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February 18th, 2017, 06:45 AM   #31
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Ya. Imagine trying to explain a formula involving logs and the likes!

1st thing I do when I try to solve a problem given here is change
the damn given example to something as simple as possible.
Like, in this case: target = 28830, r = .10, term = 27 :
Code:
01   1000   1000
02   1000   2000
...
12   1000  12000
13   1100  13100
...
24   1100  25200
25   1210  26410
26   1210  27620
27   1210  28830
Now my formula need to yield an even 1000 bucks!
Get my drift
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February 18th, 2017, 07:26 AM   #32
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Quote:
Originally Posted by Denis View Post
Ya. Imagine trying to explain a formula involving logs and the likes!

1st thing I do when I try to solve a problem given here is change
the damn given example to something as simple as possible.
Now my formula need to yield an even 1000 bucks!
Get my drift
What I actually did in my spread sheet was to work out the sum of the payments starting with an initial payment of 1 dollar (the number of years, months, indexation percentage, and total to be disbursed were input variables). Then I divided that sum into the total to be disbursed.

Same idea except I started with a simple payment rather than a simple total. I think skipjack may have been there from the start, but I really did not get what the problem even was until about page 2.
Thanks from Denis
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February 18th, 2017, 09:50 AM   #33
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Mine was using future value of annuity formula
(treating the percentage increase as an interest rate)
for the complete years, then adjusting by subtracting
above from the target amount...
as you say, same general idea...
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February 18th, 2017, 06:33 PM   #34
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Quote:
Originally Posted by SDK View Post
Finally, if a particular value for $S$ is specified one can factor and solve for $P_0$ explicitly which is
$$ P_0 = \frac{S}{12(\frac{1-(1+r)^Y)}{r}) + M(1+r)^Y}.$$
Just noticed that +M(1+r)^Y should be -M(1+r)^Y.
After that change plus inserting the missing half bracket,
it works fine. Plus simpler than mine
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February 18th, 2017, 07:28 PM   #35
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It should be $P_0 = \dfrac{S}{12((1 + r)^Y - 1)/r + M(1 + r)^Y}$.
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February 18th, 2017, 07:59 PM   #36
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Skip, doesn't that simply change the result from negative to positive?
Anyway, I prefer positive!
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March 31st, 2017, 02:17 AM   #37
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Quote:
Originally Posted by mshaughn View Post
I am trying to find a formula to solve for the monthly starting amount below:

Monthly Starting Amount = ?
Payment Term = 10 years, 5 months (125 monthly payments)
Indexation = 2% per year, compounded
Total Future Payout = \$200,000.00

That is, what is the required monthly starting amount you must invest if you want a payout of \$200,000.00 after 125 monthly payments, if payments increase by 2% per year, compounded?

I know the answer is \$1,454.63 with \$0.45 remaining at the end of the term, but I am not sure how to get there. Any help would be appreciated.

I also know that without indexation, it's simply \$1,600.00 (\$200,000.00 / 125).
Using the =PMT formula, I got about 1500 as payment but that's a fixed 2% APR per year (not counting it as increasing by 2% each time - i'm not sure how to enter this in excel).
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March 31st, 2017, 09:56 PM   #38
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Beer soaked comment follows.

I must have been sober at the time when this thread started out.
Good stuff.
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