6.2: Compound Interest
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In this section, you will learn to:
- Find the future value of a lump-sum.
- Find the present value of a lump-sum.
- Find the effective interest rate.
Compound Interest
In the last section, we examined problems involving simple interest. Simple interest is generally charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, if the interest is paid (or charged) not only on the principal, but also on the past interest, then we say the interest is compounded.
Suppose we deposit $200 in an account that pays 8% interest. At the end of one year, we will have $200 + $200(.08) = $200(1 + .08) = $216.
Now suppose we put this amount, $216, in the same account. After another year, we will have $216 + $216(.08) = $216(1 + .08) = $233.28.
So an initial deposit of $200 has accumulated to $233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of $16(.08) = $1.28, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest.
Now suppose we leave this amount, $233.28, in the bank for another year, the final amount will be $233.28 + $233.28(.08) = $233.28(1 + .08) = $251.94.
Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.
After one year, we had $200(1 + .08) = $216
After two years, we had $216(1 + .08)
But $216 = $200(1 + .08), therefore, the above expression becomes
\[\$ 200(1+.08)(1+.08) = \$ 200(1+.08)^2=\$ 233. 28 \nonumber \]
After three years, we get
\[\$ 233.28(1+.08)=\$ 200(1+.08)(1+.08)(1+.08) \nonumber \]
which can be written as
\[\$ 200(1+.08)^{3}=\$ 251.94 \nonumber \]
Suppose we are asked to find the total amount at the end of 5 years, we will get
\[200(1+.08)^{5}=\$ 293.87 \nonumber \]
We summarize as follows:
| The original amount | $200 | = $200 |
| The amount after one year | $200(1 + .08) | = $216 |
| The amount after two years | $200(1 + .08)2 | = $233.28 |
| The amount after three years | $200(1 + .08)3 | = $251.94 |
| The amount after five years | $200(1 + .08)5 | = $293.87 |
| The amount after t years | $200(1 + .08)t |
COMPOUNDING PERIODS
Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to one-fourth of 8%, or 2%.
Now if we deposit $200 in the bank, after one quarter we will have \(\$ 200\left(1+\frac{.08}{4}\right)\) or $204.
After two quarters, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{2}\) or $208.08.
After one year, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{4}\) or $216.49.
After three years, we will have \(\$ 200\left(1+\frac{.08}{4}\right)^{12}\) or $253.65, etc.
| The original amount | $200 | = $200 |
| The amount after one quarter | \(\$ 200\left(1+\frac{.08}{4}\right)\) | = $204 |
| The amount after two quarters | \(\$ 200\left(1+\frac{.08}{4}\right)^{2}\) | = $208.08 |
| The amount after one year | \(\$ 200\left(1+\frac{.08}{4}\right)^{4}\) | = $216.49 |
| The amount after two years | \(\$ 200\left(1+\frac{.08}{4}\right)^{8}\) | = $234.31 |
| The amount after three years | \(\$ 200\left(1+\frac{.08}{4}\right)^{12}\) | = $253.65 |
| The amount after five years | \(\$ 200\left(1+\frac{.08}{4}\right)^{20}\) | = $297.19 |
| The amount after t years | \(\$ 200\left(1+\frac{.08}{4}\right)^{4t}\) |
Therefore, if we invest a lump-sum amount of \(P\) dollars at an interest rate \(r\), compounded \(n\) times a year, then after \(t\) years the final amount is given by
\[A=P\left(1+\frac{r}{n}\right)^{n t} \nonumber \]
The following examples use the compound interest formula \(A=P\left(1+\frac{r}{n}\right)^{n t}\)
If $3500 is invested at 9% compounded monthly, what will the future value be in four years?
Solution
Clearly an interest of .09/12 is paid every month for four years. The interest is compounded \(4 \times 12 = 48\) times over the four-year period. We get
\[\mathrm{A}=\$ 3500\left(1+\frac{.09}{12}\right)^{48}=\$ 3500(1.0075)^{48}=\$ 5009.92 \nonumber \]
$3500 invested at 9% compounded monthly will accumulate to $5009.92 in four years.
