Comments (3)
After some thinking and experimenting I've finally found the function that fits
the bill (limit when x -> inf = constant, integral has inverse, etc.):
f(x) = a + b/(x+1)^2
where
x - dropoff distance (see below)
f(x) - food per farmer at dropoff distance of x
a - minimal food per farmer
b - maximal food per farmer minus "a"
Let's denotate with D population that can work as farmers without penalty. D +
1st farmer has dropoff distance of 1/D, D + 2nd has distance of 2/D, etc. Now,
how to calculate number of necessary farmers (excluding first D farmers)? Well,
that's why I was looking for inverse of integral function.
Integral (with condition F(0) = 0) of the function above is
F(x) = ax + b/(x+1) + b
Inverse of that integral is
F^-1(y) = (sqrt(g^2 + 4ay) + g) / 2a
where
y - food units demand (excluding demand covered by first D farmers) divided by D
g = y - a - b
a, b - same as above
Original comment by [email protected]
on 15 Nov 2011 at 2:16
- Changed state: Started
from stareater.
Original comment by [email protected]
on 15 Nov 2011 at 10:42
- Changed state: Done
from stareater.
Forumal correction:
F(x) = ax - b/(x+1) + b
F^-1(y) = (sqrt(g^2 + 4ay) + g) / 2a
where
y - food units demand (excluding demand covered by first D farmers) divided by D
g = a + b - y
a, b, x, D - same as above
Original comment by [email protected]
on 23 Aug 2013 at 2:10
from stareater.
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