1. Interest Formulas (Gradient Series)
Lecture No.8
Chapter 3
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th edition, © 2007

2. Linear Gradient SeriesP
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3. Contemporary Engineering Economics, 4th edition, © 2007
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

4. Example – Present value calculation for a gradient series
$2,000
$1,750
$1,500
$1,250
$1,000
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?
P =?
Contemporary Engineering Economics, 4th edition, © 2007

5. Contemporary Engineering Economics, 4th edition, © 2007
Method 1: Using the (P/F, i, N) Factor
$2,000
$1,750
$1,500
$1,250
$1,000
$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
P =?
Contemporary Engineering Economics, 4th edition, © 2007

6. Method 2: Using the Gradient Factor
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7. Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N)
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8. Example 3.21 – Find the Equivalent Uniform Deposit Plan
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9. Contemporary Engineering Economics, 4th edition, © 2007
Solution:
Contemporary Engineering Economics, 4th edition, © 2007

10. Example 3.22 Declining Linear Gradient Series
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11. Contemporary Engineering Economics, 4th edition, © 2007
Solution:
Contemporary Engineering Economics, 4th edition, © 2007

12. Types of Geometric Gradient Series
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13. Contemporary Engineering Economics, 4th edition, © 2007
Present Worth Factor
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14. Example 3.23 Annual Power Cost if Repair is Not Performed
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15. Solution – Adopt the new compressed-air system
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16. Example 3.24 Jimmy Carpenter’s Retirement Plan – Save $1 Million
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17. What Should be the Size of his first Deposit (A1)?
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