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I have two problems that I am having trouble with

  1. Using an interest rate of 5%, find the present value of 5000 payable in 10 years and the effective rate of discount between the 7th and 8th year.

My attempt:

$1/(1+(0.05)(10)) * 5000 = 3333.33$

So the present value is 3333.33, and the effective rate of discount is $[a(7) - a(6)]/(a7)$ = 3.7%.

I get confused on the effective rate of discount, is it the same as the effective rate of interest in terms of computing it?

  1. An amount 1000 is deposited into an account, earns 6% convertible quarterly for two years and earns continuous interest with δ(t) = $2/(1+t)^2$ . Find the accumulated amount after 5 years.

I have after two years: a(2) = $1000(1+0.06)^8$ = 1593.85

And then δ(3) = .125, so for the next 3 years, $1593.85(1.125)^3$ = 2,269.37

So the accumulated amount is 2,269.37? Or is the accumulated amount 2,269.37 - 1000?

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  1. Let $C=5000, \,i=5\%, n=10$. The present value is: $$ PV=\frac{C}{(1+i)^n}=\frac{5000}{1.05^{10}}\simeq 3069.57 $$ and the discount rate $d$ is found by $$ 1+i=\frac{1}{1-d}\Rightarrow d=\frac{i}{1+i}=\frac{0.05}{1.05}\simeq 4.76\% $$
  2. Let $C=1000,\,i^{(4)}=6\%,\,\delta(t)=\frac{2}{(1+t)^2}$. The future value after 5 years is $$ \begin{align*} FV&=C\cdot\overbrace{\left(1+\frac{i^{(4)}}{4}\right)^{4\times2}}^{\text{after 2 years}}\cdot \overbrace{\exp\left({\int_0^3\delta(t)\mathrm{d}t}\right)}^{\text{after 3 years}}\\ &=1000\cdot 1.015^8\cdot\exp\left(\frac{3}{2}\right)\\ &\simeq 5048.59 \end{align*} $$
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