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#### Geometric Gradient Series

Go to questions covering topic below

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

A_{1} = payment at EOY 1

The Present Worth (P) is
given by:

P = A_{1} ( P/ A_{1}, g, i, n ) = A_{1} [ 1 – ( 1 + g ) ^{n}
( 1 + i ) ^{–n} ] / ( i – g )

Example 1: A_{1} =
$100; g = 15%; i = 9%; n = 5

P = A_{1} ( P/ A_{1},
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: A_{1} =
$100; g = 5%; i = 8%; n = 10

P = A_{1} ( P/ A_{1},
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