Ordinary Annuities

Section 5.4 Ordinary Annuities

An annuity is a sequence of equal payments made at equal time periods (unlike one-time deposits or payments like we previously considered). An ordinary annuity is an annuity in which the payments are made to the account at the end of each period.

Because annuities involve a sequence of payments, determining the future value of an annuity can involve adding a large number of terms together, which can be extremely tedious. Fortunately, using some algebra, we can find a nice formula to make our work much easier. So we begin with some albegra.

As a an example, consider the following equality. For any real number $x\text{,}$ we have,

\begin{equation*} (x-1)(1+x+x^2+x^3+x^4) = x^5 - 1. \end{equation*}

When $x \neq 1$ this is equivalent to

\begin{equation*} 1+x+x^2+x^3+x^4 = \dfrac{x^5-1}{x-1} \end{equation*}

Similarly

\begin{equation*} 1+x+x^2+x^3+x^4 + x^5 = \dfrac{x^6-1}{x-1}, \: \textrm{and} \end{equation*}

\begin{equation*} 1+x+x^2+x^3+x^4 + x^5 +x^6= \dfrac{x^7-1}{x-1} \end{equation*}

and so on. What will be useful to us here, is that the sum on the left can be computed using division. In general, the following is true:

Fact 5.4.1.

For any natural number $m$ and real number $x \neq 1\text{,}$

\begin{equation*} 1 + x +x^2+x^3+x^4+\cdot + x^{m-1} = \dfrac{x^m-1}{x-1}. \end{equation*}

Now, suppose we deposit $$P$ at the end of each period for $m$ total compounding periods at an interest rate of $i$ each period. Note that $i$ is NOT necessarily the annual rate. To compare with our previous work, if $r$ were the annual interest rate and there were $n$ compounding periods per year, then we would use $i = \dfrac{r}{n}$ as the interest rate per period. In other words, using our new notation, we can rewrite our compound interest model as follows. For principal $P\text{,}$ compounded $n$ times a year for $t$ years at an annual interest rate $r$

\begin{equation*} P \left( 1+ \dfrac{r}{n} \right)^{nt} = P\left(1+ i \right)^{m} \end{equation*}

Now, using our compound interest model, with $$P$ deposited at the end of each period,

Accumulated amount from deposit 1 = $P(1+i)^{m-1}$
Accumulated amount from deposit 2 = $P(1+i)^{m-2}$
Accumulated amount from deposit 3 = $P(1+i)^{m-3}$

$\vdots$

Accumulated amount from deposit m-3 = $P(1+i)^{3}$
Accumulated amount from deposit m-2 = $P(1+i)^{2}$
Accumulated amount from deposit m-1 = $P(1+i)$
Accumulated amount from deposit m = $P$

Let $F$ represent the future value of this ordinary annuity. Then $F$ will equal the sum of all these accumulated amounts. Therefore, if we let $x=1+i$ and use Fact 5.4.1

\begin{align*} F \amp = P + Px + Px^2 + Px^3 + \cdots + Px^{m-2} + Px^{m-1} \\ \\ F \amp = P(1+x+x^2+x^3 + \cdots x^{m-1}) \\ \\ F \amp = P\left(\dfrac{x^m - 1}{x-1} \right) \end{align*}

Substituting $x = 1 + i$ back into this expression gives the following ordinary annuity formula.

Fact 5.4.2.

Suppose $$P$ is deposited (or paid) at the end of each of $m$ total periods at an interest rate of $i$ per period. Then the future value, $F\text{,}$ of this ordinary annuity is

\begin{equation*} F = P\left(\dfrac{(1+i)^m - 1}{i} \right) \end{equation*}

Under these same conditions, we can rewrite the above formula in terms of $P$ to determine the periodic payments needed to reach a future value, $F\text{.}$ The payments would be determined by

\begin{equation*} P = F\left(\dfrac{i}{(1+i)^m - 1} \right). \end{equation*}

This expression is often referred to as a sinking fund formula.

Example 5.4.3.

