1 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

2 IRR Example Project A costs $10,000 and will last for 10 years. Annual, end of year revenues will be $3000, and expenses will be $1000. There is no salvage value. Project B costs $20,000 and will also last for 10 years. Annual revenues will be $4000 with annual expenses of $1500. Salvage value is $14,500. What is each project’s IRR?

3 Numerical Method Example: IENG 302 IRR.xls Table Example i*(P/A, i*, 10)(P/F, i*, 10) 6% % % % %

4

5 Decisions with IRR When applied to project selection among independent projects – i.e. there are enough funds such that any or all of the projects may be selected – then investing based on IRR is easy: Select all projects with IRR > MARR! Note: Make sure that there is only one IRR (root)!

6 Decisions with IRR When applied to project selection among mutually exclusive projects – i.e. there is not enough money to do them all – IRR can produce the same result as NPW and EAW. However, incremental analysis MUST be used! REASON: IRR is a relative measure of project merit

7 Why Must Incremental Analysis be Used for Competing Projects? Assume that an MARR of 16% per year is required, and $ is available to invest: Project A requires $ upfront to obtain an IRR of 35% per year. Project B requires an $ first cost and returns an IRR of 29% per year. What could we do with the un-invested money from Project A? ($35 000)

8 Why Must Incremental Analysis be Used for Competing Projects? It would be reasonable to invest the remaining $ at the MARR: Overall IRR A = (.35) (.16) = 27.2% per year Project B returns an IRR of 29% per year on ALL the money available to invest.

9 Incremental Analysis A technique or approach that can be used with NPW, EAW, and later with IRR and Cost/Benefit to determine if an incremental expenditure should be made. Note: If using NPW or IRR, lifetimes must be equal – so use Least Common Multiple or Study Period!

10 Incremental Analysis Incremental Analysis can used by putting the options in the order of increasing initial investment. Check the feasibility of the first investment. If it is feasible, then you consider the increment to the next level of investment. If it is not feasible, then you check the second investment alternative for feasibility, and so on until all options are considered.

11 Incremental Analysis Incremental Analysis can also be used when you’re already incurring an expense (e.g., DSL internet service) and you are trying to determine if it is a good decision to spend additional funds (e.g., satellite or cable?).

12 Steps of the Process 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.

13 Steps of the Process 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 If all alternatives have been considered, select the “current best” alternative as the preferred alternative.

14 i = 18%Net Cash Flows for Alternatives A 0 - A 3 End of Year, t A 0 A 1 A 2 A 3 0$0 -$50,000-$75, ,50020, ,50020,00025, ,50020,00030, ,50020,00035, ,50020,00040,000

15 IRR Incremental Analysis Rank projects from lowest to highest initial cost Eliminate any projects with IRR < MARR Starting from the least expensive project to the next most expensive, justify each incremental investment IRR HC-LC < MARRAccept LC Project & Reject HC Project IRR HC-LC = MARRIndifferent IRR HC-LC > MARRReject LC Project & Accept HC Project