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Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2006

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Page 1: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Interest Formulas(Gradient Series)

Lecture No.8Chapter 3Contemporary Engineering EconomicsCopyright © 2006

Page 2: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Linear Gradient Series

P Gi i iN

i i

G P G i N

N

N

LNM

OQP

( )

( )

( / , , )

1 1

12

P

Page 3: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

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.

Page 4: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

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?

Example – Present value calculation for a gradient series

Page 5: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Method 1: Using the (P/F, i, N) Factor

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

$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

Page 6: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Method 2: Using the Gradient Factor

P P G2 12%,5

599 20

$250( / , )

$1, .P

$3, . $1, .

$5,204

604 08 599 20

P P A1 000 12%,5

604 80

$1, ( / , )

$3, .

Page 7: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N)

Page 8: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Example 3.21 – Find the Equivalent Uniform Deposit Plan

Page 9: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Solution:

1Given : $1,000, $300, 10%,and, 6

Find :

$1,000 $300( / ,10%,6)

$1,000 $300(2.22236)

$1,667.08

A G i N

A

A A G

Page 10: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Example 3.22 Declining Linear Gradient Series

Page 11: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Solution:

1 2

Equivalent Present Worth at = 0

1( / ,10%,5) $200( / ,10%,5) ( / ,10%,5)

$1,200(6.105) $200(6.862)(1.611)

$5,115

n

F F F

A F A P G F P

Page 12: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Types of Geometric Gradient Series

0g

Page 13: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Present Worth Factor

P Ag i

i gi g

NA i i g

N N

LNM

OQP

RS|T|

1

1

1 1 1

1

( ) ( )

/ ( ),

, if

if

Page 14: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Example 3.23 Annual Power Cost if Repair is Not Performed

Page 15: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Solution – Adopt the new compressed-air system

5 51 (1 0.07) (1 0.12)$54,440

0.12 0.07

$222,283

$54,440(1 0.23)( / ,12%,5)

$41,918.80(3.6048)

$151,109

Old

New

P

P P A

Page 16: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

Example 3.24 Jimmy Carpenter’s Retirement Plan – Save $1 Million

Page 17: Contemporary Engineering Economics, 4 th edition, © 2007 Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics

What Should be the Size of his first Deposit (A1)?

20 20

20 1

1

1

1 (1 0.06) (1 0.08)

0.08 0.06

(72.6911)

$1,000,000

$1,000,000

72.6911$13,757

F A

A

A