Reliability Cgg2

Author: Carlos González González 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 order over 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 (β). 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 or repaired. 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. 1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.-

2200 2400 3125 3300 3400 3500 3550 3650 4000 6000 8000 8000 8000 8000 8000 8000 8000 8000 8000 8000

Step 2.- Now we are going to use Median Ranks. Data are paired with data shown previously for n = 20. Failure order j 1.2.3.4.-

Time of failure t 2200 2400 3125 3300

Mdn. Rank from table 3.4 8.3 13.1 18.1 1

Kapur & Lamberson Mdn. Rank 3.43 8.33 13.23 18.13

5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.-

3400 3500 3550 3650 4000 6000 8000 8000 8000 8000 8000 8000 8000 8000 8000 8000

23.0 27.9 32.8 37.7 42.6 47.5 52.5 57.4 62.3 67.2 72.1 77.0 81.9 86.9 91.7 96.6

23.04 27.94 32.84 37.74 42.64 47.54 52.45 57.35 62.25 67.15 72.05 76.96 81.86 86.76 91.66 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)

j

=

Failure order

n

=

Sample size

--( i )

Where:

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 each order 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 2

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 μ.

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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 δ = 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/(1–F(t))] = Ln[Ln(1/(1–0.0343)] = Ln[Ln(1/0.9657)] = Ln[Ln(1.03552] = Ln(0.03490) = – 3.3550 Second Median Rank = 0.0833 Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.0833)] = 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/[1–F(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

11.-

8000 Censored 5

12.· · · 20.-

8000 · · · 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 – δ) on the x Axis and Ln [Ln(1/(1 – (t – δ))] on the y axis. An alternative to plotting Ln (t – δ) on the x – axis and Ln [Ln(1/(1 – (t – δ))] on the y – axis is to use Weibull 6

Probability Paper and plot (t – δ) versus F(t – δ). 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 β, 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, δ may not be equal to zero. The time to fail must be transformed by subtraction of an estimate of δ. 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, θ, can be calculated using the expression θ = exp(– yo /β) where: yo = the y intercept The y-intercept, yo may be found by extrapolating the line in figures 1.13 and 1.17. The yintercept for the data in table 1.5, is –25.7485. Thus, the scale parameter is θ = exp(– yo /β) = exp [(– ( – 25.7485))/2.989] = 5510.549 1.5.2.- Applications The shape parameter β 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 β, when β has other values the shape can be approach to other distributions par example:

Figure 1.18.- Values of Beta for different Probability Distributions. ß=1

The Weibull distribution is exactly an Exponential Distribution.

ß=2

The Weibull distribution is exactly a Rayleigh Distribution.

ß = 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. ß = 3.6

The Weibull distribution approaches to Normal Distribution. 7

ß=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 c 10.- 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 where: 8

---(a)

Ij n Op r

= The increment for the jth failure = The total number of data points, both censored and uncensored = The order of the previous failure = 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 δ = 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 – δ) = Ln(2400 – 0) = Ln(2400) = 7.7832 Ln{Ln[ 1 / (1 – F[t - δ])]} = 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 – δ) = Ln(5600 – 0) = Ln(5600) = 8.63052 Ln{Ln[ 1 / (1 – F[t - δ])]} = 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. 9

6.- For each uncensored observation plot the Ln(t – δ) on the X axis And the Ln{Ln( 1 / [1 – F(t -δ) ] )} on the Y axis is to use Weibull probability paper and plot (t – δ) versus F(t – δ). 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 are shown 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, β, is equal to the slope of the best straight line fit to the plotted data. The slope in Figure 1.21 is β = 2.4753 9.- The scale parameter of the Weibull distribution, θ, can be calculated using equation: θ = exp[(-Yo) / β] 10

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 θ = 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.- González G. Carlos RELIAB_EN.exe Software, México City México, 2007 Author: Carlos González González ASQ Fellow Master Black Belt ASQ Press Reviewer MBA National University san Diego CA. eMail: [email protected]

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