CTC 475 Review Is a certain cash flow economically feasible?

Methods for Determining if an Alternative is 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

Example (MARR=10%) EOY Cash Flow -$40K 1 $5K 2 $8K 3 $11K 4 $14K 5 6 7 -$40K 1 $5K 2 $8K 3 $11K 4 $14K 5 6 7 8

Cash flow breakdown-show on board 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) 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 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

Future Worth-Alternate Method “Snail”Method Using Single Sums FW= -40K(F/P10,8)+5K(F/P10,7)+8K(F/P10,6)+11K(F/P10,5) 14K(F/P10,4)+14K(F/P10,3)+11K(F/P10,2)+8K(F/P10,1) +5K FW=-$85,744+$9,744 +$14,172+ $17,716 +$20,497+$18,634+$13,310+$8,800+$5,000 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 i(%) PW 10 +$10,324 12 +$6,723 15 +$1,994 18 -$2,064

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(-)=$40,000 SIR=$50,324/$40,000=1.26 PW(+)=5K(P/A10,4)+3K(P/G10,4) +[14K(P/A10,4)- 3K(P/G10,4)](P/F10,4)=$50,324 PW(-)=$40,000 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

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