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I am trying to find the present value of an annuity, where we have a $20$-year annuity immediate where the first payment is $1000$, and each subsequent payment is increased by either $50$ for the first $10$ payments or by $100$ for the last $10$ payments.

The 1st payment= $1000$ (at time 1)

Y2 TO Y10 payment = $1000 + 50$ = $1050$

Y11 TO Y20 payment = $1000 + 50 + 100$ = $1150$

I did the present value of each individual payment using the principle $PV = Pν + (P + Q)ν^{2} + (P + 2Q)ν^{3} + ... + (P + (n − 1)Q)ν^{n}$

$PV = 1000v +1050v^{2} + 1100v^{3} + ... + 1450v^{10} + 1600v^{11} + 1750v^{12} + 1900v^{13} +...+ 4300v^{20}$

then

$$\require{enclose} PV = 1000 [ 50a_{\enclose{actuarial}{10}i} + 150v^{10} a_{\enclose{actuarial}{10}i}] \tag{1}$$

where $i = 10$%

I wasn't sure if I derived the formula correctly. I appreciate your help.

comp890
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1 Answers1

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The way you describe the cash flow does not match the equation of value you wrote.

You say that the payments are increased by $50$ for each year for the first ten years, and then by $100$ for each year for the last ten years. But the way you wrote it, you're increasing the payments by $150$ in the last ten years. Which is it?

My interpretation of the question would have the payments be: $$1000, 1050, 1100, 1150, 1200, 1250, 1300, 1350, 1400, 1450, \\ 1550, 1650, 1750, 1850, 1950, 2050, 2150, 2250, 2350, 2450.$$ Note that the total increase from the first payment to the last is $50(9) + 100(10) = 1450 = 2450 - 1000$.

In such a case, there are multiple ways to write the equation of value. One way is to take the level payments of $1000$ out, and then what is left are two increasing annuities, one deferred by one year, and the other deferred by $10$ years. Refer to the following table:

$$\begin{array}{c|c|c|c|c} \text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\ \hline 1 & 1000 & 0 & 0 & 1000 \\ 2 & 1000 & 50 & 0 & 1050 \\ 3 & 1000 & 100 & 0 & 1100 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 10 & 1000 & 450 & 0 & 1450 \\ 11 & 1000 & 500 & 50 & 1550 \\ 12 & 1000 & 550 & 100 & 1650 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 20 & 1000 & 950 & 500 & 2450 \\ \end{array}$$ This gives us the equation of value $$\require{enclose} \begin{align} PV &= 1000(v + v^2 + \cdots + v^{20}) + (50v^2 + 100v^3 + \cdots + 950v^{20}) \\ &\qquad + (50v^{11} + 100v^{12} + \cdots + 500v^{20}) \\ &= 1000 a_{\enclose{actuarial}{20} i} + 50 v (Ia)_{\enclose{actuarial}{19}i} + 50 v^{10}(Ia)_{\enclose{actuarial}{10}i}. \end{align}$$

Alternatively, we can structure the payments as follows: $$\begin{array}{c|c|c|c|c} \text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\ \hline 1 & 950 & 50 & 0 & 1000 \\ 2 & 950 & 100 & 0 & 1050 \\ 3 & 950 & 150 & 0 & 1100 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 10 & 950 & 500 & 0 & 1450 \\ 11 & 1450 & 0 & 100 & 1550 \\ 12 & 1450 & 0 & 200 & 1650 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 20 & 1450 & 0 & 1000 & 2450 \\ \end{array}$$ This gives the equation of value $$PV = 950 a_{\enclose{actuarial}{10}i} + 1450 v^{10} a_{\enclose{actuarial}{10}i} + 50 (Ia)_{\enclose{actuarial}{10}i} + 100v^{10} (Ia)_{\enclose{actuarial}{10}i}.$$ You might ask, why would we write it this way instead of the previous? Well, notice that even though there is one more annuity symbol, there's actually only two different ones, rather than three in the first equation of value. In other words, things factor more nicely in the second equation: $$PV = 50\left( (19 + 29v^{10})a_{\enclose{actuarial}{10}i} + (1 + 2v^{10})(Ia)_{\enclose{actuarial}{10}i}\right).$$ In both cases, the answer is the same, approximately $11843.639297$. But no matter how you write the equation of value, you must use an increasing annuity symbol $(Ia)_{\enclose{actuarial}{n}i}$.

heropup
  • 135,869
  • thank you for clarifying and find where my mistake was. For some reason I'm getting 11299.51. – comp890 Mar 14 '21 at 19:22
  • I did $50[ (19+29(1/1.1)^{10}) ((1-(1.1)^{-10})/0.1) + (1+2(1/1.1)^{10}) (22.89134211)]$. For (Ia) I did $[(1-(1.1)^{-10})/0.1) - 10(1/1.1)^{10}]/0.1 = 22.89134211$ – comp890 Mar 14 '21 at 19:41
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    wait, I made a mistake I use a and not "a double dot" for the formula (Ia) – comp890 Mar 14 '21 at 19:51
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    @comp890 Your formula for $\require{enclose}(Ia){\enclose{actuarial}{n}i}$ is not correct. you can check by manually calculating $v + 2v^2 + \cdots + 10v^{10}$ and seeing it does not equal $22.8913$. The correct formula is $$(Ia){\enclose{actuarial}{n}i} = \frac{\ddot a_{\enclose{actuarial}{n}i} - nv^n}{i}.$$ I see you figured it out, nice job! – heropup Mar 14 '21 at 19:51