Reliability CGG1, Weibull, Exponential, Log-Normal Distributions

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Author: Carlos González González Author: Carlos González González Reliability

Reliability

Software RELIAB_EN.exe Download from site

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

1.1.- Exponential Distribution 1.1.- Exponential Distribution

Exponential distribution figure 1.1 is the most commonly used distribution in reliability, and Exponential distribution figure 1.1 is the most commonly used distribution in reliability, and is generally used to predict the probability of surviving at a (t) time.

is generally used to predict the probability of surviving at a (t) time.

Figure 1.1 Exponential distribution is the most commonly used i

Figure 1.1 Exponential distribution is the most commonly used i n reliability.n reliability. The probability density function (pdf) of the exponential distribution is: The probability density function (pdf) of the exponential distribution is: f(t) = f(t) =

λλ

ee --λλtt , , tt

≥≥

0 0 or or f(t) f(t) = = (1/(1/

θθ

) e) e --λλtt where twhere t

≥≥

00 MTBF = MTBF =

θθ

λλ

= rate of failure = 1/= rate of failure = 1/

θθ

R(t) = e

R(t) = e --λλtt or or ee –t/ –t/θθ where twhere t

≥≥

00 F(t)

F(t) = = Unreliability Unreliability = = Failure Failure = = 1 1 – – R(t)R(t)

The hazard function for the exponential distribution =

λλ

, and is constant throughout the f, and is constant throughout the function.unction. Therefore, the exponential distribution should be used for reliability prediction during the rate of Therefore, the exponential distribution should be used for reliability prediction during the rate of constant failure or at random cause of failure or period of operation.

constant failure or at random cause of failure or period of operation. Some unique failures to the exponential distribution include:

Some unique failures to the exponential distribution include: 1.- The mathematical mean and standard deviation are equal. 1.- The mathematical mean and standard deviation are equal.

2.- Of all the values 63.21% fall below the mean value, which translates into only a 36.79% 2.- Of all the values 63.21% fall below the mean value, which translates into only a 36.79% probability of surviving past the time period of one MTBF.

probability of surviving past the time period of one MTBF.

3.- The Reliability R(t) as the time t approaches zero, approaches to one as a limit. 3.- The Reliability R(t) as the time t approaches zero, approaches to one as a limit.

Previously we saw, the reliability for a given (t) time during the constant failure rate period can be Previously we saw, the reliability for a given (t) time during the constant failure rate period can be calculated with the formula:

calculated with the formula:

R(t) = e R(t) = e --λλtt

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Where: e = base of the natural logarithms which is 2.718281828... Where: e = base of the natural logarithms which is 2.718281828...

λλ

= failure rate= failure rate t

t = = timetime Example:

Example: The equipment in a manufacturing plant has a failure rate of 0.0015/hr. (MTBF = 1500The equipment in a manufacturing plant has a failure rate of 0.0015/hr. (MTBF = 1500 hr.). What is the probability of operating for a period of 750 hr. without failure?

hr.). What is the probability of operating for a period of 750 hr. without failure?

λλ

= = 0.000666, 0.000666, t t = = 750750

ee --λλtt = e= e – – (0.000666)(750(0.000666)(750)) = e= e – 0.5 – 0.5 = 0.6065= 0.6065

A 60.65% of probability of operating for a period of 750 hr. without failure exists when the MTBF A 60.65% of probability of operating for a period of 750 hr. without failure exists when the MTBF = 1500 hr (

= 1500 hr (

λλ

= 0.000666).= 0.000666). Note:

Note: _{MTBF and}_{MTBF and}

λλ

_{do not need to be a function of time in hours. The characteristic of “time” or}_{do not need to be a function of time in hours. The characteristic of “time” or} usage can be such units as cycles instead of hours. In this case, MCBF (Mean Cycles Between usage can be such units as cycles instead of hours. In this case, MCBF (Mean Cycles Between Failures) could be the appropriate measure.

Failures) could be the appropriate measure.

Using the software RELIAB_EN.exe you can get the following result: Using the software RELIAB_EN.exe you can get the following result:

Figure 1.2

Figure 1.2 Results obtained with program RELIAB_EN.exe fromResults obtained with program RELIAB_EN.exe from

www.spc-inspector.com/cgg

Example:

Example: One cycle of the machine completes the assembly of 20 units. A study of this machineOne cycle of the machine completes the assembly of 20 units. A study of this machine predicted an MCBF of 14,000 cycles (

predicted an MCBF of 14,000 cycles (

λλ

= 0.00007143/cycle).= 0.00007143/cycle). What is the probability of operating 15,000 cycles without failure? What is the probability of operating 15,000 cycles without failure? Note:

Note:

λλ

is also a function of cycles.is also a function of cycles.

