Question
My monthly mortgage payments are $4,369.66 for 30 years at 8%, what is the original cost of the house?
Answers
MathMate
We need to know how often interest is accrued. It is not always equal to the payment frequency. Some banks compound every 3 months, some six months, and some a year.
For lack of information, we will assume that interest is compounded monthly, which simplifies the calculations.
To do the calculations, we assume:
A=amount of payment per period (month) = $4369.66
P=principal, amount borrowed
i=interest per period (month)=0.08/12
n=number of periods (month) = 30*12=360
We equate the future value of the amount borrowed and the future value of the payments, as follows:
The first payment is assumed to be made at the end of the first period.
P(1+i)^n
=A(1+i)^(n-1)+A(1+i)^(n-2)...+A(1+i)^1+A(1+i)^0
The last term represents the last payment.
The right hand side factorizes to:
A((1+i)^n -1)/(1+i-1)
=A((1+i)^n -1)/i
So the whole equation becomes:
P(1+i)^n=A((1+i)^n -1) /i
Which means that
P=A((1+i)^n -1)/[i×(1+i)^n]
For lack of information, we will assume that interest is compounded monthly, which simplifies the calculations.
To do the calculations, we assume:
A=amount of payment per period (month) = $4369.66
P=principal, amount borrowed
i=interest per period (month)=0.08/12
n=number of periods (month) = 30*12=360
We equate the future value of the amount borrowed and the future value of the payments, as follows:
The first payment is assumed to be made at the end of the first period.
P(1+i)^n
=A(1+i)^(n-1)+A(1+i)^(n-2)...+A(1+i)^1+A(1+i)^0
The last term represents the last payment.
The right hand side factorizes to:
A((1+i)^n -1)/(1+i-1)
=A((1+i)^n -1)/i
So the whole equation becomes:
P(1+i)^n=A((1+i)^n -1) /i
Which means that
P=A((1+i)^n -1)/[i×(1+i)^n]