Appendix A — Time Value of Money
This material was originaly published HERE by Department of Mathematics, Penn State University Park.
Learning Outcomes:
- Understand the concept of the time value of money (TVM).
- Calculate the present value (PV) of both single and multiple future cash flows using appropriate discount rates.
- Calculate the future value (FV) of both single and ongoing investments using given interest rates.
- Apply TVM formulas to various financial scenarios, including loans, savings, and investments, to make informed decisions.
- Understand the effects of compounding frequency on FV and PV calculations.
- Critically analyze TVM problems, taking into account the impact of rate, time, and cash flows on financial decisions.
A.1 Notation and Terminology
A.1.1 Basic Notation and Terminology
- \(P =\) Principal (i.e., value of initial deposit)
- \(A =\) Accumulated amount (i.e., sum of the principal and interest)
- \(r =\) Nominal interest rate
- \(m =\) Number of conversion periods per year, (a conversion period is the interval of time between successive interest payments)
| Annually | Semiannually | Quarterly | Monthly | Weekly | Daily |
|---|---|---|---|---|---|
| \(m=1\) | \(m=2\) | \(m=4\) | \(m=12\) | \(m=52\) | \(m=365\) |
- $t = $ Term of investment (in years)
A.1.2 Simple Interest
Interest is always computed based on the original principal.
| Interest Earned | Accumulated Amount |
|---|---|
| \(I = Prt\) | \(A = P(1 + rt)\) |
A.1.3 Discrete Compound Interest
Interest payments are added to the principal at the end of each conversion period and therefore earn interest during future conversion periods.
| Accumulated Amount | Present Value Formula |
|---|---|
| \(A = P \left(1 + \frac{r}{m}\right)^{mt}\) | \(P = A\left(1 + \frac{r}{m}\right)^{-mt}\) |
A.1.4 Continuous Compound Interest
Continuous compounding of interest is equivalent to a discrete compounding of interest where \(m\), the number of conversion periods per year, goes to infinity.
| Accumulated Amount | Present Value Formula |
|---|---|
| \(A = Pe^{rt}\) | \(P = Ae^{-rt}\) |
A.1.5 Effective Rate of Interest
The effective interest rate, \(r_{\text{eff}}\), is the simple interest rate that produces the same accumulated amount in 1 year as the nominal rate, \(r\), compounded \(m\) times a year.
\[r_{\text{eff}}=\left(1+\frac{r}{m}\right)^m-1\]
A.2 Future Value Examples
A.2.1 Example 1
Suppose $1,000 is deposited into an account with an interest rate of 16% compounded annually. How much money is in the account after 3 years?
Step 1: Since interest is compounded annually, use the accumulated amount for discrete compound interest.
\[A=P\left(1+\frac{r}{m}\right)^{mt}\]
Step 2: Plug in the given values: \(P = 1000\), \(r = 0.16\), \(m=1\), and \(t = 3\).
\[\begin{align*} A &= 1000\left(1 + \frac{0.16}{1}\right)^{1\cdot 3} \\ \\ &= 1000\left(1 + 0.16\right)^{3} \\ \\ &= 1000\left(1.16\right)^{3} \approx \$1,560.90 \end{align*}\]
Therefore, after 3 years of accumulating interest, the original investment of $1,000 is worth $1,560.90.
A.2.2 Example 2
Suppose $1,000 is deposited into an account with an interest rate of 16% compounded quarterly. How much money is in the account after 3 years?
Step 1: Since interest is compounded quarterly, use the accumulated amount for discrete compound interest.
\[A=P\left(1+\frac{r}{m}\right)^{mt}\]
Step 2: Plug in the given values: \(P = 1000\), \(r = 0.16\), \(m=4\), and \(t = 3\).
\[\begin{align*} A &= 1000\left(1 + \frac{0.16}{4}\right)^{4\cdot 3} \\ \\ &= 1000\left(1 + 0.04\right)^{12} \\ \\ &= 1000\left(1.04\right)^{12} \approx \$1,601.03 \end{align*}\]
Therefore, after 3 years of accumulating interest, the original investment of $1,000 is worth $1,601.03.
Compare the accumulated amounts in the above two examples. Both examples have the same principal, interest rate, and term. But since interest is compounded more frequently in Example 2 (4 times a year) than in Example 1 (1 time a year), the accumulated amount is higher in Example 2.
A.2.3 Example 3
Find the interest rate required for an investment of $3,000 to double in value after 5 years if interest is compounded quarterly.
Step 1: Since interest is compounded quarterly, use the accumulated amount for discrete compound interest.
\[A=P\left(1+\frac{r}{m}\right)^{mt}\]
Step 2: Plug in the given values: \(P=3000\), \(A=6000\) (since the investment is to double in value), \(m=4\), and \(t=5\).
\[\begin{align*} 6000 &= 3000\left(1 + \frac{r}{4} \right)^{4\cdot 5}\\ &= 3000\left(1 + \frac{r}{4} \right)^{20} \end{align*}\]
Step 3: Solve for the interest rate, \(r\).
