8/16/2019 Ebeling Ch2-3

1/18

(

II A

It

 I

J<

H 2

 

4iii; i t

t

, ~ j . , , ' M I I t f . I : ~ ~ ' I ' : . . .

l hc Failure

Distribution

Thl initial focus of this book is on lhe developme nt of mathematical models that de

~ n i h e

the reliability of components and systems. This chapter develops four related

plOhability functions: the reliability function, the cumulative distribution function,

11ll probability density function, and the hazard rate function.

Each

of these func

I ions can be used

to

compute reliabililies, but they offer four different perspectives.

Specifying anyone of these functions will uniquely and completely characterize the

lailure process. Various summary measures

of

reliability, such as the mean time to

failure, the variance of the failure distribution, and the median time to failure, may

then be determined. Subsequent chaplers will define several common and highly

useful theoretical reliability models.

2.1

THE RELI BILITY FUNCTION

Reliability is defined as the probability that a system component) will function over

-;ome time period t. To express this relationship mathematically we define the contin

uous random variable to be the time

to

failure of the system component);

O.

Then reliability can be expressed as

R t) = Pr{T t}

2.1 )

where R t) 2: 0, R O) =

1,

and limHoo

R t) O.

For a given value of

t R t)

is the

probability that the time to failure is greater than or equal

to t

If we define

F t) R t)

= Pr{T

<

t}

2.2)

23

8/16/2019 Ebeling Ch2-3

2/18

---------------------

ARl I: Bask

Reliability Models

where

£(0)

= 0

and

l imF t) =

I

I

'"

then F t) is the probability that a failure occurs before time I.

We will refer to R(t) as the reliability function and

F(t)

as the cumulative dis-

tributionfunction COP) of

the failure distribution.

A

third function, defined by

f(t) = dF(t) = _ dR(t)

2.3)

dt dl

is called the

probability density fun tion

(PDP).

This

function describes the shape

of

the

failure distribution. These three functions are illustrated in Pig. 2.1.

The

PDP,

I(t).

bas these two properties:

oo

f(t) : :

0 and

f(t) dt

= I

Given the PDP, f(t), then

F(t) = f(t ) dt

(2.4)

and

R(t) = f f(t ) dt

2.5)

In other words, both the reliability function and the COP represent areas under the

curve defined

by f(t).

Therefore. since the area beneath the entir e curve is equal to

one, both the reliability and the failure probability will be defined so that

o s; R(t) s;

lOs F t)

s; 1

The function

R(t)

is normally used when reliabilities are being computed, and the

function

F t)

is normally used

when

failure probabilities are

being

computed. Graph

ing the PDP,

f(t),

provides a visual representation of the failure distribution.

EXAMPLE 2.1. Given the following PDP for the random variable T, the time (in op

erating hours) to failure of a compressor, what is its reliability for a lOO-hr operating

life?

Solution

0.001

[ I) =

ci0 OOlt +

1)2

{

otherwise

Solution

R f' 0.001 d I

1

00

t) t (O.OOlt + 1)2 t '= (O.OOlt + 1) t O.OOlt +

1

I O.OOlr

and

F t) = 1 -

R t)

= I

O.OOlt

+ I 0.0011 + 1

hen

ROOO) =

0.1

1

+ I = 0.909

1M

a)

1"(1)

-

III -

 

-

 b)

( I)

Area

=

1.0

e)

FIGURE

2.1

a) The reliability function. b) The cumulative distribution function.

c)

The probability density function.

8/16/2019 Ebeling Ch2-3

3/18

26 "AM'!' I. BusicReliuhililyModels

A designlifeisdefinedto

be

thetimeto failure tR thatcorresponds to aspecifiedrehabi!

ityR. ThatisR tR) =

R.

Tofindthedesignlifeifareliability

of

0.95isdesired,wesel

1

R tR) = =0.95

Solvingfor

td,

tR 1 0 0 0 0 . ~ 5 - 1 52.6hr

Theprobability

of

afailureoccurringwithinsomeinterval

of

time

[a,

b) maybe

foundusingany

of

thethreeprobabilityfunctions,since

Prf a T

s

b}

=

F(b) - F(a)

=

R(a) R(b)

r

f ( t )dt 2.6)

Fromthepreviousexample,

Pr{

10

s

T

s lOO}

=R(lO) R lOO)

0.1 + I = 0.081

2.2

MEANTIMETOFAILURE

Themeantimetofailure(MTTF)isdefinedby

'

MTTF=E T) =

tf(t)dt

(2.7)

whichis themean,orexpectedvalue,

of

theprobabilitydistributiondefinedby f 1).

It canalsobeshown(Appendix2A)that

MTTF=tOO R(t) dt 2.8)

Equation(2.8)isofteneasierto applythan(2.7).

Themeanof thefailuredistributionisonlyoneof severalmeasures

of

central

tendencyof thefailuredistribution.Anotheristhemediantime to failure,definedby

R tmed)

=

0.5

=

Prf T ;;:::

tmed}

(2.9)

Themediandivides thedistributioninto two halves, with50 percentof the fail

uresoccurringbeforethemediantimeto failureand

50

percentoccurringafterthe

median.Themedian maybepreferredto themeanwhen thedistributionishighly

skewed.

Athirdfrequentlyusedaverageisthemode,ormostlikelyobservedfailure

time,definedby

f(tmode) = max f(t)

(2.10)

OSI

8/16/2019 Ebeling Ch2-3

4/18

Ind (or R2(t),

R2(400) = l ( X ~ ~ 4 0 0 ==

0.60

Obviously,

the

MITF

alone win not uniquely characterize a failure distribution.

Other measures are necessary. One measure that is often used to further describe

a

failure distribution is its variance (1'2, defined

by

(1'2

=

L"'(t-MITF)2f(t)dt (2.11)

The variance represents an average squared distance a failure time will be from the

MITF.

It is a measure

of

the spread, or dispersion,

of

the failure times about the

mean.

It

is shown in Appendix

2B

that the variance can also be written as

(1'2

=

fa' t

2

f(t)dt - (MTTF)2 (2.12)

Equation

(2.12)

is computationally simpler than the definitional Eq.

(2.11).

\

The

square root of the variance is the standard .deviation. Since the standard deviation

will be measured in the same time units as the random variable, T, it. is

easier

to

interpret than the variance (which is m easured in square units).

