# Calculate the mortagage he could assume for each amortization period

1. May 13, 2005

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? 2. May 13, 2005 ### JamesU Do you wanna cheat 3. May 14, 2005 ### Icebreaker What have you done so far? 4. May 14, 2005 ### robert Ihnot Actually most people who do this possess 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: May 14, 2005 5. May 16, 2005 ### answerseeker the formula used is A=Ao (1+ i) ^n A= amount after Ao=amount before ( principle amount) i=interest rate n=time period 6. May 16, 2005 ### 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). 7. May 17, 2005 ### 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.

Last edited: May 17, 2005
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