Regular Compound Interest

Section 2.2 Regular Compound Interest

When using simple interest, one only considers the value of the resource at the beginning of time and then again at the end. However, the lender does without use of the funds at all times during the period of the investment.

Indeed, if the original investment were to only be lent for 1/2 of the original time period, then the borrower would have to pay the accrued interest at that time (1/2 as much as planned) and then perhaps re-borrow the money plus the paid interest for the remaining time. However, for the second half, the amount borrowed would include the additional interest that had to be paid. Continuing this by shortening the period (say, 1/4's or 1/12's) would allow for the amount borrowed at the end of each period to grow. The net effect of this recalculation over many periods causes the interest upon interest to grow at a rate that is more than what might be expected using simple interest.

Definition 2.2.1. Compound Interest.

When simple interest is allowed to accure for a short period of time and then recursively reconsidered as part of a slightly increased principle for all subsequent time periods, the resulting increase in value is known as compound interest.

Theorem 2.2.2. Future Value of Compound Interest.

Using Compound interest over t years, compounding m times per year, yields

\begin{equation*} A = P(1+\frac{i}{m})^{mt}. \end{equation*}

Proof.

To determine the future value of a deposit of size \(P\) over several periods, utilize the simple interest formula 2.1.2 for a single period to determine the value of the deposit after that one period. Then, take that new (slightly larger) amount as a starting point for a new simple interest calculation covering the next period. Repeat for n periods to get the ending value over all periods.

To do this, let's denote the present value after k periods as \(P_k\) and note that \(P_0 = P\text{.}\)

To simplify notation here, consider using \(n=mt\) and \(r = i/m\text{.}\) Then looking at the timeline of money

\begin{equation*} \begin{matrix} \text{Money Amounts} & P_0 & P_1 & P_2 & \ldots & P_n \\ \hline \\ \text{Time} & 0 & \frac{t}{n} & \frac{2t}{n} & \ldots & t \\ \end{matrix} \end{equation*}

Using simple interest over each single time period

\begin{align*} P_1 & = P_0(1+r)\\ P_2 & = P_1(1+r) = P_0(1+r)^2 \\ P_3 & = P_2(1+r) = P_0(1+r)^3\\ P_4 & = P_3(1+r) = P_0(1+r)^4\\ & = ...\\ P_{n-1} & = P_{n-2}(1+r) = P_0(1+r)^{n-1}\\ P_n & = P_{n-1}(1+r) = P_0(1+r)^n \end{align*}

Noting that \(A = P_n \) and \(P = P_0 \) gives

\begin{align*} A & = P(1+r)^n \end{align*}

Corollary 2.2.3. Compound Interest gives Exponential Growth.

Note that \(1+\frac{i}{m} > 1\text{.}\) So, raising a number larger than one to an increasing power is by definition exponential growth.

It should also be noted that

\begin{equation*} P(1+\frac{i}{m})^n = P e^{n \cdot ln(1+\frac{i}{m})} \end{equation*}

which is a growing natural exponential since \(ln(1+\frac{i}{m})\) is positive because \(\frac{i}{m} > 0.\text{.}\)

Corollary 2.2.4. Present Value of Compound Interest.

Given a future value \(A\text{,}\) the present value using compound interest is given by

\begin{equation*} P = A(1+\frac{i}{m})^{-n} \end{equation*}

Proof.

Take the Compound Interest formula 2.2.2 and solve for P. This gives

\begin{equation*} P = A(1+\frac{i}{m})^{-n} \end{equation*}

Definition 2.2.5. Nominal Interest Rates.

The published yearly interest rate \(i\) is also known as the "nominal" rate.

As one considers investments (and especially annuities) one may desire compounding the value of money over different compounding time frames. So, comparing raw interest rates 2.2.5 becomes muddled since the same nominal rate can result in wildly different actual growth amounts. To be able to compare two different investments "apples-to-apples" you need a standard rate that takes into account the various factors involved in dealing with compound interest.

Definition 2.2.6. Effective Interest Rates.

The effective interest rate for any investment is determined by computing the simple interest 2.1.1 rate needed in order for an investment \(P\) to grow to a future value of \(A\) over a one year time frame.

