stracener_emis 7305/5305_spr08_02.28.08 1 reliability data analysis and model selection dr. jerrell...

Stracener_EMIS 7305/5305_Spr08_02.28.08

Reliability Data Analysis and Model Selection

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7305/5305Systems Reliability, Supportability and Availability Analysis

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

Reliability Model Selection

•Estimation of Reliability Model Parameters

•Probability Plotting

Estimation of Reliability Model Parameters

• Estimation of Binomial Distribution Parameters

• Estimation of Normal Distribution Parameters

• Estimation of Lognormal Distribution Parameters

• Estimation of Exponential Distribution Parameters

• Estimation of Weibull Distribution Parameters

Estimation - Binomial Distribution

Estimation of a Proportion, p

• X1, X2, …, Xn is a random sample of size n fromB(n, p), where

• Point estimate of p:

where fs = # of successes

n ..., 1,ifor

failure if 0

success if 1

Estimation - Normal Distribution

Estimation of the Mean - Normal Distribution

• X1, X2, …, Xn is a random sample of size n from N(, ), where both µ & σ are unknown.

• Point Estimate of

• Point Estimate of s

1ii XX

Estimation - Lognormal Distribution

Estimation of Lognormal Distribution

• Random sample of size n, X1, X2, ... , Xn from LN (, )

• Let Yi = ln Xi for i = 1, 2, ..., n

• Treat Y1, Y2, ... , Yn as a random sample from N(, )

• Estimate and using the Normal DistributionMethods

Estimation of the Mean of a Lognormal Distribution

• Mean or Expected value or MTBF

• Point Estimate of MTBF

where and are point estimates of and respectively.

•Median time to failure

•Point estimate of median time To Failure

σμ μ σ

5.0 et

Estimation - Exponential Distribution

Estimation of Exponential Distribution

• Random sample of size n, X1, X2, …, Xn, from E(), where is unknown.

Estimation - Weibull Distribution

Estimation of Weibull Distribution

• Random sample of size n, T1, T2, …, Tn, from W(, ), where both & are unknown.

• Point estimates

• is the solution of g() = 0

lnTTβg

Estimation of the Mean of a Weibull Distribution

• Mean or Expected value or MTBF

• Point Estimate of MTBF

where and are point estimates of and respectively.θβ

1ΓθMTBF

1θΓMTBF

Example

The following data represents a random sample from the normal distribution N(,) :

94 74 105 126 124 135 56 95 122 78 86 66 63 80 85 58 89 92 103 93

Estimate the population parameters. Then estimate the 90% percentile, and plot estimates of the probability density and distribution functions.

Example solution

T ~ N(, ), then the estimates of are is:

2.91T20

Example solution

Normal Model N(, ):

2.91ˆt0.5

3.120ˆ28.1ˆt0.9

7.22ˆ

2.91MTBF^

Standard Deviation

Example solution

- Probability Plotting

Probability Plotting

• Data are plotted on special graph paper designed for a particular distribution

- Normal - Weibull- Lognormal - Exponential

• If the assumed model is adequate, the plotted points will tend to fall in a straight line

• If the model is inadequate, the plot will not be linear and the type & extent of departures can be seen

• Once a model appears to fit the data reasonably well, parameters and percentiles can be estimated.

Probability Plotting Procedure

• Step 1: Obtain special graph paper, known asprobability paper, designed for each of the followingdistributions:Weibull, Exponential, Lognormal and Normal.http://www.weibull.com/GPaper/index.htm

• Step 2: Rank the sample values from smallest to largest in magnitude i.e., X1 X2 ..., Xn.

Probability Plotting General Procedure

• Step 3: Plot the Xi’s on the probability paper versus

depending on whether the marked axis on the paper refers to the % or the proportion of observations. The axis of the graph paper on which the Xi’s are plotted will be referred to as the observational scale, and the axis for as the cumulative probability scale.

3.0i100)x(F

i 4.0n

3.0i)x(F

)( ixF

The formula

is an approximation that can be used to estimate median ranks, called Benard’s approximation.

where n is the sample size and i is the sample order number. Tables of median ranks can be found in may statistics and reliability texts.

%)100(0.4n

0.3iMRxF ii

Median ranks represent the 50% confidence level (“best guess”) estimate for the true value of F(t), based on the total sample size and the order number (first, second, etc.) of the data.

•Step 4: If a straight line appears to fit the data, draw a line on the graph, ‘by eye’.

• Step 5: Estimate the model parameters from the graph.