The reliability for this example can be calculated from either equation noted before. The reliability for this example can be calculated from either equation noted before. (1).-

(1).- R(t) R(t) = = ee –t/ –t/θθ or or ee –c/ –c/θθ; where c = cycles,; where c = cycles, or

or (2).-

(2).- R(t) R(t) = = ee--λλtt or or ee --λλcc

For this example the equation (1) will be used. For this example the equation (1) will be used. R

R (15,000)(15,000)= e= e – 15,000/14000 – 15,000/14000= e= e – 1.0714 – 1.0714 = 0.3425= 0.3425

A 34.25% probability exists that the machine will run 15,000 cycles without failure. A 34.25% probability exists that the machine will run 15,000 cycles without failure. Using software RELIAB_EN.exe we can get:

Using software RELIAB_EN.exe we can get:

Figure 1.3

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An interesting note of prediction during the chance cause failure period is that the probability of An interesting note of prediction during the chance cause failure period is that the probability of functioning for a given time period (t) is totally independent of previous operation. Therefore, as functioning for a given time period (t) is totally independent of previous operation. Therefore, as long as operation remains in the chance cause failure mode, the probability of failure is the same long as operation remains in the chance cause failure mode, the probability of failure is the same during the first 100 hr of operation or for the period of 10,000 hr to 10,100 hr.

during the first 100 hr of operation or for the period of 10,000 hr to 10,100 hr. Fundamentals of reliability statistics

Fundamentals of reliability statistics

Among the many reliability statistical applications, there are relatively few mathematical Among the many reliability statistical applications, there are relatively few mathematical relationships that provide a large part of reliability calculations. Some of these include: hazard relationships that provide a large part of reliability calculations. Some of these include: hazard function, survival, series systems, parallel systems, perfect and imperfect switching for standby function, survival, series systems, parallel systems, perfect and imperfect switching for standby redundancy, confidence intervals for MTBF, and others.

redundancy, confidence intervals for MTBF, and others.

Although the concepts of reliability theory will not be explored completely, the following definition Although the concepts of reliability theory will not be explored completely, the following definition of reliability will be used throughout:

of reliability will be used throughout:

Reliability: The probability that an item will perform its intended function for a specified interval Reliability: The probability that an item will perform its intended function for a specified interval understated environmental conditions.

understated environmental conditions.

When making predictions using the exponential distribution, it is imperative that the failure rate in When making predictions using the exponential distribution, it is imperative that the failure rate in this period be constant; failures are random in nature. This is also known as the chance cause or this period be constant; failures are random in nature. This is also known as the chance cause or useful life period. Other periods of failure are early 0r infant mortality and wearout.

useful life period. Other periods of failure are early 0r infant mortality and wearout. 1.2.- Bathtub Curve

1.2.- Bathtub Curve

Over the life of a complex system, three distinct failure rate phases may become apparent. Over the life of a complex system, three distinct failure rate phases may become apparent. The first phase or period is referred to as the infant mortality period, which is shown as the The first phase or period is referred to as the infant mortality period, which is shown as the decreasing failure rate on the left portion of figure 1.4. The second phase is the random or constant decreasing failure rate on the left portion of figure 1.4. The second phase is the random or constant cause failure period, which is the period of time encompassing the flat portion of the curve, where cause failure period, which is the period of time encompassing the flat portion of the curve, where the failure rate remains constant. The last phase is the wearout period, which is shown on the right the failure rate remains constant. The last phase is the wearout period, which is shown on the right side of figure 1.4 as an increasing failure rate. The wearout phase is more predominant in side of figure 1.4 as an increasing failure rate. The wearout phase is more predominant in mechanical systems than in electronic system.

mechanical systems than in electronic system.

Figure 1.4

Figure 1.4 Reliability bathtub curve showing the infant mortality period as the decreasing failureReliability bathtub curve showing the infant mortality period as the decreasing failure rate (left), random constant cause failure period (middle), and wearout period as an increasing rate (left), random constant cause failure period (middle), and wearout period as an increasing failure rate (right).

failure rate (right).

Infant mortality failures are generally the result of manufacturing errors that are not caught Infant mortality failures are generally the result of manufacturing errors that are not caught in inspection prior to burn-in or placing in service. Failures resulting from time/stress dependent in inspection prior to burn-in or placing in service. Failures resulting from time/stress dependent errors may occur in this period.

