2 ms·
Ok I think I've got a counter example. First of all, I'm only going to consider values where N approaches infinity && r < 1 && r > 0 to make the math easier.
by gibybo 14y ago
Ok I think I've got a counter example. First of all, I'm only going to consider values where N approaches infinity && r < 1 && r > 0 to make the math easier.
lim [n-> inf, r < 1, r > 0 ] of S_0 = E_0 (1-r^n)/(1-r) ->
S_0 = E_0 * 1 / (1-r)
I propose that the actual relationship is:
S_0 = E_0 / (i-f)
E_0=1,f=.07,i=.10 - Original: S_0 = 36.666..., My proposal: S_0 = 33.333...
E_0=1,f=.80,i=1.00 - Original: S_0 = 10, My proposal: S_0 = 5
So it seems your formula approximates mine at low values of i and f, but diverges strongly at higher values.
I believe I can show mine is the correct one with a small amount of intuition when we let f=0:
If I make 10% annually in interest, then $1mil will provide $100k/year forever. The same thing more concisely: i * S_0 = E_0
r = F/I = (1+0/1+i)
After substitution and division, your formula is E_0 = S_0 / ((i+1)/i)
In words: If I make 10% annually in interest, then $1mil will provide $90.9k/year forever, which is intuitively false.
- nileshtrivedi 14y agoIt is not. As I mentioned in the post, you need to set aside the money for the year's expenses at the beginning of the year. $1mil provides $90.9k/year because (1m - 90.9k)*0.1 = 90.9k. Interest is received at the end of the period but we need cash in hand throughout the year to pay for the expenses.
- gibybo 14y agoAlright, so I spent another hour playing with these numbers and I think I understand it correctly now and agree with your original math. A combination of mistakes (F/I vs f/i, taking money out at the start of the year) and my imperfect intuition of inflation led me astray. Thanks for teaching me something new, and for your patience :)
- nileshtrivedi 14y agoThank you for helping me find weak points in the post and clarify it. :)