Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Reliability Testing

Status
Not open for further replies.

meverett85

Aerospace
Sep 23, 2013
5
Hi all,

Short background:

In my line of work, we must be able to destroy an errant launch vehicle if it is determined to be outside of the predetermined acceptable corridor.

We do this by having transmitter sites that send RF commands to the vehicle.

In any case, when a developer is put on contract to modernize a site, one of the requirements is a reliability of .999 for a 5 minute window.

One of the questions that recently came up, was how long one would have to test the new site in order to prove this reliability. The final component to this problem is that there must be 95% confidence associated with the test.

The confidence part is what is stumping me, since I have no idea how you attribute a z-value equivalent to such a problem? Can anybody explain some theories or approaches on how to determine the amount of test time necessary to achieve these reliability requirements at a system level?
 
Replies continue below

Recommended for you

The reliability number presumably results in some sort of failure rate, say, fails/million hours. If we assume that it's a Poisson process, then the standard deviation might be the square of the number of fails in a million hours. This would then given you a variance of failure rates that could be applied to the confidence level. Anyway, that's my guess for something similar that I'm working at the moment.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Interesting....so let's say I took a stab at this.

Reliability = e^-(failure rate*time)

Therefore, for R = .999 and a time of 5 minutes, solving for failure rate = .01201 or Mean Time Between Failure of approximately 83 hours (1/.0121).

If I said test the system for 83 hours and see how many failures you get...I believe that seems too simplistic.

Now the question becomes, how long do I have to test a large system (with many individual components) to satisfy the reliability with a 95% confidence level for a 5 minute window?

I understand how to theoretically calculate the reliability based off of individual component production data...and rolling up the standard deviations into an overall system error, but I don't know how to attribute a confidence level to a system that does not possess individual component testing or a large amount of system testing (with failure data included).

My background in stats is pretty sophomoric...
 
You should consider the period prior to the window as the critical issue. Assuming, say, a period of 24-hr or more prior to the 5-minute window that the system must not fail, since it might cause a launch delay. This would make the required failure rate more like 42 per million hrs --> 4282-hr MTBF. Obviously, if the pre-launch period is allowed to be much shorter, say 8 hr, since there would be some finite repair time that must be covered, the MTBF required would be more like 8000 hr.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Ugh, that first value should have been 24000-hr MTBF.


TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
In any case, using the second failure rate, you'd assume that it needs to be the required failure rate minus 1.645 times its standard deviation, you'd wind up with 105 failures per million hours or 9560-hr MTBF

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
dang it. 108 per million and 9260 hr

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
You are on to something....you must have range experience!

You are referring to what we call launch mandatory reliability. And the window is T-6 hours before launch to T-0. From T+0 to T+5 minutes we call this Flight Termination reliability.

You are correct that if we have a certain amount of failures before the launch we will scrub the mission (mega bucks in some cases), so this is certainly an important variable. It is addressed in the requirements specification by stating the Mean Time to Restore the system must be no greater than 30 minutes. Additionally there is much redundancy allowing us to absorb certain failures and still feel confident to press on.

However, I'm still not sure if we are on track to answer my original question or if it even can be answered. Let me reword it:

Suppose you were developing a new transmitter site (sends termination commands to the vehicle) and wanted R=.999 for a time window of 5 minutes (flight termination reliability). If you were trying to identify a set of requirements that had to be met before the range accepted the new system from the developer, is it possible to identify an amount of test time necessary to claim that you meet this reliability with a 95% confidence. Basically some sort of operational test once DT&E has been completed?

Keep in mind that testing the site for thousands and thousands of hour is impractical in this situation. With that in mind, is the only way to prove this reliability through a table top theoretical analysis using individual component data?

Thanks for the replies!
 
Testing will depend on cost per unit and number of units. If this is typical stuff, then there's only one deliverable unit, and one would have to build additional qualification units, which is not uncommon. If so, then one can do accelerated life testing using vibe, temperature, etc. to accelerate failures. Since this is mostly electronic, the both vibe and temperature might be used. The issue then, is whether there exists a plausible activation energy for the failures, so that one could plug into the Arrhenius equation and determine the acceleration factor. This might require some literature searching to develop a basis for claiming, say, 1 eV as the activation energy.

There are a bunch of companies selling highly accelerated life testing (HALT) and highly accelerated stress testing (HAST) services and equipment. Obviously, your customer will have to buy into either pure testing or pure analysis. I would guess that they would require at least the analysis, and some level of accelerated testing, just to dot the i's. This might be coupled with some sort of reliability improvement program, to cross the t's.

The analysis presumably would be to MIl-STD-217 or Bellcore. There are programs that do that sort of thing, given a parts list, by pulling failure rate data from their databases and aggregating the failure rates into a composite rate that is used to calculate the expected MTBF. Given the reliability requirements, the 108 fails per million accounts for some plausible statistical variance, but your reliability software, like RELEX, would have ranges of failure rates, which you could probably use to generate an aggregate failure rate variance value.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
So I think I've figured out a good way to do this:

1) Identify the theoretical reliability for the system for a given amount of time.
2) Calculate the MTBF using equation R = e^(-t/MTBF)
3) Figure out the Chi-Square value for 95% confidence and zero failures:

To do this use equation Testing time,T = [MTBF*Chisquare(prob,dof)]/2].

Within the Chisquare characteristic, the probability is .95 corresponding to a 95% confidence requirement and the dof formula is v = 2*failures + 2 = 2

Excel has a function called CHIINV which calculates this value.

So with the aforementioned .999 reliability requirement, after some plug and chug you determine that one needs to test the system for 372 hours with zero failures to be 95% certain that it meets its intended reliability.

Thanks for you help IRstuff.
 
Your hours seem low to me. I'm not particularly familiar with this line of calculation, so I used Method 2, with 0.995 reliability for 6.08 hr, zero fails, and 95% confidence, which returns 3634 hr test and implied MTBF of 1213 hr.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
oops, I used 0.995 instead of 0.999

MTBF = 6077
test hours = 18205, which is a VERY long time.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor