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Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations Lecture No. 9 Chapter 3 Contemporary Engineering Economics.

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Presentation on theme: "Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations Lecture No. 9 Chapter 3 Contemporary Engineering Economics."— Presentation transcript:

1 Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2006

2 Contemporary Engineering Economics, 4 th edition, © 2007 Equivalent Present Worth Calculation – Brute Force Approach using Only P/F Factors

3 Contemporary Engineering Economics, 4 th edition, © 2007 $50 $100 $150 $150 $200 0 1 2 3 4 5 6 7 8 9 Equivalent Present Worth Calculation – Grouping Approach

4 Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations – A Personal Savings Problem Situation 1: If you make 4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount (A) can be withdrawn over 4 subsequent years?

5 Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations – An Economic Equivalence Problem Situation 2: What value of A would make the two cash flow transactions equivalent if i = 10%?

6 Contemporary Engineering Economics, 4 th edition, © 2007 Method 1: Establish the Economic Equivalence at n = 0

7 Contemporary Engineering Economics, 4 th edition, © 2007 Method 2: Establish the Economic Equivalence at n = 4

8 Contemporary Engineering Economics, 4 th edition, © 2007 Multiple Interest Rates $300 $500 $400 5% 6% 4% Find the balance at the end of year 5. 0 1 2 3 4 5 F = ?

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

10 Contemporary Engineering Economics, 4 th edition, © 2007 Cash Flows with Missing Payments P = ? $100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Missing payment i = 10%

11 Contemporary Engineering Economics, 4 th edition, © 2007 Solution P = ? $100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Pretend that we have the 10 th Payment in the amount of $100 i = 10% $100 Add $100 to offset the change

12 Contemporary Engineering Economics, 4 th edition, © 2007 Approach P = ? $100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 i = 10% $100 Equivalent Cash Inflow = Equivalent Cash Outflow

13 Contemporary Engineering Economics, 4 th edition, © 2007 Equivalence Relationship

14 Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Regularity in Cash Flow Pattern $10,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C i = 10% Payments are made every other year

15 Contemporary Engineering Economics, 4 th edition, © 2007 Approach 1: Modify the Original Cash Flows $10,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 i = 10% A A A A A A A

16 Contemporary Engineering Economics, 4 th edition, © 2007 Relationship Between A and C $10,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 i = 10% A A A A A A A $10,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C i = 10%

17 Contemporary Engineering Economics, 4 th edition, © 2007 C A =$1,357.46 AA i = 10% Solution

18 Contemporary Engineering Economics, 4 th edition, © 2007 Approach 2: Modify the Interest Rate Idea: Since cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period. How: If interest is compounded 10% annually, the equivalent interest rate for two- year period is 21%. (1+0.10)(1+0.10) = 1.21

19 Contemporary Engineering Economics, 4 th edition, © 2007 Solution $10,000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C i = 21% 1 2 3 4 5 6 7

20 Contemporary Engineering Economics, 4 th edition, © 2007 Example 3.25 – At What Value of C would Make the Two Cash Flows Equivalent?

21 Contemporary Engineering Economics, 4 th edition, © 2007 Example 3.26 Establishing a College Fund

22 Contemporary Engineering Economics, 4 th edition, © 2007 Solution: Establish the Economic Equivalence at n = 18

23 Contemporary Engineering Economics, 4 th edition, © 2007 Example 3.27 Calculating an Unknown Interest Rate with Multiple Factors

24 Contemporary Engineering Economics, 4 th edition, © 2007 Establish an economic Equivalence at n =7

25 Contemporary Engineering Economics, 4 th edition, © 2007 Linear Interpolation to Find an Unknown Interest Rate

26 Contemporary Engineering Economics, 4 th edition, © 2007 Linear Interpolation

27 Contemporary Engineering Economics, 4 th edition, © 2007 Using the Goal Seek Function to Find the Unknown Interest rate


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