Interest Formulas

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Interest Formulas
(Gradient Series)
Lecture No.8
Chapter 3
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th
edition, © 2007
Linear Gradient Series
L
i (1  i )  iN  1 O
P  GM
P
N i (1  i ) Q
N
2
 G ( P / G , i, N )
P
Contemporary Engineering Economics, 4th
edition, © 2007
N
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, 4th
edition, © 2007
Example – Present value calculation for a gradient
$2,000
series
$1,250 $1,500
$1,750
$1,000
0
1
P =?
2
3
4
5
How much do you have to deposit
now in a savings account that
earns a 12% annual interest, if
you want to withdraw the annual
series as shown in the figure?
Contemporary Engineering Economics, 4th
edition, © 2007
Method 1: Using the (P/F, i, N) Factor
$2,000
$1,250 $1,500
$1,750
$1,000
0
1
P =?
2
3
4
5
$1,000(P/F, 12%, 1) = $892.86
$1,250(P/F, 12%, 2) = $996.49
$1,500(P/F, 12%, 3) = $1,067.67
$1,750(P/F, 12%, 4) = $1,112.16
$2,000(P/F, 12%, 5) = $1,134.85
$5,204.03
Contemporary Engineering Economics, 4th
edition, © 2007
Method 2: Using the Gradient Factor
P1  $1, 000 ( P / A ,12% ,5 )
 $3, 604 .80
P2  $250( P / G ,12% ,5 )
 $1, 599 .20
P  $3, 604 .08  $1, 599 .20
 $5,204
Contemporary Engineering Economics, 4th
edition, © 2007
Gradient-to-Equal-Payment Series
Conversion Factor, (A/G, i, N)
Contemporary Engineering Economics, 4th
edition, © 2007
Example 3.21 – Find the Equivalent
Uniform Deposit Plan
Contemporary Engineering Economics, 4th
edition, © 2007
Solution:
Given : A1  $1, 000, G  $300, i  10% , and, N  6
Find : A
A  $1, 000  $300( A / G ,10% , 6)
 $1, 000  $300(2.22236)
 $1, 667.08
Contemporary Engineering Economics, 4th
edition, © 2007
Example 3.22 Declining Linear Gradient
Series
Contemporary Engineering Economics, 4th
edition, © 2007
Solution:
F  F1  F2
Equivalent Present W orth at n = 0
 A1 ( F / A ,10% , 5)  $200( P / G ,10% , 5) ( F / P ,10% , 5)
 $1, 200(6.105)  $200(6.862)(1.611)
 $5,115
Contemporary Engineering Economics, 4th
edition, © 2007
Types of Geometric Gradient Series
g  0
Contemporary Engineering Economics, 4th
edition, © 2007
Present Worth Factor
R
L
1  (1  g ) (1  i ) O
|
AM
, if i  g
P
PS
ig
N
Q
|TN A / (1  i ),
if i  g
N
N
1
1
Contemporary Engineering Economics, 4th
edition, © 2007
Example 3.23 Annual Power Cost if
Repair is Not Performed
Contemporary Engineering Economics, 4th
edition, © 2007
Solution – Adopt the new compressed-air
system
PO ld
 1  (1  0.07 ) 5 (1  0.12)  5 
 $54, 440 

0.12  0.07


 $222, 283
PN ew  $54, 440(1  0.23)( P / A ,12% , 5)
 $41, 918.80(3.6048)
 $151,109
Contemporary Engineering Economics, 4th
edition, © 2007
Example 3.24 Jimmy Carpenter’s
Retirement Plan – Save $1 Million
Contemporary Engineering Economics, 4th
edition, © 2007
What Should be the Size of his first
Deposit (A1)?
 1  (1  0 .0 6 ) (1  0 .0 8)
 A1 
0 .0 8  0 .0 6

20
F2 0
 A1 (7 2 .6 9 1 1)
 $ 1, 0 0 0, 0 0 0
A1 
$ 1, 0 0 0, 0 0 0
7 2 .6 9 1 1
 $ 1 3, 7 5 7
Contemporary Engineering Economics, 4th
edition, © 2007
20



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