Divide both sides by 3000 \[ 2 = \left(1 + \frac{r}{4} \right)^{20} \]
Take the natural logarithm of both sides.
\[\begin{align*} \ln(2) &= \ln\left[\left(1 + \frac{r}{4} \right)^{20}\right] \\ &= 20\ln\left(1 + \frac{r}{4} \right) && \hbox{since $\ln(m^n) = n\ln(m)$} \end{align*}\]
Divide both sides by 20.
\[\ln(2)/20 = \ln\left(1 + \frac{r}{4} \right)\]
Take the exponential of both sides.
\[\begin{align*} e^{\ln(2)/20} &= e^{\ln\left(1 + \frac{r}{4} \right)} \\ &= 1 + \frac{r}{4} && \hbox{since $e^{\ln(x)} = x$} \end{align*}\]
Subtract 1 from both sides.
\[e^{\ln(2)/20} - 1 = \frac{r}{4} \]
And finally, multiply both sides by 4.
\[r = 4(e^{\ln(2)/20} - 1) \approx 0.1411.\]
Here is an alternate method for solving for the interest rate \(r\). We start with the following equation.
\[2 = \left(1 + \frac{r}{4} \right)^{20}\]
Instead of taking the natural logarithm of both sides as we did before, now take the 20th root of both sides (i.e., raise both sides to the power of \(1/20\)).
\[2^{1/20} = 1 + \frac{r}{4}\]
Subtract 1 from both sides.
\[2^{1/20} - 1 = \frac{r}{4}\]
And finally, multiply both sides by 4.
\[r = 4(2^{1/20} - 1) \approx 0.1411\]
Note that this value of \(r\) is numerically equal in both methods since
\[\begin{align*} e^{\ln(2)/20} &= e^{\ln(2^{1/20})} && \hbox{since $n\ln(m) = \ln(m^n)$} \\ &= 2^{1/20} && \hbox{since $e^{\ln(x)} = x$} \end{align*}\]
Therefore, an interest rate of approximately 14.11% compounded quarterly is required for an investment of $3,000 to double in value in 5 years.
A.2.4 Example 4
Find the interest rate required for an investment of $3,000 to double in value after 5 years if interest is compounded continuously.
Step 1: Since interest is compounded continuously, use the accumulated amount for continuous compound interest.
\[A=Pe^{rt}\]
Step 2: Plug in the given values: \(P=3000\), \(A=6000\) (since the investment is to double in value), and \(t=5\).
\[\begin{align*} 6000 &= 3000e^{5r} \end{align*}\]
Step 3: Solve for the interest rate, \(r\).
Divide both sides by 3000.
\[2 = e^{5r}\]
Take the natural logarithm of both sides.
\[\begin{align*} \ln(2) &= \ln(e^{5r}) \\ &= 5r && \hbox{since $\ln(e^x) = x$} \end{align*}\]
Divide both sides by 5.
\[r = \ln(2) / 5 \approx 0.1386\]
Therefore, an interest rate of approximately 13.86% compounded continuously is required for an investment of $3,000 to double in value in 5 years.
Compare the last two examples. Since continuous compounding of interest earns interest faster than discrete compounding, a lower interest rate is needed for an investment to double in value over a fixed term if interest is compounded continuously. In our examples, an interest rate of 13.86% was needed for the investment with continuous compound interest to double in value in 5 years, while an interest rate to 14.11% was needed for the investment with quarterly compound interest.
A.2.5 Example 5
How long will it take for $5,000 to grow to $8,000 if the investment earns interest at 6% per year compounded monthly?
Step 1: Since interest is compounded monthly, use the accumulated amount for discrete compound interest.
\[A=P\left(1+\frac{r}{m}\right)^{mt}\]
Step 2: Plug in the given values: \(P=5000\), \(A=8000\), \(m=12\), and \(r=0.06\).
\[\begin{align*} 8000 &= 5000\left(1 + \frac{0.06}{12} \right)^{12\cdot t}\\ &= 5000\left(1 + 0.005 \right)^{12t} \end{align*}\]
Step 3: Solve for the unknown term \(t\).
Divide both sides by 5000.
\[8/5 = 1.005^{12t}\]
Take the natural logarithm of both sides.
\[\begin{align*} \ln(8/5) &= \ln(1.005^{12t}) \\ &= 12t\ln(1.005) & \hbox{since $\ln(m^n) = n\ln(m)$} \end{align*}\]
Divide both sides by \(12\ln(1.005)\).
\[\begin{align*} t &= \frac{\ln(8/5)}{12\ln(1.005)} \approx 7.85 \end{align*}\]
Therefore, it will take approximately 7.85 years for $5,000 to grow to $8,000 if the investment earns interest at 6% per year compounded monthly.
