reliability lecture notes

7/30/2019 Reliability Lecture Notes

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STATISTICS OVERVIEW

This section overviews some methods used to determine the parameters of pdfs or

directly the R(t), F(t) and (t) from the given data. Both point estimation and confi-

dence interval estimation are discussed. A confidence interval provides a range of values

within which the distribution parameter is included with a specified confidence. These

methods to be overviewed are:

Curve Fitting

Moment Estimation

Maximum Likelihood Estimator

Maximum Entropy Estimator

Empirical methods

In curve fitting the parameters of the chosen distribution are selected to fit the data, usu-

ally using the least-squares method. In the curve fitting of data using few data points,

median rank tables obtained from Order Statistics are generally used. Since we do not

know the true rank of each failure (i.e. percent population failing before), these tables

help to assign probability values top failure times. Median rank tables are constructed so

that half of the time the probability is high and half of the time the probability is low. It is

assumed that the errors would cancel each other in the long run.

Example 1

Ten identical devices are tested until failure at times tn = 1.7, 3.5, 5.0, 6.5, 8.0, 9.6, 11.0,

13.0, 18.0, 22.0 (100) h (n = 1, . . . , 10). Assuming that failures are random, find the reli-ability function and MTTF for the device and 95% confidence interval for the MTTF.

Solution

n 1 2 3 4 5 6 7 8 9 10

tn (h) 170 350 500 650 800 960 1100 1300 1800 2200

F(tn) 0.066 0.162 0.258 0.355 0.451 0.548 0.644 0.741 0.837 0.933

Since failures are random,

F(t) = 1 e t => t= ln

1

1 F(t)

= ln

1

R(t)

d(t).

Let

AAT = [170 350 500 650 800 960 1100 1300 1800 2200]


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and

ddT = [d(t1). . . . d(t10)]

= [0. 0683 0. 1767 0. 2984 0. 4385 0. 5997 0. 7941 1. 0328 1. 3509 1. 8140 2. 7031].

Then AA = ddand can be estimated from the least-squares fitting

= AATAA

1

AATdd= 0. 0010 h1.

In general, least-squares fitting fits a set of data pairs (yn, tn) (n = 1. . . N) to a specifiedfunctional form y(t) such that

N

n=1[yn y(tn)]2

is minimum. The above expression for fits natural log of 1/[1 F(tn)] or 1/R(tn) to astraight line passing through the origin. The least-squares fitting could be also performed

using the matlab function polyfit in which case the line may not pass through the origin

and have a different slope.

0 500 1000 1500 2000 25000

0.5

1

1.5

2

2.5

3

t(hours)

ln[1/R(t)]

** Data

__ Leastsquares fitting with matlab polyfit (=0.0013/h__ Leastsquares fitting with matlab polyfit (=0.0013/h)

Leastsquares fitting with the expression in the notes (=0.0010/h)


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0 500 1000 1500 2000 25000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t(hours)

R(t)

__=0.0013/h

=0.0010/h

Using = 0. 0010 h1, the reliability function for the device is

R(t) = e 0.0010t

where tis hours. The MTTF is

MTTF=1

= 1000 hours

Whenever tables are not available median ranks can be estimated from

(n 0. 3)/(N+ 0. 4) where n is the failure order and N is the sample size.

A 100(1- alpha ) percent confidence interval for the MTTF obtained from the exponential

distribution based on complete data is

2T

2/2,2r) MTTF

2T

21/2,2r

where T is the total test time, r is the number of failures and 2/2,2(r+1) and 21/2,2r are

found from Chi-square distribution tables. For this problem,

= 0. 05 T=10

i=1 = tn = 9830 hour r= 10

=> 20.025,22) = 36. 8 0.975,20 = 9. 59

=> 267 MTTF 1025 hour


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For moment estimation, the two most commonly used estimators for two-parameter

distributions f(t) of the stochastic variable tare the mean and variance estimators:

mean t=1

N

N

n=1

tn

variance 2 =1

N 1

N

n=1 (tn t)2.

Example 2

For the failure times ofExample 1, estimate the parameters and of the Weibull distri-

bution using moment estimators.