Suppose $200 is deposited into an account at the end of each year at 4% compounded annually. How much will be in the account in 5 years? Solution.

We use the future value formula for an ordinary annuity (Item ) given in Fact 5.4.2 with $P=200\text{,}$ $i = .04\text{,}$ and $m=5\text{.}$

\begin{align*} F \amp = P\left(\dfrac{(1+i)^m - 1}{i} \right)\\ \\ F \amp = 200\left(\dfrac{(1+.04)^5 - 1}{.04} \right) \\ \\ F \amp \approx 1083.26 \end{align*}

In 5 years the account will grow to $1,083.26

Example 5.4.4.

ADTF Distributors needs $250,000 to purchase equipemnt for their new distribution center which will be open in 3 and half years. If ADTF Distributors earns 6% compounded monthly, how much does ADTF Distributors need to deposit at the end of each month to have $250,000 in the account at the end of 3 and a half years? Solution.

We use the sinking fund formula (Item ) given in Fact 5.4.2 with $F=250,000\text{,}$ $i = \dfrac{.06}{12} = .005\text{,}$ and $m=3.5 \cdot 12 = 42\text{.}$

\begin{align*} P \amp = F\left(\dfrac{i}{(1+i)^m - 1} \right)\\ \\ P \amp = 250,000\left(\dfrac{.005}{(1+.005)^{42} - 1} \right) \\ \\ P \amp \approx 5,364.05 \end{align*}

ADTF Distributors need to deposit $5,364.05 at the end of each month for 42 months to have $250,000 for equipment at the end of 3 and a half years.

Example 5.4.5.

To build their new distribution center, ADTF Distributors had to borrow $5 million. They recently issued bonds to raise money to help pay back the $5 million they borrowed. The bonds have a face value of $5 million and are due and payable in 12 years. Assuming they earn 7.4% quarterly on their deposits, how much should ADTF Distributors deposit at the end of each quarter in order to retire the bond issue at the end of year 12. Solution.

We use the sinking fund formula (Item ) given in Fact 5.4.2 with $F=5,000,000\text{,}$ $i = \dfrac{.074}{4} = .0185\text{,}$ and $m=12 \cdot 4 = 48\text{.}$

\begin{align*} P \amp = F\left(\dfrac{i}{(1+i)^m - 1} \right)\\ \\ P \amp = 5,000,000\left(\dfrac{.0185}{(1+.0185)^{48} - 1} \right) \\ \\ P \amp \approx 65,573.78 \end{align*}

ADTF Distributors needs to deposit $65,573.78 at the end of each quarter for 48 quarters to retire the bond issue at the end of year 12.

Example 5.4.6.

Cindy opens a money market at her bank that offers an annual rate of 1.11% compounded monthly. She contributes $950 a month.

How much will Cindy have in the money market at the end of year 10 years? Solution.
We use the future value formula for an ordinary annuity (Item ) given in Fact 5.4.2 with $P=950\text{,}$ $i = \dfrac{.011}{12}\text{,}$ and $m=10 \cdot 12 = 120\text{.}$
\begin{align*} F \amp = P\left(\dfrac{(1+i)^m - 1}{i} \right)\\ \\ F \amp = 950\left(\dfrac{\left(1+\dfrac{.011}{12}\right)^{120} - 1}{\dfrac{.011}{12}} \right) \\ \\ F \amp \approx 120,508.87 \end{align*}
By the end of year 10, Cindy will have $120,508.87.
After the 10 years is up, Cindy decides to invest the amount from her money market annuity into another account with annual interest rate 1.24% compounded monthly. If she continues to make her monthly deposits of $950, how much will she have in her account in another 5 years? Solution.
We need to do this in two steps:
- First, we use the compound interest formula Item 1 to determine the future value of $120,508.87. With $P=120,508.87\text{,}$ $r=.0124\text{,}$ and $n=12\text{,}$ the model we use is
  \begin{equation*} A(t) = 120,508.87\left(1+\dfrac{.0124}{12}\right)^{12t} \end{equation*}
  Evaluating at $t=5$ gives
  \begin{align*} A(5) \amp = 120,508.87\left(1+\dfrac{.0124}{12}\right)^{12 \cdot 5} \\ A(5) \amp \approx 128,212.80 \end{align*}
  The amount from Cindy's previous account will grow to $128,212.80 in another 5 years.
- Next we calculate the amount that will be generated by Cindy's continued deposits using the ordinary annuity formula (Item given in Fact 5.4.2) with $P=950\text{,}$ $i = \dfrac{.0124}{12}\text{,}$ and $m=5 \cdot 12 = 60\text{.}$
  \begin{align*} F \amp = P\left(\dfrac{(1+i)^m - 1}{i} \right)\\ \\ F \amp = 950\left(\dfrac{\left(1+\dfrac{.024}{12}\right)^{60} - 1}{\dfrac{.0124}{12}} \right) \\ \\ F \amp \approx 58,772.78 \end{align*}
  The amount generated from Cindy's continued deposits is $58,772.78
Finally, we add the two dollar amounts above to find the final value in Cindy's account:
\begin{equation*} $128,212.80 + $58,772.78 = $186,985.58 \end{equation*}

