Math in Games

Gauss is credited with this story but I don't think he was the first person to come up with this... anyway, Gauss as a child was in school and his teacher asked the students to add up all numbers from 1 to 100. Most of Gauss' classmates started furiously adding everything by hand. Gauss just thought about the problem for a minute and then reasoned it out this way: suppose you write all the integers from 1 to 100 in order from left to right. Now suppose you write them all again from 100 to 1, left to right, directly under the first row. Now, he thought, add each column. In each of the 100 columns you get a sum of 101. 100 cases of 101 adds up to 10100. 10100 is the sum of 1 to 100 TWICE. Therefore, cutting the number in half, 5050, is the sum from 1 to 100 of integers. His teacher was of course impressed. From this we see that the sum of consecutive integers is the lowest integer plus the highest integer times half the total quantity of integers being added.

It's just a chain of the first attack with half the number of attacks, (odd = rounded up) + a chain of the second attack with half the number of attacks, (odd = rounded down)

example: 40+70+42+72+44... = 40+42+44... + 70+72...

You figured out the regular chain formula already.

(T_c - T_f) * c > T_c

Where T_c is number of turns to grow, if a new cottage is built
T_f is number of turns to grow, if a new farm is built
c is the remaining number of cottages to be built

So, finding the values of T_c and T_f:

T_c = G / (B + f)

T_f = G / (B + f + 1)

Where G is the amount of Food required for growth,
B is the food bonus of the single Food Resource
f is the current number of Farms

So,

(G/(B+f) - G/(B+f+1) ) * (C - (n-f-1-1) ) > G/(B+f)

where n is the current pop count
C is the final number of cottages

Now, divide both sides by G and multiply by (B+f):

(1 - (B+f)/(B+f+1)) * (C-n+f+2) > 1

or

(1 - (1- 1/(B+f+1) ) * (C-n+f+2) > 1

or

1/(B+f+1) * (C-n+f+2) > 1

or

C - n + f + 2 > B + f + 1

Subtracting (f+1) from both sides gives us:

C - n + 1 > B

or

n < C - B + 1

Resulting in the the same formula arrived at by vicawoo.

So, this means, with C=15, B=5, you should Farm when:

n < 15 - 5 + 1

n < 11


Note that this result is independent of both G (the food cost of pop growth) and f (the current number of farms), even though they were used in the inequality.

Let M(n,d[n+1]) be the increase in empirewide maintenance by adding an n+1st city at a distance of d[n+1] from the capital (In this notation d[i] is the distance from the capital of the ith city).

In the long run, if we increase empire wide maintenance by 4, we would have to switch two tiles to 2 food 2 commerce tiles to maintain equilibrium.
However, we get 1 commerce from the new city itself; you may even get an additional 1 if you have a trade route.

So new city negatively impacts empire wide production by
( M(n,d[n+1]) - 1 )/tile_c tiles, which would have otherwise yielded tile_y (tile yield). So the cost of the new city in food/hammers is
M(n,d[n+1])*tile_y/tile_c
Note:
[spoiler]If we give up a riverside mine for a riverside cottage, we are only netting one commerce, so in practice tile_c is increase in commerce from switching from our production oriented tile to our commerce tile. If by chance we have a domestic trade route to the new city, we can just pretend the new city costs the empire one less gold coin in maintenance[/spoiler]

Profit = Revenue - Cost
Profit = tile_y[n+1] - ( (M(n,d[n+1] ) -1 )*tile_y/tile_c

We divide our profit by our initial investment, that is 100 production.

Now we compare this to growing our ith city from size p[i] to p[i]+1.

Again Profit = Revenue - Cost
Cost = d[i]/29 + n/60
Note that in most cases this cost will usually be much less than 1.
Profit = yield'_fh + (yield'_c- d[i]/29 + n/60)*tile_y/tile_c
In most of these cases we're debating working a mine or it's commerce equivalent.
And the investment is 2*(10+p[i]) food, with an opportunity cost of
2*(10+p[i])*(f[i]+h[i])/f[i] food and hammers, where f[i] is the food surplus and h[i] is the hammer production of the ith city.

Ok, we have tons of variables. Let's simplify. Since the increase in maintenance from growing is much less than 1 and we're dealing with integer production values, we'll ignore that. We'll assume we are growing to work a mine, that is, a net +2 food/hammer production.

(tile_y[n+1] + ( 1+tile_c[n+1]- M(n,d[n+1] )*2/tile_c)/100 > 2/( 2*( 10+p[i] )*( f[i]+h[i] )/f[i] )

So what's the point of all this, since sometimes we may only care about maximizing hammers and other times we may care more about our tech rate? It provides a numerical value in the commerce adjusted value of new cities.