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== Precise Mathematical Analysis ==
== Precise Mathematical Analysis ==
In some cases you will need to know how many [[centrifuge]]s will you need exactly for each [[reactor]] stability without [[Kovarex enrichment process]]. For this purpose man can derive some formulas for estimation: probability of obtaining the required amount of [[uranium-235]] or even number of needed [[centrifuge]]s.
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.


The main goal of these calculus is to derive formula that can be easily applied and sustainable for predictions despite advancement along [[technologies]], game version updates, or ingame development and modifications.  
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.


=== Variables ===
=== Variables ===
Line 54: Line 54:
* ''<code>&alpha;</code> - desired probability of having success number [[uranium-235]] to sustain all <code>T</code> time of [[reactor]] stable work**''
* ''<code>&alpha;</code> - desired probability of having success number [[uranium-235]] to sustain all <code>T</code> time of [[reactor]] stable work**''


<nowiki>*</nowiki> In seconds or you can measure in in-game [[tick]]s, it does not detracts from the generality of reasoning
<nowiki>*</nowiki> Measured in seconds or in-game [[tick]]s, this does not affect the general reasoning
 
<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]].


<nowiki>**</nowiki> Known as [[https://en.wikipedia.org/wiki/Confidence_interval confidence level]]. Due to probability nature this variable can be only in <code>(0; 1)</code> interval. To preserve scientific papers treatment of significance we will use: 0,9; 0,95; 0,99 values. This value is opposite to [[https://en.wikipedia.org/wiki/Statistical_significance significance level]] for convinience and ease of interpretation.
=== Assumptions ===
=== Assumptions ===
;Assumption 1
;Assumption 1
: All <code>k</code> centrifuges are working without stop and simultaneously during each <code>n</code> cycles and have the same <code>&tau;</code>
: All <code>k</code> centrifuges work continuously and simultaneously for each <code>n</code> cycle with the same <code>τ</code>
;Assumption 2
;Assumption 2
: There are always enough resources for transforming [[uranium-235]] in reactor fuel *
: There are always enough resources to convert [[uranium-235]] into reactor fuel *
;Assumption 3
;Assumption 3
: Transformation of [[uranium-235]] in fuel for reactors can be negleted **
: The conversion of [[uranium-235]] into fuel for reactors can be neglected **
;Assumption 4 (for longrun play)
;Assumption 4 (for long-term play)
: Game can be played infinitely (this is technical assumption, for practical usage it's enough average playtime)
: The game can be played indefinitely (technically assumed for practical use, average playtime suffices)


<nowiki>*</nowiki> For instance, base game [[uranium-238]] has very high probability (<code>0.993</code>) so there are always enough of it for [[uranium fuel cell]]
<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.


<nowiki>**</nowiki> For instance, base game [[uranium-235]] is transforming into 10 [[uranium fuel cell]]s for <code>12 sec</code>, but [[reactor]] cycle is <code>200 sec</code> - that's insignificant and you can always build more [[assembler]]s to sustain production throughput time.
<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.


=== Exact approach ===
=== Exact Approach ===
The main question we're going to answer:
The main question we aim to answer:


<blockquote>  
<blockquote>
How many <code>k</code> [[centrifuge]]s do we need during <code>n</code> cycles of [[centrifuge]] production to sustain [[reactor]] working continuously for <code>T</code> time with probability <code>&alpha;</code>
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>&alpha;</code>?
</blockquote>
</blockquote>
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]]).


