## 1.1 The Time Value of Money

This section reviews basic time value of money calculations. The concepts of future value, present value and the compounding of interest are defined and discussed.

### 1.1.1 Future value, present value and simple interest

Consider an amount $$\V$$ invested for $$n$$ years at a of $$R$$ per annum (where $$R$$ is expressed as a decimal). If compounding takes place only at the end of the year, the after $$n$$ years is: $\begin{equation} FV_{n}=\V(1+R)\times\cdots\times(1+R)=\V\cdot(1+R)^{n}.\tag{1.1} \end{equation}$ Over the first year, $$\V$$ grows to $$\V(1+R)=\V+\V\times R$$ which represents the initial principal $$\V$$ plus the payment of simple interest $$\V\times R$$ for the year. Over the second year, the new principal $$\V(1+R)$$ grows to $$\V(1+R)(1+R)=\V(1+R)^{2},$$ and so on.

Example 1.1 (Future value with simple interest)

Consider putting 1000 in an interest checking account that pays a simple annual percentage rate of 3%. The future value after $$n=1,5$$ and 10 years is, respectively, \begin{align*} FV_{1} & =\1000\cdot(1.03)^{1}=\1030,\\ FV_{5} & =\1000\cdot(1.03)^{5}=\1159.27,\\ FV_{10} & =\1000\cdot(1.03)^{10}=\1343.92. \end{align*} Over the first year,30 in interest is accrued; over three years, $$\159.27$$ in interest is accrued; over five years, $343.92 in interest is accrued. These calculations are easy to perform in R. V = 1000 R = 0.03 FV1 = V*(1 + R) FV5 = V*(1 + R)^5 FV10 = V*(1 + R)^10 c(FV1, FV5, FV10) ##  1030.000 1159.274 1343.916 $$\blacksquare$$ The future value formula (1.1) defines a relationship between four variables: $$FV_{n},$$ $$V$$, $$R$$ and $$n$$. Given three variables, the fourth variable can be determined. For example, given $$FV_{n}$$, $$R$$ and $$n$$ and solving for $$V$$ gives the formula: $\begin{equation} V=\frac{FV_{n}}{(1+R)^{n}}.\tag{1.2} \end{equation}$ Here, $$V$$ represents the present or current value of the future value $$FV_{n}$$ that is to be received $$n$$ years from today. The future value is discounted by the factor $$(1+R)^{-n}$$ to reflect the fact that dollars to be received in $$n$$ years are worth less than current dollars because current dollars can be invested today returning $$R$$ per year. The discount factor $$(1+R)^{-n}$$ is less than $$1$$, and represents the price today of $$\1$$ to be received in $$n$$ years when the interest rate is $$R$$ per year. Taking $$FV_{n}$$, $$n$$ and $$V$$ as given, the annual interest rate on the investment is defined as: $\begin{equation} R=\left(\frac{FV_{n}}{V}\right)^{1/n}-1.\tag{1.3} \end{equation}$ Here, $$R$$ represents a type of average rate of return on an investment known as the compound annual return. Finally, given $$FV_{n}$$, $$V$$ and $$R$$ we can solve for $$n$$: $\begin{equation} n=\frac{ln(FV_{n}/V)}{ln(1+R)}.\tag{1.4} \end{equation}$ The expression (1.4) can be used to determine the number of years it takes for an investment of $$\V$$ to double. Setting $$FV_{n}=2V$$ in (1.4) gives: $n=\frac{\ln(2)}{\ln(1+R)}\approx\frac{0.7}{R},$ which uses the approximations $$\ln(2)=0.6931\approx0.7$$ and $$\ln(1+R)\approx R$$ for $$R$$ close to zero. The approximation $$n\approx0.7/R$$ is called the rule of 70. Example 1.2 (Using the rule of 70) The rule of 70 gives a good approximation as long as the interest rate is not too high. R = seq(0.01, 0.10, by=0.01) nExact = log(2)/log(1 + R) nRule70 = 0.7/R cbind(R, nExact, nRule70) ## R nExact nRule70 ## [1,] 0.01 69.660717 70.000000 ## [2,] 0.02 35.002789 35.000000 ## [3,] 0.03 23.449772 23.333333 ## [4,] 0.04 17.672988 17.500000 ## [5,] 0.05 14.206699 14.000000 ## [6,] 0.06 11.895661 11.666667 ## [7,] 0.07 10.244768 10.000000 ## [8,] 0.08 9.006468 8.750000 ## [9,] 0.09 8.043232 7.777778 ## [10,] 0.10 7.272541 7.000000 For $$R$$ values between 1% and 10% the approximation error in the rule of 70 is never more than half a year. $$\blacksquare$$ ### 1.1.2 Multiple compounding periods If interest is paid $$m$$ times per year, then the future value after $$n$$ years is: $FV_{n}^{m}=\V\cdot\left(1+\frac{R}{m}\right)^{m\cdot n}.$ $$R/m$$ is often referred to as the periodic interest rate . As $$m$$, the frequency of compounding, increases, the rate becomes continuously compounded and it can be shown that the future value becomes: $FV_{n}^{c}=\lim_{m\rightarrow\infty}\V\cdot\left(1+\frac{R}{m}\right)^{m\cdot n}=\V\cdot e^{R\cdot n},$ where $$e^{(\cdot)}$$ is the exponential function and $$e^{1}=2.71828$$. Example 1.3 (Future value with different compounding frequencies) For a simple annual percentage rate of 10%, the value of$1000 at the end of one year ($$n=1)$$ for different values of $$m$$ is calculated below.
V = 1000
R = 0.1
m = c(1, 2, 4, 356, 10000)
FV = V*(1 + R/m)^(m)
print(cbind(m, FV), digits=7)
##          m       FV
## [1,]     1 1100.000
## [2,]     2 1102.500
## [3,]     4 1103.813
## [4,]   356 1105.155
## [5,] 10000 1105.170

