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I am working on the following problems which should not involve a difficult polynomial, but I tried a couple of times without success.

Victor invests $300$ into a bank account at the beginning of each year for $20$ years. The account pays out interest at the end of every year at an effective annual interest rate of $i\%$. The interest is reinvested at an effective annual rate of $i/2 \%$. The yeild rate on the entire investment over the $20$ year period is $8\%$ effective annual. Determine $i$.

The following is what I have in mind.

1), For the first bank account, Victor invests $300$ every year and reinvests his interest into the second account, so the accumulated amount at the beginning of year $20$ must simply be

$$300 + 300 + \cdots +300 = 6000$$

2), For the second bank account, Victor reinvests the interest as follows. At the end of the first year he earns $i\%$ of $300$ thus $3i$. At the end of the second year his interest is $i\%$ of $600$ thus $6i$ ... up to his last investment $60i$.

I was not too comfortable going to the accumulated amount right away, so I used the present value as

$$X=3i(v+2v^2+ \cdots+20v^{20})$$

where $v=(1+(\frac{i}{2}\%))^{-1}$

I identified this as an increasing annuity present for 20 payments with rate $i/2\%$, so

$$\begin{align} X &= 3i(Ia)_{\overline{20}\rceil {\frac{i}{2}\%}}\\ & =3i \left(\frac{\ddot{a}_{\overline{20}\rceil {\frac{i}{2}\%}}-20v^{20}}{\frac{i}{2}\%}\right)\\ \end{align}$$

Multiplying this by $v^{-20}$ gives the accumulated value, so

$$6000+v^{-20}X=300(1.08)^{20}$$

is what I think we should be solving for, but since this is an Exam FM problem I don't think I should be solving for such a complex polynomial... is there a trick behind this problem?

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My interpretation of the question is as follows: the accumulated value of the reinvested interest plus the principal is equal to the accumulated value of a corresponding level payment annuity at 8%. Your last equation clearly cannot be what was intended, since the right-hand side of that equation merely corresponds to the accumulated value of a single payment of 300 after 20 years at 8%.

If each payment of $300$ is invested at an effective annual interest rate of $i$, then the total interest received at the end of each year is $300i, 600i, 900i, \ldots, 6000i$, and in particular, the $k^{\rm th}$ year-end interest received is $300ki$. Once reinvested at a rate of $i/2$, these payments have accumulated value $$300i(1+i/2)^{19} + 600i(1+i/2)^{18} + \cdots + 6000i = 300i(Is)_{\overline{20}\rceil i/2}.$$ The total principal is $20(300) = 6000$. Finally, the accumulated value of $300$ paid at the beginning of each year is $$300 \ddot s_{\overline{20}\rceil 0.08}.$$ Thus our equation of value is $$6000 + 300i(Is)_{\overline{20}\rceil i/2} = 300 \ddot s_{\overline{20}\rceil 0.08},$$ because on both sides of the equation, we are assuming the same investment schedule of $300$ at the beginning of each year. However, one could make the argument that the right-hand side should be $6000(1.08)^{20}$, since this represents the accumulated value of the total money Victor invested into the original account. The question is not absolutely clear on this.

Now recalling that $$(Is)_{\overline{n}\rceil} = \frac{\ddot s_{\overline{n}\rceil} - n}{i},$$ we get $$20 + 2\left(\ddot s_{\overline{20}\rceil i/2} - 20\right) = \ddot s_{\overline{20} \rceil 0.08},$$ or $$\ddot s_{\overline{20} \rceil i/2} = 10 + \tfrac{1}{2} \ddot s_{\overline{20}\rceil 0.08}.$$ This equation doesn't have an elementary closed form solution; the way to do it on the exam is to use numerical approximation or a solving method available on a financial calculator. This type of question is within the scope of Exam 2/FM. If you don't have a calculator that can solve for the interest rate, then one way is to use guess-and-check. First calculate the right-hand side, which is $34.7115$. Then if your calculator has an Ans button, you can do the following: guess a value for the interest rate and enter that into the calculator, say $i = 0.09$. Then type in

(1+Ans/2)((1+Ans/2)^20 - 1)/(Ans/2)

which will give you $\ddot s_{\overline{20}\rceil 0.045} = 32.7831$. This number is too small, so we need to try something larger, say $i = 0.12$, by entering this value into the calculator. Then scroll up twice to retrieve the previous formula you used, and recall that expression (so you don't have to retype everything). Since the formula uses the Ans button, it will automatically calculate $\ddot s_{\overline{20}\rceil 0.06} = 38.9927$ for $i = 0.12$. This is too large, so we guess $i = 0.10$ to get $\ddot s_{\overline{20}\rceil 0.06} = 34.7193$. This is very close to what we want, and we can refine the result further by using a few more guesses, to give a final answer of approximately $i = 0.09996$.

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  • $\begingroup$ Thank you, that makes things much clearer. Regarding the use of calculators, the book I am using mentions the use of a finance calculator so I am assuming that something like that might be available on the computer screen that I will be working on. Even if it is not, the solution for this particular question definitely is something that we can work with by guess and check and I am familiar with the use of my calculator so it actually sounds feasible. Especially since the exam will be a multiple choice. $\endgroup$ – hyg17 Nov 5 '14 at 4:18

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