A.2.6 Example 6
How long will it take for $5,000 to grow to $8,000 if the investment earns interest at 6% per year compounded continuously?
Step 1: Since interest is compounded continuously, use the accumulated amount for continuous compound interest.
\[A=Pe^{rt}\]
Step 2: Plug in the given values: \(P=5000\), \(A=8000\), and \(r=0.06\).
\[\begin{align*} 8000 &= 5000e^{0.06t} \end{align*}\]
Step 3: Solve for the unknown term \(t\).
Divide both sides by 5000.
\[8/5 = e^{0.06t}\]
Take the natural logarithm of both sides.
\[\begin{align*} \ln(8/5) &= \ln(e^{0.06t}) \\ &= 0.06t & \hbox{since $\ln(e^x) = x$} \end{align*}\]
Divide both sides by \(0.06\).
\[\begin{align*} t &= \frac{\ln(8/5)}{0.06} \approx 7.83 \end{align*}\]
Therefore, it will take approximately 7.83 years for $5,000 to grow to $8,000 if the investment earns interest at 6% per year compounded monthly.
Compare the last two examples. Both examples have the same principal, accumulated amount, and interest rate. But since continuous compounding of interest earns interest faster than discrete compounding, it should take less time for the investment to grow to $8,000 if interest is compounded continuously.
A.2.7 Example 7
Find the effective interest rate corresponding to a nominal interest rate of 10% compounded semiannually.
Step 1: Recall the formula for effective interest rate, \(r_{\text{eff}}\).
\[r_{\text{eff}}=\left(1+\frac{r}{m}\right)^m-1\]
Step 2: Plug in the given values: \(r = 0.1\) and \(m=2\).
\[\begin{align*} r_{\text{eff}} &= \left(1+\frac{0.1}{2}\right)^2-1 \\ &= 1.05^2 - 1 \\ &= 0.1025 \end{align*}\]
Therefore, an investment earning interest compounded semiannually at 10% earns the same amount of interest after 1 year as an investment earning simple interest at 10.25%.
A.2.8 Example 8
Suppose you have $12,000 in the bank earning interest at a rate of 12% compounded quarterly. Your cousin calls you and needs $12,000 to buy a new car. You are willing him to loan him the money, but you’d hate to lose out on the interest you would gather by simply leaving your money alone. If you charge your cousin an interest rate compounded continuously, what rate should you charge in order to earn the same amount of interest you otherwise would have?
Step 1: Assume your cousin is prepared to pay you back after \(t\) years.
We’ll use \(t\) as the term in each of the following calculations. Eventually, we’ll see that the interest rate you charge does not depend on the specific value of \(t\).
Step 2: Compute the accumulated amount of the $12,000 after \(t\) years assuming you leave your money in the bank.
\[\begin{align*} A &= P\left(1 + \frac{r}{m}\right)^{m\cdot t} \\ &= 12000 \left( 1 + \frac{0.12}{4} \right)^{4t} \\ &= 12000 \left( 1.03 \right)^{4t} \end{align*}\]
Step 3: Compute the accumulated amount of the $12,000 after \(t\) years assuming you let your cousin borrow the money.
This would be the amount that your cousin repays you after \(t\) years.
\[\begin{align*} A &= Pe^{rt}\\ &= 12000e^{rt} \end{align*}\]
Step 4: Equate the two accumulated amounts and solve for \(r\).
\[12000\left(1.03\right)^{4t} = 12000e^{rt}\]
Divide both sides by 12000.
\[1.03^{4t} = e^{rt}\]
Take the natural logarithm of both sides.
\[\ln(1.03^{4t}) = \ln(e^{rt})\]
Simplify using properties of logarithms (\(\ln(m^n) = n\ln(m)\) and \(\ln(e^x) = x\)).
\[4t\ln(1.03) = rt\]
And finally, divide both sides by \(t\). Here is where we see that the time it would take your cousin to repay you does not affect the interest rate you would charge.
\[r = 4\ln(1.03) \approx 0.1182\]
Therefore, charging your cousin 11.82% interest compounded continuously earns the same amount of interest as leaving your money in the bank earning interest at a rate of 12% compounded quarterly.
A.3 Present Value Examples
A.3.1 Example 1
How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years, the accumulated amount will be $20,000?
