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| == Initiating the [[Kovarex enrichment process]] == | | == Initiating the [[Kovarex enrichment process]] == |
| Unlike all other crafting processes in the game, uranium processing creates U-235 and U-238 based on probability, rather than in guaranteed deterministic amounts. Additionally, the Kovarex process requires a bulk stockpile of 40 units of the rare uranium-235 isotope to initiate in a single centrifuge. Therefore, planning uranium mining and processing for the Kovarex process takes certain considerations not encountered elsewhere in the game. | | Unlike most other crafting processes, uranium processing creates U-235 and U-238 based on probability, rather than in guaranteed deterministic amounts. Additionally, the Kovarex process requires a bulk stockpile of 40 units of the rare uranium-235 isotope to initiate in a single centrifuge. Therefore, planning uranium mining and processing for the Kovarex process takes certain considerations not encountered elsewhere in the game. |
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| === Expected value === | | === Expected value === |
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| == Precise Mathematical Analysis == | | == Trivia == |
| To determine the exact number of [[centrifuge]]s required for maintaining [[reactor]] stability without using the [[Kovarex enrichment process]], we need to employ probability theory. This analysis will help us estimate the probability of obtaining the required amount of [[uranium-235]] and the number of [[centrifuge]]s needed.
| | * Excluding Space Age-exclusive recipes, this is the only recipe in the base game to have random products. |
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| The primary goal of these calculations is to derive a formula that can be easily applied and sustainable for predictions despite advancements in [[technologies]], game version updates, or in-game developments and modifications.
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| === Variables ===
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| Let's introduce some variables that we'll need in further calculations
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| * ''<code>T</code> - overall time of a [[reactor]] working (in seconds*)'' | |
| * ''<code>t</code> - time of a [[reactor]] cycle (in seconds*)''
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| * ''<code>c</code> - consumption of [[uranium-235]] per reactor cycle''
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| * ''<code>k</code> - [[centrifuge]]s number (our target value for calculations)''
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| * ''<code>p</code> - probability of [[uranium-235]] production per centrifuge cycle''
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| * ''<code>n</code> - overall number of [[centrifuge]] cycles''
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| * ''<code>τ</code> - [[centrifuge]] time per cycle to produce 1 recipe''
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| * ''<code>α</code> - desired probability of having success number [[uranium-235]] to sustain all <code>T</code> time of [[reactor]] stable work**''
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| <nowiki>*</nowiki> Measured in seconds or in-game [[tick]]s, this does not affect the general reasoning
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| <nowiki>**</nowiki> Known as the [[https://en.wikipedia.org/wiki/Confidence_interval confidence level]]. Due to the nature of probability, this value can only be in the interval <code>(0; 1)</code>. We will use the values 0.9, 0.95, and 0.99 to maintain scientific rigor and ease of interpretation, opposite to the [[https://en.wikipedia.org/wiki/Statistical_significance significance level]].
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| === Assumptions ===
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| ;Assumption 1
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| : All <code>k</code> centrifuges work continuously and simultaneously for each <code>n</code> cycle with the same <code>τ</code>
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| ;Assumption 2
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| : There are always enough resources to convert [[uranium-235]] into reactor fuel *
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| ;Assumption 3
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| : The conversion of [[uranium-235]] into fuel for reactors can be neglected **
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| ;Assumption 4 (for long-term play)
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| : The game can be played indefinitely (technically assumed for practical use, average playtime suffices)
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| <nowiki>*</nowiki> For instance, base game [[uranium-238]] has a very high probability (<code>0.993</code>), ensuring a sufficient supply for [[uranium fuel cell]] production.
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| <nowiki>**</nowiki> For instance, converting base game [[uranium-235]] into 10 [[uranium fuel cell]]s takes <code>12 sec</code>, but the [[reactor]] cycle is <code>200 sec</code> - a negligible time difference, easily managed by building more [[assembler]]s.
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| === Exact Approach ===
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| The main question we aim to answer:
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| <blockquote>
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| How many <code>k</code> [[centrifuge]]s are needed during <code>n</code> cycles of [[centrifuge]] production to sustain a [[reactor]] working continuously for <code>T</code> time with probability <code>α</code>?
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| </blockquote>
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| Assume each [[centrifuge]] in each production cycle is represented by a discrete random variable '''''X<sub>ij</sub>''''', where '''''i''''' is the [[centrifuge]] number and '''''j''''' is the production cycle number. If '''''X<sub>ij</sub>''''' produces [[uranium-235]], its value is <code>X<sub>ij</sub>=1</code>. If not, <code>X<sub>ij</sub>=0</code>. Thus, <code> P(X<sub>ij</sub> = 1) = 0.007</code> and <code> P(X<sub>ij</sub> = 0) = 0.993</code> (This is a [[https://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli distribution]]).
