Formula for combined efficiency of a Breeder and reactor

  • If you just want to know the formula scroll down if you want to see whats behind it read it all.



    So now some maths on breeder efficiency. Let say we take this design: http://www.talonfiremage.pwp.b…i=1m101010114010101001019 (note: this is the max eff you can get with a breeder)


    Whats the max efficiency boost in percents @6000 and @9000 heat taking into account breeders also produce eu.


    1 uranium ore = 8 Near depleted cells


    @6000 heat
    To get the reactor started we need 1 uranium cell and 4 depleted cells. This cost 1,5 uranium in total. In return you get 4 uranium cells and 2million eu. Do note the fact breeder always nets a 2 million eu for each cycle means lower efficiency reactors will get a higher boost in %.


    Lets say we use this reactor to make eu:
    http://www.talonfiremage.pwp.b…1k101010037ps011111101110
    Specs:
    3 eff = 6 million eu/uranium cell
    4 uranium cells/cycle (ok this does not change the maths but easier example)


    With the breeder you make 4 uranium with 1,5 uranium ore. This means the 1,5 uranium ore made 4*6+2= 26 million eu. This means for each uranium ore you find you get 17,33 million eu. This is a efficiency of 8,67.


    BUT as most here will know uranium cells have 25% chance to turn into a near depleted uranium cell. So the actual efficiency is even higher. Whats the worth of 1 near depleted uranium cell? Its 1/8 of a uranium ore. In this example its 17,33/8 = 2,16625 million eu. Its easy to figure out what the gain is after 1 cycle. But whats the gain after 10 cycles? or 100 cycles, or endless cycles? The chance that a uranium cell turns into a near depleted cell is 25%. The chance this happens again with the same uranium cell is (0.25^2)*100
    = 6,25%. The formula to calculate the chance is 0.25^cycle. After 2 cycles the gain is (0.25^1 + 0.25^2)*100 = 31,25%. Now we want to know what the gain is after x cycles. As you repeat this with more cycles the gain will come closer and closer to 33,33333333%. Thus the gain is 33,33333333%. So this gives us a extra 2,16625/3 = 0,722 million eu or 0.36 more eff making the total eff 9.03.


    In the end the formula for total eff @6000 heat is this: Total eff = ((8*Reactor eff+2)/1,5)*(0.5+(1/48 ))
    If you want eff boost in percents do this (Total eff/Reactor eff)*100



    @9000 heat
    This formula is almost the same but you only need 1 uranium ore to make 4 uranium cells thus the 1,5 in the formula changes to 1. Also the uranium cell used to breed only produces half the eu per breeding cycle (9000 heat means half the time). So the formula is now: Total eff = (8*Reactor eff+1)*(0.5+(1/48 )). In this example it would mean total eff is 13,02.
    If you want eff boost in percents do this (Total eff/Reactor eff)*100


    I hope i didnt do any miscalculations but the answers seem to be realistic.


    Conclusion:
    Total eff@6000 heat = ((8*Reactor eff+2)/1,5)*(0.5+(1/48 ))
    Total eff@9000 heat = (8*Reactor eff+1)*(0,5+1/14)
    Eff boost in percent = (Total eff/Reactor eff)*100

  • I like this. Also cooldown time when switching between breeding and energy production can be taken into account to calculate Effective EU/t for this process.

  • The formula can easily be changed to work with any temperature and any breeding setup. You just need to know how many uranium cells and eu you get per uranium ore from the breeding. Havent really tested the formula so iam not 100% i didnt made some error somewhere in the calculations. I do not have much time to do this since iam rather busy with school lately.

  • I did my own calculations but they disagree with yours:


    One uranium ore makes 8 near-depleted cells, x1.33333 = 10 2/3 near-depleted cells per ore total.
    A good breeder can use one uranium cell to recharge 8. (Let's call this breeder efficiency B = 8).


