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CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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Page 1: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS

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Page 2: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

A SHORT INTRODUCTION TO A CASH FLOW STATEMENT

Definition: a financial statement showing information on cash inflows and outflows occurring during the fiscal year.

It is prepared for either a enterprise wide basis or a specific investment project basis.

Other types of the financial statements are a balance sheet, an income statement, and a statement of changes in owner’s equity.

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Page 3: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Types of the Financial Statements

- balance sheet- income profit/loss(PL) statement- cash flow statement- statement of changes in owner’ equity

Cash flow Basis

Depositor Tong_Il Bank

10 million won(cash outflow)

11million won(cash inflow)

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Page 4: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Cash Flow Statement prepared with MS_Excel

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Page 5: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.1]

Year Net Cash Flows

0 -300,000

1 90,000

2 90,000

3 90,000

4 90,000

5 90,000

(Unit: 000 won)

An investment project for purchasing the equipment

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Page 6: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

- Evaluation Techniques

(1) net present value: NPV

(2) annual equivalent worth: AE

(3) net future value: NFV

(4) internal rate of return: IRR

Ex) NPV using MS-Excel

=NPV(D1,D5:D9)+D4

An investment project for purchasing the equipment

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Page 7: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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We pay interest only when we borrow money. Borrowing is a matter of obtaining purchasing power that we have not yet earned. Borrowers want money now though they currently have no valuable service to offer in exchange for it. They persuade lenders to provide them with money bow by promising to pay later. Then enter a mutually agreed-upon contract. The ratio between what is given back later and what is obtained now determined the interest rate Interest is thus the price that people pay to obtain resources now rather than wait until they have earned the purchasing power with which to buy the resources. The best way to think about interest is to view it as the premium paid to obtain current command of resources. It surely is not the price of money. Current resources are generally more valuable than future resources because having resources now usually expands one’s opportunity. Present command of resources will often enable us to do things that cause our earning capacity to increase over time, so that we will have more resources at some future date than we would otherwise have had. When we see such a prospect, we want to borrow. And we are willing to pay, if we have to, a premium-interest-as long as the interest is less than what we expect to gain as a result of borrowing

What Is Interest?

Page 8: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

THE TYPES OF INTEREST

1. A SIMPLE INTEREST2. A COMPOUND INTEREST

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Page 9: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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• A Simple Interest Calculation Interest is applied to the initial principal only

(1 )F P I P iN

( )I iP N

Page 10: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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• A Compound Interest Calculation Interest is applied to the total amount

accumulated by so far.

22 (1 ) [ (1 )] (1 )(1 ) (1 )F P i i P i P i i P i

1 (1 )F P Pi P i

2 2 33 (1 ) [ (1 ) ] (1 )F P i i P i P i

(1 )NF P i

Page 11: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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A Simple Interest

• P = principal• i = interest rate• N = a number of

interest periods• Ex 2.2]:

P = 200milliomn won

i = 6%N = 3 years

N

Balance at the

beginning

Interest

Balance at the ending

0 200,000

1 200,000 12,000 212,000

2 212,000 12,000 224,000

3 224,000 12,000 236,000

(unit: 000 won)

Page 12: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

12

A Compounding Interest

• Ex 2.3]:P = 200 million

woni = 6%N = 3 years

(unit: 000 won)

N

Balance at the

beginning

interest

Balance at the ending

0 200,000

1 200,000 12,000 212,000

2 212,000 12,720 224,720

3 224,72013,483

.2238,203.

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Page 13: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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(1 )NF P i

The Fundamental of EE

Page 14: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

A COMPARISON OF A SIMPLE AND COMPOUND INTEREST

A white man bought Manhattan Island for $24 in 1626Given: P=$24 i=8% N=383yearsSimple: F =24(1+0.08×383) = $759.4

Compound: F =24(1+0.08)383 = $151trillion 883.1billion 14

Page 15: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

WHO IS THE WEALTHIEST IN THE WORLD?

15

Rothschild FamilyGiven:P: $600 million in 1840i: 8%N: 169 years

F=6*(1+0.08)169=$267trillion 156.7billion(the value as of 2009

Bill Gates: $57 bl in 2008Warren Buffett: $50 bl in 2008

Page 16: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

THE FUTURE VALUE OF THE INVESTMENT PROJECT FOR THE EQUIPMENT PURCHASE

Ex 2.5]

Given: cash flows for each year, i=12%, N=5 years

16

N C.F. Future Value Factor

F. V.

0 -300,000 (1+0.12)5=1.762 -528,703

1 90,000 (1+0.12)4=1.574 141,617

2 90,000 (1+0.12)3=1.405 126,444

3 90,000 (1+0.12)2=1.254 112,896

4 90,000 (1+0.12)1=1.120 100,800

5 90,000 (1+0.12)0=1.000 90,000

Total 43,054

(unit: 000 won)

Page 17: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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Inflation ? The value of Money

Earning Power

Purchasing Power

Earning PowerPurchasing power

Investment Opportunities

Decrease in purchasing power (inflation)Increase in purchasing power (deflation

Page 18: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Calculate Inflation Rate

- Calculate the inflation rate by dividing the difference between the previous and current year’s price by the previous year’s price.