Step 1: Notice that this is a present value problem since we’re given the accumulated amount and we’re asked to find the principal. And since interest is compounded monthly, we’ll use the present value formula for discrete compounding of interest.
\[P = A\left(1 + \frac{r}{m}\right)^{-mt}\]
Step 2: Plug in the given values: \(A = 20000\), \(r = 0.06\), \(m=12\), and \(t=3\).
\[\begin{align*} P &=20000\left(1+\frac{0.06}{12}\right)^{-(12)(3)}\\ \\ &=20000(1.005)^{-36} \approx \$16,712.90 \end{align*}\]
Therefore, $16,712.90 invested at 6% interest compounded monthly will be worth $20,000 in 3 years.
A.3.2 Example 2
Use the accumulated amount for discrete compound interest to solve the previous example.
Step 1: Start with the formula for accumulated amount for discrete compounding of interest.
\[A = P\left( 1 + \frac{r}{m}\right)^{mt}\]
Step 2: Plug in the given values: \(A = 20000\), \(r = 0.06\), \(m=12\), and \(t=3\).
\[\begin{align*} 20000 &= P\left(1+\frac{0.06}{12}\right)^{(12)(3)}\\ &=P(1.005)^{36} \end{align*}\]
Step 3: Solve for \(P\).
\[\begin{align*} P = \frac{20000}{1.005^{36}} \approx \$16,712.90 \end{align*}\]
A.3.3 Example 3
Parents wish to establish a trust fund for their child’s education. If they need $170,000 in 7 years, how much should they set aside now if the money is invested at 20% compounded continuously?
Step 1: Notice that this is a present value problem since we’re given the accumulated amount and we’re asked to find the principal. And since interest is compounded continuously, we’ll use the present value formula for continuous compounding of interest.
\[P = Ae^{-rt}\]
Step 2: Plug in the given values: \(A = 170000\), \(r = 0.2\), and \(t=7\).
\[\begin{align*} P &=170,000e^{-(0.2)(7)}\\ \\ &=170,000e^{-1.4} \approx {\$41,921.48} \end{align*}\]
Therefore, $41,921.48 invested at 20% interest compounded continuously will be worth $170,000 in 7 years.
A.3.4 Example 4
Use the accumulated amount for continuous compound interest to solve the previous example.
Step 1: Start with the formula for accumulated amount for continuous compounding of interest.
\[A = Pe^{rt}\]
Step 2: Plug in the given values: \(A = 170000\), \(r = 0.2\), and \(t=7\).
\[\begin{align*} 170000 &= Pe^{(0.2)(7)}\\ &= Pe^{1.4} \end{align*}\]
Step 3: Solve for \(P\).
\[\begin{align*} P &= \frac{170000}{e^{1.4}} \approx \$41,921.48 \end{align*}\]
A.4 Try It Yourself
A.4.1 Exercise 1
If $6,00 is invested at 7% compounded continuously, what will be the accumulated amount after 6 years?
\(A = 6000e^{0.42}\)
A.4.2 Exercise 2
If $7,000 is invested at 16% compounded quarterly, what will be the accumulated amount after 3 years?
\(A = 7000(1.04)^{12}\)
A.4.3 Exercise 3
Find the interest rate \(r\) needed for an investment of $2,000 to grow to $8,000 in 7 years if compounded continuously.
\(r = \ln(4)/7\)
A.4.4 Exercise 4
Find the interest rate \(r\) needed for an investment of $7,000 to grow to $12,000 in 21 years if compounded monthly.
\(r = 12\left[(12/7)^{1/252} - 1\right]\)
A.4.5 Exercise 5
Find the time it would take for an investment of $1,000 to grow to $100,000 if interest is compounded quarterly at an annual rate of 8%.
\(t = \frac{\ln(100)}{4\ln(1.02)}\)
A.4.6 Exercise 6
Find the time it would take for an investment of $2,500 to grow to $6,000 if interest is compounded continuously at an annual rate of 24%.
\(t = \frac{25}{6}\ln(12/5)\)
A.4.7 Exercise 7
Calculate the effective rate of interest corresponding to a nominal interest rate of 52% compounded weekly.
\(r_{eff} = 1.01^{52} - 1\)
A.4.8 Exercise 8
Your grandma would like to establish a trust fund for your education. How much should she set aside now if she wants $50,000 in 9 years and interest is compounded monthly at an annual rate of 12%?
\(P = 50000(1.01)^{-108}\)
A.4.9 Exercise 9
You are preparing to run for president and want to have $100,000 in 6 years to start your campaign. How much money do you need now if interest is compounded continuously at an annual rate of 15%?
\(P = 100000e^{-0.9}\)
A.4.10 Exercise 10
You have $50,000 in the bank earning 7% interest compounded quarterly. However, your cousin needs a $50,000 investment to start up his new financial consulting business. In order to get the same total return as leaving your money in the bank, what interest rate \(r\) should you request from your cousin if interest is compounded continuously?
\(r = 4\ln(1+0.07/4)\)