# Loan Repayments and Yield Rates

Jose loans Martin $12000$. Martin repays the loan by paying $5000$ at the end of two years and $10000$ at the end of $4$ years. The money received at time $t=2$ is immediately reinvested at an annual effective interest rate of $2.4$%. Find Martin's rate of interest and Jose's annual yield rate.

Well assuming the amount is compounded once every two years, then we can figure it out. The first payment will go partly towards interest and partly towards the principle (usually the interest is paid first). So let's assume a rate of $x$ compounded every two years. The interest due would then be $A_0x$ so the final balance after the initial payment of $p_0$ would be:

$$A_2 = A_0(1 + x) - p_0$$

Note that we know $A_0$ and $p_0$ here (but not $A_2$). Next we make another payment (two years later) which just so happens to be twice the initial payment meaning the new balance would be:

\begin{align} A_4 =& A_2(1 + x) - 2p_0\\ =& (A_0(1 + x) - p_0)(1 + x) - 2p_0 \\ =& A_0(1 + x) - p_0 + A_0(1 + x)x - p_0x - 2p_0 \\ =& A_0(1 + x + x + x^2) - p_0(1 + x + 2) \\ =& A_0(1 + 2x + x^2) - p_0(3 + x) \\ =& A_0x^2 + x(2A_0 - p_0) + (A_0 - 3p_0) \end{align}

Since the loan is payed off after 4 years, $A_4 = 0$ (there is no remaining balance). So this is a quadratic in the interest rate $x$ which can be solved using the quadratic equation:

$$x = \frac{p_0 - 2A_0 \pm \sqrt{(2A_0 - p_0)^2 - 4A_0(A_0 - 3p_0)}}{2A_0}$$

We know $A_0 = 12000$ and $p_0 = 5000$:

$$p_0 - 2A_0 = 5000 - 24000 = -19000 \\ A_0 - 3p_0 = 12000 - 15000 = -3000 \\ 4A_0(A_0 - 3p_0) = -48000 * 3000 = -144 * 1000^2 \\ (2A_0 - p_0)^2 = 361 * 1000^2 \\ (2A_0 - p_0)^2 - 4A_0(A_0 - 3p_0) = (361 + 144)*1000^2 = 505*1000^2 \\ \sqrt{(2A_0 - p_0)^2 - 4A_0(A_0 - 3p_0)} \approx 22472 \\ \frac{p_0 - 2A_0 + \sqrt{(2A_0 - p_0)^2 - 4A_0(A_0 - 3p_0)}}{2A_0} \approx \frac{-1900 + 22472}{24000} \approx 0.1447$$

So Martin's interest rate is approximately $14.47\%$ per two years, which would mean an annual interest rate of half of that: $\approx 7.23\%$.

We can check to make sure that's right:

$$A_2 \approx 12000 \cdot 1.1447 - 5000 = 8736.4 \\ A_4 \approx 8736.4 \cdot 1.1447 - 10000 = 0.55708$$

Notice that there is $55$ cents left out of a \$12,000 loan (because the interest rate of$14.47\%\$ isn't exact).

All interest rates are annual.

The borrower’s rate:

1. The residual debt after paying 5000 is $$12000(1+b)^2-5000$$
2. The residual debt after paying the remaining 10000 is zero:

$$[12000(1+b)^2-5000](1+b)^2=10000$$

this is a quadratic in $$x=(1+b)^2$$:

$$12x^2-5x-10=0, x=\frac{5+\sqrt{5^2-4\cdot12\cdot(-10)}}{2\cdot12}=\frac{5}{24}(1+\sqrt{21})$$

from here we deduce the borrower’s rate: $$b=\sqrt{\frac{5}{24}(1+\sqrt{21})}-1\approx 7\%$$

The lender’s return rate:

$$\left(\frac{5000(1+2.4\%)^2+10000}{12000}\right)^{\frac{1}{4}}-1\approx 6\%$$