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Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005
16

Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Dec 22, 2015

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Page 1: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Interest Formulas(Gradient Series)

Lecture No.6Professor C. S. ParkFundamentals of Engineering EconomicsCopyright © 2005

Page 2: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Linear Gradient Series

P

Page 3: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Present Value of Perpetuities

Perpetual Cash flow with equal payment series

P=A/i A=Annual perpetual cash stream i=Interest rate

Page 4: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Gradient Series as a Composite Series

Page 5: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

$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 6: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Method 1:

$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 7: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Method 2:

P P A1 000 12%,5

604 80

$1, ( / , )

$3, .

P P G2 12%,5

599 20

$250( / , )

$1, .

P

$3, . $1, .

$5,204

604 08 599 20

Page 8: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

0 1 2 3 4 5 6 7 25 26

$3.44 million

0 1 2 3 4 5 6 7 25 26

Cash Option

$175,000$189,000

$357,000

$196,000G = $7,000

Annual Payment Option

Example 2.16 Supper Lottery

Page 9: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Equivalent Present Value of Annual Payment Option at 4.5%

[$175,000 $189,000( / , 4.5%,25)

$7,000( / , 4.5%,25)]( / , 4.5%,1)

$3,818,363

P P A

P G P F

Page 10: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

ExcelSolution

Page 11: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Geometric Gradient SeriesMany engineering economic problems, particularly those relating to construction costs, involve cash flows that increase over time, not by a constant amount, but rather by a constant percentage (geometric), called compound growth.

PA

g i

i gi g

NA i i g

N N

1

1

1 1 1

1

( ) ( )

/ ( ),

, if

if

Page 12: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Alternate Way of Calculating P

1

Let '1

( / , ', )(1 )

i gg

g

AP P A g N

g

Page 13: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Example 2.17: Find P, Given A1,g,i,N Given:g = 5%

i = 7%

N = 25 years

A1 = $50,000

Find: P

25 251 (1 0.05) (1 0.07)$50,000

0.07 0.05

$940,696

P

Page 14: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Required Additional Savings

$50,000( / ,7%,25)

$582,679

$940,696 $582,679

$358,017

P P A

P

Page 15: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

The Five Types of Cash Flows

(a) Single cash flow

(b) Equal (uniform) payment series

(c) Linear gradient series

(d) Geometric gradient series

(e) Irregular payment series

Page 16: Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

Summary

Money has a time value because it can earn more money over time.

Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same value. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.

The purpose of developing various interest formulas was to facilitate the economic equivalence computation.