Chapter 7 An Economic Appraisal II: NPV, AE, IRR Technique

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

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

NPV... - Equation NPV ( MARR) N t 0 CFt (1 MARR) t CF ( P / F, MARR,0) CF ( P / F, MARR,1) CF ( P / F, MARR,2) 0 1 2 CF N ( P / F, MARR, N) Where, CF t : cash flow at time t, MARR: minimum attractive rate of return on a project - Decision Rule NPV(MARR) > 0 Accept it NPV(MARR) = 0 Indifferent NPV(MARR) < 0 Reject it Powerpoint Templates Page 3

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. Powerpoint Templates Page 4

Ex 7.1] 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 1 2 3 48 49 50 n 50 60 80 50 100 100 60 150 (unit: Million won) Powerpoint Templates Page 5

Sol] Continued.. - 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. A Present Value of The construction Costs 50( P/ A,8%, 2) 60( P/ F,8%,3) 80( P/ F,8%,4) 50( P/ F,8%,5) 100( P/ A,8%, 2)( P/ F,8%,5) 150( P/ F,8%,8) 100( P/ F,8%,9) 60( P/ F,8%,10) 509.85M (2) a present value of all the net cash flows incurred during the commercialization stage of 50 years A Present Value of All the Cash Inflows 120( P/ A,8%,50)( P/ F,8%,10) 680M Powerpoint Templates Page 6

Continued.. - Step 5: Make an investment decision on the project with NPV(8%) NPV (8%) 680 509.85 170.15M Since NPV(8%)=170.15 M >0, accept the project (NPV Unit: $M 100 50 50 100 150 200 Break_Even Interest Rate=IRR=9.78% 0.5 1.0 1.5 2.0 MARR( 100%) The Sensitivity Analysis of the NPV with A Varying Interest Rate Powerpoint Templates Page 7

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

An Investment Decision with A NFW for Ex. 7.1 Sol] (1) A future value of all the cash flows incurred under construction Futue Value of All the Cash Outflows 60( F / P,8%,0) 100( F / P,8%,1) 150( F / P,8%, 2) 100( F / A,8%, 2)( F / P,8%,3) 50( F / P,8%,5) 80( F / P,8%,6) 60( F / P,8%,7) 50( F / A,8%, 2)( F / P,8%,8) 1,100.7M (2) A present value of all the cash flows incurred during the commercialization stage Present Value of All the Cash Flows 120( P / A,8%,50) 1,468M Since NFV (8%) 1, 468 1,100.7 367.3 M ),accept the project. Powerpoint Templates Page 9

A Project Balance Concept - 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 (unit: 000 won) n 0 1 2 3 4 5 NCF (62,500) 33,982 33,726 33,205 33,135 82,013 PB PB PB PB PB PB 0 1 2 3 4 5 62,500,000 62,500(1 0.15) 33,982 37,893,000 37,893(1 0.15) 33,726 9,851,000 9,851(1 0.15) 33,205 21,876,000 21,876(1 0.15) 33,135 58,292,000 58,292(1 0.15) 82,013 149,049,000 Powerpoint Templates Page 10

Continued (unit: 000 won) N 0 1 2 3 4 5 Beginning Bal. -62,500-37,893-9,851 +21,876 +58,292 Interest -9,375-5,684-1,478 +3,281 +8,744 Ending Bal. -62,500 +33,982 +33,726 +33,205 +33,135 +82,013-62,500-37,893-9,851 +21,876 +58,292 +149,049 NPV(15%) = 149,049 (P/F, 15%, 5) =74,107,000 9,851 DPBP 2 2.3years 33,205 NFV(15%) Powerpoint Templates Page 11

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

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 CE() i P A( P / A,, i ) A i A n 1 1 Note P A n i P = CE(i) : i i 1 Powerpoint Templates Page 13

Continued. 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 2 CE(8%) A( P / A,8%, ) 25( 억원) 0.08 200M P = CE(8%)=2.5B Powerpoint Templates Page 14

