# Loan repayment and level amortization

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.

• 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 suspicious close to $500,000$. You might check if that also gives $1860.$ Apr 30, 2021 at 20:40
• @saulspatz If you could put your alternative method as an answer below, I can upvote it... Apr 30, 2021 at 20:57
• Actually, I now realize that there's a mistake in this calculation. You are ignoring the growth of the loan balance between payments. You can't say that the reduction in the balance is $(P - S_{29}) - (P - S_{28})$; it's actually $(P - S_{29}) - (1+j)(P - S_{28})$. I should have noticed this immediately, but it's been many, many years since I last practiced as an actuary. You always have to refer all cash flows to a single point in time. Apr 30, 2021 at 22:22
• Oh yes! My mistake, I forgot to account for the fact that $(P - S_{29}) - (P - S_{28})$ includes interest. Thank you for the correction. Apr 30, 2021 at 22:24

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

• Could you could go over your second method? I tried it with the numbers in OPs post but I cannot seem to make it work.
– user902292
Apr 30, 2021 at 21:05
• @Turnipsavocados Sure, just give me a few minutes. Apr 30, 2021 at 21:06
• I'm getting a wildly different answer than you did. I posted it, so as not to lose the MathJax, but I need to figure out what's causing the discrepancy. Apr 30, 2021 at 21:44
• I think I see it; why exactly does your formula denominator have $d$ rather than $i$?
– user902292
Apr 30, 2021 at 21:48
• It's an annuity due, because the first payment will be made immediately. Have you not learned about that yet? Apr 30, 2021 at 21:50

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

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

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

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

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

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

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

Therefore,

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

Therefore, the 61st loan reduction is

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