AI wiki: Energy Efficiency: Error

[Ranakastrasz]

Going to change the chart, to show how it actually works. its quite obvious when you look at it how you calculated the values, and the fact that it is x% less than the previous amount, rather than -x% of the original amount is not obvious. Even more funny is that you put twice the efficiency than would exist if it worked like this. the T1 reactor uses 4 resources (2 metal and 2 crystal) and you are working under the assumption that it is a linear drop, when it is a quadratic drop.

Resource Ratio: Energy/Resource         
# of reactors   Mark 1   Mark 2   Mark 3
                  1   1250.0  1333.3   1000.0
                  2   1000.0    800.0     200.0
                  3     800.0    480.0       40.0
                  4     640.0    288.0   
                  5     512.0    172.8
                  6     409.6    103.7
                  7     327.7      62.2
                  8     262.1
                  9     209.7
                10     167.7
                11     134.2
                12     107.4
                13       85.9
Energy
Produced   :        5000   40000   80000
Resources
Used:                  4         30       80
Eff Reduction:      0.2      0.4      0.8

[Nalgas]

Going to change the chart, to show how it actually works.

And the moral of the story is that mk3 energy reactors are absolutely terrifying.  I've actually built a second one at a couple planets in the same game in an emergency instead of just stacking a dozen mk1s and mk2s back in the day before realizing just how sharp the dropoff is compared to the others, but ever since learning that I almost feel like the AI deserves a special voice taunt to mock you for hosing your economy to that degree.  Heh.

[Ranakastrasz]

Going to change the chart, to show how it actually works.

And the moral of the story is that mk3 energy reactors are absolutely terrifying.  I've actually built a second one at a couple planets in the same game in an emergency instead of just stacking a dozen mk1s and mk2s back in the day before realizing just how sharp the dropoff is compared to the others, but ever since learning that I almost feel like the AI deserves a special voice taunt to mock you for hosing your economy to that degree.  Heh.

Note however that the loss for making enough Mark1 or 2 reactors to match that of two Mark 3 reactors is compatible to the loss of 80%.

I probably ought to calculate how many Mark 1 and 2 reactors it takes to match 1 or 2 Mark 3 reactors.

[Nalgas]

Note however that the loss for making enough Mark1 or 2 reactors to match that of two Mark 3 reactors is compatible to the loss of 80%.

Sure, but you don't have to match two of them in practice, because if you only use one running at full efficiency, you don't have to worry about the second, absurdly inefficient one.  If you already have one each of mk1/mk2/mk3 plus a second each of mk1/mk2, a third each of mk1/mk2 will produce more power than a second mk3 would for less than half the operating cost.  You basically never, ever want to have low efficiency mk3 reactors for any reason.  They should come with a "build this and lose the game" warning label on them or something if you try to build a second one in a system that already has one.

[Ranakastrasz]

Note however that the loss for making enough Mark1 or 2 reactors to match that of two Mark 3 reactors is compatible to the loss of 80%.

Sure, but you don't have to match two of them in practice, because if you only use one running at full efficiency, you don't have to worry about the second, absurdly inefficient one.  If you already have one each of mk1/mk2/mk3 plus a second each of mk1/mk2, a third each of mk1/mk2 will produce more power than a second mk3 would for less than half the operating cost.  You basically never, ever want to have low efficiency mk3 reactors for any reason.  They should come with a "build this and lose the game" warning label on them or something if you try to build a second one in a system that already has one.

True, you would only have to build *That many* reactors if you were, say, turtling up on 1-2 planets for some reason.

[Nalgas]

Note however that the loss for making enough Mark1 or 2 reactors to match that of two Mark 3 reactors is compatible to the loss of 80%.

Sure, but you don't have to match two of them in practice, because if you only use one running at full efficiency, you don't have to worry about the second, absurdly inefficient one.  If you already have one each of mk1/mk2/mk3 plus a second each of mk1/mk2, a third each of mk1/mk2 will produce more power than a second mk3 would for less than half the operating cost.  You basically never, ever want to have low efficiency mk3 reactors for any reason.  They should come with a "build this and lose the game" warning label on them or something if you try to build a second one in a system that already has one.

True, you would only have to build *That many* reactors if you were, say, turtling up on 1-2 planets for some reason.

I was going to say that I wasn't sure where you'd even get enough stuff to use up that much energy with that few planets/that little knowledge to unlock things, but then someone would go hide behind a wall of fortresses and viral shredders just to prove a point.  Heh.

