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Chapter 7An Economic Appraisal II:NPV, AE, IRR Technique
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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
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NPV....
- Equation
- Decision RuleNPV(MARR) > 0 Accept itNPV(MARR) = 0 IndifferentNPV(MARR) < 0 Reject it
0
0 1 2
( )(1 )( / , ,0) ( / , ,1) ( / , , 2)
( / , , )
Nt
tt
N
CFNPV MARRMARR
CF P F MARR CF P F MARR CF P F MARRCF P F MARR N
Where, CFt: cash flow at time t, MARR: minimum attractive rate of return on a project
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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.
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Ex 7.1] An Investment Decision on A Construction Project of A Small Power Plant with A NPVNPV
Given] Cash Flows Diagram, MARR=8%
5060
80
50
100
150
100
60
120 120
0-1-2-3-4-5-6-7-8-9-101 2 3 504948
……………
n
(unit: Million won)
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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……..
A Present Value of The construction Costs50( / ,8%, 2) 60( / ,8%,3) 80( / ,8%,4) 50( / ,8%,5)100( / ,8%, 2)( / ,8%,5) 150( / ,8%,8) 100( / ,8%,9)60( / ,8%,10) 509.85
P A P F P F P FP A P F P F P F
P F M
120( / ,8%,50)( / ,8%,10) 680P A P F M A Present Value of All the Cash Inflows
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- Step 5: Make an investment decision on the project with NPV(8%)
Since NPV(8%)=170.15 M >0, accept the project
Continued……..
(8%) 680 509.85 170.15NPV M
0.5 1.0 1.5 2.0
200
150
100
50
50
100
MARR(100%)
(NPV
Unit:$M
Break_Even Interest Rate=IRR=9.78%
The Sensitivity Analysis of the NPV with A Varying Interest Rate
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Net Future Value TechniqueGiven: Cash Flows and MARR (i)
Find: A Net Equivalent Value at the End of a Project Life
75,000
24,400 27,34055,760
01 2 3
Project Life
(unit: 000 won)
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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
60( / ,8%,0) 100( / ,8%,1) 150( / ,8%, 2)100( / ,8%, 2)( / ,8%,3) 50( / ,8%,5)80( / ,8%,6) 60( / ,8%,7) 50( / ,8%, 2)( / ,8%,8)1,100.7
F P F P F PF A F P F P
F P F P F A F PM
Futue Value of All the Cash Outflows
120( / , 8% , 50) 1, 468P A M Present Value of All the Cash Flows
(8%) 1, 468 1,100.7 367.3 ) ,accept the project.Since NFV M
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- Project Balance (PB): Cash Flows Left inside A Project- Equation
PB0=PB0 PBn=PBn-1(1+MARR)+CFn
Ex 7.2] An Economic Meaning of An NPV Based on A PB
A Project Balance Concept
n 0 1 2 3 4 5
NCF (62,500) 33,982 33,726 33,205 33,135 82,013
0
1
2
3
4
5
6 2 , 5 0 0 , 0 0 06 2 , 5 0 0 (1 0 .1 5 ) 3 3 , 9 8 2 3 7 , 8 9 3 , 0 0 03 7 , 8 9 3 (1 0 .1 5 ) 3 3 , 7 2 6 9 , 8 5 1, 0 0 09 , 8 5 1(1 0 .1 5 ) 3 3 , 2 0 5 2 1, 8 7 6 , 0 0 0
2 1, 8 7 6 (1 0 .1 5 ) 3 3 , 1 3 5 5 8 , 2 9 2 , 0 0 05 8 , 2 9 2 (1 0 .1 5 ) 8 2 , 0 1 3 1 4 9 , 0 4 9 , 0 0 0
P BP BP BP BP BP B
(unit: 000 won)
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Continued…
NPV(15%) = 149,049 (P/F, 15%, 5) =74,107,0009,8512 2.3
33, 205DPBP years
(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
-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
-62,500
-62,500
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-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
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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
( ) ( / , , ) ACE i P A P A ii
A
P = CE(i)
1 1:
1
niNote P A ni i
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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 ….
2(8%) ( / ,8%, ) 25( )0.08
CE A P A 억원
200M
P = CE(8%)=2.5B ∞
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Annual Equivalent
2 3 4 5 N1
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
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Ex 7.4] Convert the irregular cash flows into an equivalent worth
Continued…..
2 3 4 5 61
A=$1.835
0 1 2 3 4 5 60
Unit:: $M
0
(15%) $6.946NPV M(15%) $6.946( / ,15%,6) $1.835AE A P M
15
3.5
59
12 108
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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
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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 toprovide 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 servicesuch as a purchase cost and sales taxes.
