I was wondering if someone could help me confirm the answer for the following problem regarding force of interest.
X deposits $1$ at time $t=0$ with force of interest $\delta_t=\frac{t^2}{k}$. Y deposits the same amount at the same time with nominal interest of $8\%$ per annum convertible semi annually. At $t=5$, their accumulated values are equal. What is $k$?
I have two questions.
1), "nominal interest of $8\%$ per annum convertible semiannually"
What does this mean? I want to say that interest is compounded $8\%$ per year so
$$(1+ \frac{i^{(2)}}{2})^2=1.08$$
But could it mean
$$i^{(2)}=8\%$$
?
If that is not the case, how do you describe the other case?
2), I tried the calculation by
$$\exp \left[ {\int_0^5\frac{t^2}{k}dt} \right] =(1.08)^5$$
which yields
$$k=\frac{125}{15 \ln (1.08)}$$
The book tells me the answer is $102$, which I think is not an approximation deviation.
Can someone explain me what's going on?
To get the result you have to use this definition of future value:
$FV_n=\frac{\text{DPV}}{(1-i)^n}$
DPV: Discounted present value
You have semi anually intervals. Thus the formula is
$FV_{5}=\frac{\text{1}}{(1-\frac{i}{2})^{5\cdot 2}}=\frac{\text{1}}{(1-\frac{0.08}{2})^{10}}$
Hence the equation is: $$\large{e^{\int_0^5 x^2/k}=\frac{1}{(1-0.04)^{10}}}$$
$\int_0^5 x^2/k=ln \left( \frac{1}{(1-0.04)^{10}} \right)$
$\frac{125}{3k}=-10*ln(0.96)$
$\frac{1}{k}=-\frac{30}{125}\cdot ln \left( 0.96 \right)$
$\Rightarrow k=-\frac{125}{30\cdot ln(0.96)}\approx 102$