mmm4444bot said:
Hello Jonah:
It looks like you think that i = 0.05/12
Exactly, that’s what my conditional “
if P = $100,000;
i = .05/12…” and my (actual) sleepy comment was for. My bad. I guess temporarily quitting the bottle has its undesirable effect of making me miss the obvious.
mmm4444bot said:
The interest rate is 5%. (That much is standard.)
I’d have to disagree with you on that one.
i is almost always defined as nominal rate divided by the interest period in finance math jargon.
mmm4444bot said:
PS: I would have liked to have retrieved the scrap paper from the recycle bin to be able to upload an image of my work for you (I had already pulled it out once to type my version for Denis; I would gladly have dug through the bin a second time), but, alas, the recycling truck came earlier today.
Can you perhaps recreate the process of your work that led to this formula? Impractical as it seems (my opinion, as in 12^360) and monstrous as it was (as Denis called it), I’m almost certain I’m not the only one curious to see how you managed to derive it.
Denis said:
Jonah, Mark's monstrosity does work:
100000(.05)/12 * 12.05^360 / (12.05^360 - 12^360) = 536.821623...
The division by 12 in 1st term effectively makes the interest .05/12
So it does.
By the way, I’d like to thank you for that insightful formula that you cited at:
http://www.freemathhelp.com/forum/viewtopic.php?f=17&t=26743
It is essentially a generalization or clever manipulation (depending on how you look at it) of the so-called retrospective method for determining the outstanding liability just after any particular payment.
Had I not gotten used to obtaining the same results from spreadsheet amortization schedules, I might have come up with it in like a few years after retiring. If I might inquire with you: Did you perhaps learned it from a textbook? Or did you perhaps “discovered” it while you were still active in the workplace? Either way, I am definitely green with envy. To refresh your memory:
Denis said:
So as example: a $3000 loan at rate 12% cpd monthly is being repaid at $100 per month;
how much interest is paid between payment number x and payment number y (y > x) ?
100(y - x) - {3000(1 + .01)^x - 100[(1 + .01)^x - 1] / .01 - 3000(1 + .01)^y - 100[(1 + .01)^y - 1] / .01}
A slight correction would be:
100(y - x) - {3000(1 + .01)^x - 100[(1 + .01)^x - 1] / .01 - 3000(1 + .01)^y
+ 100[(1 + .01)^y - 1] / .01}
which then simplifies to
100(y - x) + (100/.01 – 3,000)[ (1 + .01)^x – (1 + .01)^y]
Let's see, how many "perhaps" did I use?