Built to Fade
Woodpecker
When You Know The Math, Your Options Expand
Bumping this thread by posting a full mathematical formula of a compound interest account. This assumes that rates and amounts are the same. It is not the case in the real markets, but it gives an understanding about the power of exponentials.
Say you put in 10% of your salary into this account. Let's say that your annual salary is $60 000 before taxes.
You would put in $6 000 a year into this account or $500 per month into a compound interest account.
Assume that your interest rate is at 2% per annum.
The return formula is (1 + [period return rate]/[no. of intervals per year])^[no. of intervals per year]
Monthly interest will be 1 + (0.02/12) = 1/600% or 0.166[recurring 6s]% per month. It gives a slightly higher rate than an annual investment.
The maximum return occurs at an instantaneous investments (0 seconds). It is represented by (1+x/[infinite intervals])^[infinite intervals]. That is one way to derive the numeral e when using 100% interest per annum.
The full compound interest formula is:
_____ M(X^Y - 1)_____________A(X - 1)___________log([A(X - 1) / M] + 1)________M
A = ______________ or, M = ______________ or, Y = ____________________ or, X = ____ + 1
_______ X - 1 _______________(X^Y - 1)_________________log(X)_______________ P
Where:
Example with $500 per month (M), 2% interest paid monthly (X) & 600 months (50 years) (Y):
___ 500([1 + 1/600]^600 - 1)
_ ________________________ = $514 806.01 after 50 years.
_______ 1 + 1/600 - 1
Using the substitution 500([1 + 1/600]^600 - [1 + 1/600]^480), the first 120 instalments (10 years) will go from $12 000 to $147 588.20, a whopping 28.67% of the total amount for just the first 10 years.
Substituting 500([1 + 1/600]^480 - [1 + 1/600]^360) during Years 11-20 will go from $12 000 to $120 855.12 (23.48% total)
Substituting 500([1 + 1/600]^360 - [1 + 1/600]^240) during Years 21-30 will go from $12 000 to $98 964.28 (19.22% total)
Substituting 500([1 + 1/600]^240 - [1 + 1/600]^120) during Years 31-40 will go from $12 000 to $81 038.59 (15.74% total)
Substituting 500([1 + 1/600]^120 - [1 + 1/600]^0) during Years 41-50 will go from $12 000 to $66 359.83 (12.89% total)
Conversely, the full loan formula is:
______________ M(X^Y - 1)
A = P x X^Y - ______________
_________________ X - 1
Where:
Bumping this thread by posting a full mathematical formula of a compound interest account. This assumes that rates and amounts are the same. It is not the case in the real markets, but it gives an understanding about the power of exponentials.
Say you put in 10% of your salary into this account. Let's say that your annual salary is $60 000 before taxes.
You would put in $6 000 a year into this account or $500 per month into a compound interest account.
Assume that your interest rate is at 2% per annum.
The return formula is (1 + [period return rate]/[no. of intervals per year])^[no. of intervals per year]
Monthly interest will be 1 + (0.02/12) = 1/600% or 0.166[recurring 6s]% per month. It gives a slightly higher rate than an annual investment.
The maximum return occurs at an instantaneous investments (0 seconds). It is represented by (1+x/[infinite intervals])^[infinite intervals]. That is one way to derive the numeral e when using 100% interest per annum.
The full compound interest formula is:
_____ M(X^Y - 1)_____________A(X - 1)___________log([A(X - 1) / M] + 1)________M
A = ______________ or, M = ______________ or, Y = ____________________ or, X = ____ + 1
_______ X - 1 _______________(X^Y - 1)_________________log(X)_______________ P
Where:
- A is the future amount
- M is the periodic investment amount
- P is the total amount invested (= M x no. of investments)
- X is the return rate per period (yearly, monthly, or weekly)
- Y is the amount of periods (yearly, monthly or weekly)
Example with $500 per month (M), 2% interest paid monthly (X) & 600 months (50 years) (Y):
___ 500([1 + 1/600]^600 - 1)
_ ________________________ = $514 806.01 after 50 years.
_______ 1 + 1/600 - 1
Using the substitution 500([1 + 1/600]^600 - [1 + 1/600]^480), the first 120 instalments (10 years) will go from $12 000 to $147 588.20, a whopping 28.67% of the total amount for just the first 10 years.
Substituting 500([1 + 1/600]^480 - [1 + 1/600]^360) during Years 11-20 will go from $12 000 to $120 855.12 (23.48% total)
Substituting 500([1 + 1/600]^360 - [1 + 1/600]^240) during Years 21-30 will go from $12 000 to $98 964.28 (19.22% total)
Substituting 500([1 + 1/600]^240 - [1 + 1/600]^120) during Years 31-40 will go from $12 000 to $81 038.59 (15.74% total)
Substituting 500([1 + 1/600]^120 - [1 + 1/600]^0) during Years 41-50 will go from $12 000 to $66 359.83 (12.89% total)
Conversely, the full loan formula is:
______________ M(X^Y - 1)
A = P x X^Y - ______________
_________________ X - 1
Where:
- A is the amount owed
- P is the principal (original loan amount)
- M is the periodic payment amount
- X is the owing interest rate per period (yearly, monthly, or weekly)
- Y is the amount of periods (yearly, monthly or weekly)