reliability and Six Sigma
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 This topic has 8 replies, 4 voices, and was last updated 18 years, 1 month ago by Hemant Gham.

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April 26, 2004 at 4:09 pm #35349
We are carrying out an reliability test under one six sigma project. out small y’s were as follows. Please help me to find the test hours for the test.
A test is needed to verify an chip for mobile phone to 300,000 – hour MTBF product criterion. The product is typically used 24 hours per day by customersa. If five failures are allowed and a 90% confidence interval statement is desired, determine the total number of test hours0April 26, 2004 at 9:31 pm #99152In 1997 RAMS conference, phadke explained that you don’t have to run 40,000 tests for a copy machine to guarantee that there will be no paper jam. It had something to do with Robust Design. They came up with less than 100 tests. You can contact him
Manee0April 27, 2004 at 7:51 am #99169
Hemant GhamParticipant@HemantGham Include @HemantGham in your post and this person will
be notified via email.Bono,
You have incomplete information posted in your question.
MTBF/MTTF = 300,000 hours;
Failures ‘f’ = 5;
Confidence Level (%) = 90 %.
How many units do you have? OR How many units you would like to be tested? This is also required to get the “Test time – t”.
Secondly, in order to design and conduct such a test, something about the behaviour of the product will need to be known. For example, the shape parameter of the chip’s life distribution. You need to study the financial tradeoffs between the no. of units and how much test hours is needed to demonstrate the desired goal (MTBF=300,000 hours). Use cumulative binomial distribution and find out the test hours required. But you will require the no. of units also in addition to the above data you have.
In your case, we have the value of MTBF. Assume you have Weibull distribution with shape parameter = 1.5. Calculate the value of the scale parameter.
MTBF = h.G(1+1/b)………………..(I)h = Scale parameter
G(x) = Gamma function of x (Ask your MBB or Stats/Reliability expert to obtain this)
b = Shape parameter = 1.5 (assumed)
From (I),
h = MTBF / G (1+1/b)
Let me give the cumulative binomial equation, which for Weibull distribution appears as,
1 CL = iS f (n!/(i!(ni)!)) .(1 e(t/h)b)i . ( e(t/h)b)(ni)) …(II)
Now, knowing the values of n (no. of units), CL (Confidence level), f (no. of failures allowed),h (scale parameter), and b (shape parameter), all that remains is to solve the above equation (II) for test hours (‘t’).
Hope this helps. Alternatively, you may discover that instead of finding test hours, you require to find the no. of units required since you have at your disposal only specified test time given by the customer or dictated by the market. In this case, you can get the no. of units required to test for specified amount of time for a given % reliability (or MTBF). I think software tools are available in the market to do these calculations. If you haven’t got any of them, then you need to solve the equations manually.
Good luck !0April 27, 2004 at 10:54 am #99176Congratulation for your excellent demonstration.You are one of the few experts in this Forum who go for details,kind regards
0April 27, 2004 at 1:15 pm #99194Thx a lot hemant for your sincere effort, but I was wondering if it can be solved with the pascal distribution.
0April 27, 2004 at 1:19 pm #99195please revert Mr Hemant ASAP
0April 27, 2004 at 5:26 pm #99229I will be highly obliged if somebody throw some light on how reliability engineering get mapped with Six Sigma so that we may thin k of implementing the same in our organisation.
0April 27, 2004 at 7:35 pm #99254Please comment on this formulae
The chisquared equation for test time is
This method only returns the necessary accumulated test time for a demonstrated reliability or MTTF, but not a specific time/test unit combination that is obtained using the cumulative binomial method described above. The accumulated test time is equal to the total amount of time experienced by all of the units on test. Assuming that the units undergo the same amount of test time, this works out to be,
where,
= the chisqaured distribution,
= the necessary accumulated test time,
CL = the confidence level, and
f = the number of failures.
0April 28, 2004 at 5:09 am #99282
Hemant GhamParticipant@HemantGham Include @HemantGham in your post and this person will
be notified via email.Good finding Bono.
But again, remember, here you are only assuming constant failure rate, an exponential distribution instead of using cumulative binomial equation. Another difference is if you are interested in only accumulated test time then this will hold true.
For financial tradeoff, it is always better to go with cumulative binomial method if we have specified test time or no. of units. In your case if this is not the criteria then accumulated test time will solve your purpose.
One important thing. We discussed two cases with assumptions of underlying life distributions. You may wish to find the appropriate life distribution using Goodness of Fit and corresponding AndersonDarling values.A smaller AndersonDarling statistic means that the distribution provides a better fit. You may as well do it using Reliability software packages. Once you have the life distribution you can always choose a method to find the test time.
Best Regards,
Hemant0 
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