Estimating weibull parameters

[Anonymous]

[Originally Posted: 10/24/00--Transferred by ReliaSoft Moderator]

How I can estimating the three-parameter of weibull distribution?

[RS Tech Support]

[Originally Posted: 10/24/00-Transferred by ReliaSoft Moderator]

The most common ways to proceed are to use rank regression or Maximum Likelihood Estimation. These estimation methods are presented in many of the books listed at the following site, http://www.weibull.com/knowledge/books.htm. In addition, we also have our Weibull++ 5 software package which is designed for Life Data Analysis and the 3-parameter Weibull distribution is included. If you would like more information regarding Weibull++, you can contact us at support@reliasoft.com.

[Antoine]

You can also use the AMSAA model. There is an easy procedure which gives you the results with the appropriate risk rate.

[Usha]

How to calculate Bx life of a product?

[RS Support]

BX life is defined as the time by which X% of the units in a population will have failed. For example, if an item has a B10 life of 100 hours, that means that 10% of the population will have failed by 100 hours of operation. This is also the same as the time at which (100-X%) of the units have survived (or B10 is the same as time at 90% reliability)
Thus, once the parameters of the model have been determined the BX life can be obtained directly from the reliability equation (or the probability plot). For the weibull distribution this is given in the section “The Weibull Reliable Life” on page http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm

Related references:
http://www.weibull.com/hotwire/issue13/relbasics13.htm
http://www.weibull.com/basics/lifedata.htm

[Jengo]

In your presentation about the Maximum Likelihood Estimators for the parameters of
NHPP with Power Law intensity function you have given the formule of the joint
probability density distribution function and the likelihood function. But I do not
understand where the T* come from.Would you please give some explanation? Thanks!

[RS Support]

What presentation are you referring to?

[Jengo]

http://www.weibull.com/RelGrowthWeb/parameter_estimation(nh).htm

[RS Support]

T* represents the termination time. If the test is failure terminated then the termination time is equal to the last time-to-failure. If the test is time terminated then the termination time will end up being greater than the last time-to-failure.

[DonGiacomo]

Hello,

I have a great problem, as I am not a statistic or engineer, only economic student ;-(

Well, I'll try to explain my problem.
I need to test if following data are weibull-distributed:

x1 5
x2 9
x3 3
x4 1
x5 11
x6 2
x7 3
x8 11
x9 6
x10 7
x11 1
x12 4

I read about the Kolmogorov- Smirnov- Test (K-S-Test), but to say it clear: I have not really understood what to do exactly.
Another problem here is, that e.g. X4 and X11 are same, so there is some kind of relation.
Because I have to test a lot of such functions as seen above, I ask for help! Can somebody explain me how to do such a test???
If I know it for one example, I can do the rest alone!
Please help a student, who lost all nerves!

Thank you, in advance

George

[Pantelis]

George,

To begin with a KS is only one test – and there are other goodness-of-fit tests and methods. Also the fact that two of the values are the same does not matter.

Now what you are asking is to determine if Weibull is a better fit than another model … (e.g. you will never know what the right model is just that it is a better fit than the ones considered – or another way to say it is you “fail reject the hypothesis” at some level that the data is distributed using a Weibull distribution.)

Now to answer this you will need to fit the parameters to the model(s) (say Weibull) and then use GOF tests. Software packages (such as Weibull++, http://weibull.reliasoft.com can easily do this).

If you need specific info as to how to do this manually, you can start out at http://www.weibull.com/lifedatawebcontents.htm and then continue from there. It is a well documented process and widely available -- you can also search on Google and you will find additional references.

Now, for the data presented I run a quick analysis (using software), and testing against [ Lognormal – Normal & Exponential ] Weibull is a better candidate distribution.

[Raul Suarez]

Back to calculating B10. What if my life tests show no failures? We can use standard techniques to calculate c=90 lower limit for R, but this is not Weibull Statistics? Please advise.

[Pantelis]

You can do that using a 1-parameter Weibull formulation and MLE (with an assumed beta value).

[manoel Henrique]

hi,
my question is about the same mater.
i´m trying to calculate the weibull three parameters, using rank regression, and i tested it, using a simulation of random exponencial distribution, the problem is:
when i calculate the two parameters, the rank regression result values are near the ideal, but when i calculate three parameters using less square regression and rank regression the parameters are far from the ideal.

my doubt is , am i wrong, using less square regression?

my e-mail:manoel.henrique@globo.com

[Pantelis]

You can use regression, now is your algorithm right?

