Chapter 4 Finance
4.2 Annuities
Learning Objectives
By the end of this section, you will be able to:
- Find the future value of an annuity
- Find deposits needed to fund an annuity
- Find the interest earned
Savings Annuities
Most of us aren’t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a savings annuity. Most retirement plans like 401(k) plans or IRA plans are examples of savings annuities.
Suppose we will deposit $100 each month into an account paying 6% interest. How much will we have after a year? We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. In this example:
[latex]r = 0.06 \text{ (6%)}[/latex]
[latex]k=12 \text{ (12 compounds/deposits per year)}[/latex]
[latex]d = \$100 \text{ (our deposit per month)}[/latex]
[latex]t=1 \text{ year}[/latex]
With ordinary annuities we assume the payment is made at the end of the period. The $100 we deposit at the end of the first month will earn interest for 11 months and at the end of the year will be worth [latex]A=100(1+\frac{0.06}{12})^{11}=100(1.005)^{11}[/latex].
The $100 deposited at the end of the second month will have 10 months to grow and will be worth [latex]A=100(1.005)^{10}[/latex] at the end of the year. This pattern continues down to the last deposit, which has no time to compound and will be worth [latex]A = 100[/latex].
In total, we will have accumulated:
[latex]A=100(1.005)^{11}+100(1.005)^{10}+...+100(1.005)^{2}+100(1.005)^{1}+100[/latex]
This equation leaves a lot to be desired, though—it doesn’t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:
[latex]1.005A=1.005(100(1.005)^{11}+100(1.005)^{10}+...+100(1.005)^{2}+100(1.005)+100)[/latex]
Distributing on the right side of the equation gives
[latex]1.005A=100(1.005)^{12}+100(1.005)^{11}+...+100(1.005)^{3}+100(1.005)^{2}+100(1.005)[/latex]
Now we’ll line this up with like terms from our original equation and subtract each side:
[latex]\begin{array}{rcll} 1.005A & = & 100(1.005)^{12} + & 100(1.005)^{11}+...+100(1.005)\\ A & = & & 100(1.005)^{11} +...+ 100(1.005) + 100 \\ \end{array}[/latex]
Almost all the terms cancel on the right-hand side when we subtract, leaving
[latex]1.005A-A=100(1.005)^{12}-100[/latex]
Now we solve this equation for [latex]A[/latex].
[latex]A=\frac{100((1.005)^{12}-1)}{0.005}[/latex]
Recall 0.005 was [latex]\frac{r}{k}[/latex], 100 was the deposit [latex]d[/latex], and 12 was the number of months, [latex]kt[/latex]. Generalizing this result, we get the saving annuity formula.
Annuity Formula
[latex]A=\frac{d((1+\frac{r}{k})^{kt}-1)}{(\frac{r}{k})}[/latex]
[latex]A[/latex] is the balance in the account after t years.
[latex]d[/latex] is the regular deposit (the amount you deposit each year, each month, etc.).
[latex]r[/latex] is the annual interest rate in deci