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## Geometric Gradient Series

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#### Geometric Gradient Series

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Suppose that there is a series of "n" payments uniformly spaced, but differing from one period to the next by a constant multiple. The change or "gradient" multiple from one period to the next is denoted "g." There will, of course, also be an interest rate “i” that applies.

g = gradient percent (as a decimal in calculations)

i = interest percent (as a decimal in calculations)

n = number of periods

A1 = payment at EOY 1

The Present Worth (P) is given by:

P = A1 ( P/ A1, g, i, n ) = A1 [ 1 – ( 1 + g ) n ( 1 + i ) –n ] / ( i – g )

Example 1: A1 = \$100; g = 15%; i = 9%; n = 5

P = A1 ( P/ A1, g, i, n ) = 100 [ 1 – ( 1 + 0.15 ) 5 ( 1 + 0.09 ) –5 ] / ( 0.09 – 0.15 )

= 100 [ 1 – 2.0114 * 0.6499] / (-0.06)

= 100 [-0.3072] / (-0.06)

= 512.1

Example 2: A1 = \$100; g = 5%; i = 8%; n = 10

P = A1 ( P/ A1, g, i, n ) = 100 [ 1 – ( 1 + 0.05 ) 5 ( 1 + 0.08 ) –5 ] / ( 0.08 – 0.05 )

= 100 [ 1 – 1.628894627 * 0.463193488] / (0.03)

= 100 [ 0.2455] / (0.03)

= 818.3554

### More Interest Formulas

#### Geometric Gradient Series

Question 1

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Question 1.

Suppose that the maintenance for a piece of equipment costs \$300 EOY1 and increases by 15% every year for 5 years. The value of money is 9%. What is the equivalent present cost over the time horizon?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A P = \$2022.71

B P = \$1175.16

C P = \$1536.22

D P = \$1166.90

Review topic

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