Case study: compounding and continuous rates

Compounding $n$ times a year

A quoted annual rate $r$ paid $n$ times a year pays a fraction $r/n$ of the balance at the end of each period. The interest then earns interest itself. That is all compounding means.

So each period multiplies the balance by the same factor, $1 + r/n$. This is percentage growth from lesson 1.4.2, with $g = r/n$. Throughout, $r$ is a yearly rate written as a decimal (5% is 0.05) and $t$ is a time in years.

1. 1
   $$\begin{gathered} \text{after 1 period:} \\ PV\left(1+\frac{r}{n}\right) \end{gathered}$$

   Why?
2. 2
   $$\begin{gathered} \text{after 1 year:} \\ PV\left(1+\frac{r}{n}\right)^n \end{gathered}$$

   Why?
3. 3
   $$FV = PV\left(1+\frac{r}{n}\right)^{nt}$$

   Why?

Here $PV$ is the present value, the amount you start with, and $FV$ is the future value after $t$ years.

Continuous compounding

In lesson 1.4.2 you saw $(1 + x/n)^n$ settle on $e^x$ as $n$ grows. Here $x = r$, so the yearly factor settles on $e^r$, and $t$ years gives $e^{rt}$. That is continuous compounding, and running it backwards is continuous discounting:

Try the explorer below. Step through the presets from annual to hourly and watch the growth factor climb towards the dashed line at $e^{rt}$. Then raise the rate or the years: the gap between the conventions widens.

Over many years

Over one year the conventions differ by cents. Over ten years at 10%, they differ by dollars.

Simple interest (case study 0.3.5) pays interest on the original $100 only, so it adds $10 a year. Compounding pays interest on interest, so each year adds more than the last. The more often it compounds, the faster it pulls away.

Comparing rates fairly

Rates quoted with different compounding cannot be compared directly. Convert each to its effective annual rate (EAR): the plain growth it produces over one year.

import math


def test_annual():
    """$100 at 5% compounded annually for 1 year is $105"""
    expect_close(future_value(100, 0.05, 1, n=1), 105.0, what="future_value(100, 0.05, 1, n=1)")


def test_monthly():
    """$100 at 5% compounded monthly for 1 year"""
    expect_close(future_value(100, 0.05, 1, n=12), 100 * (1 + 0.05 / 12) ** 12, what="future_value(100, 0.05, 1, n=12)")


def test_many_years():
    """Years multiply the number of periods: 10 years quarterly is 40 periods"""
    expect_close(future_value(250, 0.04, 10, n=4), 250 * 1.01**40, what="future_value(250, 0.04, 10, n=4)")


def test_continuous():
    """n=None compounds continuously: $100 at 5% for 2 years is 100·e^0.1"""
    expect_close(future_value(100, 0.05, 2), 100 * math.exp(0.1), what="future_value(100, 0.05, 2)")


def test_zero_rate():
    """A zero rate leaves the money unchanged, for any convention"""
    expect_close(future_value(80, 0.0, 7, n=12), 80.0, what="future_value(80, 0, 7, n=12)")
    expect_close(future_value(80, 0.0, 7), 80.0, what="future_value(80, 0, 7)")


def test_frequent_approaches_continuous():
    """Compounding a million times a year is within 1e-6 of continuous"""
    frequent = future_value(100, 0.05, 1, n=1_000_000)
    expect_close(frequent, 100 * math.exp(0.05), what="n=1,000,000 vs continuous", rel=1e-8, abs_tol=1e-6)

import math


def test_annual_is_unchanged():
    """Annual compounding: the EAR equals the quoted rate"""
    expect_close(effective_annual_rate(0.05, n=1), 0.05, what="effective_annual_rate(0.05, n=1)")


def test_monthly():
    """5% compounded monthly has an EAR of about 5.1162%"""
    expect_close(effective_annual_rate(0.05, n=12), (1 + 0.05 / 12) ** 12 - 1, what="effective_annual_rate(0.05, n=12)")


def test_continuous():
    """5% continuous has an EAR of e^0.05 - 1, about 5.1271%"""
    expect_close(effective_annual_rate(0.05), math.exp(0.05) - 1, what="effective_annual_rate(0.05)")


def test_returns_a_decimal():
    """The answer is a decimal, not a percentage: 5% annual gives 0.05, not 5"""
    ear = effective_annual_rate(0.05, n=1)
    assert ear is not None and abs(ear) < 1, f"expected a decimal like 0.05, got {ear}"

Converting between conventions

Two quotes are equivalent when they grow $1 by the same amount in a year. Write $r_n$ for a rate compounded $n$ times a year and $r_c$ for a continuous rate. Equivalence says

Solving for each rate takes one log or one root (lessons 1.4.1 and 1.4.4).

Take 6% compounded twice a year. A year multiplies $1 by $1.03^2 = 1.0609$, so the EAR is 6.09%. The equivalent quarterly rate is 5.9557%, the monthly rate 5.9263% and the continuous rate 5.9118%. The more often a rate compounds, the lower it must be quoted to give the same growth.

Year fractions and day counts

Every formula here measures $t$ in years. A desk measures time in days, so each quote comes with a rule for turning days into a year fraction: its day-count convention.

Dollar and euro money markets divide the actual number of days by 360, written ACT/360. Sterling divides by 365, written ACT/365. Many bonds use 30/360, which counts every month as 30 days.