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Random Failures and wearout failures are generally a factor of design. Random Failures and wearout failures are generally a factor of design.

No distinct break off from infant mortality to random to wearout failure has been No distinct break off from infant mortality to random to wearout failure has been established. Random failures can occur anywhere in the three periods, as can infant mortality established. Random failures can occur anywhere in the three periods, as can infant mortality failures. A failure caused by a cold solder joint may occur well into the service life, but this is really failures. A failure caused by a cold solder joint may occur well into the service life, but this is really an infant mortality type failure. Wearout of mechanical parts also begins the moment the product is an infant mortality type failure. Wearout of mechanical parts also begins the moment the product is put into service.

put into service.

The probability distributions occurring most often during the infant mortality period are The probability distributions occurring most often during the infant mortality period are Weibull, gamma, and decreasing exponential. Probability distributions of value in the constant Weibull, gamma, and decreasing exponential. Probability distributions of value in the constant failure rate period are exponential and Weibull. During the wearout period, the curves generally failure rate period are exponential and Weibull. During the wearout period, the curves generally follow the normal or Weibull distributions.

follow the normal or Weibull distributions.

Figure 1.5 is the same as shown in Figure 1.4, except that the nonrandom failure period has Figure 1.5 is the same as shown in Figure 1.4, except that the nonrandom failure period has been designated as the constant failure rate period.

been designated as the constant failure rate period.

Figure 1.5

Figure 1.5 Reliability bathtub curve showing nonrandom failure period as the constant failure rateReliability bathtub curve showing nonrandom failure period as the constant failure rate period.

period.

1.3.- Log-Normal Distribution 1.3.- Log-Normal Distribution

When working with continuous distributions that show positive skewness, calculations using When working with continuous distributions that show positive skewness, calculations using the Gaussian (normal) distribution prove inadequate. When this situation occurs, and all the values the Gaussian (normal) distribution prove inadequate. When this situation occurs, and all the values are (or transformed to be) > 0, taking the natural logs of the values may result in a normal are (or transformed to be) > 0, taking the natural logs of the values may result in a normal distribution.

distribution.

For the log-normal distribution: For the log-normal distribution:

__ __ f(x) = {1 / [

f(x) = {1 / [

σ

xx

√√

22

ππ

] } ] } ee [ -1/2 ([ln x -[ -1/2 ([ln x -μμ] /] /σσ)^2 ])^2 ] , , x x > > 00 The mean and variance of the log-normal distribution are: The mean and variance of the log-normal distribution are:

^ ^

Equation

Equation (i): (i): Mean Mean = = ee[[μμ+ (S^2)/ 2)]+ (S^2)/ 2)]

∧ ∧

Equation

Equation (ii): (ii): Variance = Variance = [e[e(2(2μμ+ + S^2 S^2 )) ] [e] [eS^2S^2 –1]–1] Where:

Where:

μμ

estimatedestimated= mean of the natural logs of individuals.= mean of the natural logs of individuals.

S

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Example:

Example: Twenty Twenty five five measurements measurements are are taken taken of of time time to to failure failure of of a a component component in in hours. hours. TheThe natural logarithms are found to be normally distributed with an estimated mean natural logarithms are found to be normally distributed with an estimated mean

μμ

estimatedestimated = 3.5 and a y variance (S= 3.5 and a y variance (S22) = 1.30 (remember that) = 1.30 (remember that

μμ

estimatedestimated and sand s22 are for are for

natural log values). Find the untransformed mean and variance in hours. natural log values). Find the untransformed mean and variance in hours. Mean Mean = = ee [3.5 + 1.30 / 2][3.5 + 1.30 / 2]= 63.434= 63.434 Variance = [e Variance = [e2(3.5) + 1.302(3.5) + 1.30] [e] [e1.301.30

−−

1] =1] = = = [6634.24][2.6669[6634.24][2.6669] ] = = 17692.8517692.85 = 17692.85 = 17692.85

To calculate tail area probabilities, translated log-normal values are used. To calculate tail area probabilities, translated log-normal values are used.

∧ ∧ Zln Zln = = [ln(x) [ln(x) --

μμ

estimatedestimated] / s] / s ∧ ∧ Where:

Where:

μμ

estimatedestimated and and s are s are log-normal log-normal values.values.

∧ ∧

Example:

Example: In In the the previous previous example example it it was was found found thatthat

μμ

estest= 3.5 and s = = 3.5 and s = 1.14017.1.14017.

What percentage will survive 100 hr.? What percentage will survive 100 hr.?

