# Reliability CGG1, Weibull, Exponential, Log-Normal Distributions

June 21, 2018 | Author: cgg1115572 | Category: Normal Distribution, Reliability Engineering, Logarithm, Median, Probability Theory

#### Short Description

Basic Studies of Exponential, Log-Normal and Weibull Distributions for Reliability, applications with Software RELIAB_EN...

#### Description

Author: Carlos González González Reliability Software RELIAB_EN.exe Download from site www.spc-inspector.com/cgg 1.1.- Exponential Distribution

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.

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

f(t)

= λe - t ,

t≥0

f(t)

= (1/θ) e -

λt

or

where t ≥ 0

MTBF = θ

λ = rate of failure = 1/θ λt

R(t) = e F(t)

θ

or e  –t/

where t ≥ 0

= Unreliability = Failure = 1 – R(t)

The hazard function for the exponential distribution = λ, and is constant throughout the function. f unction. 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. Some unique failures to the exponential distribution include: 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%  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. Previously we saw, the reliability for a given (t) time during the constant failure rate period can be calculated with the formula: λt

R(t) = e -

1

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

λ

= failure rate

t

= time

Example: The 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? λ = 0.000666, t = 750 -λt (0.000666)(750) e = e  – (0.000666)(750) = e  – 0.5 = 0.6065

A 60.65% of probability of operating for a period of 750 hr. without failure exists when the MTBF = 1500 hr ( λ = 0.000666).  Note: MTBF and λ 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 Failures) could be the appropriate measure. Using the software RELIAB_EN.exe you can get the following result:

Figure 1.2 Results obtained with program RELIAB_EN.exe from www.spc-inspector.com/cgg Example: One cycle of the machine completes the assembly of 20 units. A study of this machine  predicted an MCBF of 14,000 cycles ( λ = 0.00007143/cycle). What is the probability of operating 15,000 cycles without failure?

Note: λ is also a function of cycles. The reliability for this example can be calculated from either equation noted before. (1).-

θ

R(t)

= e  –t/

R(t)

=e-

θ

e  –c/ ; where c = cycles,

or

or  (2).-

λt

λc

or e -

For this example the equation (1) will be used.   – 15,000/14000 R (15,000) = e  – 1.0714 (15,000)= e

= 0.3425

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

Figure 1.3 Results obtained with software RELIAB_EN.exe

2

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 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. Fundamentals of reliability statistics 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 function, survival, series systems, parallel systems, perfect and imperfect switching for standby redundancy, confidence intervals for MTBF, and others. Although the concepts of reliability theory will not be explored completely, the following definition of reliability will be used throughout: Reliability: The probability that an item will perform its intended function for a specified interval understated environmental conditions. 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  useful life period. Other periods of failure are early 0r infant mortality and wearout. 1.2.- Bathtub Curve 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 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 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 mechanical systems than in electronic system.

Figure 1.4 Reliability 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 failure rate (right).

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 errors may occur in this period. 3

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 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 an infant mortality type failure. Wearout of mechanical parts also begins the moment the product is  put into service. The probability distributions occurring most often during the infant mortality period are 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 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  been designated as the constant failure rate period.

Figure 1.5 Reliability bathtub curve showing nonrandom failure period as the constant failure rate  period. 1.3.- Log-Normal Distribution

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 are (or transformed to be) > 0, taking the natural logs of the values may result in a normal distribution. For the log-normal distribution:  __  μ σ f(x) = {1 / [ σ x √2π ] } e [ -1/2 ([ln x - ] / )^2 ] ,

x >0

The mean and variance of the log-normal distribution are: Equation (i): Mean =

e

^ [ μ + (S^2)/ 2)]

Equation (ii): Variance = [e

(2μ + S^2 )

] [eS^2 –1]

Where:

μestimated = mean of the natural logs of individuals. S2 = Variance of natural logs of individuals. 4

Example:

Twenty five measurements are taken of time to failure of a component in hours. The natural logarithms are found to be normally distributed with an estimated mean μestimated = 3.5 and a y variance (S 2) = 1.30 (remember that μestimated and s2 are for  natural log values). Find the untransformed mean and variance in hours.