Assume, that each [[centrifuge]] in each production cycle is represented by '''''X<sub>ij</sub>''''' [[https://en.wikipedia.org/wiki/Random_variable#:~:text=%5Bedit%5D-,Discrete%20random%20variable,-%5Bedit%5D discrete random variable]], where '''''i''''' is number of [[centrifuge]] and '''''j''''' is number of production cycle.
To find out how long [[centrifuge]]s should produce [[uranium-235]] to sustain a [[reactor]]:
We say, that if '''''X<sub>ij</sub>''''' has produced [[uranium-235]], then it's value <code>X<sub>ij</sub>=1</code>, if '''''X<sub>ij</sub>''''' hasn't, it's value <code>X<sub>ij</sub>=0</code>. That's why, <code> P(X<sub>ij</sub> = m) = 0.007, if m=1</code> and <code> P(X<sub>ij</sub> = m) = 0.993, if m=0</code> (This is so called [[https://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli distribution]]).
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>.
From the other side, the [[centrifuge]]s work continuously for '''''n''''' cycles during time '''''T''''': <code>T = n*τ</code>.


Let's find out how many time [[centrifuge]]s should produce [[uranium-235]] in all cycles to sustain a [[reactor]]:
Combining these, our required quantity of '''''u''''' is <code>u=n*τ*c/t</code> for the entire [[reactor]] operation time '''''T'''''.
<br>
From one side, there are <code>T/t</code> of [[reactor]] working cycles and as a result <code>T*c/t</code> overall [[uranium-235]] quantity need for working continuously all '''''T''''' time. Let's denote this value as '''''u''''', so <code>u=T*c/t</code>.
<br>
From other side, our [[centrifuge]]s during '''''T''''' time will work without stop and simultaneously for '''''n''''' cycles, that means that <code>T = n*&tau;</code>
 
Wrapping all up, we get that our needed quantity of '''''u''''' is <code>u=n*&tau;*c/t</code> for all '''''T''''' time [[reactor]] works.
 
1) Let's consider simple case when we need only for '''''n''''' cycles and '''''k''''' [[centrifuge]]s to produce exactly '''''u''''' amount of [[uranium-235]] (technically, you need ceil <code>&LeftCeiling;u&RightCeiling;</code>, but we will omit this for simplicity of calculations). As we figured out earlie each '''''X<sub>ij</sub>''''' has Bernoulli distribution, but for all '''''T''''' we produce <code>X<sub>all</sub> = &sum;<sub>i</sub><sup>n</sup>&sum;<sub>j</sub><sup>k</sup>X<sub>ij</sub></code> quantity of [[uranium-235]] (this is sum for '''''n''''' cycles and '''''k''''' [[centrifuge]]s). But as all '''''X<sub>ij</sub>''''' are independent Bernoulli trials, so <code>X<sub>all</sub></code> is distributed by definition by [[https://en.wikipedia.org/wiki/Binomial_distribution binomial distribution]] <code>X<sub>all</sub> ~ B(nk, p)</code>. This leads us to probability of producing exactly '''''u''''' [[uranium-235]] for '''''n''''' cycles with '''''k''''' [[centrifuge]]s <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>, where <code>C<sub>nk</sub><sup>u</sup></code> is [[https://en.wikipedia.org/wiki/Binomial_coefficient binomial coefficient]]. As long as we know <code>u=n*&tau;*c/t</code> we can get full probability value of getting exact amount [[uranium-235]] for sustaining [[reactor]]:


Consider the simple case where we need '''''n''''' cycles and '''''k''''' [[centrifuge]]s to produce exactly '''''u''''' [[uranium-235]]. 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:
<blockquote>
<blockquote>
<code>P(X<sub>all</sub>=n*&tau;*c/t) = C<sub>nk</sub><sup>n*&tau;*c/t</sup> * p<sup>n*&tau;*c/t</sup> * (1-p)<sup>nk-n*&tau;*c/t</sup></code>
<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>
</blockquote>
</blockquote>
 