EXAMPLE

l.4.From the first failure distribution in Example 2.3,

we

have

2

17 ==

fo' t

2

(0.002e-

o

.

O2J

)dt

-

(500)2

=

250,000

and 17

=

500.

From the second. with f(t)

=

-dR(t)Jdt = 111000,

17

2

fOOO 2 l ~

)dt -

(500f

1000

3

1

.

= 0 -

(500)2

=

83,333

and

17 =

288.67.

Therefore, although their

MTIFs

are identical, they have considerably different

standard deviations. from which we would conclude that their reliability distributions

should

be

inherently different. We would generally prefer the distribution having the

smaller variance. Why?

HAZARD RATE FUNCTION

In addition to the probability functions defined earlier, another function, called the

failure rate

or

hazard rate Junction,

is often used

in

reliability. It provides an instan

taneous (at time t) rate of failure. From Eq. (2.6),

Pr{ t

s

T

:::.:;

t

+

Ill} = R(t) - R(t

+

Ilt)

an(.!

the conditional probability of a failure in the time interval from

t

to t

+

Ilt given

t h a ~ the system nas survIved to time t is

Pr\

t

s

l

s

t + At 1T 2: I -

R(I)

  t ~ . -

Thon

R(t) - R(t +

Ilt)

R(t) III

III the conditional probability

of

failure per unit

of

time (failure rate).

Set

A(O = lim =[R(t

+

Ilt)

- R(tli .

_1_

Al

-..0

Ilt R(t )

(2.13)

-dR(t) 1 f(t)

  ;

---;rr .

R(t)

=::

R ij

Then

A(t) is known as the instantaneous hazard rate

or

failure rate function.

The

I'llilure rate function A(t) provides an alternative way

of

describing a failure distribu

(ion. Failure rates

in

some cases may be characterized as increasing (IFR), decreas

ing DFR), or constant

eFR)

when A I) is an increasing, decreasing,

or

constant

funclion.

A

particular hazard rate function will uniquely determine a reliability function.

To see this, let

-dR(t)

1

A(t)

dt

,-, . R(t):

-dR(t)

A(t)dt

==

I f ( i )

or

lntegrating,

t

,  

JR t)

: S ..')

J

A(t )dt - I R(t')

1 establishes the lower limit

in the

integral on

the

right-hand side.

where R O)

==

Then

-L(t )dr =

In

R(t)

(2.1,t)

.ot

R(I)

exp

l

L

(I')

d

t

1

Equation (2.14) can then

be

used to derive the reliability function from a known

hazard rate function.

EXAMPLE 2.5. Given the linear hazard rate function

A(t)

= 5

X

O 6

t

where t is mea

sured in operating hours, what is the design life

if

a 0.98 reliability is desired?

Solution

R(t) P

l l:

5

x

W .

d 1

«P

[ - 2.5 x

\0"','

1

0.98

2.3

8/16/2019 Ebeling Ch2-3

5/18

J

In 0.98 = 89.89' 90

hr

or

10 98

=

umulative and average failure rate

The cumulative failure rate over a period

of

time t is defined

by

L(t)

=

f; A(t')

dt

(2.15)

A related and useful concept is the average failure rate, defined between two times

tl and

t2:

AFR(tl,

t2)

1

J'

2

A(t )dt =

InR(tI) - nR( t2)

(2.16)

t2 -

tIl l

t2 - t)

FromEq.(2.l4)andlettingtl

=

Oandt2

=

T,Eq.(2.l6)maybewritten

AFR(t) =

In R O) - In

R t) =

--

In R t)

L t)

since In

R O)

In

1

= O.

If

AFR t)

is a nondecreasing function, the failure distribution is characterized

as having an increasing failure rate average (IFRA). I f the function is nonincreasing,

the distribution has a DFRA. Obviously IFR (DFR) systems are also IFRA (DFRA),

but the Converse is not necessarily true.

EXAMPLE 2.5 (CONTINUED). The cumulative failure rate is given by

L(t) = {' 5

x

to-

6

t dt 2.5

x

lO6 (2

Jo

.

and the average failure rate from time 0 to t is

6

2

AFR t) 2.5

X

10--

t =

2.5 x 10-

6

t

t

EXA

MPLE

2.6. A component has a reliability function gi ven by

t

2

R(t) 1 - 2 for 0 t a

a

where a is a parameter of the distribution representing the component's maximum life.

Then

2

f(t) = 2t

A(t) = 2tla = _2_t_

and

forO t a

a

2

t

2

 a

2

t

2

)l a

2

2 3

t ) t IQ

= aMTTF

=

i

Q

 

1 - - dt =

t

-

a

2

o 3a

2

0 3

t

2

m ~ d = 0.5; tmed

"r0.5a

2

0.707a

a

AFR t)

=

-In(1

-

t

2

1a

2

)

t

The average failure rate up

to

the MTTF is

AFR(MTTF) = AFRG a =

In 1

-  4I:1

a

)/

  ~ a )

= 0.8:17

).(1)

Wearou[

Uscfullifc

Hum-in

I ;

do

m failures I ;

Ran

I

Early \

I

'"

...  failures

I

:_ Wearout

:

- - , f ,l i, re '

• •

I

 

............................

_.;

  ..........

FIGURE 2.3

The bathtub curve.

2 4

n THTUB URVE

All important form

of

the hazard rate function is shown in Fig. 2.3. Because of its

shape, it is commonly referred to as the bathtub

curve.

Systems having this haz

ard rate function experience decreasing failure rates early in their life cycle (infant

mortality), followed by a nearly constant failure rate (useful life), followed by an

increasing failure rate (wearout). This curve may

be

obtained, as shown later, as a

composite of several failure distributions, or as shown in the following example, as

u function

of

piecewise linear and constant failure rates. Table 2.1 summarizes some

lit' the distinguishing features of the bathtub curve.

EXAMPLE 2.7. A simplified form ofthe bathtub curve is based upon lin ear and constant

hazard rates:

o

t Co

Cit + A

CI

Co

< t to

CI

A t)

C2 t

- to)

+

A

to <

t

Then

o t s Co

exp - (co + A)t Cj(P/2)j

CI

Co

< t

to

c2 )

CI

R(t)

= \ exp At

+

exp [( {)(t - to> + At + ] t < f

where Co' CI,

C2.

and to are constants to

be

determined. Figure 2.4 portrays this hazard

rate graphically. Appendix 2D derives R(t) from A(I).