How much should be invested in an account paying 9% compounded daily for it to accumulate to $5,000 in five years?
Solution
We know the future value, but need to find the principal.
\[\begin{array}{l}
\$ 5000=P\left(1+\frac{.09}{365}\right)^{365 \times 5} \\
\$ 5000=P(1.568225) \\
\$ 3188.32=P
\end{array} \nonumber \]
$3188,32 invested into an account paying 9% compounded daily will accumulate to $5,000 in five years.
If $4,000 is invested at 4% compounded annually, how long will it take to accumulate to $6,000?
Solution
\(n = 1\) because annual compounding means compounding only once per year. The formula simplifies to \(A=(1+r)^{t}\) when \(n = 1\).
\[\begin{aligned}
\$ 6000 &=4000(1+.04)^{t} \\
\frac{6000}{4000} &=1.04^{t} \\
1.5 &=1.04^{t}
\end{aligned} \nonumber \]
We use logarithms to solve for the value of \(t\) because the variable \(t\) is in the exponent.
\[t=\log _{1.04}(1.5) \nonumber \]
Using the change of base formula we can solve for \(t\):
\[t=\frac{\ln (1.5)}{\ln (1.04)}=10.33 \text { years } \nonumber \]
It takes 10.33 years for $4000 to accumulate to $6000 if invested at 4% interest, compounded annually
If $5,000 is invested now for 6 years what interest rate compounded quarterly is needed to obtain an accumulated value of $8000.
Solution
We have \(n = 4\) for quarterly compounding.
\[\begin{aligned}
\$ 8000 &=\$ 5000\left(1+\frac{r}{4}\right)^{4 \times 6} \\
\frac{\$ 8000}{\$ 5000} &=\left(1+\frac{r}{4}\right)^{24} \\
1.6 &=\left(1+\frac{r}{4}\right)^{24}
\end{aligned} \nonumber \]
We use roots to solve for \(t\) because the variable \(r\) is in the base, whereas the exponent is a known number.
\[\sqrt[24]{1.6}=1+\frac{\mathrm{r}}{4} \nonumber \]
Many calculators have a built in “nth root” key or function. In the TI-84 calculator, this is found in the Math menu. Roots can also be calculated as fractional exponents; if necessary, the previous step can be rewritten as
\[1.6^{1 / 24}=1+\frac{\mathrm{r}}{4} \nonumber \]
Evaluating the left side of the equation gives
\[\begin{array}{l}
1.0197765=1+\frac{\mathrm{r}}{4} \\
0.0197765=\frac{\mathrm{r}}{4} \\
\mathrm{r}=4(0.0197765)=0.0791
\end{array} \nonumber \]
An interest rate of 7.91% is needed in order for $5000 invested now to accumulate to $8000 at the end of 6 years, with interest compounded quarterly.
Effective Interest Rate
Banks are required to state their interest rate in terms of an “effective yield” ” or “effective interest rate”, for comparison purposes. The effective rate is also called the Annual Percentage Yield (APY) or Annual Percentage Rate (APR).
The effective rate is the interest rate compounded annually would be equivalent to the stated rate and compounding periods. The next example shows how to calculate the effective rate.
To examine several investments to see which has the best rate, we find and compare the effective rate for each investment.
Example \(\PageIndex{5}\) illustrates how to calculate the effective rate.
If Bank A pays 7.2% interest compounded monthly, what is the effective interest rate?
If Bank B pays 7.25% interest compounded semiannually, what is the effective interest rate? Which bank pays more interest?
Solution
Bank A: Suppose we deposit $1 in this bank and leave it for a year, we will get
\[\begin{array}{l}
1\left(1+\frac{0.072}{12}\right)^{12}=1.0744 \\
\mathrm{r}_{\mathrm{EFF}}=1.0744-1=0.0744
\end{array} \nonumber \]
We earned interest of $1.0744 - $1.00 = $.0744 on an investment of $1.
The effective interest rate is 7.44%, often referred to as the APY or APR.
Bank B: The effective rate is calculated as
\[\mathbf{r}_{\mathrm{EFF}}=1\left(1+\frac{0.072}{2}\right)^{2}-1=.0738 \nonumber \]
The effective interest rate is 7.38%.
