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Incremental Analysis Graphical Technique in Solving problems with Mutually Exclusive Alternatives Using Spreadsheets in Incremental Analysis Chapter Outline 2

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Define Incremental Analysis Apply Graphical Technique in Solving Problems with Mutually Exclusive Alternatives Use Spreadsheets in Incremental Analysis Learning Objectives 3

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4 By definition in Chapter 7: Given a cash flow stream, IRR is the interest rate i at which the benefits are equivalent to the costs. This can be expressed mathematically in five different ways. NPW=0 PW of benefits - PW of costs =0 PW of benefits = PW of costs PW of benefits / PW of costs=1 EUAB-EUAC=0 Internal Rate of Return (IRR)

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Example Cash flows for an investment are shown in the following figure. What is the IRR to obtain these cash flows? YEARCASH FLOW 0($500) 1$100 2$150 3$200 4$250 5

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QUESTION CONTINUES -8.85 YEARCASH FLOW 0($500) 1$100 2$150 3$200 4$250 6

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QUESTION CONTINUES -8.85 YEARCASH FLOW 0($500) 1$100 2$150 3$200 4$250 7

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INTERPOLATION 5% 15% X% 30.95 -8.85 5-X 30.95 1039.80 0 8

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INTERPOLATION 5% 15% X% 30.95 -8.85 5-X 30.95 1039.80 0 9

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EXCEL solution IRR = irr(a1:a5) = 12.83% 10

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Only one alternative may be implemented All alternatives serve the same purpose Objective of incremental analysis is to select the best of these mutually exclusive alternatives Mutually Exclusive Alternatives 11

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When there are two alternatives, rate of return analysis is performed by computing the incremental rate of return, ΔIRR, on the difference between the two alternatives, as discussed in Chapter 7. Incremental Analysis 12

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Incremental Analysis The cash flow for the difference between alternatives is calculated by taking the higher initial-cost alternative minus the lower initial-cost alternative. The below decision path is made for incremental rate of return (ΔIRR) on difference between alternatives: Two -Alternative Situations Decision ΔIRR≥MARRChoose the higher-cost alternative ΔIRR

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The cash flows for four different alternatives are given in table below. If MARR 10%, which is the best alternative? Using the incremental analysis, we need to repeat 3 times, by comparing 2 alternatives at a time. Example 14

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EUAB =EUAC (Increment) ΔIRR≥MARRChoose the higher-cost alternative MARR = 10% ΔIRR

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Copyright Oxford University Press 2009 Incremental Analysis Could be applied to rate of return (IRR), present worth (PW), equivalent uniform annual cost (EUAC), or equivalent uniform annual worth (EUAW) approaches. [Higher-cost alternative] = [Lower-cost alternative] + [Increment between them] The “defender” is the best alternative identified so far in the process, and “challenger” is the next higher-cost alternative to be evaluated. For a set of N mutually exclusive alternatives, (N - 1) “challenger/defender” comparisons must be made. 16

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Given the alternatives below: Select the one best alternative if MARR = 8%. Use incremental rate of return analysis. Example 17

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MARR = 8% Since the MARR is 8%, Alt. D may be eliminated, as the ROR is less than 8% Among the remaining alternatives A, B, and C, the two lower cost alternatives are A and B. (A - B) increment: PW of benefit = PW of cost (623 - 531)(P/A, i, 10) = (4,000 - 3,000) (P/A, i, 10) = 1,000/92 = 10.86 PW of benefit = PW of cost (1,020 - 531)(P/A, i, 10) = (6,000 - 3,000) (P/A, i, 10) = 3,000/489 = 6.13 (C - B) increment: ∆ROR is greater than 8%. Therefore, choose the higher-cost alternative, Alt. C 18

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Example 8-1 High CapacityLow CapacityIncrement Cost$13,400$10,310$3090 Benefit$4000/year$3300/year$700/year Life5 years IRR Low IRR High IRR Increment = 4.3% PW LOW = -$10,310 + $3300(P/A,i,5) PW HIGH = -$13,400 + $4000(P/A,i,5) 19

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Example 8-1 – Continued In column B, input formula: PW LOW = –$10,300 + $3300(P/A,i,5) = –$10300 + pv(A3, 5, –3300) In column C, input formula: PW HIGH = –$13,400 + $4000(P/A,i,5) = –$13400 + pv(A3, 5, –4000) From a3 to a24, input interest from 0 to 0.21 In column D, input formula for incremental cost PW HIGH–LOW : = C3 – B3 Then draw a line chart! Or use EXCEL function npv(i, value range) -10300+npv(A3, $A$28:$A$32) -13400+npv(A3, $A$36:$A$40)

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Example 8-1 – Continued IRR Low IRR High IRR Increment = %4.3 PW LOW = -$10,300 + $3300(P/A,i,5) PW HIGH = -$13,400 + $40000(P/A,i,5) Interest RateBest Choice 0% ≤ i ≤ 4.3% High Capacity 4.3 % ≤ i ≤ 18% Low Capacity 18% ≤ i Do Nothing PW high-low = -$3090 + $700(P/A, i, 5)IRR Increment = %4.3 21