8/16/2019 Ebeling Ch2-3

6/18

TA

B t

It

2. '

The bathtubcurve

Characterizedby

Causedby Reducedby

Bum-in

DFR

Manufacturingdefects: Burn-intesting

weldingflaws,cracks, .

Screening

defective parts,

Quality control

poorqualitycontrol, Acceptancetesting

contamination,poor

workmanship

Usefullife

CFR

Environment

Redundancy

Random loads Excessstrength

Human error

"Acts of God"

Chanceevents

Wear-out IFR

Fatigue

Derating

Corrosion Preventivemaintenance

Aging

Partsreplacement

Friction Technology

Cyclicalloading

2.5

CONDITIONALRELIABILITY

Conditionalreliabilityisusefulindescribingthereliability

of

acomponentorsystem

followingabum-inperiod

To

orafterawarrantyperiodTo. Wedefineconditional

reliabilityasthereliabilityof asystemgiventhat

it

hasoperatedfortimeTo:

5

4

3

2

2.41 3.21 4.01

4.81 5.61

Time

FIGURE2.4

Apiecewiselinearbathtubcurve.

R(I\

To PI (

7'

,.

1'0

+ t iT ;> Tol

R(To + t)

Prl

T:>To+

t\

- ~ , , - :

R(To)

Pr \T>To

\

(2.17)

:::

ex

p

[ -  OTo+t

A(tl)dtIJ

ex [

ro A(t )dt ]

"'+  ::" W)dt 1

p

EXAMPLE 2.8.

Let

-0.5

0.5

t

t inyears

A t) = 1000Cooo

)

whichis \l. DFR.Thenfrom

Eq.

(2.14),forareliabilityof

0.90

t 112

R(t) == exp - 1000

=

0.90

(

andthedesignlifeisfoundfrom

t

== lOOO(-1nO.90)2 ILl yr

If

we letTo = 0.5,asix-monthbum-inperiod,then

R IT

=R(t

+ 0.5)

t

0

R 0.5)

=

exp- I +

0 . 5 ) 1 H ~ 1 ° . 5 _

0.90

exp- (0.511

oooj'

r  05 \ . 51

t

=

lOOOl-lno.90+ yX {» -0.5

=

~ s y r

and

Thisis

an

increaseofover4yearsinthedesignlifeasaresultofasix-monthbum-in

period.This improvementin reliabilityresultingfrom a bum-in periodTo willonly

be realized fora DFR asillustrated in the following example and shown in Ap

pendix2C.

EXAMPLE 2.9.

Let

A(t)

=

At,

an

IFR

for

A> O.

Then

R(t)

= e-(lI2)M

2

-(lJ2»).

(1+

TO)2

e

R(t \To) = -(II2)Uij

and

e

which.canbesimplifiedto

2

R(t I

To). ;:;

e -

ATo

'e-(II2)Al

Sinceexp(_ ATo t for)..> 0isadecreasingfunctionof To. theconditionalreliability

willdecreaseasthebum-inperiod

To increases.

http:///reader/full/lOOOl-lno.90http:///reader/full/lOOOl-lno.90http:///reader/full/lOOOl-lno.90

8/16/2019 Ebeling Ch2-3

7/18

EX M1'1.E Z. 10. Forthl.' reliabilityfunction giyeninExample2.6.thecondit IlInlll ..l:'Ii-

abilityis

Nit

I

To) = I

(t +

To)2/a

2

a

2

-

(t

+

I

T51a2 T2

o

2

Residual

MTTF

Since R(t ITo) isareliabilityfunction,aresidualMTTF maybeomained from

x R( t ) . ,

I

MTTF(To) r R(t ITo)dt =

R(T ) dt = R(To)

I R(t')

dt' (2.18)

.10

o

where

t'

=

I

+

To.

For

thoseunitshavingsurvivedtotime

To,

Eq.(2.18)determines

theirmeanremaininglifetime.Forcomponentshavingan IFR (DFR). one would

expect

the

MTIF(To) to

beadecreasing(increasing)functionof

To.

asshowninthe

followingexamples.

EXAMPLE 2.11. Thereliabilityt"unctionR t) IJ

I)lbforO:s

t :s

bandzemelse-

wherehasanlFR(seeExercise

2.8).

lis residual MTTF isgivenby

h b 1')2

(b To)

MTTF(T

o

)   ~ h ~ O )

- To

forO

°

saparameter(constant)of thedistribution,has[hehazardrate lun,:

;'

..,

ACt =

a

+ f

whichisdecreasing. I'heresidualMTTF is

-a

2

. T 2 I ·

,

MTTF(T )

(a

T 0)

dt'

a

t

To

o

To

a

+ t ):

a + t'

whichhastheinterestingproperty!!lattheresidualmeanincreaseshytheamount

of

the

currentage.If

To

= 0,theunconditionalmean,MTTF =a, isohtained.

2.6

S(JMMARY

A

failureprocess.rep resented bythe random variableT (time tofailure),may be

1Illi'luely

characterizedbyanyof thefollowingfourfunctions:

I. lUI. theprobahilitydensityfunction(PDF)

2. ' (1). the

cumulative

distributionfunction

(CDF)

J.

/((1),

thereliabilityfunction

4. ).(1),thehazardratefunction

~ ~ ~ ~ ~

 

. .

Tht

following

relutionllhlllM hili".

R(t) =

r'

f(t') dt '

'.'(1) \./ f(t ' )dt '

.0

f(t) = -dR(t ) = dF(t)

R(t) =1 - F(t) dt

A t)

=  J l

R t)

R(t) exp [ i: A(t')dt'l

MTIF

J:

R(t)dt

oo

(T

2

t

2

f(t) dt (MTTF)2

R(t I To)

!!(t + T o ~

R(To)

L(t) = 1/ A(t') dt'

o

12 A(t)dt In In

R(h)

AFR(t\.

t2)

t2

- t\

t2

-

tl

R(/') dt '

MTTF(To)

= R(To)

APPENDIX2A

DERIVATION

OF

EQUATION(2.8)

By

definition.