Also when using simulation, how many data points are you using?

[Manoel Henrique]

My algorithm just takes the values from less square regression and calculates the parameters, i changed the usual form of the equation of weibull in a linear form like y=ax+b, this method has a name, but i forgot now.
I get the parameters using the regression and this equation.

About the points, i'm using a hundred point, i can use any amounts, i get them using a software called R, a statistical data analysis software, that operates very well.

if you want, i can send my data points to you and my algorithm that is an excel file.

thanks.

[Pantelis]

You can linearize a 2 parameter Weibull - but how exactly do you linearize a 3 parameter Weibull? You cannot use standard linear regression for 3P Weibull... your x will contain the term (T-LP) but you need to estimate LP (Location Parameter).
Its actually not that simple.

[manoel Henrique]

there´s a book here in Brazil, that brings one method that do this.

just test the parameter To with many values, and analyse the regression coefficient, when it´s used the linear regression, the To value the maximize the regression coefficient is the final To value, the others parameters of weibull will be obtained at same process, bacause the regression coefficient refers to one set of parameters.

take a look in this home page, it´s in portuguese, but there´s an explain in english about weibull, where this method is completely describe.

http://www.qualytek.com.br/page13.html

[Pantelis]

Manoel,

You started the discussion by saying that you did not like the results that you were getting using your algorithm. My first answer was to check your algorithm/method. Solving for a three parameter Weibull is much more complex than the trivial case of the 2-parameter Weibull. Now you mention a search for the best value --- and a reference to some method posted on the web.

First lets discuss the web reference you provided. It is a good example but I have some issues with it. First it looks like the data is interval data yet the example treats them as complete data (using the endpoint) (see http://www.weibull.com/LifeDataWeb/data_classification.htm for more info on datatypes). If we do treat them as complete data -- then I also have an issue with the example because using the cumulative frequency method presented, the last data point is ignored. Median Ranks (or other ranks) should be utilized.

Having said that, yes I agree that an answer for gamma can be found using that method– but again it’s a matter of implementation. Searching for a gamma using the correlation coefficient as a guide can get you close to a value of gamma, if implemented correctly -but it is not the most efficient way to do it. In other words the method presented there is at best crude and not the best way to do it.

As an example and for the data set given (assuming complete data and using the end points as the example did) yields the following results:


Beta=3.58
Eta=623.12
Gamma (To)=925.75

(Results using Weibull++ http://www.reliasoft.com/Weibull/index.htm . From Brazil should you require any more help you can contact http://www.reliasoft.com.br/ )

[manoel henrique]

thanks Pantelis,

first thing,
In the book reference, it´s used median ranks, and in the web reference it´s used cumulative frequency method.
i made one versions of each one.

one good notice is that, i tested the results using Kolmogorov-Smirnov test, the median ranks get the better results than cumulative frequency method.

But about the three parameters,
can i use this method discribed in the book reference.

you´re telling me that is not the best way, i took a look in the MLE method, but i think that i can´t implement it at an easy way in the excel.

i read the home page.

i think that using median rank is better, but in these descriptions, i saw some about using median ranks with suspensions.

In the book reference, these treatment is done using the equation of median ranks lefting the N (number of equipments) equal to the total of equipments and just calculating the median ranks with the amount of failed equipment data, that it´s less then the number os equipments.

In the weibull.com is showed other method.

What of these can i use?


and to end:

let´s discurse just here, or do you prefer to do it in the other list?

it´s the same think!!

thanks for your patience.

[Pantelis]

Manoel,

As I said earlier the method in the book is not really the best way to do it, but will get you close to an answer.

Now wrt to median ranks, I obviously would say that what is on weibull.com is what is correct (since I wrote some of that).

Now as to how do you implement in Excel, well there is no easy way for 3P Weibull. We have numerical methods to do well and quickly in our software, but I hope that you understand that this is not something we would provide on a discussion forum for obvious reasons.

[Manoel Henrique]

thanks Pantelis,

Don't worry, i understood what you're trying to tell me.

but you already helped me.

Now i'm using the method indicated for you and it's implemented in excel with 3P weibull and numerical methods, i'm using one after test with some data.

it's true that i'm not worry about my rigths for done an software or excel worksheet developed by me, spending my time.

but in this list, this is very important.

you helped me, orientate me to keep on the right way.

this month is the 17th month that i'm working in my after-graduate work, i don't know how this type of work or course is called in your country.

thanks a lot.