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Example 8-2 Machine XMachine Y Initial Cost$200$700 Uniform Annual Benefit$95$120 End-of-Useful-Life Salvage Value$50$150 Useful Life, in Years612 MARR = 10% RateMachine XMachine Y 0%$840.00$890.00 1.322752.24 2710.89687.31 4604.26519.90 6515.57380.61 8441.26263.90 10378.56165.44 12325.3081.83 14279.7710.37 16240.58-51.08 18206.65-104.23 20177.10-150.47 Net Present Worth 22 In column C, –700 + pv(A3, 12, –120, –150) In column B, –200 + pv(A2, 6, -95, -50) + pv(A2,6, 0, 200 – pv(A2, 6, -95, -50))

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RateMachine XMachine Y 0%$840.00$890.00 1.322752.24 2710.89687.31 4604.26519.90 6515.57380.61 8441.26263.90 10378.56165.44 12325.3081.83 14279.7710.37 16240.58-51.08 18206.65-104.23 20177.10-150.47 IRR Y ∆ IRR Increment =1.3% Example 8-2 Net Present Worth For MARR ≤ 1.3%, Machine Y is the right choice For MARR ≥1.3%, Machine X is the right choice Machine X Machine Y 23

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Example 8-3 ABC Initial Cost$2000$4000$5000 Uniform Annual Benefit410639700 Consider the three mutually exclusive alternatives: Each alternative has a 20 year life and no salvage value. If the MARR is 6%, which alternative should be selected? 24

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ABC Initial Cost$2000$4000$5000 Uniform Annual Benefit410639700 If MARR ≥ 9.6%, Choose Alt. A If 9.6% ≥ MARR≥2%, Choose Alt. B If 2% ≥ MARR ≥ 0%, Choose Alt. C ∆ IRR B-A =9.6% ∆ IRR C-B = 2% Net Present Worth Graph of Alternatives A, B, and C. Alt. C Alt. B Alt. A 25

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If MARR ≥ 9.6%, Choose Alt. A If 9.6% ≥ MARR≥2%, Choose Alt. B If 2% ≥ MARR ≥ 0%, Choose Alt. C ∆ IRR B-A =9.6% ∆ IRR C-B = 2% How to find the intersection points: Alt. C Alt. B Alt. A NPW(C-B) = -$5000+$4000+($700-$639)(P/A,i,20) = 0 ∆IRR(C-B) = 2% NPW(B-A) = -$4000+$2000+($639- $410)(P/A,i,20) = 0 ∆IRR(B-A) = 9.6% 26

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Example 8-4 BrassStainlessTitanium Cost$100,000$175,000$300,000 Life41025 If 6.3% ≥ MARR ≥ 0%, Choose Titanium If 15.3% ≥ MARR≥ 6.3%, Choose Stainless If MARR ≥ 15.3%, Choose Brass IRR Stainless - Brass =15.3% IRR Titanium - Stainless = 6.3% 27

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Example 8-5 Incremental Analysis (with Do-Nothing option) Machine XMachine YMachine Z Initial Cost$200$700$425 Uniform Annual Benefit65110100 Useful Life, in years6128 If MARR≥23%, Choose “Do-Nothing” If 23% ≥ MARR≥11%, Choose X If 11%≥MARR≥3.5%, Choose Z If 3.5%≥MARR≥0%, Choose Y IRR Z-X =11% IRR Y-Z =3.5% IRR X =23% X Z Y 28

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Example 8-5 Incremental Analysis (without Do-Nothing option) Machine XMachine YMachine Z Initial Cost$200$700$425 Uniform Annual Benefit65110100 Useful Life, in years6128 If MARR≥11%, Choose X If 11%≥MARR≥3.5%, Choose Z If 3.5%≥MARR≥0%, Choose Y IRR Z-X =11% IRR Y-Z =3.5% IRR X =23% X Z Y 29

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Example 8-6 Incremental Analysis using Graphical Comparison ABCDE Initial Cost$4000$2000$6000$1000$9000 Uniform Annual Benefit639410761117785 IRR A-B =9.6% IRR C-A =2% IRR B =20% 30 Life = 20 yrs

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Example 8-6 Incremental Analysis using Graphical Comparison ABCDE Initial Cost$4000$2000$6000$1000$9000 Uniform Annual Benefit639410761117785 Calculating Incremental Interest ∆IRR(C-A) = $6000-$4000 = ($761 - $639)(P/A, i, 20) = 2% ∆IRR(A-B) = $4000-$2000 = ($ 639 - $410)(P/A, i, 20) = 9.6% And to find where the NPW of B crosses the 0 axis IRR (B) = $2000 = $410(P/A, i, 20) = 20% 31

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Example 8-6 Incremental Analysis using Graphical Comparison ABCDE Initial Cost$4000$2000$6000$1000$9000 Uniform Annual Benefit639410761117785 If MARR≥20%, Choose Do-Nothing If 20%≥MARR≥9.6%, Choose B If 9.6%≥MARR≥2%, Choose A If 2%≥MARR≥0%, Choose C IRR A-B =9.6% IRR C-A =2% IRR B =20% 32

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Spreadsheet and Incremental Analysis Excel FunctionsPurpose Rate (n, A, -P, [F], [Type], [guess])To find rate of return or incremental rate of return given n, P, and A IRR (range, [guess])To find internal rate of return (or incremental rate of return) of a series of cash flow (or incremental cash flow) Excel ToolsPurpose Goal SeekIt varies the value in one specific cell until a formula that's dependent on that cell returns the wanted result. SolverSolver adjusts the values in the changing cells to produce the result from the target cell formula. Constraints are applied to restrict the values Solver can use in the model. 33

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End of Chapter 8 34

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