algebra-precalculusfinance
a) A $\$50,000$ mortgage is to be repaid by monthly payments for $20$ years. Determine the monthly payments if the nominal rate is $12\%$ converted monthly
b) an extra payment of $\$1,000$ is made at the end of each year. Determine the monthly payment.
I have solved (a) but am not sure on how to solve (b)
a) monthly interest rate = $1\%$ & period = $240$ months
$$\begin{align}payment &= \frac{PVr}{1-(1+r)^{-n}}\\ &= \frac{50,000\cdot.01}{1 – 1.01^{-240}}\\ &= $550.54\end{align}$$
Firstly we need a reference date for your payments and the payments of the bank. I´ve made a time line. The converted montly interest rate is $i_{12}=\frac{0.04}{12}$
Your payments start at the last day of January. We calculate the future value of your 12 payments at 31.12. We pretend that you start at 01.01, but make the payments at the end of each month. Then the future value is at $\color{red}{t=0} \ (31.12)$ This is the left hand side of the equation. Then we calculate the present value of the payments made by the bank. Since the payments ($+x$) start one month after $t=0$ we just discount the future value 6 times. The equation is
$$250\cdot\frac{(1+\frac{0.04}{12})^{12}-1}{\frac{0.04}{12}}=x\cdot\frac{(1+\frac{0.04}{12})^{6}-1}{\frac{0.04}{12}}\cdot\frac1{\left(1+\frac{0.04}{12}\right)^6}$$
$$3055.6157=x\cdot 5.930618 \Rightarrow x=\large{\color{grey}{£}} \ \normalsize 515.23 \ \ $$
As I said in a comment, the amount of the loan should be calculated as $$\require{enclose} L = 600 a^{(12)}_{\enclose{actuarial}{10} i} = 600\left(1-(1/1.01)^{120}\right)/0.01 = 41820.31 $$
You calculation of the total interest is correct. It's just the total payments less the amount of the loan.
The interest paid in the $10$ payment is $1\%$ of the amount of the loan outstanding after $9$ payments. Since there are then $111$ payments left, you do it it just as you calculated the amount of the loan, but with $111$ in the exponent instead of $120$.
For the amount of principal paid in the $20$ payment, calculate the amount of interest in the payment as above. The amount of principal in the payment is $600$ minus the amount of interest in the payment.
Best Answer
b) Without the yearly payment, the equation you used was derived using the present value of money:
$$P(1+r)^n = m \sum_{k=0}^{n-1} (1+r)^k$$
where $P$ is the principal and $r$ the monthly interest rate. If we allow a fixed yearly payment $f$ in addition, the equation becomes
$$P(1+r)^{12 y} = m' \sum_{k=0}^{12 y-1} (1+r)^k + f \sum_{k=0}^{y-1} (1+r')^k$$
where $y$ represents the number of years on the loan ($y=20$), and $r'$ is the effective annual rate of interest:
$$r' = (1+r)^{12}-1$$
Evaluating the geometric sums, we may solve for the new monthly payment $m'$:
$$m' = P \frac{r}{1-(1+r)^{-12 y}} - f \frac{r}{(1+r)^{12}-1}$$
Plugging in the numbers given above (i.e., $r=0.01$, $y=20$, $f=1000$,$P=50000$), we get a monthly payment of $m=471.69$.