    But that uranium cell can itself be a recharged near-depleted cell, so if we ignore the one piece of uranium to start with, asymptotically you spend 1/B (as a proportion) of your near depleted cells on breeding, with 2M power production from the breeder as well for each of these cells. That leaves 4B/3 * (1 - 1/B) uranium cells + (4B/3 * 1/B * 2 = 8/3) million EU per uranium ore.
    The highest efficiency reactor you can get is 4.44 (reactor effiency R = 4.44)
    Each of these remaining uranium cells can give you 2R million EU. So from a single uranium ore you get (8BR / 3 * (1 - 1/B) + 8/3) million EU
    simplifying:
    8BR/3 * (1 - 1/B) + 8/3
    = 8BR/3 - 8R/3 + 8/3
    = 8/3 (BR - R + 1)


    = 8/3 (B(R - 1) + 1) million EU per ore.
    With the best possible breeder (B=8) and reactor (R=4.44) that's 76.053 million EU per ore.

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  • huh? 38 eff? Thats seems extremely high. I think you maths could be a bit off. Why the x4/3? How you get over 10 near depleted cells from 1 ore? You get only 8 then that ore is gone.


    EDIT: I see what you done now. You use the fact that 33% of your uranium get turned back into near depleted cells and you apply that to the 8 near depleted cells you get from 1 ore making 10+( 1/3). Then later you apply this 33% again resulting in a double dip which also explains the extremely high eff you had.

  • I see what you done now. You use the fact that 33% of your uranium get turned back into near depleted cells and you apply that to the 8 near depleted cells you get from 1 ore making 10+( 1/3). Then later you apply this 33% again resulting in a double dip which also explains the extremely high eff you had.

    Wait.. Isn't that 33% actually 25% ?

  • Wait.. Isn't that 33% actually 25% ?

    For 1 cycle its 25% but for endless cycles it 33+(1/3)%. Its because there is a 25% chance that a uranium cell turns into a near depleted cell but there is also a chance that this happens again with the same cell.


    Also since you are only getting 8 near depleted cells per ore the absolute maximum efficiency boost (meaning with no uranium wasted during the breeding proces or using a 1 eff reactor) is 8x. It is not possible to go higher than this simply because you only get 8 cells per uranium ore. immibis had like 8,5x efficiency boost.

  • I did indeed multiply 8*4/3 twice, but discarded the result the second time. It wasn't 8*4/3*4/3. Fixed anyway.


    No it does not seem extremely high. Let's do the same calculation assuming the chance of near-depleted cells is 0, and recharging is free, which makes it much simpler. (and those two simplifcations partially cancel each other out)


    1 uranium ore -> 1 refined uranium -> 8 near-depleted cells -> 8 uranium cells


    8 uranium cells in an eff 4.44 reactor = 35.52 eff.


    The absurdly high figure is probably because that was using the most efficient possible breeder and reactor.


    The 10 2/3 is deliberately double-counting 25%, triple-counting 25%^2, quadruple-counting 25%^3, etc. It's the average total number of near depleted cells you will eventually get from one uranium ore piece.

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  • It takes 2 uranium to make and breed 8 near depleted cells to uranium cells. The ratio is 1:4. Its impossible to get 8 uranium cells from just 1 ore. You always need 1 uranium cell to breed. Like i told you 8 eff is the highest you can get with free breeding. You had 8.5x eff so it can never be correct. If that would be true breeding would be op :)


    So lets take a 5 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 40 million eu. So after this cycle have ended you end up with 41 million eu. Resulting in 20,5 efficiency. This is a increase of 4.1x in efficiency


    So lets take a 4 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 32 million eu. So after this cycle have ended you end up with 33 million eu. Resulting in 16,5 efficiency. This is a increase of 4.125x in efficiency.


    So lets take a 3 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 24 million eu. So after this cycle have ended you end up with 25 million eu. Resulting in 12,5 efficiency. This is a increase of 4.16x in efficiency.


    So lets take a 2 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 16 million eu. So after this cycle have ended you end up with 17 million eu. Resulting in 8,5 efficiency. This is a increase of 4.25x in efficiency.


    So lets take a 1 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 8 million eu. So after this cycle have ended you end up with 9 million eu. Resulting in 4,5 efficiency. This is a increase of 4.5x in efficiency.


