MHB Future Value of Savings Account: y Years, j Rate, f Deposit

[Wilmer]

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:

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]

I find:

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

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

 

[Wilmer]

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]

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]

...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]

I gather you want a closed form...well, let's look at the difference equation:

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

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

The homogeneous solution will take the form:

$$h_y=c_1(1+j)^y$$

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]

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]

     d[(1 + i)^n - in - 1] 
FV = --------------------- 
             i^2

 

[MarkFL]

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]

...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]

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 $n$:

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

We see that the homogeneous solution is given by:

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

And we seek a particular solution of the form:

$$p_n=An+B$$

Substituting, we find:

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

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

Equating coefficients, we find:

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

Thus, we have:

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

And so the general solution is:

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

Using $FV_0=0$, we find:

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

Thus, we have (after some simplification):

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

 

[Wilmer]

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%.

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