The price index changes every 5 years in Korea

Ex) calculate the inflation rate with the price of a sports footwear

Given] 100,000 won in 2008 110,000 won in 2009Find] f

Sol.]

18

%101.010

1011

f

Page 19: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Market Interest Rate (i): It is an interest rate for which the inflation and deflation effects are reflected.

Inflation_Free Interest Rate(i’): it is an interest rate for which the inflation and deflation effects are completely removed

Fisher’s Equation: The Relationship among the interest and inflation (deflation)rates

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A Type of Interest Rates Based on the Inflation Concept

(1 ) (1 ')(1 )

' '

i i f

i i f i f

Page 20: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Actual Cash Flow(An): It is a cash flows for which the inflation and deflation effects are reflected.

Constan Cash flow(A’n): It is a cash flows for which the inflation and deflation effects are not reflected.

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A Type of Cash Flows Based on The Inflation Effect

A.C.F

C.C.F Inflation_Free. I.R(i’)

M.I.R(i)

Page 21: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

21

A Nominal and Real Rate of Return on Investment

Ex 2.6]Given] P=100million won, I=8 million won, f=5.5%

Find] A Nominal ROR(r) and A Real ROR(r’)

Sol]

-

-

min ROR

8000.08 8%

100,000

No al

r

Re ( ') ' Equation( , ' ')

' '

0.08 ' 0.055 0.055 '

( ) 0.08 0.055' 0.0237 2.37%

(1 0.055)

al ROR r Fisher s i r i r

r r f r f

r r

r fr

i f

Page 22: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.7]Given] Cash flows, inflation rate(f)=5.4%, market interest rate(i)=10%

%364.404364.0)054.01(

054.01.0)(',''

fi

fiififii

( 단위 : 천원 )

An Economic Evaluation Based on The Constant Cash Flow

n C. C. FCompounding Factor(i=4.364%)

F. V.

0 -300,000 (1+0.4364)5=1.238 -371,428

1 90,000 (1+0.4364)4=1.186 106,769

2 90,000 (1+0.4364)3=1.137 102,304

3 90,000 (1+0.4364)2=1.089 98,027

4 90,000 (1+0.4364)1=1.044 93,928

5 90,000 (1+0.4364)0=1.000 90,000

total 119,600

(unit: 000 won)

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Page 23: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

n C. C. F Inflation Factor A. C. F.

0 -300,000 (1+0.054)0=1.000 -300,000

1 90,000 (1+0.054)1=1.054 94,860

2 90,000 (1+0.054)2=1.111 99,982

3 90,000 (1+0.054)3=1.171 105,381

4 90,000 (1+0.054)4=1.234 111,072

5 90,000 (1+0.054)5=1.301 117,07023

(unit: 000 won)

Inflation Rate=5.4%, Market Interest Rate=10%

An Economic Evaluation Based on A Actual Cash Flow

♦Step 1: convert the constant cash flows to the actual cash flows

Page 24: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

n A. C. F. Compounding Interest Factor

F. V.

0 -300,000 (1+0.1)5=1.611 -483,153

1 94,860 (1+0.1)4=1.464 138,885

2 99,982 (1+0.1)3=1.331 133,076

3 105,381 (1+0.1)2=1.210 127,511

4 111,072 (1+0.1)1=1.100 122,179

5 117,070 (1+0.1)0=1.000 117,070

Total 155,568 24

(unit: ooo won)

An Economic Evaluation Based on A Actual Cash Flow

♦ step 2: calculate the future value of each actual cash flow using the market interest rate

Inflation Rate=5.4%, Market Interest Rate=10%

Page 25: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

The Characteristics of the Fixed Asset: its value decreases as time goes onDefinition: decrease in the value of the fixed assetsA Depreciation Cost: the amount of decrease in the value of the fixed assets over the specific period of time

Ex: The worth of 15million won for an automobile

Dep

recia

tion

n Market value

Depreciation

0

1

2

3

4

5

15,00010,0008,0006,0005,0004,000

5,0002,0002,0001,0001,000

(unit: thousand won

25

Depreciation

Page 26: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

THE FACTORS CONSIDERED WITH DEPRECIATION

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A Depreciable LifeA Salvage ValueA Depreciation MethodCost Basis

Page 27: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

THE CONDITIONS FOR A DEPRECIABLE ASSETS

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The fixed assets must be used longer than one year with the limited time.They must be used for production, not for sale.Their value must decreases over the time due to erosion, worn out, and so on.

Considering the three conditions, land can not be a depreciable asset.

