# Accumulated Value of Investment

So I have been given this questions

€500 is invested over a period of 4-years. In year 1 a nominal rate of interest of 6% p.a. convertible quarterly applies. In year 2 a nominal rate of discount of 10% p.a. convertible 4-monthly applies. In year 3 a force of interest of 0.75% per month applies. In year 4 a compound rate of interest of 10% per 2-years applies. Calculate the accumulated value of the investment at the end of the 4-year period and the equivalent level annual force of interest over the 4-year period.

I started with year 1 $$i^{(4)} = 6\text{%}$$ $$p=4$$ from this $$i^{*} = \frac{i^{(p)}}{p}$$ $$i^{*} = \frac{i^{(4)}}{4} = 0.015$$ $$AV_{1} = 500(1+i^{*})^{4} = 530.6817753$$

We then move to year 2

- In year 2 a nominal rate of discount of 10% p.a. convertible 4-monthly applies.

$$p = 3$$

this is where my confusion arises, as it says discount rate. I assume I must discount back some payments, however i am unsure of which payments that it is I am supposed to discount back

Let $$A(n)$$ the accumulated value at the end of year $$n$$, starting from $$A(0)=€ \,500$$.
• $$i_1^{(4)}=6\%$$, so we have $$A(1)=A(0)\left(1+\frac{i_1^{(4)}}{4}\right)^4= 530.68$$
• $$d_2^{(3)}=10\%$$, so we have $$A(2)=\frac{A(1)}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}=587.49$$
• $$\delta_3=0.75\%$$, so we have $$A(3)=A(2)\;\textrm{e}^{\delta_3 \times 12}=642.82$$
• $$i_4^{(1/2)}=10\%$$, so we have $$A(4)=A(3)\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}=704.18$$ So we have $$A(4)=A(0)\underbrace{\frac{\left(1+\frac{i_1^{(4)}}{4}\right)^4}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}\;\textrm{e}^{\delta_3 \times 12}\;\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}}_{1.408351517}=704.18$$ and we find the equivalent force of interest $$\delta$$ as $$A(4)=A(0)\;\textrm{e}^{\delta \times 4}$$, i.e. $$\delta=\frac14\times\log\left(\tfrac{\left(1+\frac{i_1^{(4)}}{4}\right)^4}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}\;\textrm{e}^{\delta_3 \times 12}\;\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}\right)=\frac14\times\log(1.408351517)\approx 8.56\%$$
• In your answer you have $A(2) e^{\delta_3 \times 12} = 605.39$ When I attempted this calculation assuming you used $A(2) = 587.49$ and $\delta_3 =0.0075$ the answer I got was $A(3) = 642.82$. Just wondering if I have missed something or am I using the wrong figure somewhere? Thanks – Rito Lowe Oct 8 '18 at 15:35