Loan repayment and level amortization

Question

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.

Answer 1

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 $$P=K\frac{1-v^{100}}i=649019.62$$

Answer 2

I used an alternative approach to arrive at the same answer.

$L = $ amount of the loan.
$P = $ amount of the payment.

$j = $ the interest rate charged each quarter, where
$(1 + j)^4 = (1.01)^{12} \implies (1 + j) = (1.01)^3 = 1.030301.$

Let $v = \frac{1}{1+j}.$

Then $L$ represents the present value of the 100 payments.

Therefore $\displaystyle L = P (v + v^2 + \cdots + v^{100}) = Pv~\frac{1 - v^{100}}{1 - v}.$

$\displaystyle (1 - v^{100}) = 1 - \frac{1}{(1 + j)^{100}} = \frac{(1 + j)^{100} - 1}{(1 + j)^{100}}.$

$\displaystyle (1 - v) = 1 - \frac{1}{1 + j} = \frac{j}{1 + j}.$

Therefore

$$L = P \left(\frac{1}{1 + j}\right) \left[\frac{\frac{(1 + j)^{100} - 1}{(1 + j)^{100}}} {\frac{j}{1 + j}} \right] = P \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right]. \tag{1}$$

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At the time of the $(k)$-th payment, the loan has grown to $\displaystyle L(1 + j)^k.$

This has been offset by payments, whose value at the time of the $(k)$-th payment is

$\displaystyle P[(1+j)^{k-1} + (1+j)^{k-2} + \cdots + 1] = P\left[\frac{(1+j)^k - 1}{(1 + j) - 1}\right] = P\left[\frac{(1+j)^k - 1}{j}\right].$

This means that the loan balance, immediately after your $(k)$-th payment is

$\displaystyle L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right].$

During the period between the payments $(k)$ and $(k+1)$, the interest on this loan balance is

$\displaystyle \left\{L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right]\right\} \times j$

$\displaystyle = \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}.$

Therefore, the principal reduction for payment $(k+1)$ is

$\displaystyle P - \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}$

$\displaystyle = P - L(1 + j)^k j + P\left[(1+j)^k - 1\right]$

Using equation (1) above, this equals

$\displaystyle = P - P \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right](1 + j)^k j + P\left[(1+j)^k - 1\right]$

$\displaystyle = P\left\{1 - \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right](1 + j)^k j + \left[(1+j)^k - 1\right]\right\}$

$\displaystyle = P\left\{(1+j)^k - \left[\frac{(1 + j)^{100} - 1}{(1 + j)^{100 - k}} \right]\right\}$

$$= P(1+j)^{k-100}. \tag{2}$$

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Therefore,

$\displaystyle P(1 + j)^{[28-100]} = (1860) \implies P = (1860)(1+j)^{72}.$

Therefore, the 61st loan reduction is

$\displaystyle (1860)(1+j)^{72} \times (1+j)^{[60-100]} = (1860)(1+j)^{32} = 4834.647641.$