Average MTBF

[dvaidr]

I have twenty indentical units, (the same type), and have calculated MTBFfor all.

How do I calculate the 'average' so that I quote an MTBF in the manual.

Might sound a bit of a strnage question but it was debated hotly in a meeting yesterday.

Cheers.

Dave.

[Pantelis]

Can you elaborate...

How did you calculate each one?

[dvaidr]

2 parameter weibull using linear regression and the weibull function in MS excel. I 'checked' it using the weibull sectin of RCMCost software and both were the same.

The regression bit give me the shape parameter and the characterstic life and hence can determine MTBF from the using the WEIBULL(a,b,c) function.

[Pantelis]

Ok... so if I understand you right:

· You computed a Weibull model for the times to failure (and/or suspensions) of 20 identical units using rank regression.
· From the model parameters you then computed the average time to failure or MTTF using equations like the ones here http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm (http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm) (Note that in this case it is really not an MTBF, since if you are using a Weibull model and you are assuming non-repairable units, but the MTTF).

Now that MTTF (Mean Time To failure) is the average time to failure for these units. What other average are you seeking?

[dvaidr]

Ok... so if I understand you right:

· You computed a Weibull model for the times to failure (and/or suspensions) of 20 identical units using rank regression.
· From the model parameters you then computed the average time to failure or MTTF using equations like the ones here http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm (http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm) (Note that in this case it is really not an MTBF, since if you are using a Weibull model and you are assuming non-repairable units, but the MTTF).

Now that MTTF (Mean Time To failure) is the average time to failure for these units. What other average are you seeking?

I am seeking the average of the MTTF for all units. i.e. say the twenty units yielded the following MTTF - 220, 276, 210, 290, 310, 203, 405 etc etc. These are the MTTF calculated for all twnty units. Now that I have these would it just be a case of taking the median or would I have to repeat the rank regression?

I know it seems like a stupid question but I want to be sure.

Thanks for your help.

Dave.

[Pantelis]

I am still not getting it!

You got twenty units… each one has a time to failure T1, T2, … T20. You then computed the MTTF … that’s a single value, how did you get multiple MTTFs? In other words, if I got 3 data points 10, 20, 30 then the normal mean is 20…. You compute a mean from the data, a single mean!


Now I may have totally missed your question, and you are dealing with repairable units, i.e. unit 1 has multiple failure times T1_1, T1_2 … and so forth, in which case the Weibull analysis you mentioned is not applicable.

[dvaidr]

OK, I'll try to make it clearer.

The units are repairable - I should have said MTBF!

Each unit has a MTBF calculated from the failure intervals, i.e. during its life, a unit might fail at 100, 167, 231, 342 etc hours. From this I determined MTBF for twenty units. There is no censored data here for arguments sake.

I end up with twenty separate values of MTBF for twenty units. I wish to give the user an average MTBF for that unit type.

Do I just use the median or is it not that simple?

Hope I managed to make myself clear.

Cheers.

Dave.

[Pantelis]

Ok… Now I got you …

First you are looking at each unit individually. Ok…

Earlier you said you fit a Weibull to each one (I am assuming to the inter-failure times). Now this is not correct for repairable units unless each repair makes the unit as good as new. If the units are as good as new after each repair then each failure time is independent of any past history, therefore each data point is an iid (independently and identically distributed) that can be used in Weibull analysis. Thus, what you would do is fit all the data in a single date set and get your MTTF (in other words if each unit failed three times and you have twenty units, 60 failure points would be fitted to a single Weibull obtaining a single MTTF).

Unfortunately the as good as new assumption for repairable systems is not usually valid. I don’t know whet you are testing so I can’t comment.

In this case you would analyze all the data again but using different modeling techniques. In the case of as good as old an NHPP model can be utilized or in the more general case a General Renewal Process (GRP) model should be used. Both of these models would utilize all the data giving you a single MTBF metric for the analysis.

For details on Recurrent Events Data Analysis see
http://www.weibull.com/LifeDataWeb/recurrent_events_data_analysis.htm (http://www.weibull.com/LifeDataWeb/recurrent_events_data_analysis.htm)

For general repairable systems analysis see also:
http://www.weibull.com/RelGrowthWeb/Repairable_Systems.htm (http://www.weibull.com/RelGrowthWeb/Repairable_Systems.htm)

Hope this finally answers your question.
------------------------------------------
P.S. With respect to how the analysis is done the aforementioned links are textbooks explaining this.
Also ReliaSoft Software (http://www.reliasoft.com/ (http://www.reliasoft.com/)) automate this type of analysis.
Specifically for the GRP process see Weibull++ 7 http://weibull.reliasoft.com/
For repairable systems see RGA http://rga.reliasoft.com/

[dvaidr]

Thanks fo your help.

I thought this might be the case but I had to check.

It just gets more and more involved.

[Pantelis]

Good luck David.

I am now closing this thread.