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Hi all,
I am in search of a formula that will provide the number of test units required to demonstrate a reliability figure with allowance for 'x' failures.
I know of this formula which allows for zero failures:
N=(ln(1-confidence)/ln(R)), where N is the number of units (round up this answer), R is the reliability from 0-1. Confidence level is also 0-1. Straight forward.
Then, I found this formula which is confusing and also solves for confidence.
______________f_____n!
(1-confidence)= Σ ----------- x (1-R)^i x R^(n-i)
_____________i=0_ i! x (n-i)!
Yikes! Brain hurting....
Need a similar formula to solve for number of units.
Thanks. Robert
I am in search of a formula that will provide the number of test units required to demonstrate a reliability figure with allowance for 'x' failures.
I know of this formula which allows for zero failures:
N=(ln(1-confidence)/ln(R)), where N is the number of units (round up this answer), R is the reliability from 0-1. Confidence level is also 0-1. Straight forward.
Then, I found this formula which is confusing and also solves for confidence.
______________f_____n!
(1-confidence)= Σ ----------- x (1-R)^i x R^(n-i)
_____________i=0_ i! x (n-i)!
Yikes! Brain hurting....
Need a similar formula to solve for number of units.
Thanks. Robert