Z = [ln(100) – 3.5]

Z = [ln(100) – 3.5] / 1.14017 = (4.60517 – 3.5) / 1.14017 = 1.10517/1.14017 = 0.9693/ 1.14017 = (4.60517 – 3.5) / 1.14017 = 1.10517/1.14017 = 0.9693 Z = 0.9693. Using the normal Z tables a value of 0.1662 ó 16.62% will exceed 100 hr.

Z = 0.9693. Using the normal Z tables a value of 0.1662 ó 16.62% will exceed 100 hr. Using the software RELIAB_EN.exe we obtain:

Using the software RELIAB_EN.exe we obtain:

Figure 1.6 Results calculated using Software RELIAB_EN.exe site:

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1.4.- Weibull Distribution 1.4.- Weibull Distribution

One of the most versatile for use in reliability applications is the Weibull distribution. With One of the most versatile for use in reliability applications is the Weibull distribution. With its many changing shapes the Weibull

its many changing shapes the Weibull can be made to fit can be made to fit many distributions. Among these Gaussianmany distributions. Among these Gaussian (normal) and exponential distributions, the calculations for the scale and shape parameters which (normal) and exponential distributions, the calculations for the scale and shape parameters which will be described later are extremely tedious along with be complex. These values of the shape and will be described later are extremely tedious along with be complex. These values of the shape and scale parameters generally estimated using Weibull probability paper to made a graphical scale parameters generally estimated using Weibull probability paper to made a graphical application.

application.

In using the Weibull distribution several parameters will be used, which are defined as In using the Weibull distribution several parameters will be used, which are defined as follows:

follows:

ββ

= = shape shape parameter, parameter, which which determines determines distribution distribution shapeshape

η

= = scale scale parameter; parameter; 63.21% 63.21% of of the the values values fall fall below below this this parameter parameter

θθ

= = estimated estimated mean mean (MTBF (MTBF estimate estimate in in reliability) reliability) (Characteristic (Characteristic Life).Life).

Γ

(x) (x) = = gamma gamma function function of of a a variable variable (x). (x). Values Values of of

Γ

(x) are listed in the gamma(x) are listed in the gamma function tables along with the equation for calculating the values of

function tables along with the equation for calculating the values of

Γ

(x) for large values of x.(x) for large values of x. t

t = = noted noted time time o o fan fan individual individual characteristic.characteristic. The probability density function for Weibull is:

The probability density function for Weibull is:

ββ

F(t)

F(t) = = { { ((

ββ

//

η

) ( t/) ( t/

η

))ββ-1-1 ee [-(t/[-(t/ηη) ) ]] for tfor t

≥≥

00 0

0 for t < 0for t < 0 and the survival function P(s) is: and the survival function P(s) is:

ββ

P(s)

P(s) = = ee[-(t/[-(t/ηη) ) ]] Equation (iii): Mean Weibull

Equation (iii): Mean Weibull

μμ

ww==

ηΓ

(1 + 1/(1 + 1/

ββ

))

Equation (iv): standard deviation standard Weibull Equation (iv): standard deviation standard Weibull

σ

ww==

_________________ _________________

σ

==

η√

η√ Γ

Γ

(1+2/(1+2/

ββ

) -) -

Γ

22(1+1/(1+1/

ββ

)) Example:

Example: A A unit unit was was tested tested and and the the following following were were the the results results of of the the test:test:

η

= 25,000 y= 25,000 y

ββ

= 2.0.= 2.0. Calculate the Weibull mean, standard deviation, and P

Calculate the Weibull mean, standard deviation, and P(s)(s) for 10,000 hr.for 10,000 hr.

∧ ∧

Weibull mean estimate

μμ

ww= 25,000 {= 25,000 {

Γ

(1 + 1/2.0) = (1 + 1/2.0) = 22,155.6822,155.68

∧ ∧

Weibull standard deviation estimate Weibull standard deviation estimate

σ

ωω == ∧ ∧ ____________________ ____________________

σ

ww= 20,000= 20,000

√√ Γ

Γ

(1+2/2.0) -(1+2/2.0) -

Γ

22(1+1/2.0) (1+1/2.0) = = 11581.2811581.28 2.0 2.0 P(s) = e P(s) = e –(10,000/25,000) –(10,000/25,000) = = ee−−0.160.16= 0.852143= 0.852143

Using software RELIAB_EN.exe to calculate Weibull we get: Using software RELIAB_EN.exe to calculate Weibull we get:

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Figure 1.7

Figure 1.7 Statistical calculation found with RELIAB_EN.exe.Statistical calculation found with RELIAB_EN.exe. Review of Weibull functions.

