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I invest $\$750$ in an annuity for eight years at $6\%$pa with interest compounded annually.

Find the present value of annuity to nearest dollar.

Now, this can simply be solved using an interest factor table so $750 \times 6.2098$

However, I tried an approach without interest factor:

Let $B_n$ be the balance after $n$ years.

Then $B_1 = 750$

$B_2 = 750(1.06)+750$

$B_3 = B_2 (1.06) + 750$ and so on up to $B_8$, which when plugged in a calculator loop gives incorrectly $7423$ dollars.

Is my approach at least in the right direction?

user71207
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1 Answers1

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More or less. Your approach is for the future value at $t=8$. To obtain the present value $B_0$ you have discount every n-th payment by $(1+i)^{n-1}$. I assume that every payment is made at the beginning of the year.

  • First payment: $B_0=750$
  • First two payments: $B_0=750+\frac{750}{1.06}$
  • First three payments: $B_0=750+\frac{750}{1.06}+\frac{750}{1.06^2}$
  • All 8 payments: $B_0=750+\frac{750}{1.06}+\frac{750}{1.06^2}+\ldots \frac{750}{1.06^7}$

In general the present value for the sum of all payments $C_0$ is

$$C_0=r+\frac{r}{ (1+i)}+\frac{r}{ (1+i)^2}+\ldots +\frac{r}{ (1+i)^{n-1}} \quad (1)$$

This is a kind of geometric series. See the next steps show how a closed formula can be derived.

  1. Multiplying the equation by $\frac{1}{ (1+i)}$

$$\frac{1}{ (1+i)}\cdot C_0=\frac{r}{ (1+i)}+\frac{r}{ (1+i)^2}+\ldots +\frac{r}{ (1+i)^{n-1}}+\frac{r}{ (1+i)^{n}} \quad (2)$$

  1. Subtracting (2) from (1):

$C_0-\frac{1}{ (1+i)}\cdot C_0=r-\frac{r}{ (1+i)^{n}}$

  1. Factoring out $C_0$ and $r$

$C_0\cdot \left(1-\frac{1}{ (1+i)}\right)=r\left(1-\frac{1}{ (1+i)^{n}}\right)$

$C_0=r\cdot \frac{1-\frac{1}{ (1+i)^{n}}}{1-\frac{1}{ (1+i)}}$

  1. Expanding the fraction by $(1+i)$

$C_0=r\cdot (1+i)\cdot \frac{1-\frac{1}{ (1+i)^{n}}}{1+i-1}=r\cdot (1+i)\cdot \frac{1-\frac{1}{ (1+i)^{n}}}{i}$

  1. Expanding the fraction by $(1+i)^n$

$$C_0=r\cdot (1+i)\cdot \frac{(1+i)^n-1}{i\cdot (1+i)^n}$$

This is the closed formula. Putting in the values we obtain:

$$C_0=750\cdot (1+0.06)\cdot \underbrace{\frac{(1+0.06)^8-1}{0.06\cdot (1+0.06)^8}}_{=6.2098}$$

Remark:

$(1+0.06)\cdot \frac{(1+0.06)^8-1}{0.06\cdot (1+0.06)^8}=6.5824$, because of the assumption that the payments are made at the beginning of each year.

If you assume that the payments are made at the end of each year, then you have $\frac{(1+0.06)^8-1}{0.06\cdot (1+0.06)^8}=6.2098$

So if the interest factor table is for payments which are made at the end of each period (year), then you have to multiply the factor $f(i;n)$ by (1+i) to obtain the right factor for your calculation.

In your case: $1.06\cdot f(0.06;8)=1.06\cdot 6.2098=6.5824$

callculus42
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  • hmm ok. Why do you discount every n-th payment? What does discount even mean - are you decreasing the yearly deposit? – user71207 May 08 '21 at 12:05
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    Not really. The payments are all $750$. The first payment at $t=0$ is $750$ and has a value at $t=0$ of 750. The second payment at $t=1$ is $750$ as well. But it has an equivalent value at $t=0$ of $\frac{750}{1.06}$. To see the equivalent you could imagine that you deposit this equivalent value at $t=0$ for one year you would get $\frac{750}{1.06}\cdot 1.06=750$ Conclusion: So all payments have a value of $750$ at the period where they are made. But to get the corresponding value at $t=0$ we discount all payments. The later the payments are made the more they have to be discounted. – callculus42 May 08 '21 at 13:37
  • @user71207 (continued) To sum all payments you need a reference date. In this case it is $t=0$ (now). – callculus42 May 08 '21 at 13:39
  • ok i see. So you are considering each deposit independently. But I am also interested as why my method doesn't work. Why can't we base each year off the previous year (as with my relation)? This would mean each deposit + its interest is accounted for in each period - as in compound interest. I can't understand why it woudln't work here. – user71207 May 08 '21 at 14:15
  • @user71207 Your reference date is $t=7$. So you have to divide your result by $(1.06)^7$. That´s it. See here the result for the factor. – callculus42 May 08 '21 at 14:24
  • To make clear what you have calculated. Your $B_1$ is the eighth payment at $t=7$. Then at B_2 you add the seventh payment at $t=6$ of 750, which has a value of $750\cdot 1.06$ at $t=7$ Then you add the sixth payment of 750 which has an equivalent value of $750\cdot 1.06^2$. And so on. At the end you have to discount the sum by $(1.06)^7$ to obtain the present value. – callculus42 May 08 '21 at 14:40
  • perhaps I'm not understanding what the "present value" is - you stated $t=0$ or $B_0$. I am thinking that this is the "initial" time ie no interest has been obtained even! The present value to me is the amount after eight years right? Why do I divide by $1.06^7$? – user71207 May 08 '21 at 14:44
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    @user71207 The present value to me is the amount after eight years right? No, at the present value and the the so called future value (t=7) all 8 payments are regarded. At the present value your reference date is $t=0$. That means that is the value of the 8 payments at t=0. At the future value ($\approx 7423$) you calculate the value of the 8 payments at $t=7$. To obtain the present value you divide the future value by $(1.06)^7$. – callculus42 May 08 '21 at 14:51
  • Ohhhh I see. So what I have calculated was indeed the future value. Thank you so much for helping me out. I understand now – user71207 May 08 '21 at 14:56
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    @user71207 It pleases me, that all is clear now. In general a timeline is very helpful for the imagination. With the follwing keywords you will obtain many hits at google pictures: timeline future present value – callculus42 May 08 '21 at 15:05