Bank A pays slightly higher interest, with an effective rate of 7.44%, compared to Bank B with effective rate 7.38%.
Continuous Compounding
Interest can be compounded yearly, semiannually, quarterly, monthly, and daily. Using the same calculation methods, we could compound every hour, every minute, and even every second. As the compounding period gets shorter and shorter, we move toward the concept of continuous compounding.
But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts? When interest is compounded "infinitely many times", we say that the interest is compounded continuously. Our next objective is to derive a formula to model continuous compounding.
Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be \(\$ 1(1+1)=\$ 2\).
- If the interest is compounded semiannually, in one year we will have \(\$ 1(1+1 / 2)^{2}=\$ 2.25\)
- If the interest is compounded quarterly, in one year we will have \(\$ 1(1+1 / 4)^{4}=\$ 2.44\)
- If the interest is compounded monthly, in one year we will have \(\$ 1(1+1 / 12)^{12}=\$ 2.61\)
- If the interest is compounded daily, in one year we will have \(\$ 1(1+1 / 365)^{365}=\$ 2.71\)
We show the results as follows:
| Frequency of compounding | Formula | Total amount |
|---|---|---|
| Annually | \(\$ 1(1 + 1)\) | $2 |
| Semiannually | \(\$ 1(1+1 / 2)^{2}\) | $2.25 |
| Quarterly | \(\$ 1(1+1 / 4)^{4}=\$ 2.44\) | $2.44140625 |
| Monthly | \(\$ 1(1+1 / 12)^{12}\) | $2.61303529 |
| Daily | \(\$ 1(1+1 / 365)^{365}\) | $2.71456748 |
| Hourly | \(\$ 1(1+1 / 8760)^{8760}\) | $2.71812699 |
| Every minute | \(\$1(1+1 / 525600)^{525600}\) | $2.71827922 |
| Every Second | \(\$ 1(1+1 / 31536000)^{31536000} \) | $2.71828247 |
| Continuously | \(\$ 1(2.718281828 \ldots)\) | $2.718281828... |
We have noticed that the $1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name "e" after the great mathematician Euler.
In mathematics, we say that as \(n\) becomes infinitely large the expression equals \(\left(1+\frac{1}{n}\right)^{n}\) e.
Therefore, it is natural that the number e play a part in continuous compounding.
It can be shown that as \(n\) becomes infinitely large the expression \(\left(1+\frac{r}{n}\right)^{n t}=e^{r t}\)
Therefore, it follows that if we invest $\(P\) at an interest rate \(r\) per year, compounded continuously, after \(t\) years the final amount will be given by
\[ A = P \cdot e^{rt} \nonumber \]
$3500 is invested at 9% compounded continuously. Find the future value in 4 years.
Solution
Using the formula for the continuous compounding, we get \(A = Pe^{rt}\).\begin{aligned}
A &=\$ 3500 e^{0.09 \times 4} \\
A &=\$ 3500 e^{0.36} \\
A &=\$ 5016.65
\end{aligned}
If an amount is invested at 7% compounded continuously, what is the effective interest rate?
Solution
If we deposit $1 in the bank at 7% compounded continuously for one year, and subtract that $1 from the final amount, we get the effective interest rate in decimals.\[\begin{array}{l}
\mathrm{r}_{\mathrm{EFF}}=1 \mathrm{e}^{0.07}-1 \\
\mathrm{r}_{\mathrm{EFF}}=1.0725-1 \\
\mathrm{r}_{\mathrm{EFF}}=.0725 \text { or } 7.25 \%
\end{array} \nonumber \]
If an amount is invested at 7% compounded continuously, how long will it take to double?
We offer two solutions.
Solution 1 uses logarithms to calculate the exact answer, so it is preferred. We already used this method in Example \(\PageIndex{3}\) to solve for time needed for an investment to accumulate to a specified future value.
Solution 2 provides an estimated solution that is applicable only to doubling time, but not to other multiples. Students should find out from their instructor if there is a preference as to which solution method is to be used for doubling time problems.
