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Geometrically increasing annuities

Payments growing by a fixed PERCENTAGE each period - inflation-linked cash flows. One substitution turns it back into a level annuity.

The formulas

Present value

$$PV = \dfrac{1 - \left(\dfrac{1+g}{1+i}\right)^{n}}{i - g}$$

first payment 1 at time 1, growing at rate g

The trick rate

$$1 + j = \dfrac{1+i}{1+g}, \qquad PV = \dfrac{a_{\overline{n}|j}}{1+g}$$

When i = g

$$PV = n\,v = \dfrac{n}{1+i}$$

Growing perpetuity

$$PV = \dfrac{1}{i-g}, \quad g < i$$

Where it comes from

1. The $k$th payment is $(1+g)^{k-1}$ at time $k$, with present value $(1+g)^{k-1}v^k$.
2. The series is geometric with ratio $\frac{1+g}{1+i}$, so it sums to $\frac{1 - \left(\frac{1+g}{1+i}\right)^n}{i-g}$.
3. Defining $j$ by $1+j = \frac{1+i}{1+g}$ makes the same sum $\frac{1}{1+g}a_{\overline{n}|j}$ - a level annuity at the 'real' rate $j$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• The denominator is i − g, not i. If the payments grow, the effective discount is only the EXCESS of i over g.
• When i = g every term is v, so the answer is just n·v - no division by zero, a limit.

Traps

• Dividing by i − g when the growth starts at the FIRST payment rather than the second - check whether payment 1 is 1 or (1+g).
• Using this for arithmetic increases.

Related

• Increasing annuities
• Perpetuities

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