1. 1 1. Order alternatives from lowest to highest initial investment. 2. Let Alternative A 0 (do nothing) be considered the current best. 3. Consider next Alternative ( j = j+ 1) 4. Determine cash flows for “current best” and Alternative j. 5. Determine incremental cash flows between “current best” and Alternative j. 6. Calculate PW, AW, FW, IRR (or Benefit/Cost) of only the incremental cash flows. Steps of the Incr. Analysis Process

2. 2 7. If incremental investment yields NPW, EAW, or a NFW > 0*, then the new “current best” becomes Alternative j. * (B/C ratio > 1, or IRR > MARR) 8.If there are remaining alternatives, go to Step 3. 9. If all alternatives have been considered, select the “current best” alternative as the preferred alternative. Steps of the Incr. Analysis Process

3. 3 Internal Rate of Return Internal Rate of Return (IRR): The interest rate i* at which NPW = 0 Note: This is the same as finding the roots of a polynomial equation. If there is more than one sign change in the net annual cash flows over the life of the project, then there is more than one internal rate of return (root)! We may find the IRR by either the manual method we used for the bond yield, or we may use the computer to find the roots by either plots or numerical methods

4. Incremental Example Investment12345 Initial Investment $15,000$25,000$40,000$50,000$70,000 Annual Return$3750$5000$9250$11,250$14,250 Salvage Value$15,000$25,000$40,000$50,000$70,000 IRR25.00%20.00%23.13%22.5%22.36% 4 Another way to find the IRR is to set the Annual worth to Zero: 0 = (P-F)(A/P,i,n) + Fi – A Capital Cost Equation So- since A = F in example i= A/P Looking @ IRR – Option 1 looks best!

5. Incremental Example Cont. MARR = 18% Using a 10 year Horizon 5 Investment12345 Initial Investment $15,000$25,000$40,000$50,000$70,000 Annual Return$3750$5000$9250$11,250$14,250 Salvage Value$15,000$25,000$40,000$50,000$70,000 Present Worth$4718.79$2247.04$9212.88$10,111.69$7415.24 Using NPV Option 4 looks best!

6. Incremental Example Cont. Why don’t they match??? 6 Reference: White, Case, Pratt. “Principles of Engineering Economic Analysis” Fifth Edition. 355-357.

7. Incremental Example Cont. 7 Investment12 - 13 - 14 - 35 Initial Investment $15,000$10,000$25,000$10,000$20,000 Annual Return$3750$1250$5500$2000$3,000 Salvage Value$15,000$10,000$25,000$10,000$20,000 IRR25.00%12.50%22.00%20.00%15.00% > MARR or 18% YesNoYes No Defender11344

8. 8 IRR Incr. Analysis Example 15