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Reliability Application
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Systems Engineering Program
Department of Engineering Management, Information and Systems
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An Application of Probability toReliability Modeling and Analysis
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• Figures of merit
• Failure densities and distributions
• The reliability function
• Failure rates
• The reliability functions in terms of the failure rate
• Mean time to failure (MTTF) and mean time between failures (MTBF)
Reliability Definitions and Concepts
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• Reliability is defined as the probability that an item will perform its intended function for a specified interval under stated conditions. In the simplest sense, reliability means how long an item (such as a machine) will perform its intended function without a breakdown.
• Reliability: the capability to operate as intended, whenever used, for as long as needed.
Reliability is performance over time, probability that something will work when you want it to.
What is Reliability?
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• Basic or Logistic Reliability
MTBF - Mean Time Between Failures
measure of product support requirements
• Mission Reliability
Ps or R(t) - Probability of mission success
measure of product effectiveness
Reliability Figures of Merit
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“If I had only one day left to live, I would live it in my statistics class --it would seem so much longer.”
From: Statistics A Fresh ApproachDonald H. SandersMcGraw Hill, 4th Edition, 1990
Reliability Humor: Statistics
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The Reliability of an item is the probability that the item willsurvive time t, given that it had not failed at time zero, when used within specified conditions, i.e.,
)tT(PtR
t
)t(F1dt)t(f
The Reliability Function
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Relationship between failure density and reliability
tRdt
dtf
Reliability
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Remark: The failure rate h(t) is a measure of proneness to failure as a function of age, t.
tF-1
tf
tR
tfth
Relationship Between h(t), f(t), F(t) and R(t)
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The reliability of an item at time t may be expressed in termsof its failure rate at time t as follows:
where h(y) is the failure rate
t
0dy)y(ht
0
edy)y(hexp)t(R
The Reliability Function
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Mean Time to Failure (or Between Failures) MTTF (or MTBF)is the expected Time to Failure (or Between Failures)
Remarks:
MTBF provides a reliability figure of merit for expected failure free operation
MTBF provides the basis for estimating the number of failures ina given period of time
Even though an item may be discarded after failure and its mean life characterized by MTTF, it may be meaningful tocharacterize the system reliability in terms of MTBF if thesystem is restored after item failure.
Mean Time to Failure and Mean Time Between Failures
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If T is the random time to failure of an item, themean time to failure, MTTF, of the item is
where f is the probability density function of timeto failure, iff this integral exists (as an improperintegral).
0
dtttfMTTFTE
Relationship Between MTTF and Failure Density
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Relationship Between MTTF and Reliability
0
dttRMTTFMTBF
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Reliability “Bathtub Curve”
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Reliability Humor
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DefinitionA random variable T is said to have the ExponentialDistribution with parameters , where > 0, if the failure density of T is:
, for t 0
, elsewhere
t
e1
)t(f
0
The Exponential Model: (Weibull Model with β = 1)
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• Weibull W(, )
, for t 0
Where F(t) is the population proportion failing in time t
• Exponential E() = W(1, )
t
e-1 )t(F
t
e-1 )t(F
Probability Distribution Function
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RemarksThe Exponential Model is most often used in Reliability applications, partly because of mathematical convenience due to a constant failure rate.
The Exponential Model is often referred to as the Constant Failure Rate Model.
The Exponential Model is used during the ‘Useful Life’ period of an item’s life, i.e., after the ‘Infant Mortality’period before Wearout begins.
The Exponential Model is most often associated withelectronic equipment.
The Exponential Model
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Probability Distribution Function• Weibull
• Exponential
t
e )t(R
t
e )t(R
Reliability Function
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Reliability Functions
R(t)
t
t is in multiples of
β=5.0
β=1.0
β=0.5
1.0
0.8
0.6
0.4
0.2
00 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
The Weibull Model - Distributions
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Weibull
Exponential
MTBF
1
1 MTBF
Mean Time Between Failure - MTBF
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The Gamma Function
0
1ax dxxe)a(
)a(a)1a(
y=a (a) a (a) a (a) a (a)1 1 1.25 0.9064 1.5 0.8862 1.75 0.9191
1.01 0.9943 1.26 0.9044 1.51 0.8866 1.76 0.92141.02 0.9888 1.27 0.9025 1.52 0.887 1.77 0.92381.03 0.9836 1.28 0.9007 1.53 0.8876 1.78 0.92621.04 0.9784 1.29 0.899 1.54 0.8882 1.79 0.92881.05 0.9735 1.3 0.8975 1.55 0.8889 1.8 0.93141.06 0.9687 1.31 0.896 1.56 0.8896 1.81 0.93411.07 0.9642 1.32 0.8946 1.57 0.8905 1.82 0.93691.08 0.9597 1.33 0.8934 1.58 0.8914 1.83 0.93971.09 0.9555 1.34 0.8922 1.59 0.8924 1.84 0.94261.1 0.9514 1.35 0.8912 1.6 0.8935 1.85 0.9456
1.11 0.9474 1.36 0.8902 1.61 0.8947 1.86 0.94871.12 0.9436 1.37 0.8893 1.62 0.8959 1.87 0.95181.13 0.9399 1.38 0.8885 1.63 0.8972 1.88 0.95511.14 0.9364 1.39 0.8879 1.64 0.8986 1.89 0.95841.15 0.933 1.4 0.8873 1.65 0.9001 1.9 0.96181.16 0.9298 1.41 0.8868 1.66 0.9017 1.91 0.96521.17 0.9267 1.42 0.8864 1.67 0.9033 1.92 0.96881.18 0.9237 1.43 0.886 1.68 0.905 1.93 0.97241.19 0.9209 1.44 0.8858 1.69 0.9068 1.94 0.97611.2 0.9182 1.45 0.8857 1.7 0.9086 1.95 0.9799
1.21 0.9156 1.46 0.8856 1.71 0.9106 1.96 0.98371.22 0.9131 1.47 0.8856 1.72 0.9126 1.97 0.98771.23 0.9108 1.48 0.8858 1.73 0.9147 1.98 0.99171.24 0.9085 1.49 0.886 1.74 0.9168 1.99 0.9958
2 1
Values of theGamma Function
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• Weibull
and, in particular
• Exponential
1
P p)-ln(1- t
t 632.0
p)-ln(1- Pt
Percentiles, tp
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• Failure Rate
a decreasing function of t if < 1Notice that h(t) is a constant if = 1
an increasing function of t if > 1
• Cumulative Failure Rate
• The Instantaneous and Cumulative Failure Rates, h(t) and H(t), are straight lines on log-log paper.
