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Fundamentals of ReliabilityFundamentals of Reliability Engineering and Applications
Part 2 of 3
E. A. Elsayed©2011 ASQ & Presentation ElsayedPresented live on Dec 07th, 2010
http://reliabilitycalendar.org/The_Reliability Calendar/Short Courses/Shliability_Calendar/Short_Courses/Short_Courses.html
ASQ Reliability DivisionASQ Reliability Division Short Course SeriesShort Course SeriesThe ASQ Reliability Division is pleased to present a regular series of short courses
featuring leading international practitioners, academics and consultantsacademics, and consultants.
The goal is to provide a forum for the basic andThe goal is to provide a forum for the basic and continuing education of reliability
professionals.
http://reliabilitycalendar.org/The_Reliability Calendar/Short Courses/Shliability_Calendar/Short_Courses/Short_Courses.html
Fundamentals of Reliability Engineering and Applications
E. A. Elsayedelsayed@rci rutgers [email protected]
Rutgers University
December 7, 20101
,
OutlineOutlinePart 1. Reliability Definitions
Reliability Definition Time dependent Reliability Definition…Time dependent characteristics
Failure Rate Availability MTTF and MTBF Time to First Failure Mean Residual Life
C l i Conclusions
2
OutlineOutlinePart 2. Reliability Calculations
1 U f f il d t1. Use of failure data 2. Density functions 3 Reliability function3. Reliability function 4. Hazard and failure rates
3
OutlineOutlinePart 3. Failure Time Distributions
1 C t t f il t di t ib ti1. Constant failure rate distributions 2. Increasing failure rate distributions 3 Decreasing failure rate distributions3. Decreasing failure rate distributions 4. Weibull Analysis – Why use Weibull?
4
OutlineOutlinePart 2. Reliability Calculations
1 U f f il d t1. Use of failure data a) Interval data (no censoring)b) Exact failure times are knownb) Exact failure times are known
2. Density functions 3. Reliability function y4. Hazard and failure rates
5
Basic Calculations
Suppose n0 identical units are subjected to aSuppose n0 identical units are subjected to a test. During the interval (t, t +∆t ), we observed nf (t ) failed components Let n (t ) be thenf (t ) failed components. Let ns (t ) be the surviving components at time t , then we define:
Failure density function 0
( )ˆ( ) fn tf tn t
Failure rate function
0
( )ˆ( ) , ( )fn th t
n t t
Reliability function
( )sn t t
( )ˆ ( ) ( ) sn tR t P T t6
Reliability function60
( )( ) ( ) srR t P T t
n
Basic Definitions Cont’dThe unreliability F(t) is
︵ ︶1 ︵ ︶F t R t ︵ ︶1 ︵ ︶F t R t
Example: 200 light bulbs were tested and the failures in 1000-hour intervals are
Time Interval (Hours) Failures in theinterval
0-10001001-20002001 3000
10040202001-3000
3001-40004001-5000
201510
5001-60006001-7000
87
Total 200
7
Total 200
7
Calculations
Time I t l
Failure Density( )f t x 410
Hazard rate( )h t x 410Interval ( )f t x 410 ( )h t x 410
0-1000 3
1 0 0 5 .02 0 0 1 0
3
1 0 0 5 .02 0 0 1 0
Time Interval Failures
1001 2000
4 0 2 0
4 0 4 0
(Hours) in theinterval
0-1000 100 1001-2000
2001-3000
3 2 .02 0 0 1 0
2 0 1 .0
3 4 .01 0 0 1 0
3
2 0 3 .3 3
1001-20002001-30003001-4000
402015 2001 3000
……
3 1 .02 0 0 1 0
……..
