
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?
Solution: [Spoiler] It's a trick question!
9 mxes 14 solars vs 10 mexes 14 solars.
the grids I gave in the post are very confusing, but what it amounts to is in the 9 mex sinario 8 mexes have access to 1 energy in it's grid, and in the 10 mex scenario 2 mexes have access to 4 energy in their grids.
But wait! what you're saying is that both scenarios have only 8 excess energy, but when you look at the math it's actually 10 excess energy and 12 excess energy! The answer to that is confusing, isn't it? if all the excess energy were converted to metal, you would not have any energy to SPEND that metal. what this means is that the excess energy is separated into 2 sections, the energy spent to overdrive into metal, and the energy saved to spend that metal. And these values end up equivalent, where the metal value of the energy for overdrive equals the energy value left over.
Here's a good time to remind you the formula for overdrive. overdrive% = (sqrt(energy into mex)/4)*100% that's a lot to unpack, so here are some simpler values to remember: 1e =25% overdrive 4e = 50% overdrive 9e = 75% overdrive 16e = 100% overdrive 64e = 200% overdrive 256e = 300% overdrive having the first 2 key values is important to us. so let's go back to the problem.
in case 9 mexes, you have 12 excess energy. take my word for it, split that into 8 overdrive energy and 4 spend energy. The grids are confusing, but the energy can be evenly distributed to all 8 mexes at 1e/mex, and that translates to 25% of 2 or 0.5 m/e. 8*0.5=4, equal to our spend energy, so 12 excess energy converts to 4 spendable metal in overdrive.
Note that the 2 caretakers and strider hub produce .9 metal and energy.
in case 10 mexes, you have 10 excess energy. only 2 mexes are in grid, so we can only overdrive those 2 mexes. split the energy into 8 overdrive energy and 2 spend energy. the 8 OD energy goes 4e per mex. that's 50% overdrive on a 2.0 mex, or 0.25m/e. .25*8= 2 metal, equal to our spend energy. In this set up, the 10 excess energy converted into 2 usable metal in overdrive.
so, calculating the sum total of everything: 9 mex scenario, you have 22 metal base and 34 energy, excess energy converts into 4 metal for 26 total metal. 10 mex scenario, you have 24 metal base and 34 energy, excess energy converts into 2 metal for 26 total metal.
So the answer is that it's a trick question, production is the same and the detriments come out at the same time! silly lobsters, building detriments. Why don't you build a paladin in half the time and win the game?
+4 / 0



im gonna guess the one with 10 mexes
+1 / 0



Reclaim detri and spam ht
+3 / 0



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?
:p
Edit: It's a good question, though. I just grid whenever I can and hope my team does the same.
+3 / 0



building that unit is detrimental
+5 / 0



Without excess energy overdrive does not occur. Therefore the Detriment is completed sooner with 10 metal extractors.
Edit:
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



Case 1: 10×2+4=24 M/s (base) 14×2+6=34 E/s excess E/s=3424=10 E/s
Need to distribute 10 E/s over 2 mex: 5 E/s per mex.
Total: 2×sqrt(5)/4=2×0.56=1.12 mexequivalents from overdrive
Case 2: 9×2+4=22 M/s (base) 14×2+6=34 E/s excess E/s=3422=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 overdriveinefficient are used first by constructors.)
Total extra mexequivalents: 2×sqrt(1)/4=2×0.25=0.5 5×sqrt(2.2)/4=5×0.37=1.85 Total=2.35 mexequivalents 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 mexequivalents, and case 2 has about 9+2.35=11.35 mexequivalents. 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. [Spoiler] It's a trick question! Haha, got you! with careful overdrive grids, the 9 mex scenario makes up that 2 m/s!
cathartes and Rhade did an attempt at solving it in this thread, good on you! But, cathartes overlooked a key problem, if you overdrive all the excess energy to metal, you have no energy to SPEND that metal. Going over the scenarios again but only overdriving 8 energy each, you get much cleaner numbers. In 10 mex scenario, 8e= 4e/mex * 2mex then 2(4e= 50%OD)= 2 * 1= 2m, and you can spend that 2m with the left over 2 energy. In the 9 mex scenario, split 12 excess e into 8 and 4, 8e= 1e/mex *8mex then we get 8(1e= 25%OD)= 8 * .5= 4 metal. the 4 metal can be spent with that 4 energy I reserved from earlier.the 9 mex scenario generates an extra 2m/s, gapping the bridge between the mex difference generating the same income.
video demonstration here for completeness. you need to pause to see everything because I move kind of fast and have no mic and don't edit my videos, sorry.
+0 / 0



cant we just make it simple... the cost of a new mex vs the same cost in overdriving 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 Steel_Blue. It gave me a much better perspective on Overdrive and how effective it is (or isn't).
[Spoiler]The two situations will take exactly the same amount of time to make the Detriment and in fact both give 22em/s. One producing 2m/s overdrive and the other producing 4m/s overdrive.
EDIT: Checking your solution it looks like I forgot to include the commander in my 22em/s, even though I took into account the fact he produces 2 more energy than metal! XD
+1 / 0



Steel_Blue: Not a bad trick question. Normally, I'd have personally gone for the 9 mexes with more evenlydistributed 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 earlygame.) 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.
SmokeDragon: 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



but what if they are all in the grid
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quote: 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 ;)
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