1-t )t(h
)t(ht )t(H
1-
Failure Rates - Weibull
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• Failure Rate
• Note:
Only for the Exponential Distribution
•Cumulative Failure
1
)t(h
)t(H
rate failure
1MTBF
Failure Rates - Exponential
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Failure Rates
h(t)
t
t is in multiples of h(t) is in multiples of 1/
3
2
1
0
0 1.0 2.0
β=5
β=1
β=0.5
The Weibull Model - Distributions
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Problem -
Four Engine Aircraft
Engine Unreliability Q(t) = p = 0.1
Mission success: At least two engines survive
Find RS(t)
The Binomial Model - Example Application 1
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Solution -
X = number of engines failing in time t
RS(t) = P(x 2) = b(0) + b(1) + b(2)
= 0.6561 + 0.2916 + 0.0486 = 0.9963
The Binomial Model - Example Application 1
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• Simplest and most common structure in reliability analysis.
• Functional operation of the system depends on the successful operation of all system components Note: The electrical or mechanical configuration may differ from the reliability configuration
Reliability Block Diagram
• Series configuration with n elements: E1, E2, ..., En
• System Failure occurs upon the first element failure
E1 E2 En
Series Reliability Configuration
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• Reliability Block Diagram
•Element Time to Failure Distribution
with failure rate , for i=1, 2,…, n
• System reliability
where
tS
Se)t(R
SS
S θλ
1MTTF
is the system failure rate
• System mean time to failure
n
1iiS )t(
ii θE~T
E1 E2 En
ii θ
1λ
Series Reliability Configuration with Exponential Distribution
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• Reliability Block Diagram Identical and independent Elements Exponential Distributions
• Element Time to Failure Distribution
with failure rate
• System reliability
tnS e)t(R
E1 E2 En
θE~T θ
1λ
Series Reliability Configuration
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• System mean time to failure
Note that /n is the expected time to the first failure, E(T1), when n identical items are put into service
nMTTFS
Series Reliability Configuration
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Parallel Reliability Configuration – Basic Concepts
• Definition - a system is said to have parallel reliability configuration if the system function can be performed by any one of two or more paths
• Reliability block diagram - for a parallel reliability configuration consisting of n elements, E1, E2, ... En
E1
E2
En
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Parallel Reliability Configuration
• Redundant reliability configuration - sometimes called a redundant reliability configuration. Other times, the term ‘redundant’ is used only when the system is deliberately changed to provide additional paths, in order to improve the system reliability
• Basic assumptions
All elements are continuously energized starting at time t = 0
All elements are ‘up’ at time t = 0
The operation during time t of each element can be describedas either a success or a failure, i.e. Degraded operation orperformance is not considered
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Parallel Reliability Configuration
System success - a system having a parallel reliability configuration operates successfully for a period of time t if at least one of the parallel elements operates for time t without failure. Notice that element failure does not necessarily mean system failure.
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Parallel Reliability Configuration
• Block Diagram
• System reliability - for a system consisting of n elements, E1, E2, ... En
n
jiij
ji
n
1iiS )t(R)t(R)t(R)t(R
n
ii
nk
n
kjiijk
ji tRtRtRtR1
1 )()1...()()()(
if the n elements operate independently of each other and where Ri(t) is the reliability of element i, for i=1,2,…,n
E1
E2
En
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System Reliability Model - Parallel Configuration
• Product rule for unreliabilities
n
iiS tRtR
1
)(11)(
•Mean Time Between System Failures
0
SS (t)dtRMTBF
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Parallel Reliability Configuration
s
p=R(t)
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Element time to failure is exponential with failure rate
• Reliability block diagram:
•Element Time to Failure Distribution
with failure rate for I=1,2.
• System reliability
• System failure rate
t
t
S e2
e12)t(h
ttS eetR 22)(
E1
E2
θE~Ti θ
1λ
Parallel Reliability Configuration with Exponential Distribution
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• System Mean Time Between Failures:
MTBFS = 1.5
Parallel Reliability Configuration with Exponential Distribution
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A system consists of five components connected as shown.Find the system reliability, failure rate, MTBF, and MTBM if Ti~E(λ) for i=1,2,3,4,5
E1
E2
E3
E4 E5
Example
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This problem can be approached in several different ways. Here is one approach:There are 3 success paths, namely,Success Path EventE1E2 AE1E3 BE4E5 C
Then Rs(t)=Ps= =P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC) =P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+
P(A)P(B)P(C) =P1P2+P1P3+P4P5-P1P2P3-P1P2P4P5
-P1P3P4P5+P1P2P3P4P5
assuming independence and where Pi=P(Ei) for i=1, 2, 3, 4, 5
)( CBAP
Solution
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Since Pi=e-λt for i=1,2,3,4,5
Rs(t)
hs(t)
tttt
ttttt
tttttttt
ttttttttt
5λ-4λ-3λ-2λ-
λ-λ-λ-λ-λ-
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