Mean =

e [3.5 + 1.30 / 2] = 63.434

Variance

=

[e 2(3.5) + 1.30] [e1.30

=

[6634.24][2.6669] [6634.24][2.6669] = 17692.85

=

17692.85

−1] =

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

Zln

=

[ln(x) - μestimated] / s

Where:μestimated

and

s are log-normal values. ∧

Example:

In the previous example it was found that μest = 3.5 and s = 1.14017.

What percentage will survive 100 hr.? Z = [ln(100) – 3.5] / 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. Using the software RELIAB_EN.exe we obtain:

Figure 1.6 Results calculated using Software RELIAB_EN.exe site: www.spc-inspector.com/cgg 5

1.4.- Weibull Distribution

One of the most versatile for use in reliability applications is the Weibull distribution. With its many changing shapes the Weibull can be made to fit many distributions. Among these Gaussian (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 scale parameters generally estimated using Weibull probability paper to made a graphical application. In using the Weibull distribution several parameters will be used, which are defined as follows:

β

=

shape parameter, which determines distribution shape

η

=

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

θ

=

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

Γ(x)

= gamma function of a variable (x). Values of  Γ(x) are listed in the gamma function tables along with the equation for calculating the values of  Γ(x) for large values of x. t

=

noted time o fan individual characteristic.

The probability density function for Weibull is: F(t)

β-1

= { (β / η) ( t/η ) 0 for t < 0

e

[-(t/η) ]

β

for t ≥ 0

and the survival function P(s) is: β

P(s)

=e

[-(t/η) ]

Equation (iii): Mean Weibull μw = ηΓ(1 + 1/β) Equation (iv): standard deviation standard Weibull  _________________  σ = η√ Γ(1+2/β) - Γ 2 (1+1/β)

σw =

Example: A unit was tested and the following were the results of the test: η = 25,000 y Calculate the Weibull mean, standard deviation, and P (s) for 10,000 hr. ∧

Weibull mean estimate

μw = 25,000 { Γ (1 + 1/2.0) = 22,155.68 ∧

Weibull standard deviation estimate σω = ____________________  ∧ σw = 20,000√ Γ(1+2/2.0) - Γ 2 (1+1/2.0) P(s) = e  –(10,000/25,000)

2.0

= 11581.28

= e−0.16 = 0.852143

Using software RELIAB_EN.exe to calculate Weibull we get:

6

β = 2.0.

Figure 1.7 Statistical calculation found with RELIAB_EN.exe.

Review of Weibull functions. Cumulative Distribution Function (CDF):

F(t) = 1 – e

Reliability Function

R(t) = e

β

–(t/θ)

–(t/θ)

β

β

Probability Density Function

f(t) = (β/t) (t/θ) e

Rate of Failure Function

h(t) = (β/t) (t/θ)

–(t / θ)

β

β

This formula or equations can be used once the parameters value are known to obtain the Reliability predictions, percentiles and failure rate. There are procedures to estimate the Weibull parameters from data using the maximum likelihood estimation method [MLE] 1.4.1- Shape Parameter

The shape parameter or slope (β (β) is the main influence infl uence over the distribution shape. When θ = 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  β = 2.0 this is a Raleigh distribution. 1.4.2.- Weibull Fit

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 ( 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 10 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. Step 1.- Data were ordered in ascendant manner by column for the 10 units.

1.2.3.-

2200 2400 3125 7

4.5.6.7.8.9.10.-

3300 3400 3500 3550 3650 4000 6000

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

Failure order j 1.2.3.4.5.6.7.8.9.10.-

Time of failure t 2200 2400 3125 3300 3400 3500 3550 3650 4000 6000

Mdn. Rank from table 6.7 16.2 25.9 35.5 45.2 54.8 64.5 74.1 83.8 93.3

Kapur & Lamberson Mdn. Rank  6.73 16.34 25.96 35.57 45.19 54.80 64.42 74.03 83.65 93.26

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 n

= =

Failure order  Sample size

Example: if the failure j = 5, n = 10 when you substitute substitute those values en la formula ( i ) you obtain: F(t) = (5 – 0.3) / (10 + 0.4) 0.4) = 4.7 / 10.4 10.4 = 45.19 and so forth every calculation for each order of failure.

