Resources are the basis on which everything in OGame is based. Simply put: whatever you want to do, you need resources to do it. It is only fitting, then, that OMath starts with the basics.
How does one acquire resources?
Strategies to obtain resources can be classified in two categories: offensive strategies (using fleets to take resources from other players) and defensive strategies (building mines to produce your own). This specific article, as its name suggest, focuses on the latter.
Say you want to produce your own resources. You'll need mines for that (for simplicity's sake, we take "mines" to include the Deuterium Synthesizer). And developing mines cost resources.
All mines follow the following pattern. If L is a mine's current level, the cost to improve it to level L+1 will be given by C = B * R^L, where B is the base cost (the cost to build level 1 of a mine), R is a certain ratio (1.5 for a Metal Mine or Deuterium Synthesizer and 1.6 for a Crystal Mine), and C is the final cost. This formula is actually applied twice: once for the metal cost, and the other for the crystal cost. Although both use the same ratio R, each have a different base cost B, which yields different final costs C. Consider now the cost C' to upgrade a level L+1 Metal Mine. We have
C' = B * 1.5^(L+1) = B * 1.5^L * 1.5 = C * 1.5 = 1.5 * C
This simple calculation gives us an interesting result: each time a Metal Mine is upgraded, the cost for the next upgrade is increased by 50% of the old value (this also holds for a Deuterium Synthesizer, but the increase is actually 60% for a Crystal Mine). And this applies to both metal and crystal.
Let's put that aside and look at production for a moment.
The formula for a mine's production is trickier. P = B * L * 1.1^L Here, we see that all three types of mines use the same ratio (1.1). Again, L is the current level of the mine and B is the base amount produced (the amount produced by a level 1 mine; note that the value of B depends on a planet's average temperature for a Deuterium Synthesizer). The key difference is that the production is multiplied by the current level of the mine. Think about this. Mines take place: each upgrade takes up 1 field. In other words, mines use up a number of fields equal to their levels. So, what is the average production per field of a mine? Well, we already know the sum (the total production P), so we only need to divide by the number of fields (the current level L). We then find that the production-per-field Pf is Pf = B * 1.1^L Isn't that interesting? This means that both the upgrade cost of a mine and its production per field used create geometric sequences. This would allow us, for example, to calculate the total amount of resources required to upgrade any mine to any arbitrary level (and the increase in production that represents, but the game can already to that for us).
Trick question: should I improve a level 18 mine on my home planet or a level 15 mine on a colony? To answer it, we will consider the upgrade cost not as a per-level cost, but as a per-additionnal-unit-of-resource-per-hour cost; put simply, we want to know which of these two upgrades gives the most production for the least resources. Of course, this will be dependent on the current level of the mine. This special cost is simply obtained by dividing the production increase by the upgrade cost. (We will replace the B in the cost equation by D in order to avoid ambiguity, and add some fancy colours that help us read it all.)
(B * (L+1) * 1.1^(L+1) - B * L * 1.1^L) / (D * R^L)
Okay, that looks like gibberish. We have a constant (read: independent of the current level) term, B/D, that we can safely ignore. What's left is (1.1/R)^L and (0.1 L + 1.1). We know that either R = 1.5 (Metal Mine or Deuterium Synthesizer) or R = 1.6 (Crystal Mine). This has the effect that 1.1/R is always smaller than 1. This is very important. How? Because we then have that the limit of (1.1/R)^L is 0 (multiply a number between 0 and 1 with itself enough times and it will essentially become 0) [we use a limit here because we wish to calculate a long-term behaviour as the value of L increase]. However, the limit of (0.1 L + 1.1) is infinity (0.1 times infinity is still infinity, and adding 1.1 to that doesn't make any real difference). What happens when you have 0 * infinity? First you scratch your head. Then you make some weird calculations that imply finding some derivatives. Finally, you find that (in this particular case), 0 * infinity = 0. This isn't always true, and depends greatly on the context. In our current situation, this means that (1.1/R)^L will decrease much faster than (0.1 L + 1.1), so that their product will decrease as L increase.
So what does this all means? Well, each time you increase the level of a mine, you spend a certain number of units of resources to increase the production of the mine by a certain number of units per hour. However, we just calculated that the production increase divided by the cost comes closer to zero as the level increase. In simpler terms, the higher a mine's level, the more expensive each unit-per-hour becomes when you improve that mine's level.
Thus, the increase in production obtained from upgrading a higher-level mine is less than the increase in production obtained from spending the same amount of resources to upgrade lower-level mines. In other words, investing in a lot of small mines gives you more resources than have a few big ones.