Let's now look at the present value of an ordinary annuity. The question is, how much should be invested presently at a periodic interest rate of $i$ to ensure a payment of $$P$ can be made from the account at the end of each of $m$ periods. To ensure $$P$ is in the account for the first payment (here, we are viewing $P$ as the future value; in other words $P$ is playing the role of $F$ in Item 3),

\begin{equation*} \dfrac{P}{(1+i)} \end{equation*}

must be depsoited presently. To ensure the second payment of $$P$ (again, $P$ is the future value) is in the account by the end of the second period

\begin{equation*} \dfrac{P}{(1+i)^2} \end{equation*}

must be depsoited presently. To ensure the third payment of $$P$ (again, $P$ is the future value) is in the account by the end of the third period

\begin{equation*} \dfrac{P}{(1+i)^3} \end{equation*}

must be depsoited presently. This pattern continues up to and including the final payment, for which

\begin{equation*} \dfrac{P}{(1+i)^m} \end{equation*}

must be depsoited presently to ensure the last payment can be made from the account. What we need to deposit, then, to ensure $m$ periodic payments of $P$ (at the end of each period), is

\begin{equation*} \dfrac{P}{(1+i)} + \dfrac{P}{(1+i)^2} + \dfrac{P}{(1+i)^3} + \cdots + \dfrac{P}{(1+i)^m}. \end{equation*}

Lertting $x=1+i$ and factoring a $\dfrac{P}{x}$ from each term, the present value, $PV\text{,}$ of this ordinary annuity has the form

\begin{equation*} PV = \dfrac{P}{x} \left( 1 + \dfrac{1}{x} + \left(\dfrac{1}{x}\right)^2 + \cdots + \left(\dfrac{1}{x}\right)^{m-1} \right) \end{equation*}

where $P$ is the periodic pay-out amount. Using Fact 5.4.1 this can be rewritten as

\begin{equation*} PV = \dfrac{P}{x} \left( \dfrac{ 1 - \left( \frac{1}{x}\right)^m }{1-\frac{1}{x}}\right) = P\left(\dfrac{1-x^{-m}}{x-1}\right). \end{equation*}

Substituting $1+i$ back in for $x$ gives the following:

Fact 5.4.7.

Suppose $$P$ is paid (or deposited) at the end of each of $m$ total periods at an interest rate of $i$ per period. Then the present value, $PV\text{,}$ of this ordinary annuity is

\begin{equation*} PV = P\left(\dfrac{1-(1+i)^{-m}}{i} \right) \end{equation*}

Under these same conditions, we can rewrite the above formula in terms of $P$ to determine the periodic payments associated with an ordinary annuity with present value, $PV\text{.}$ The payments would be determined by

\begin{equation*} P = PV \left(\dfrac{i}{1-(1+i)^{-m}} \right). \end{equation*}

Loan amortization is the process of paying off the interest and principal balance on a loan with regular payments over time. If the payments are made at the end of the period, then the payment formula, Item , for $P$ above will provide the payment amount needed to pay off a loan of $$PV$ at periodic interest rate $i$ over $m$ periods. Also observe that if both sides of the expression for $PV$ above (Item ) are multiplied by $(1+i)^m$ to find the future value $F$ of the loan under compound interest, we arrive back at the future value formula,Item , for an oridinary annuity.