MTTF = t

f(t)

dt

FromEq.(2.3),

MTIF J oo dR(t)

-T t

tdt

o

Using integrationby parts,

MTTF -tR(t)1' +

foo

R(n

d

!

(2.8)

10 0

MTIF =

J:

R(t)

dt

· ---- ''''''',T

...... ~ , . .

8/16/2019 Ebeling Ch2-3

8/18

~ l I l c e

lim

t

R(t) = lim

te x

p

[

rA(t')dt'j

0

I-+::n l - ~ Jo

and

O'R(O)

==

0

APPENDIX 28

DERIVATION OF EQUATION

(2.12)

[J.2

f (t -MITFi

J(t)dt

== LX> [p

-

  t

.

MITF +

(MITFi]

J(t) dt

2

r t

f( t)dt 2MTTFJ: t f t )d t

+ (MTIF)'

J: f(t)dt

=

10'

P

J(t)dt

- 2(MITF)2

+

(MTTF)2

since

LX> tJ(t)dt == MTTF

and

LX> JCt)dt

I

Therefore

(T2 == L t

2

J(t)dt

- (MITF)2

(2.12)

APPKNUfX 2C

CONUI'f'IONAL RELIABILITY

ND

ItAltURE RATES

_1Iq, (2.17),

R(T, + t) [(t+TO 1

H(t I To)

R(°"j;;) ==

exp -

JTo

.

A(t

'

)dt'

l'IIAI'II'1l ."

'I'hl:

1101111111:

l l i ~ l l h l l l l o l l 1'1

d IW

I

To) I J J I

j

d

(: t

Mt')

dr'

j

IIWH,ton' /' , cxp

I

AU

I

)dr /'"

l 10

1' (

0

N(t ITo)

AU

+ To) + A(To)]

III o,der

r r

the reliability

to

improve as a function of

To,

the derivative, or slope.

nl

I\(

r I 'I 'l l ) with respect to Ttl must be positive. From the above result, this will occur

I I l Ih II A(fl l ) .

AU

i {o). ln

other words, the failure rate must be decreasing,

r\PIJENDIX 2D

INTI':R!\lEDIATE CALCULATIONS

!iOa TilE

LINEAR

BATHTUB

ClJRVE

•.

__ _ . u ; ' ; ' ~

h" r .

rT

R(t\ =-

exp

[

Jo

A)dt']

2

exp-(cot

-

2

l

t

+

[

hlll ' I l < I "

'II:

R(I)

R cxp

(

('1)

c, .

A

(

xp -

( I

+

j

co)

  cxp

CI

(

At Aco)

CI

cxp

(At + 2('1 )

For

to

t:

' [c2(t' (0) + A1dlt1

= exp - (Ato

+

) exp

;

r 10)2

+ At

- Ato)

R(t) R(lo)exp

C1

EXP -

i (t ,

+ At 2e )

(

l

M n t I ~ i

. .

......

........ ..,

 

.,

..

8/16/2019 Ebeling Ch2-3

9/18

.. _, .

 

APPENDIX2E

TABLEOFINTEGRALS

2E.l

INDEFINITEINTEGRALS

l.

J

adt

=at

2.

R

1

J

t +

¥

ndt

=

  -

n + 1

3,

J

Int

eat

.

Je

l

[

dl

a

5.

(a + bt)n+l

J

a + bt)R dt =

n ¥

n + l)b

6.

J

ear

teat dt = a

2

(at

1)

7.

J

udv

=

uv -

J

vdu

(integrationbyparts)

2E.2

DEFINITEINTEGRALS

8.

J(,' te-at

dt =

a>

0

9.

f

-aid _ n.

 

a>

0

 

t e

. . .

o a

n

+

I n aposItiveInteger

£XERCISES

2.1

COllSiderthefollowingreliabilityfunction,where

t

is inhours:

1

R t) O.OOlt +

1

t

;;= 0

(a) Find thereliabilityafter 100operatinghours;after1000operatinghours.

(b)Derivethehazardratefunction.

Is

itan increasingoradecreasingfailurerate?

_

ACOIIII)UIlt'nt hasIhefllllowlIIglinellrhazard!"dIe, where

t

isinyears:

A t) OAt 0

(II)

Find

N t) anddetennine theprobabilityof acomponent f

8/16/2019 Ebeling Ch2-3

10/18

(d) Is thefailurerateOFR,CFR,orIFR'!

(e) Willaone-yearbum-inperiodimprovethereliabilityin part(a)?If so,whatisthe

newreliability?

2.8

A umfonn failuredistributionhasthecharacteristicthatequalintervalsof timehave

equalfailureprobabilities.Thedensityfunctionisgiven by

I

f t)

= - 0   =; t =;

b

b

Analyzethisgeneralfailuredistributionbyfinding

F t \

R t), A t), MTIF, Tmod,Tmode,

andu.

2.9

RepeatExercise

2.8

assumingthatb

=

20

yr.

2.10 Thereliabilityof aturbinebladecartbe representedbythefollowing:

R t) =

 I - r 0  =;

t

=;

t

where

t

isthemaximumlifeoftheblade.

(a) Showthatthebladesareexperi encingwearout.

(b) ComputetheMTIF as afunctionof themaximumlife.

(c)

If

themaximumlifeis2000operatinghours,whatisthedesignlifeforareliability

of 0.901

2.11 Anew fuelinjectionsystemisexperiencinghighfailurerates. The reliabilityfunction

hasbeenfound to be

  t + 0-

3

R t) =

t

2

0

where

t

ismeasuredin years.The reliabilityoveritsintendedlifeof 2

yr

is0.19,which

isunacceptable.Willabum-inperiod

of

6monthssignificantlyimproveuponthisreli

ability?

If

so,byhowmuch?

2.12 UsingEq. (2.14),derivethereliabilityfunctionforasystemhavingalinearlyincreasing

hazardratefunction,i.e., A=

at, a > 0, t

2

O

Findthedensityfunction, and derive

expressionsforthemediantimetofailureandthemode.

2.13 ComputetheaveragefailurerateforthefailuredistributioninExercise2.4.Whatisthe

averagefailurerateoverthefirst500 hours?

1.14 A

householdappliance

is

advertisedash avingmorethan alO-yrlife.If thefollowing

is

itsPDF,determineitsreliabilityforthenext10

yr

if it hassurviveda l-yr warranty

period:

f t) 0.10 + 0.05t)-3 t

2

0

Whatisits

MTIF

beforethewarrantyperiod,andwhatisits

MTIF

afterthewarranty

IIcriodassumingithasstillsurvived?

