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I'm having trouble solving this FM problem.

For $3000$, Nick purchases an perpetuity-immediate paying $100$ at the end of each $6$ months period. For the same amount and for the same effective annual rate Paul purchase an annuity-immediate with $80$ quarterly payments that begin at amount $P$ and decreases by $1.1$ each quarter. Find $P$.

I honestly have no clue where to start.

So far I have:

Nick:

$3000 = 200*/i$

$i = 0.067$

For Paul:

Paul's effective rate is $j = (1.067)^{(1/4)} - 1 = 0.0163$

Is it $3000 = P*a_{80|0.0163}-(1.1*79)$

user270494
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  • First of all, maybe write the problem in a more structured form, say, a table? – Andrew Mar 18 '16 at 02:58
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    Also, it is better to not use the numbers right away, but instead write down the formulas first, together with the assignment of the coefficients to the given values. (reason in terms of symbols not numbers) – Andrew Mar 18 '16 at 04:10
  • Note on math typesetting: use curve brackets to group elements, like "a^{bc}" to get $a^{bc}$. – Andrew Mar 18 '16 at 04:13

1 Answers1

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Write out the cash flows.

Let $i$ be the effective annual interest rate; then Nick's cash flow is $$3000 = 100(v^{1/2} + v + v^{3/2} + v^2 + \cdots) = 200 a_{\overline{\infty}\rceil i}^{(2)} = 100 \cdot \frac{1}{v^{-1/2}-1}$$ where $v = 1/(1+i)$ is the effective annual present value discount factor. Therefore, $v = 900/961 \approx 0.936524$ and $i = 61/900 \approx 0.0677778$.

For Paul, the present value of his cash flow is $$\begin{align*}3000 &= Pv^{1/4} + (P-1.1)v^{2/4} + (P-2.2)v^{3/4} + \cdots + (P-86.9)v^{20} \\ &= (P+1.1)a_{\overline{80}\rceil j} - 1.1(Ia)_{\overline{80} \rceil j}, \end{align*}$$ where $$j = (1+i)^{1/4} - 1 \approx 0.01653$$ is the effective quarterly interest rate. Since $$a_{\overline{n}\rceil i} = \frac{1 - v^n}{i}, \quad (Ia)_{\overline{n}\rceil i} = \frac{\ddot a_{\overline{n}\rceil i} - nv^n}{i},$$ we obtain $$\begin{align*} P &= \frac{3000 + 1.1 (Ia)_{\overline{80}\rceil j}}{a_{\overline{80}\rceil j}} - 1.1 \\ &\approx \frac{3000 + (1.1)(1414.29)}{44.1989} - 1.1 \\ &\approx 101.973. \end{align*}$$

heropup
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  • @alexjo No. Check the math yourself: your equation $$3000 = 200/i_s$$ at $i_s = 0.067$ would correspond to an perpetuity-immediate of $200$ every six months. That clearly is not what the question stated. – heropup Mar 18 '16 at 19:32