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]]:
2) However, practically there is a probability to get more than '''u''' amount of [[uranium-235]], this corresponds to our purpose of sustaining [[reactor]] working continuously too. The main goal of [[centrifuge]]s production is not producing less than <code>u=n*τ*c/t</code>. That's why we need to calculate the probability of such event won't never occur in future, in other words our probability of always being success: <code>P(X<sub>all</sub> ≥ u)</code>. On the other side, <code>P(X<sub>all</sub> ≥ u) = 1 - P(X<sub>all</sub> < u)</code>, but <code>P(X<sub>all</sub> < u)</code> is [[https://en.wikipedia.org/wiki/Binomial_distribution#:~:text=0.059535.-,Cumulative%20distribution%20function,-%5Bedit%5D Cumulative distribution function]] and has exact formula for computation -- as sum of all probabilities for each <code>m < u</code> or <code>P(Xall < u) = &sum;<sub>i</sub><sup>u-1</sup>P(X<sub>all</sub>=m)</code>, '''m''' is integer. The most important part is that we would like the probability <code>P(X<sub>all</sub> ≥ u)</code> to be high enough, that's why we need to set '''''&alpha;''''' variable (as for nature of probability it's impossible for <code>P(X<sub>all</sub> ≥ u)</code> to equal exactly '''''1'''''). That's how we can derive equation for variables selection:
 
<blockquote>
<blockquote>
<code>P(X<sub>all</sub>≥u) ≥ &alpha;</code> ⇒ <code>1 - &sum;<sub>m</sub><sup>u-1</sup>P(X<sub>all</sub>=m) ≥ &alpha;</code> ⇒ {using results of previous paragraph 1} ⇒ <code>&sum;<sub>m</sub><sup>n*&tau;*c/t-1</sup>C<sub>nk</sub><sup>m</sup> * p<sup>m</sup> * (1-p)<sup>nk-m</sup> ≤ 1 - &alpha;</code>
<code>P(X<sub>all</sub>≥u) ≥ &alpha;</code> ⇒ <code>1 - &sum;<sub>m</sub><sup>u-1</sup>P(X<sub>all</sub>=m) ≥ &alpha;</code> ⇒ {using results from paragraph 1} ⇒ <code>&sum;<sub>m</sub><sup>n*&tau;*c/t-1</sup>C<sub>nk</sub><sup>m</sup> * p<sup>m</sup> * (1-p)<sup>nk-m</sup> ≤ 1 - &alpha;</code>
</blockquote>
</blockquote>
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.


So now we have formula (technically speaking, hypothesis test verification) to figure out how many '''''k''''' [[centrifuge]]s we need to sustain [[reactor]] with consumption '''''c''''' during '''''n''''' cycles of [[centrifuge]]s production.
==== Example Base Game Short-Term ====
This is the most general formula and not very comfort to use. There is no pure analitics form of sum, that's why we need to calculate each probability in some separetly. Moreover, the main disadvantage is in using high value of '''''n''''' and '''''u'''''. It is impossible due to binomial coefficiant will become an enormous number! So it can not be used to predict figures for longrun playtime. We will consider this in future calculations. However, this formula will be extremely useful if you are considering playing game speedrunning and prepare some blueprints.
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]]:
 
==== Example base game shortrun ====
Assume, you would like to find out how many [[centrifuge]]s you need to fully work to sustain 1 [[reactor]] with 0.99 confidence without any modules. Many are mistaken in confusing that 0.007 is speed of production, but that's untrue, as there is always a probability that even during whole game [[centrifuge]] won't produce any [[uranium-235]] (this probability is of course drastically small). Let's find out, for example, that we would like keep working reactor for 2000 [[centrifuge]]'s production cycles (neary ''2000*12/3600 = 6.7'' hours of [[reactor]] working) if we set 1 [[centrifuge]]:
 


{| class="wikitable"
{| class="wikitable"
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|}
|}


As you can see we need to produce 12 '''''u''''' for 2000 cycles. So due to our formula we need sum up all probabilities of production less than 12 [[uranium-235]] and compare with our desired confidence level:
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:
 
<code>&sum;<sub>m</sub><sup>2000*12*0.1/200-1</sup>C<sub>2000*1</sub><sup>m</sup> * 0.007<sup>m</sup> * (1-0.007)<sup>2000*1-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>
<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>


We can interpret this value as "Probability to produce less than needed [[uranium-235]] for 2000 [[centrifuge]]s cycles". And we know that this probability should be less than <code>1-&alpha; = 0.01</code>. However, this is wrong, so we can not be sure that 1 [[centrifuge]] will produce enough [[uranium-235]] to sustain [[reactor]] work during 2000 cycles. That's why we need to increase [[centrifuge]] quantity:
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.


