Weibull and Exponential Distributions

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RELIABILITY ENGINEERING UNIT

ASST4403

Lecture 8: WEIBULL AND EXPONENTIAL DISTRIBUTIONS

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earn ng outcomes

Familiarity with the properties of Weibull and

exponential distributions

Ability to find characteristics of a random variablewith Weibull distribution

Ability to find characteristics of a random variablewith ex onential distribution

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: shape parameter1 t

t

: characteristic life

(scale parameter)

e

(Hyper-exponential)

f(t)

=3.5 xponen a

(Normal)

(Rayleigh)

3t


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What is behind

Used first to model fatigue/s reng

An extreme value distribution

(the smallest extreme value)

Most flexible

(Waloddi Weibull:1887-1979, Swedishengineer)

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: shape parameter1 tt

: c arac er s c et e

t

1( ) 1 0.632F t e

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Weibull Distribution

0.8

0.9

, c arac er s c e: me a w c

cumulative probability = 63.2%

0.6

0.7

0.3

0.4

.

63.2%

0.1

0.2

t0

Note: when =1 =

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: shape parameter1

1

The mean value

2

2 2 2 1 The standard

deviation

The median (the random variable has 50% chance of being

lower and 50% greater than this value) /15.0 )5.0ln( mediantt

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Gamma function

,

Otherwise the value (z) can be found in a table

o e a so or a z

Example: (4)=(4-1)!=3!=321=6 2.6 =1.42962

(3.7)=(2.7+1) =2.7 (2.7)= 2.7 1.54469=4.1707 8


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0.

0.

ame growing

0.

0.Same &growing

0.

0.

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Effect of and on the shape of the Weibull pdf


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3-Parameter

We u str ut on

t1

etf )(

1f(t)

f(t)

0

63.2%

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Weibull distribution : examples

A weibull distributed random variable has shapeparame er an sca e parame er , r. n

The mean an ar eva on The median The robabilit the random variable takes a value

between 800 and 1,200 hours The time at which the random variable has 90%

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The mean)

11(1000)

11(

hr23.88688623.01000)5.1(1000

)2

11()11(1000)

11()

21(

22

The median hr25.46388623.011000 2

hrtt median 55.832)5.0ln(1000)5.0ln(2/1/1

5.0

e pro a y e ran om var a e a es a va uebetween 8,000 and 24,000 hours

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e me a w c e ran om vara e as

chance to be greater than


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The probability the random variable takes a value,

)800()1200()1200800Pr(

228001200

FFT

2904.011 10001000

ee

e me a w c e ran om vara e aschance to be greater than

tTtT

t

1.0)Pr(9.0)Pr(

1000

9.09.0

2

9.0

14

... 9.09.0


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Exponential Distribution

A special case when =1 in Weibull distribution Very important in reliability engineering

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Exponential Distribution

1

f(t)

0.8

0.9

0.5

0.6

.

f(t)

0.3

0.4

0.1

0.2

0

t, time

16

t

etf

)(


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x

t 0, 0( ) ,t

tf t e

( ) { }F t P X t

01

tx te dx e

63.2% = t17


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t Q: Why 63.2%?1( ) 1 1 0.632t

tF t e e

1 mean

Standard1

63.2%

= t18


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Exponential distribution : examples

Redo the example for Weibull distributed randomvara e w s ape parame er an sca eparameter 1,000 hr.

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Exponential distribution :

examp es The mean)

11(1000)

11(

hr100011000)2(1000

)1

11()21(1000)

11()

21(

22

The median hr1000!1!21000

2

hrtt median 15.693)5.0ln(1000)5.0ln(1/1/1

5.0

e pro a y e ran om var a e a es a va uebetween 8,000 and 24,000 hours

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e me a w c e ran om vara e as

chance to be greater than


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Exponential distribution :

examp es The probability the random variable takes a value,

)800()1200()1200800Pr(

8001200

FFT

1481.011 10001000

ee

e me a w c e ran om vara e aschance to be greater than

tTtT

t

1.0)Pr(9.0)Pr(

1000

9.09.0

9.0

... ..

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Weibull and Exponential Distributions