IIJI Mhllw thatif thehazardratefunctionisdecreasing,thePDF, f t) , isalsoadecreasing

l\anctionanditsmodemustthereforeoccurat

t =

O.

l

' I II

' II

f'(O and R t) whenthehazardratefunctionisexponentiaLThat is,A(t)

=

ae

,

wh.".

/I

is aconstantanda

>

O. Thisisonefonn of anextremevaluedistribution

  hili heenused to modelwearoutduetocorrosion. It isalsoknown as Gumbel's

.' hudlln.

( II

,,','1': It .\

'M I •

.......

  -

, . . . , . . , . .

..... - ...

, , "

ConstantFailureRateModel

These nexttwochapterswilldevelopseveralprobabilitymodelsusefulindescrib

ingafailureprocess.Thesemodelsarebasedupontheexponential,Weibull,normal,

andlognormalprobabilitydistributions.Theyareoftenreferred toastheoreticaldis

tributionssincethey arederivedmathematicallyandnotempirically.Animportant

question

to

beansweredisthatoftheabilityofaparticulartheoreticaldistribution

to u.escribe the failures andreliability ofacomponentora system.The answerto

thatquestion

is

acentralthemeinPart

11

ofthetext.The immediateconcern,how

ever, is thedevelopmentanduseofthesetheoreticalmodelsinanalyzingafailure

process. .

Afailure distribution thathas aconstant failure rate

is

calledan

exponential

probability distribution, Theexponentialdistributionisoneof themostimportant

reliabilitydistributions.Manysystemsexhibitconstantfailurerates,and,theexpo

nential distribution

is

inmanyrespects thesimplestreliabilitydistribution

to

ana

lyze. Another usefulconcept,failuremodes,

is

also discussed.If thefailure rates

of

allfailuremodesofacomponentareconstantandindependent,thentheoverall

failurerateofthecomponentisalsoconstant.

3.1

THE

EXPONENTIALRELIABILITYFUNCTION

Oneofthemostcommonfailuredistributionsinreliabilityengineeringistheexpo

nential,

Of

CFR,model.failures due to completelyrandom orchance eventswill

followthisdistribution.

It

shoulddominateduringtheusefullifeofasystemorcom

ponent.It isalsooneoftheeasiestdistributionsto analyzestatistically.Forexample.

4l

'_111

8/16/2019 Ebeling Ch2-3

11/18

 

gOlld parametcr cstimators and cxact conlidcnce intervals call bc cm,ily computed

under differcnt testing conditions (e.g., in the presence of censoring). r

We begin the development of the CFR model by assuming that

A t) =

A,

t

0,

A> O. Then from Eq. (2.14)

R(t)

=

exp/-

fa Adtl] =

e-

At

, t

2:

0

(3.1 )

and

F(t) = 1 - e-

At

Then

J( t )

.

.

-

dR(t)

=

Ae -

At

dt

The

three probability functions are illustrated graphically in Fig. 3.1 for several dlf ..

ferent values of

A.

To find the

MITF.

we use

Eq.

(2.8):

f

a

e-

At

dt

e-AtJ'oo 1

MTTF

=

= _

  _

o A

A

3.2)

o

The variance is given by Eq. (2.11), or

2

(7 2

I -At

- -

Ae

dt

-

A

)

A2

(3.3)

and the standard deviation is

A MITF.

This is an interesting result since it im

plies that the variabilityof failure time increases as the reliability MITF) increases.

R(t)

1 2 ~

I

1.0

0.8

0 [

0.4

0.2

o

0.5

1.0

r

,

\

"

till

.25 -

1.00

5.00 -

 

1.5

2.0

2.5

3.0

3.5

4.0

(a)

FIGURE

3.1

(a) The exponential reliability function.

I

Censoring will

be

discussed later. It requires the analysis of incomplete failure dam (not all items tested

have failed).

F .)

1.2

10

0.8

0.6

'

0.4

0.2

0.25

1.00

5.0 1

A.

o

0.5 1.0

1.5 2.0 2.5 3.0 3.5

(b)

f(t)

6

5

A

.0

2.5

 

r

5.0

3

2

'

 

.

......

': -

  .-

······ •..7.··:-:.':":.":".:" '.':'":..'"r"._._._.

o

0.5

1.5

(e)

FIGURE 3.1

continued)

The exponential cumulative distribution function. c) The exponyntial

probability densitv function.

High variability in failure times is often observed in practice. It should also be noted

that the mean time to failure is the reciprocal of the failure rate. Although

AU)

is

always in units of time per (between) failures, it is the mean ofthe failure distribution

for the

CFR

model only.

A second observation concerning this distribution is that

R MTTF) e MTIFIMTIF = e-

I

0.368

2

...

8/16/2019 Ebeling Ch2-3

12/18

A

component having

a

CFR has

a slightly better than one-third

chance

of

surviving

to its

mean time

to failure.

The design

life of a

component having an

exponentially

distributed failure

time may be obtained

by solving for

the

inverse of

the

reliability

function. That is, for a

given

reliability

R,

e-

 t

R(tR)

=

R

then

t«

1

  ln

A

(3.4)

When R

=

0.5, the

median of

the distribution

is

obtained from Eq. (3.4):

Imed

-

-

_ 1

I

05

An .

0.69315

(3.5)

= 0.69315 MTTF

The

median is always

less

than

the

mean

since

the exponential

distribution is skewed

to the right.

EXAMPLE 3 t A microwave transmitter has exhibited a constant failure rate of

0.00034 failure per operating hour. Therefore MTIF

1/0.00034 =

2941

hr

and

tmcd

0.69315

X

2941

=

2039 hr. The reliability function

is

given by

R(t)

= e-O.OOO34t.

The reliability over a 30-day continuous operating period is

R(30

X

24)

=

0.78286. The

design life for a reliability of 0.95 specification is (-In 0.95)/0.00034 = 150.86 hr

Memorylessness

A

well-known

characteristic

of the CFR

model,

one

not

shared

by

other

failure

distributions, is its lack of memory. That is, the time to failure of a component is

not

dependent

on

how

long

the component

has been operating. There is no

aging

or

wearout effect. The probability that the

component

will

operate

for

the next

1000 hi-

is the

same

regardless of

whether

the

component

is

brand

new, has

been

operating

for several

hundred

hours, or

has been

operating for several

thousand

hours.

