Download Failure Rate - Space Sciences Laboratory

Failure Rate Estimation
M.Lampton UCB SSL
• An upper limit on failure rate, or a lower limit on MTTF, is
required to establish system reliability
• The limit is to be obtained by measurement
• How many failure-free operations, or hours of operation,
will establish a given limit?
• What is meant by the confidence level of a limit?
• General reference: I.Bazovsky “Reliability Theory &
Practice” Prentice-Hall 1961, 1982.
Simplest case: independent random failures
ref: I.Bazovsky Chapters 3,4
• Failures are discrete, complete, and unambiguous: a failure
happens, or it doesn’t.
• Failures are independent and they happen at random. They
are not correlated. They obey Poisson statistics.
• Failure rate is statistically constant: no infant mortality, no
wear-out, no limited life effects.
m N
e m
P ( N , m) 
N!
m
P (0, m)  e
How to test for failure rate?
• Any test will involve measuring some
random events
• The test, when repeated, can/will give
various answers
• Yikes
• How to quantify a test result when the
events being observed are random?
• Answer is Confidence Level
What is Confidence Level?
• Confidence = probability of making a
correct conclusion, given your test data.
• Pick a confidence level, say 90%
• Then adjust the wording of your conclusion
so that it is correct at least 90% of the time,
for every possible true parameter value.
What would we expect?
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Let the true failure rate = R failures/hour
Let a test have duration T hours.
Mean number of expected failures is RT.
Probability of zero failures is exp(-RT).
The confidence region for R given zero observed
failures and given any desired confidence C is...
log e (1 /(1  C ))
0 R
T
or
0  R  2.3/T for 90% confidence .
Why does this work?
• There are just two possibilities:
R  2.3/T
then t he claim is, in fact, correct.
or
R  2.3/T
then t he claim is wrong, but the
observed result (zero failures! ) is unusual :
only a 10% chance, if R  2.3/T
only a 9% chance, if R  2.4/T
only an 8% chance, if R  2.5/T
only a 1% chance, if R  4.6/T
only a 0.1% chance, if R  6.9/T,
etc.
Example
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Have a 40000 hour mission (5 years).
Want MTTF>100000 hours with 90% confidence.
Therefore want R<1E-5/hour, 90% confidence.
Therefore need T>2.3E5 unit hours with zero failures.
This is a lot! One unit=29 years; 10 units=2.9 years.
Probably need “accelerated testing” i.e. faster cycling, or
more frequent thermal stresses, or higher RPMs, or
whatever makes for accelerated failure rates. But then, you
need to estimate how much acceleration you are actually
obtaining -- a problem in itself.