Solution

From the data given,

t=1

10

10

n=1 tn = 983 h

and

2 =1

10

10

n=1 (tn 983)2 = 4. 114 105 h2

For the Weibull distribution

t=

0

dt t t

1

e(t/) = (1 + 1/ ) = 983

2 =

0

dt(t t)2 t

1

e(t/) =

2{(1 + 2/ ) [(1 + 2/ )]2} = 4. 114 105

Numerical solution of these equations yields = 1. 58 and

= 1095 h. If f(t) has morethan two parameters, higher moment estimators are needed.

The maximum likelihood estimator obtains the M parameters m , (m = 1, . . . , M)of the pdf f(t, 1, 2, . . . , M) from

m

ln L = 0 , (m = 1, . . . , M) and,


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where L is the likelihood function defined as

L N

n=1 f(tn,1, 2, . . . , M).

Example 3

Derive the maximum likelihood estimators for the lognormal distribution

Solution

f(t) =1

2 texp

ln2(t/)

22

lnL

= N

1

2 ln 2 + ln

N

n=1 [lnt

n +

ln2(tn/)

22 ]

ln L = 0 => 2 =1

N

N

n=1 ln2(tn/)

ln L = 0 => ln =1

N

N

n=1 ln tn

Example 4

Estimate and for the data given in Example 1, if the failure times satisfy the lognor-

mal distribution.

Solution

ln =1

N

N

n=1 ln tn = 110 [ln(170) + ln(350) + ln(500) + ln(650) + ln(800)

+ ln(960) + ln(1100) + ln(1300) + ln(1800) + ln(2200)] = 6. 66 => = 780. 691

2 =1

N

N

n=1 ln2(tn/) = 110 [ln2(0. 218) + ln2(0. 448) + ln2(0. 641) + ln2(0. 769) + ln2(1. 025)

+ ln2(1. 230) + ln2(1. 409) + ln2(1. 665) + ln2(2. 306) + ln2(2. 818)] = 0. 539 => = 0. 734274

The maximum entropy estimator differs from the maximum likelihood estimator mainly

due to the weighting of the chosen distribution in the function to be maximized. For this


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purpose, we first define the entropy function

H= N

n=1 f(tn) ln[f(tn)]

where f(tn) is the chosen distribution and rewrite f(tn) as

f(tn) = exp[ 0 1T1(tn) 2T2(tn) MTM(tn)]

where m (m = 0, 1, . . . , M) are functions of the parameters to be estimated and Tm(tn)(n = 0, 1, . . . , N) are functions of time. Then the estimates are obtained from

0m

=1

N

N

n=1Tm(tn) m = 1, . . . , M

Example 5

Derive the maximum entropy estimators for the gamma distribution

Solution

f(t) =(t)r1

(r)et= eln() + (r 1) ln() + (r 1) ln(t) ln[(r)] t

=> 0 = ln[(r)] rln() 1 = 2 = 1 r

T1(t) = t T2(t) = ln(t)

=> 0

1=

[ln[(r)] rln()] =

r

=

1

N

N

n=1 tn

01

= 0

(1 r)=

0r

{ln[(r)] rln()} =(r) ln() =1

N

N

n=1 ln(tn)

where

(x) =d(x)

dx

1

(x)

is called the digamma function.

Empirical methods are also referred to as non-parametric methods or distribution-free

methods because R(t), F(t) and (t) are obtained directly from the data without assum-

ing a functional form. They are used when no parametric distribution fits the data.


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Ungrouped Complete Data

Let N n be the number of units surviving out of a population ofN at time tn. A possi-ble set of estimates for the reliability function and the Cdf are, respectively,

R(tn) = N nN =1 nN

and F(tn) = 1 R(tn) = nN. (I)

Such a choice of estimates predicts R(tN) = 0 which is not realistic. Tw o other possiblechoices are

F(tn) =n

N+ 1and R(tn) = 1 F(tn) =

N+ 1 nN+ 1

(II)

and

F(tn) =n 0. 3

N+ 0. 4and R(tn) = 1 F(tn) =

N + 0. 1 n

N+ 0. 4. (III)

The pdf can be obtained from

f(tn) = R(tn+1) R(tn)

tn+1 tn

and the failure rate can be obtained from

(tn) =f(tn)

R(tn).

