Annuity and Loan Repayment Question. Show the amount of Loan.

Question

A loan was taken out on 1 September 1998 and was repayable by the following scheme: The first repayment was made on 1 July 1999 and was £1000. Thereafter, repayments were made on 1 November 1999, 1 March 2000, 1 July 2000, 1 November 2000, etc until 1 March 2004, inclusive (note that the loan was fully repajd on 1 March 2004). Each payment was 5% greater than its predecessor. The effective rate of interest throughout the period was 6% per annum.

1)Calculate the effective rate of interest per month j;

2)Show that the amount of loan was £17692 to the nearest pound.

3)Calculate the amount of capital repaid on 1 July 1999.

4)Calculate both the capital component and the interest component of the seventh repayment (1 July 2001)

My attempts:

Take $\triangle t=4 months$

1)$1+0.06=(1+i)^{12}$ , so the effective rate is $i=0.486755$%

2)The first payment at $t=10$ months (1 July 1999) is $1000 u^{2.5}$ where $u=1/(1+i)$ and we take $\triangle t=4$ months

This means that the next payments at $t=14$ is $1000(1+0.05)\times u^{3.5}$

In total we have to make 15 payments and the 15th payment is $1000(1+0.05)^14\times u^{16.5}$

Factorize to get Present Value: $$PV=1000u^{2.5}(1+1.05u+1.05u^2+...+1.05^{14}u^{14})$$

I can rewrite $(1+1.05u+1.05u^2+...+1.05^{14}u^{14})$ as $\sum (1.05u)^k=\frac{1-(1.05u)^{k+1}}{1-10.5u}$

But when I do the computations I seem to get $PV=20520$

3) Simply Payment(1000) - the interest($i^{10}\times 17692$) , correct?

4) Find Loan Outstanding after 6th payment. This is $17692-1000u^{2.5}(1+1.05u+1.05u^2....1.0.5u^5)$. Calculate interest paid by multiplying by $i$ and deduct this from the 7th payment which is $1.05^61000$

Answer 1

Because of the timing of repayments relative to the time of the loan, let's look at the cash flow in monthly increments rather than 4-month increments. We note that if $PV$ is the present value of the loan, and $K(m) = 1000(1.05)^{m-1}$ is the amount of the $m^{\rm th}$ repayment, and there are a total of $n = 15$ repayments from the first payment 1 Jul 1999 to the last payment 3 Mar 2004, then the cash flow is $$PV = 1000v^{10} + 1000(1.05)v^{14} + 1000(1.05)^2 v^{18} + \cdots + 1000(1.05)^{14} v^{66},$$ where $v = (1+j)^{-1}$ is the present value discount factor for one month, and $j = (1+0.06)^{1/12} - 1 \approx 0.00486755$ is the monthly effective interest rate. This is because the first payment occurs $10$ months after the loan is taken, and each repayment occurs at a 4-month interval. Now we simplify: $$PV = 1000v^{10}\left( \sum_{m=0}^{14} (1.05)^m v^{4m} \right) = 1000v^{10} \sum_{m=0}^{14} (1.05 v^4)^m = 1000v^{10} \ddot a_{\overline{15}\rceil j^*}$$ where $j^*$ satisfies the relationship $$\frac{1}{1+j^*} = 1.05v^4.$$ This is because $1.05v^4 = v^*$ can be viewed as a "modified" present value discount factor taking into account the increasing size of the repayment over 4-month periods, thus corresponds to a similarly modified per-period interest rate $j^*$ for the purposes of using the annuity-due formula; thus $$PV = 1000v^{10} (1 + j^*) \left( \frac{1 - (1.05v^4)^{15}}{j^*} \right) \approx 17691.768479.$$

Answer 2

1. From $\left(1+\frac{i^{(12)}}{12}\right)^{{12}}=1+i$, we find $j=\frac{i^{(12)}}{12}=\left(1+0.06\right)^{\frac{1}{12}} -1=0.486755\%$
2. The loan on 1 September 1998 is the present value of annuity payments: $$ \begin{align*} L&=1000\left(v^{\frac{10}{12}}+(1.05)v^{\frac{14}{12}}+(1.05)^2 v^{\frac{18}{12}}+\cdots+(1.05)^{14}v^{\frac{66}{12}}\right)\\ &=1000 v^{\frac{10}{12}}\left[\frac{1-\left(1.05v^{\frac{4}{12}}\right)^{15}}{1-1.05v^{\frac{4}{12}}}\right]=17691.77\approx 17692 \end{align*} $$
3. The loan outstanding on 30/6/99 is $$L'=L\times(1+i)^{10}=17691.77\times (1.06)^{\frac{10}{12}}=18,572.04.$$ Thus the interest in first instalment is $I=L-L'=18,572.04-17,691.77=880.27$ and the amount of capital repaid on 1 July 1999 is $$K=1000-880.27=119.73.$$
4. The capital outstanding after 6 repayments is the present value of payments at 1/3/2001 $$ \begin{align*} K' &=1000\cdot (1.05)^6\left(v^{\frac{4}{12}}+(1.05)v^{\frac{8}{12}}+\cdots+(1.05)^{8}v^{\frac{36}{12}}\right)\\ &=1000\cdot (1.05)^6\cdot v^{\frac{4}{12}}\left[\frac{1-\left(1.05v^{\frac{4}{12}}\right)^{9}}{1-1.05v^{\frac{4}{12}}}\right]= 13,341.57 \end{align*} $$ The interest in 7th payment is $I'= 13,341.57 \left(1.06^{\frac{4}{12}}-1\right)= 261.67$ and the loan repaid in 7th instalment is $L''=1.05^6\times 1000-261.67= 1078.43.$