6.4 Compound Interest

Since the interest is compounded annually, the interest will be computed at the end of each year and added to the CD’s value. The interest at the end of the following year will be based on the value found form the previous year.

Step 1: After the first year, the interest in Abena’s CD is computed using the interest formula $I=P\times r\times t$. The principal is $P$ = 1,000, the rate, as a decimal, is 0.04, and the time is one year, so $t$ = 1. Using that, the interest earned in the first year is $I=P\times r\times t=\mathrm{1,000}\times 0.04\times 1=40$, so the interest earned in the first year was $40.00. This is added to the value of the CD, making the CD worth $\text{\$}\mathrm{1,000}+\text{\$}40=\text{\$}\mathrm{1,040}$.

Step 2: At the end of the second year, interest is again computed, but is computed based on the CD’s new value, $1,040. Using this new value and the interest formula ($r$ and $t$ are still 0.04 and 1, respectively), we see that the CD earned $I=P\times r\times t=\mathrm{1,040}\times 0.04\times 1=41.6$, or $41.60. This is added to the value of the CD, making the CD now worth $\text{\$}\mathrm{1,040.00}+\text{\$}41.60=\text{\$}\mathrm{1,081.60}$.

Step 3: At the end of the third year, interest is again computed, but is computed based on Abena’s CD’s new value, $1,081.60. Using this value and the interest formula ($r$ and $t$ are still 0.04 and 1, respectively), we see that the CD earned $I=P\times r\times t=\mathrm{1,081.60}\times 0.04\times 1=43.264$, or $43.26 (remember to round down). This is added to the value of the CD, making the CD now worth $\text{\$}\mathrm{1,081.60}+\text{\$}43.26=\text{\$}\mathrm{1,124.86}$.

After 3 years, Abena’s CD is worth $1,124.86.

Had Abena invested $1,000 in a 4% simple interest CD for 3 years, her CD would have been worth $P+P\times r\times t=\mathrm{1,000}+\mathrm{1,000}\times 0.04\times 3=\mathrm{1,120}$, or $1,120.00. With interest compounded annually, Abena’s CD was worth $1,124.86. The difference between compound and simple interest is $\text{\$}\mathrm{1,124.86}-\text{\$}\mathrm{1,120.00}=\text{\$}4.86$. So compound interest earned Abena $4.86 more than the simple interest did.

1. The principal is $P$ = $5,000, interest rate, in decimal form, $r$ = 0.038, compounded monthly so $n$ = 12, and for $t$ = 5 years. Substituting these values into the formula, we find
   $$\begin{array}{ccc}\hfill A& \hfill =\hfill & P{\left(1+\frac{r}{n}\right)}^{nt}=5,000{\left(1+\frac{0.038}{12}\right)}^{12\times 5}\hfill \\ \hfill & \hfill =\hfill & 5,000{\left(1+0.0031\overline{6}\right)}^{60}\hfill \\ \hfill & \hfill =\hfill & 5,000{\left(1.0031\overline{6}\right)}^{40}\hfill \\ \hfill & \hfill =\hfill & 5,000\times 1.20888663572\hfill \\ \hfill & \hfill =\hfill & 6,044.4332\hfill \end{array}$$
   
   The future value of the investment is $6,044.43.
   

2. The principal is $P$ = $18,500, interest rate, in decimal form, $r$ = 0.0625, compounded quarterly so $n$ = 4, and for $t$ = 17 years. Substituting these values into the formula, we
   $$\begin{array}{ccc}\hfill A& \hfill =\hfill & P{\left(1+\frac{r}{n}\right)}^{nt}=18,500{\left(1+\frac{0.0625}{4}\right)}^{4\times 17}\hfill \\ \hfill & \hfill =\hfill & 18,500{\left(1+0.015625\right)}^{68}\hfill \\ \hfill & \hfill =\hfill & 18,500{\left(1.015625\right)}^{68}\hfill \\ \hfill & \hfill =\hfill & 18,500\times 2.86992151999\hfill \\ \hfill & \hfill =\hfill & 53,093.5481\hfill \end{array}$$
   The future value of the investment is $53,093.54.

Cody’s initial investment is $7,500, so $P$ = $7,500. The annual interest rate is 4.5%, which is 0.045 in decimal form. Compounding quarterly means there are four periods in a year, so $n$ = 4. He invests the money for 10 years. Substituting those values into the formula, we calculate

After 10 years, Cody’s initial investment of $7,500 is worth $11,732.82.

