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

2. Powerpoint Templates Page 2 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 2 3 4 5 0 1 Cash Inflows Cash Outflows 0 PV(i) CIF PV(i) COF NPV NPV(i) > 0

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

4. Powerpoint Templates Page 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. Powerpoint Templates Page 5 Ex 7.1] An Investment Decision on A Construction Project of A Small Power Plant with A NPVNPV Given] Cash Flows Diagram, MARR=8% 50 60 80 50 100 150 100 60 120 0-2-3-4-5-6-7-8-9-10 123504948 …………… n (unit: Million won)

6. Powerpoint Templates Page 6 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 Continued……..

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

8. Powerpoint Templates Page 8 8 Net Future Value Technique  Given: Cash Flows and MARR (i)  Find: A Net Equivalent Value at the End of a Project Life 75,000 24,400 27,340 55,760 0 12 3 Project Life (unit: 000 won)

9. Powerpoint Templates Page 9 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 An Investment Decision with A NFW for Ex. 7.1

10. Powerpoint Templates Page 10 -Project Balance (PB): Cash Flows Left inside A Project -Equation  PB 0 =PB 0  PB n =PB n-1 (1+MARR)+CF n Ex 7.2] An Economic Meaning of An NPV Based on A PB A Project Balance Concept n 012345 NCF (62,500)33,98233,72633,20533,13582,013 (unit: 000 won)

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

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

13. Powerpoint Templates Page 13  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 Capitalized Equivalent A P = CE(i)

14. Powerpoint Templates Page 14 Ex 7.3] 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 Continued …. 200M P = CE(8%)=2.5B ∞

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

16. Powerpoint Templates Page 16 Ex 7.4] Convert the irregular cash flows into an equivalent worth Continued….. 2 3 4 5 6 1 A=$1.835 0 1 2 3 4 5 6 0 Unit:: $M 0 15 3.5 5 9 12 10 8

17. Powerpoint Templates Page 17 1.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. 2.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. 3.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. The Advantages of An AE Technique

18. Powerpoint Templates Page 18  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. Capital Cost and Operating Cost

19. Powerpoint Templates Page 19  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. Annual Equivalence Analysis CC OC + AE Cost

20. Powerpoint Templates Page 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. 0 N I S 0 1 2 3 N CR(i) ………………

21. Powerpoint Templates Page 21 Calculate a CR(i) - CR(i) for Mini Cooper Unit: 000 won TypeModel Purchase Cost SV at the end of year 3 SmallMini Cooper19,80012,078

22. Powerpoint Templates Page 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 nBegin. Inv. CostInterest with i= 6%PV of the interest 119.8001.1881.188(P/F,6%,1)=1.12075 217.2261.033561.03356(P/F,6%,2)=0.919.6 314.6520.87912o.87912(P/F,6%,3)=0.73813 총액 2.77874 AE of the Interest = 2,778.74(A/P, 10%, 3)=1.03995M

23. Powerpoint Templates Page 23 Ex 7.5] 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 - Method 1: first obtain the NPV and transform it into AE - Method 2: Make a decision with CR(i) AE-CR(i)

24. Powerpoint Templates Page 24 Ex 7.6] 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 24,400 0 1 2 3 55,760 27,340 75,000 Operating hrs 2,000hrs

25. Powerpoint Templates Page 25 - 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. Rate of Return or Internal Rate of Return 01230123 Unit:000 n Begin. Unrecovered Bal. -10,000 -6,979 -3,656 -1,000 -698 -366 Interest on the Unrec. Bal.(10%) Recov. Money +4,021 Ending. Unrec.Bal -10,000 -6,979 -3,656 0

26. Powerpoint Templates Page 26 - 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 ROR

27. Powerpoint Templates Page 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 01230123 Unit:000 n Begin. Unrecovered Bal. -10,000 -6,979 -3,656 -1,000 -698 -366 Interest on the Unrec. Bal.(10%) Recov. Money +4,021 Ending. Unrec.Bal -10,000 -6,979 -3,656 0

28. Powerpoint Templates Page 28 28 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. The Types of Projects

29. Powerpoint Templates Page 29 Ex 7.7] Given] cash flows(refer to the table below) Simple and Nonsimple Investment Project Period (N) Project AProject BProject C 0-225-270-450 1135158270 22,02515890 3 4 -45 590 667 Unit: 000

30. Powerpoint Templates Page 30 Sol] Identify a number of sign changes in the net cash flows Continued…. n0123456 # of sign changes S or NS Project A-++1S Project B-++++1S Project C -+++-++3NS

31. Powerpoint Templates Page 31 Ex 7.8] 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 Concept 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 IRR=29%

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