Interest calculation determines how money balances deposited into an interest earning account grow with time. Simple interest lets funds grow in a linear manner. In contrast, compound interest results in an exponential increase that slightly speeds up when compounding periods shorten.
This introduction starts with simple interest and then explains the evolution of money balances with compound interest. It concludes with compound interest converging to exponential growth when reinvestment periods become infinitesimally small.
Simple Interest
The practice of charging interest in return for a loan appears to be older than coined money. In ancient Assyria, loans of grain were paid back with larger amounts of grain at harvest[2]. Just like nature is reproductive, money creates more of its kind. It multiplies with an interest rate and the passing of time. But in the case of simple interest, only the original amount bears fruit, not the interest itself:
\begin{aligned} B(\tau) &= B_0 \cdot ( 1 + \tau \cdot r)\\ \\ B(\tau) &: \mathrm{amount \space at \space time} \space \tau\\ B_0 &: \mathrm{original \space amount}\\ \tau &: \mathrm{time \space in \space years}\\ r &: \mathrm{interest \space rate} \end{aligned}
So the original amount B0 grows to B(τ) in a linear manner with interest rate r and time-span τ. In financial mathematics, time is usually measured in years and thus interest rates are annual. For instance, an amount of $1000 invested for one month at a rate of 10% grows to $1008.33.
B(1/12) = B0 * (1 + 1/12 * 10 / 100)
B(0.0833) = B0 * (1 + 0.00833)
Compound Interest
As in natural reproduction, in compound interest calculation the interest accruing from an original amount in turn generates interest. However, the reinvestment of interest income does not happen continually, but at regular time intervals. That is, interest accrues linearly until it adds to the total balance at the end of compounding intervals. Investing accordingly for two years at interest r with annual compounding yields the following end balance B(2):
\begin{aligned} B(2) & = B_0 \cdot (1 + r) \cdot (1 + r)\\ B(2) & = B_0 \cdot (1+r) ^ 2 \end{aligned}
Investing for N years with annual compounding is a straightforward extension:
\begin{aligned} &B(N) = B_0 \cdot (1 + r)^N\\ &N \in \mathbb{N} \end{aligned}
Note that the above formula works with a positive integer number of years. If the last compounding period is incomplete the last multiplication factor should be computed like in the case of simple compounding.
Of course, compounding can happen any number of times per year. In the general case of compound interest with m compounding steps per year the interest rate r will be applied for a year fraction of 1/m. So in compound interest each multiplication factor is computed like in the case of simple compounding with τ=1/m. Investing over a time span of t years with m compounding periods per year then results in the following formula for compound interest calculation:
\begin{alignat*}{2} B(t)&&=&B_0 \cdot (1+\frac{1}{m}\cdot r)^{t \cdot m}\\ \\ B(t):&& &\mathrm{amount \space at \space time} \space t\\ B_0: && &\mathrm{original \space amount}\\ t: && & \mathrm{time \space in \space years}\\ r: && & \mathrm{interest \space rate}\\ m: && & \mathrm{compounding \space periods}\\ && & \mathrm{per \space year}\\ \end{alignat*}
The above is the general formula for compound interest calculation. Alternatively, the number of compounding periods per year is often called compounding frequency.
To give an example, investing $1000 for 10 years at a rate 4% with semi-annual compounding yields $1485.95. A web calculator initialized to these numbers can be found here.
B(10) = 1000 * (1 + 4 / 100 / 2)^(2*10)
B(10) = 1485.95
The below table illustrates how the annual compounding frequency influences the resulting final amount. Again, $1000 were invested for 10 years at 4% interest. Note that the compounding frequency of zero implies simple interest:
Comp. frequency | Final amount |
---|---|
0 | $1400.00 |
1 | $1480.24 |
2 | $1485.95 |
4 | $1488.86 |
12 | $1490.83 |
365 | $1491.79 |
Continuous Compounding
The case of continuous compounding arises when the compounding frequency tends to infinity and consequently reinvestment periods become infinitesimally small. In Table 1 of the compound interest section the final amount only grew from $1490.83 to $1491.79 when moving from monthly to daily compounding. This suggests that results converge to a finite amount as the compounding frequency goes to infinity.
\begin{aligned} \tag{1}B(t+\tau) &= B(t)\lim_{m \to \infty}(1 + r/m)^{m \cdot \tau}\\ &= B(t)~exp(r \cdot \tau) \end{aligned}
Indeed, with infinitely short compounding periods compound interest converges to an exponential function of base e≈2.71828 and instantaneous rate of return r. Consequently, in the limit the investment from Table 1 would grow to $1491.82 moving up from daily compounding.
While continuous compounding is next to irrelevant in private banking, it’s of crucial importance in financial calculus when pricing derivatives.
A derivation of Equation (1) is available here.
References
[1] Compound Interest: Wikipedia.org
[2] The Price of Time: The Real Story of Interest, Edward Chancellor, 2022, Allen Lane