Review of Weibull functions.

ββ

Cumulative

Cumulative Distribution Distribution Function Function (CDF): (CDF): F(t) F(t) = = 1 1 – – ee –(t/ –(t/θθ))

ββ

Reliability

Reliability Function Function R(t) R(t) = = ee –(t/ –(t/θθ))

ββ

Probability

Probability Density Density Function Function f(t) f(t) = = ((

ββ

/t) (t//t) (t/

θθ

))ββ ee –(t / –(t /θθ)) Rate

Rate of of Failure Failure Function Function h(t) h(t) = = ((

ββ

/t) (t//t) (t/

θθ

))ββ

This formula or equations can be used once the parameters value are known to obtain the This formula or equations can be used once the parameters value are known to obtain the Reliability predictions, percentiles and failure rate.

Reliability predictions, percentiles and failure rate.

There are procedures to estimate the Weibull parameters from data using the maximum There are procedures to estimate the Weibull parameters from data using the maximum likelihood estimation method [MLE]

likelihood estimation method [MLE] 1.4.1- Shape Parameter

1.4.1- Shape Parameter

The shape parameter or slope (

The shape parameter or slope (ββ) is the main infl) is the main influence over the distribution shape. Whenuence over the distribution shape. When θθ == 1, the data fit a Weibull distribution with a shape parameter of 3.44, the distribution has a Gaussian 1, the data fit a Weibull distribution with a shape parameter of 3.44, the distribution has a Gaussian distribution. A shape parameter of 1.0 is an Exponential distribution, while if

distribution. A shape parameter of 1.0 is an Exponential distribution, while if ββ = 2.0 this is a= 2.0 this is a Raleigh distribution.

Raleigh distribution. 1.4.2.- Weibull Fit 1.4.2.- Weibull Fit

When using a Weibull Probability Paper, the failure times are plotted in ascending order 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 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).

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 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 (

line is transferred and drawn to determine the shape parameter (ββ). The percentage of units that is). The percentage of units that is expected to survive at a given time it is established along to the vertical axis corresponding to the 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.

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 10 units Example: When you ran a life test of a mechanical system, with a sample of 10 units

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

hours. Step

1.-Step 1.- Data were ordered in ascendant manner by column for the 10 units.Data were ordered in ascendant manner by column for the 10 units. 1.- 2200 1.- 2200 2.- 2400 2.- 2400 3.- 3125 3.- 3125

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4.- 3300 4.- 3300 5.- 3400 5.- 3400 6.- 3500 6.- 3500 7.- 3550 7.- 3550 8.- 3650 8.- 3650 9.- 4000 9.- 4000 10.- 6000 10.- 6000 Step

2.-Step 2.- Now we are to use Median Ranks. Data are paired with data shown previously for n = 10.Now we are to use Median Ranks. Data are paired with data shown previously for n = 10. Failure

Failure Time Time of of Mdn. Mdn. Rank Rank Kapur Kapur & & LambersonLamberson order

order j j failure failure t t from from table table Mdn. Mdn. Rank Rank 1.- 1.- 2200 2200 6.7 6.7 6.736.73 2.- 2.- 2400 2400 16.2 16.2 16.3416.34 3.- 3.- 3125 3125 25.9 25.9 25.9625.96 4.- 4.- 3300 3300 35.5 35.5 35.5735.57 5.- 5.- 3400 3400 45.2 45.2 45.1945.19 6.- 6.- 3500 3500 54.8 54.8 54.8054.80 7.- 7.- 3550 3550 64.5 64.5 64.4264.42 8.- 8.- 3650 3650 74.1 74.1 74.0374.03 9.- 9.- 4000 4000 83.8 83.8 83.6583.65 10.- 10.- 6000 6000 93.3 93.3 93.2693.26

If you do not have a median rank table that you can use then you can calculate the median 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 ranks with a formula developed by authors Kapur & Lamberson, 1977, then those values are going to be paired with column t of failure

to be paired with column t of failure F(t)

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

Where: j

j = = Failure Failure order order n

n = = Sample Sample sizesize Example:

Example: if the failure j = 5, n = if the failure j = 5, n = 10 10 when you subswhen you substitute those values en la formula ( titute those values en la formula ( i ) youi ) you obtain: F(t) = (5 –

obtain: F(t) = (5 – 0.3) / (10 + 0.40.3) / (10 + 0.4) = 4.7 / 10) = 4.7 / 10.4 = 45.19 and .4 = 45.19 and so forth every calculation for eachso forth every calculation for each order of failure.

order of failure.

