Calculate the mortagage he could assume for each amortization period

  • Thread starter Thread starter answerseeker
  • Start date Start date
  • Tags Tags
    Period
AI Thread Summary
Bob can afford a monthly mortgage payment of $575 at an interest rate of 6.75%. For a 15-year amortization period, the monthly payment per $1,000 is $8.85, leading to an estimated mortgage amount of approximately $64,972. Calculations involve using the formula for present value of future payments, factoring in the interest rate compounded monthly. Participants discussed various methods, including financial calculators and specific formulas to derive the mortgage amount. The consensus suggests that the calculated mortgage amount is around $64,978.40.
answerseeker
Messages
27
Reaction score
0
Bob estimates he can afford a monthly mortgage payment of $575. Current interest rates are 6.75%. Calculate the mortagage hecould assume for each amortization period.
A) 15 years

the extra info is that the monthly payments per $1000 for this percentage + 15years is $8.85.

How would you do this problem? :confused:
 
Mathematics news on Phys.org
Do you want to cheat
 
What have you done so far?
 
Actually most people who do this possesses a financial calculator. The formula is derived at http://www.moneychimp.com/articles/finworks/fmmortgage.htm

P(z^n)-a((z^n)-1)/(z-1) = debt remaining.

Here, z=(1+i), where i is interest per payment. a is the payment, P is the principal borrowed, and n is the number of payments made.
 
Last edited:
the formula used is A=Ao (1+ i) ^n
A= amount after
Ao=amount before ( principle amount)
i=interest rate
n=time period
 
How about trying 575/8.5*1000=64972. Otherwise, amount assumed equals present value of future payments where interest interest is compounded monthly.
Mortgage Amount= 575* Sum[((1+(.0675/12))^(-k)),k=1,2,...,12*15]=
575* [1-(1+(.0675/12))^(-15*12)]/(.0675/12).
 
answerseeker: the formula used is A=Ao (1+ i) ^n. That is not correct because we regularly subtract the monthly payment from the principal.

LittleWolf: How about trying 575/8.5*1000=64972. Sounds pretty good. Sum[((1+(.0675/12))^(-k)), As far as that goes, I don't think it does any good to sum.

My calculator, HP15C, takes it straight across from the formula I put previously. That is for this case:

P=\frac{a(z^n-1)}{z^n(z-1)}

Putting in $575 for a, 1.005625 for z, n = 180, since it is a monthly payment. Then, I arrive at $64978.40. REMEMBER: 6% is not 6, it is .06 in decimal form. Thus 6.75%/12 = .005625.
 
Last edited:
Back
Top