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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

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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. $$ \delta=\frac14\times\log\left(\tfrac{\left(1+\frac{i_1^{(4)}}{4}\right)^4}{\left(1-\frac{d_2^{(3)}}{3}\right)^3}\;\textrm{e}^{\delta_3 \times 12}\;\left(1+\frac{i_1^{(1/2)}}{1/2}\right)^{1/2}\right)=\frac14\times\log(1.408351517)\approx 8.56\% $$
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  • $\begingroup$ The force of interest is constant $\endgroup$ – alexjo Oct 8 '18 at 15:26
  • $\begingroup$ 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 $\endgroup$ – Rito Lowe Oct 8 '18 at 15:35
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    $\begingroup$ You are right...my mistake! now corrected. $\endgroup$ – alexjo Oct 8 '18 at 16:13

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