Such choices are called plotting positions because they yield the ordinate values in plot-

ting R(tn), F(tn) or f(tn).

Example 5

For the failure times tn given below, plot f(tn) and R(tn) for the plotting position options

I, II and III.

n 1 2 3 4 5 6 7 8 9 10

tn (h) 170 350 500 650 800 960 1100 1300 1800 2200


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Solution

0 500 1000 1500 2000 25000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10

3

Time (hours)

f(t)

f(tn)=n/Nf(tn)=n/(N+1)

f(tn)=(n0.3)/(N+0.4)

with quadratic fittings

0 200 400 600 800 1000 1200 1400 1600 1800 20000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (hours)

R(t)

F(tn)=n/N

F(tn)=n/(N+1)

F(tn)=(n0.3)/(N+0.4)

For large N, the difference between different options is negligible.


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Grouped Complete Data

If for Nnumber of units at the start of the test, the failure data are reported in terms of the

number nk of units surviving prior to failures at ordered times tk (k= 0, 1, . . . , K) insteadof individual failure times, the data are referred to as grouped data. Then for tk

1 t< tk

(k= 1, 2, . . . , K)

R(t) =nk

N(I)

F(t) = 1 R(t) = 1 nk

N

f(t) = R(tk) R(tk1)

tk tn1=

nk1 nk(tk tk1)N

,

(tk) =f(tk)

R(tn) =

nk1 nk(tk tk1)nk

.

Example 6

Plot f(tk) and R(tk) for the data ofExample 5 for grouping in time intervals tk1 t< tkwith tk = 0, 200 , 400, 700, 1000, 1500, 2500.

Solution

tk (hours) nk

200 8

400 7

700 5

1000 3

1500 1

2500 0


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0 500 1000 1500 2000 25000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10

3

Time (hours)

f(t)

Grouped Data

f(tn)=n/N

f(tn)=n/(N+1)f(tn)=(n0.3)/(N+0.4)


Ungrouped Censored Data

Sometimes it is necessary to remove points from test data due to wearout, systematic

errors etc. The removal process is called sensoring. One way to account for sensoring is

to use the Lewis product limit estimator:

R(tk) =nk+ 1N+ 1

=> R(tk) =nk+ 1

nk1 + 1R(tk1).

Here nk is again the number of units surviving prior to failures at ordered times tk. The

interpretation of the quantitynk+ 1

nk1 + 1is that it is the conditional probability that the unit

will survive from tk1 to tk, giv en that it has survived until tk1.

Another product limit estimator is the Kaplan-Meier estimator:

R(tk) =

1 for t0 = 0

uncensored k 1

1

nk+ 1

otherwise.

Again,

f(t) = R(tk) R(tk1)

tk tk1(tk1 t< tk for k uncensored).


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Example 7

Plot f(tk) and R(tk) for the data ofExample 5 for sensoring at times t = 350, 800 and

1300 h using the Lewis and Kaplan-Meier productlimit estimators.

k 1 2 3 4 5 6 7 8 9 10

tk (h) 170 350+ 500 650 800+ 960 1100 1300+ 1800 2200

Solution

Using the Lewis estimator

k tk R(tk)

0 0 1

1 170 10/11 = 0.9091

2 350+

3 500 (8/9)*0.9091 = 0.8081

4 650 (7/8)*0.8081 = 0.7071

5 800+

6 960 (5/6)*0.7071 = 0.5892

7 1100 (4/5)*0.5892 = 0.4714

8 1300+

9 1800 (2/3)*0.4714 = 0.3143

10 2200 (1/2)*0.3143 = 0.1572

Using the Kaplan-Meier estimator

k tk R(tk)

0 0 1

1 170 1-1/10 = 0.9000

2 350+

3 500 (1-1/8)*0.9000 = 0.7875

4 650 (1-1/7)*0.7875 = 0.6750

5 800+

6 960 (1-1/5)*0.6750 = 0.5400

7 1100 (1-1/4)*0.5400 = 0.4050

8 1300+

9 1800 (1-1/2)*0.4050 = 0.2025

10 2200 (1-1/1)*0.3143 = 0


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0 500 1000 1500 2000 25000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10

3

Time (hours)

f(t)

Lewis

KaplanMeier

f(tn)=(n0.3)/(N+0.4)


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