Capital Cost and Operating Cost
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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 C
ost
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Capital Recovery Cost(CR) Definition: to be an annulequivalent of capital costsItems 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)
………………
iSNiPASINiFASNiPAIiCR
),,/)((),,/(),,/()(
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Calculate a CR(i)
- CR(i) for Mini Cooper
Unit: 000 won
Type Model Purchase Cost
SV at the end of year
3
Small Mini Cooper 19,800 12,078
(6%) ( )( / ,6%, )(19,800 12,078)( / ,6%,3) 12,078(0.06) 3.61355
R I S A P N iSP M
CA
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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=12.078MA Depreciation Method: A SL Method
19.800 12.078 2.5743
D M
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
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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)
(10%) 20,000 4, 400( / ,10%,5) 4,000( / ,10%,5)0.83688(10%) 836.88( / ,10%,5) 0.22076
NPV P A P F
AE A P
1
2
1 2
( ) ( )[(20,000 4,000)( / ,10%,5) (0.10)(4,000)]4.62076
( ) 4.400(10%) ( ) ( )
4,620.76 4, 400 0.22076
AE i CR iA P
MAE i M
AE AE i AE iM
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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
55,76027,340
75,000
Operating hrs
2,000hrs 2,000hrs 2,000hrs
(15%) 3,553( / ,15%,3) 1.556AE A P M
1.556 / 2,000780 /
M hrshr
Saving/MCH
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- 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
0
1
2
3
Unit:000
nBegin.
Unrecovered Bal.
-10,000
-10,000
-6,979
-3,656
-1,000
-698
-366
Interest on the Unrec. Bal.(10%)
Recov. Money
+4,021
+4,021
+4,021
Ending. Unrec.Bal
-10,000
-6,979
-3,656
0
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- 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
* * *
**
0
31 20 * * 2 * 3
1* 1 *
( ) ( ) ( ) 0
( )(1 )
(1 ) (1 ) (1 )
(1 ) (1 )0
Nt
tt
N NN N
NPV i PV i PV iCFNPV i
iCFCF CFCF
i i iCF CF
i i
cash inflows cash outflows
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ROR=IRRInternal 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
0
1
2
3
Unit:000
nBegin.
Unrecovered Bal.
-10,000
-10,000
-6,979
-3,656
-1,000
-698
-366
Interest on the Unrec. Bal.(10%)
Recov. Money
+4,021
+4,021
+4,021
Ending. Unrec.Bal
-10,000
-6,979
-3,656
0
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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
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Ex 7.7]
Given] cash flows(refer to the table below)
Simple and Nonsimple Investment Project
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
Unit: 000
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Sol] Identify a number of sign changes in the net cash flows
Continued….
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
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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}]
1.0 0.5 0.5 1.0
50 000
50 000
100 000
150 000
200 000
250 000
300 000
i
** * 2 * 3
18,500 18,500 18,500( ) 49,950(1 ) (1 ) (1 )
NPV ii i i
0)1(
500,18)1(
500,18)1(
500,186*5*4*
iiiIRR=29%
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- Check up IRR=29% with the PB concept
Prove it with a PB
nnn CFIRRPBPB )1(1
PB at “0” :
PB at “1” :
PB at “2”:
PB at “3” :
0 0 49.950PB CF M
1 49,950(1 0.29) 18,50045.939
PBM
2 45,939(1 0.29) 18,50040.757
PBM
3 40,757(1 0.29) 18,50034.076
PBM
PB at “4” :
PB at “5” :
PB at “6”:
5 25, 458(1 0.29) 18,50014.341
PBM
6 45,939(1 0.29) 18,50040.757
PBM
4 34,076(1 0.29) 18,50025.458
PBM
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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*.
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Ex 7.9] Determine the IRR of the project
Given] Cash Flow Diagrams
Calculate IRR
(a) investing project
1 2 3 4년
5,000
5
1,318.99
0 1 2
3 4 5 6 7년
90,000
180,000
90,000
80,000120,000
150,000
(b) borrowing project
0
Unit: 000
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Sol] 1) Determine the IRR of Project (a)
- Obtain a numberr of i*
- Determine the IRR
Continued…..
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}]
1.0 0.5 0.5 1.0
200 000
100 000
100 000
200 000
300 000
400 000
500 000
i
NPV
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}
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1) Determine the IRR of Project (b)
- Obtain a number of i*.
