Pay back loan with an annual withdraw

Question

I was given question 7b as homework:

I am guessing that there are numerous ways of approaching this.

The one method I have tried was to calculate the effective interest year for the year.

Then come up with something like this:

$180000 = (x-2500)\cdot(1.1157^4-\frac{1}{1.1157})\cdot1.1157$

$X$ would represent the annual-payement thus I divide it by $12$ to get the monthly payment.

Yet my answer does not match that of the book being $4046.34$.

Answer 1

Let $r = 1 + \frac{0.11}{12} = 1.009166666$ be the monthly interest rate plus $1$ and, to simplify the equations below, let

$$R = \dfrac{r^{12} - 1}{r - 1}$$

Let $P_i$ be the principal after $i$ years, and let $X$ be the monthly repayment.

$$P_0 = 182500$$

$$P_1 = P_0 r^{12} - RX + 2500$$

$$P_2 = P_1 r^{12} - RX + 2500$$

$$P_3 = P_2 r^{12} - RX + 2500$$

$$P_4 = P_3 r^{12} - RX$$

$$0 = P_5 = P_4 r^{12} - RX$$

Therefore, repeatedly substituting these $P_i$ values:

\begin{eqnarray*} 0 &=& (P_3 r^{12} - RX) r^{12} - RX \\ && \\ &=& P_3 r^{24} - RX r^{12} - RX \\ && \\ &=& (P_2 r^{12} - RX + 2500) r^{24} - RX r^{12} - RX \\ && \\ &=& P_2 r^{36} - RXr^{24} + 2500 r^{24} - RX r^{12} - RX \\ && \\ &=& (P_1 r^{12} - RX + 2500) r^{36} - RXr^{24} + 2500 r^{24} - RX r^{12} - RX \\ && \\ &=& P_1 r^{48} - RXr^{36} + 2500 r^{36} - RXr^{24} + 2500 r^{24} - RX r^{12} - RX \\ && \\ &=& (P_0 r^{12} - RX + 2500) r^{48} - RXr^{36} + 2500 r^{36} - RXr^{24} + 2500 r^{24} - RX r^{12} - RX \\ && \\ &=& P_0 r^{60} - RXr^{48} + 2500 r^{48} - RXr^{36} + 2500 r^{36} - RXr^{24} + 2500 r^{24} - RX r^{12} - RX \\ && \end{eqnarray*}

Re-arranging this to solve for $X$ we get \begin{eqnarray*} RX(r^{48} + r^{36} + r^{24} + r^{12}) &=& P_0 r^{60} + 2500 r^{48} + 2500 r^{36} + 2500 r^{24} \\ && \\ X\dfrac{r^{60} - 1}{r-1} &=& P_0 r^{60} + 2500 r^{48} + 2500 r^{36} + 2500 r^{24} \\ && \\ X &=& \dfrac{(P_0 r^{60} + 2500 r^{48} + 2500 r^{36} + 2500 r^{24})(r-1)}{r^{60} - 1} \\ && \\ &=& \$4045.08 \end{eqnarray*}

There could be some rounding error explaining the discrepancy between this and the book's answer. There might be a simpler method than this or some shortcuts I'm not aware of - my background is in math rather than finance.

Answer 2

The idea is to treat the additional withdrawals as a separate cash flow, and to compute its present value: if $2500$ is borrowed immediately and at the beginning of each year for four years, then its present value is $$2500 \ddot a_{\overline{4}|i} = 2500 (1+i) \frac{1 - (1+i)^{-4}}{i},$$ where $i$ is the effective annual interest rate. Then the present value of the total loan is $180000 + 2500 \ddot a_{\overline{4}|i}$. The present value of the total series of payments is $K a_{\overline{60}|j}$, where $K$ is the monthly level payment at the end of each month, and $j$ is the monthly interest rate.