    There is no way that the fact that every uranium cell has 25% chance to turn into a near depleted cell will raise this efficiency to anywhere near your results. That would mean it has to double the efficiency i just calculated. Its just impossible. I dont know what you exactly did wrong but the answer cannot be correct.

  • 'Simulation' of a 3 efficiency reactor


    1th Cycle:
    So lets take a 3 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 24 million eu. So after this cycle have ended you end up with 25 million eu. Resulting in 12,5 efficiency. This is a increase of 4.16x in efficiency.


    2th Cycle
    1 of those cells turns into a near depleted cell. I takes 0.125 uranium cells to recharge this. This proces makes 0.25 million eu. The recharged cell will make 6 million eu. So thats 6.25 million eu total. Combined with the 1th cycle you now made 30.25 million eu at the cost of 1.125 uranium cells. This is a efficiency of 13.444444 (ok i realise that the near depleted cell part of my formula isnt working correctly). This is a increase of 4.48x in efficiency. Still nowhere near your 8.5x. Also this is a increase of 1.075555556x over 12.5 eff.


    3th Cycle
    0.25 of those cells turn into a near depleted cell. It takes 0.03125 cells to recharge this. This proces makes 0.0625 million eu. The recharged cell will make 1.5 million eu. So thats 1.5625 million eu total. Combined with the 1th and 2th cycle you now made 31.8125 million eu at the cost of 1.15625 uranium cells. This is a efficiency of 13.75675676. This is a 4.59x increase in efficiency. Also this is a increase of 1.100540541x over 12.5 eff


    4th Cycle
    0.0625 of those cells turn into a near depleted cell. It takes 0.0078125 cells to recharge this. This proces makes 0.015625 million eu. The recharged cell will make 0.375 million eu. So thats 0.390625 million eu total. Combined with the 1th and 2th cycle you now made 32.203125 million eu at the cost of 1.1640625 uranium cells. This is a efficiency of 13.83221477. This is a 4.61x increase in efficiency. Also this is a increase of 1.106577181x over 12.5 eff

    Endless cycles:

    After running for endless cycles you will have 33+(1/3) million eu at the cost of 1+(1/6) uranium cells this is a efficiency of 14+(2/7). This is a increase of 4+(16/21) efficiency.
    Also is a increase of 1+(1/7)x over 12,5 efficiency.

  • So, each ore can become 8 near-depleted cells. Each of those, once processed and used, has a 25% chance to become a near-depleted cell. Of those 25%, 1/4 will become a near-depleted cell when used, again. It continues on like this, for a total of 100% + 25% + 6.25% + 1.5625% + 0.390625% (and so on, though I won't bother with them). With just those 5 cycles, you get a total of 1.33203125%, and this continues to approach 1 + 1/3 the longer it goes.


    So, each uranium ore can potentially provide 8 + (8/3) = 10 + 2/3 cells, with processing.


    Now, assuming that one uranium cell can recharge 4 near-depleted cells, it would take (10 + 2/3)/4, or 32/12, or 8/3, or 2 + 2/3 uranium cells to recharge the full output of a single uranium ore into 10 + 2/3 uranium cells. Now, if you were to use 2 + 2/3 of those in order to recharge the next ore's cells, that leaves exactly eight uranium cells that can be used elsewhere.


    So, as long as you aren't stupid, and use recharged cells to recharge later near-depleted cells (rather than making a fresh uranium cell each time), each uranium ore you gather can produce a full 8 cells to be used in whatever non-breeder you choose. As mentioned by people previously, that could be as high as 4.44, giving 35.52 for each uranium ore you mine.


    Oh, and that's *completely* ignoring the energy produced by the breeder reactor.


    If you use a 9k temperature breeder (if I understand correctly halving the recharge time, so a single uranium cell can charge two sets of four near-depleted cells), you get an extra 1 + 1/3 cells output, since you only need half as many.


    So, with the extra output, and then assuming that each uranium cell in the breeder gets at least an efficiency of 1, you get ((9 + 1/3) * 4.44) + ((1 + 1/3) * 1) = 41.44 + 1.3333... = 42.7733 per ore of input.



