1 Special Continuous Probability Distributions -Exponential Distribution -Weibull Distribution Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering

1

Special Continuous Probability Distributions-Exponential Distribution

-Weibull Distribution

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08


2

Exponential Distribution


3

A random variable X is said to have the ExponentialDistribution with parameters , where > 0, if the probability density function of X is:

, for 0

, elsewhere

x

e1

0

x )( xf

The Exponential Model - Definition


4

• Probability Distribution Function

for < 0

for 0

*Note: the Exponential Distribution is said to be without memory, i.e.

• P(X > x1 + x2 | X > x1) = P(X > x2)

P(X )(xFx

e

-1

0)x

x

x

Properties of the Exponential Model


5

• Mean or Expected Value

• Standard Deviation

)(XE

Properties of the Exponential Model


6

Suppose the response time X at a certain on-line computer terminal (the elapsed time between the end of a user’s inquiry and the beginning of the system’s response to that inquiry) has an exponential distribution with expected response time equal to 5 sec. The E(X) = 5=θ, so λ = 0.2.

(a) What is the probability that the response time is at most 10 seconds?

(b) What is the probability that the response time is between 5 and 10 seconds?

(c) What is the value of x for which the probability of exceeding that value is 1%?

Exponential Model - Example


7

The probability that the response time is at most 10 sec is:

)10( XP

865.0

135.01

1

)2.0,10()10)(2(.

e

F

)105( XP

233.0

)1()1(

)2.0;5()2.0;10(12

ee

FF

The probability that the response time is between 5 and 10 sec is:



8

The value of x for which the probability of exceeding x is 1%:

)( xXP

sec 025.232.0

605.4

)01.0ln(λx

01.0

99.01

x

x

e

ex

x



9

Weibull Distribution


10

•Definition - A random variable X is said to have the Weibull Probability Distribution with parameters and , where > 0 and > 0, if the probability density function of is:

, for 0

, elsewhere

Where, is the Shape Parameter, is the Scale Parameter. Note: If = 1, the Weibull reduces to the Exponential Distribution.

x

ex 1

0

x

x

)( xf

The Weibull Probability Distribution Function


11

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



12

for x 0

β

θ

x

e-1 xXPF(x)

F(t) for various and = 100

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

x

pro

bab

ilit

y, p

F(x)



13

• Derived from double logarithmic transformation of

the Weibull Distribution Function.

• Of the form

where

•Any straight line on Weibull Probability paper is a Weibull

Probability Distribution Function with slope, and intercept, - ln , where the ordinate is ln{ln(1/[1-F(t)])} the abscissa is ln t.

)/t(e1)t(Fbaxy

)t(F11lnlny

a tlnx lnb

Weibull Probability Paper (WPP)


14

Weibull Probability Paper links

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

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

Weibull Probability Paper (WPP)


15

F(x)in %

x

Cum

ulat

ive

prob

abili

ty in

per

cent

1.8 in.

1 in.

Use of Weibull Probability Paper


16

• 100pth Percentile

and, in particular

• Mean or Expected Value

Note: See the Gamma Function Table to obtain values of (a)

1

p)-ln(1- px

632.0 x

1

1 )X(E

Properties of the Weibull Distribution


17

• Standard Deviation of X

where

2

1

2 11

12

22 )()( aa



18

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

The Gamma Function


19

• Mode - The value of x for which the probability density function is masimum

i.e.,

1mode 11x

)(maxxmode xff

xmode

0

f(x)

x

Max f(x)=f(xmode)



20

Let X = the ultimate tensile strength (ksi) at -200 degrees F of a type of steel that exhibits ‘cold brittleness’ at low temperatures. Suppose X has a Weibull distribution with parameters = 20, and = 100. Find:

(a) P( X 105)

(b) P(98 X 102)

(c) the value of x such that P( X x) = 0.10

Weibull Distribution - Example


21

(a) P( X 105) = F(105; 20, 100)

(b) P(98 X 102) = F(102; 20, 100) - F(98; 20, 100)

930.0070.01120)100/105( e

287.0226.0513.0

2020 )02.1()98.0( ee

Weibull Distribution - Example Solution


22

(c) P( X x) = 0.10

P( X x)

Then

10.0120)100/( xe

90.020)100/( xe

90.0ln)100/( 20 x

90.0ln)100/( 20 x

20/190.0ln100/ x

20/190.0ln100 x

36.89x

Weibull Distribution - Example Solution


23

The random variable X can modeled by a Weibull distribution with = ½ and = 1000. The spec time limit is set at x = 4000. What is the proportion of items not meeting spec?



24

The fraction of items not meeting spec is

4000XP

That is, all but about 13.53% of the items will not meet spec.