Another formula from White 1969 also has a good approach. Another formula from White 1969 also has a good approach.

F(t)

F(t) = = [j [j – – (3/8)] (3/8)] / / [n [n + + (1/4)] (1/4)] --( --( ii ii )) Where:

Where: j

n = = Sample Sample sizesize One more formula f

One more formula from Dovich 1990 also has a good approach.rom Dovich 1990 also has a good approach. MR

MR (j)(j) = [n – j + (0.5)= [n – j + (0.5) 1/n1/n(2j – n – 1)] / (n-1)(2j – n – 1)] / (n-1)

Where: Where:

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j

n = = Sample Sample sizesize

Example for Median Rank order 7 then: j = 7, n = 10 Example for Median Rank order 7 then: j = 7, n = 10

MR MR (7)(7)= [10 – 7 + (0.5)= [10 – 7 + (0.5) 1/101/10 {2(7) – 10 – 1}] / {2(7) – 10 – 1}] / (10-1)(10-1) = [3 + 0.5 = [3 + 0.5 0.10.1 {14 – 10 – 1} ] / 9{14 – 10 – 1} ] / 9 = [3 + .9330(3)] / 9 = [3 + .9330(3)] / 9 = (3 + 2.799) / 9 = (3 + 2.799) / 9 = 5.799 / 9 = 5.799 / 9 = 64.43 = 64.43 Step 3.-.

Step 3.-. Using Weibull Probability Paper. Exist different types of forms of Weibull ProbabilityUsing Weibull Probability Paper. Exist different types of forms of Weibull Probability Paper with horizontal or vertical shape.

Paper with horizontal or vertical shape. Step

4.-Step 4.- Once you select the form and shape of the Weibull Probability Paper, horizontal axis hasOnce you select the form and shape of the Weibull Probability Paper, horizontal axis has three logarithmic cycles horizontal then you attach and scale the data.

three logarithmic cycles horizontal then you attach and scale the data. Step

5.-Step 5.- Now you register pairs of values one at a time over the Weibull Probability Paper, plottingNow you register pairs of values one at a time over the Weibull Probability Paper, plotting each point where correspond hours and percentages.

each point where correspond hours and percentages. Step

6.-Step 6.- Always you should find the best fitting line, the drawing of an straight line helped with aAlways 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 transparent rule made of plastic is recommended with preference of the tendency of the majority of points.

points. Step

7.-Step 7.- You need to transport in parallel with the same tilt, and slope as the line already drawn.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 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 4.1 which indicates a shape of until cuts the graduated arc. (in this case, the value that you get is 4.1 which indicates a shape of almost a normal distribution skewed a little bit to the right).

almost a normal distribution skewed a little bit to the right). Step

8.-Step 8.- Then, now you should prolong the best fitting line until you cut the horizontal axis at theThen, 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 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 100% of population at least survive a minimum of 550 minimum life of the product this means that 100% of population at least survive a minimum of 550 hours. (Really 550 hours is reached by 99.97% of the product or just 0.03% can not get 550 Hours, hours. (Really 550 hours is reached by 99.97% of the product or just 0.03% can not get 550 Hours, according to the design of the Weibull Probability Paper).

according to the design of the Weibull Probability Paper). Step

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

Example: What What percentage percentage of of product product will will fail fail at at 2400 2400 hours hours of of life? life? The The procedure procedure to to knowknow that is as follows: You can go vertically starting in an horizontal value of 2400 until you get the 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 fi

slopped line plotted previously, then you go horizontally to the left until you fi nd the vertical axis of nd the vertical axis of percentages where you read the searched value. (In this case approximately 14%).

percentages where you read the searched value. (In this case approximately 14%). Step

10.-Step 10.- In this particular case the estimator is called the characteristic life.In this particular case the estimator is called the characteristic life. (1 – 1/e) for this case too 3750 or 3800 hours is the value of the life

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Weibull Weibull Slope Slope

Weibull Probability

P

Paa eer

r

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Weibull Weibull Slope Slope

Hours

Weibull

Weibull Probability

Probability Paper

Paper

Figure 1.8

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Figure 1.9

Figure 1.9 Time of failure for the sample of 10 (Condition)Time of failure for the sample of 10 (Condition) A).- Assume

A).- Assume δδ = 0 and order the observations from smallest to largest.= 0 and order the observations from smallest to largest. B).- Compute the median rank as we do in step 2.