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| To find out how long [[centrifuge]]s should produce [[uranium-235]] to sustain a [[reactor]]:
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| From one side, the [[reactor]] has <code>T/t</code> working cycles, needing a total of <code>Tc/t</code> [[uranium-235]] to run continuously for time '''''T'''''. Let's denote this value as '''''u''''': <code>u=Tc/t</code>.
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| From the other side, the [[centrifuge]]s work continuously for '''''n''''' cycles during time '''''T''''': <code>T = n*τ</code>.
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| Combining these, our required quantity of '''''u''''' is <code>u=n*τ*c/t</code> for the entire [[reactor]] operation time '''''T'''''.
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| Consider the simple case where we need '''''n''''' cycles and '''''k''''' [[centrifuge]]s to produce exactly '''''u''''' [[uranium-235]] (technically, we need ceil of '''''u''''', but it won't be practically useful and doesn't detract generality of our calculations). Each '''''X<sub>ij</sub>''''' is a Bernoulli trial, so <code>X<sub>all</sub> = ∑<sub>i</sub><sup>n</sup>∑<sub>j</sub><sup>k</sup>X<sub>ij</sub></code> is the total production, distributed as a [[https://en.wikipedia.org/wiki/Binomial_distribution binomial distribution]] <code>X<sub>all</sub> ~ B(nk, p)</code>. This gives the probability of producing exactly '''''u''''' [[uranium-235]] in '''''n''''' cycles with '''''k''''' [[centrifuge]]s:
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| <blockquote>
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| <code>P(X<sub>all</sub>=u) = C<sub>nk</sub><sup>u</sup> * p<sup>u</sup> * (1-p)<sup>nk-u</sup></code>
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| </blockquote>
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| Practically, we are interested in producing at least '''''u''''' [[uranium-235]], corresponding to <code>P(X<sub>all</sub> ≥ u)</code>. This can be calculated as <code>P(X<sub>all</sub> ≥ u) = 1 - P(X<sub>all</sub> < u)</code>, where <code>P(X<sub>all</sub> < u)</code> is the [[https://en.wikipedia.org/wiki/Binomial_distribution#:~=0.059535.-,Cumulative%20distribution%20function,-%5Bedit%5D cumulative distribution function]]:
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| <blockquote>
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| <code>P(X<sub>all</sub>≥u) ≥ α</code> ⇒ <code>1 - ∑<sub>m</sub><sup>u-1</sup>P(X<sub>all</sub>=m) ≥ α</code> ⇒ {using results from paragraph 1} ⇒ <code>∑<sub>m</sub><sup>n*τ*c/t-1</sup>C<sub>nk</sub><sup>m</sup> * p<sup>m</sup> * (1-p)<sup>nk-m</sup> ≤ 1 - α</code>
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| </blockquote>
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| So, we now have a formula to determine the number of '''''k''''' [[centrifuge]]s needed to sustain a [[reactor]] with consumption '''''c''''' during '''''n''''' cycles. This formula is general and not always practical for long-term play due to the high values of '''''n''''' and '''''u'''''. It is best suited for short-term or speedrun calculations.
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| ==== Example Base Game Short-Term ====
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| To illustrate, let's determine how many [[centrifuge]]s are needed to sustain 1 [[reactor]] with 0.99 confidence without modules. It's a common mistake to think that 0.007 is the production speed, but it represents the probability of producing [[uranium-235]] per cycle. Suppose we want a reactor to work for 2000 [[centrifuge]] production cycles (about 6.7 hours of reactor operation) with 1 [[centrifuge]]:
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| {| class="wikitable"
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| ! n !! tau !! c !! t !! m !! p !! u !! ceil(u-1) !! alpha !! 1-alpha !! !! k !! C(nk, m) !! P(m) !! ∑<sub>m</sub>P
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| |-
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| | 2000 || 12 || 0,1 || 200 || 0 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 1 || 7,92E-07 || 7,92E-07
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| |-
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| | 2000 || 12 || 0,1 || 200 || 1 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 2000 || 1,12E-05 || 1,20E-05
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| |-
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| | 2000 || 12 || 0,1 || 200 || 2 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 2,0E+06 || 7,86E-05 || 9,06E-05
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| |-
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| | 2000 || 12 || 0,1 || 200 || 3 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 1,3E+09 || 0,0004 || 0,0005
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| |-
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| | 2000 || 12 || 0,1 || 200 || 4 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 6,6E+11 || 0,0013 || 0,0018
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| |-
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| | 2000 || 12 || 0,1 || 200 || 5 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 2,7E+14 || 0,0037 || 0,0054
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| |-
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| | 2000 || 12 || 0,1 || 200 || 6 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 8,8E+16 || 0,0086 || 0,0140
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| |-
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| | 2000 || 12 || 0,1 || 200 || 7 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 2,5E+19 || 0,0172 || 0,0312
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| |-
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| | 2000 || 12 || 0,1 || 200 || 8 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 6,3E+21 || 0,0302 || 0,0302
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| |-
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| | 2000 || 12 || 0,1 || 200 || 9 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 1,4E+24 || 0,0472 || 0,0472
