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C. Ebeling, Intro to Reliability & Maintainability Engineering,
2nd ed. Waveland Press, Inc. Copyright 2010
Chapter 3
The Constant Failure Rate Model
Exponential Probability
Distribution
Chapter 3 1
Hurry, come here!
Chapter 3 is starting now.
You dont want to miss any of this.
-
The Exponential Distribution
Chapter 3 2
0t , = (t)
0t ,e =e = R(t)t-dt-
t0 '
Then
Let
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-The Reliability Function
Chapter 3 3
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-The CDF and PDF
Chapter 3 4
e-1 = F(t)t-
e = dt
F(t) d=f(t) t-
-
-Probability Density Function (PDF)
Chapter 3 5
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-The MTTF
Chapter 3 6
.368 = e = e = R(MTTF)1-
MTTF
-MTTFnote that
or
0|- t
- t
0
1eMTTF = dt = = e
-
0
1tMTTF te dt
= =
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-Variance & Standard Deviation
Chapter 2 7
definitional form:
computational form:
22
0 = (t - MTTF f(t)dt)
22 2
0 = f(t)dt - (MTTF )t
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-The Standard Deviation
Chapter 3 8
2
= 0
2
t -1
-te dt =
12
and MTTF
= =1
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-The Median Time to Failure
Chapter 3 9
.69315 = .5
1- = tmed ln
= .69315 MTTF
R t e t( ) .= = 5
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-The Design Life
Chapter 3 10
For a given reliability R, let
lnthenR
1= - Rt
R- tRR( ) = = Rt e
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-Example #1 - CFR model
An electrical transmission line has been tested and found to have a CFR of .001 defects per foot. Find:
a. R(100)
b. R(1000)
c. MTTF
d. Standard Deviation
e. tmedf. 90 percent design life
Chapter 3 11
-
-Example #1 - solution
Chapter 3 12
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-Memoryless Property
Chapter 3 13
e
e =
)TR(
)T+R(t = )T|R(t
T-
)T+(t-
0
00
0
0
R(t) = e = e
e e = t-
T-
T-t-
0
0
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-Failure Modes
R(t) =
n
=
R (t)
i
i
1
Chapter 3 14
Ri(t) is the reliability function for the ith failure mode, then, assuming independence among the failure modes,the system reliability, R(t) is found from
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-More on Failure Modes
iR (t) = - 0
t i ( t ) d te
Chapter 3 15
Let
R(t) =
n
i=1
- 0t i ( t )dt 'e
Then
i
i
n
=1
(t) = (t) where
01
exp ( ') 'nt
i
i
= t dt=
t0
- (t )d t= e
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-Failure Modes and CFR
Chapter 3 16
in
=1i
==(t)
If a system consists of n independent,
serially related components each with
CFR, then
and
t0
- d t - tR(t) = = e e
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-Example - Failure Modes
An engine tune-up kit consists of 3 parts each having CFRs (in failures per mile) of .000034, .000017, and .0000086.
Find the MTTF, median time to failure, standard deviation, and reliability of the tune-up kit at 10,000 miles.
Chapter 3 17
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-Example - solution
Chapter 3 18
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-Parts Count Approach
An integrated circuit board consists of the following components each having a CFR.
Component a-Failure Rate(10-5) b- Quantity (a) x (b)
Diodes, silicon .00041 10 .0041
Resistors .014 25 .3500
Capacitors .0015 12 .0180
Transformer .0020 2 .0040
Relays .0065 6 .0390
Inductive devices .0004 12 .0048
total .4199 x 10-5
Chapter 3 19
Therefore Rsys(t) = e- .000004199 t and MTTF = 1/.4199 x 105
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-All Failure Modes are CFR
MTTF = 1 =
1
1
MTTF
where MTTF = 1
i=1
n
i
i=1
n
i
i
i
= 1
Chapter 3 20
= n MTTF = 1
n 1
1
and
If all components have identical failure rates, then:
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-3.5 Poisson Process
If a component having a constant failure rate 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 Poisson probability mass function pn(t):
Chapter 3 21
( ) ( )
2
0, 1, 2, !
