MTBF calculation

J

Thread Starter

Jeremy W

I have 3 products I tested for 1000 hours each with no failures. I will be using 8 of these products for 800 hours a year (each) for 5 years, I need to know the probability of 1, 2, or 3 failing, I also need to know the overall MTBF. Everytime I try to calculate it I get infinity so I am trying to find the lower bound with a certain degree of confidence, 95%, but I am also having difficulty finding a calculation that suites my test data.
Can anyone help me out there?

 
G
You don't have a large enough sample size, and haven't experienced any
failures. All you know is your MTBF is at least 3,000 hours. I suggest you
buy a basic reliability engineering text. There's simply too much to try and
cover in e-mails.
 
Jeremy W:
> I have 3 products I tested for 1000 hours each with no failures. I will
> be using 8 of these products for 800 hours a year (each) for 5 years, I
> need to know the probability of 1, 2, or 3 failing, I also need to know
> the overall MTBF.

If you find any answers, can you send me a copy, please? I'd be interested
in this also.

I've been thinking about a similar problem this week, and getting various
formulae which I don't really have the inclination to integrate right now
if I can help it...

> Everytime I try to calculate it I get infinity so I am trying to find the
> lower bound with a certain degree of confidence, 95%, but I am also
> having difficulty finding a calculation that suites my test data. Can
> anyone help me out there?

One possibility is to use Bayes formula, but to do that you need to specify
what you expected the MTBF to be before you ran your test (or rather the
expected probabilities of getting the various MTBF figures).

I've the advantage that my problem involves a limited range, so I could
simply say ``all answers are equally likely''.

If you can get that, the calculation is relatively simple - basically for
each possible MTBF figure you multiply how likely you think it is with how
likely it would be to give you 3 sets of 1000 hours no failure, add them
all up, and then find the 95th percentile (or 5th, depending on which way
you set it up).

You could integrate it, but it's probably not worth the bother: a numerical
calculation in a spreadsheet should be good enough.

HTH - and if you get a better answer, please do fwd me a copy

Jiri
--
Jiri Baum <[email protected]>
http://www.csse.monash.edu.au/~jiribvisit the MAT LinuxPLC project at http://mat.sourceforge.net
 
As others have already advised, there will be a myriad number of solutions. However, I will attempt to provide you with a starting point... simply because engineers love numbers.

You're problem is not unique. Usually one tests a new model to its initial failure, fixes it, tests to the second failure, fixes it, etc, until the "n"th failure. For your purposes assume that the failure-rate of your product is constant with respect to time. Additionally, assume that test termination occurs at a time that doesn't correspond with the "n"th failure, but instead terminates with the 1st failure, say 1 second
after the 1,000 hour. And, since you have 3 identical(?) units the cumulative time (T) is 3,000 hours.

Two Cases are now possible, one optimistic (o), the other pessimistic (p). For Case 1, the n failures have occurred in time T; and for Case 2
that n+1 failures have occurred in time T. Thus, there are two mean time to failures to deal with, Ho=T/n, and Hp=T/(n+1)

From here on you apply the 'chi-square' distribution formula for both Hp and Ho, using degrees of freedom, Fp =2n, and Fo =2(n+1), respectively. For the optimistic case, I calculated that for a 90% confidence level the lower value is about 380 hrs, while the higher value is about 8,500 hrs. The pessimistic case calculation is left to the reader!

Good hunting!

Regards,
Phil Corso, PE
(Boca Raton, FL)
 
There is a method for an exponential failure distribution using chi-square
distribution that provides a lower limit estimation the MTBF for a time
truncated test with no failures. It is given in the following books with
the book by Dorvich also providing a simplified equation.

Reliability Statistics, Section 8.3 pg 26
Robert A. Dovich,
ASQC Quality Press, Milwaukee, WI
ISBN 0-87389-086-8

Reliability Maintainability and Risk, 5th edition, Section 5.3 pg. 47
David J. Smith
Butterworth Heinemann
ISBN 0 7506 3752 8

If the unit's failure distribution is exponential, the failure rate is
the inverse of the MTBF from which the probability of failure can then be
easily calculated. Either of the above books will have the methods to
calculate the probability.

Bill Mostia
===========================================
William(Bill) L. Mostia, Jr. PE
Independent I &E Consultant
WLM Engineering Co.
P.O. Box 1129
Kemah, TX 77565
[email protected]
281-334-3169
These opinions are my own and are offered on the basis of Caveat Emptor.
 
R

R R Stephens

Take a look at this web site. You can use Weibull paper( a log-log type of
paper) to plot your date and read beta/MTBF/Characteristic Life from the
curve. Now days software does the work I believe they have the software you
need to calculate the MTBF/Characteristic Life is located on this web site.

http://www.weibull.com/
 
Correction to my Thu, Jul 12, 2:53pm post on the subject. I inadvertently interchanged Fp and Fo. The correct notation should be:

Fp = 2(n+1) and Fo = 2n.

Optimistic Case
The lower and upper MTBF values at a 90% confidence level are 475 hrs and 3,670 hrs, respectively.

Pessimistic Case
The lower and upper MTBF values at a 90% confidence lever are 515 and 2,960 hrs.

Regards,
Phil Corso, PE
(Boca Raton, FL)
 
G

George Condur

Better, you ask this information from the products producer. Normaly, exept the asian sources, every R&D laboratory calculate this parameter.
 
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