Case study: annuities, perpetuities and bond prices

Cash flows on a timeline

An annuity pays the same amount $C$ once a period for $n$ periods. On a desk the first question is when each payment arrives.

A loan repayment, a bond coupon or a pension payment usually comes at the end of each period. So the first one is a full period away.

Each payment is discounted back to today, as in case study 0.2.4: a payment at time $t$ is worth $\frac{C}{(1 + r)^t}$ now. Here $r$ is the rate per period, compounded once per period, and $t$ counts periods.

Rate and period must match. A rate quoted as 6% a year and paid monthly means $r = 0.06 / 12 = 0.005$ per month, over $n = 12 \times 30 = 360$ months (case study 1.4.6). Mixing a yearly rate with monthly payments is the most common error on a desk.

An annuity is a geometric series

Add the present values of the $n$ payments:

Write $v = \frac{1}{1 + r}$, the value today of $1 paid in one period. Each term is the one before times $v$, so this is a geometric series (lesson 1.5.2) with first term $Cv$, ratio $v$ and $n$ terms.

Its sum is $Cv\frac{1 - v^n}{1 - v}$. Since $1 - v = \frac{r}{1 + r}$, the front part $\frac{Cv}{1 - v}$ is just $\frac{C}{r}$, and $v^n = (1 + r)^{-n}$.

When the rate is zero

At $r = 0$ the formula reads $\frac{0}{0}$: the ratio $v$ is exactly 1, the case lesson 1.5.2 warned about. Nothing is discounted, so the value is simply $nC$. Twelve payments of $100 at 0% are worth $1,200.

Zero and even slightly negative rates do happen, so pricing code needs a branch for $r = 0$. A negative rate above $-100\%$ still works in the formula: $v$ is then above 1, so each payment is worth more today than its amount.

Payments at the start: the annuity due

Rent, leases and insurance premiums are usually paid in advance, at the start of each period. A stream whose first payment is today is an annuity due.

Every payment arrives one period earlier than in the ordinary annuity, so each is worth $1 + r$ times as much. The total is too.

In code, the two timings differ only in where the sum starts. The end of each period is range(1, n + 1); the start is range(0, n). An off-by-one there is a real pricing error, not a rounding one.

Perpetuities: the annuity that never stops

A perpetuity pays $C$ every period for ever. Its value is the limit of the annuity as $n$ grows. When $r > 0$, the ratio $v$ is below 1, so $v^n = (1 + r)^{-n} \to 0$ (lesson 1.5.1) and the infinite geometric series of lesson 1.5.3 converges.

How many years does it take to get within 1% of the perpetuity? The annuity is $\frac{C}{r}$ times $1 - (1.04)^{-n}$, so it falls short by the fraction $(1.04)^{-n}$. Solve $(1.04)^{-n} < 0.01$ as in lesson 1.5.3: $n > \frac{\ln 100}{\ln 1.04} \approx 117.4$, so 118 years.

Turning this round explains the annuity formula. An $n$-period annuity is a perpetuity starting now minus one whose first payment is in period $n + 1$.

The second is worth $\frac{C}{r}$ at time $n$, so $\frac{C}{r}(1 + r)^{-n}$ today. For $5 at 4% over 10 years, $125 - 125 \times 1.04^{-10} = 40.55$.

A growing perpetuity, and when it breaks

If the payment grows by a fraction $g$ each period, the terms are $\frac{C}{1 + r}, \frac{C(1 + g)}{(1 + r)^2}, \ldots$: a geometric series with ratio $\frac{1 + g}{1 + r}$. It converges only when that ratio is below 1, that is when $g < r$.

When $g \ge r$ the series diverges, exactly as lesson 1.5.3 found for a ratio of 1 or more. The formula does not warn you: with $g = 7\%$ it returns $\frac{5}{-0.01} = -500$, a negative price for positive payments.

Analysts using this formula to value shares check $g < r$ first. For the same reason, a level perpetuity has no finite value at a zero or negative rate.

Bond prices

A bond pays a coupon $c$ (in dollars, once a year) and repays its face value $F$ after $T$ years. So it is an annuity of coupons plus one final payment. Case study 1.3.5 added the payments one by one; the annuity factor does the coupons in one step.

Discount both at the bond's yield $y$:

The 5-year 4% bond pays $4 a year on $100 of face value. At a 5% yield its coupons are worth 17.32 and its face value 78.35, so its price is 95.67, as in case study 1.3.5. Prices are quoted per 100 of face value.