{| class="wikitable"
{| class="wikitable"
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|}
|}


And now you can see that our condition is fulfilled. So now we can say, that without any [[module]]s, [[kovarex enrichment process]] we can achieve stability of 1 [[reactor]] during 2000 [[centrifuge]]s cycles using at least 2 [[centrifuge]]s! However, does this mean that we need 4 [[centrifuge]]s for 2 [[reactor]]s, 6 [[centrifuge]]s for 3 [[reactor]]s and etc? I suggest you think about this question yourself, I'm sure you will be very surprised by results, applying the approach outlined above.
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.


=== Distribution approximation ===
=== Distribution Approximation ===
If you are going to play long enough while your [[reactor]]s setup is working, this will mean that we can pretend as <code>n→∞</code>. In this case as long as we know that our [[centrifuge]]s overall production <code>X<sub>all</sub> ~ B(nk, p)</code>, we can use [[https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem De Moiver-Laplace theorem]]. In short, it says that if our random variable has binomial distribution and we have large trials, then <code>(X<sub>all</sub>-nkp)/(sqrt(nkp(1-p)) ~ N(0, 1)</code>. That's why if we would like to estimate <code>P(X<sub>all</sub> ≥ u)</code> we should reduce it to this expression:
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:


<code>P((X<sub>all</sub>-nkp)/(sqrt(nkp(1-p)))) ≥ (u-nkp)/(sqrt(nkp(1-p))) ≥ &alpha;</code> ⇒ <code>{1-P<sub>dist</sub> ≥ &alpha;}</code> ⇒ <code>(nkp - u)/(sqrt(nkp(1-p)) ≥ ɸ<sup>-1</sup>(1-&alpha;)</code>, where ɸ<sup>-1</sup> is inverse standart distribution and can be find by [[https://en.wikipedia.org/wiki/Standard_normal_table standart normal table]]
<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]].


Let's denote <code>ɸ<sup>-1</sup>(&alpha;)</code> as &lambda;. For each confidencle level &alpha; we can calculate it from standart normal table:  
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:
<code>&lambda;<sub>0.99</sub>=2.3263</code>, <code>&lambda;<sub>0.95</sub>=1.6449</code>, <code>&lambda;<sub>0.9</sub>=1.2816</code>
<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>.


Now we need to solve the equation above on a variable '''''k''''':
Now we need to solve the equation above for the variable '''''k''''':


<code>(nkp-u)/(sqrt(nkp(1-p)) ≥ &lambda;</code> ⇒ <code>(nkp-nτc/t)/(sqrt(nkp(1-p)) ≥ &lambda;</code>
<code>(nkp-u)/sqrt(nkp(1-p)) ≥ λ</code> ⇒ <code>(nkp-nτc/t)/sqrt(nkp(1-p)) ≥ λ</code>.


After some basic arithmetic operations and solution to quadratic equation you can get '''''k''''' value:
After some basic arithmetic operations and solving the quadratic equation, you can determine the value of '''''k''''':


<blockquote>
<blockquote>
Line 200: Line 191:
<code>χ<sub>1</sub> = &lambda;*sqrt(np(1-p))</code>,
<code>χ<sub>1</sub> = &lambda;*sqrt(np(1-p))</code>,
<code>χ<sub>2</sub> = n&tau;c/t</code>,
<code>χ<sub>2</sub> = n&tau;c/t</code>,
<code>&lambda;</code> is inverse standart normal distribution (popular values: 2.3263; 1.6449; 1.2816 for corresponding *alpha;: 0.99; 0.95; 0.9)
<code>&lambda;</code> is the inverse standard normal distribution (popular values: 2.3263; 1.6449; 1.2816 for corresponding &alpha; values: 0.99; 0.95; 0.9).
</blockquote>
</blockquote>


