Reliability CGG2

8/14/2019 Reliability CGG2

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Author: Carlos Gonzlez Gonzlez

Reliability

Software RELIAB_EN.exe Download from site www.spc-inspector.com/cgg

1.5 WEIBULL TEST, SINGLE AND MULTIPLE CENSORED

1.5.1.- Weibull Fit with Single Censored Data

When using a Weibull Probability Paper, the failure times are plotted in ascending orderover the the Weibull Probability Paper using a median ranks table matched with the proper sample

size or you can use the formula developed by (Kapur & Lamberson).

The best fitting line is drawn through the plotted points, and a parallel line to the best fitting

line is transferred and drawn to determine the shape parameter (). The percentage of units that is

expected to survive at a given time it is established along to the vertical axis corresponding to the

best fitting line. The horizontal axis is to establish the time or number of cycles to fail.

Example: When you ran a life test of a mechanical system, with a sample of 20 units

The obtained results were collected in hours per failure for each unit, they were not replaced orrepaired. Numbers are as follows: 2200, 2400, 3125, 3300, 3400, 3500, 3550, 3650, 4000, 6000.

hours. The other 10 were censored all together at once when accomplished 8000 hours

Step 1.- Data were ordered in ascendant manner by column for the 20 units.

1.- 2200

2.- 2400

3.- 3125

4.- 3300

5.- 3400

6.- 35007.- 3550

8.- 3650

9.- 4000

10.- 6000

11.- 8000

12.- 8000

13.- 8000

14.- 8000

15.- 8000

16.- 8000

17.- 8000

18.- 8000

19.- 8000

20.- 8000

Step 2.- Now we are going to use Median Ranks. Data are paired with data shown previously for n

= 20.

Failure Time of Mdn. Rank Kapur & Lamberson

order j failure t from table Mdn. Rank

1.- 2200 3.4 3.432.- 2400 8.3 8.33

3.- 3125 13.1 13.23

4.- 3300 18.1 18.13
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5.- 3400 23.0 23.04

6.- 3500 27.9 27.94

7.- 3550 32.8 32.84

8.- 3650 37.7 37.74

9.- 4000 42.6 42.64

10.- 6000 47.5 47.54

11.- 8000 52.5 52.45

12.- 8000 57.4 57.35

13.- 8000 62.3 62.2514.- 8000 67.2 67.15

15.- 8000 72.1 72.05

16.- 8000 77.0 76.96

17.- 8000 81.9 81.86

18.- 8000 86.9 86.76

19.- 8000 91.7 91.66

20.- 8000 96.6 96.56

If you do not have a median rank table that you can use then you can calculate the median

ranks with a formula developed by authors Kapur & Lamberson, 1977, then those values are going

to be paired with column t of failure

F(t) = ( j 0.3) / (n + 0.4) --( i )

Where:

j = Failure order

n = Sample size

Example: if the failure j = 5, n = 20 when you substitute those values in the formula ( i ) you

obtain: F(t) = (5 0.3) / (20 + 0.4) = 4.7 / 20.4 = 23.04 % and so forth every calculation for eachorder of failure the only change as you can see is that you are using n = 20.

Step 3.-. Using Weibull Probability Paper. There are different types of forms of Weibull Probability

Paper with horizontal or vertical shape and one, two, three or four cycles (we are going to use a two

cycle paper).

Step 4.- Once you select the form and shape of the Weibull Probability Paper, in this case two

cycles, horizontal axis has two logarithmic cycles, then you attach and scale the data.

Step 5.- Now you register pairs of values one at a time over the Weibull Probability Paper, plotting

each point where correspond hours and percentages.

Step 6.- Always you should find the best fitting line, the drawing of an straight line helped with a

transparent rule made of plastic is recommended with preference of the tendency of the majority of

points.

Step 7.- You need to transport in parallel with the same tilt, and slope as the line already drawn.