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| |-
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| | 2000 || 12 || 0,1 || 200 || 10 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 2,8E+26 || 0,0662 || 0,1133
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| |-
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| | 2000 || 12 || 0,1 || 200 || 11 || 0,007 || 12 || 11 || 0,99 || <span style="color: red">0,01</span> || || 1 || 5,0E+28 || 0,0844 || <span style="color: red">0,1977</span>
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| |-
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| | 2000 || 12 || 0,1 || 200 || 12 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 8,3E+30 || 0,0986 ||
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| |-
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| | 2000 || 12 || 0,1 || 200 || 13 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 1 || 1,3E+33 || 0,1063 ||
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| |}
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| As you can see, we need to produce 12 '''''u''''' for 2000 cycles. According to our formula, we need to sum up all the probabilities of producing less than 12 [[uranium-235]] and compare this with our desired confidence level:
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| <code>∑<sub>m</sub><sup>2000120.1/200-1</sup>C<sub>20001</sub><sup>m</sup> * 0.007<sup>m</sup> * (1-0.007)<sup>20001-m</sup> = 7.92E-07 + 1.12E-05 + 7.86E-05 + 0.0004 + 0.0013 + 0.0037 + 0.0086 + 0.0172 + 0.0302 + 0.0472 + 0.0662 + 0.0844 = 0.1977</code>
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| We can interpret this value as the "Probability of producing less than the required amount of [[uranium-235]] for 2000 [[centrifuge]] cycles." This probability should be less than <code>1-α = 0.01</code>. However, since it is not, we cannot be sure that 1 [[centrifuge]] will produce enough [[uranium-235]] to sustain [[reactor]] operation during 2000 cycles. Therefore, we need to increase the number of [[centrifuge]]s.
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| {| class="wikitable"
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| ! n !! tau !! c !! t !! m !! p !! u !! ceil(u-1) !! alpha !! 1-alpha !! !! k !! C(nk, m) !! P(x=m) !! sum(P(m<=ceil(u-1)))
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| |-
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| | 2000 || 12 || 0,1 || 200 || 0 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 1 || 6,3E-13 || 6,27E-13
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| |-
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| | 2000 || 12 || 0,1 || 200 || 1 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 4000 || 1,8E-11 || 1,83E-11
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| |-
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| | 2000 || 12 || 0,1 || 200 || 2 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 8,0E+06 || 2,5E-10 || 2,67E-10
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| |-
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| | 2000 || 12 || 0,1 || 200 || 3 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 1,1E+10 || 2,3E-09 || 2,61E-09
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| |-
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| | 2000 || 12 || 0,1 || 200 || 4 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 1,1E+13 || 1,6E-08 || 1,91E-08
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| |-
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| | 2000 || 12 || 0,1 || 200 || 5 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 8,5E+15 || 9,3E-08 || 1,12E-07
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| |-
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| | 2000 || 12 || 0,1 || 200 || 6 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 5,7E+18 || 4,4E-07 || 5,48E-07
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| |-
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| | 2000 || 12 || 0,1 || 200 || 7 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 3,2E+21 || 1,8E-06 || 2,30E-06
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| |-
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| | 2000 || 12 || 0,1 || 200 || 8 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 1,6E+24 || 6,2E-06 || 6,17E-06
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| |-
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| | 2000 || 12 || 0,1 || 200 || 9 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 7,2E+26 || 1,9E-05 || 1,93E-05
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| |-
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| | 2000 || 12 || 0,1 || 200 || 10 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 2,9E+29 || 0,0001 || 0,0001
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| |-
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| | 2000 || 12 || 0,1 || 200 || 11 || 0,007 || 12 || 11 || 0,99 || <span style="color: red">0,01</span> || || 2 || 1,0E+32 || 0,0001 || <span style="color: red">0,0002</span>
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| |-
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| | 2000 || 12 || 0,1 || 200 || 12 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 3,4E+34 || 0,0003 ||
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| |-
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| | 2000 || 12 || 0,1 || 200 || 13 || 0,007 || 12 || 11 || 0,99 || 0,01 || || 2 || 1,1E+37 || 0,0007 ||
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| |}
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| And now you can see that our condition is fulfilled. Therefore, we can say that without any [[module]]s or [[Kovarex enrichment process]], we can achieve the stability of 1 [[reactor]] during 2000 [[centrifuge]] cycles using at least 2 [[centrifuge]]s! However, does this imply that we need 4 [[centrifuge]]s for 2 [[reactor]]s, 6 [[centrifuge]]s for 3 [[reactor]]s, and so on? I encourage you to think about this question yourself. You may be surprised by the results when applying the approach outlined above.