Another formula from White 1969 also has a good approach. F(t)

=

[j – (3/8)] / [n + (1/4)]

--( ii )

Where:

j n

= =

Failure order  Sample size

One more formula from f rom Dovich 1990 also has a good approach. MR (j) (j) = [n – j + (0.5)

1/n

(2j – n – 1)] / (n-1)

Where: 8

j

=

Failure order

n

=

Sample size

Example for Median Rank order 7 then: j = 7, n = 10 MR (7) (7) = [10 – 7 + (0.5) = [3 + 0.5

0.1

1/10

{2(7) – 10 – 1}] / (10-1)

{14 – 10 – 1} ] / 9

= [3 + .9330(3)] / 9 = (3 + 2.799) / 9 = 5.799 / 9 = 64.43 Step 3.-. Using Weibull Probability Paper. Exist different types of forms of Weibull Probability Paper with horizontal or vertical shape. Step 4.- Once 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. 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 4.1 which indicates a shape of  almost a normal distribution skewed a little bit to the right). 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 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, according to the design of the Weibull Probability Paper). Step 9.- This straight line give us an instantaneously relation of % of failure and hours.

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 fi nd the vertical axis of   percentages where you read the searched value. (In this case approximately 14%). Step 10.- 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

9

μ.

Weibull Slope

Weibull Probability Pa er

10

Weibull Slope

Hours

Weibull Probability Paper

Figure 1.8 Weibull Probability Paper plotting paired points

11

Figure 1.9 Time of failure for the sample of 10 (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 = 0.0673 Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.0673)] = Ln[Ln(1/0.9327)] = Ln[Ln(1.072156] Ln[Ln(1.072156] = Ln(0.06967) = – 2.6639 Second Median Rank = 0.1635 Ln[Ln(1/(1–F(t))] = Ln[Ln(1/(1–0.1635)] = Ln[Ln(1/0.8365)] = Ln[Ln(1.19545] = Ln(0.17852) = – 1.7230 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 Order Time to Fail t – D F(t) K&L Natural Log # (t – D) Median Rank (t – D)

LnLn (1/[1–F(t – D)])

1.-

2200

6.73

7.6962

–2.6638

2.-

2400

16.34

7.7832

–1.7233

3.-

3125

25.96

8.0472

–1.2020

4.-

3300

35.57

8.1017

–.8217

5.-

3400

45.19

8.1315

–.5086

6.-

3500

54.80

8.1605

–.2304

7.-

3550

64.42

8.1747

.0329

8.-

3650

74.03

8.2025

.2990

9.-

4000

83.65

8.2940

.5940

10.-

6000

93.26

8.6995

.9927

12

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 –  δ) on the x – axis and Ln [Ln(1/(1 – (t –  δ))] on the y – axis is to use Weibull Probability Paper and plot (t – δ – δ) versus F(t – δ – δ). As we do some pages before instead of a computer   program as RELIAB_EN.exe

Figure 1.10 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  be used. The data Table 1.4 is plotted in figure 1.11. The slope of this line, which is equal to β, is is 3.774. 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 of δ. A discussion of how to handle a nonzero parameter is made using RELIAB_EN.exe.

Figure 1.11 Best fitting line and calculation of Delta = 0, Beta = 3.774, Teta = 3896, and R 2 =0.859

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(– y o /β) where: yo = the y intercept 13

The y-intercept, yo may be found by extrapolating the line in figures 1.8 and 1.11. The yintercept for the data in table 1.4, is –31.2023. Thus, the scale parameter is θ = exp(– y o /β) = exp [(– [( – ( – 31.2023))/3.774] = 3895.98 1.4.3.- Applications

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

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.

ß=5

The Weibull distribution approaches to Normal Distribution Lepto-Kurtic.

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

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). Bibliography 1.- Reliability Toolkit: Commercial Practices Edition, Rome Laboratory RAC (Reliability Analysis Center 1988, 1993 revision 14

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 Reliabil ity 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 i n 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]

15