaneous failure rate) at that time?

(c) What is the Mean Time To Failure?

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The Weibull Model – Example 2 Solution

(a)

(b)

9608.0

)200(2)1000/200(

eR

1t

)t(h

/te)t(R

0004.0

1000

2002)200( 2

12

h

failures per hour

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The Weibull Model – Example 2 Solution continued

(c)

From the Gamma Function table:

2

111000

11

MTTF

23.886

5.11000

MTTF

88623.0)5.1(