36 0 1 0
……
4001-50005001-60006001-7000
1087
6001-7000
3
7 0 .3 52 0 0 1 0
3
7 1 07 1 0
Total 200
88
Failure Density vs. Time
×10-
4
1 2 3 4 5 6 7 x 103
Ti i h9
Time in hours9
Hazard Rate vs. Time
-4×1
0-
1 2 3 4 5 6 7 × 103
Time in Hours
1010
Reliability CalculationsReliability Calculations
Time Interval Reliability ( )R t 0 1000 5/5=1 0
Time Interval(Hours)
Failures in theinterval 0-1000
1001-2000
5/5=1.0 2.0/4.0=0.5
interval
0-10001001-20002001 3000
1004020
2001-3000
/ 1/3.33=0.33
2001-30003001-40004001-50005001 6000
2015108
…… ……
5001-60006001-7000
87
Total 200
6001-7000 0.35/10=.035
1111
Reliability vs. Timey
1 2 3 4 5 6 7 x 103
Time in hours
1212
Exponential DistributionExponential Distribution
Definition (t)
( ) 0, 0h t t Time
( ) exp( )f t t
( ) exp( ) 1 ( )R t t F t ( ) p( ) ( )
1313
Exponential Model Cont’dExponential Model Cont d
Statistical Properties
1
MTTF 6 Failures/hr5 10
MTTF=200,000 hrs or 20 years
2
1Variance 6 Failures/hr5 102 Standard deviation of MTTF is
200,000 hrs or 20 years
12Median life ︵ln ︶ Median life =138,626 hrs or 14
years
1414
Exponential Model Cont’dExponential Model Cont d
Statistical Properties
6F il /h5 101
MTTF 6Failures/hr5 10
MTTF=200,000 hrs or 20 years
It is important to note that the MTTF= 1/failureIt is important to note that the MTTF 1/failure is only applicable for the constant failure rate case (failure time follow exponential distribution.( p
1515
Empirical Estimate of F(t) and R(t)Empirical Estimate of F(t) and R(t)When the exact failure times of units is known, weuse an empirical approach to estimate the reliabilitymetrics. The most common approach is the RankppEstimator. Order the failure time observations (failuretimes) in an ascending order:times) in an ascending order:
1 2 1 1 1... ...i i i n nt t t t t t t
16
Empirical Estimate of F(t) and R(t)p ( ) ( )
is obtained by several methods( )iF t y
1 Uniform “naive” estimatori
1. Uniform naive estimator
2 Mean rank estimator
ni
2. Mean rank estimator
3 M di k ti t (B d)
1n0 3.i
3. Median rank estimator (Bernard) 0 4.n
3 8/i4. Median rank estimator (Blom) 3 81 4
//
in
17
Empirical Estimate of F(t) and R(t)Assume that we use the mean rank estimator
iˆ ( )11
iiF t
ni
11ˆ( ) 0,1, 2,...,
1i i in iR t t t t i n
n
Since f (t ) is the derivative of F(t ), then
11
ˆ ˆ( ) ( )ˆ ( )( 1)
i ii i i i
F t F tf t t t tt
.( 1)1ˆ ( )
i
i
t n
f t
1818
( ).( 1)i
i
ft n
Empirical Estimate of F(t) and R(t)1ˆ( )
( 1 )it
( ).( 1 )
ˆ ˆ( ) ln ( ( )
ii
i i
t n i
H t R t
( ) ( ( )i i
Example:
Recorded failure times for a sample of 9 units are observed at t=70 150 250 360 485 650 855observed at t=70, 150, 250, 360, 485, 650, 855, 1130, 1540. Determine F(t), R(t), f(t), ,H(t)︵ ︶t
1919
Calculations
i t (i) t(i+1) F=i/10 R=(10-i)/10 f=0.1/t =1/(t.(10‐i)) H(t)i t (i) t(i+1) F i/10 R (10 i)/10 f 0.1/t 1/(t.(10 i)) H(t)
0 0 70 0 1 0.001429 0.001429 0
1 70 150 0.1 0.9 0.001250 0.001389 0.10536052
2 150 250 0.2 0.8 0.001000 0.001250 0.22314355
3 250 360 0.3 0.7 0.000909 0.001299 0.35667494
4 360 485 0.4 0.6 0.000800 0.001333 0.51082562
5 485 650 0.5 0.5 0.000606 0.001212 0.69314718
6 650 855 0.6 0.4 0.000488 0.001220 0.91629073
7 855 1130 0.7 0.3 0.000364 0.001212 1.2039728
8 1130 1540 0.8 0.2 0.000244 0.001220 1.60943791
9 1540 - 0.9 0.1 2.30258509
2020
Reliability Functiony
1.2
1
1.2
R li bilit
0.6
0.8Reliability
0.4
0
0.2
0 200 400 600 800 1000 1200
Time
2121
Probability Density FunctionProbability Density Function
0.001600
0.001200
0.001400
0.000800
0.001000
Density Function
0.000400
0.000600
Function
0 000000
0.000200
0.000400
0.0000000 200 400 600 800 1000 1200
Time
2222
Constant Failure Rate
0 001500
0.001700
0.001900
0.001100
0.001300
0.001500
Failure Rate
0.000700
0.000900
Failure Rate
0.000300
0.000500
0.0001000 200 400 600 800 1000 1200
Time
2323
Exponential Distribution: Another ExampleExponential Distribution: Another Example
Gi f il d tGiven failure data:
Plot the hazard rate, if constant then use the exponential distribution with f (t), R (t) and h (t) as defined before.
We use a software to demonstrate these steps.
2424
Input Data
2525
Plot of the Data
2626
Exponential Fit
2727
Exponential Analysis
SummarySummary
In this part, we presented the three most importantrelationships in reliability engineering.relationships in reliability engineering.
We estimated obtained estimate functions for failureWe estimated obtained estimate functions for failure rate, reliability and failure time. We obtained these function for interval time and exact failure timesfunction for interval time and exact failure times.
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