In general, equity is defined to be total assets minus total liabilities. For a homeowner or property owener, equity would be the (market) value of the home or property less the amount still owed. To calculate the amount still owed (the outstanding debt), we view the remaining payments left on the laon as an ordinary annuity, and find its present value. This amount would be the amount still owed on the loan. Consider the following example.

H.E. Pennypacker, a wealthy American industrialist, is looking to purchase a sliver mine in the mountains of Peru. In need of loan, Pennypacker finances the entire value of the mine and borrows $500,000,000 from Varnsen Savings and Loan. The terms of the loan stipulate that Pennypacker pay an annual interest rate of 2.25% compounded monthly for 10 years.

Let's determine Pennypacker's monthly payments to Varnsen Savings and Loan. We can think of this two ways:

We can view $500,000,000 as the present value of an ordinary annuity formed by Pennypacker's sequence of payments.. In this case, using Item to find the amortization payments,
\begin{equation*} P = 500,000,000 \left( \dfrac{\frac{.0225}{12}}{1-\left(1+\dfrac{.0225}{12}\right)^{-120}}\right) \approx 4,656,868.59 \end{equation*}
so that Pennypacker's monthly payments are $4,656,868.59.
Using the compound interest model
\begin{equation*} A(t) = 500,000,000\left(1+\frac{.0225}{12}\right)^{12t} \end{equation*}
we find the future value of the loan to be
\begin{equation*} F = A(10) = 500,000,000\left(1+\frac{.0225}{12}\right)^{12 \cdot 10} \approx 626,029,455.95, \end{equation*}
or $626,029,455.95. Viewing this as the future value of an ordinary annuity and using Item to find the payments,
\begin{equation*} P = 626,029,455.95 \left( \frac{ \frac{.0225}{12} }{ \left(1+\frac{.0225}{12}\right)^{120} - 1} \right) \approx 4,656,868.59, \end{equation*}
and again we find Pennypacker's payments are $4,656,868.59.

So, if Pennypacker pays the loan in 120 monthly payments of $4,656,868.59, he will end up paying

\begin{equation*} 4,656,868.59(120) - 500,000,000 = 58,824,230.92 \end{equation*}

or $58,824,230.92 in interest.

Suppose that Pennypacker wants to pay off his loan halfway through the term, with 60 payments remaining. How much does Pennypacker owe Varnsen Savings and Loan? To determine the amount owed, we find the present value of the annuity defined by these 60 remaining payments using Item :

\begin{equation*} PV = 4,656,868.59 \left(\frac{1-\left(1+\frac{.0225}{12}\right)^{-60}}{ \frac{.0225}{12}}\right) \approx 264,034,561.60 \end{equation*}

If Pennypacker wants to pay off the laon at the end of the fifth year (after 60 payments), he owes Varnsen Savings and Loan $264,034,561.60. In this case, Pennypacker paid

\begin{equation*} (4,656,868.59(60)+264,034,561.60)- 500,000,000 = 43,446,677.06 \end{equation*}

or $43,446,677.06 in interest.

We can also use the present value, $PV\text{,}$ above to calculate Pennypacker's equity after the fifth year of the laon. Assuming that the value of the silver mine remained constant, Pennypacker's equity after his 60th payment is

\begin{align*} \textrm{Equity} \: \amp = \: \textrm{Value} \: - \: \textrm{Amount Owed} \\ \textrm{Equity} \: \amp = 500,000,000 - 264,034,561.60 \\ \textrm{Equity} \: \amp = 235,965,438.40 \end{align*}

Pennypacker's equity after the fifth year of the laon is $235,965,438.40 or about 47%.

Business Mathematics: Applications Using Algebra and Calculus

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