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Old July 23rd, 2002, 06:17 PM
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DirectorTsaarx DirectorTsaarx is offline
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Default Re: Tech Progression Formula?

Quote:
Originally posted by Baron Munchausen:
Those formulas were posted by PvK, and he was just describing what he saw without thinking very much about the underlying math. I don't think PvK is a mathematician or even a programmer. I think the new formulas posted in this thread are cleaner and probably closer to the truth of what's going on in the game code. Just goes to show you that there are many ways to describe the same thing.
Well, DavidG's formulas are mathematically equivalent to Andrés' formulas (yeah, I'm a geek; I did a quick arithmetic proof). DavidG's formulas are just the recursive form (i.e., the total cost is the previous level's cost plus the difference calculated from his formulae), while Andrés' formulas are the absolute form (i.e., you just need to know the level cost and the level being researched to find the total cost).

For those who are interested, the proof involves taking the "absolute" formula for level "L" and subtracting the formula for level "L-1", then reducing algebraically to come up with the difference between each level.

Full proof follows:

Low tech cost: absolute formula is LC*L

LC*L - LC*(L-1) = LC*L - LC*L + LC = LC

Therefore, the difference in cost for each level is equal to the level cost.

-----------------------------------
Medium tech cost: absolute formula is LC*(L˛/2)

LC*(L˛/2) - LC*[(L-1)˛/2] = LC*[L˛/2 - (L˛/2 - 2L/2 + 1/2)] = LC*[2L/2 - 1/2] = LC*(L - 0.5)

Therefore, the difference in cost for each level is equal to the level cost times "Level - 0.5"

-----------------------------------
High tech cost: absolute formula is LC*(L˛)

LC*(L˛) - LC*(L-1)˛ = LC*[L˛ - (L˛ - 2*L + 1)] = LC*(2L - 1) = 2*LC*(L - 1/2)

Therefore, the difference in cost for each level is equal to the level cost times 2, times "Level - 0.5"

[ July 23, 2002, 17:18: Message edited by: DirectorTsaarx ]
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