On Sat, Oct 08, 2016 at 12:58:00PM -0400, Patrick 'P. J.' McDermott wrote: > If you want the amount a project receives to be exactly equal to the > amount a patron intends to donate, here's the math to determine the > extra fee to add. > > Remember that Stripe's fee is 2.9% + $0.30, and begin with an equation > that yields the final donation after fees ($0.30 less than 100%-2.9% > of the payment): > > donation = total_payment * (1 - 0.029) - $0.30 > > Do some algebra, first adding $0.30 to both sides: > > donation + $0.30 = total_payment * (1 - 0.029) > > And then divide both sides by 100%-2.9% to isolate total_payment and > get a formula to calculate it: > > donation + $0.30 > total_payment = ---------------- > 1 - 0.029
Awesome, thanks! I was caught up calculating the fee, but this way of looking at it is easier to implement. I spent some time trying to show that the rounding works out analytically, but in the end I just brute-forced the calculation on every possible value. It works (until integer truncation errors occur up at $20M). Plus, I made some cool graphs: https://www.desmos.com/calculator/xtzs3n4lxq So, I'll just go ahead with charging enough so that crowdmatch = net_transaction_balance I don't really like the lack of an analytical solution, but I'm satisfied that it actually works. Brute force code attached. -Bryan
import Data.Int
stripeFee :: Int32 -> Int32
stripeFee = round . (+ 30) . (* 0.029) . fromIntegral
payment :: Int32 -> Int32
payment = round . (/ (1 - 0.029)) . (+ 30) . fromIntegral
bad c =
let p = payment c
f = stripeFee p
in p - f /= c
main = do
putStr "First bad at: "
print (take 1 (filter bad ([1..])))
#include <stdio.h>
#include <limits.h>
#include <math.h>
long int stripe_fee (int c) {
return lrint(c * 0.029 + 30);
}
long int payment (int c) {
return lrint( (c + 30) / (1 - 0.029) );
}
int test (long int c) {
int p, f;
p = payment(c);
f = stripe_fee(p);
return (p - f) != c;
}
int main () {
long int i = 0;
int max = INT_MAX;
for (i = 0; i < max; i++) {
if (test(i)) {
printf("First error at %ld\n", i);
return 0;
}
}
}
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