Formula for compounded contributions

[joema]

Homework Statement

What is the formula for compounded annual contributions which increase at a fixed % rate, yet where the principal interest rate is 0%. IOW the classic "money under the mattress" savings where the yearly contributed amount increases at a fixed %.

E.g, year one contribution is $1,000, each year the contributed amount increases by x%, but with zero % interest on the principal. How much total saved after n years?

Homework Equations

All the equations I can find assume compounded principal and fixed annual contributions. What I want is the opposite - non-compounded principal and annual contributions which increase at a fixed % each year.

 

[Jilang]

Would it not just be the original amount multiplied by (1 + x%)^n where n is the number of years?

 

[SammyS]

Would it not just be the original amount multiplied by (1 + x%)^n where n is the number of years?

The interest rate issued by a mattress is 0% .

 

[SammyS]

Homework Statement

What is the formula for compounded annual contributions which increase at a fixed % rate, yet where the principal interest rate is 0%. IOW the classic "money under the mattress" savings where the yearly contributed amount increases at a fixed %.

E.g, year one contribution is $1,000, each year the contributed amount increases by x%, but with zero % interest on the principal. How much total saved after n years?

Homework Equations

All the equations I can find assume compounded principal and fixed annual contributions. What I want is the opposite - non-compounded principal and annual contributions which increase at a fixed % each year.

How much is contributed yearly?

How many years?


(Use reason rather than some meaningless formula.)

 

[xiavatar]

Sorry. My roommate was messing around with my computer and posted that.

 

[joema]

How much is contributed yearly?

How many years?

(Use reason rather than some meaningless formula.)

I'm not looking at a specific case of x dollars per year for n years. Rather a formula which could be applied to various cases.

To state one case, what if $1000 per year was saved under the mattress the first year, then 3% more the 2nd year ($1030), then 3% more the 3rd year ($1060.90). After three years the total saved would be 1000+1030+1060.90 = $3090.90.

What is the formula to calculate this using as variables the # of years and % increase of annual contributions? I know the answer is very simple but I can't find it.

 

[Ray Vickson]

I'm not looking at a specific case of x dollars per year for n years. Rather a formula which could be applied to various cases.

To state one case, what if $1000 per year was saved under the mattress the first year, then 3% more the 2nd year ($1030), then 3% more the 3rd year ($1060.90). After three years the total saved would be 1000+1030+1060.90 = $3090.90.

What is the formula to calculate this using as variables the # of years and % increase of annual contributions? I know the answer is very simple but I can't find it.

First you need to find the formula for the contribution in year n for n = 2,3,4,...; this will be a geometric sequence. Then, to find the N-year total you need to add up all the individual yearly contributions; this will be a finite geometric sum, and you can find the needed formulas in books or on-line.

 

[ChrisVer]

you just have to sit down and write, what you wrote above by logic, in maths...
So you have some amount of input money X...
what will happen in 1 year? It will become A depending on X and the % rate...

Now this A will be your new input (which will be interested)

what will happen to it in another year? It will become A', which will depend on A and the % rate... But you know how A depends on X and %, so then you can write A' as a function of X, and % rate...

Keep doing that for as many years you want- but you will already be able to deduce the result for k years...

 

[Jilang]

The interest rate issued by a mattress is 0% .

X% refers to the contribution rate not the interest rate of the mattress, lol . ; )

 

[Mentallic]

I'm not looking at a specific case of x dollars per year for n years. Rather a formula which could be applied to various cases.

x and n would be variables which means you can exchange it for any dollar value and number of years respectively, hence it's not a specific case as you put it.

So you start with x dollars under the mattress, then after 1 year it's
x+x*(1+r) where r is the interest rate, then after 2 years it's
x+x*(1+r)+(x*(1+r))*(1+r) = x+x*(1+r)+x*(1+r)²

Are you following this sequence? Is there anything you don't understand? Can you continue the sequence for n years?

 

[joema]

OK after several more hours of searching (beyond those I did before my initial post), I finally found the formula.

It is FV = PMT/i * ((1 + i)^n - 1), where

FV = final value
PMT = initial payment
i = % increase of each payment
n = number of periods

E.g: Initial payment = $1000,
% increase of each subsequent payment = 3%,
number of periods (years) = 3

1000/0.03*((1+0.03)^3 - 1) = 3090.90

This is a very difficult area to search for because keywords like "interest", "compound interest", etc will deflect the search results to common formulas.

The above formula describes the future value of an annuity, but it has wide application -- any scenario where the annual expenditure increases at a fixed % and you want to calculate the total expenditure over time.

E.g, if global energy consumption is 550 quadrillion btu per year and increases at an average annual rate of 2% per year, by 2040 the cumulative energy consumed over that period will be 19,439 quadrillion BTUs.

 

[Mentallic]

Alright, well, seeing as how you preferred to research for hours to find the formula rather than deriving it yourself, I'll continue with where I was going in my previous post to show you how much easier it could have been:

S = sum of money in total
x = initial amount
r = rate of increase per year
n = number of years

After n years, the total amount would be

[tex]S= x+x(1+r)+x(1+r)^2+...+x(1+r)^n[/tex]

[tex]S=x\left(1+(1+r)+(1+r)^2+...+(1+r)^n\right)[/tex]

Using the geometric sum formula on the part in the large brackets

[tex]1+p+p^2+...+p^n=\frac{p^{n+1}-1}{p-1}[/tex]

where in our case, p=1+r, this gives us

[tex]S=x\frac{(1+r)^{n+1}-1}{(1+r)-1}[/tex]
[tex]=\frac{x}{r}((1+r)^n-1)[/tex]

as you found.