Variations in Present Worth Analysis Lecture No. 17 Chapter 5 Contemporary Engineering Economics Copyright © 2016

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. $47,309 $37,360 $35,560 $31,850 $34,400 1 2 3 $76,000 Project life

Excel Solution A B C 1 Period Cash Flow 2 ($76,000) 3 $35,650 4 ($76,000) 3 $35,650 4 $37,360 5 $31,850 6 $34,400 7 PW(12%) $30,145 8 FW(12%) $47,434 =FV(12%,4,0,-B7)

FW Calculation with the Cash Flow Analyzer Payback Period Project Cash Flows Net Present Worth Net Future Worth

Example 5.6: Future Equivalent at an Intermediate Time Figure: 05-09EXM

Example 5.8: Project’s Service Life is Extremely Long 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 Q1: Was Bracewell's $800,000 investment a wise one? Q2: How long does he have to wait to recover his initial investment, and will he ever make a profit?

Mr. Bracewell’s Hydroelectric Project Figure: 05-11EXM

Find P for a Perpetual Cash Flow Series, A Figure: 05-10

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 n P=CE(i)

Practice Problem $2,000 $1,000 ∞ P = CE (10%) = ? Given: i = 10%, N = ∞ Find: P or CE (10%) $2,000 $1,000 10 ∞ P = CE (10%) = ?

Solution $2,000 $1,000 10 ∞ P = CE (10%) = ?

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%

Cash Flow Diagram for the Bridge Construction Project Years 15 30 45 60 $50,000 $500,000 $500,000 $500,000 $500,000 $2,000,000

Solution Construction Cost P1 = $2,000,000 Maintenance Costs Renovation Costs P3 = $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 = P1 + P2 + P3 = $3,463,423

Alternate Way to Calculate P3 Concept: Find the effective interest rate per payment period. Interest rate: Find the effective interest rate for a 15-year cycle. i = (1 + 0.05)15 − 1 = 107.893% Capitalized equivalent worth P3 = $500,000/1.0789 = $463,423 Effective interest rate for a 15-year period 15 30 45 60 $500,000