[ ]
nt
n
e tp t n
nE n t
= =
= =
K
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The Poisson and Exponential
Chapter 3 22
( ) ( ) ( )0
00!
t
te tp t e R t
= = =
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-The Gamma Distribution
LetTi = time between failure i 1 and failure i having an
exponential distribution with parameter .
Yk = the time of the kth failure.
The sum of k independent exponential random variables has a
gamma distribution with parameters and k.
If k is an integer (i.e. the Erlang distribution),
Chapter 3 23
1
k
k i
i
Y T=
=
( )( ) ( )
1
( ) for , , 0
1 !, positive int.
k k t
Y
t ef t k t
k
n n n
=
=
-
-The Gamma Distribution
The CDF is the probability that the kth failure will occur by
time t.
The mean value for Yk is k/, and the variance is k/2.
The mode is (k 1)/(why?).
Chapter 3 24
{ } ( ) ( ) ( ) ( )
( ) ( ) ( )
0 1 1
0 11
0
Pr 1 ...
1 ... 10! 1! !
it can be obtained by integration on Gamma function too.
kk Y k
k ik
t t t
i
Y t F t p t p t p t
t t te e e
k i
=
= = =
=
-
?
Related Failure Distributions
Chapter 3 25
Random Variable Distribution Parameter(s)
T, time to failure Exponential Yk, time of the k
th failure Gamma (Erlang) , kN, number of failures in time t Poisson
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EXAMPLE 3.9
A specially designed welding machine has a nonrepairable motor
with a constant failure rate of 0.05 failures per year. The company
has purchased two spare motors. If the design life of the welding
machine is 10 yr, what is the probability that the two spares will be
adequate?
Chapter 3 26
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EXAMPLE 3.9
Let Y3 be the time of the third failure. Y3 has a gamma
distribution with k = 3 and = 0.05.
The expected, or mean, time to obtain 3 failures is 3/0.05 =
60 yr.
The probability that the third failure will occur within 10 yr is
Chapter 3 27
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-Redundancy2 identical components with CFR
( )2
1 Pr one fails
2- t
R(t) =
1-(1- )e
=
Chapter 3 28
= 1-(1-2e +e )- t -2 t
= 2e - e- t -2 t
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Redundancy - Hazard Rate Function
[ ]1 - t -2 t- t -2 t
R(t)f(t) 2 - 2e e(t) = =
R(t) R(t) 2 -e e
=
Chapter 3 29
= (1 - e )
(1-.5 e )
- t
- t
Not a CFR process!
-
Chapter 3 30
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Redundancy, CFR & MTTF
Chapter 3 31
= 2-1
2 =
3
2 =
1.5
- t -2 t
0 0MTTF = R(t)dt = (2 - )dte e
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-Two-parameter Exponential
Chapter 3 32
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More 2-parameter Exponential
Chapter 3 33
R t e
t t t
t tR
med
t t
med
R
med o( ) .
ln . .
ln
( )= =
= +
= +
= +
5
05 0 693150 0
0
Note that the variance does not change and the mode occurs at t0.
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Example CFR Model
A certain type of surge protector is observed to fail at the constant rate of .0005 failures per day. Find:
a. The MTTF, median, and the standard deviation.
b. Find the reliability during the first year; the second year; the second year given it has survived the first year.
c. The 90% design life.
Chapter 3 34
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Example - solution
Chapter 3 35
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Bonus Round - Return Period
An asteroid 2/3 of a mile across smashing into the Earth at a high speed which would cause a blast with the power of thousands of hydrogen bombs could happen within the next 300,000 years.
- Dave Morrison (NASA)
Chapter 3 36
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The End
Chapter 3 37
Our next assignment
will be in Chapter 4.
First, however, you will
want work the
problems identified for
chapter 3.