1. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Variations in Present Worth Analysis Lecture No. 17 Chapter 5 Contemporary Engineering Economics Copyright © 2016

2. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Future Worth Criterion  Given Cash flows and MARR (i)  Find The net equivalent worth at a specified period other than the “present,” commonly at the end of the project life  Decision Rule Accept the project if the equivalent worth is positive. $76,000 $35,560 $37,360 $34,400 0 1 2 3 Project life $31,850 $47,309

3. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Excel Solution ABC 1PeriodCash Flow 20($76,000) 31$35,650 42$37,360 53$31,850 64$34,400 7PW(12%)$30,145 8FW(12%)$47,434 =FV(12%,4,0,-B7)

4. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved FW Calculation with the Cash Flow Analyzer Net Present Worth Net Future Worth Payback Period Project Cash Flows

5. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 5.6: Future Equivalent at an Intermediate Time

6. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 5.8: Project’s Service Life is Extremely Long o Q1: Was Bracewell's $800,000 investment a wise one? o Q2: How long does he have to wait to recover his initial investment, and will he ever make a profit? Built a hydroelectric plant using his personal savings of $800,000 Power generating capacity of 6 million kwhs Estimated annual power sales after taxes − $120,000 Expected service life of 50 years

7. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Mr. Bracewell’s Hydroelectric Project

8. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Find P for a Perpetual Cash Flow Series, A

9. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Capitalized Equivalent Worth  Principle: PW for a project with an annual receipt of A over infinite service life  Equation CE(i) = A(P/A, i, ) = A/i A 0 P =CE(i) n

10. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Practice Problem 10 $1,000 $2,000 P = CE (10%) = ? 0 Given: i = 10%, N = ∞ Find: P or CE (10%) ∞

11. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution 10 $1,000 $2,000 P = CE (10%) = ? 0 ∞

12. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved A Bridge Construction Project Construction cost = $2,000,000 Annual maintenance cost = $50,000 Renovation cost = $500,000 every 15 years Planning horizon = infinite period Interest rate = 5%

13. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Cash Flow Diagram for the Bridge Construction Project $500,000 $2,000,000 $50,000 0 15 30 4560 Years

14. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution Construction Cost P 1 = $2,000,000 Maintenance Costs P 2 = $50,000/0.05 = $1,000,000 Renovation Costs P 3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60) : = {$500,000(A/F, 5%, 15)}/0.05 = $463,423 Total Present Worth P = P 1 + P 2 + P 3 = $3,463,423

15. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Alternate Way to Calculate P 3 o Concept: Find the effective interest rate per payment period. o Interest rate: Find the effective interest rate for a 15-year cycle. i = (1 + 0.05) 15 − 1 = 107.893% o Capitalized equivalent worth P 3 = $500,000/1.0789 = $463,423 15 30 4560 0 $500,000 Effective interest rate for a 15-year period