1. Interest Formulas – Equal Payment Series
Lecture No.7
Chapter 3
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th edition, © 2007

2. Equal Payment Series Equivalent Future Worth F P 1 2 N A A A N 1 2 NA A A P N 1 2 N
Contemporary Engineering Economics, 4th edition, © 2007

3. Contemporary Engineering Economics, 4th edition, © 2007
Equal Payment Series – Compound Amount Factor F A A A 1 2 N N 1 2 F = 1 2 N A A A
Contemporary Engineering Economics, 4th edition, © 2007

4. Contemporary Engineering Economics, 4th edition, © 2007
Process of Finding the Equivalent Future Worth, F F A
A(1+i)N-2 A A A
A(1+i)N-1 N 1 2 1 2 N
Contemporary Engineering Economics, 4th edition, © 2007

5. Another Way to Look at the Compound Amount Factor
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6. Contemporary Engineering Economics, 4th edition, © 2007
Equal Payment Series Compound Amount Factor (Future Value of an Annuity) F N A
Example:
Given: A = $5,000, N = 5 years, and i = 6%
Find: F
Solution: F = $5,000(F/A,6%,5) = $28,185.46
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7. Contemporary Engineering Economics, 4th edition, © 2007
Validation
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8. Contemporary Engineering Economics, 4th edition, © 2007
Finding an Annuity Value F N
A = ?
Example:
Given: F = $5,000, N = 5 years, and i = 7%
Find: A
Solution: A = $5,000(A/F,7%,5) = $869.50
Contemporary Engineering Economics, 4th edition, © 2007

9. Handling Time Shifts in a Uniform Series
First deposit occurs at n = 0
i = 6%
$5,000 $5, $5, $5,000 $5,000
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10. Contemporary Engineering Economics, 4th edition, © 2007
Annuity Due
Excel Solution
Beginning period
=FV(6%,5,5000,0,1)
Contemporary Engineering Economics, 4th edition, © 2007

11. Contemporary Engineering Economics, 4th edition, © 2007
Sinking Fund Factor F N A
Example: College Savings Plan
Given: F = $100,000, N = 8 years, and i = 7%
Find: A
Solution:
A = $100,000(A/F,7%,8) = $9,746.78
Contemporary Engineering Economics, 4th edition, © 2007

12. Contemporary Engineering Economics, 4th edition, © 2007
Excel Solution
Given:
F = $100,000
i = 7%
N = 8 years
Find:
=PMT(i,N,pv,fv,type)
=PMT(7%,8,0,100000,0)
=$9,746.78
Contemporary Engineering Economics, 4th edition, © 2007

13. Contemporary Engineering Economics, 4th edition, © 2007
Example 3.15 Combination of a Uniform Series and a Single Present and Future Amount
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14. Solution: A Two-Step Approach
Step 1: Find the required savings at n = 5.
Step 2: Find the required annual contribution (A) over 5 years.
Contemporary Engineering Economics, 4th edition, © 2007

15. Comparison of Three Different Investment Plans – Example 3.16
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16. Contemporary Engineering Economics, 4th edition, © 2007
Solution:
Investor A:
Investor B:
Investor C:
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17. How Long Would It Take to Save $1 Million?
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18. Contemporary Engineering Economics, 4th edition, © 2007
Loan Cash Flows
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19. Contemporary Engineering Economics, 4th edition, © 2007
Capital Recovery Factor P N
A = ?
Example: Paying Off an Educational Loan
Given: P = $21,061.82, N = 5 years, and i = 6%
Find: A
Solution: A = $21,061.82(A/P,6%,5) = $5,000
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20. Example 3.17 Loan Repayment
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21. Contemporary Engineering Economics, 4th edition, © 2007
Solution:
Using Interest Factor:
Using Excel:
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22. Example 3.18 – Deferred Loan Repayment
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23. Contemporary Engineering Economics, 4th edition, © 2007
A Two-Step Procedure
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24. Present Worth factor – Find P, Given A, i, and N N A
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25. Example 3.19 Louise Outing’s Lottery Problem
Given:
A = $280,000
i = 8%
N = 19
Find: P
Using interest factor:
P =$280,000(P/A,8%,19)
= $2,689,008
Using Excel:
=PV(8%,19, )
Contemporary Engineering Economics, 4th edition, © 2007