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| === Distribution Approximation ===
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| If you plan to play for an extended period while your [[reactor]] setup is operating, we can consider <code>n→∞</code>. In this case, since our [[centrifuge]]s' overall production <code>X<sub>all</sub> ~ B(nk, p)</code>, we can use the [[https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem De Moivre-Laplace theorem]]. In short, it states that if our random variable follows a binomial distribution with a large number of trials, then <code>(X<sub>all</sub>-nkp)/sqrt(nkp(1-p)) ~ N(0, 1)</code>. Thus, to estimate <code>P(X<sub>all</sub> ≥ u)</code>, we simplify it to:
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| <code>P((X<sub>all</sub>-nkp)/sqrt(nkp(1-p))) ≥ (u-nkp)/sqrt(nkp(1-p)) ≥ α</code> ⇒ <code>{1-P<sub>dist</sub> ≥ α}</code> ⇒ <code>(nkp - u)/sqrt(nkp(1-p)) ≥ ɸ<sup>-1</sup>(1-α)</code>, where ɸ<sup>-1</sup> is the inverse standard normal distribution, found using the [[https://en.wikipedia.org/wiki/Standard_normal_table standard normal table]].
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| Let's denote <code>ɸ<sup>-1</sup>(α)</code> as λ. For each confidence level α, we can calculate it from the standard normal table:
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| <code>λ<sub>0.99</sub>=2.3263</code>, <code>λ<sub>0.95</sub>=1.6449</code>, <code>λ<sub>0.9</sub>=1.2816</code>.
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| Now we need to solve the equation above for the variable '''''k''''':
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| <code>(nkp-u)/sqrt(nkp(1-p)) ≥ λ</code> ⇒ <code>(nkp-nτc/t)/sqrt(nkp(1-p)) ≥ λ</code>.
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| After some basic arithmetic operations and solving the quadratic equation, you can determine the value of '''''k''''':
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| <blockquote>
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| <code>k ≥ (χ<sub>1</sub> + sqrt(χ<sub>1</sub><sup>2</sup> + 4χ<sub>0</sub>χ<sub>2</sub>))/(2*χ<sub>0</sub>)</code>
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| where
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| <code>χ<sub>0</sub> = np</code>,
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| <code>χ<sub>1</sub> = λ*sqrt(np(1-p))</code>,
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| <code>χ<sub>2</sub> = nτc/t</code>,
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| <code>λ</code> is the inverse standard normal distribution (popular values: 2.3263; 1.6449; 1.2816 for corresponding α values: 0.99; 0.95; 0.9).
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| </blockquote>
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| This formula can be calculated for very large '''''n'''''. That's why it's best choice for planning average playtime of experienced player.
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| === Infinite Approximation ===
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| If we consider the formula for '''''k''''' [[centrifuge]]s above, we can derive an analytical form for <code>n→∞</code> by substituting all '''''χ''''' into the formula and finding the limit:
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| <blockquote>
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| <code>lim<sub>n→∞</sub>(χ<sub>1</sub> + sqrt(χ<sub>1</sub><sup>2</sup> + 4χ<sub>0</sub>χ<sub>2</sub>))/(2*χ<sub>0</sub>)</code> ⇒ {substitute all '''''χ'''''} ⇒ <code>k ≥ τ*c/(p*t)</code>
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| </blockquote>
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| Thus, to determine the minimal number of [[centrifuge]]s needed to achieve [[reactor]] stability, we use the final formula:
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| <blockquote>
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| <code>k = ⌈τ*c/(p*t)⌉, where ⌈x⌉ is the ceiling function</code>
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| </blockquote>
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| Notice that this formula does not depend on '''''λ'''''. The main reason is that when <code>n</code> is infinitely large, the cumulative probability tends to 1, and at infinity, we can be 100% sure that this number of [[centrifuge]]s will sustain our [[reactor]]'s stability forever!
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| This formula provides the minimal number of [[centrifuge]]s needed to sustain [[reactor]] stability indefinitely. To determine this number, we need only 4 parameters.
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| == History == | | == History == |
