Variations in Present Worth Analysis Lecture No 17 Chapter Contemporary Engineering Economics Copyright © 2016 Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Future Worth Criterion Given $47,309 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 $35,560 $37,360 $31,850 $34,400 Decision Rule Accept the project if the equivalent worth is positive $76,000 Project life Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Excel Solution A Period PW(12%) FW(12%) B Cash Flow ($76,000) $35,650 $37,360 $31,850 $34,400 $30,145 $47,434 Contemporary Engineering Economics, th edition Park C =FV(12%,4,0,-B7) Copyright © 2016 by Pearson Education, Inc All Rights Reserved FW Calculation with the Cash Flow Analyzer Payback Period Project Cash Flows Net Present Worth Net Future Worth Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 5.6: Future Equivalent at an Intermediate Time Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 5.8: Project’s Service Life • Built a hyd roelectric is Extremely Long plant u pers 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, onal savin sing his gs of $800 ,000 • Powe r generati ng capacit kwhs y of mill ion • Estim ated annu al power s taxes − $1 ales afer 20,000 • Expec ted serv ice life of years and will he ever make a profit? Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Mr Bracewell’s Hydroelectric Project V1 V2 $1,101K $1, 468K $367 K V2 120 K ( P / A,8%,50) $1, 468 K V1 $50 K ( F / P,8%,9) $50 K ( F / P,8%,8) L $100 K ( F / P,8%,1) 60 K $1,101K Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Find P for a Perpetual Cash Flow Series, A Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Capitalized Equivalent Worth A Principle: PW for a project with an annual receipt of A over infinite service life n Equation CE(i) = A(P/A, i, ) = A/i Contemporary Engineering Economics, th edition Park P =CE(i ) Copyright © 2016 by Pearson Education, Inc All Rights Reserved Practice Problem Given: i = 10%, N = ∞ Find: P or CE (10%) $2,000 $1,000 10 ∞ P = CE (10%) = ? Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution $2,000 $1,000 10 ∞ $1,000 $1,000 (P / F ,10%,10) 0.10 0.10 $10,000(1 0.3855) CE(10%) P = CE (10%) = ? $13,855 Contemporary Engineering Economics, 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% Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Cash Flow Diagram for the Bridge Construction Project 15 Years 30 45 60 $500,000 $500,000 $500,000 $500,000 $50,000 $2,000,000 Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution • • • Construction Cost P1 = $2,000,000 Maintenance Costs P2 = $50,000/0.05 = $1,000,000 • Total Present Worth Renovation Costs P = P1 + P2 + P3 P3 = $500,000(P/F, 5%, = $3,463,423 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60) : Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Alternate Way to Calculate P3 o Concept: Find the effective interest rate per payment period o Interest rate: Find the effective interest rate for a 15-year cycle Effective interest rate for a 15-year period 30 15 45 60 i = (1 + 0.05)15 − = 107.893% $500,000 $500,000 $500,000 $500,000 o Capitalized equivalent worth P3 = $500,000/1.0789 = $463,423 Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved ... $500,000 every 15 years Planning horizon = infinite period Interest rate = 5% Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Cash... Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Find P for a Perpetual Cash Flow Series, A Contemporary Engineering Economics,... Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 5.6: Future Equivalent at an Intermediate Time Contemporary Engineering Economics,