The origin where start such line is located over the vertical axis left side, near the top, prolonged

until cuts the graduated arc. (in this case, the value that you get is 3.0 which indicates a shape

between Log-Normal and a Normal distribution skewed a little bit to the left).

Step 8.- Then, now you should prolong the best fitting line until you cut the horizontal axis at the

top and at the bottom. In this last case the cut is around 550 hours which corresponds with the

minimum life of the product this means that 99.90% of population at least survive a minimum of


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550 hours. (Really 550 hours is reached by 99.90% of the product or just 0.10% can not get 550

Hours, according to the design of the Weibull Probability Paper).

Figure 1.13 Weibull Probability Paper plotting paired points

Step 9.- This straight line give us an instantaneously relation of % of failure and hours of life.

Example: What percentage of product will fail at 2400 hours of life? The procedure to know

that is as follows: You can go vertically starting in an horizontal value of 2400 until you get the

slopped line plotted previously, then you go horizontally to the left until you find the vertical axis of

percentages where you read the searched value. (In this case approximately 9%).

Step 10.- In this particular case the estimator is called the characteristic life.

(1 1/e) for this case too 5000 or 5500 hours is the value of the life .

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Figure 1.14 Weibull Probability Paper 3 Cycles

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Figure 1.15 Time of failure and censored for the sample of 20 (Condition)

A).- Assume = 0 and order the observations from smallest to largest.

B).- Compute the median rank as we do in step 2.

C).- Compute the natural logarithm of the time to fail for each observation.

D).- Compute the Natural Logarithm of the Natural Logarithm of the inverse of 1 the median rank

for each uncensored observation.

As an example for the first two median ranks calculated:

First Median Rank = 0.0343

Ln[Ln(1/(1F(t))] = Ln[Ln(1/(10.0343)] = Ln[Ln(1/0.9657)] = Ln[Ln(1.03552] = Ln(0.03490)

= 3.3550

Second Median Rank = 0.0833

Ln[Ln(1/(1F(t))] = Ln[Ln(1/(10.0833)] = Ln[Ln(1/0.9167)] = Ln[Ln(1.09087] = Ln(0.086975)

= 2.4420

Table 1.5 and figure 1.16 summarizes the calculations listed from A).- to D).-

Table 1.5 Calculations for 20 values of Failure Times 1 to 10 and 11 to 20 censored.

Order Time to Fail t D F(t) K&L Natural Log LnLn

# (t D) Median Rank (t D) (1/[1F(t D)])

1.- 2200 3.43 % 7.6962 3.3548

2.- 2400 8.33 % 7.7832 2.4417

3.- 3125 13.24 % 8.0472 1.9521

4.- 3300 18.14 % 8.1017 1.6088

5.- 3400 23.04 % 8.1315 1.3399

6.- 3500 27.94 % 8.1605 1.1157

7.- 3550 32.84 % 8.1747 .9210

8.- 3650 37.75 % 8.2025 .7467

9.- 4000 42.65 % 8.2940 .5871

10.- 6000 47.55 % 8.6995 .4381

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11.- 8000 Censored


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12.- 8000

20.- 8000 Censored

Figure 1.16 Statistical Calculation of Weibull study from software RELIAB_EN.exe.

Figure 1.17 Best fitting line and calculation of Delta = 0, Beta = 2.989, Teta = 5510, and R2 =

0.79871

E).- Plot the Ln (t ) on the x Axis and Ln [Ln(1/(1 (t ))] on the y axis. An alternative to

plotting Ln (t ) on the x axis and Ln [Ln(1/(1 (t ))] on the y axis is to use Weibull

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Probability Paper and plot (t ) versus F(t ). As we do some pages before instead of a computer

program as RELIAB_EN.exe

F).- Fit a straight line to the data points. This can be done visually, or a least square regression may

be used. The data Table 1.5 is plotted in figures 1.13 and 1.17. The slope of this line, which is equal

to , is 2.989. This is not the case but if the data in the probability plot appears to fall on a

downward or upward curve, may not be equal to zero. The time to fail must be transformed by

subtraction of an estimate of. A discussion of how to handle a nonzero parameter is made using

RELIAB_EN.exe.Probability plots are commonly used as goodness of fit tests. A straight line of the plotted

points indicates the chosen density function is acceptable.

