Calculate the mortagage he could assume for each amortization period

[answerseeker]

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?

 

[JamesU]

Do you want to cheat

 

[Icebreaker]

What have you done so far?

 

[robert Ihnot]

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.

 

[answerseeker]

the formula used is A=Ao (1+ i) ^n
A= amount after
Ao=amount before ( principle amount)
i=interest rate
n=time period

 

[LittleWolf]

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).

 

[robert Ihnot]

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.