Revision as of 15:54, 30 May 2024

Uranium processing.png
 Uranium processing

Recipe

Time.png
12
+
Uranium ore.png
10
→
Uranium-235.png
0.007
+
Uranium-238.png
0.993

Total raw

Time.png
12
+
Uranium ore.png
10

Prototype type

recipe

Internal name

uranium-processing

Required technologies

Uranium processing (research).png

Produced by

Centrifuge.png

Uranium processing is the only way to use uranium ore and the first source of uranium-235 and uranium-238 that is available to the player. The process has a 99.3% chance to produce 1 uranium-238 and a 0.7% chance to produce 1 uranium-235.

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.

Expected value

As U-235 is created in centrifuges running uranium processing with p = 0.007, the expected number of processing cycles to gain a single unit is E(1,p) = 1 ÷ p = ~143. The expected number of cycles to gain 40 units is then E(40,p) = 40 × E(1,p) = ~5,714 cycles. Note that running this many cycles requires mining 57,140 uranium ore, a non-trivial task.

However, given the nature of probability, it is impossible to guarantee that 40 units will be reached in this number of cycles; as a matter of fact, some mathematics yields that the probability that 5,714 cycles will produce at least 40 units of U-235 is only about 52%.

Confidence levels

Probability of reaching 40 U-235 from uranium processing by number of cycles (click to enlarge)

Some further mathematics enables one to calculate the number of uranium processing cycles one must run to obtain at least 40 units of U-235 (given p = 0.007) with a given level of confidence (probability of achieving the set goal).

As discussed in the previous section, the level of confidence corresponding to 5,714 cycles (for which the expected value of U-235 gained is 40) happens to be about 52% (or, in other words, with this number of cycles, one can expect to fail to reach 40 U-235 a bit less than once out of every 2 tries).

Some further example confidence levels that may be of interest to players are given in the table below. Consult the chart to the right for custom values.

Confidence level Failures Cycles required
10% 9 out of 10 4,595
50% 1 out of 2 5,667
90% 1 out of 10 6,894
95% 1 out of 20 7,272
99% 1 out of 100 8,015

Precise Mathematical Analysis

To determine the exact number of centrifuges 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 centrifuges needed.

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.

Variables

Let's introduce some variables that we'll need in further calculations

  • T - overall time of a reactor working (in seconds*)
  • t - time of a reactor cycle (in seconds*)
  • c - consumption of uranium-235 per reactor cycle
  • k - centrifuges number (our target value for calculations)
  • p - probability of uranium-235 production per centrifuge cycle
  • n - overall number of centrifuge cycles
  • τ - centrifuge time per cycle
  • α - desired probability of having success number uranium-235 to sustain all T time of reactor stable work**

* Measured in seconds or in-game ticks, this does not affect the general reasoning

** Known as the [confidence level]. Due to the nature of probability, this value can only be in the interval (0; 1). We will use the values 0.9, 0.95, and 0.99 to maintain scientific rigor and ease of interpretation, opposite to the [significance level].

Assumptions

Assumption 1
All k centrifuges work continuously and simultaneously for each n cycle with the same τ
Assumption 2
There are always enough resources to convert uranium-235 into reactor fuel *
Assumption 3
The conversion of uranium-235 into fuel for reactors can be neglected **
Assumption 4 (for long-term play)
The game can be played indefinitely (technically assumed for practical use, average playtime suffices)

* For instance, base game uranium-238 has a very high probability (0.993), ensuring a sufficient supply for uranium fuel cell production.

