You just proved my point even better than I could have with 2 pages of material.
I just finished a finite math class..... Here is what supercomputer dude does in his head
Let the original loan balance be: LB(0) {it is the $100,000.00 you borrowed}
Then the interest due ID(1) at the end of the first payment period is:
ID(1) = LB(0) * i
where i is the effective interest rate per payment period.
Our payment period is 1 month (1 year/12), so the effective interest rate is 1/12 of the note rate. In our example, it is 0.07/12.
The principal part PP(1) of the first monthly payment (the part that goes toward the loan balance) is calculated by subtracting the interest part from the monthly payment:
PP(1) = PMT - ID(1)
The loan balance after the first payment LB(1) is calculated by subtracting the principal part (it was calculated above) from the original loan balance.
LB(1) = LB(0) - PP(1)
LB(1) = LB(0) - (PMT - ID(1))
LB(1) = LB(0) - (PMT - LB(0) * i)
LB(1) = LB(0)*(1 + i) - PMT
Similarly, the loan balance after the second payment is:
LB(2) = LB(1)*(1 + i) - PMT
Replacing the LB(1) with the expression we just derived, we get:
LB(2) = (LB(0)*(1 + i) - PMT)*(1 + i) - PMT
or
LB(2) = LB(0)*(1 + i)^2 - PMT*((1 + i) + 1)
The loan balance after the third payment is:
LB(3) = LB(0)*(1 + i)^3 - PMT*((1 + i)^2 + (1 + i) + 1)
The loan balance after n payments is:
LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^(n-1) + ... + (1 + i) + 1)
The sum of the finite series: 1 + a + (a^2) + (a^3) + ... + (a^n) is (1-a^(n+1))/(1-a)
As it can be easily seen:
if SUM(n) is the series sum, then SUM(n) - a * SUM(n) = 1 - a^(n+1)
solving for SUM(n):
SUM(n) = (1-a^(n+1))/(1-a)
Now, with a simple re-arrangement, our equation for loan balance after n payments becomes:
LB(n) = LB(0)*(1 + i)^n - PMT*(1-(1 + i)^n)/(1-(1 + i))
LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^n-1)/i
The loan balance after 360 payments is: $0.00 (the loan is paid off). So:
LB(0)*(1 + i)^360 = PMT*((1 + i)^360-1)/i
Here is the solution for PMT:
PMT = i * LB(0)*(1 + i)^360 / ((1 + i)^360-1)
Your monthly payment will be:
PMT = (0.07/12) * 100000*(1 + 0.07/12)^360 / ((1 + 0.07/12)^360-1) = 665.302495179183
It must be rounded to the nearest cent: 665.30.
Now, determining the interest part of each monthly payment is simple. It is computed by multiplying the current loan balance by the effective interest rate per payment period.
Your initial loan balance is $100,000.00.
When your first monthly payment is due, the interest part is: $100,000.00 * 0.07/12 = $583.33.
Let's calculate the remaining part of the payment (the principal part), that goes toward the amount borrowed:
$665.30 - $583.33 = $81.97.
When your second monthly payment is due, your loan balance is: $100,000.00 - $81.97 = $99,918.03.
The interest part is: $99,918.03 * 0.07/12 = $582.86.
The remaining part of the payment (the principal part) is: $665.30 - $582.86 = $82.45.
The total interest paid is: $583.33 + $582.86 = $1,166.19.
Now we have the first two rows of the loan amortization table:
Amortization Table
Event Payment Interest Principal Total Interest Loan Balance
Loan $100,000.00
Payment # 1 $665.30 $583.33 $81.97 $583.33 $99,918.03
Payment # 2 $665.30 $582.86 $82.45 $1,166.19 $99,835.58
The process repeats each month...
Payment # 3 $665.30 $582.37 $82.93 $1,748.56 $99,752.66
Payment # 4 $665.30 $581.89 $83.41 $2,330.45 $99,669.24
Payment # 5 $665.30 $581.40 $83.90 $2,911.86 $99,585.34
...
Payment # 359 $665.30 $7.69 $657.61 $139,505.04 $661.44
Payment # 360 $665.30 $3.86 $661.44 $139,508.90 $0.00
How to compute the total interest paid over the life of the loan:
How much did you borrow?
$100.000.
How much did you pay?
360 monthly payments, $665.30 each time.
You paid: 360 * $665.30 = $239,508.
The total interest paid over the life of the loan = $239,508 - $100.000 = $139,508.
http://www.youtube.com/watch?v=Re23EpgGPQs
Here is a link on how to do it on a T-84. Look how many data point Mr. SuperComputer can work with just in his head.
Look Mr. Mensa I really think you are wasting your valuble intellect on a stoner forum.
