ebeling ch2-3
TRANSCRIPT
-
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.