Warranty Exposure

[Jason Bower]

Our customer has dictated to us a design target of B20 life at 90% confidence (R80/C90)at 750,000 miles. They also specify that these types of calculations, we should use a beta of 2.5. How can I take this information and determine what that means for a failure rate at the end of our warranty period of 100,000 miles.

I have used the free form data type to specify the two points and then checked the failure rate at 100,000 but I'm not sure if this adequately accounts for the 90% confidence in the design target.

Thanks in advance for any help you can provide.

[Dr. Dave]

First let's decide what the required eta is. If we have an infinitely large sample size (max precision), then .8 = exp (- (750000/eta)^2.5)

This means that eta must be 1.36655 10^6.

If we have a smaller sample size, then by using a LCB 90 we would have the sample eta estimated as smaller value as a hedge against sampling error.

Now we need that failure rate at 100,000 miles. The failure rate formula is lambda (t) = beta / eta * (t/eta)^(beta - 1). For t = 100000, lambda = 2.5 /1.36655E6 *(100000/1.36655E6)^1.5
= 3.62140E-8.

Hope this helps. Best wishes,

Dr. Dave Olwell