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Chapter 7 An Economic Appraisal II: NPV, AE, IRR Technique. Powerpoint Templates. NPV ( i ) > 0. Net Present Value Technique. NPV=The Sum of The Present Values of All Cash Inflows – The Sum of The Present Value of All Cash Outflows. Cash Inflows. 0 1.

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  1. Chapter 7 An Economic Appraisal II: NPV, AE, IRR Technique Powerpoint Templates

  2. NPV(i) > 0 NetPresent Value Technique NPV=The Sum of The Present Values of All Cash Inflows – The Sum of The Present Value of All Cash Outflows Cash Inflows 0 1 2 3 4 5 Cash Outflows NPV PV(i) CIF 0 PV(i) COF

  3. NPV.... • Equation • Decision Rule NPV(MARR) > 0 Accept it NPV(MARR) = 0 Indifferent NPV(MARR) < 0 Reject it Where, CFt: cash flow at time t, MARR: minimum attractive rate of return on a project

  4. The Steps to Make a Decision with the NPV Technique • Step 1: Determine an MARR. • Step 2: Estimate a project life. • Step 3: Calculate a net cash flow(all cash inflows – all cash outlfows) • Step 4: Calculate a net present value with an MARR. • Step 5: Make a Decision on the Project with the NPV Derived in Step 4.

  5. Ex7.1] An Investment Decision on A Construction Project of A Small Power Plant with A NPVNPV Given] Cash Flows Diagram, MARR=8% 120 120 …………… -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 n 1 2 3 48 49 50 50 50 60 60 80 100 100 (unit: Million won) 150

  6. Continued…….. Sol] - Step 1: MARR=8% Already Determined. - Step 2: A project life turns out be 60 years including a construction time of 10 years. - Step 3: A net cash flows are presented in the cash flow diagram. - Step 4: Calculate the net present value. (1) a present value of all the net cash flows incurred under construction. (2) a present value of all the net cash flows incurred during the commercialization stage of 50 years

  7. Continued…….. - Step 5: Make an investment decision on the project with NPV(8%) Since NPV(8%)=170.15 M >0, accept the project Break_Even Interest Rate=IRR=9.78% (NPV Unit: $M MARR(100%) The Sensitivity Analysis of the NPV with A Varying Interest Rate

  8. Net Future Value Technique • Given: Cash Flows and MARR (i) • Find: A Net Equivalent Value at the End of a Project Life (unit: 000 won) 55,760 27,340 24,400 0 3 1 2 75,000 Project Life 8

  9. An Investment Decision with A NFW for Ex. 7.1 Sol] (1) A future value of all the cash flows incurred under construction (2) A present value of all the cash flows incurred during the commercialization stage

  10. A Project Balance Concept • Project Balance (PB): Cash Flows Left inside A Project • Equation • PB0=PB0 • PBn=PBn-1(1+MARR)+CFn Ex7.2] An Economic Meaning of An NPV Based on A PB (unit: 000 won)

  11. Continued… (unit: 000 won) N 0 1 2 3 4 5 Beginning Bal. Interest Ending Bal. -62,500 -62,500 -62,500 -9,375 +33,982 -37,893 -37,893 -5,684 +33,726 -9,851 -9,851 -1,478 +33,205 +21,876 +21,876 +3,281 +33,135 +58,292 +58,292 +8,744 +82,013 +149,049 NPV(15%) = 149,049 (P/F, 15%, 5) =74,107,000 NFV(15%)

  12. A Project Balance Diagram as A Function of Time 160,000 140,000 120,000 100,000 80,000 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -80,000 149,049 PB at the End of a Project Life 58,292 PB 21,876 Discounted Payback Peirod -9,851 -37,893 -62,500 n 0 1 2 3 4 5

  13. Capitalized Equivalent • Principle: a present value of cash flows which are oriented with an equivalent amount of money of “A” over an infinite period of time. • Equation A P = CE(i)

  14. Continued …. Ex7.3] CE Given] A=200 M won, i=8%, N= ∞ Sol] Calculate a CE to prepare for an annual maintenance and repair cost of a building 200M ∞ P = CE(8%)=2.5B

  15. Annual Equivalent 0 1 …… 2 3 4 5 N A …… 0 1 2 3 4 5 N • Decision Rule • - if AE(i) > 0, accept the project • -if AE(i) = 0, remain indifferent • -if AE(i) <, reject the project 0 NPV(i) AE(i) =NPV(i)(A/P, i, N)

  16. Continued….. Unit:: $M Ex7.4] Convert the irregular cash flows into an equivalent worth 0 1 12 10 9 8 5 2 3 4 5 6 3.5 15 A=$1.835 0 1 2 3 4 5 6 0

  17. The Advantages of An AE Technique Consistency of report formats: Financial managers more commonly work with annual rather than with overall costs in any number of internal and external reports. Engineering managers may be requires to submit project analyses on an annual basis for consistency and ease of use by other members of the corporation and stockholders. Need for unit costs/profits:In many situations, projects must be broken into unit costs/profits for ease of comparison with alternatives. Make-or-buy and reimbursement analyses aree key examples. Unequal project lives: Comparing projects with unequal service lives is complicated by the need to determine the lowest common multiple life. For the special situation of an indefinite service period and replacement with identical projects, we can avoid this complication by use of AN analysis.

