How compounding and discounting move money through time

Is $1,000 in ten years worth more than $600 today?

It depends on the interest rate. Money held today can earn interest, so at a positive rate a future payment is worth less now. Dividing by the growth factor moves it back to today, and the rate where the two offers tie is the break-even rate.

Before you start · 3 quick checks on exponents and logarithms, about 2 min

Setup and definitions

Let $m$ be a positive integer. An annual rate $r$ is compounded $m$ times a year when interest of $r/m$ times the balance is added at the end of each of the $m$ equal periods in a year. After $t$ years, with $mt$ a whole number of periods, an amount $P$ grows to

\[ FV = P\Big(1 + \frac{r}{m}\Big)^{mt}. \]

Continuous compounding at rate $r$ is the limit as $m \to \infty$. Stage 3 proves that this limit gives $FV = P\,e^{rt}$.

We take $1 + r/m \gt 0$, so the growth factor is positive; every rate on this page satisfies this. The discount factor for a payment at time $t$ is the reciprocal of the growth factor: $D(t) = (1 + r/m)^{-mt}$, or $D(t) = e^{-rt}$ with continuous compounding. A payment $F$ at time $t$ has present value $PV = F\,D(t)$.

The effective annual rate is $EAR = (1 + r/m)^m - 1$, the growth over one year written as a single annual rate. Two quoted rates are equivalent when they have the same effective annual rate, so they grow every deposit by the same factor over every whole number of years.

Assumptions behind every number on this page

The interest rate is constant over the whole horizon and is the same for saving and for borrowing.
Every promised payment is certain: no default, no taxes, no fees.
Time $t$ is measured in years. With $m$ periods a year, $mt$ is a whole number of periods; every computed answer here satisfies this. The one exception is the sliding bar in stage 1, whose positions between whole years are a visual interpolation.
Cash flows are in nominal dollars. Inflation is not modeled separately.

1 See

$600 today or $1,000 in ten years?

You are offered a choice: $600 paid now, or $1,000 paid in exactly 10 years. Both payments are certain. Money you hold now can be saved at the interest rate shown, compounded once a year. Make your choice for each rate. The page then slides the $1,000 back along the time axis to today, dividing by $1 + r$ for each year it moves, and compares what is left with $600.

What the sliding bar computes

At a whole year $s$ from now, the $1,000 due at year 10 is worth $1000\,(1+r)^{-(10-s)}$: the amount that, saved at rate $r$ with interest added once a year, grows to $1,000 by year 10. Each whole year the bar moves back, its height is divided by $1+r$. At $s = 0$ it is the present value $PV = 1000\,(1+r)^{-10}$.

Between whole years the bar is drawn with the same formula at fractional $s$. That is a visual interpolation: with interest added only at year ends, annual compounding does not by itself say what the payment is worth in the middle of a year. The dots mark the whole years.

The comparison works the same way forward: $600 saved for 10 years grows to $600\,(1+r)^{10}$. That beats $1,000 exactly when $1000\,(1+r)^{-10} \lt 600$, because both sides of the inequality were multiplied by the same positive factor $(1+r)^{10}$.

Key point To compare money at different dates, move it to one date. The present value of $1,000 in 10 years falls as the rate rises, and it equals $600 at the break-even rate.

2 Try

Compounding more often, and its limit

Explorer: reach the goals

Answer the two questions above to open the explorer.

Choose how often interest is added, then set the rate and the number of years. The figure follows $1,000 deposited today. The stepped line jumps at each compounding date; the dashed curve is continuous compounding at the same quoted rate.

Key point At a positive rate, more frequent compounding grows money faster, but the growth factor for one year rises toward a ceiling, $e^r$. Compare quoted rates through their effective annual rates.

3 Prove

Why the limit is $e^r$

Theorem (continuous compounding)

For every real number $r$, with $m$ running through the integers $m \ge 2$,

\[ \lim_{m\to\infty}\Big(1 + \frac{r}{m}\Big)^{m} = e^{r}. \]

Consequently, for every $t \gt 0$, the growth factor $(1 + r/m)^{mt}$ tends to $e^{rt}$ and the discount factor $(1 + r/m)^{-mt}$ tends to $e^{-rt}$.

What the proof uses.

$\ln$ is the inverse of $\exp$ on $(0,\infty)$, and $a^b = e^{b\ln a}$ for $a \gt 0$ and real $b$.
$\ln$ is differentiable at 1 with $\ln'(1) = 1$.
$\exp$ is continuous, and $y \mapsto y^t$ is continuous on $(0,\infty)$.
A limit along $m \to \infty$ can be computed through a substitution $x = r/m$ that tends to 0 through nonzero values.

Conditions used in the proof. $r$ is real; the case $r = 0$ is settled in line 1, and from then on $r \ne 0$. $m$ is an integer with $m \ge 2$ and $m \gt |r|$, so $1 + r/m \gt 0$; a limit as $m \to \infty$ depends only on large $m$. In line 6, $t$ is any real number with $t \gt 0$. The same argument works for real $m \to \infty$.

Second proof: the present value of a level annuity (4 lines)

Theorem (level annuity)

A level annuity pays $C \gt 0$ at the end of each of $n$ periods, where $n \ge 2$ is an integer. With interest rate $i \gt 0$ per period, write $v = 1/(1+i)$ for the one-period discount factor. Its present value is

\[ PV = C\,\frac{1 - (1+i)^{-n}}{i}. \]

Where this goes next: continuous discounting

Later pages discount with $e^{-rT}$. On the Black–Scholes page, under that page's model (a European call on a stock that pays no dividends, constant rate $r$ and volatility $\sigma$), the call price is $C = S_0 N(d_1) - K e^{-rT} N(d_2)$. There $S_0$ is today's stock price, $K$ the strike, $T$ the time to expiry in years, $N$ the standard normal distribution function, and $d_1, d_2$ numbers built from those inputs.

The factor $K e^{-rT}$ is the present value, at continuously compounded rate $r$, of the strike $K$ paid at time $T$. It is the $m \to \infty$ limit of $K(1 + r/m)^{-mT}$ from the theorem above.

Key point $(1 + r/m)^m = \exp\big(m\ln(1 + r/m)\big)$, and the exponent tends to $r$ because $\ln(1+x)/x \to 1$. So continuous discounting over $T$ years multiplies by $e^{-rT}$.

4 Apply

Problems

All offers, loans and rates are hypothetical.

How compounding and discounting move money through time

$600 today or $1,000 in ten years?

Where the two offers are equal

Compounding more often, and its limit

Explorer: reach the goals

Why the limit is \(e^r\)

Problems

Further reading