|MTTF = Total Run Hours (# units x test time)/Number of failures = (200x50)/10 = 1000 hrs|
|R (100 hrs) = e^(-lamda x t) where lamda = 1/MTTF and t=100 hours and e = 2.71828 = e^(-.1) = .904841|
Thanks a lot for your replies.
Regarding the MTTF calculation, there were 10 failures during the testing, but we dont know when they failed. Do we still calculate the test time of 10 failures as 50 hours? Is it possible that these failure units failed before 50 hours testing was done?
I found the solutions for this question in question bank is:
I don't understand this solution.
Your image does not have good resolution, but I think in the question bank solution there is that minor correction to the MTTF formula given by Duke in order to take into account that some of the units failed before the 50 hous testing, notice that in practice the result from Duke and the one from the question bank are basically the same.
What confused me is that why the total run hours need to be divided by 2 as you listed above. Do we want to get an average run hour of the two extreme cases: 1. 10 units failed at the 50th hour; 2. 10 units failed at the 0 hour ? Thanks.
Because in the problem they are not giving you the exact time when each of the 10 units failed you are using an average to estimate the total running time:
- 200 units with a running time of 50 hours. (no failures)
- 190 units with a running time of 50 hours. (10 failures at time 0)