At, above and below par

The coupon rate is $\frac{c}{F}$, here 4%. When the yield equals the coupon rate, $c = yF$, so the coupon part is $F\big(1 - (1 + y)^{-T}\big)$. Adding the face part $F(1 + y)^{-T}$ leaves exactly $F$: the bond is priced at par, whatever its maturity.

At a 0% yield nothing is discounted, so the price is the plain total of the payments, $5 \times 4 + 100 = 120$. That is the zero-rate branch again.

The curve falls as the yield rises, and it bends upwards. Case study 2.2.4 measures how fast it falls, the bond's duration, using this formula.

Two desk conventions

Pull to par. As a bond nears maturity its price moves towards face value. At a 5% yield the 4% bond is worth 92.28 with 10 years left, 95.67 with 5, 98.14 with 2 and 99.05 with 1. At maturity it repays exactly 100, with its last coupon.

Semi-annual coupons. Many government and corporate bonds pay half the coupon every six months. Then the coupon is $\frac{c}{2}$, the rate per period is $\frac{y}{2}$ and there are $2T$ periods. The 5-year 4% bond at a 5% yield, priced this way, is worth 95.62 instead of 95.67.

Solving for the payment

A loan is an annuity seen from the lender's side: the amount lent is the present value of the payments. Rearrange the annuity formula for $C$ (lesson 0.3.2).

Each payment first covers the month's interest on what is still owed; the rest repays the loan. The balance at any time is the present value of the payments still to come. After 10 years, 240 payments remain, worth $251,057.17: about 84% of the loan is still owed after a third of the term.

def test_annuity_matches_sum():
    """$300 a year for 4 years at 8% matches the direct sum, about $993.64"""
    direct = sum(300 / 1.08**t for t in range(1, 5))
    expect_close(annuity_pv(300, 0.08, 4), direct, what="annuity_pv(300, 0.08, 4)")


def test_annuity_due():
    """Paid at the start of each year instead, the same $300 is worth about $1,073.13"""
    direct = sum(300 / 1.08**t for t in range(0, 4))
    expect_close(annuity_pv(300, 0.08, 4, due=True), direct, what="annuity_pv(300, 0.08, 4, due=True)")


def test_annuity_zero_rate():
    """At a 0% rate, 12 payments of $100 are worth $1,200"""
    expect_close(annuity_pv(100, 0, 12), 1200.0, what="annuity_pv(100, 0, 12)")


def test_long_annuity_near_perpetuity():
    """$5 a year for 500 years at 4% is within a cent of the perpetuity, $125"""
    expect_close(annuity_pv(5, 0.04, 500), 125.0, what="annuity_pv(5, 0.04, 500)", abs_tol=0.01, rel=0)


def test_mortgage_payment():
    """A $300,000 loan at 0.5% a month for 360 months costs about $1,798.65 a month"""
    expect_close(loan_payment(300_000, 0.005, 360), 1798.6515754582708, what="loan_payment(300000, 0.005, 360)")


def test_payment_repays_loan():
    """The payments' present value equals the amount borrowed"""
    pay = loan_payment(20_000, 0.004, 60)
    expect_close(sum(pay / 1.004**t for t in range(1, 61)), 20_000, what="present value of your payments")


def test_loan_zero_rate():
    """At a 0% rate, $12,000 over 24 months costs $500 a month"""
    expect_close(loan_payment(12_000, 0, 24), 500.0, what="loan_payment(12000, 0, 24)")


def test_bond_at_par():
    """A 4% bond at a 4% yield is priced at its face value, 100"""
    expect_close(bond_price(4, 100, 5, 0.04), 100.0, what="bond_price(4, 100, 5, 0.04)")


def test_bond_matches_sum():
    """The 5-year 4% bond at 5% matches the payment-by-payment sum, about 95.67"""
    direct = sum(4 / 1.05**t for t in range(1, 6)) + 100 / 1.05**5
    expect_close(bond_price(4, 100, 5, 0.05), direct, what="bond_price(4, 100, 5, 0.05)")


def test_bond_premium():
    """At a 3% yield the same bond trades above par, at about 104.58"""
    direct = sum(4 / 1.03**t for t in range(1, 6)) + 100 / 1.03**5
    expect_close(bond_price(4, 100, 5, 0.03), direct, what="bond_price(4, 100, 5, 0.03)")


def test_bond_zero_yield():
    """At a 0% yield the price is the plain total of the payments, 120"""
    expect_close(bond_price(4, 100, 5, 0), 120.0, what="bond_price(4, 100, 5, 0)")