Solution: Solution 1: Calculating the answer exactly: \(P e^{0.07t} = A\).
We don’t know the initial value of the prinicipal but we do know that the accumulated value is double (twice) the principal.
\[\mathrm{P}_{e}^{0.07t}=2 \mathrm{P} \nonumber \]
We divide both sides by \(\mathrm{P}\)
\[e^{.07 t}=2 \nonumber \]
Using natural logarithm:
\[\begin{array}{l}
.07 \mathrm{t}=\ln (2) \\
\mathrm{t}=\ln (2) / .07=9.9 \: \mathrm{years}
\end{array} \nonumber \]
It takes 9.9 years for money to double if invested at 7% continuous interest.
Solution 2: Estimating the answer using the Law of 70:
The Law of 70 is a useful tool for estimating the time needed for an investment to double in value. It is an approximation and is not exact and comes from our previous solution. We calculated that
\[\mathrm{t}=\ln (2) / \mathrm{r} \text{ where } \mathrm{r} \text{ was 0.07 in that solution.} \nonumber \]
Evaluating \(\ln(2) = 0.693\), gives \(t = 0.693/\mathrm{r}\). Multiplying numerator and denominator by 100 gives \(t = 69.3/ (100\mathrm{r})\)
If we estimate 69.3 by 70 and state the interest rate as a percent instead of a decimal, we obtain the Law of 70:
Law of 70: The number of years required to double money ≈ 70 ÷ interest rate
- Note that this is an approximate estimate only.
- The interest rate is stated as a percent (not decimal) in the Law of 70.
Using the Law of 70 gives us \(t\) ≈ 70/7=10 which is close to but not exactly the value of 9.9 years calculated in Solution 1.
| Annual interest rate | 1% | 2% | 3% | 4% | 5% | 6% | 7% | 8% | 9% | 10% |
| Number of years to double money | 70 | 35 | 23 | 18 | 14 | 12 | 10 | 9 | 8 | 7 |
The pattern in the table approximates the Law of 70.
With technology available to do calculations using logarithms, we would use the Law of 70 only for quick estimates of doubling times. Using the Law of 70 as an estimate works only for doubling times, but not other multiples, so it’s not a replacement for knowing how to find exact solutions.
However, the Law of 70 can be useful to help quickly estimate many “doubling time” problems mentally, which can be useful in compound interest applications as well as other applications involving exponential growth.
- At the peak growth rate in the 1960’s the worlds population had a doubling time of 35 years. At that time, approximately what was the growth rate?
- As of 2015, the world population’s annual growth rate was approximately 1.14%. Based on that rate, find the approximate doubling time.
Solution
a. According to the law of 70,doubling time = \(35 \approx 70 \div r\)
\(r \approx 2\) expressed as a percent
Therefore, the world population was growing at an approximate rate of 2% in the 1960’s.
b.. According to the law of 70,
doubling time \(t \approx 70 \div r = 70 \div 1.14 \approx 61\) years
If the world population were to continue to grow at the annual growth rate of 1.14% , it would take approximately 61 years for the population to double.
SECTION 6.2 SUMMARY
Below is a summary of the formulas we developed for calculations involving compound interest:
COMPOUND INTEREST \(\mathbf{n}\) times per year
- If an amount \(\mathrm{P}\) is invested for \(t\) years at an interest rate \(r\) per year, compounded \(n\) times a year, then the future value is given by \[A=P\left(1+\frac{r}{n}\right)^{n t} \nonumber \] \(\mathbf{P}\) is called the principal and is also called the present value.
- If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is given by \[\mathbf{r}_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^{n}-1 \nonumber \]
CONTINUOUSLY COMPOUNDED INTEREST
- If an amount \(\mathrm{P}\) is invested for \(t\) years at an interest rate \(r\) per year, compounded continuously, then the future value is given by \[\mathrm{A} = \mathrm{P}e^{rt} \nonumber \]
- If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is given by \[\mathrm{r}_{\mathrm{EFF}}=e^{\mathbf{r}}-1 \nonumber \]
- The Law of 70 states that
The number of years to double money is approximately 70 ÷ interest rate