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

Continued.. Ex 7.4] Convert the irregular cash flows into an equivalent worth Unit:: $M 0 1 3.5 12 9 10 5 8 2 3 4 5 6 15 A=$1.835 0 NPV (15%) $6.946M 0 1 2 3 4 5 6 AE(15%) $6.946( A/ P,15%,6) $1.835M Powerpoint Templates Page 16

The Advantages of An AE Technique 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. Powerpoint Templates Page 17

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. Powerpoint Templates Page 18

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. AE Cost CC + OC Powerpoint Templates Page 19

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) I( A/ P, i, N) S( A/ F, i, N) ( I S)( A/ P, i, N) is CR(i) Powerpoint Templates Page 20

Calculate a CR(i) Unit: 000 won Type Model Purchase Cost SV at the end of year 3 Small Mini Cooper 19,800 12,078 - CR(i) for Mini Cooper CR(6%) ( I S)( A/ P,6%, N) is (19,800 12,078)( A/ P,6%,3) 12,078(0.06) 3.61355M Powerpoint Templates Page 21

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 19.800 12.078 D 2.574M 3 n Begin. Inv. Cost Interest with i= 6% PV of the interest 1 19.800 1.188 1.188(P/F,6%,1)=1.12075 2 17.226 1.03356 1.03356(P/F,6%,2)=0.919.6 3 14.652 0.87912 o.87912(p/f,6%,3)=0.73813 총액 2.77874 AE of the Interest = 2,778.74(A/P, 10%, 3)=1.03995M Powerpoint Templates Page 22

AE-CR(i) 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 NPV (10%) 20,000 4, 400( P / A,10%,5) 4,000( P / F,10%,5) 0.83688 AE(10%) 836.88( A / P,10%,5) 0.22076 - Method 2: Make a decision with CR(i) AE() i 1 CR() i [(20,000 4,000)( A/ P,10%,5) (0.10)(4,000)] 4.62076M AE() i 2 4.400M AE(10%) AE( i) 1 AE( i) 2 4,620.76 4, 400 0.22076M Powerpoint Templates Page 23

Ex 7.6] profit/machine time used when the operating time is constant Given] NPV=3.553M, N=3 years, i= /year: 2,000 hrs Sol] Saving/machine time used 15%, machine time used 24,400 27,340 55,760 Operating hrs 0 2,000hrs 2,000hrs 2,000hrs 1 2 3 75,000 AE(15%) 3,553( A/ P,15%,3) 1.556M Saving/MCH 1.556 M / 2,000hrs 780 / hr Powerpoint Templates Page 24

- 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. n 0 Rate of Return or Internal Rate of Return Begin. Unrecovered Bal. -10,000 Interest on the Unrec. Bal.(10%) Recov. Money Unit:000 Ending. Unrec.Bal -10,000 1-10,000-1,000 +4,021-6,979 2-6,979-698 +4,021-3,656 3-3,656-366 +4,021 0 Powerpoint Templates Page 25

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 NPVi PVi PVi NPV i * * * ( ) ( ) cash inflows ( ) cash outflows 0 * ( ) N t * t t 0 (1 i ) CF CF CF CF (1 i ) (1 i ) (1 i ) 1 2 3 0 * * 2 * 3 CFN 1 CFN * N 1 * (1 i ) (1 i ) 0 CF N Powerpoint Templates Page 26

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 n 0 Begin. Unrecovered Bal. -10,000 Interest on the Unrec. Bal.(10%) Recov. Money Ending. Unrec.Bal -10,000 1-10,000-1,000 +4,021-6,979 2-6,979-698 +4,021-3,656 3-3,656-366 +4,021 0 Powerpoint Templates Page 27

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. Powerpoint Templates Page 28 28

Ex 7.7] Simple and Nonsimple Investment Project Given] cash flows(refer to the table below) Unit: 000 Period (N) Project A Project B Project C 0-225 -270-450 1 135 158 270 2 2,025 158 90 3 90 90 4 90-45 5 90 6 67 Powerpoint Templates Page 29

Continued. Sol] Identify a number of sign changes in the net cash flows n 0 1 2 3 4 5 6 # of sign changes S or NS Project A - + + 1 S Project B - + + + + 1 S Project C - + + + - + + 3 NS Powerpoint Templates Page 30