** For instance, converting base game uranium-235 into 10 uranium fuel cells takes 12 sec, but the reactor cycle is 200 sec - a negligible time difference, easily managed by building more assemblers.

Exact Approach

The main question we aim to answer:

How many k centrifuges are needed during n cycles of centrifuge production to sustain a reactor working continuously for T time with probability α?

Assume each centrifuge in each production cycle is represented by a discrete random variable Xij, where i is the centrifuge number and j is the production cycle number. If Xij produces uranium-235, its value is Xij=1. If not, Xij=0. Thus, P(Xij = 1) = 0.007 and P(Xij = 0) = 0.993 (This is a [Bernoulli distribution]).

To find out how long centrifuges should produce uranium-235 to sustain a reactor: From one side, the reactor has T/t working cycles, needing a total of Tc/t uranium-235 to run continuously for time T. Let's denote this value as u: u=Tc/t. From the other side, the centrifuges work continuously for n cycles during time T: T = n*τ.

Combining these, our required quantity of u is u=n*τ*c/t for the entire reactor operation time T.

Consider the simple case where we need n cycles and k centrifuges to produce exactly u uranium-235. Each Xij is a Bernoulli trial, so Xall = ∑in∑jkXij is the total production, distributed as a [binomial distribution] Xall ~ B(nk, p). This gives the probability of producing exactly u uranium-235 in n cycles with k centrifuges:

P(Xall=u) = Cnku * pu * (1-p)nk-u

Practically, we are interested in producing at least u uranium-235, corresponding to P(Xall ≥ u). This can be calculated as P(Xall ≥ u) = 1 - P(Xall < u), where P(Xall < u) is the [cumulative distribution function]:

P(Xall≥u) ≥ α ⇒ 1 - ∑mu-1P(Xall=m) ≥ α ⇒ {using results from paragraph 1} ⇒ ∑mn*τ*c/t-1Cnkm * pm * (1-p)nk-m ≤ 1 - α

So, we now have a formula to determine the number of k centrifuges 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.

Example Base Game Short-Term

To illustrate, let's determine how many centrifuges 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:

n tau c t m p u ceil(u-1) alpha 1-alpha k C(nk, m) P(m) ∑mP
2000 12 0,1 200 0 0,007 12 11 0,99 0,01 1 1 7,92E-07 7,92E-07
2000 12 0,1 200 1 0,007 12 11 0,99 0,01 1 2000 1,12E-05 1,20E-05
2000 12 0,1 200 2 0,007 12 11 0,99 0,01 1 2,0E+06 7,86E-05 9,06E-05
2000 12 0,1 200 3 0,007 12 11 0,99 0,01 1 1,3E+09 0,0004 0,0005
2000 12 0,1 200 4 0,007 12 11 0,99 0,01 1 6,6E+11 0,0013 0,0018
2000 12 0,1 200 5 0,007 12 11 0,99 0,01 1 2,7E+14 0,0037 0,0054
2000 12 0,1 200 6 0,007 12 11 0,99 0,01 1 8,8E+16 0,0086 0,0140
2000 12 0,1 200 7 0,007 12 11 0,99 0,01 1 2,5E+19 0,0172 0,0312
2000 12 0,1 200 8 0,007 12 11 0,99 0,01 1 6,3E+21 0,0302 0,0302
2000 12 0,1 200 9 0,007 12 11 0,99 0,01 1 1,4E+24 0,0472 0,0472
2000 12 0,1 200 10 0,007 12 11 0,99 0,01 1 2,8E+26 0,0662 0,1133
2000 12 0,1 200 11 0,007 12 11 0,99 0,01 1 5,0E+28 0,0844 0,1977
2000 12 0,1 200 12 0,007 12 11 0,99 0,01 1 8,3E+30 0,0986
2000 12 0,1 200 13 0,007 12 11 0,99 0,01 1 1,3E+33 0,1063

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:

∑m2000120.1/200-1C20001m * 0.007m * (1-0.007)20001-m = 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

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 1-α = 0.01. 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 centrifuges.