The result with continuous compounding is

print(V*exp(R), digits=7)
##  1105.171

$$\blacksquare$$

The continuously compounded return analogues to the present value, annual return and horizon period formulas (1.2), (1.3) and (1.4) are: \begin{align*} V & =e^{-Rn}FV_{n},\\ R & =\frac{1}{n}\ln\left(\frac{FV_{n}}{V}\right),\\ n & =\frac{1}{R}\ln\left(\frac{FV_{n}}{V}\right). \end{align*}

### 1.1.3 Effective annual rate

We now consider the relationship between simple interest rates, periodic rates, effective annual rates and continuously compounded rates. Suppose an investment pays a periodic interest rate of 2% each quarter. This gives rise to a simple annual rate of 8% (2% $$\times$$ 4 quarters). At the end of the year, 1000 invested accrues to: $\1000\cdot\left(1+\frac{0.08}{4}\right)^{4\cdot1}=\1082.40.$ The effective annual rate, $$R_{A}$$, on the investment is determined by the relationship: $\1000\cdot(1+R_{A})=\1082.40.$ Solving for $$R_{A}$$ gives: $R_{A}=\frac{\1082.40}{\1000}-1=0.0824,$ or $$R_{A}=8.24\%.$$ Here, the effective annual rate is the simple interest rate with annual compounding that gives the same future value that occurs with simple interest compounded four times per year. The effective annual rate is greater than the simple annual rate due to the payment of interest on interest. The general relationship between the simple annual rate $$R$$ with payments $$m$$ times per year and the effective annual rate, $$R_{A}$$, is: $(1+R_{A})=\left(1+\frac{R}{m}\right)^{m}.$ Given the simple rate $$R$$, we can solve for the effective annual rate using: $\begin{equation} R_{A}=\left(1+\frac{R}{m}\right)^{m}-1.\tag{1.5} \end{equation}$ Given the effective annual rate $$R_{A}$$, we can solve for the simple rate using: $R=m\left[(1+R_{A})^{1/m}-1\right].$ The relationship between the effective annual rate and the simple rate that is compounded continuously is: $(1+R_{A})=e^{R}.$ Hence, \begin{align*} R_{A} & =e^{R}-1,\\ R & =\ln(1+R_{A}). \end{align*} Example 1.4 (Determine effective annual rates) The effective annual rates associated with the investments in example 1.2 are calculated below. RA = FV/V - 1 print(cbind(m, FV, RA), digits=7) ## m FV RA ## [1,] 1 1100.000 0.1000000 ## [2,] 2 1102.500 0.1025000 ## [3,] 4 1103.813 0.1038129 ## [4,] 356 1105.155 0.1051554 ## [5,] 10000 1105.170 0.1051704 The effective annual rate with continuous compounding is print(exp(R) - 1, digits=7) ##  0.1051709 $$\blacksquare$$ Example 1.5 (Determine continuously compounded rate from effective annual rate) Suppose an investment pays a periodic interest rate of 5% every six months ($$m=2,R/2=0.05$$). In the market this would be quoted as having an annual percentage rate, $$R_{A}$$, of 10%. An investment of100 yields $$\100\cdot(1.05)^{2}=\110.25$$ after one year. The effective annual rate, $$R_{A}$$, is then 10.25%. To find the continuously compounded simple rate that gives the same future value as investing at the effective annual rate we solve: $R=\ln(1.1025)=0.09758.$ That is, if interest is compounded continuously at a simple annual rate of 9.758% then \$100 invested today would grow to $$\100\cdot e^{0.09758}=\110.25$$

$$\blacksquare$$