Fundamentals of reliability engineering and applications part2of3

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 eduelsayed@rci.rutgers.edu

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.

29