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Economy riddle

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Say you're playing on a map where the mexes all give 2.0 metal/s.

you have 10 metal extractors and 14 solar collectors, your com is alive, the 14 solar collectors are connected in a grid with 2 mexes and 8 mexes are not in grid.

you have 9 metal extractors and 14 solar collectors, your com is alive, with a much more complicated grid. 7 solars are connected in a grid with 2 mexes, 3 solars connect a grid of 2 mexes, 1 solar connects a grid of 2 close mexes, there are 2 cases of a mex connected to a single solar and a single lone mex.

Ignoring any threat of enemies and assuming you are spending all the metal you produce into a detriment assisted by 50 build power(or any number more than your income), and your energy is at max storage, which scenario will produce the detriment the fastest?

+4 / -0

8 months ago
im gonna guess the one with 10 mexes
+1 / -0
8 months ago
Reclaim detri and spam ht
+3 / -0

8 months ago
hey fellas, let's beat up this math dweeb and steal his avatar.

it's not a riddle, it's a test, so ok professor here's my guess:
overdrive is the key here and with the 10 mex scenario you have 8 of them getting nothing.
in the 9 mex scenario you have a more than a few mexes recieving hefty overdrive buffs and a couple with small overdrive buffs. at the very least, 5 of these are outperforming their former scenario counterpart by 20% - meaning at this intersection is where the overdrive is producing more metal than that extra mex would.
it's a very clean implementation diminishing returns

i may run off and test just to see though. good question
+1 / -0
Since I'm mathematically challenged, I'd rather skip that question and ask: why you are building a Detriment when you could build a Paladin in less than half the time and win the game?


Edit: It's a good question, though. I just grid whenever I can and hope my team does the same.
+3 / -0
8 months ago
building that unit is detrimental
+5 / -0
Without excess energy overdrive does not occur. Therefore the Detriment is completed sooner with 10 metal extractors.


On second tought... there is some excess energy available.
In the first case: 14 * 2 + 6 = 34 energy and 20 + 4 metal production -> 10 energy to overdrive overall
In the second case: same amount of energy and 18 + 4 metal -> 12 energy to overdrive

I did not calculate it, but I doubt that +2 energy to overdrive will yield +2 metal income.
+1 / -0
8 months ago
Case 1:
10×2+4=24 M/s (base)
14×2+6=34 E/s
excess E/s=34-24=10 E/s

Need to distribute 10 E/s over 2 mex:
5 E/s per mex.

2×sqrt(5)/4=2×0.56=1.12 mex-equivalents from overdrive

Case 2:
9×2+4=22 M/s (base)
14×2+6=34 E/s
excess E/s=34-22=12

Need to distribute 12 E/s over 7 mex:
2 mexes get 1 E/s each since they share one solar.
5 mexes and 11 energy remain.
2.2 E/s to all remaining mexes, distributed equally because they are all connected to enough energy. (The grid layout does not matter as long as there is enough energy to each mex, since energy sources that are overdrive-inefficient are used first by constructors.)

Total extra mex-equivalents:
Total=2.35 mex-equivalents from overdrive

However, this analysis is not complete! It needs to be iterated to stability with the new metal production values because metal production has increased and there will be less excess energy. The first iteration is probably good enough though, because I am too lazy to do a second iteration.

So on the first iteration, case 1 has about 10+1.12=11.12 mex-equivalents, and case 2 has about 9+2.35=11.35 mex-equivalents. More iterations should change these numbers slightly, but not enough to flip the result I think. So the second case gets more metal due to better distribution of energy, and thus gets a slightly faster detriment.
+6 / -0
OP updated to a poor video demonstration on Titan duel. Briefly explained here.
+0 / -0
cant we just make it simple... the cost of a new mex vs the same cost in over-driving a mex..

so a mex costs 75 and a solar costs 70

rough numbers now...

a mex gains + 4

an overdrive mex gains +20% = + 1 almost

so a new mex is op and like 4 times better then the best overdrive.. build more of them
+1 / -0
Oooh, that was quite an thought provoking exercise USrankSteel_Blue. It gave me a much better perspective on Overdrive and how effective it is (or isn't).

+1 / -0
8 months ago
USrankSteel_Blue: Not a bad trick question. Normally, I'd have personally gone for the 9 mexes with more evenly-distributed energy grid, since it's a bit cheaper than the 10 mexes without a good energy grid. (It's also why I prefer to run Wind Generators from one Mex to another early-game.) However, in order to spend that metal, you'd need at least an equal amount of buildpower to spend it, and at least an equal amount of energy to run all that buildpower. Therefore, we could discount at least 10 of those solar collectors from the inefficient energy grid system, and at least 9 from the more efficient grid system, since they'd most likely be busy powering the Strider Hub that's building the Detriment.

AUrankSmokeDragon: Keep in mind that you need at least as much buildpower as you have metal income, and as much energy income as you have buildpower, in order to spend all of the metal that you earn as soon as you receive it. Therefore, if you want to spend all of the metal that you earn from a Metal Extractor that produces 4 metal per second, you'll need at least 2 Solar Collectors in order to run the 4 buildpower needed to spend that 4 metal before any additional energy is spent overdriving your mexes.
+2 / -0
8 months ago
but what if they are all in the grid
+1 / -0
8 months ago
But what if they are all in a grid

Then my numbers aren't so clean and I've got no advanced formula to tell me where to put the line between ODenergy and spendennergy, you're welcome to come back with some calculus on that though ;)
+0 / -0