6.3 Annuities and Sinking Funds

There are \(60\) deposits in this account. The first payment stays in the account for \(59\) months, the second for \(58\) months, the third for \(57\) months, and so on.

The first payment of \$500 accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{59} $$

The second payment accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{58} $$

The third payment accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{57} $$

The fourth payment accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{56} $$

And so on.

Finally, the next-to-last \((59^{\text{th}})\) payment accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right) $$

The last payment is made at the same time the money is taken out, so it earns no interest.

To find the total amount after five years, we add the accumulated values of all \(60\) payments:

$$ 500\left(1 + \frac{0.08}{12}\right)^{59} + 500\left(1 + \frac{0.08}{12}\right)^{58} + 500\left(1 + \frac{0.08}{12}\right)^{57} + \dots + 500 $$

Written in the forward order, this is

$$ 500 + 500\left(1 + \frac{0.08}{12}\right) + 500\left(1 + \frac{0.08}{12}\right)^2 + \dots + 500\left(1 + \frac{0.08}{12}\right)^{59} $$

This is a geometric series with first term \(a = 500\), common ratio

$$ x = 1 + \frac{0.08}{12} $$

and number of terms \(n = 60\). So the sum is

$$ A = \frac{500\left[\left(1 + \frac{0.08}{12}\right)^{60} - 1\right]}{0.08/12} $$ $$ A \approx 500(73.47686) = 36738.43 $$

So the final amount after five years is

$$ \$36,738.43 $$

Use the ordinary annuity formula with \(m = 100\), \(r = 0.08\), \(n = 4\), and \(t = 1\).

$$ A = \frac{100\left[\left(1 + \frac{0.08}{4}\right)^4 - 1\right]}{0.08/4} $$ $$ A = \frac{100[(1.02)^4 - 1]}{0.02} \approx 412.16 $$

So the account will contain approximately \(\$412.16\).

We use the future value formula for an ordinary annuity.

Here,

• \(m = 300\)
• \(r = 0.0575\)
• \(n = 4\)
• \(t = 4\)

So,

$$ A = \frac{300\left[\left(1 + \frac{0.0575}{4}\right)^{16} - 1\right]}{0.0575/4} $$ $$ A \approx 300(17.8463) = 5353.89 $$

If Tanya deposits \$300 into a savings account earning \(5.75\%\) compounded quarterly for 4 years, then at the end of 4 years she will have

$$ \$5,353.89 $$

Suppose Robert saves \(m\) dollars per month. After three years he will have

$$ \frac{m\left[\left(1 + \frac{0.08}{12}\right)^{36} - 1\right]}{0.08/12} $$

But Robert wants this amount to be \(5,000\). Therefore,

$$ \frac{m\left[\left(1 + \frac{0.08}{12}\right)^{36} - 1\right]}{0.08/12} = 5000 $$

Evaluating the factor gives

$$ m(40.5356) = 5000 $$

So,

$$ m = \frac{5000}{40.5356} = 123.35 $$

Robert needs to deposit

$$ \$123.35 $$

at the end of each month for 3 years into an account paying \(8\%\) compounded monthly in order to have \(5,000\) at the end of 3 years.

Use \(m = 150\), \(r = 0.06\), \(n = 12\), and \(t = 2\).

$$ A = \frac{150\left[\left(1 + \frac{0.06}{12}\right)^{24} - 1\right]}{0.06/12} $$ $$ A \approx 3814.79 $$

So the account will contain approximately \(\$3,814.79\).

Suppose \(m\) dollars are deposited each quarter. After five years, the future value of the fund should be \(450,000\). This gives the equation

$$ \frac{m\left[\left(1 + \frac{0.09}{4}\right)^{20} - 1\right]}{0.09/4} = 450000 $$

Evaluating the annuity factor gives

$$ m(24.9115) = 450000 $$

So,

$$ m = \frac{450000}{24.9115} = 18063.93 $$

The business needs to deposit

$$ \$18,063.93 $$

at the end of each quarter for 5 years into a sinking fund earning \(9\%\) compounded quarterly in order to have \(450,000\) at the end of 5 years.

Use the sinking fund equation with \(A = 3000\), \(r = 0.05\), \(n = 1\), and \(t = 3\).

$$ m = \frac{3000(0.05)}{(1.05)^3 - 1} $$ $$ m \approx 951.63 $$

So each yearly deposit should be about \(\$951.63\).

There are \(60\) deposits. The first payment stays in the account for \(60\) months, the second for \(59\) months, the third for \(58\) months, and so on.

The first payment of \(500\) accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{60} $$

The second accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{59} $$

The third accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right)^{58} $$

And so on.

The last payment stays in the account for one month, so it accumulates to

$$ 500\left(1 + \frac{0.08}{12}\right) $$

To find the total amount, we add the series

$$ 500\left(1 + \frac{0.08}{12}\right)^{60} + 500\left(1 + \frac{0.08}{12}\right)^{59} + 500\left(1 + \frac{0.08}{12}\right)^{58} + \dots + 500\left(1 + \frac{0.08}{12}\right) $$

Written forward, this is

$$ 500\left(1 + \frac{0.08}{12}\right) + 500\left(1 + \frac{0.08}{12}\right)^2 + \dots + 500\left(1 + \frac{0.08}{12}\right)^{60} $$

If we add \(500\) to this series and later subtract \(500\), the value does not change. So we get

$$ 500 + 500\left(1 + \frac{0.08}{12}\right) + 500\left(1 + \frac{0.08}{12}\right)^2 + \dots + 500\left(1 + \frac{0.08}{12}\right)^{60} - 500 $$

Except for the last subtraction, this is a geometric series with first term \(500\), common ratio \(1 + \frac{0.08}{12}\), and \(61\) terms. Therefore,

$$ A = \frac{500\left[\left(1 + \frac{0.08}{12}\right)^{61} - 1\right]}{0.08/12} - 500 $$ $$ A \approx 500(74.9667) - 500 = 37483.35 - 500 = 36983.35 $$

So, when payments are made at the beginning of each month, the final amount is

$$ \$36,983.35 $$

Use the annuity due formula with \(m = 100\), \(r = 0.06\), \(n = 12\), and \(t = 1\).

$$ A = \frac{100\left[\left(1 + \frac{0.06}{12}\right)^{13} - 1\right]}{0.06/12} - 100 $$ $$ A \approx 1239.72 $$

So the account will contain approximately \(\$1,239.72\).