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

C).- Compute the natural logarithm of the time to fail for each observation. 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 D).- Compute the Natural Logarithm of the Natural Logarithm of the inverse of 1 – the median rank for each uncensored observation.

for each uncensored observation.

As an example for the first two median ranks calculated: As an example for the first two median ranks calculated: First Median Rank = 0.0673

First Median Rank = 0.0673

Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.0673)] = Ln[Ln(1/0.9327)] =

Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.0673)] = Ln[Ln(1/0.9327)] = Ln[Ln(1.072156Ln[Ln(1.072156] = ] = Ln(0.06967)Ln(0.06967) = – 2.6639

= – 2.6639 Second Median Rank = 0.1635 Second Median Rank = 0.1635

Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.1635)] = Ln[Ln(1/0.8365)] =

Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.1635)] = Ln[Ln(1/0.8365)] = Ln[Ln(1.19545] = Ln(0.17852)Ln[Ln(1.19545] = Ln(0.17852) = – 1.7230

= – 1.7230

Table 1.4 and figure 1.10 summarizes the calculations listed from A).- to Table 1.4 and figure 1.10 summarizes the calculations listed from A).- to D).-Table 1.4 Calculations for 10 values of Failure Times 1 to 10

Table 1.4 Calculations for 10 values of Failure Times 1 to 10 Order

Order Time Time to to Fail Fail t t – – D D F(t) F(t) K&L K&L Natural Natural Log Log LnLnLnLn #

# (t (t – – D) D) Median Median Rank Rank (t (t – – D) D) (1/[1–F(t (1/[1–F(t – – D)])D)]) 1.- 1.- 2200 2200 6.73 6.73 7.6962 7.6962 –2.6638–2.6638 2.- 2.- 2400 2400 16.34 16.34 7.7832 7.7832 –1.7233–1.7233 3.- 3.- 3125 3125 25.96 25.96 8.0472 8.0472 –1.2020–1.2020 4.- 4.- 3300 3300 35.57 35.57 8.1017 8.1017 –.8217–.8217 5.- 5.- 3400 3400 45.19 45.19 8.1315 8.1315 –.5086–.5086 6.- 6.- 3500 3500 54.80 54.80 8.1605 8.1605 –.2304–.2304 7.- 7.- 3550 3550 64.42 64.42 8.1747 8.1747 .0329.0329 8.- 8.- 3650 3650 74.03 74.03 8.2025 8.2025 .2990.2990 9.- 9.- 4000 4000 83.65 83.65 8.2940 8.2940 .5940.5940 10.- 10.- 6000 6000 93.26 93.26 8.6995 8.6995 .9927.9927

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E).- Plot the Ln (t –

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

plotting Ln (t – δδ) on the x – axis and Ln [Ln(1/(1 – (t – ) on the x – axis and Ln [Ln(1/(1 – (t – δδ))] on the y – axis is to use Weibull))] on the y – axis is to use Weibull Probability Paper and plot (t –

Probability Paper and plot (t – δδ) versus F(t – ) versus F(t – δδ). As we do some pages before ). As we do some pages before instead of a computer instead of a computer program as RELIAB_EN.exe

program as RELIAB_EN.exe

Figure 1.10

Figure 1.10 Statistical Calculation of Weibull study from software RELIAB_EN.exe.Statistical Calculation of Weibull study from software RELIAB_EN.exe.

F).- Fit a straight line to the data points. This can be done visually, or a least square regression may 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 1.4 is plotted in figure 1.11. The slope of this line, which is equal to be used. The data Table 1.4 is plotted in figure 1.11. The slope of this line, which is equal to ββ, i, iss

3.774. This is not the case but if the data in the probability plot appears to fall on a downward or 3.774. This is not the case but if the data in the probability plot appears to fall on a downward or upward curve,

upward curve, δδ may not be equal to zero. The time to fail must be transformed by subtraction of anmay not be equal to zero. The time to fail must be transformed by subtraction of an estimate of

estimate of δδ. A discussion of how to handle . A discussion of how to handle a nonzero parameter a nonzero parameter is made using RELIAB_EN.exe.is made using RELIAB_EN.exe.

