1. Annie Oakley is purchasing a home for $215,000. She will finance the mortgage for 15 years and pay 7% interest on the loan. She makes a down payment that is 20% of the purchase price. a. Find the monthly payment, including principal and interest. b. Calculate the total interest Annie will pay over the 15 year period. c. How much more interest would Annie pay by paying for the home in 30 years rather than 15 years? d. The annual taxes on Annie’s property come to $7100, and she pays $535 for insurance each year. Find her monthly PITI payment if she takes a 15 year loan.
P- Initial Balance W- Monthly Payment R- Monthly Interest Rate \begin{align} & 1 \text{st} \text{Month}\text{ }= P(1+R) -W\\&2 \text{nd} \text{Month}\text{ }=((P(1+R) -W)(1+R))-WP(1+R)^2-W(1+R)-W\\&3\text{rd} \text{Month}=\left(\left(P(1+R)^2-W(1+R)-W\right)(1+R)\right)-W\\&P(1+R)^3-W(1+R)^2-W(1+R)-W \end{align} \[N \text{th} \text{Month}= P(1+R)^N-W\left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)\] Using Geometric Sum \[\left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)=\frac{\left(1-(1+R)^N\right)}{1-(1+R)}=-\frac{1-(1+R)^N}{R}\] \[P(1+R)^N-W\left(-\frac{1-(1+R)^N}{R}\right)=P(1+R)^N+W\frac{\left(1-(1+R)^N\right) }{R}\]
To find a) p=215,000(1-.2)=172000 n=15*12=180 r=.07/12 \[P(1+R)^N+W\frac{\left(1-(1+R)^N\right) }{R}\] \[172000\left(1+\frac{.07}{12}\right)^{180}+W\frac{\left(1-\left(1+\frac{.07}{12}\right)^{180}\right) }{\frac{.07}{12}}=0\] Solve for W, I got W=$1545.98
b)Total Amount she will pay is 1545*180=278,277 principal was $172000 278,277-172000=$106277
c)same formula as A but 360 instead of 180 \[172000\left(1+\frac{.07}{12}\right)^{360}+W\frac{\left(1-\left(1+\frac{.07}{12}\right)^{360}\right) }{\frac{.07}{12}}=0\] Monthly payment is $1144.32 Multiply that by 360 to get total amount she paid =$411955 amount she paid for 30 years - amount she paid for 15 years $411955-$278,227=133728
15 years amortization table http://tinyurl.com/3drxnnc 30 years amortization table http://tinyurl.com/3msr3sm
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