Ex 7.8] understanding the IRR conept IRR Concept given] I= 49.950M, Cash Inflow= 18.5M, Life= 6 years Sol] Determine the IRR -obtain # of real root using Mathematica 300 000 250 000 200 000 IRR=29% 150 000 100 000 50 000 1.0 0.5 0.5 1.0 50 000 i NPV i 18,500 18,500 18,500 * * 2 * 3 (1 i ) (1 i ) (1 i ) 18,500 18,500 18,500 0 * 4 * 5 * 6 (1 i ) (1 i ) (1 i ) * ( ) 49,950 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}] Powerpoint Templates Page 31

Prove it with a PB - Check up IRR=29% with the PB concept PB n PB (1 n 1 IRR) CF n PB at 0 : PB0 CF0 49.950M PB at 1 : PB1 49,950(1 0.29) 18,500 45.939M PB at 2 : PB2 45,939(1 0.29) 18,500 40.757M PB at 3 : PB at 4 : PB4 34,076(1 0.29) 18,500 25.458M PB at 5 : PB5 25, 458(1 0.29) 18,500 14.341M PB at 6 : PB6 45,939(1 0.29) 18,500 40.757M PB3 40,757(1 0.29) 18,500 34.076M Powerpoint Templates Page 32

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

Calculate IRR Ex 7.9] Determine the IRR of the project Given] Cash Flow Diagrams Unit: 000 150,000 5,000 120,000 80,000 0 1 2 3 4 5 6 7 년 0 1 2 3 4 5 년 90,000 90,000 1,318.99 180,000 (a) investing project (b) borrowing project Powerpoint Templates Page 34

Continued.. Sol] 1) Determine the IRR of Project (a) - Obtain a numberr of i * 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}] 500 000 400 000 NPV 300 000 200 000 100 000 1.0 0.5 0.5 1.0 100 000 i 200 000 - Determine the IRR 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) 7 0, {i,0.05}] {i* 0.10473633423827057} Powerpoint Templates Page 35

Continued 1) Determine the IRR of Project (b) - Obtain a number of i *. 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}] 4000 NPV 2000 1.0 0.5 0.5 1.0 i 2000 4000 - Find out the IRR 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) 5 0, {i,0.05}]{i 0.100001} Powerpoint Templates Page 36

An Example Which Is Economically Analyzed with a Variety of The Appraisal Techniques Ex 7.11] An Economic Analysis with a Variety of The Techniques Given] Cash Flows, MARR=8% Sol] Unit: 000 n NCF of Project A NCF of Project B NCF of Project C 0-200,000-200,000-200,000 1 20,000 80,000 60,000 2 40,000 60,000 60,000 3 60,000 60,000 60,000 4 60,000 40,000 60,000 5 60,000 40,000 6 68,000 20,000 총이익 108,000 40,000 100,000 PBP(Pref. B, C,A) 4.3 yrs 3yrs 3.3 yrs DPBP(Pref.: B, C,A) More than 6yrs 3.9 yrs 4.05yrs ARR* 9% 5% 8.3% NPV(8%) 28,230 2,546 38,554 IRR 12% 8.5% 15% Powerpoint Templates Page 37

The Conditions to Compare Alternatives 1. Keep the alternatives independent or mutually exclusive one another 2. Set up a common time period 3. Perform an incremental analysis if neccessay(irr) Powerpoint Templates Page 38

The Reasons for Why NPV, AE, and IRR Techniques Are Chosen to Determine A Preference Ordering (1)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 Powerpoint Templates Page 39

Determine a preference ordering of the alternatives Ex 7.12] 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 n A B C D 0-100,000-100,000-100,000-100,000 1-100,000 140,000 50,000 470,000 2 200,000-10,000-50,000-720,000 3 200,000 200,000 360,000 Powerpoint Templates Page 40