Figure 1.11

Figure 1.11 Best fitting line and calculation of Delta = 0, Beta = 3.774, Teta = 3896, and R Best fitting line and calculation of Delta = 0, Beta = 3.774, Teta = 3896, and R 22=0.859=0.859 Probability plots are commonly used as goodness of fit tests. A straight line of the plotted Probability plots are commonly used as goodness of fit tests. A straight line of the plotted points indicates the chosen density function is acceptable.

points indicates the chosen density function is acceptable. G).- The scale parameter of the Weibull distribution,

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

θ = exp(– y= exp(– yoo//ββ))

where: where:

y

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The y-intercept, y

The y-intercept, yoo may be found by extrapolating the line in figures 1.8 and 1.11. The y-may be found by extrapolating the line in figures 1.8 and 1.11. The

y-intercept for the data in table 1.4, is –31.2023. Thus, the scale parameter is intercept for the data in table 1.4, is –31.2023. Thus, the scale parameter is

θ

θ = exp(– y= exp(– yoo//ββ) = exp [() = exp [(– ( – 31.2023))/3.774] = 3895.98– ( – 31.2023))/3.774] = 3895.98

1.4.3.- Applications 1.4.3.- Applications

The shape parameter

The shape parameter ββ and probability density function take different shapes as is shown inand probability density function take different shapes as is shown in figure 1.12. Weibull Distribution can be used in a wide variety of applications, depending on values figure 1.12. Weibull Distribution can be used in a wide variety of applications, depending on values of

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

ß

ß = = 1 1 The The Weibull Weibull distribution distribution is is exactly exactly an an Exponential Exponential Distribution.Distribution. ß

ß = = 2 2 The The Weibull Weibull distribution distribution is is exactly exactly a a Rayleigh Rayleigh Distribution.Distribution. ß

ß = = 2.5 2.5 The The Weibull Weibull distribution distribution approaches approaches to to Log-Normal Log-Normal Distribution. Distribution. ThoseThose distributions are nearly equal to Log-Normal Distribution but require a sample size bigger than 50 distributions are nearly equal to Log-Normal Distribution but require a sample size bigger than 50 to be distinguished between them.

to be distinguished between them. ß

ß = = 3.6 3.6 The The Weibull Weibull distribution distribution approaches approaches to to Normal Normal Distribution.Distribution. ß

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

Figure

1.12.-Figure 1.12.- Values of Beta for different Probability Distributions.Values of Beta for different Probability Distributions.

Due to this flexibility, there are, very few failure rates observed that can not be exactly Due to this flexibility, there are, very few failure rates observed that can not be exactly modeled by the Weibull Distribution, some examples are:

modeled by the Weibull Distribution, some examples are:

a).- The components resistance or the required stress of metal fatigue. a).- The components resistance or the required stress of metal fatigue. b).- The time of failure of electronic components.

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

c).- Failure time of items that suffers wearout, as the auto tires. 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 d).- Systems fail when the less resistant component fails (Weibull Distribution represents a distribution of extreme value).

distribution of extreme value). Bibliography

Bibliography

1.- Reliability Toolkit: Commercial Practices Edition, Rome Laboratory RAC (Reliability Analysis 1.- Reliability Toolkit: Commercial Practices Edition, Rome Laboratory RAC (Reliability Analysis Center 1988, 1993 revision

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2.- Kapur K.C., Lamberson L.R. Reliability

2.- Kapur K.C., Lamberson L.R. Reliability in Engineering Design. New York, NY: John Wileyin Engineering Design. New York, NY: John Wiley and Sons, inc., 1974.

and Sons, inc., 1974.

3.- Ireson W.G. (Editor). Reliabil

3.- Ireson W.G. (Editor). Reliability Handbook. New York, NY: McGraw-Hill, Inc. ity Handbook. New York, NY: McGraw-Hill, Inc. 1966.1966. 4.- Dodson Bryan Weibull Analysis ASQC Quality Press Milwaukee,

4.- Dodson Bryan Weibull Analysis ASQC Quality Press Milwaukee, Wisconsin. 1994.Wisconsin. 1994. 5.- Lewis, E. E. Introduction to

5.- Lewis, E. E. Introduction to Reliability Engineering, New York: John Wiley, 1987.Reliability Engineering, New York: John Wiley, 1987. 6.- Dhillon, B.S. Reliability Engineering i

6.- Dhillon, B.S. Reliability Engineering in Systems Design and Operation, Princeton, N.J. Vann Systems Design and Operation, Princeton, N.J. Van Nostrand Reinhold, 1983.

Nostrand Reinhold, 1983.

7.- González G. Carlos RELIAB_EN.exe Software, México City México, 2007 7.- González G. Carlos RELIAB_EN.exe Software, México City México, 2007 Author: Carlos González González

Author: Carlos González González ASQ Fellow

ASQ Fellow

Master Black Belt Master Black Belt ASQ Press Reviewer ASQ Press Reviewer

MBA National University san Diego CA. MBA National University san Diego CA. eMail: