26 Compounded percent change
In the previous section we used a fixed percent rate to build exponential models. In this section we consider situations where the rate is applied multiple times per year — this is called compounding.
Example 26.1 (Activity: El Prado Auto Traders) El Prado Auto Traders are offering a deal for first-time customers: no payments for the first year of their car loan. However, there is a catch: the auto dealer will continue to charge interest on the loan during that first year.
Suppose that Genevieve purchases a car from El Prado Auto Traders for $5,000. The interest rate on her loan is 6% annually, though it is compounded monthly. We would like to compute what Genevieve’s loan balance will be at the end of 12 months, when she will start making payments.
Let’s start by looking at what happens during the first month:
- The balance begins at $5,000.
- At the end of the month, the dealer charges interest: the 6% annual rate divided among 12 months gives a monthly rate of \[ \frac{6\%}{12} = \underline{\hspace{1in}} \]
- The interest charged during the first month is \[\phantom{\frac{1}{2}}\underline{\hspace{1.5in}}\]
- The new balance is 100% of the previous balance, plus the interest charged: \[ \text{New Balance} = \underline{\hspace{1.5in}} + \underline{\hspace{1.5in}} \] We rewrite this as \[ \text{New Balance} = \$5000\big(\,\underline{\hspace{1in}} + \underline{\hspace{1in}}\, \big)\]
Our conclusion is that each month, the balance is multiplied by: \[ R = 1 + \frac{6\%}{12} = \underline{\hspace{1in}} \]
- Complete the data table for Genevieve’s loan balance.
| \(x=\) time (months) | \(y=\) loan balance ($) |
|---|---|
| 0 | 5,000 |
| 1 | \(\phantom{\dfrac{1}{2}}{\hspace{0.5in}}\) |
| 2 | \(\phantom{\dfrac{1}{2}}{\hspace{0.5in}}\) |
| 3 | \(\phantom{\dfrac{1}{2}}{\hspace{0.5in}}\) |
| 4 | \(\phantom{\dfrac{1}{2}}{\hspace{0.5in}}\) |
- Plot the data from your table above.
- Construct a formula that gives the loan balance in terms of months.
Use your formula to determine what the loan balance will be after 12 months.
Solutions
Monthly rate: \(\dfrac{6\%}{12} = 0.5\%\), so \(R = 1 + 0.005 = 1.005\).
| Month (\(x\)) | Balance (\(y\)) |
|---|---|
| 0 | $5,000.00 |
| 1 | $5,025.00 |
| 2 | $5,050.13 |
| 3 | $5,075.38 |
| 4 | $5,100.75 |
Formula: \(y = 5000 \cdot (1.005)^x\).
After 12 months: \(y = 5000 \cdot (1.005)^{12} \approx 5000 \cdot 1.0617 \approx \$5{,}308.39\).
26.1 The compounded percent change framework
When a percent rate \(r\) is applied \(N\) times per unit period (e.g., monthly within a year), the rate is divided among those \(N\) compounding periods. The multiplier for each period is \[ R = 1 + \frac{r}{N}. \] With \(x\) measured in compounding periods, the formula is \[ y = A \cdot \left(1 + \frac{r}{N}\right)^x. \]
Example 26.2 (Activity: Elk hunting guides wage increase) The Taos County Elk Hunting Guide Association has also negotiated a new contract. Currently, they are being paid $20 per hour. Under their contract, this amount will increase by 5% each year, but with the increase computed quarterly (every 3 months). We want to model the hourly wage during the duration of the contract.
What is the starting amount \(A\) in this situation?
What is the annual rate \(r\)?
What is \(N\), the number of compounding periods per year? What time unit does each period represent?
What is the multiplier \(R = 1 + \dfrac{r}{N}\)?
Using \(x =\) number of quarters and \(y =\) hourly wage (dollars), write the equation for this model. \[ y = \underline{\hspace{3in}} \]
What will the guides’ hourly wage be after 6 years?
Solutions
\(A = \$20\) per hour.
\(r = 0.05\) (5% annual rate).
\(N = 4\) quarters per year; each period is one quarter (3 months).
\(R = 1 + \dfrac{0.05}{4} = 1 + 0.0125 = 1.0125\).
Formula: \(y = 20 \cdot (1.0125)^x\) where \(x\) is the number of quarters.
After 6 years = 24 quarters: \(y = 20 \cdot (1.0125)^{24} \approx 20 \cdot 1.3474 \approx \$26.95\) per hour.
26.2 Practice
Example 26.3 (Activity: Certificate of deposit) A local bank is offering a 4-year certificate of deposit (CD) at an annual rate of 3.3%. The interest is compounded monthly.
- Suppose that $1,000 is invested in this CD. Make a data table showing the value of the account during the first 4 months.
- Construct a formula for the value in terms of months.
- Sketch a plot of your model.
- What will the value be at the end of the four year period?
Solutions
\(A = 1000\), \(r = 0.033\), \(N = 12\), \(R = 1 + \dfrac{0.033}{12} = 1.00275\), \(x =\) months.
| Month (\(x\)) | Value (\(y\)) |
|---|---|
| 0 | $1,000.00 |
| 1 | $1,002.75 |
| 2 | $1,005.51 |
| 3 | $1,008.27 |
| 4 | $1,011.04 |
Formula: \(y = 1000 \cdot (1.00275)^x\).
After 4 years = 48 months: \(y = 1000 \cdot (1.00275)^{48} \approx 1000 \cdot 1.1397 \approx \$1{,}139.70\).
26.3 Homework exercises
Exercise 26.1 A company purchases a skid steer for $50,000. The equipment depreciates at a rate of 12% per year, calculated quarterly.
- Construct a data table showing the value each quarter for the first 5 quarters.
- Make a plot showing the data in your table.
- Write down the formula for the value in terms of quarters.
- What will the skid steer be worth after 3 years?
Solutions
\(A = 50000\), \(r = 0.12\) (depreciation), \(N = 4\) quarters per year, \(R = 1 - \dfrac{0.12}{4} = 1 - 0.03 = 0.97\), \(x =\) quarters.
| Quarter (\(x\)) | Value (\(y\)) |
|---|---|
| 0 | $50,000.00 |
| 1 | $48,500.00 |
| 2 | $47,045.00 |
| 3 | $45,633.65 |
| 4 | $44,264.64 |
| 5 | $42,936.70 |
Formula: \(y = 50000 \cdot (0.97)^x\).
After 3 years = 12 quarters: \(y = 50000 \cdot (0.97)^{12} \approx 50000 \cdot 0.6938 \approx \$34{,}692\).
Exercise 26.2 $8,500 is placed in a fund that grows at 7% per year, compounded quarterly.
- Construct a data table showing the value of the account each quarter for the first 5 quarters.
- Make a plot showing the data in your table.
- Write down the formula for the value in terms of quarters.
- How much will the investment be worth after 3 years?
Solutions
\(A = 8500\), \(r = 0.07\), \(N = 4\), \(R = 1 + \dfrac{0.07}{4} = 1.0175\), \(x =\) quarters.
| Quarter (\(x\)) | Value (\(y\)) |
|---|---|
| 0 | $8,500.00 |
| 1 | $8,648.75 |
| 2 | $8,799.61 |
| 3 | $8,952.59 |
| 4 | $9,107.75 |
| 5 | $9,265.14 |
Formula: \(y = 8500 \cdot (1.0175)^x\).
After 3 years = 12 quarters: \(y = 8500 \cdot (1.0175)^{12} \approx 8500 \cdot 1.2314 \approx \$10{,}467\).
Exercise 26.3 Consider the exponential model \[ y = 500 \cdot \left(1 + \frac{0.15}{12}\right)^x, \] where \(x\) is measured in months.
- What is the starting amount \(A\) for this model?
- What is the annual rate \(r\)? What is \(N\)? How often is the rate being compounded?
- Draw a sketch of the plot for this model.
- What will the amount be after 7 years? Show this point on your plot.
Solutions
\(A = 500\).
\(r = 0.15\) (15% annual rate), \(N = 12\) (compounded monthly). The monthly multiplier is \(R = 1 + \dfrac{0.15}{12} = 1.0125\).
After 7 years = 84 months: \(y = 500 \cdot (1.0125)^{84} \approx 500 \cdot 2.875 \approx \$1{,}438\).