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FV challenge...

Wilmer

In Memoriam
Mar 19, 2012
376
d is deposited in a savings account for 3 years, then 2d for 3years,
then 3d for 3 years, and so on similarly;
here's an example for 9 years, 1st deposit = $100, rate = 10% annual:
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
GIVENS:
y = number of years (multiple of 3): 9 in above example
j = annual rate / 100 : .10 in above example
f = first deposit : $100 in above example

What is the future value of the account (2,460.51 in above example) in terms of y, j and f ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I find:

\(\displaystyle FV(y,j,f)=f\sum_{k=1}^y\left(\left\lceil\frac{k}{3} \right\rceil(1+j)^{y-k} \right)\)

I simply wrote down the first few terms, observed the pattern, then checked the above formula against the provided table.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Nice, Mark, but that's not really what I was after;

I meant something like a standard financial formula, a bit like:
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g)
P=initial payment
r=discount rate or interest rate
g=growth rate
n=number of periods

Guess I should have said so, then(Bandit)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Nice, Mark, but that's not really what I was after;

I meant something like a standard financial formula, a bit like:
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g)
P=initial payment
r=discount rate or interest rate
g=growth rate
n=number of periods

Guess I should have said so, then(Bandit)
I have FV on the left of the equal sign and on the right a set of mathematical symbols representing the value of FV...it's just like what you are after. (Clapping)(Sun)
 

Wilmer

In Memoriam
Mar 19, 2012
376
...it's just like what you are after.
No it's not...the right side is not really a "formula", even if correct,
but more a definition of how the formula is arrived at...
Well, not really important anyway: won't affect price of groceries (Sun)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I gather you want a closed form...well, let's look at the difference equation:

\(\displaystyle FV_{y+1}-(1+j)FV_{y}=sf\)

where $s$ is the step-function \(\displaystyle s=\left\lceil\frac{y}{3} \right\rceil\)

The homogeneous solution will take the form:

\(\displaystyle h_y=c_1(1+j)^y\)

But, I have no idea what form the particular solution must take, given that pesky step function, which also thwarts my ability to use symbolic differencing to obtain a homogeneous recurrence. (Emo)
 

Wilmer

In Memoriam
Mar 19, 2012
376
Well, this was mine :

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] 
FV = --------------------- 
             i^2
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Your formula does work when $y$ is a multiple of 3, and if I had read your original post more carefully, I would have noticed you did state that $y$ is a multiple of 3...I was trying to find a formula that worked for any natural number $y$.

When I have more time later, I will attempt to post a derivation of your formula. (Nerd)
 

Wilmer

In Memoriam
Mar 19, 2012
376
...and if I had read your original post more carefully, I would have noticed you did state that $y$ is a multiple of 3...
Oh oh: 15 minutes in the corner (Wait)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The 15 minutes in the corner was nothing compared to the battle I had with silly errors in coming up with the formula...(Sweating)(Rofl)

Let's let $y=3n$ where $n\in\mathbb{N_0}$.

Now, consider the following difference equation obtained from analyzing what goes on in between each value of $n$:

\(\displaystyle FV_{n+1}-(1+j)^3FV_{n}=(n+1)f\left(j^2+3j+3 \right)\)

We see that the homogeneous solution is given by:

\(\displaystyle h_n=c_1(1+j)^{3n}\)

And we seek a particular solution of the form:

\(\displaystyle p_n=An+B\)

Substituting, we find:

\(\displaystyle (A(n+1)+B)-(1+j)^3(An+B)=(n+1)f\left(j^2+3j+3 \right)\)

\(\displaystyle -jAn+\frac{A}{j^2+3j+3}-jB=fn+f\)

Equating coefficients, we find:

\(\displaystyle A=-\frac{f}{j},\,B=-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}\)

Thus, we have:

\(\displaystyle p_n=-\frac{f}{j}n-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}\)

And so the general solution is:

\(\displaystyle FV_{n}=h_n+p_n=c_1(1+j)^{3n}-\frac{f}{j}n-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}\)

Using $FV_0=0$, we find:

\(\displaystyle c_1=\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}\)

Thus, we have (after some simplification):

\(\displaystyle FV_{n}=\frac{f}{j}\left(\frac{(1+j)^{3n}-1}{1-(1+j)^{-3}}-n \right)\)
 

Wilmer

In Memoriam
Mar 19, 2012
376
I simply converted a "3 periods" to an equivalent "1 period";
in my example, that's 331 deposited periodically , and
increasing by 331 each period, at 33.1%.
Code:
P  DEPOSIT INTEREST BALANCE
0                        .00
1   331.00            331.00
2   662.00  109.56  1,102.56
3   993.00  364.95  2,460.51