1353.0

e

e

)4000(F1

)4000(P1

2

1000

40001/2

X



25

An Application of Probability toReliability Modeling and Analysis


26

• Figures of merit

• Failure densities and distributions

• The reliability function

• Failure rates

• The reliability functions in terms of the failure rate

• Mean time to failure (MTTF) and mean time between failures (MTBF)

Reliability Definitions and Concepts


27

• Reliability is defined as the probability that an item will perform its intended function for a specified interval under stated conditions. In the simplest sense, reliability means how long an item (such as a machine) will perform its intended function without a breakdown.

• Reliability: the capability to operate as intended, whenever used, for as long as needed.

Reliability is performance over time, probability that something will work when you want it to.

What is Reliability?


28

• Basic or Logistic Reliability

MTBF - Mean Time Between Failures

measure of product support requirements

• Mission Reliability

Ps or R(t) - Probability of mission success

measure of product effectiveness

Reliability Figures of Merit


29


30

“If I had only one day left to live, I would live it in my statistics class --it would seem so much longer.”

From: Statistics A Fresh ApproachDonald H. SandersMcGraw Hill, 4th Edition, 1990

Reliability Humor: Statistics


31

The Reliability of an item is the probability that the item willsurvive time t, given that it had not failed at time zero, when used within specified conditions, i.e.,

)tT(PtR

t

)t(F1dt)t(f

The Reliability Function


32

Relationship between failure density and reliability

tRdt

dtf

Reliability


33

Remark: The failure rate h(t) is a measure of proneness to failure as a function of age, t.

tF-1

tf

tR

tfth

Relationship Between h(t), f(t), F(t) and R(t)


34

The reliability of an item at time t may be expressed in termsof its failure rate at time t as follows:

where h(y) is the failure rate

t

0dy)y(ht

0

edy)y(hexp)t(R

The Reliability Function


35

Mean Time to Failure (or Between Failures) MTTF (or MTBF)is the expected Time to Failure (or Between Failures)

Remarks:

MTBF provides a reliability figure of merit for expected failure free operation

MTBF provides the basis for estimating the number of failures ina given period of time

Even though an item may be discarded after failure and its mean life characterized by MTTF, it may be meaningful tocharacterize the system reliability in terms of MTBF if thesystem is restored after item failure.

Mean Time to Failure and Mean Time Between Failures


36

If T is the random time to failure of an item, themean time to failure, MTTF, of the item is

where f is the probability density function of timeto failure, iff this integral exists (as an improperintegral).

0

dtttfMTTFTE

Relationship Between MTTF and Failure Density


37

Relationship Between MTTF and Reliability

0

dttRMTTFMTBF


38

Reliability “Bathtub Curve”


39

Reliability Humor


40

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

)t(f

0

The Exponential Model: (Weibull Model with β = 1)


41

• Weibull W(, )

, for t 0

Where F(t) is the population proportion failing in time t

• Exponential E() = W(1, )

t

e-1 )t(F

t

e-1 )t(F

Probability Distribution Function


42

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.

The Exponential Model


43

Probability Distribution Function• Weibull

• Exponential

t

e )t(R

t

e )t(R

Reliability Function


44

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

The Weibull Model - Distributions


45

Weibull

Exponential

MTBF

1

1 MTBF

Mean Time Between Failure - MTBF


46

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


47

• Weibull

and, in particular

• Exponential

1

P p)-ln(1- t

t 632.0

p)-ln(1- Pt

Percentiles, tp


48

• Failure Rate

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

an increasing function of t if > 1

• Cumulative Failure Rate

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

1-t )t(h

)t(ht )t(H

1-

Failure Rates - Weibull


49

• Failure Rate

• Note:

Only for the Exponential Distribution

•Cumulative Failure

1

)t(h

)t(H

rate failure

1MTBF

Failure Rates - Exponential


50

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

The Weibull Model - Distributions


51

Problem -

Four Engine Aircraft

Engine Unreliability Q(t) = p = 0.1

Mission success: At least two engines survive

Find RS(t)

The Binomial Model - Example Application 1


52

Solution -

X = number of engines failing in time t

RS(t) = P(x 2) = b(0) + b(1) + b(2)

= 0.6561 + 0.2916 + 0.0486 = 0.9963

The Binomial Model - Example Application 1


53

• Simplest and most common structure in reliability analysis.

