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
$$\begin{align}payment &= \frac{PVr}{1-(1+r)^{-n}}\\ &= \frac{50,000\cdot.01}{1 - 1.01^{-240}}\\ &= $550.54\end{align}$$
On
Consider the series of 20 payments of 1000 made annually as a mortgage in itself. The interest rate is quoted as 12% compounded monthly. This means that the equivalent effective annual rate to be used with the 1000 dollar payments is given by $$(1+.01)^{12}-1=0.12682503$$ Re-arranging your first equation for the regular payment to yield the present value, with $n=20, payment=1000, r=0.12682503$, we obtain 7161.00 . IOW, the extra payments will pay off 7161.00 of the 50000 mortgage, leaving 42,839.00 to be paid off by the regular monthly payments. Using your first equation in its original form, with this reduced present value, produces a payment of 471.69
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 $m=471.69$.