Reliability Calculation without Failure (Success Run)

C

convivial

#1
I would like to know the reliability of 20 samples. Assume there is no failure:

Reliability R(t) = ?
Confidence Level P = 80% (in every case)
Required test duration = 100 hours
Weibull paramete b = 2 (always remains 2)

I know I can use this formula: R(t) = (1 - P)^(1/(L)^b.n) where L = life time ratio

But before applying this formula I divide these 20 samples in to two groups having 10 samples each.

For first group (10 samples) I take Lifetime ratio (L) = 0.7, I apply the above formula and get R(t) = 65% (say)

For Second group (10 samples) I take Lifetime ratio (L) = 1.3, I apply the above formula and get R(t) = 70% (say)

Now I have got two reliabilities for these 2 groups. Can I combine these two reliabilities to get a final reliability? If yes, then how shall I combine these two? Shall I multiply or simple add these two?
 
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Marc

Hunkered Down for the Duration
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#2
Any Reliability Engineers out there who can help with this? My Thanks! in advance!
 

Miner

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#3
Why did you divide the samples into two groups?

Are these from different time periods, suppliers, different components in series, in parallel?

The answer depends on your response.
 
C

convivial

#4
Thanks for your reply!

No these are not components instead 20 identical systems. Therefore the concept of parallel or series does not apply here. Sometimes it happens that when you have to test them for a particular period of time and no failure occurs then you say ok I would like to test for a longer time to see whether if the samples still survive or not. E.g. if these survive then their reliability would increase.
In order to make the problem clear lets say there were 20 identical samples (not components) in a test chamber and first ten samples were tested for 100 hours (exactly equal to required test time) and were taken out but the remaining ten were tested for longer time say 200 hours.
I can calculate the reliability for each group of samples as shown in my first post but someone told me that in this case I need to find the total reliability i.e. have to combine the reliabilities of these two groups and here is the point where I don't know how to combine the reliabilities?

P.S. You can see the formula of calculating reliability of each group in the first post.
 

Miner

Forum Moderator
Staff member
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#5
If you can attach or list the actual data, I can calculate this for you. This is just a censoring issue. The complete data set can be used.
 
C

convivial

#6
The complete data is:

Group "A" Samples
1-10 80 hours

Group "B" Samples
11-20 200 hours

I want to find the total reliability
 

Miner

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#7
I performed a Bayes analysis with a Weibul shape parameter = 2. With zero failures, I had to perform a Maximum Likelihood estimate for the lower 95% confidence bound.

The Minitab analysis is attached. The asterices appear because there are zero failures upon which to estimate the values.
 

Attachments

T

topguy

#8
I performed a Bayes analysis with a Weibul shape parameter = 2. With zero failures, I had to perform a Maximum Likelihood estimate for the lower 95% confidence bound.

The Minitab analysis is attached. The asterices appear because there are zero failures upon which to estimate the values.
Could you kindly provide the step by step guidence on how to calculate in MINITab? thanks a lot
 
V

vida99

#9
hi
could you please tell me where did you get this formula( R(t) = (1 - P)^(1/(L)^b.n) ).
im searching for it everywhere and i didnt find it.
thanks
 

Miner

Forum Moderator
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#10
The OP is no longer active on this forum.

I am not familiar with that approach. I did track it down to section 8.3, equation 8.5 in Reliability in Automotive and Mechanical Engineering: Determination of Component and System Reliability written by Bernd Bertsche and published by Springer.
 
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