Finding present value without an intrest factor

Question

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?

Answer 1

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 $\frac{1}{ (1+i)}$

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

2. Subtracting (2) from (1):

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

3. Factoring out $C_0$ and $r$

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

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

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

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

5. 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: $1.06\cdot f(0.06;8)=1.06\cdot 6.2098=6.5824$