Transcript Interest Formulas – Gradient Series

Lecture No.8
Chapter 3
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Linear Gradient Series
A Strict Gradient Series
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.
Contemporary Engineering Economics, 5th edition, © 2010
Example3.21 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
 Excel Solution:
Contemporary Engineering Economics, 5th edition, © 2010
Gradient-to-Equal-Payment Series Conversion Factor,
(A/G, i, N)
 Cash Flow Series
 Given: G = $1,000, N = 10
years, i = 12%
 Find: A
 Solution:
A  $1,000(A / G ,12%,10)
 $1,000(3.5847)
 Factor Notation
 $3,584.70
Contemporary Engineering Economics, 5th edition, © 2010
Example 3.22 – 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
Contemporary Engineering Economics, 5th edition, © 2010
Example 3.23 Declining Linear Gradient Series
 Given: A1 = $1,200, G = -
$200, N = 5 years, and i = 10%
per year
 Find: F
 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.
Contemporary Engineering Economics, 5th edition, © 2010
Present Worth of Geometric Gradient Series
 Formula:
 Factor Notation:
Contemporary Engineering Economics, 5th edition, © 2010
Example 3.24 –Geometric Gradient Series
 Given: A1 = $54,600, g = 7%,
N = 5 years, and i = 12% per
year
 Find: P
 1  (1  0.07)5 (1  0.12)5 
POld  $54,600 

0.12

0.07


 $222,937
PNew  $54,600(1  0.23)(P / A,12%,5)
 $42,042(3.6048)
 $151,552
Contemporary Engineering Economics, 5th edition, © 2010
Example 3.25 Retirement Plan – Saving $1 Million
 Given: F
= $1,000,000, g =
6%, i = 8%, and N = 20
 Find: A1
 Solution:
Contemporary Engineering Economics, 5th edition, © 2010