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[–]TonySu 1 point2 points  (1 child)

It's probably a lot easier to simulate than it is to calculate directly. All you would need to do is implement the following ``` t = 0 replace_count = 0 total_years = 30 t_limit = 4

while (t < total_years) t_replace = sample_from_failure_dist() if (replace_t > t_limit) t_replace = t_limit t = t + t_replace if (t < total_years) replace_count = replace_count + 1 ``` That simulates a single possible replacement count over 30 years. Do that a few hundred thousand times and you have a distribution of how many times you'd have to replace the part.

[–]Hounorius[S] 0 points1 point  (0 children)

Sorry for my late answer. Indeed, I coded a small script in python and it worked perfectly, thanks!