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CTC 475 Review. Is a certain cash flow economically feasible?. CTC 475. Methods for Determining if an Alternative is Economically Feasible. Objectives. Know the various methods for determining if an alternative is economically feasible
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CTC 475 Review • Is a certain cash flow economically feasible?
CTC 475 Methods for Determining if an Alternative is Economically Feasible
Objectives • Know the various methods for determining if an alternative is economically feasible • Be able to use any method for economic feasibility studies
Methods for Economic Feasibility Studies • Present Worth (PW) • Annual Worth (AW) • Future Worth (FW) • Internal Rate of Return (IRR) • External Rate of Return (ERR) • Savings/Investment Ratio (SIR) or • Benefit/Cost Ratio (B/C) • Payback Period Method (PBP) • Capitalized Worth Method (CW)
Equivalent Methods • PW • AW • FW • IRR • ERR • SIR or B/C
Nonequivalent Methods • PBP • CW
When is an alternative feasible? • PW > 0 • AW > 0 • FW > 0 • IRR > MARR • ERR > MARR • SIR or B/C > 1
Net Cash Flows • It’s a good idea to use net cash flows (one cash flow at each period). • It doesn’t matter with respect to whether a project is feasible or not; however, absolute numbers (ERR and SIR) may differ
Cash flow breakdown • Years 1-4: Uniform ($5K) + Gradient ($3K) • n=4 • P will occur at t=0 • F will occur at t=4 • Years 5-8: Uniform ($14K) - Gradient ($3K) • n=4 • P will occur at t=4 • F will occur at t=8
Present Worth • PW= -40K+5K(P/A10,4)+3K(P/G10,4) +[14K(P/A10,4)- 3K(P/G10,4)](P/F10,4) • PW= -40K+5K(3.1699)+3K(4.3781) +[14K(3.1699)-3K(4.3781)]*0.6830 • PW= +$10,324 • PW>0 ; therefore, cash flow is economically feasible
Annual Worth • Find A given P • AW=PW(A/P10,8) • AW=$10,324(.1874) • AW=$1,935 • AW>0; therefore, cash flow is economically feasible
Future Worth • FW=PW(F/P10,8) or PW(1.1)8 • FW=$10,324(2.1436) • FW=$22,130 OR • FW= AW(F/A10,8) • FW=$1,935(11.4359) • FW=$22,128 • FW>0; therefore, cash flow is economically feasible
Future Worth-Alternate Method • FW= -40K(F/P10,8)+[5K+3K(A/G10,4)](F/A10,4)(F/P10,4) +[14K-3K(A/G10,4)](F/A10,4) • FW=-$85,744+($9,144)(6.7949)+($9,856)(4.6410) • FW=+$22,129
IRR-Find i that gives a PW=0 • PW= -40K+5K(P/Ai,4)+3K(P/Gi,4) +[14K(P/Ai,4)- 3K(P/Gi,4)](P/Fi,4) Interpolate to get an IRR = 16.5% IRR>MARR
ERR: Set FW of + using MARR = FW of – using ERR; solve for ERR • FW(+) = 5K+3K(A/G10,4)](F/A10,4)(F/P10,4) +[14K-3K(A/G10,4)](F/A10,4) = $107,873 • FW(-) = 40K(1+ERR)8 • 40K(1+ERR)8 = $107,873 • (1+ERR)8 = 2.6968 • ERR=13.2% • ERR>MARR • Check: MARR=10%; ERR=13.2%; IRR=16.5%
SIR or B/C • SIR=PW(+)/PW(-) • PW(+)=5K(P/A10,4)+3K(P/G10,4) +[14K(P/A10,4)- 3K(P/G10,4)](P/F10,4)=$50,324 • SIR=$50,324/$40,000=1.26 • SIR>1
PBP-Payback Period • If MARR=0 how many periods does it take to get your investment back? • At 1 year; $5K<$40K • At 2 years: $13K<$40K • At 3 years: $24K<$40K • At 4 years: $38K<$40K • At 5 years: $52K>$40K • PBP is 5 years
Next lecture • Bonds