G).- The scale parameter of the Weibull distribution, , can be calculated using the expression

= exp( yo /)

where:

yo = the y intercept

The y-intercept, yo may be found by extrapolating the line in figures 1.13 and 1.17. The y-

intercept for the data in table 1.5, is 25.7485. Thus, the scale parameter is

= exp( yo /) = exp [( ( 25.7485))/2.989] = 5510.549

1.5.2.- Applications

The shape parameter and probability density function take different shapes as is shown in

figure 1.18. Weibull Distribution can be used in a wide variety of applications, depending on values

of, when has other values the shape can be approach to other distributions par example:

Figure 1.18.- Values of Beta for different Probability Distributions.

= 1 The Weibull distribution is exactly an Exponential Distribution.

= 2 The Weibull distribution is exactly a Rayleigh Distribution.

= 2.5 The Weibull distribution approaches to Log-Normal Distribution. Those

distributions are nearly equal to Log-Normal Distribution but require a sample size bigger than 50

to be distinguished between them.

= 3.6 The Weibull distribution approaches to Normal Distribution.

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= 5 The Weibull distribution approaches to Normal Distribution Lepto-Kurtic.

Due to this flexibility, there are, very few failure rates observed that can not be exactly

modeled by the Weibull Distribution, some examples are:

a).- The components resistance or the required stress of metal fatigue.

b).- The time of failure of electronic components.

c).- Failure time of items that suffers wearout, as the auto tires.

d).- Systems fail when the less resistant component fails (Weibull Distribution represents a

distribution of extreme value).

1.5.3.- Weibull Fit with Multi-Censored Data

When you build a probability plot using multiple censored data, a modified failure order

must be calculated, considering the data in Table 1.6.

Table 1.6. Multi-censored data

Time to Fail and Censored

1.- 2200 c

2.- 2400 f

3.- 3100 c

4.- 3500 c

5.- 4000 f

6.- 4500 c

7.- 5000 f

8.- 5600 f

9.- 6300 c10.- 6600 c

11.- 7000 f

12.- 7600 f

13.- 8000 c

14.- 8400 f

15.- 8800 f

16.- 9200 f

17.- 9600 c

18.- 10100 f

19.- 11000 f

20.- 12000 c

In such table letters f and c mean f = failure and c = censored, censored indicates that

the item was removed from testing without failing; it was censored.

The second item could have been the second item to fail, the third item failed and the fourth

item was censored, because of that the item that failed at time = 3100 can not be considered as the

second ordered failure, and the item that failed and the item that failed at time = 4000 can not be

considered as the third ordered failure either; a weighted average ordered failure must be calculated

for each failure after the first censored data point. Equations (a) and (b) can be used to determine

the modified failure order for multiple-censored data.

Ij = [(n+1) Op] / 1 + r ---(a)

where:


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Ij = The increment for the jth failure

n = The total number of data points, both censored and uncensored

Op = The order of the previous failure

r = The number of data points

Oj = Op Ij ---(b)

where:

Oj = The order of the jth failure

Steps to plot the probability when multiple-censored data

1.- Assume = 0 and order the observations from smallest to largest. This has already been

done in table 1.6.

2.- For each uncensored observation compute the increment, Ij, using equation (a), and the

modified order number, Op using equation (b).

For the second ordered observation the first failure occurred at 2400 hours, (the first

uncensored observation), using equation, (a).