Weibull Probability Plotting Paper

the cumulative probability distribution function is

We now need to linearize this function into the form y = ax +b:

et 1)(F

θβ,W~T

which is the equation of a straight line of the form y = ax +b,

lnln)T(F1

ln)T(F1lnln

)T(F1ln

ln)T(F1ln

,)t(F1

i.e., ,ln b

which is a linear equation with a slope of b and an intercept of Now the x- and y-axes of the Weibull probability plotting paper can be constructed. The x-axis is simply logarithmic, since x = ln(T) and

,ln xy

,)t(F1

cumulativeprobability

(in %)

Probability Plotting - example

To illustrate the process let 10, 20, 30, 40, 50, and 80 be a random sample of size n = 6.

Probability Plotting – Example Solution

Based on Benard’s approximation, we can now calculate F(t)

for each observed value of X. These are shown in the following

table:

For example, for x2=20,

%100*0.46

0.3220F

i xi F(xi)

1 10 10.9%2 20 26.6%3 30 42.2%4 40 57.8%5 50 73.4%6 80 89.1%

cumulativeprobability

(in %)

Probability Plotting- example

Now that we have y-coordinate values to go with the x-coordinate sample values so we can plot the points on Weibull probability paper.

F(x)(in %)

Probability Plotting- example

The line represents the estimated relationship between x and F(x):

F(x)(in %)

In this example, the points on Weibull probability paper fall in a fairly linear fashion, indicating that the Weibull distribution provides a good fit to the data. If the points did not seem to follow a straight line, we might want to consider using another probability distribution to analyze the data.

Probability Paper - Normal

Probability Paper - Lognormal

Probability Paper - Exponential

Example - Probability Plotting

Given the following random sample of size n=8,which probability distribution provides the best fit?

1 79.409682 88.120933 91.063944 98.730945 104.15366 105.10197 106.50368 112.0354

40 specimens are cut from a plate for tensile tests. The tensile tests were made, resulting in Tensile Strength, x, as follows:

i x i x i x i x1 48.5 11 55.0 21 53.1 31 54.62 54.7 12 55.7 22 49.1 32 49.93 47.8 13 49.9 23 55.6 33 44.54 56.9 14 54.8 24 46.2 34 52.95 54.8 15 49.7 25 52.0 35 54.46 57.9 16 58.9 26 56.6 36 60.27 44.9 17 52.7 27 52.9 37 50.28 53.0 18 57.8 28 52.2 38 57.49 54.7 19 46.8 29 54.1 39 54.8

10 46.7 20 49.2 30 42.3 40 61.2

Perform a statistical analysis of the tensile strength data and estimate the probability that tensile strength on a new design will be less than 50, i.e, reliability at 50.

Example: 40 Specimens

04/21/23

Time Series plot:

By visual inspection of the scatter plot, there seems to be no trend.

40 Specimens

0 5 10 15 20 25 30 35 40

40 Specimens

Descriptive Statistics

Count 40Sum 2104.82Mean 52.62Median 53.03Standard Deviation 4.45Range 18.84Minimum 42.35Maximum 61.18

Using the descriptive statistics function in Excel, the following were calculated:

40 Specimens

From looking at the Histogram and the Normal Probability Plot, we see that the tensile strength can be estimated by a normal distribution.

Using the histogram feature of excel the following data was calculated:

and the graph:

Bin Frequency40 045 350 1055 1660 9

More 2

Histogram of Tensile Strengths

40 45 50 55 60 More

40 SpecimensBox Plot

40 45 50 55 60 65

The lower quartile 49.45The median is 53.03The mean 52.6The upper quartile 55.3The interquartile range is 5.86

lowerextreme upper

extreme

lowerquartile

upperquartile

medianmean

40 Specimens

Normal Probability Plot

99.90%

40 45 50 55 60 65

40 Specimens

LogNormal Probability Plot

99.90%

10 100

40 Specimens

Weibull Probability Plot

40%50%60%70%80%

90%95%

99.90%

41 44 48 52 56 61

40 Specimens

The point estimates for μ and σ are:

The tensile strength distribution can be estimated by

40 Specimens

45.4ˆ,62.52μN~X

Norm al Dis tributions f(x)

49 50 51 52 53 54 55

Normal Distribution F(x)

40 Specimens

Estimate of Probability that P(x<50) is:

325.040

13)50(ˆ

2776.0)5888.0(

62.5250()50(ˆ)50(ˆ

stracener_emis 7305/5305_spr08_02.28.08 1 reliability data analysis and model selection dr. jerrell...

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