- Find out the IRR
Continued……
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}]
1.0 0.5 0.5 1.0
4000
2000
2000
4000
i
NPV
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}]{i0.100001}
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Ex 7.11] An Economic Analysis with a Variety of The TechniquesGiven] Cash Flows, MARR=8%Sol]
An Example Which Is Economically Analyzed with a Variety of The Appraisal Techniques
n NCF of Project A NCF of Project B NCF of Project C
0
1
2
3
4
5
6
-200,000
20,000
40,000
60,000
60,000
60,000
68,000
-200,000
80,000
60,000
60,000
40,000
-200,000
60,000
60,000
60,000
60,000
40,000
20,000
총 이익 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%
Unit: 000
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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)
The Conditions to Compare Alternatives
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(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
The Reasons for Why NPV, AE, and IRR Techniques Are Chosen to Determine A Preference Ordering
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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
Determine a preference ordering of the alternatives
n A B C D
0
1
2
3
-100,000
-100,000
200,000
200,000
-100,000
140,000
-10,000
-100,000
50,000
-50,000
200,000
-100,000
470,000
-720,000
360,000
Unit: 000
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- 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 toundertake 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
A Preference Ordering with the NPV
(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%,3) 200,000( / ,10%, 4)124.640
(10%) 100,000 50,000( / ,10%,1) 50,000( / ,10%,3) 200,000( / ,10%, 4)53.950
(10%) 100,000 470,000( / ,1
A
C
D
NPV P F P F P FM
NPV P F P F P FM
NPV P F
0%,1) 720,000( / ,10%,3) 360,000( / ,10%, 4)
2.705P F P F
M
(10%) 124.640 (10%) 53.950 (10%) 2.705A C DNPV M NPV M NPV M
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(1) Obtain the AE with a least common multiple of 6 years
Cash Flow of “A” over 6 years
n0 1
2 3
100,000
200,000 200,000
100,000
45 6
n0
1 2 3
100,000
140,000
4 5 6
140,000 140,000
Cash Flow of “B”
10,000
Determine a preference ordering with the AE
10,00010,000
100,000100,000
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- A LCM of 6 years
Calculate the AE of “A”
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
(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%, 2) 100,000( / ,10%,3)100,000( / ,10%, 4) 200,000( / ,10%,5) 200,000( / ,10%,6)218.289
ANPV P F P F P FP F P F P F
M
(10%) (10%)( / ,10%,6)218, 289( / ,10%,6)50.120
A AAE NPV A PA P
M
A Comparison with the AE Technique
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For “B”
For “C”
(10%) 19,008 19,008( / ,10%, 2) 19,008( / ,10%,4)19,008{(1 ( / ,10%,2) ( / ,10%, 4)}47.700
BNPV P F P FP F P F
M
(10%) (10%)( / ,10%,6)47,700( / ,10%,6)10.952
B BAE NPV A PA P
M
(10%) 54,395 54,395( / ,10%,3)54,395{(1 ( / ,10%,3)}95.262
CNPV P FP F
M
(10%) (10%)( / ,10%,6)95, 262( / ,10%,6)21.873
C CAE NPV A PA P
M
Continued…
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For “D”
Way to Calculate the AE of the Alternatives with a single cycle of cash flows For “ A”
(10%) 2,705 2,705( / ,10%,3)2,705{(1 ( / ,10%,3)}4.737
DNPV P FP F
M
(10%) (10%)( / ,10%,6)4,737( / ,10%,6)1.088
D DAE NPV A PA P
M
(10%) (10%)( / ,10%,3)124,643( / ,10%,3)50.120
A AAE NPV A PA P
M
(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%, 2) 200,000( / ,10%,3)124.643
ANPV P F P F P FM
Continued…
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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
(10%) (10%)( / ,10%, 2)19,008( / ,10%, 2)10.952
B BAE NPV A PA P
M
(10%) (10%)( / ,10%,3)2,705( / 10%,3)1
,.088
D DAE NPV A PA P
M
for “C”
(10%) (10%)( / ,10%,3)54,395( / ,10%,3)21.873
C CAE NPV A PA P
M
Continued…
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-Question: Can we determine the preference ordering of the alternatives according to the seize of the IRR?
n
0
1
IRR
NPV(10%)
Unit: 000
A1
-1,000
2,000
100%
818
A2
-5,000
7,000
40%
1,364
>
<
A Comparison with the IRR
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Incremental Analysis
• 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.
n A1 A2 (A2 – A1)
01
-1,0002,000
-5,0007,000
-4,0005,000
IRRNPV(10%)
100%818
40%1,364
25%546
Unit: 0000
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Step 1: Subtract the cash flows of Project (A) whose initial investment cost isless than the other from the cash flows of Project(B) whose initialinvestment cost is greater
-Step 2: Calculate the the IRRB-A of the incremental project.
Step 3: Select the project based the following decision rules.
If IRR B-A > MARR, undertake Project BIf IRR B-A = MARR, remain indifferentIf IRR B-A < MARR, undertake Project A.
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The Steps of The Incremental Analysis
Powerpoint Templates Page 50
Ex 7.13] The Example for the Incremental Analysis with the IRR Technique
Given] Cash Flows for two projects, MARR=10%
Example
Unit: 000
n B1 B2 B2 - B10123
-3,0001,3501,8001,500
-12,0004,2006,2256,330
-9,0002,8504,4254,830
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