Interest Formulas
(Gradient Series)
Lecture No. 8
Chapter 3
Contemporary Engineering Economics
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Linear Gradient Series
A Strict Gradient Series
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Gradient Series as a Composite Series of a
Uniform Series of N Payments of A1 and the
Gradient Series of Increments of Constant
Amount G
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Example 3.18: Linear Gradient: Find P, Given A1,
G, N, and i
Given: A1 = $1,000, G = $250, N = 5 years, and i = 12%
per year
Find: P
Contemporary Engineering Economics, 6 th edition
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Solution
Excel Solution
Contemporary Engineering Economics, 6 th edition
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Gradient-to-Equal-Payment Series
Conversion Factor, (A/G, i, N)
Given:
G = $1,000,
N = 10 years, i =
12%
Find: A
Solution
• Cash Flow Series
• Factor Notation
Contemporary Engineering Economics, 6 th edition
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Example 3.19: Linear Gradient: Find A,
Given A1, G, i, and N
Given: A1 = $1,000, G = $300, N = 6 years, and i =
10% per year
Find: A
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Solution
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Example 3.20: Declining Linear Gradient Series:
Find F, Given A1, G, I, and N
Given: A1 = $1,200,
G = -$200, N = 5 years,
and i = 10% per year
Find: F
Contemporary Engineering Economics, 6 th edition
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Solution
Strategy: Since we have no
interest formula to compute
the future worth of a linear
gradient series directly, we first
find the equivalent present
worth of the gradient series
and then convert this P to its
equivalent F.
Solution
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Present Worth of Geometric
Gradient Series
Formula
Factor Notation
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Example 3.21: Geometric Gradient Series
Given: A1 = $54,600,
g = 7%, N = 5 years,
and i = 12% per year
Find: P
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Solution
Contemporary Engineering Economics, 6 th edition
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Example 3.22: Retirement Plan: Saving $1
Million
Given:
o F = $1,000,000,
o g = 6%,
o i = 8%, and
o N = 20
Find: A1
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Solution
Contemporary Engineering Economics, 6 th edition
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