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)))
2000 12 0,1 200 0 0,007 12 11 0,99 0,01 2 1 6,3E-13 6,27E-13
2000 12 0,1 200 1 0,007 12 11 0,99 0,01 2 4000 1,8E-11 1,83E-11
2000 12 0,1 200 2 0,007 12 11 0,99 0,01 2 8,0E+06 2,5E-10 2,67E-10
2000 12 0,1 200 3 0,007 12 11 0,99 0,01 2 1,1E+10 2,3E-09 2,61E-09
2000 12 0,1 200 4 0,007 12 11 0,99 0,01 2 1,1E+13 1,6E-08 1,91E-08
2000 12 0,1 200 5 0,007 12 11 0,99 0,01 2 8,5E+15 9,3E-08 1,12E-07
2000 12 0,1 200 6 0,007 12 11 0,99 0,01 2 5,7E+18 4,4E-07 5,48E-07
2000 12 0,1 200 7 0,007 12 11 0,99 0,01 2 3,2E+21 1,8E-06 2,30E-06
2000 12 0,1 200 8 0,007 12 11 0,99 0,01 2 1,6E+24 6,2E-06 6,17E-06
2000 12 0,1 200 9 0,007 12 11 0,99 0,01 2 7,2E+26 1,9E-05 1,93E-05
2000 12 0,1 200 10 0,007 12 11 0,99 0,01 2 2,9E+29 0,0001 0,0001
2000 12 0,1 200 11 0,007 12 11 0,99 0,01 2 1,0E+32 0,0001 0,0002
2000 12 0,1 200 12 0,007 12 11 0,99 0,01 2 3,4E+34 0,0003
2000 12 0,1 200 13 0,007 12 11 0,99 0,01 2 1,1E+37 0,0007

And now you can see that our condition is fulfilled. Therefore, we can say that without any modules or Kovarex enrichment process, we can achieve the stability of 1 reactor during 2000 centrifuge cycles using at least 2 centrifuges! However, does this imply that we need 4 centrifuges for 2 reactors, 6 centrifuges for 3 reactors, 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.

Distribution Approximation

If you plan to play for an extended period while your reactor setup is operating, we can consider n→∞. In this case, since our centrifuges' overall production Xall ~ B(nk, p), we can use the [De Moivre-Laplace theorem]. In short, it states that if our random variable follows a binomial distribution with a large number of trials, then (Xall-nkp)/sqrt(nkp(1-p)) ~ N(0, 1). Thus, to estimate P(Xall ≥ u), we simplify it to:

P((Xall-nkp)/sqrt(nkp(1-p))) ≥ (u-nkp)/sqrt(nkp(1-p)) ≥ α ⇒ {1-Pdist ≥ α} ⇒ (nkp - u)/sqrt(nkp(1-p)) ≥ ɸ-1(1-α), where ɸ-1 is the inverse standard normal distribution, found using the [standard normal table].

Let's denote ɸ-1(α) as λ. For each confidence level α, we can calculate it from the standard normal table: λ0.99=2.3263, λ0.95=1.6449, λ0.9=1.2816.

Now we need to solve the equation above for the variable k:

(nkp-u)/sqrt(nkp(1-p)) ≥ λ ⇒ (nkp-nτc/t)/sqrt(nkp(1-p)) ≥ λ.

After some basic arithmetic operations and solving the quadratic equation, you can determine the value of k:

k ≥ (χ1 + sqrt(χ12 + 4χ0χ2))/(2*χ0) where χ0 = np, χ1 = λ*sqrt(np(1-p)), χ2 = nτc/t, λ is the inverse standard normal distribution (popular values: 2.3263; 1.6449; 1.2816 for corresponding α values: 0.99; 0.95; 0.9).

History

See also

Intermediate products

Resources

Materials

Crafting components

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