Ij = [(20 +1) 0] / (1 + 19) = 19/20 = 1.05

For the eighth ordered observation, (the fourth uncensored observation),

Using the equation Ij = [(n+1) Op] / 1 + r ---(a)

Ij = [(20 +1) 3.4753] / (1 + 13) = 17.5247 / 14 = 1.251764 1.2518

Oj = 3.4753-1.2518 = 2.2235

3.- For each uncensored observation compute the median rank using equation

^F(t) = [1.05 0.3] / (20 + 0.4) = 0.0368

For the eighth ordered observation (the fourth uncensored observation),

^

F(t) = [2.2235 0.3] / (20 + 0.4) x 100 = 9.4289 % 9.43%

4.- Compute the natural Logarithm of the natural logarithm of the inverse of (1 Mdn Rank) for

each uncensored 0bservation.

For the second ordered observation, (the first uncensored observation),

Ln(t ) = Ln(2400 0) = Ln(2400) = 7.7832

Ln{Ln[ 1 / (1 F[t - ])]} = Ln{Ln[ 1 / (1 0.0368)]} = Ln {Ln(1 / [0.9632])}= Ln{Ln(1.03820)

Ln(0.03749) = -3.2835

For the eighth ordered observation (the fourth uncensored observation),

Ln(t ) = Ln(5600 0) = Ln(5600) = 8.63052

Ln{Ln[ 1 / (1 F[t - ])]} = Ln{Ln[ 1 / (1 0.2170)]} = Ln {Ln(1 / [0.783])}= Ln{Ln(1.2771)Ln(0.2446) = -1.4080

Table 1.7 summarizes the calculations listed in steps 1 through 5.


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6.- For each uncensored observation plot the Ln(t ) on the X axis

And the Ln{Ln( 1 / [1 F(t -) ] )} on the Y axis is to use Weibull probability paper and plot (t )

versus F(t ). However, this is a tedious task and less accurate without using a computer program.

7.- Draw the best fitting line to the data points. This can be done using the software

RELIAB_EN.exe that can be downloaded from site www.spc-inspector.com/cgg results areshown in Figures 1.19, 1.20, 1.21.

Figure 1.19 Data of constructed file shown on screen.

Table 1.7 Calculations

Figure 1.21 Calculation of each failure and censored items

8.- The shape parameter of the Weibull distribution, , is equal to the slope of the best straight line

fit to the plotted data. The slope in Figure 1.21 is = 2.4753

9.- The scale parameter of the Weibull distribution, , can be calculated using equation:

= exp[(-Yo) / ]

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where Yo = y-intercept

y-intercept may be found by extrapolating the line in Figure 1.21 or using calculations shown in

same figure. Yo = 22.7643. Thus, the scale parameter is

= exp [ ( 22.7643) / 2.4753] = exp(9.1965) = 9863.36

There are some tiny differences between the manual calculations and computer calculations due to

more accurate calculations made by computer programs

Figure 1.21. Plotted line of Weibull distribution

Bibliography

1.- Reliability Toolkit: Commercial Practices Edition, Rome Laboratory RAC (Reliability Analysis

Center 1988, 1993 revision

2.- Kapur K.C., Lamberson L.R. Reliability in Engineering Design. New York, NY: John Wiley

and Sons, inc., 1974.

3.- Ireson W.G. (Editor). Reliability Handbook. New York, NY: McGraw-Hill, Inc. 1966.

4.- Dodson Bryan Weibull Analysis ASQC Quality Press Milwaukee, Wisconsin. 1994.

5.- Lewis, E. E. Introduction to Reliability Engineering, New York: John Wiley, 1987.

6.- Dhillon, B.S. Reliability Engineering in Systems Design and Operation, Princeton, N.J. Van

Nostrand Reinhold, 1983.

7.- Gonzlez G. Carlos RELIAB_EN.exe Software, Mxico City Mxico, 2007

Author: Carlos Gonzlez Gonzlez

ASQ Fellow

Master Black Belt

ASQ Press Reviewer

MBA National University san Diego CA.

eMail: [email protected]

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mailto:[email protected]:[email protected]

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Reliability CGG2