Help Mortgage payment

• Feb 6th 2008, 02:54 AM
crazykitty
Help Mortgage payment
Hi!!
In one of my assignments, we had to figure out monthly mortgage payments.

I used a calculator because I do know know how to show the work.

The question was getting a 30year mortgage@ 100,000 with a 6.5% rate. What would be monthly payment. Answer is $632.07 Next question is using a 7.125% interest rate, answer is$673.72
Next question is a 15 yr loan, at 6.125%

My text book does not show a formula but i went online and found one but dont know how to figure it out Im pasting it below. THANKYOU :)
• Feb 6th 2008, 02:55 AM
crazykitty
formula i said I would include
Im losing my mind!!

Here is the formula I found, if anyone knows how to work it

First you must define some variables to make it easier to set up:
# P = principal, the initial amount of the loan
# I = the annual interest rate (from 1 to 100 percent)
# L = length, the length (in years) of the loan, or at least the length over which the loan is amortized.

The following assumes a typical conventional loan where the interest is compounded monthly. First I will define two more variables to make the calculations easier:
# J = monthly interest in decimal form = I / (12 x 100)
# N = number of months over which loan is amortized = L x 12

Okay now for the big monthly payment (M) formula, it is:

J
M = P x ------------------------

1 - ( 1 + J ) ^ -N

where 1 is the number one (it does not appear too clearly on some browsers)

So to calculate it, you would first calculate 1 + J then take that to the -N (minus N) power, subtract that from the number 1. Now take the inverse of that (if you have a 1/X button on your calculator push that). Then multiply the result times J and then times P. Sorry, for the long way of explaining it, but I just wanted to be clear for everybody.

The one-liner for a program would be (adjust for your favorite language):
• Feb 6th 2008, 05:26 AM
Soroban
Hello, crazykitty!

Sorry, I don't understand that formula you found.

This is an Amortization problem.
This is the standard formula: . $\displaystyle A \;=\;P\cdot\frac{i(1+i)^n}{(1+i)^n-1}$

where: .$\displaystyle \begin{Bmatrix} A &=&\text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=&\text{number of periods} \end{Bmatrix}$

Quote:

The question was getting a 30-year mortgage for $100,000 with a 6.5% rate. What would be monthly payment? Answer:$632.07

We have: .$\displaystyle P \:=\: \$100,000,\quad i \:=\: \frac{0.065}{12}\:\approx\:0.0054167\,\quad n \:=\: 360$Then: .$\displaystyle A \;=\;\$100,000\cdot\frac{0.0054167\,(1.0054167)^{3 60}}{(1.0054167)^{360} - 1} \;=\;632.070... \;\approx\;\$632.07$Quote: Next question is using a 7.125% interest rate. Answer:$673.72

We have: .$\displaystyle P \:=\:\$100,000,\quad i \:=\:\frac{0.7124}{12}\:=\:0.0059375,\quad n \:=\:360$Then: .$\displaystyle A \;=\;\$100,000\cdot\frac{0.0059375\,(1.0059375)^{3 60} } {(1.0059375)^{360} - 1} \;=\;673.718... \;\approx\;\$673.72 $Quote: Next question is a 15 yr loan, at 6.125% We have: .$\displaystyle P \:=\:\$100,000,\quad i \:=\:\frac{0.06125}{12} \:\approx\:0.005104,\quad n \:=\:180$

Then: .$\displaystyle A \;=\;\$100,000\cdot\frac{0.005104\,(1.005104)^{180 }}{(1.005104)^{180}-1} \;=\;850.641... \;\approx\;\$850.64$

• Feb 6th 2008, 07:45 PM
crazykitty
Hey!
THANKYOU SO MUCH!!! : )

I dont know how to thank you all!!! This forum (Rock)!!!