CE 203 Rate of Return Analysis (EEA Chapter 7) ISU CCEE Rate of Return Analysis • “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 ISU CCEE Rate of Return Analysis, RoR • P = F (P/F, i, n) or P = A(P/A, i, n) • A = P (A/F, i, n) EEA 7: i? for benefits = costs ISU CCEE EEA 5 EEA 6 Rate of Return Analysis • Internal RoR, i*, solve for i in : – NPW – EUAW = = PWB – PWC = 0 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) ISU CCEE In-class example 7-1 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% ISU CCEE OR: Use SOLVER In-class example 7-1 Use plotting: ISU CCEE Rate of Return Analysis • 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?) ISU CCEE What is easier to understand? • NPW = $5000 • EUAW = $800 • RoR = 8% ISU CCEE Investment vs. Borrowing Situation • Investment: subsequent inflow > initial amount • Borrowing: subsequent outflow > initial amount • Usually (but not always) investigate initial cash flow – Investment if negative – Borrowing if positive ISU CCEE Investment vs. Borrowing Example Investment ISU CCEE Borrowing Year 0 Cash Flow #1 -$5,000 Cash Flow #2 $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 Minimum Attractive Rate of Return • 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.) ISU CCEE Rate of Return Analysis • 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 (DIRR) on the difference between the alternatives and compare to MARR If DIRR > MARR, choose higher-cost alternative If DIRR < MARR, choose lower-cost alternative ISU CCEE In-class Example 7-2 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Year 0 Alternative #1 - $5,000 Alternative #2 - $5,000 1 $4,500 $500 2 $1,400 $5,700 (RoR = 14.5%) (RoR = 11.9%) Which is the better alternative? To answer, consider both RoR and MARR ISU CCEE In-class Example 7-2 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Year 0 Alternative #1 - $5,000 Alternative #2 - $5,000 1 $4,500 $500 2 $1,400 $5,700 Total C. F. +$900 + $1200 Both total cash flows are positive, so both are “investments”; Alternative 2 has larger total, so use Alt. 2 – Alt. 1 for DIRR ISU CCEE In-class Example 7-2 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Alt. #2 – Alt. #1 $0 Year 0 Alt. #1 - $5,000 Alt. #2 - $5,000 1 $4,500 $500 - $4000 2 $1,400 $5,700 +$4,300 Total C. F. +$900 + $1200 + $300 Note: total or net cash flow for difference is positive ISU CCEE In-class Example 7-2 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Year Alt. #1 Alt. #2 Alt. #2 – Alt. #1 0 - $5,000 - $5,000 $0 1 $4,500 $500 - $4000 2 $1,400 $5,700 +$4,300 Total C. F. +$900 + $1200 + $300 Need i such that NPW = 0 = - 4000 (1 + i) -1 + 4300 (1 + i) -2 For this simple case, i = 300/4000 = .075 = 7.5% = DIRR ISU CCEE In-class Example 7-2 Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%. Year 0 Alt. #1 - $5,000 Alt. #2 - $5,000 Alt. #2 – Alt. #1 $0 1 $4,500 $500 - $4000 2 $1,400 $5,700 +$4,300 Total C. F. +$900 + $1200 + $300 Since DIRR = 7.5% > MARR = 6%, choose alternative #2 ISU CCEE In-class Example 7-2 Or, to look at it another way: Alternative #1 Year Action 0 Invest $5,000 1 Receive $4,500 and invest it at 6% (MARR) 2 Receive $1,400 + $4,500 (1 + .06) = $6,170 Alternative #2 Year Action 0 Invest $5,000 ISU CCEE 1 Receive $500 and invest it at 6% (MARR) 2 Receive $5,700 + $500 (1 + .06) = $6,230 In-class Example 7-2 with MARR = 9% Suppose MARR = 9% for payback alternatives for an initial investment of $5000: Year 0 Alternative #1 - $5,000 Alternative #2 - $5,000 1 $4,500 $500 2 $1,400 $5,700 Since DRoR = 7.5% < MARR = 9%, choose alternative #1 ISU CCEE In-class Example 7-2 with MARR = 9% Alternative #1 Year Action 0 Invest $5,000 1 Receive $4,500 and invest it at 9% (MARR) 2 Receive $1,400 + $4,500 (1 + .09) = $6,305 Alternative #2 Year Action 0 Invest $5,000 ISU CCEE 1 Receive $500 and invest it at 9% (MARR) 2 Receive $5,700 + $500 (1 + .09) = $6,245 In-class Example 7-2 but we are the borrower Year 0 Alternative #1 - $5,000 Alternative #2 - $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. ISU CCEE In-class Example 7-3 What is the internal RoR (i*) for the cash flow shown in the table below? Year Cash flow 0 $1,020 1 - $2,000 2 $500 3 $500 0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F, i, 3). Solve for i. 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% NPW vs. Interest Rate 20 Net Present Worth 15 10 5 0 1 3 5 7 9 11 13 15 17 19 -5 -10 -15 Interest Rate ISU CCEE 21 23 25 27 29 Two answers! 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) ISU CCEE