  18. Capital Cost and Operating Cost • Operating Costs : to be costs which are incurred repeatedly over the life of a project by the operation of physical plant and or equipment needed to provide servicee suck as labor and raw materials. • Capital Costs : to be costs which are incurred only one time over the life of a project by purchasing assets to be used in production and service such as a purchase cost and sales taxes.

  19. Annual Equivalence Analysis • When only costs are involved, an AE cost analysis may be useful. • A profit must exceed a sum of operating and capital costs such that a project be economically viable. CC AE Cost + OC

  20. Capital Recovery Cost(CR)  Definition: to be an annul equivalent of capital costs Items of costs (1) Initial cost being the same as the cost basis(I) (2) Salvage value(S)  CR(i) : Considering two costs above, we obtain the following expression. S 0 N I ……………… 0 1 2 3 N CR(i)

  21. Calculate a CR(i) Unit: 000 won • CR(i) for Mini Cooper

  22. A Relationship between a CR(i) and a Depreciation Cost In a practical term, a CR(i) cost consists of (1)a depreciation cost and (2) an interest on the investment cost. I=19.8M, N = 3 years, i=6%, S=12.078M A Depreciation Method: A SL Method AE of the Interest= 2,778.74(A/P, 10%, 3)=1.03995M

  23. AE-CR(i) Ex7.5] Given] I=20M, S=4M, A=4.4M, N=5 years, i=10% Sol] An investment decision-making with AE, and make a decision with the AE - Method1: first obtain the NPV and transform it into AE - Method2: Make a decision with CR(i)

  24. Ex7.6] profit/machine time used – when the operating time is constant Given] NPV=3.553M, N=3 years, i= 15%, machine time used /year: 2,000 hrs Sol] Saving/machine time used 55,760 27,340 24,400 Operating hrs 0 1 2 3 75,000 2,000hrs 2,000hrs 2,000hrs

  25. Rate of Return or Internal Rate of Return - Def 1: ROR is the interest rate earned on the unpaid balance of an amortized loan - Ex: A bank lend 10 million won and receives 4.021 million won each year over the next 3 years. Then, it can be said that this bank earns 10% of ROR on the loan. Unit:000 Interest on the Unrec. Bal.(10%) Begin. Unrecovered Bal. Ending. Unrec.Bal Recov. Money n 0 1 2 3 -10,000 -10,000 -6,979 -3,656 -1,000 -698 -366 +4,021 +4,021 +4,021 -10,000 -6,979 -3,656 0

  26. ROR - Def 2: ROR is the break-even interest rate, i*, which equates the present value of a project’s cash outflows to the present value of its cash inflows. - Equation

  27. ROR=IRR Internal rate of return is the interest rate charged on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project balance will be zero Unit:000 Interest on the Unrec. Bal.(10%) Begin. Unrecovered Bal. Ending. Unrec.Bal Recov. Money n 0 1 2 3 -10,000 -10,000 -6,979 -3,656 -1,000 -698 -366 +4,021 +4,021 +4,021 -10,000 -6,979 -3,656 0

  28. The Types of Projects • Simple Investment • Def: one in which the initial cash flows are negative, and only one sign change occurs in the net cash flow series. • Example: -100, 250,300 (-, +, +) • i* : Only one unique i* i* becomes the IRR • Nonsimple Investment • Def: one in which more than one sign change occurrs in the cash flow series • Example: -100, 250,300(-, +, +,-) • i* : the real i* may exist as many as a number of sign changes in the cash flow series. • So, any i* can not be the IRR. 28

  29. Simple and Nonsimple Investment Project Ex7.7] Given] cash flows(refer to the table below) Unit: 000

  30. Continued…. Sol] Identify a number of sign changes in the net cash flows

  31. IRR Concept Ex7.8] understanding the IRR conept given] I= 49.950M, Cash Inflow= 18.5M, Life= 6 years Sol] Determine the IRR -obtain # of real root using Mathematica IRR=29% Plot[-49950+18500/(1+i)1+18500/(1+i)2+18500/(1+i)3 +18500/(1+i)4+18500/(1+i)5+18500/(1+i)6,{i, -1, 1}, PlotRange  {-50000,300000}] i

  32. Prove it with a PB - Check up IRR=29% with the PB concept  PB at “0” :  PB at “1” : • PB at “2”: •  PB at “3”:  PB at “4” :  PB at “5” : •  PB at “6”:

  33. A Decision Rule for the case in which there exists only a unique i*. ifIRR > MARR, accept the project ifIRR = MARR, remain indifferent ifIRR > MARR, reject the project Note that this decision rule can not be applied for the case in which there exist more than one i*.

