Uranium processing expected times
Posted: Fri Nov 29, 2024 4:46 am
(Posting here since wiki discussion pages are dead) I added a section at https://wiki.factorio.com/Uranium_proce ... cted_value about how one centrifuge continuously running should (in theory) be able to power one reactor.
There is some math on the distribution of cycles required to reach 40 U-235, although that should probably be in the Kovarex enrichment page, and I don't think it's useful for most players as it's not critical for running the power supply. It wasn't mentioned but I believe it comes from the binomial distribution, specifically the probability of X >= k = 40 successes in n trials, where probability of success is p = 0.007. This is 1 - P(X <= 39) = 1 - CDF(k = 39, n, p = 0.007).
Maybe more useful is the probability that you will be able to run your reactor successfully with just one centrifuge. 1 U-235 produces 10 uranium cells, each of which lasts 200 seconds. To find the distribution of at least 1 U-235 produced over n cycles, it is the same binomial distribution but with a shortcut, as it is P(X >= 1) = 1 - P(X = 0) = 1 - (1-p)^n. 2000 seconds is 2000 / 12 = 166 cycles, so there's only a 1 - 0.993^166 = 68.8% chance that you'll get 1 U-235 in that time. With two centrifuges, you have 332 cycles so 90.3%, and for three centrifuges you have 498 cycles so 97%.
Is this worth adding? Hopefully I did the math right.
There is some math on the distribution of cycles required to reach 40 U-235, although that should probably be in the Kovarex enrichment page, and I don't think it's useful for most players as it's not critical for running the power supply. It wasn't mentioned but I believe it comes from the binomial distribution, specifically the probability of X >= k = 40 successes in n trials, where probability of success is p = 0.007. This is 1 - P(X <= 39) = 1 - CDF(k = 39, n, p = 0.007).
Maybe more useful is the probability that you will be able to run your reactor successfully with just one centrifuge. 1 U-235 produces 10 uranium cells, each of which lasts 200 seconds. To find the distribution of at least 1 U-235 produced over n cycles, it is the same binomial distribution but with a shortcut, as it is P(X >= 1) = 1 - P(X = 0) = 1 - (1-p)^n. 2000 seconds is 2000 / 12 = 166 cycles, so there's only a 1 - 0.993^166 = 68.8% chance that you'll get 1 U-235 in that time. With two centrifuges, you have 332 cycles so 90.3%, and for three centrifuges you have 498 cycles so 97%.
Is this worth adding? Hopefully I did the math right.