1. ISU CCEE CE 203 Rate of Return Analysis (EEA Chapter 7)

2. ISU CCEE “Equivalent” cash flows: same value at some given time for a given interest rate Internal rate of return (definitions): – interest rate such that, for given payment schedule, loan is paid off with final payment – interest rate such that, for given payment schedule, unrecovered investment = 0 at final payment – interest rate such that benefits = costs Rate of Return Analysis

3. ISU CCEE P = F (P/F, i, n) or P = A(P/A, i, n) EEA 5 A = P (A/F, i, n)EEA 6 Rate of Return Analysis, RoR EEA 7: i? for benefits = costs

4. ISU CCEE Internal RoR, i*, solve for i in : – NPW = PWB – PWC = 0 – EUAW = EUAB – EUAC = 0 To solve for i* : – Iterative solution (get close, interpolate) – Use “solver” – Plot NPW or EUAW, “read” i* at NPW = 0 – Spreadsheet (Excel or ???) » RATE (N, A, P, F, Type, guess) » IRR (value, guess) Rate of Return Analysis

5. ISU CCEE If you invest $10,000 now and are paid $5,200 at the end of each of the next two years, what is the internal rate of return? Use iteration, then interpolation to find i NPW = $ 5,200(1+i) -1 + $ 5,200(1+i) -2 - $10k = 0 Try 2% = $ 5,098 + $ 4,998 - $ 10,000 = $ 96 Try 3% = $ 5,045 + $ 4,902 - $ 10,000 = -$ 53 Interest rate, from linear interpolation 2% + 96/(96+53)(3-2) = 2.64% OR: Use SOLVER In-class example 7-1

6. ISU CCEE Use plotting: In-class example 7-1

7. ISU CCEE Chapter 7: compare two alternatives Chapter 8: compare three + alternatives Advantages of RoR analysis: – More widely understood – Single value of “merit” – Most widely used (but maybe not in CE?) Rate of Return Analysis

8. ISU CCEE NPW = $5000 EUAW = $800 RoR = 8% What is easier to understand?

9. ISU CCEE Investment: subsequent inflow > initial amount Borrowing: subsequent outflow > initial amount Usually (but not always) investigate initial cash flow – Investment if negative – Borrowing if positive Investment vs. Borrowing Situation

10. ISU CCEE Investment vs. Borrowing Example YearCash Flow #1Cash Flow #2 0-$5,000$5,000 1$1,000-$1,000 2$3,000-$3,000 3$2,000-$2,000 Sum = $1,000 Investment Sum = - $1,000 Borrowing Investment Borrowing

11. ISU CCEE Minimum Attractive Rate of Return (MARR) Rate of return (RoR) below which we will not invest (because we can invest elsewhere at MARR or simply decide not to invest if RoR is < MARR) MARR is the highest of – Interest rate for borrowing money – Average interest rate for the cost of capital (loans, bonds, stock, etc.) Minimum Attractive Rate of Return

12. ISU CCEE RoR criterion: If internal rate of return (i*) P > MARR, the investment is considered acceptable (but not necessarily the best) RoR analysis for two alternatives – Determine the cash flow for the difference between alternatives (highest total cash flow alternative minus lower total cash flow alternative) – Determine the incremental rate of return (  IRR) on the difference between the alternatives and compare to MARR  If  IRR > MARR, choose higher-cost alternative  If  IRR < MARR, choose lower-cost alternative Rate of Return Analysis

13. ISU CCEE In-class Example 7-2 YearAlternative #1Alternative #2 0- $5,000 1$4,500$500 2$1,400$5,700 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. (RoR = 14.5%)(RoR = 11.9%) Which is the better alternative? To answer, consider both RoR and MARR

14. ISU CCEE In-class Example 7-2 YearAlternative #1Alternative #2 0- $5,000 1 $4,500 $500 2 $1,400 $5,700 Total C. F. +$900+ $1200 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Both total cash flows are positive, so both are “investments”; Alternative 2 has larger total, so use Alt. 2 – Alt. 1 for  IRR

15. ISU CCEE In-class Example 7-2 YearAlt. #1Alt. #2Alt. #2 – Alt. #1 0- $5,000 $0 1 $4,500 $500- $4000 2 $1,400 $5,700+$4,300 Total C. F. +$900+ $1200 + $300 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Note: total or net cash flow for difference is positive

16. ISU CCEE In-class Example 7-2 YearAlt. #1Alt. #2Alt. #2 – Alt. #1 0- $5,000 $0 1$4,500$500- $4000 2$1,400$5,700 +$4,300 Total C. F. +$900+ $1200+ $300 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Need i such that NPW = 0 = - 4000 (1 + i) -1 + 4300 (1 + i) -2 For this simple case, i = 300/4000 =.075 = 7.5% =  IRR

17. ISU CCEE In-class Example 7-2 YearAlt. #1Alt. #2Alt. #2 – Alt. #1 0- $5,000 $0 1$4,500$500- $4000 2$1,400$5,700 +$4,300 Total C. F. +$900+ $1200+ $300 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Since  IRR = 7.5% > MARR = 6%, choose alternative #2

18. ISU CCEE In-class Example 7-2 YearAction 0Invest $5,000 1Receive $4,500 and invest it at 6% (MARR) 2Receive $1,400 + $4,500 (1 +.06) = $6,170 Or, to look at it another way: Alternative #1 YearAction 0Invest $5,000 1Receive $500 and invest it at 6% (MARR) 2Receive $5,700 + $500 (1 +.06) = $6,230 Alternative #2

19. ISU CCEE In-class Example 7-2 with MARR = 9% YearAlternative #1Alternative #2 0- $5,000 1$4,500$500 2$1,400$5,700 Suppose MARR = 9% for payback alternatives for an initial investment of $5000: Since DRoR = 7.5% < MARR = 9%, choose alternative #1

20. ISU CCEE In-class Example 7-2 with MARR = 9% YearAction 0Invest $5,000 1Receive $4,500 and invest it at 9% (MARR) 2Receive $1,400 + $4,500 (1 +.09) = $6,305 Alternative #1 YearAction 0Invest $5,000 1Receive $500 and invest it at 9% (MARR) 2Receive $5,700 + $500 (1 +.09) = $6,245 Alternative #2

21. ISU CCEE In-class Example 7-2 but we are the borrower YearAlternative #1Alternative #2 0- $5,000 1$4,500$500 2$1,400$5,700 First, if MARR = 6%, we would choose neither alternative and go to bank to get $$$. If forced to choose, selection criterion is reversed: we would choose Alternative #1.

22. ISU CCEE In-class Example 7-3 YearCash flow 0$1,020 1- $2,000 2$500 3 What is the internal RoR (i*) for the cash flow shown in the table below? 0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F, i, 3). Solve for i.

23. ISU CCEE In-class Example 7-3 0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F, i, 3). Graphing solution (using EXCEL): NPW = 0 at i values of 5.24% and 27.4% Two answers!

24. ISU CCEE Multiple values for ROR possible! …there may be as many positive values for i* as there are sign changes in cash flow table (in example, +1020 to -2000 to +500) …try the modified internal rate of return (p. 238 of the textbook)