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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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Chapter 7

An Economic Appraisal II:

NPV, AE, IRR Technique

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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


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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


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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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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


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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


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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


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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


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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


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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)


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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%)


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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


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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)


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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


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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)


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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


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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.


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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.


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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


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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)


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Calculate a CR(i)

Unit: 000 won

  • CR(i) for Mini Cooper


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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=12.078M

A Depreciation Method: A SL Method

AE of the Interest= 2,778.74(A/P, 10%, 3)=1.03995M


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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)


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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


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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


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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


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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


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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


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Simple and Nonsimple Investment Project

Ex7.7]

Given] cash flows(refer to the table below)

Unit: 000


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Continued….

Sol]

Identify a number of sign changes in the net cash flows


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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


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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”:


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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*.


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Calculate IRR

150,000

5,000

120,000

80,000

1

2

1

2

3

4

5

0

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


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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}


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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}


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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


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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)


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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


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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


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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


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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”


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A Comparison with the AE Technique

  • A LCM of 6 years

     Calculate the AE of “A”


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Continued…

 For “B”

 For “C”


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Continued…

 For “D”

Way to Calculate the AE of the Alternatives with a single cycle of cash flows

 For “A”


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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”


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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

>

<


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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.


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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


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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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