Tyson wants to pay off some debt and decides to borrow money from James. He takes a $25$-year loan that is to be paid off in level amortization payments at the end of each quarter. If the nominal annual interest rate is $12$%, convertible monthly, and the principal reduction in the $29$th payment is $1860$, find the amount of principal reduction in the $61$st payment.
My attempt:
The nominal yearly interest is $12$% so the month effective interest rate is $\frac{12}{12} = 1$%. The quarterly effective interest, $j$ is
$$(1 + j)^{4} = (1 + 0.01)^{12}$$
$$j = 0.030301$$
Let us let $P$ be the amount borrowed in the beginning. This must be satisfied:
$$K \cdot a_{100, j} = P$$
, where $K$ is the level payment amount.
The payment amount from the $29$th payment is $1860$ so we have
$(P – S_{29}) – (P – S_{28}) = S_{28} – S_{29} = U_{29}$, where $S_n$ is the outstanding loan balance after the $n$th payment and $U_n$ is the reduction from the payment.
$K \cdot a_{72, j} – K \cdot a_{71, j} = 1860$
$K (\frac{1 – 1.030301^{-72}}{0.030301} – \frac{1 – 1.030301^{-71}}{0.030301}) = 1860$
$K = \frac{1860}{29.1552 – 29.0386}$
$K = 15956.20656$
Using $K$ we can find $P$:
$P = K \cdot a_{100, j} = 15956.20656 \cdot \frac{1 – 1.030301^{-100}}{0.030301} = 499979.1373$
The amount of principal reduction in the $61$st payment would be
$U_{61} = (P – S_{61}) – (P – S_{60}) = S_{60} – S_{61} = K \cdot a_{40, j} – K \cdot a_{39, j}$
$U_{61} = 15956.20656(\frac{1 – 1.030301^{-40}}{0.030301} – \frac{1 – 1.030301^{-39}}{0.030301})$
$U_{61} = 15956.20656(23.0027 – 22.6997)$
$U_{61} = 4834.6472$
Is this correct? I'm fairly new to loan repayment and amortization questions so any assistance is much appreciated.
Best Answer
I haven't checked the arithmetic, but the logic looks right to me. I'd point out that the amount of interest in any payment is $jB$ so the amount of principal is $B(1−j)$, where $B$ is the loan balance just before the payment. This might be a little easier than computing two successive balances. Also, the answer is suspiciously close to $500,000$. You might check if that also gives $1860$.
EDIT
Let $j$ be the periodic interest rate, $j=.030301$. Let $v=\frac1{1+j}=.970590.$ Let $d$ be the discount rate, $d=\frac j{1+j}=0.02940985$ Now, just before the $29$th payment, there are $72$ payments remaining, and the first of them is due at once, so the loan balance is $$K\frac{1-v^{72}}{d}=30.03860564K$$ The interest in the payment is $j$ times the loan balance, or $.91019978$K, so that the principal in the payment is $.089800210K$ This gives $K=20712.646349$ and $$P=K\frac{1-v^{100}}i=649019.62$$