Deposits shorter than a year usually earn simple interest (case study 0.3.5). Put $1,000,000 on deposit for 91 days at 5%:

• ACT/360: $t = \frac{91}{360} \approx 0.252778$, so the interest is $12,638.89.
• ACT/365: $t = \frac{91}{365} \approx 0.249315$, so the interest is $12,465.75.

The same quoted 5% pays $173.14 more under ACT/360. So 5% on ACT/360 is worth $5\% \times \frac{365}{360} \approx 5.0694\%$ on an ACT/365 basis.

How long until money doubles?

Set the growth factor equal to 2 and solve for $t$, exactly as in lesson 1.4.4:

With continuous compounding, $e^{rt} = 2$ gives the simpler $t = \ln 2 / r$. At 7%, doubling takes 10.24 years with annual compounding and 9.90 years continuously.

The formula breaks where the maths says it should. At a rate of 0 it divides by zero, and at a negative rate it returns a negative time. Both mean the money never doubles: $e^{rt} = 2$ has no positive solution unless $r > 0$, the no-solution case of lesson 1.4.4.

What rate does a price imply?

A bond you hold rose from 92.50 to 100 in 1.5 years. The continuous rate $r$ satisfies $100 = 92.50\,e^{1.5r}$. Divide by 92.50, take $\ln$, and divide by 1.5:

A fall in value gives a negative rate. Negative rates are not only for losses: some central banks have set them, so desk code must accept them. The formulas still work, as long as each period's factor $1 + r/n$ stays positive.

Rates over several periods

Continuous rates add over time, because exponents add (lesson 1.4.1). Two years at 4% and then one year at 6%, both continuous, grow $1 to $e^{0.08} e^{0.06} = e^{0.14} \approx 1.150274$. The average rate over the three years is the time-weighted average, $0.14 / 3 \approx 4.6667\%$.

With annual compounding the plain average is slightly off: the equivalent steady rate is $(1.04^2 \times 1.06)^{1/3} - 1 \approx 4.6624\%$. This is one reason risk systems store rates in continuous form. Case study 1.4.7 meets the same idea as log returns that add up.

Units and bad inputs

Most rate bugs on a desk are unit bugs. Check these before trusting a number.

• Percent or decimal. Formulas take decimals: 5% is 0.05. A basis point (bp) is 0.01%, or 0.0001, so a rise of 25 bp takes 5% to 5.25%.
• Per year. $r$ is per year and $t$ is in years. A monthly rate of 0.5% is 6% a year compounded monthly, not 0.5% a year.
• Day counts. Turn days into years with the quote's convention, not always 365.
• Factors stay positive. $\ln(1 + r/n)$ needs $1 + r/n > 0$, so a rate of $-100\%$ or below breaks every formula here.

import math


def test_doubling_annual():
    """Doubling at 7% annual takes ln 2 / ln 1.07, about 10.24 years"""
    expect_close(years_to_grow(2, 0.07, n=1), math.log(2) / math.log(1.07), what="years_to_grow(2, 0.07, n=1)")


def test_doubling_continuous():
    """Doubling at 7% continuous takes ln 2 / 0.07, about 9.90 years"""
    expect_close(years_to_grow(2, 0.07), math.log(2) / 0.07, what="years_to_grow(2, 0.07)")


def test_monthly_tripling():
    """Tripling at 6% monthly: (1 + 0.005)^(12t) = 3"""
    t = years_to_grow(3, 0.06, n=12)
    expect_close((1 + 0.06 / 12) ** (12 * t), 3.0, what="growth after years_to_grow(3, 0.06, n=12) years")


def test_round_trip():
    """Growing for the computed time gives back the multiple"""
    t = years_to_grow(1.5, 0.04, n=4)
    expect_close(1.01 ** (4 * t), 1.5, what="1.01^(4t) with t = years_to_grow(1.5, 0.04, n=4)")


def test_falling_value():
    """At -5% continuous, a value falls to 90% after ln 0.9 / -0.05, about 2.107 years"""
    expect_close(years_to_grow(0.9, -0.05), math.log(0.9) / -0.05, what="years_to_grow(0.9, -0.05)")


def _raises_value_error(*args, **kwargs):
    try:
        years_to_grow(*args, **kwargs)
    except ValueError:
        return True
    except ZeroDivisionError:
        raise AssertionError(f"years_to_grow{args} divided by zero: raise ValueError instead")
    return False


def test_zero_rate_never_doubles():
    """At a 0% rate money never doubles, so raise ValueError"""
    assert _raises_value_error(2, 0.0), "expected ValueError for years_to_grow(2, 0.0)"
    assert _raises_value_error(2, 0.0, n=12), "expected ValueError for years_to_grow(2, 0.0, n=12)"


def test_negative_rate_never_doubles():
    """At -2% a year money never doubles, so raise ValueError (not a negative time)"""
    assert _raises_value_error(2, -0.02), "expected ValueError for years_to_grow(2, -0.02)"
    assert _raises_value_error(2, -0.02, n=1), "expected ValueError for years_to_grow(2, -0.02, n=1)"


def test_implied_rate_bond():
    """92.50 to 100 in 1.5 years implies about 5.197% continuous"""
    expect_close(implied_continuous_rate(92.5, 100, 1.5), math.log(100 / 92.5) / 1.5, what="implied_continuous_rate(92.5, 100, 1.5)")


def test_implied_rate_loss():
    """A fall in value gives a negative rate"""
    r = implied_continuous_rate(100, 90, 2)
    assert r is not None and r < 0, f"expected a negative rate for a loss, got {r}"
    expect_close(100 * math.exp(2 * r), 90.0, what="100·e^(2r)")