8.4: Annuities
Sometimes it makes better financial sense to put small amounts of money away (to save for a period of time time), in order to purchase a large item in the future instead of taking out a loan now with a high interest rate.
Annuity
In the previous sections of this chapter, we examined problems where an amount of money was deposited lump sum in an account and was left there for the entire time period. Now we will solve problems where a series of payments of the same size are made into an account. When a series of payments of some fixed amount are made into an account at equal intervals of time, we call that an annuity . When payments are made at the end of each period rather than at the beginning, we call it an ordinary annuity .
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The Future Value, F, of an ordinary annuity is the amount in the account, including interest, after making all payments. \[F=PMT\cdot \dfrac{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}{\left(\dfrac{r}{m}\right)}\] where, \(F\) = Future value \(PMT\) = Periodic payment/deposit \(r\) = Annual percentage rate (APR) changed to a decimal \(t\) = Number of years \(m\) = Number of compounding periods per year |
Tanya deposits $300 at the end of each quarter in her savings account. If the account earns 5.75% compounded quarterly, how much money will she have in 4 years?
Solution
PMT =300, r =0.0575, t =4, m =4
The future value of this annuity can be found using the above formula.
\(F=300\dfrac{\left[\left(1+\dfrac{0.0575}{4}\right)^{4(4)}-1\right]}{\left(\dfrac{0.0575}{4}\right)}\)
\(F=300\dfrac{\left[\left(1+\dfrac{0.0575}{4}\right)^{16}-1\right]}{\left(\dfrac{0.0575}{4}\right)}\)
\(F=300(17.8463)\)
\(F=5353.89\)
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.
A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?
Solution
\(\begin{array}{ll} PMT= 100 & \text{the monthly deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ t = 20 & \text{we want the amount after 20 years} \\ m = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \end{array}\)
\(F=100\dfrac{\left[\left(1+\dfrac{0.06}{12}\right)^{20(12)}-1\right]}{\left(\dfrac{0.06}{12}\right)}\)
\(F=100\dfrac{\left((1.005)^{240}-1\right)}{(0.005)}\)
\(F=100\dfrac{(3.3102-1)}{(0.005)}\)
\(F=100\dfrac{(2.3102)}{(0.005)}=\$ 46204\)
The account will grow to $46,204 after 20 years.
Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned . In this case it is \(\$46,204 - \$24,000 = \$22,204\).
A more conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?
- Answer
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\(\begin{array}{ll} PMT= 5 & \text{the daily deposit} \\ r = 0.03 & 3\% \text{ annual rate} \\ t = 10 & \text{we want the amount after 10 years} \\ m = 365 & \text{since we’re doing daily deposits, we’ll compound daily} \end{array}\)
\(F=5\dfrac{\left[\left(1+\dfrac{0.03}{365}\right)^{365(10)}-1\right]}{\dfrac{0.03}{365}}=\$ 21,282.07\)
We would have deposited a total of \(\$ 5 \cdot 365 \cdot 10=\$ 18,250\), so $3,032.07 is from interest.
Sinking Fund
You may want to save regularly to have a fixed amount available in the future. The account that you establish for your deposits is called a sinking fund . Calculating the sinking fund deposit uses the same method as the previous problem. Observe that we are using the words deposit and payment synonymously.
Robert needs $5,000 in three years. How much should he deposit each month in an account that pays 8% compounded monthly in order to achieve his goal?
Solution
We need to find how much Robert should deposit, PMT, each month.
\(\begin{array}{ll} F= 5000 & \text{the future value} \\ r = 0.08 & 8\% \text{ annual rate} \\ t = 3 & \text{we want the amount after 3 years} \\ m = 12 & \text{compounded monthly} \end{array}\)
\(5000=PMT\dfrac{\left[\left(1+\dfrac{0.08}{12}\right)^{12(3)}-1\right]}{\left(\dfrac{0.08}{12}\right)}\)
Therefore,\[\begin{align*} PMT\frac{\left[\left(1+\dfrac{.08}{12}\right)^{36)}-1\right]}{\dfrac{.08}{12}} &=\ 5000 \\[4pt] \mathrm{PMT}(40.5356) &=\ 5000 \\[4pt] \mathrm{PMT} &=\frac{5000}{40.5356}\\[4pt] &=\ 123.3483 \end{align*}\]
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 5 years.
Solve the Annuity Formula for Payment, PMT
In the example above, notice we were given the future value and asked to find the payment, PMT . It would be helpful to solve the annuity formula for the payment first.
\(F=PMT\cdot \dfrac{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}{\left(\dfrac{r}{m}\right)}\)
---Since the PMT is multiplied by a fraction, solve for PMT by multiplying both sides of the equation by the reciprocal of that fraction.