A Preference Ordering with the NPV - In case which the alternatives are independent( A, C, D) NPV (10%) 100,000 100,000( P / F,10%,1) 200,000( P / F,10%,3) 200,000( P / F,10%, 4) A Conclusion: 124.640M NPV (10%) 100,000 50,000( P / F,10%,1) 50,000( P / F,10%,3) 200,000( P / F,10%, 4) C 53.950M NPV (10%) 100,000 470,000( P / F,10%,1) 720,000( P/ F,10%,3) 360,000( P/ F,10%,4) D 2.705M 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) NPV (10%) 124.640 M NPV (10%) 53.950 M NPV (10%) 2.705M A C D Conclusion: It is required to undertake alternative A only because its NPV is greater than others Powerpoint Templates Page 41

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

A Comparison with the AE Technique - A LCM of 6 years n A B C D 0 1 2 3 4 5 6-100,000-100,000 200,000 100,000-100,000 200,000 200,000-100,000 140,000-110,000 140,000-110,000 140,000-10,000-100,000 50,000-50,000 100,000 50,000-50,000 200,000-100,000 470,000-720,000 260,000 470,000-720,000 360,000 Calculate the AE of A NPV (10%) 100,000 100,000( P / F,10%,1) 200,000( P / F,10%,2) 100,000( P / F,10%,3) A 100,000( P/ F,10%, 4) 200,000( P/ F,10%,5) 200,000( P/ F,10%,6) 218.289M AE (10%) NPV (10%)( A / P,10%,6) A A 218,289( A/ P,10%,6) 50.120M Powerpoint Templates Page 43

Continued For B NPV (10%) 19, 008 19, 008( P / F,10%, 2) 19, 008( P / F,10%, 4) B 19,008{(1 ( P/ F,10%,2) ( P/ F,10%, 4)} 47.700M AE (10%) NPV (10%)( A / P,10%,6) B B 47,700( A/ P,10%,6) 10.952M NPV C (10%) 54,395 54,395( P / F,10%,3) For C 54,395{(1 ( P/ F,10%,3)} 95.262M AE (10%) NPV (10%)( A / P,10%,6) C C 95,262( A/ P,10%,6) 21.873M Powerpoint Templates Page 44

Continued NPV For D D (10%) 2,705 2,705( P / F,10%,3) Way to Calculate the AE of the Alternatives with a single cycle of cash flows For A 2,705{(1 ( P/ F,10%,3)} 4.737M AE (10%) NPV (10%)( A / P,10%,6) D D 4,737( A/ P,10%,6) 1.088M NPV (10%) 100,000 100,000( P / F,10%,1) 200,000( P / F,10%,2) 200,000( P / F,10%,3) A 124.643M AE (10%) NPV (10%)( A / P,10%,3) A A 124,643( A/ P,10%,3) 50.120M Powerpoint Templates Page 45

Continued For B AE (10%) NPV (10%)( A / P,10%, 2) B B 19,008( A/ P,10%, 2) 10.952M for C AE (10%) NPV (10%)( A / P,10%,3) C C 54,395( A/ P,10%,3) 21.873M For D AE (10%) NPV (10%)( A / P,10%,3) D D 2,705( A/ P, 10%,3) 1.088M 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 Powerpoint Templates Page 46

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 A1 A2 0-1,000-5,000 1 2,000 7,000 IRR 100% > 40% NPV(10%) 818 < 1,364 Powerpoint Templates Page 47

Incremental Analysis Unit: 0000 n A1 A2 (A2 A1) 0-1,000-5,000-4,000 1 2,000 7,000 5,000 IRR 100% 40% 25% NPV(10%) 818 1,364 546 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. Powerpoint Templates Page 48

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 IRR B-A of the incremental project. Step 3: Select the project based the following decision rules. If IRR B-A > MARR, undertake Project B If IRR B-A = MARR, remain indifferent If IRR B-A < MARR, undertake Project A. Powerpoint Templates 49 Page 49

Example Ex 7.13] The Example for the Incremental Analysis with the IRR Technique Given] Cash Flows for two projects, MARR=10% Unit: 000 n B1 B2 B2 - B1 0 1 2 3-3,000 1,350 1,800 1,500-12,000 4,200 6,225 6,330-9,000 2,850 4,425 4,830 IRR 25% 17.43% 15% 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 Powerpoint Templates Page 50