This

property

is consistent

with

the

completely

random and independent nature

of

the

failure process. For

example,

when external,

random environmental

stresses

are the

primary

cause

of failures, the failure or operating history of

the component

will not

he relevant.

This property

can

be demonstrated mathematically using conditional reliability.

I;rom Eq. (2.1

R(t iTo) = R(t + To) _ e--).(t+T

o

)

-

R(To)

e-

M

•

t

e- = R(t)

In

other words, a

bum-in period 10 has

no

subsequent effect on

reliability

and

will

not improve the component's reliability.

T:me

to failure

depends only

on

the

length

01

h l ~ observed operating

time (I)

and not

on

its

current

age

(To).

\ 1

1,'i\lUJRE MOll1:'"

,"ll1l1linl!

th\.· tU/; l!d ;,Je Itlllt '\! '

11

all failure modes, as shown below.

Let

Ai t)

be the failure rate function for the

ilh

failure I11()J.:

TI1l.'11

1

RI(t) ex

p

[ A,1/)l l t ' \

(I

I

)

8/16/2019 Ebeling Ch2-3

13/18

I:X ·.u ...U : J.2.

BATHTUB

8/16/2019 Ebeling Ch2-3

14/18

1

exp [ -

[ :0

A(t )dt ]

]

r

T

+

1

R(t To) = = exp -

A(t )dt

exp [ -   [°

A(t')

dt ] ,To

R(t) = ex

p

[ - f

A(t )dt ]

To+t

il

Then

A(t )dt

=

A(t )dt

J

o

0

which requires

A(t) =

A, a constant. Therefore if failures are detennined by com

pletely random independent events not associated with the age of the system, a con

stant failure rate model will result.

3.3.1 Renewal Process

Assume that a system is comprised of many components acting independently and in

such a way that an individual component failure causes a system failure. A renewal

process, in which a failed component is immediately replaced with a new one, will

cause the system to reach a steady-state, constant number

of

failures per unit

of

time.

This is best illustrated by a numerical example.

EXAMPLE 3.5. Consider the following discrete failure distribution for each of 1000

identical components.

n

Pr{T n} 15

for n = 1,2,3,4,5

where T represents the number

of

operating cycles until failure. The number

of i l u r c ~

(and hence replacements

each

cycle) is shown in Table 3.1. Since each component is

identical,

MTTF;

15 + 2 fs )+

+

+ 5 J..)

5 3

where MTTF; is the mean time

(number of

operating cycles) to failure for the ith comp('

nent.

The

system mean time between failures is (Il/3)/lOOO

=

0.0036667, and a steady·

state failure rate

of

110.0036667

=

272.7 failures per cycle is observed. In general, for

a renewal process in which failed components are replaced as they fail, a steady-srate

c')nstant failure rate A is obtained such that

A=

;=1 MTTF;

Renewal processes will be discussed further in Chapter 9.

3.3.2 Repetitive Loading

If there is a small constant probability of failure

p

as a result of a load or stress being

placed on the system and

if

independent loads are applied at constant, fixed intervals

'I'.\IU.\I.

l.1

SYHtem renewalll

. . . . . . . . . . . . . . . . . . . . . . . . . .

__ - - - - - - - - - - ~ - - . ~ 4 J - - - - - - ~ m l.. -.---

( yde

Number of l l u ~ s

fs(lOOO) = 67

is(lOOO) + -(s(67) "" 138

.

2

rs(lOOO) +

fs(67)

+

-(s(l38) = 218

4

~ ( l O O O ) + fs(67) + is   I38) + -(s(218)

=

313

5

~ ( 1 0 0 0 ) +

~ ( 6 7 )

+ rs(138) + i s ~ 2 1 8 ) + -(s(313) 429

6

~ ( 6 7 )

+

~ ( 1 3 8 )

+

1«218)

+

1\(313)

+

-/s(429) '"

173

7

rs(138) +

~ ( 2 1 8 )

+ f:i(313) + 1\(429) + M173) 235

Continuing in this manner:

Number of failures

Cycle

Number

of failures

Cycle

271

18

8 281

274

19

9 303

271

20

10 294

272

21

II 236

275

22

12 269

274

23

13 283

273

24

14 281

273

25

15

271

27326

16 263

17 275

A steady-state constant rate

of

273 failures per cycle is reached by the 24th

III

lime, an approximate exponential reliability function r e s u l t s ~ Let

R

l

p),

thl' reliability

of

a single load. Then

Rn l t en1n 1-p)

i'i

the reliability given n loads. Since

6

1nO - p) = p for very small p,

np

Rn

= e-

Lei n =

tlat, with at being the fixed time between loads. Then

R(t) = e- plD..I)1 = e -

AI

with A

=

plat, a constant failure rate.

A similar result is obtained if the loads are applied at random if thl: number

of

loads per time period

hilS a Poisson distribution.

"This follows from a first-order Taylor series approximation around the origin;

i

e., fix) In(l

x)

I(()+ xf (Q) = O+x - I )

=

-x .

DM

: . ioe ualifit

M h h r ~

8/16/2019 Ebeling Ch2-3

15/18

.LLl Reliability Rounds

Even when the CFR model does not apply, it can still be used to provide bounds

on system reliability provided the system hazard rate can

be

bounded. Assume

that

0<

AL ; A(t) ; Au

Then L

L dt' ;

J A(t')

dl' ;

f;

Au dt'

and expl -

J

ALdt']

expl

L

(t')dt'] expl

L

~ d t l ]

e-

ALt Aut

or

R t)

(3. 11)

Therefore if the hazard rate function is bounded, the reliability function can also be

bounded.

