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During their useful life, 200 units were tested for 50 hours and not replaced when they failed. At the end of the test were 10 failures. What is the 100-hour reliability of this unit? How to calculate the MTTF? Thanks in advance.

9 Replies

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Can anyone help look into this question? Thanks in advance.

606 Posts

Duke Okes - can you help answer this question?

Also check out the Reliability and Risk division

Also, through the search function, I found these other discussions on CRE, which may be of help https://my.asq.org/search?query=cre

Also check out the Reliability and Risk division

Also, through the search function, I found these other discussions on CRE, which may be of help https://my.asq.org/search?query=cre

125 Posts

Wow it's been decades since I studied this but I googled the terms and here is what I came up with:

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 |

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Duke, Trish,

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.

Thanks,

Vincent

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.

Thanks,

Vincent

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Hi Vincent,

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.

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.

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Miroslav,

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.

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.

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Yes,

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)

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)

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I also had concerns about not knowing the actual run time for units that failed, but didn't find anything referencing it. Thanks for your addition to the solution. I am not a CRE!

606 Posts

Thanks Duke Okes and Miroslav Vulinovic ! I am certainly no math wiz. Glad I know people much smarter than me who can help :D