1 an application of probability to reliability modeling and analysis dr. jerrell t. stracener, sae...
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An Application of Probability toReliability Modeling and Analysis
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Systems Engineering ProgramDepartment of Engineering Management, Information and Systems
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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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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
and, in particular
• Exponential
1
P p)-ln(1- t
t 632.0
p)-ln(1- Pt
Percentiles, tp
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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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• 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 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 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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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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Weibull
Exponential
MTBF
1
1 MTBF
Mean Time Between Failure - MTBF
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Problem -
Four Engine Aircraft
Engine Reliability R(t) = p = 0.9
Mission success: At least two engines survive
Find RS(t)
The Binomial Model - Example
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Solution –
1) X1 = number of engines surviving in time t
Then X1 ~ B (4,0.9)
RS(t) = P(x1 ≥ 2) = b(2) + b(3) + b(4)
= 0.0486 + 0.2916 + 0.6561 = 0.9963
2) X2 = number of engines failing in time t
Then X2 ~ B (4,0.1)
RS(t) = P(x2 2) = b(0) + b(1) + b(2)
= 0.6561 + 0.2916 + 0.0486 = 0.9963
The Binomial Model - Example
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The Weibull Model
Time to failure of an item follows a Weibull distribution with = 2 and = 1000 hours.
(a) What is the reliability, R(t), for t = 200 hours?
(b) What is the hazard rate, h(t), (instantaneous failure rate) at that time?
(c) What is the Mean Time To Failure?
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The Weibull Model –Solution
(a)
(b)
9608.0
)200(2)1000/200(
eR
1t
)t(h
/te)t(R
0004.0
1000
2002)200( 2
12
h
failures per hour
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The Weibull Model –Solution continued
(c)
From the Gamma Function table:
2
111000
11
MTTF
23.886
5.11000
MTTF
88623.0)5.1(
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The design life of a given type of pump for a given operating environment has a Lognormal distribution. If t0.10 = 2000 hours and the median life is 3748 hours. What is the mean life and the the 50th & 90th percentile? A system uses five pumps of this type. What is the probability of at least one of these pumps failing in 3000 hours?
Example – Pump life
Pump Life Solution
Given that t0.10 = 2000 hours and the median is 3748, we need to first find the values for and .
Since the median life is
=ln 3748
=8.2290
And since t0.1= =2000,
-1.28 = ln 2000
1.28 = 8.2290 – 7.6009
= 0.4907
μe3748
1.28σ-μe
Pump Life Solution
Now t0.9 is the value of t for which
But
So that
and t0.9=7023.8
Probability of one pump failing within 3000 hours
326.00.491
229.83000ln Φ3000)P(TF(3000)
90.00.491
229.8ln t)F(t)tP(T 0.90
0.90.90
90.0)28.1(tF 0.90
1.28,0.491
229.8ln t0.9
Pump Life - solution
Now t0.5 is the value of t for which
But
So that
And t0.5=3748
Mean life is
6.4227eeE(t) 8.34939μ 2
2σ
50.0σ
μln t)F(t)tP(T 0.50
0.50.50
50.0)0(tF 0.50
00.491
229.8ln t0.5
Pump Life Solution
Probability of at least one pump failing in 3000 hours
Y = # of pumps failing in 3000 hoursy = 0, 1, 2, 3, 4, 5
Y has a binomial distribution withn = 5 and p = 0.326
or you could work this using a probability tree
)0(1)1( YPYP
50 674.0326.00
51
,861.0
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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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• 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
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
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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
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
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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
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
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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.
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
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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
Stracener_EMIS 7370/STAT 5340_Fall 08_09.30.08
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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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Compare the following reliability configurations I, II, and III in terms of (a) system reliability, (b) system failure rate, (c) system mean time between failures and (d) system mean time between maintenance, assuming that a failure requires maintenance.Element E has an exponential distribution of time to failure, T, with failure rate = 0.01. State all ground rules and assumptions, show all work and present the results graphically to convey your results.
I. II. III.
EE
E
E
E
Example - 3 Configurations
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Example - 3 Configurations - Solution
For the baseline system:
and
,etR λtS
,λthS
θ,MTBFS
θMTBMS
E
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Example - 3 Configurations – Solution
For alternative A:
and
E
E
,e-e2tR λt2λtSA
,e2
e12λth
0.01t
0.01t
SA
1.5θ.MTBFSA
0.5θ2
θ
2λ
1MTBMSA
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Example - 3 Configurations – Solution
For alternative B:
and
E
E
,λt1
tλth
2
SB
2θMTBFSB
,eλt1tR λtSB
θMTBMSB
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Now we compare Alternatives A & B to the baseline system.
In terms of reliability,
and
λtλt
λt2λt
S
SA e-2e
e-2e
tR
tR
λt1
e
eλt1
tR
tRλt
λt
S
SB
Example - 3 Configurations – Solution
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In terms of failure rate,
and
0.01t
0.01t0.01t
0.01t
S
SA
e2
e12
λe2e1
2λ
th
th
λt1
1
1
λt1
λt
λλt1tλ
th
th
2
S
SB
Example - 3 Configurations – Solution
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In terms of MTBF,
and
In terms of MTBM,
and
1.5θ
1.5θ
MTBF
MTBF
S
SA
2θ
2θ
MTBF
MTBF
S
SB
5.0θ
.5θ0
MTBM
MTBM
S
SA
1θ
θ
MTBM
MTBM
S
SB
Example - 3 Configurations – Solution
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0
1
2
3
4
5
6
7
8
0 100 200 300 400 500 600 700
Rs
Rsa
Rsb
Rsa/Rs
Rsb/Rs
Example - 3 Configurations – Solution
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0
0.2
0.4
0.6
0.8
1
1.2
0 100 200 300 400 500 600 700
hs
hsa
hsb
hsa/hs
hsb/hs
Example - 3 Configurations – Solution
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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-
λ-λ-λ-λ-λ-λ-λ-λ-
-λ-λ-λ-λ-λ-λ-λ-λ-λ
ee2e3e
))(e)(e)(e)(e(e
))(e)(e)(e(e-))(e)(e)(e(e-
))(e)(e(e-))(e(e))(e(e))(e(e
ttt
ttt
tttt
tttt
s
sdtd
e
tR
tR
3λ-2λ-λ-
3λ-2λ-λ-
3λ-2λ-λ-λ2
5λ-4λ-3λ-2λ-
ee2e3
e5e8e36λ
)ee2e3(
λe5λe8λe36e
)(
)(
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MTBFs
0.87θ θ30
6151045λ5
1
λ2
1
λ3
1
λ2
3
λ5
e
λ2
e
λ3
e
λ2
3e
)(
0
5λ-4λ-3λ-2λ-
0
tttt
s dttR
θ2.0λ5
1SMTBM