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mr.stahl
April 17th, 2008, 09:16 PM
I'm evaluating a version of weibull++ and I have the following question:

I have 8 specimen with 3 failure modes S,V,B.
Here ist the time and x means failure
Time,S,V,B
7481xx-
5989x--
4360-xx
5875x--
6249xx-
5220--x
3791x--
6616x--

when you have failure mode one(S) the test machine will stop - and it's possible to detect one of the other two modes(V,B) also.
When you have mode B the test machine will stop and it's possible to have failure two(V). The machine will take a while to detect that B has happened which means the failure initiation is a while ago.

Thanks for your help,
Stef

David
April 18th, 2008, 04:26 PM
Hi Stef,

Welcome to the Forum.

I have a couple of questions. First, what specifically are you trying to model? Are the failure modes independent and are B and V dormant failures?

You indicate that if S occurs the test machine will stop. After this it is possible that you may also be able to detect V or B. But the next paragraph states "When you have mode B the test machine will stop...The machine will take a while to detect that B..." So it appears that the B mode is not dormant in one line, but then dormant in the next. Can you please clarify your situation?

mr.stahl
April 18th, 2008, 10:57 PM
Hi David

Many thanks for your post.

So it's like this. V is a dormant failure. B is a "domant" failure for a long time of the test untill this failure (crack) has grown so big and the machine recognizes. So somehow the time to failure did not mean that we know when the failure occured. And we don't have an inspection interval.
But maybe we have to say that B is not dormant.
For the test data this means:
7481 - detected through S - V is dormant
5989 - detected through S
4360 - detected through B - crack initiation a while ago
5875 - detected through S
6249 - detected through S - V is dormant
5220 - detected through B - crack initiation a while ago
3791 - detected through S
6616 - detected through S

I modeled this in three Folios for every single event - just time and F/S. Example for S:
F 7481
F 5989
S 4360
F 5875
F 6249
S 5220
F 3791
F 6616

then I used a block, where I said. When S happens - failed, when B happens - failed (both in line) and put the V in parallel to these 2 blocks because it never happens the the machine will stop because of V.

I hope it's clear how I wrote this,
Stef

David
April 21st, 2008, 01:36 PM
Hi Stef,

Thanks for the information. You are correct in how you would model S event. Events V and B would be handled in a similar manner. Just be aware that the failures for the V event would be interval (left censored) data points. You only know that it occurred sometime between 0 and 7481 hours. For example, the first V failure would be entered as a failure with the last inspection time equal to zero and the state end time equal to 7481.

The B event can be modeled similarly as the S event. The crack may have started at a time previously, but you are modeling when it brought the system down. Since you are assuming B is not a dormant failure then you can put this block in series with the S event. Now the V event (dormant failure) would be put in parallel with a block that does not fail.

I hope this helps.

mr.stahl
April 22nd, 2008, 11:23 AM
Hi David,

many thanks for your answer. Your description is very clear and was very helpful also I have some problems with the solution for a model of V because of the small number of failure points.

Many thanks again,
Stef

David
April 23rd, 2008, 06:32 PM
Unfortunately, there is not much to go on with only two left censored data points. I would recommend MLE in this case (I tried a 2-parameter Weibull distribution), although you will still get a notification regarding the calculations. The notification applies to the plotting, but the estimated parameters via MLE are correct.