\(F\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}=PMT \dfrac{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}{\left(\dfrac{r}{m}\right)}\cdot {\dfrac{\left(\dfrac{r}{m}\right)}{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}}\)
---Simply the right side of the equation, then rewrite as,
\(PMT=F\dfrac{\left(\dfrac{r}{m}\right)}{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}\)
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Use this formula when you know the future value and you want to find the payment, PMT . \(PMT=F\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}\) where, \(F\) = Future value \(PMT\) = Periodic payment/deposit \(r\) = Annual percentage rate (APR) changed to a decimal \(t\) = Number of years \(m\) = Number of compounding periods per year |
*Notes:
The compounding frequency is not always explicitly given, but is determined by how often you make payments.
In making payments into a sinking fund, we will always round the payment
up
to the next cent.
You want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?
Solution
We need to find the amount of the monthly deposit, \(PMT\).
\(\begin{array}{ll} F=\ 200,000 & \text{The amount you want to have in 30 years}\\ r = 0.08 & 8\% \text{ annual rate} \\ t = 30 & \text{30 years} \\ m = 12 & \text{compounded monthly} \end{array}\)
\(\begin{aligned}
&PMT=200,000\dfrac{\left(\dfrac{0.08}{12}\right)}{\left[\left(1+\dfrac{0.08}{12}\right)^{30(12)}-1\right]} \\
&PMT=200,000(6.7043\times 10^{-4}) \\
&PMT=134.086 \\
\end{aligned}\)
You would need to deposit \(\$134.09\) each month to have \(\$200,000\) in 30 years if your account earns 8% interest.
A business needs $450,000 in five years. How much should be deposited each quarter in a sinking fund that earns 9% compounded quarterly to have this amount in five years?
Solution
Again, suppose that \(PMT\) dollars are deposited each quarter in the sinking fund. After five years, the future value of the fund should be $450,000. This suggests the following relationship:
\(\begin{aligned}
&PMT=450,000\dfrac{\left(\dfrac{0.09}{4}\right)}{\left[\left(1+\dfrac{0.09}{4}\right)^{4(5)}-1\right]} \\
&PMT=450,000(0.040142) \\
&PMT=18063.9 \\
\end{aligned}\)
The business needs to deposit $18,063.90 at the end of each quarter for 5 years into a sinking fund earning 9% interest compounded quarterly in order to have $450,000 at the end of 5 years.
Solving for time, t
Sara wants to save $10,000 for a down payment on a new car. If she deposits $300 each month into an account earning 6.8% interest, how long will it take her to save up the $10,000 she needs?
F = $10,000, PMT = $300, r = 0.068, m = 12
Note: We will use the annuity formula and solve for t.
\(F=PMT\cdot \dfrac{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}{\left(\dfrac{r}{m}\right)}\)
\(10000=300\dfrac{\left[\left( 1+\dfrac{0.068}{12}\right)^{12 t}-1\right]}{\left(\dfrac{0.068}{12}\right)}\)
\(33.333333=\dfrac{(1.005667)^{12 t}-1}{0.005667}\)
\(1.188900=(1.005667)^{12 t}\)
To solve for t , time, take the logarithm (log) of both sides. Recall the “Power Rule” of logs from section 7.4 which allows us to bring the exponent out front.
\[\log x^{r} = r \log x\]
\(\log(1.188900)=12t\cdot \log(1.005667)\)
\(t=\dfrac{log(1.188900)}{12t\cdot \log(1.005667)}\)
It will take Sara about 2.6 years to save up the $10,000.
At the end of each quarter a 50-year old woman puts $1200 in a retirement account that pays 7% interest compounded quarterly. When she reaches age 60, she withdraws the entire amount and places it into a mutual fund that pays 9% interest compounded monthly. From then on she deposits $300 in the mutual fund at the end of each month. How much is in the account when she reaches age 65?
First, calculate the future value for the first 10 years.
PMT
= $1200,
r
= 0.07,
t
= 10,
m
= 4
\(F=PMT\cdot \dfrac{\left[\left( 1+\dfrac{r}{m}\right)^{m t}-1\right]}{\left(\dfrac{r}{m}\right)}\)
\(F=1200\dfrac{\left[\left( 1+\dfrac{0.07}{4}\right)^{4(10)}-1\right]}{\left(\dfrac{0.07}{4}\right)}\)
Second, she puts this lump sum plus $300 a month for 5 years at 9%. Think of the lump sum and the new monthly deposits as separate things. The lump sum just sits there earning interest so use the compound interest formula. The monthly payments are a new payment plan, so use the savings plan formula again.
Total = (lump sum + interest) + (new deposits + interest)
Total = 107,532.48 + 22,627.24 = 130,159.72
She will have $130,159.72 when she reaches age 65.