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January 23rd, 2017, 10:56 AM   #1
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Financial formula

To de-mistify(!) a bit:
f = future value, d = periodic deposit, i = interest rate, u = 1+i,
m = positive integer constant, n = number of periods:
in other words, a complicated annuity problem!

As a series, future value is:
f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i + ..... + d(u^(nm))/i

I've been trying to work out a closed form formula: any suggestions?
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January 23rd, 2017, 02:44 PM   #2
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$f =\dfrac d i \displaystyle{\sum_{k=1}^n}~u^{k m}$

$f = \dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^m\right)^k =$

$u^m \dfrac {d}{i} \dfrac{1 - u^{nm}}{1-u^m}$
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January 23rd, 2017, 05:29 PM   #3
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Don't think so, Romsek; your formula gives ~5460.51
for the example below...an EXACT 3000 too much...
Code:
year  deposit    interest      balance 
 0                                           .00 
 1     100.00          .00        100.00 
 2     100.00       10.00       210.00 
 3     100.00       21.00       331.00 
 4     200.00       33.10       564.10 
 5     200.00       56.41       820.51 
 6     200.00       82.05    1,102.56 
 7     300.00      110.26   1,512.82 
 8     300.00      151.28   1,964.10 
 9     300.00      196.41   2,460.51
d = 100, i = .10, u = 1.1, m = 3, n = 3

f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i = ~2460.51
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January 23rd, 2017, 06:58 PM   #4
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Quote:
Originally Posted by Denis View Post
Don't think so, Romsek; your formula gives ~5460.51
for the example below...an EXACT 3000 too much...
Code:
year  deposit    interest      balance 
 0                                           .00 
 1     100.00          .00        100.00 
 2     100.00       10.00       210.00 
 3     100.00       21.00       331.00 
 4     200.00       33.10       564.10 
 5     200.00       56.41       820.51 
 6     200.00       82.05    1,102.56 
 7     300.00      110.26   1,512.82 
 8     300.00      151.28   1,964.10 
 9     300.00      196.41   2,460.51
d = 100, i = .10, u = 1.1, m = 3, n = 3

f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i = ~2460.51
can you write your formula in latex?

I get 5460.51 for that using the sum formula directly and using the closed form formula
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January 23rd, 2017, 08:55 PM   #5
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My bad; forgot the "-1"'s

d=100, i = .1, u=1.1

d*(u^3 - 1)/i + d*(u^6 - 1)/i + d*(u^9 - 1)/i = 2460.51

What it boils down to is 3 annuities of 100 annually,
one for 3 years, one for 6 years and one for 9 years.
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January 23rd, 2017, 09:38 PM   #6
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Quote:
Originally Posted by Denis View Post
My bad; forgot the "-1"'s

d=100, i = .1, u=1.1

d*(u^3 - 1)/i + d*(u^6 - 1)/i + d*(u^9 - 1)/i = 2460.51

What it boils down to is 3 annuities of 100 annually,
one for 3 years, one for 6 years and one for 9 years.
so this?

$f =\dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^{k m}-1\right)$

$f = \dfrac d i \left(\displaystyle{\sum_{k=1}^n}~u^{k m}-\displaystyle{\sum_{k=1}^n}~1\right)$

$f = \dfrac {d}{i} \left(u^m\dfrac{1 - u^{nm}}{1-u^m} - n\right)$

This gives this correct results for that set of parms you posted.
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January 24th, 2017, 08:04 AM   #7
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YES! Works fine.
I didn't like mine, as I was converting each "period of 3"
into a "period of 1", as per these contortions:

y = number of years (multiple of 3) [9 in my example]
j = annual rate / 100 [.10 in example]
f = first deposit [$100 in example ]

n = FLOOR(y/3) + 1 [4 in example]
i = (1 + j)^3 - 1 [.331 in example]
d = f(j^2 + 3j + 3) [331.00 in example]
Code:
       d[(1 + i)^n - in - 1] 
f =  ----------------------- 
                 i^2
Thanks again.
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January 26th, 2017, 06:21 AM   #8
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Beer soaked ramblings follow.
Accumulated value of a simple increasing annuity (payments are in arithmetic progression). The given example can be viewed as an annuity whose payments are made at the end of 4 months for 3 years; year end payment for 1st year is 100(1.10^3-1)/.10=331, year end payment for 2nd year is 200(1.10^3-1)/.10=662,
year end payment for 3rd year is 300(1.10^3-1)/.10=993; thus giving a common difference of 331. With R = 331 and an effective rate of i = 1.10^3-1 and n = 3 years, the problem is reduced to a blind application of the aforementioned formula of
S = (R/i){(1+i)[(1+i)^n-1]/i-n}
which bears some resemblance to Sir Romsek's formula.
Thanks from Denis

Last edited by jonah; January 26th, 2017 at 06:24 AM.
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