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Contemporary Engineering Economics, 4 th edition, ©2007 Interest Formulas for Single Cash Flows Lecture No.6 Chapter 3 Contemporary Engineering Economics.

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Presentation on theme: "Contemporary Engineering Economics, 4 th edition, ©2007 Interest Formulas for Single Cash Flows Lecture No.6 Chapter 3 Contemporary Engineering Economics."— Presentation transcript:

1 Contemporary Engineering Economics, 4 th edition, ©2007 Interest Formulas for Single Cash Flows Lecture No.6 Chapter 3 Contemporary Engineering Economics Copyright © 2006

2 Contemporary Engineering Economics, 4 th edition, © 2007 Types of Common Cash Flows in Engineering Economics

3 Contemporary Engineering Economics, 4 th edition, © 2007 Equivalence Relationship Between P and F

4 Contemporary Engineering Economics, 4 th edition, © 2007 Single Cash Flow Formula- Compound-Amount Factor Single payment compound amount factor (growth factor) Given: Find: F P F N 0

5 Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years? $2,000 F = ? 8 0 i = 10%

6 Contemporary Engineering Economics, 4 th edition, © 2007 Solution

7 Contemporary Engineering Economics, 4 th edition, © 2007 Single Cash Flow Formula – Present-Worth Factor Single payment present worth factor (discount factor) Given: Find: P F N 0

8 Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem You want to set aside a lump sum amount today in a savings account that earns 7% annual interest to meet a future expense in the amount of $10,000 to be incurred in 6 years. How much do you need to deposit today?

9 Contemporary Engineering Economics, 4 th edition, © 2007 Solution 0 6 $10,000 P

10 Contemporary Engineering Economics, 4 th edition, © 2007 Solving for i

11 Contemporary Engineering Economics, 4 th edition, © 2007 Solution EXCEL Solution using RATE Function =RATE(5,0,-10,20)=14.87%

12 Contemporary Engineering Economics, 4 th edition, © 2007 Rule of 72 – Number of Years Required to Double Your Investment

13 Contemporary Engineering Economics, 4 th edition, © 2007 Solution – Analytical Approach P 2P2P 0 N = ?

14 Contemporary Engineering Economics, 4 th edition, © 2007 Solution - Rule of 72 Approximating how long it will take for a sum of money to double

15 Contemporary Engineering Economics, 4 th edition, © 2007 Number of Years Required to Double an Initial Investment at Various Interest Rates

16 Contemporary Engineering Economics, 4 th edition, © 2007 Uneven Payment Series How much do you need to deposit today (P) to withdraw $25,000 at n =1, $3,000 at n = 2, and $5,000 at n =4, if your account earns 10% annual interest? 0 1 2 3 4 $25,000 $3,000 $5,000 P

17 Contemporary Engineering Economics, 4 th edition, © 2007 0 1 2 3 4 $25,000 $3,000 $5,000 P 0 1 2 3 4 $25,000 P1P1 0 1 2 3 4 $3,000 P2P2 0 1 2 3 4 $5,000 P4P4 ++ Uneven Payment Series

18 Contemporary Engineering Economics, 4 th edition, © 2007 Check 01234 Beginning Balance 028,6226,484.204,132.624,545.88 Interest Earned (10%) 02,862648.42413.26454.59 Payment+28,622-25,000-3,0000-5,000 Ending Balance $28,6226,484.204,132.624,545.880.47 Rounding error It should be “0.”


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