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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Chapter 7
An Economic Appraisal II:
NPV, AE, IRR Technique
Powerpoint Templates
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
NPV....
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
The Steps to Make a Decision with the NPV Technique
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
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
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
Net Future Value Technique
(unit: 000 won)
55,760
27,340
24,400
0
3
1
2
75,000
Project Life
8
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
A Project Balance Concept
Ex7.2] An Economic Meaning of An NPV Based on A PB
(unit: 000 won)
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%)
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
Capitalized Equivalent
A
P = CE(i)
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
Annual Equivalent
0
1
……
2 3 4 5 N
A
……
0 1 2 3 4 5 N
0
NPV(i)
AE(i) =NPV(i)(A/P, i, N)
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
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. Makeorbuy 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.
Capital Cost and Operating Cost
: 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.
: 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.
Annual Equivalence Analysis
CC
AE Cost
+
OC
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)
Calculate a CR(i)
Unit: 000 won
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
AECR(i)
Ex7.5]
Given] I=20M, S=4M, A=4.4M, N=5 years, i=10%
Sol] An investment decisionmaking 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)
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
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
ROR
 Def 2: ROR is the breakeven interest rate, i*, which equates the present value of a project’s cash outflows to the present value of its cash inflows.
 Equation
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
The Types of Projects
i* becomes the IRR
100, 250,300(, +, +,)
28
Simple and Nonsimple Investment Project
Ex7.7]
Given] cash flows(refer to the table below)
Unit: 000
Continued….
Sol]
Identify a number of sign changes in the net cash flows
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
Prove it with a PB
 Check up IRR=29% with the PB concept
PB at “0” :
PB at “1” :
PB at “4” :
PB at “5” :
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*.
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
Continued…..
Sol] 1) Determine the IRR of Project (a)
 Obtain a numberr of i*
 Determine the IRR
Plot[90000180000/(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[90000180000/(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)70, {i,0.05}]
{i*0.10473633423827057}
Continued……
1) Determine the IRR of Project (b)
 Obtain a number of i*.
 Find out the IRR
Plot[50001318.99/(1+i)1318.99/(1+i)21318.99/(1+i)31318.99/(1+i)41318.99/(1+i)5, {i,1,1}, PlotRange {5000,5000}]
NPV
i
FindRoot[50001318.99/(1+i)1318.99/(1+i)21318.99/(1+i)31318.99/(1+i)41318.99/(1+i)50, {i,0.05}]{i0.100001}
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
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)
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
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
A Preference Ordering with the NPV
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
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”
A Comparison with the AE Technique
Calculate the AE of “A”
Continued…
For “B”
For “C”
Continued…
For “D”
Way to Calculate the AE of the Alternatives with a single cycle of cash flows
For “A”
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”
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
>
<
Incremental Analysis
Unit: 0000
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 IRRBAof the incremental project.
Step 3: Select the project based the following decision rules.
If IRRBA > MARR, undertake Project B
If IRRBA = MARR, remain indifferent
If IRRBA < MARR, undertake Project A.
49
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 B2B1 >15% > 10% which is greater than MARR=10%,,it is recommended to undertake Project B2