Theorem 2.2.7. Computing Effective Interest Rate.

Given a nominal interest rate \(i\) compounded \(m\) times in a year, the effective interest rate \(i_e\) is given by

\begin{equation*} i_e = (1+\frac{i}{m})^m - 1 \end{equation*}

Proof.

Consider computing the increase compounded over one year's time span with \(n = m\) compounding periods per year. Equate the net proceeds using the compound interest 2.2.2 formula with rate \(r\) and set it equal to the formula for simple interest 2.1.2 with the effective rate \(i_e\text{.}\)

\begin{equation*} P(1 + i_e \cdot t) = P(1+i/m)^m \end{equation*}

Cancelling the common term

\begin{equation*} 1 + i_e \cdot 1 = (1+i/m)^m \end{equation*}

and easily solving for \(i_e\)gives the effective rate

\begin{equation*} i_e = (1+i/m)^m - 1 \end{equation*}

Example 2.2.8. Comparing different compounding periods.

Determine the effective rate of compounding daily that is equivalent to compounding monthly at an annual rate of \(i_{\text{monthly}} = 4 \%.\)

To do so, first compute the effective interest rate of the monthly compounding (since you know its stated rate) using

\begin{equation*} (1+ \frac{0.04}{12})^{12} - 1. \end{equation*}

This is the nominal rate needed to equal the compounded rate of \(4 \%\) monthly. On the other hand, when compounding daily, you would get an effective rate of

\begin{equation*} (1+ \frac{r_{365}}{365})^{12} - 1, \end{equation*}

where \(r_{365}\) is the effective rate when compounding daily. Setting these equal and working your way backward allows you to solve for the annual rate \(r_{365}\) associated with compounding daily.

Using this approach,

\begin{align*} (1+ \frac{0.04}{12})^{12} - 1 & = (1+ \frac{r_{365}}{365})^{365} - 1 \\ (1+ \frac{0.04}{12})^{12} & = (1+ \frac{r_{365}}{365})^{365} \\ \ln(1+ \frac{0.04}{12})^{12} & = \ln(1+ \frac{r_{365}}{365})^{365} \\ 12 \ln(1+ \frac{0.04}{12}) & = 365 \ln(1+ \frac{r_{365}}{365}) \\ \frac{12}{365} \ln(1+ \frac{0.04}{12}) & = \ln(1+ \frac{r_{365}}{365}) \\ e^{\frac{12}{365} \ln(1+ \frac{0.04}{12})} & = 1+ \frac{r_{365}}{365} \\ 365(e^{\frac{12}{365} \ln(1+ \frac{0.04}{12})} - 1) & = r_{365} \end{align*}

and the rest is arithmetic.

Theorem 2.2.9. Comparing Effective Rates for Different Compounding Rates.

If one compounding scheme utilizes nominal rate of \(r_1\) compounded \(m_1\) times per year and another compounds \(m_2\) times per year, then the nominal rate needed for the second scheme would require a nominal rate \(r_2\) given by

\begin{equation*} m_2(e^{\frac{m_1}{m_2} \ln(1+ \frac{r_1}{m_1})} - 1) = r_2 \end{equation*}

Proof.

Using similar algebra to the example above

\begin{align*} (1+ \frac{r_1}{m_1})^{m_1} - 1 & = (1+ \frac{r_2}{m_2})^{m_2} - 1 \\ (1+ \frac{r_1}{m_1})^{m_1} & = (1+ \frac{r_2}{m_2})^{m_2} \\ \ln(1+ \frac{r_1}{m_1})^{m_1} & = \ln(1+ \frac{r_2}{m_2})^{m_2} \\ m_1 \ln(1+ \frac{r_1}{m_1}) & = m_2 \ln(1+ \frac{r_2}{m_2}) \\ \frac{m_1}{m_2} \ln(1+ \frac{r_1}{m_1}) & = \ln(1+ \frac{r_2}{m_2}) \\ e^{\frac{m_1}{m_2} \ln(1+ \frac{r_1}{m_1})} & = 1+ \frac{r_2}{m_2} \\ m_2(e^{\frac{m_1}{m_2} \ln(1+ \frac{r_1}{m_1})} - 1) & = r_2 \end{align*}