Kathy’s initial investment is $10,000, so $P$ = $10,000. The annual interest rate is 5.6%, which is 0.056 in decimal form. Compounding daily means there are 364 periods in a year, so $n$ = 365. She invests the money for 20 years, so $t$ = 20. Substituting those values into the formula, we calculate

After 20 years, Kathy’s initial investment of $10,000 is worth $30,645.90.

1. To reach a final account value of $A$ = $250,000, invested at 6.75% interest, in decimal form $r$ = 0.0675 (decimal form!), compounded monthly, so $n$ = 12, for 30 years, substitute those values into the formula for present value. Calculating, we find the present value of the $250,000.
   
   $$\begin{array}{ccc}\hfill PV& \hfill =\hfill & \frac{A}{{\left(1+\frac{r}{n}\right)}^{n\times t}}=\frac{\mathrm{250,000}}{{\left(1+\frac{0.0675}{12}\right)}^{12\times 30}}\hfill \\ \hfill & \hfill =\hfill & \frac{\mathrm{250,000}}{{\left(1+0.005625\right)}^{360}}\hfill \\ \hfill & \hfill =\hfill & \frac{\mathrm{250,000}}{{\left(1.005625\right)}^{360}}\hfill \\ \hfill & \hfill =\hfill & \frac{\mathrm{250,000}}{7.5332454772}\hfill \\ \hfill & \hfill =\hfill & \mathrm{33,186.2277}\hfill \end{array}$$
   
   In order for this account to reach $250,000 after 30 years, $33,186.23 needs to be invested.
   

2. To reach a final account value of $A$ = $500,000, invested at 7.1% interest, in decimal form $r$ = 0.071, compounded quarterly, so $n$ = 4, for 40 years, substitute those values into the formula for present value. Calculating, we find the present value of the $500,000.
   
   $$\begin{array}{ccc}\hfill PV& \hfill =\hfill & \frac{A}{{\left(1+\frac{r}{n}\right)}^{n\times t}}=\frac{500,000}{{\left(1+\frac{0.071}{4}\right)}^{4\times 40}}\hfill \\ \hfill & \hfill =\hfill & \frac{500,000}{{\left(1+0.01775\right)}^{160}}\hfill \\ \hfill & \hfill =\hfill & \frac{500,000}{{\left(1.01775\right)}^{160}}\hfill \\ \hfill & \hfill =\hfill & \frac{500,000}{16.6946672846}\hfill \\ \hfill & \hfill =\hfill & 29,949.6834\hfill \end{array}$$
   
   In order for this account to reach $500,000 after 40 years, $29,949.69 needs to be invested.

Knowing how much to deposit at age 23 to reach a certain value later is a present value question. The target value for Pilar is $1,500,000. The interest rate is 6.35%, which in decimal form is 0.0635. Compounded monthly means $n$ = 12. She’s 23 and will leave the money in the account until the age of 67, which is 44 years, making $t$ = 44. Using this information and substituting in the formula for present value, we calculate

Pilar will need to invest $92,442,51 in this account to have $1,500,000 at age 67.

Here, $n$ = 4 (quarterly) and $r$ = 0.06 (decimal form). Substituting into the formula we find that the effective annual yield is

Therefore, a rate of 6% compounded quarterly is equivalent to a simple interest rate of 6.14%.

In this case, the rate is $r$ = 0.05 and $n$ = 365 (daily). Using the formula $Y={\left(1+\frac{r}{n}\right)}^n-1$, we have

This tells us that an account earning 5% compounded daily is equivalent to earning 5.13% as simple interest.

To compare these directly, Minh could change each interest rate to its effective annual yield, which would allow direct comparison between the rates. Computing the effective annual yield for all three choices gives:

ABC Bank: $Y={\left(1+\frac{0.0208}{12}\right)}^{12}-1=0.0210=2.10\%$

123 Bank: $Y={\left(1+\frac{0.0209}{1}\right)}^1-1=0.0209=2.09\%$

XYZ Bank: $Y={\left(1+\frac{0.0205}{365}\right)}^{365}-1=0.0207=2.07\%$

ABC Bank has the highest effective annual yield, so Minh should choose ABC bank.