[keith.lamothe]

Sorry, didn't see this discussion.  I don't want y'all to have to guess at the math

The efficiency malus is applied multiplicatively.  For, example, the actual output of a mkI is calculated basically like this:

int baseProduction = 5000;
int actualProduction = baseProduction;
for ( int i = 1; i < numberOfReactorsOfTheSameMarkAndPlayerAndPlanet; i++ )
      actualProduction = ( actualProduction * 8 ) / 10;

if( actualProduction < (baseProduction / 10) )
      actualProduction = baseProduction / 10;

(for mkII, replace the 8 with a 6; for mkIII replace the 8 with a 2; they also start with different baseProduction, of course)

[Ranakastrasz]

Sorry, didn't see this discussion.  I don't want y'all to have to guess at the math

The efficiency malus is applied multiplicatively.  For, example, the actual output of a mkI is calculated basically like this:

int baseProduction = 5000;
int actualProduction = baseProduction;
for ( int i = 1; i < numberOfReactorsOfTheSameMarkAndPlayerAndPlanet; i++ )
      actualProduction = ( actualProduction * 8 ) / 10;

if( actualProduction < (baseProduction / 10) )
      actualProduction = baseProduction / 10;

(for mkII, replace the 8 with a 6; for mkIII replace the 8 with a 2; they also start with different baseProduction, of course)

So they cant get any worse than 1/10th of the normal production eh? interesting. So at least if you are already taking 2 mark 3 reactors, then you can add a third one at slightly less efficiency, and then the rest will have the same efficiency as the third. Ofc 80 resources for 8k energy is quite steep, but still.

[keith.lamothe]

So they cant get any worse than 1/10th of the normal production eh? interesting. So at least if you are already taking 2 mark 3 reactors, then you can add a third one at slightly less efficiency, and then the rest will have the same efficiency as the third. Ofc 80 resources for 8k energy is quite steep, but still.

Yea, basically if you're willing to pay 1 (m+c)/s for every 100 energy you can have as much of it as you want   But it's somewhat more efficient to do it with mkIIs (which give a minimum efficiency of 133.3~ per 1(m+c)/s ; mkIs give 125).
But the words "brutally expensive" come to mind.

[Ranakastrasz]

So they cant get any worse than 1/10th of the normal production eh? interesting. So at least if you are already taking 2 mark 3 reactors, then you can add a third one at slightly less efficiency, and then the rest will have the same efficiency as the third. Ofc 80 resources for 8k energy is quite steep, but still.

Yea, basically if you're willing to pay 1 (m+c)/s for every 100 energy you can have as much of it as you want   But it's somewhat more efficient to do it with mkIIs (which give a minimum efficiency of 133.3~ per 1(m+c)/s ; mkIs give 125).
But the words "brutally expensive" come to mind.

Let us build more Zenith power generators plz.

[BobTheJanitor]

So the correct build efficiency order is still 2, 1, 1, 2, 3?

[keith.lamothe]

Ok, after a bit of excellery, I think these numbers are correct for the order in which to build reactors to get max efficiency (preferring larger output in case of efficiency ties) :

Mark	Number	Actual	Cost	Efficiency
2	1	40000	30	1333.333333
1	1	5000	4	1250
3	1	80000	80	1000
1	2	4000	4	1000
2	2	24000	30	800
1	3	3200	4	800
1	4	2560	4	640
1	5	2048	4	512
2	3	14400	30	480
1	6	1638.4	4	409.6
1	7	1310.72	4	327.68
2	4	8640	30	288
1	8	1048.576	4	262.144
1	9	838.8608	4	209.7152
3	2	16000	80	200
2	5	5184	30	172.8
1	10	671.08864	4	167.77216
1	11	536.870912	4	134.217728
2	6	4000	30	133.3333333

"Number" being the order of processing within that mark.

If you really need more than 19 reactors on a single planet and you've already built all those, Mk2s are the most efficient at 133.3~.

Anyway, there are actually 2 (theoretically 4) equally "optimal" orders  for the first 5 reactors:

If you prefer the highest output first, then:
2, 1, 3, 1, 2

If you prefer the lowest output first (and thus lowest overall resource cost), then:
2, 1, 1, 3, 1

[Mayjori]

not to belittle anyone, but I'm curious, were either Ranak or Blasphemy right? (been to long since i did math and that stuf is giving me a headache trying to figure it out )

Anyway, there are actually 2 (theoretically 4) equally "optimal" orders  for the first 5 reactors:

If you prefer the highest output first, then:
2, 1, 3, 1, 2

If you prefer the lowest output first (and thus lowest overall resource cost), then:
2, 1, 1, 3, 1

those are after you build the initial 1 of each? or include??

[Ranakastrasz]

I was closer, But I didn't know about the 10% Cap.

[Mayjori]

I was closer, But I didn't know about the 10% Cap.

so you did all the math right, except for not taking into account the floor?
See my edit above if you missed it