Finance Unit

1.8 Savings Plans and Investments

An annuity is a sequence of equal payments made at equal time periods.

\(\begin{equation} A=P M T \times \frac{\left[\left(1+\frac{R}{n}\right)^{n T}-1\right]}{\left(\frac{R}{n}\right)} \end{equation} \)

Where A is the accumulated value of the annuity,

PMT is the deposit made at the end of each compounding period,

R is the annual interest rate written in decimal form,

n is the number of compounding periods per year,

and T is the number of years.

Ben and Arthur were friends who grew up together. They both knew that they needed to start thinking about the future. At age 19, Ben decided to invest $2,000 every year. He picked investment funds that averaged a 7% interest rate. Then, at age 30, Ben stopped putting money into his investments.

Now Arthur didn’t start investing until age 40. Just like Ben, he put $2,000 into his investment funds every year until he turned 65. He got the same 7% interest rate as Ben.

When both Ben and Arthur turned 65, they decided to compare their investment accounts. Who do you think had more?

  1. How much did Ben invest?
  2. $2,000 per year for 11 years = $22,000

  3. How much did Arthur invest?
  4. $2,000 per year for 25 years = $50,000

  5. How much did Ben have in his account at age 30?
  6. \(\begin{equation} A=\frac{2000 \cdot\left[\left(1+\frac{.07}{1}\right)^{11 \cdot 1}-1\right]}{(.07 / 1)}= \$ 31567.20 \end{equation} \)

  7. How much did Ben have in his account at age 65?
  8. \(\begin{equation} A=31567.20\left(1+ \frac{.07}{1}\right)^{35}= \$ 337029.78 \end{equation} \)

  9. How much did Arthur have in his account at age 65?
  10. \(\begin{equation} A=\frac{2000 \cdot\left[\left(1+\frac{.07}{1}\right)^{25 \cdot 1}-1\right]}{(.07 / 1)} = \$126498.08 \end{equation} \)

  11. At age 25, to save for retirement, you decide to deposit $300 at the end of each month into an IRA (Individual Retirement Account) that pays 6.5% compounded monthly. How much will you have in the IRA when you retire at age 65?
  12. \(\begin{equation} A=300 \cdot \frac{\left[\left(1+\frac{.065}{12}\right)^{12 \cdot 40}-1\right]}{(.065 / 12)} \end{equation}= \$ 685085.68\)

  13. How much would you need to deposit each month beginning at your child’s birth to have $20,000 in a college fund when your child turns 18 years old if you pay into an annuity that pays 7% interest compounded monthly?
  14. \(\begin{equation} 20000= \text{ PMT} \cdot \frac{\left[\left(1+\frac{.07}{12}\right)^{12 \cdot 18}-1\right]}{(.07 / 12)} \end{equation} \)

    \(\begin{equation} 20000= \text{ PMT} \cdot 430.7210266 \end{equation} \)

    \(\begin{equation} P M T=\$ 46.43 \end{equation} \)

  15. Which is better when saving for a graduation beach vacation 4 years from now? (Determine the future value in each case.)
    1. Deposit $100 per month under the mattress.
    2. \(\begin{equation} A=100 \cdot 4 \cdot 12= \$ 4800 \end{equation} \)
    3. Deposit $100 per month in an annuity paying 3.5% interest compounded monthly.
    4. \(\begin{equation} A=100 \cdot \frac{\left[\left(1+\frac{.035}{12}\right)^{12 \cdot 4}-1\right]}{\left(\frac{.035}{12}\right)} \end{equation} \)

      \(\begin{equation} A=\$ 5144.21 \end{equation} \)

    5. Deposit a lump sum of $4800 now into a savings account paying 3.5% interest compounded monthly.
    6. \(\begin{equation} A=4800\left(1+\frac{.035}{12}\right)^{12 \cdot 4} \end{equation} \)

      \(\begin{equation} A= \$ 5520.19 \end{equation} \)

  16. How much should you deposit at the end of each year into an IRA (Individual Retirement Account) that pays 5% compounded yearly to have $1 million when you retire in 50 years?
  17. \(\begin{equation} 1000000=P M T \cdot \frac{\left[\left(1+\frac{.05}{1}\right)^{1 \cdot 50}-1\right]}{\left(\frac{.05}{1}\right)} \end{equation} \)

    \(\begin{equation} 1000000= PMT \cdot 209.3479957 \end{equation} \)

    \(\begin{equation} P M T= \$ 4776.74 \end{equation} \)

    How much interest will you have earned?

    \(\begin{equation} 4776.74 \cdot 50= \$ 238837 \end{equation} \)

    \(\begin{equation} 1000000-238837= \$ 761163 \end{equation} \)