Math in Games

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.