Annuities and Loans

Question

A loan for $8\, 000$ must be repaid with $6$ year end payments at an annual rate of $11 \%$. What is the annual payment? I know that the present value of an annuity with end payments is $\frac{1-v^n}{i}$ where $v = 1/(1+i)$. Likewise, the future value is $(1+i)^{n}\frac{1-v^n}{i}$. How do I use this to solve this problem?

Answer 1

We have to find the value of constant (I assume, since nothing is specified on the contrary) payments $A$ given the principal $P$ during $n$ yearly periods and the interest rate $i$. The value of $A$ in the period $k$ is equivalent to the present value $A/\left( 1+i\right) ^{k}$ monetary units, where $i$ is the interest rate in each capitalization period. Summing in $k$, from 1 to $n$, we get the sum

$$\displaystyle\sum_{k=1}^{n}\dfrac{A}{\left( 1+i\right) ^{k}}$$

This is a geometric progression with ratio $r=1/(1+i)$ and first term $u_{1}=A/\left( 1+i\right)$ whose sum is:

$$\dfrac{A}{1+i}\dfrac{\left( \dfrac{1}{1+i}\right)^{n}-1}{\dfrac{1}{1+i}-1}=A\dfrac{\left( 1+i\right) ^{n}-1}{i\left( 1+i\right) ^{n}}=P.$$

Hence

$$A=P\dfrac{i\left( 1+i\right) ^{n}}{\left( 1+i\right) ^{n}-1}.$$

For the given problem, the payments will be made during $n=6$ years, with $i=11\%=\dfrac{11}{100}$ and $P=8000$:

$$A=8000\times \dfrac{0.11(1.11)^{6}}{(1.11)^{6}-1}\approx 1891$$

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Sum of a geometric progression:

$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{n}$$

$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}$$

$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$

$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{n}\right) -\left( ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}\right) $$

$$(1-r)S=u_{1}-ru_{n}$$

$$S=\dfrac{u_{1}-ru_{n}}{1-r}=\dfrac{u_{1}-u_{1}r^{n}}{1-r}=u_1\times\dfrac{1-r^{n}}{1-r}=u_1\times\dfrac{r^{n}-1}{r-1}.$$