    Now, the very first time you run the breeder, you'll need to supply a uranium cell to bootstrap the whole process, but since you can use the output of that very first cycle to provide the next uranium cells for later cycles, all other uranium you use can have that nice 8x multiplier.

  • 'Simulation' of a 5 efficiency reactor


    1th Cycle:
    So lets take a 5 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 40 million eu. So after this cycle have ended you end up with 41 million eu. Resulting in 20.5 efficiency. This is a increase of 4.1x in efficiency.


    2th Cycle
    1 of those cells turns into a near depleted cell. I takes 0.125 uranium cells to recharge this. This proces makes 0.25 million eu. The recharged cell will make 10 million eu. So thats 10.25 million eu total. Combined with the 1th cycle you now made 51.25 million eu at the cost of 1.125 uranium cells. This is a efficiency of 22,777778. This is a increase of 4.55556x in efficiency. Still nowhere near your 8.5x. Also this is a increase of 1.1111x over 20.5 eff.


    3th Cycle
    0.25 of those cells turn into a near depleted cell. It takes 0.03125 cells to recharge this. This proces makes 0.0625 million eu. The recharged cell will make 2.5 million eu. So thats 2.5625 million eu total. Combined with the 1th and 2th cycle you now made 53.8125 million eu at the cost of 1.15625 uranium cells. This is a efficiency of 23,27027027. This is a 4.654x increase in efficiency. Also this is a increase of 1.135x over 20,5 eff


    4th Cycle
    0.0625 of those cells turn into a near depleted cell. It takes 0.0078125 cells to recharge this. This proces makes 0.015625 million eu. The recharged cell will make 0.625 million eu. So thats 0.640625 million eu total. Combined with the 1th and 2th cycle you now made 54,453125 million eu at the cost of 1.1640625 uranium cells. This is a efficiency of 23,38926174. This is a 4.678x increase in efficiency. Also this is a increase of 1.140939597x over 20.5 eff

    Endless cycles:

    After running for endless cycles you will have 54+(2/3) million eu at the cost of 1+(1/6) uranium cells this is a efficiency of 23,428. This is a increase of 1+(1/7) over 20.5 efficiency.

    So far:

    The 25% of uranium cells turn into near depleted mechanic (@9000 heat) increase the efficiency by:
    1+(1/7)x for a 3 efficiency reactor
    1+(1/7)x for a 4 efficiency reactor
    1+(1/7)x for a 5 efficiency reactor


    I think its pretty safe to asume that this mechanic increases the output by 1/7 for any eff reactor. So now the formula for a 9000 heat breeder is:
    Total eff@9000 heat = (8*Reactor eff+1)*(0,5+1/14)


    I will update the 6000 heat formula later. Currently marked it red since it isnt completely correct (through its only about 10% off).

  • I see your point. If you reuse the recharged cells you lose less uranium to breeding. But on the other hand getting a almost 10x multiplier over your reactors efficiency sounds op. Still doubting who is correct. Its so easy to overlook something.


    A few simple rules and you need so much maths just to get a eff formula XD

  • A breeder with over 9000 heat will recharge two sets of isotope cells in a cycle. (or one tick off it, which is negligible)

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  • For 1 cycle its 25% but for endless cycles it 33+(1/3)%. Its because there is a 25% chance that a uranium cell turns into a near depleted cell but there is also a chance that this happens again with the same cell.


    Also since you are only getting 8 near depleted cells per ore the absolute maximum efficiency boost (meaning with no uranium wasted during the breeding proces or using a 1 eff reactor) is 8x. It is not possible to go higher than this simply because you only get 8 cells per uranium ore. immibis had like 8,5x efficiency boost.


    Where did you get the 25% rate from? I'm just asking because when I was going through the IC2 source code I have (from MC 1.81, there's a thread on decompiiation somewhere) and the code I see is:



    Now I have close to 0 experience with java, but the documentation I found says the method nextInt(int n) for rand is inclusive for 0 but exclusive for n, i.e. it picks a number from 0 to n-1, in this case 0, 1, or 2. From this I would assume, provided I am not insane, that the probability is 33%?