  34. Calculate IRR 150,000 5,000 120,000 80,000 1 2 1 2 3 4 5 0 년 년 3 4 5 6 7 90,000 90,000 1,318.99 180,000 (b) borrowing project (a) investing project Ex7.9] Determine the IRR of the project Given] Cash Flow Diagrams Unit: 000 0

  35. Continued….. Sol] 1) Determine the IRR of Project (a) - Obtain a numberr of i* - Determine the IRR Plot[-90000-180000/(1+i)-90000/(1+i)2+80000/(1+i)3+ 80000/(1+i)4+120000/(1+i)5+120000/(1+i)6+150000/(1+i)7,{i,-1,1},PlotRange {-200000,500000}] NPV i FindRoot[-90000-180000/(1+i)-90000/(1+i)2+80000/(1+i)3+80000/(1+i)4+120000/(1+i)5+120000/(1+i)6+150000/(1+i)7Š0, {i,0.05}] {i*0.10473633423827057}

  36. Continued…… 1) Determine the IRR of Project (b) - Obtain a number of i*. - Find out the IRR Plot[5000-1318.99/(1+i)-1318.99/(1+i)2-1318.99/(1+i)3-1318.99/(1+i)4-1318.99/(1+i)5, {i,1,1}, PlotRange {-5000,5000}] NPV i FindRoot[5000-1318.99/(1+i)-1318.99/(1+i)2-1318.99/(1+i)3-1318.99/(1+i)4-1318.99/(1+i)5Š0, {i,0.05}]{i0.100001}

  37. An Example Which Is Economically Analyzed with a Variety of The Appraisal Techniques Ex7.11] An Economic Analysis with a Variety of The Techniques Given] Cash Flows, MARR=8% Sol] Unit: 000

  38. The Conditions to Compare Alternatives Keep the alternatives independent or mutually exclusive one another 2. Set up a common time period 3. Perform an incremental analysis if neccessay(IRR)

  39. The Reasons for Why NPV, AE, and IRR Techniques Are Chosen to Determine A Preference Ordering NPV - It is most widely used by companies - Its final result is expressed in an absolute value (2) AE - Its format is consistent with a fiscal year - It provides a unit cost and profit - It makes us convenient to compare alternatives whose lives are different (3) IRR - Its final result is expressed as percentage such that managers easily understand its meaning - It is also one of the techniques which are most widely used by companies

  40. Determine a preference ordering of the alternatives Ex7.12] Determine a preference ordering of the alternatives with a variety of the techniques Given] Cash Flow, MARR=10%, It is assumed that a capital budgett can be provided without limit Unit: 000

  41. A Preference Ordering with the NPV • In case which the alternatives are independent(A, C, D) Conclusion: Since all the alternatives are independent and their NPVs are greater than 0, it is better to undertake all of them. - In case which they are mutually exclusive (A, C, D) Conclusion: It is required to undertake alternative “A” only because its NPV is greater than others

  42. 200,000 200,000 0 1 4 n 6 2 3 5 100,000 100,000 Determine a preference ordering with the AE (1) Obtain the AE with a least common multiple of 6 years 140,000 140,000 140,000 0 n 1 4 5 6 2 3 10,000 10,000 10,000 100,000 100,000 100,000 Cash Flow of “A” over 6 years Cash Flow of “B”

  43. A Comparison with the AE Technique • A LCM of 6 years  Calculate the AE of “A”

  44. Continued…  For “B”  For “C”

  45. Continued…  For “D” Way to Calculate the AE of the Alternatives with a single cycle of cash flows  For “A”

  46. Continued… For “B”  For “D”  it is recommended to use the AE techniques for which the project lives are different - When the alternatives are independent - Select the alternative with a highest AE  for “C”

  47. A Comparison with the IRR -Question: Can we determine the preference ordering of the alternatives according to the seize of the IRR? Unit: 000 n 0 1 IRR NPV(10%) A1 -1,000 2,000 100% 818 A2 -5,000 7,000 40% 1,364 > <

  48. Incremental Analysis Unit: 0000 • If MARR=10%,it implies that the project earns a profitability of 10% is guaranteed. That is to say, the investment of 4 million won will grow up to 4.4 million won. • We can earn 5 million won one year after once we invest 4 million won in A2. Since the IRR of 25% is greater than the MARR, it is desirable to undertake the project.

  49. The Steps of The Incremental Analysis Step 1: Subtract the cash flows of Project (A) whose initial investment cost is less than the other from the cash flows of Project(B) whose initial investment cost is greater -Step 2: Calculate the the IRRB-Aof the incremental project. Step 3: Select the project based the following decision rules. If IRRB-A > MARR, undertake Project B If IRRB-A = MARR, remain indifferent If IRRB-A < MARR, undertake Project A. 49

  50. Example Ex7.13] The Example for the Incremental Analysis with the IRRTechnique Given] Cash Flows for two projects, MARR=10% Unit: 000 If MARR = 10%, which one is the better in an economic sense? Since IRR B2-B1 >15% > 10% which is greater than MARR=10%,,it is recommended to undertake Project B2

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