The

exponential, because

of

its ~ o n s t a n t failure rate, provides a boundary be

tween distributions having decreasing failure rates and those having increasing fail

ure rates. Barlow and Proschan [1967] develop the following upper and lower relia

bility bounds for when the hazard rate is decreasing and increasing, respectively. In

addition to knowing that the process has an

IFR

or DFR, one must know the mean

time

to

failure. For a

DFR

process,

e-t lMTfF

t MTTF

(3.12)

R(t):::;

{

M T T F e ~ t > MTTF

For

an

IFR process,

R t) { ~ - t I M T I F

t < MTTF

(3.13)

t

MTTF

They also develop the following upper bound for when the hazard rate is in

creasing.

t

; MTTF

(3.14)

R t) :::;

{ -WIMTfF

t>

MTTF

where

1 - wMTTF

=

e-

wt

In order for the above equation to be satisfied, a different (tJ must be found for

each t

EXA

MPLE 3.6.

A government specification calis for a mechanical pump to have an

MTIF of

1000 operating hours. These pumps exhibit continuous wearout and there

fore have an IFR. As a result, lower and upper bounds on the pump's reliability after

200 operating hours are given by 0.8187 ;

R 2oo)

1. For 2000 operating "ours,

o R 2000):::; 0.2032, where w was found to be 0.0007968 by solving Eq. (3.14)

numerically with I = 2000.

It XA tot ru : l .7.

The l'oUuwinll J\\lIllhlhly functilln' has DFR and has

IIIl MTI P

of 400.

R(t) t' jti2(-.J

('umpllring the aclual reliability III the upper bound computed using

Eq.

(3.12):

Upper bound

R(t)

0.9753

0.7996

0.8825

0

0.6060

0.7780

0

0.4930

0.6065

00

0.3678

0.4723

00

0.2938

0.3679

00

400 0.243

0.294

0.2057

0C

0.245

0.1769

600

J.4

TilE

TWO-PARAMETER EXPONENTIAL DISTRIBUTION

If a failure will never occur prior to some specified time

to,

then to is a minimum,

or threshold, time. t is also known as the guaranteed lifetime. The parameter

to

is

II

IOl:ation

parameter that shifts the distribution

an

amount equal to to to the right

nil

the time (horizontal) axis. This is equivalent to rewriting the density function by

Il'placing t with

t

_

to,

with the domain of the random variable now t to· Fo ' the

('xponential distribution, the probability density function becomes

(3.15)

dR t -

-

  (t):=

=

Ae A t to) 0 < to

;

t < 00

dt

alltl the reliability function will take

on

the following form:

(3.16)

R t) = e-A(t-

tu)

t to

From

Eqs.

(3.15)

and

(3.16)

the failure rate is

A(t) =

(t)/R(t)

A

However, the

mean of the distribution is no longer

II

A but is shifted a distance to along the taxis.

l Jsing integration by parts or formula 6 in Appendix 2E:

1

(3.17)

MTIF =

A t e - · ~ t - t o )

dt to + -

A

The median of the distribution is obtained by solving Eq. (3.16) for tmed,

(3.18)

R(tmed) = e -A tmed-10)

=

0.5

anu obtaining

InO.5

0.69315 (3.19)

tmed = to + to +

A

IThis is a Weibuil i "liability function. The Wcibull distribution will be discussed in following chaptel.

8/16/2019 Ebeling Ch2-3

16/18

Tilt' tit'sign lift' tH for II spl'l.:ifIL'd desigll reliability R can he obtainl'd III the SHllIl'

lllaUller as Ihl' median lime. Therefore,

InR

fR =

to

+ (3.20)

-A

The

variance and standard deviation

of

the two-parameter exponential distribution

are not affected

by

the location parameter. Therefore [f =

IIA. The

mode

OCCurs

at to.

EXAMPLF.

3.8.

Let A

0.00]

and to

= 200.

Then

R t)

2

and

MIT

- I _ 0.69315 893

F -

200 + 0.00

I

1200 tmed - 200 + o:om _

J.

0.95 3 I 1000

to.95 =

200 In 0.001 =

25L

u 6.001

3.5

POISSON PROCESS

If a component having a constant failure rate A is immediately repaired or replaced

upon failing, the

number of

failures observed over a time period

t

has a Poisson

distribution.

The

probability

of

observing n failures in time

t

is given by the Poi:;son

probability mass function

Pn t):.

e-At(AW

Pn(t)

n.

r n

0,1,2,.. .

(3.21)

Unlike the failure distribution that is continuous

over

time, the POIsson distribution

1s

discrete.

The

mean or expected

number of

failures over time

t

is given by At, and

the variance of

the distribution is also

At.

Equation (3.21) can be derived by first letting

Y

k

Ti

where

T

i ,

a random variable, is the time between failure I and failure

i

and

has

an exponential distribution with

parameter A.

Therefore

Yk

is a random variable,

the time of the

kth

failure. Since the sum

of k

independent exponential random vari

ables has a

gamma

distribution

8

with parameters k and

A

the cumulative distribution

function for Y

k

can be written as

S

A proof of this result

may

be

found in

many statistics texts, e.g.,

Ross

[1987, pp. 114-1l7

J.

Since k is

:In

integer, this

is

a special ease of the gamma distribution known a"

the

Erlang distribution.

II

i, only

Ih,s

special case that results in a closed-form

expression

for the

CDF.

FyY = 1

t

At   (A!l!

(3.22)

Prj YA .

tl

''---'

·0 /.

The

cumulative rrobabilit y given by Eq. (3.22) is the probabilhy that

the

kth failure

WIll occur

by

time

t. The

mean value for Y

k

is kIA, and the variance is k1A2.

The

mode is k 1)/,1. We have

Pn(t)

= Pr{Y

n

$

t}

-

Pr{Y

nJ

$

t} Fyn(t) -- Fyn+,(t)

I ~ u a t i o n (3.21) follows from the above and from Eq. (3.22).

The

relationship between the two probability distributions can also be seen by

determining the probability

of

no failures occurring in time

t,

which is equivalent to

Pr{T

2::

t}.

That

is,

e-At At)O

Po t)

=

e-

A

= R t)

The Poisson process is often used in inventory analysis to determine the number

of

spare components when the time between failures is exponential.

For

exallll:\e.

if S spare components are available to support a continuous operation over a time

period

t

then

s

Rs t)

= L

Pn t)

(3.23)

11=0

is the cumulative probability of

S Of

fewer failures occurring during time

t.

Equa

tion (3.23) therefore represents the probahility

of

satisfying all demands

fOf

spare

components during time

t.

Therefore, Rs t) is the component reliability if there are

S

spares available for immediate replacement when a failure occurs.

EXAMPLE 3.9.