• Functional operation of the system depends on the successful operation of all system components Note: The electrical or mechanical configuration may differ from the reliability configuration

Reliability Block Diagram

• Series configuration with n elements: E1, E2, ..., En

• System Failure occurs upon the first element failure

E1 E2 En

Series Reliability Configuration


54

• Reliability Block Diagram

•Element Time to Failure Distribution

with failure rate , for i=1, 2,…, n

• System reliability

where

tS

Se)t(R

SS

S θλ

1MTTF

is the system failure rate

• System mean time to failure

n

1iiS )t(

ii θE~T

E1 E2 En

ii θ

1λ

Series Reliability Configuration with Exponential Distribution


55

• Reliability Block Diagram Identical and independent Elements Exponential Distributions

• Element Time to Failure Distribution

with failure rate


tnS e)t(R

E1 E2 En

θE~T θ

1λ



56

• System mean time to failure

Note that /n is the expected time to the first failure, E(T1), when n identical items are put into service

nMTTFS



57

Parallel Reliability Configuration – Basic Concepts

• Definition - a system is said to have parallel reliability configuration if the system function can be performed by any one of two or more paths

• Reliability block diagram - for a parallel reliability configuration consisting of n elements, E1, E2, ... En

E1

E2

En


58

Parallel Reliability Configuration

• Redundant reliability configuration - sometimes called a redundant reliability configuration. Other times, the term ‘redundant’ is used only when the system is deliberately changed to provide additional paths, in order to improve the system reliability

• Basic assumptions

All elements are continuously energized starting at time t = 0

All elements are ‘up’ at time t = 0

The operation during time t of each element can be describedas either a success or a failure, i.e. Degraded operation orperformance is not considered


59


System success - a system having a parallel reliability configuration operates successfully for a period of time t if at least one of the parallel elements operates for time t without failure. Notice that element failure does not necessarily mean system failure.


60


• Block Diagram

• System reliability - for a system consisting of n elements, E1, E2, ... En

n

jiij

ji

n

1iiS )t(R)t(R)t(R)t(R

n

ii

nk

n

kjiijk

ji tRtRtRtR1

1 )()1...()()()(

if the n elements operate independently of each other and where Ri(t) is the reliability of element i, for i=1,2,…,n

E1

E2

En


61

System Reliability Model - Parallel Configuration

• Product rule for unreliabilities

n

iiS tRtR

1

)(11)(

•Mean Time Between System Failures

0

SS (t)dtRMTBF


62


s

p=R(t)


63

Element time to failure is exponential with failure rate

• Reliability block diagram:

•Element Time to Failure Distribution

with failure rate for I=1,2.


• System failure rate

t

t

S e2

e12)t(h

ttS eetR 22)(

E1

E2

θE~Ti θ

1λ

Parallel Reliability Configuration with Exponential Distribution


64

• System Mean Time Between Failures:

MTBFS = 1.5

Parallel Reliability Configuration with Exponential Distribution


65

A system consists of five components connected as shown.Find the system reliability, failure rate, MTBF, and MTBM if Ti~E(λ) for i=1,2,3,4,5

E1

E2

E3

E4 E5

Example


66

This problem can be approached in several different ways. Here is one approach:There are 3 success paths, namely,Success Path EventE1E2 AE1E3 BE4E5 C

Then Rs(t)=Ps= =P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC) =P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+

P(A)P(B)P(C) =P1P2+P1P3+P4P5-P1P2P3-P1P2P4P5

-P1P3P4P5+P1P2P3P4P5

assuming independence and where Pi=P(Ei) for i=1, 2, 3, 4, 5

)( CBAP

Solution


67

Since Pi=e-λt for i=1,2,3,4,5

Rs(t)

hs(t)

tttt

ttttt

tttttttt

ttttttttt

5λ-4λ-3λ-2λ-

λ-λ-λ-λ-λ-

λ-λ-λ-λ-λ-λ-λ-λ-

-λ-λ-λ-λ-λ-λ-λ-λ-λ

ee2e3e

))(e)(e)(e)(e(e

))(e)(e)(e(e-))(e)(e)(e(e-

))(e)(e(e-))(e(e))(e(e))(e(e

ttt

ttt

tttt

tttt

s

sdtd

e

tR

tR

3λ-2λ-λ-

3λ-2λ-λ-

3λ-2λ-λ-λ2

5λ-4λ-3λ-2λ-

ee2e3

e5e8e36λ

)ee2e3(

λe5λe8λe36e

)(

)(


68

MTBFs

0.87θ θ30

6151045λ5

1

λ2

1

λ3

1

λ2

3

λ5

e

λ2

e

λ3

e

λ2

3e

)(

0

5λ-4λ-3λ-2λ-

0

tttt

s dttR

θ2.0λ5

1SMTBM

Documents

1 Special Continuous Probability Distributions -Exponential Distribution -Weibull Distribution Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering