27 ms·

I also left out explanation for the edge cases. For low enough values of r, savings will never decrease and never reach zero. For practical values of r, the equ

I also left out explanation for the edge cases. For low enough values of r, savings will never decrease and never reach zero. For practical values of r, the equation is correct.

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 ago
  It 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 ago
    Alright, 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 ago
      Thank you for helping me find weak points in the post and clarify it. :)

I also left out explanation for the edge cases. For low enough values of r, savings will never decrease and never reach zero. For practical values of r, the equation is correct.

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 ago
It 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 ago
  Alright, 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 ago
    Thank you for helping me find weak points in the post and clarify it. :)