A specially designed welding machine has a nonrepairable motor with

a constant failure rate of

0.05

failure per year. The company has purchased two spare

motors. f the design life

of

the welding machine is

10

yr, what is the probability that the

two spares will be adequate?

Solution.

The expected number

of

failures over the life

of

the machine

is At

0.05(10) = 0.5. From Eq. (3.23),

R (l0) ± - O = ? 5 1 l e

0.5

I + 0.5 + 0.25 .= 0.9856

2

n O

n.

is

the probability of 2 or fewer failures occurring over the

10

yr.

Let Y, be the time

of

the third failure. Y

3

has a gamma distribution with k

and

A

0.05.

Therefore, the expected. or mean, time

10

obtain

3

failures

is 3/0.05

60 yr. The probability that the third failure will occur within

10

yr is obtained ffilm

Eq. (3.22):

I

+

0.05 x

10 +

(0.05 x

1 0 } C , ~

0.0144

FY, IO) = I

C

.•

serve that 0.0144 0.9S56 since the prnbabi

lilY of

two or fewer

f a i l u r e ~

i'l 10

yr is cornplc1TltCntary 10 the event that the third failure occurs within 10 yr.

3

.1.6

8/16/2019 Ebeling Ch2-3

17/18

----------

Rf

8/16/2019 Ebeling Ch2-3

18/18

3.6

3.7

3.8

3.9

A landing gear system has repetitive stresses placed on it twice a day as a result of

landings. The probability of a failure during landing is 0.0028. Determine the

of the landinl I ear system over a 30-day contingency operation. What is the probability

10

and 20 of the operation?

A system contains 20 identical and critical components that will

be

replaced on failure

(renewal process). As a result, a constant failure rate for the system will be observed.

f

a design life of

IO

yr with a reliability of 0.99 is required, what should the system MTTF

and median time to failure be?

If

each component has a CFR, what will the component

MTTF and median time to failure be?

Two identical and CFR comput ers are

reliability is to be 0.95 at 3000

for each computer.

Specifications for a power unit consisting of three independent and serial ly related com

ponents(failure modes) require a design life of 5 yr with a

0.95 reliability.

Let each component have a constant failure rate such that the fir,t component's rate

is

twice that of the second and the third component's rate

is

three times that of the

second. What should be the MTTF of each component and the system?

(b) If two identical power units are placed in parallel, what is the system reliability at

5

yr,

and what

is

the system MTTF?

3.10 The time to failure of fluorescent lights in a large office building is expunentiady di,,· .

tributed with a failure rate

of

0.03125

hr.

How many spare tubes must the hullding

custodian maintain to have at least a 0.95 probability of replacing

11::

failures

on

a given

Assume continuous 24-hour use of the lights.

3.11

J.12

A l1ashlight contains

two

batteries each having an MITF of 5 operating hours (assume

CFR).

(a) What

is

the probability of battery failure occurring within the first 2 hr

of

operation'

(b) If failed batteries are immediately replaced, what

is the probability

of

mor than

one failure occurring during the first 5 hr

of

operation?

(c) Would

you

expect batteries to have a constant failure rate?

Consider two redundant components having constant but different failure rates.

Derive the reliability function and the MITE

(b) Find the reliability at 1000 hr and the MTTF o f a two-component redundant systc

where

A1(t)

0.000356 per hour andA2(t) 0.00156 per hour.

3.13 An electronic circuit board with

A t)

0.00021 per hour replaced on failure. What

is the probability that the third failure will occur by 10000 hours?

3.14 In reliability testing it is of interest to know how long the test must run in order to

generate a specified number of failures. A new condenser fan motor is believed

to

have

a constant failure rate of

3.4 failures per 100 operating hours. A single test stand is

to

be used

in

which a motor is operated until failure and then replaced with a new motor

from oroduction. What is the expected test time if

IO

failures are desired?

month" and t r.llul'l

r t

IIr 0.1 f"i1l1rr

prr

yellr. Whlll i the N y ~ l e l l l rdinhililY fur

10,000 hr of continuaUII 0JHIrllllnn'/

~ 1 6

\)erive u general

expreNlIlll1l for

R t) and the MTIF for the two-component system de

N ~ ~ r i h c d

in Exercise

? t I ~

~ 1 7 A microwave link in a communications network has a high failure rate. Although sev

eral pieces of equipment have been used

in

the link, the link always seems to fail at

IIhout the same rate regardless of the age of the equipment and its prior maintenance

In general, microwave transmissions are subject to fading. Selective fading oc

curs when atmospheric conditions bend a transmission

to

the extent that signals reach

Ihe receiver in slightly different paths. The merging paths can cause interference and

acate

data errors. Other channels in the microwave transmission are not affected

by

selective fading. Selective fading occurs when there is an electrical storm. Flat fading

occurs during

fog

and when the surrounding ground is very moist. t is more serious

since

it

may last several hours and affect surrounding channels. f during the current

season, electrical storms occur about once every week and fog alerts are issued at the

rate of one every two months, what ig the reliability of the link over a 24-hour period?

What assumptions, if any, are necesgary?

.\. 11 1

A 60-watt outdoor lightbulb is advertised as navmg an average life MITF of 1000

(operating) hours. However, experience has shown that it will also fail on demand an

average of once every 120 cycles. A particular bulb is turned 011 once each evening for

an average of

10 hr.

f it

is

desired to have a reliability of

90

percent, what is its design

life in days?

.ut A more general exponential reliability model may be defined by

R t) = where a > \, b > 0

and a and b are parameters to be determined. Find the hazard rate function, and show

how this model is equivalent to R t)

.UO Repet itive loading, A packaging machine (cartoner) in a food processing facility will

with a constant probability of 0.005 per application (per carton). Twelve cans of

coffee are combined into a single case for shipment to buyers. The production rate is

30 cans of coffee every minute. What is the probability (reliability) of no jams during

a I-hr production run?

.'-21 For the reliability function R t) e ~ I I I O O O ) 2

,

use Eqs. (3.13) and (3.14) and I;ompare the

upper and lower bounds with the actual reliabiIities at

100,200,500,800,

1000,2000,

5000, and 10,000 hr. This failure distribution has an IFR with an MITF of 1772.46.

".12

Show for the exponential distribution that the residual mean life is II A egardless of the

of time the system has been operating.