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Geometric Gradient Present Worth

P = A_{1} \frac{1 - ( 1 + g ) ^{n} ( 1 + i ) ^{- n}}{i - g} (i \neq = g); P = \frac{n A _{1}}{1 + i} (i = g)

A geometric gradient grows (or shrinks) by a constant rate g each period: A_t = A_1 (1+g)^{t−1}. The Handbook closed form discounts the series to t = 0. Do not confuse with the arithmetic gradient G (0, G, 2G, …).

A_{1}

First-period cash flow

g

Growth rate per period

i

Interest rate per period

n

Number of periods

Step-by-step solved example

A_1 = $1,000, g = 5%, i = 10%, n = 2. Find P.

Each A grows by (1+g) per period.

1. Cash flows
Year 1: 1000. Year 2: 1000(1.05) = 1050.
2. Closed form
P = 1000 [1 − (1.05)²(1.10)^{−2}] / (0.10−0.05) = 1000[1 − 1.1025/1.21]/0.05 = 1000(0.08884)/0.05 = $1,777.
$P = 1777$

Answer: P ≈ $1,777

10 practice questions

0/10 correct

1.If g = 0 the geometric-gradient P reduces to

2.When i = g the Handbook formula uses

3.A_t = A_1 (1+g)^{t−1}. For t = 1, A_1 equals

4.g = −0.04 means the cash flow

5.A_1 = $500, g = 0, i = 10%, n = 1. P =

6.Arithmetic gradient G vs geometric g: year-1 gradient cash is

7.If g > i > 0 and n is large, P of a growing series

8.Units of g and i must be

9.A_1 = 100, g = 10%, n = 2 (no interest). Sum of cash flows =

10.The factor [1 − (1+g)^n (1+i)^{−n}] / (i−g) is