Page 28: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

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Revenue-Expense:(COGS)( Depreciation)(Operating)

Taxable Income

- Taxes

Net Income

Managerial Expense:

A depreciation cost decreases a taxable income and finally results in taxes

The Reason Why Considering A Depreciation Cost

Page 29: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

- Straight-Line Depreciation Method: SL) The value of the fixed assets decreases by the same amount of money per year

Equation

- Declining-Balance Method: DB) : The value of the fixed assets decreases by the same rate of mony per year

where, I: a cost basis N: a depreciable life S: a salvage value at the end of the depreciable life

N

SID

Depreciation Methods

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Page 30: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.8] Given: cost basis(I)=300 million won, depreciable life(N)=5 years,

salvage value(S)=50M won, depreciation method= SL

Sol.:

300,000 50,00050

3

I SD Mwon

N

n 도 depreciation Book value

0 300,000

1 50,000 250,000

2 50,000 200,000

3 50,000 150,000

4 50,000 100,000

5 50,000 50,000

(unit: ooo won)

Calculate the Depreciation Cost Using the SL Method

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Page 31: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

RevenueExpenses: Cost of Goods Sold Depreciation Operating CostTaxable IncomeCorporate Taxes

Net Income

Item

Taxable Income and Taxes

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Page 32: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.9]

Given] Taxable Income=350M won, Corporate Tax Rate

Sol.]Taxable Income

Tax Rate Tax

100,000 13% 13,000

250,000 25% 62,500

Total 75,500

Calculate A Corporate Tax

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Page 33: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

- Those which do not consider interest payback period: PB accounting rate of return: ARR

- Those which consider interest net present value: NPV net future value: NFV annual equivalent worth: AE internal rate of return: IRR

Appraisal Techniques

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Page 34: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Definition Time to completely recover the initial investment cost with the cash flows generated by the underlying project.

Method: Point in time which equates the initial investment cost

with the accumulated cash flow 수 Mathematical Expression:

Payback Period

n

ttCFCF

10

Where t: time at which a cash flow occurs CF0: the initial investment cost incurred at time= “0” CFt: a cash flows occurring at time = “t”

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Page 35: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.10]

n Cash

FlowAcc. C. F.

0 -300,000 -300,000

1 90,000 -210,000

2 90,000 -120,000

3 90,000 -30,000

4 90,000 60,000

5 90,000 150,000

(unit: 000 won)

300,000(initial cost)=90,000(C. F. for the 1st year)+ 90,000(C.F. for the 2nd year)

+ 90,000(C.F. for the 3rd year) +

PB=3~4 years = years

3

13

000,90

)000,270000,300(

Determine The Payback Period of The Project for Purchasing The Tooling

Equipment

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Page 36: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

This technique does not consider interest like the PB This technique depends on a net(accounting) income instead of a cash flow Mathematical Expression

-After Tax Average Net IncomeARR

Initial Investment Cost

An Accounting Rate of Return

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Page 37: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.11](unit: 000 won)

nNet Income for Each

Alternative

A B C D

0 -1,000 -1,000 -1,000 -1,000

1 100 0 100 200

2 900 0 200 300

3 100 300 300 500

4 -100 700 400 500

5 -400 1,300 1,250 600

ARR -8% 26% 25% 22%

The After-Tax Average Accounting Profit for Alt. “A”

The ARR for Alt. “A”

805

400100100900100000,1

- Pr 808%

1,000A

After Tax Average ofitARR

Initial Cost

Calculate The Accounting Rate of Return

37

Page 38: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Definition: It is a technique which allows us to convert the value of all cash flows occurring at any time to the value of the cash flows at time=0 using an interest rate given and subtract the sum of the present value of the cash outflows from that of the cash inflows.

Decision Rule: if NPV(i)> 0, accept, if NPV(i)<0, reject *minimum attractive rate of rate: MARR

2 3 4 5

0 1

Cash inflow

Cash outflow

0

PW(i)

PW(i)

NPV

PW(i) > 0

Net Present Value

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Page 39: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.12] Make an investment decision based on the NPV

Given] cash flows for the tooling equipment, MARR=15%

n C.F. Discounting Factor (MARR=15%)

PV

0 -300,000 (1+0.15)0=1.000 -300,000

1 90,000 (1+0.15)-1=0.870 78,260

2 90,000 (1+0.15)-2=0.756 68,052

3 90,000 (1+0.15)-3=0.658 59,176

4 90,000 (1+0.15)-4=0.572 51,458

5 90,000 (1+0.15)-5=0.497 44,746

Total 1,692

(unit: 000 won)

Since NPV(15%)=1.692M won>0 , it is desirable for a company to undertake the project.

NPV for the Tooling Equipment Project

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Page 40: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Given: Cash Flows MARR (i)

Find: To find the future value at the end of the project life

40

Net Future Value

Page 41: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.13] Make an investment decision based on the NFV

Given] Cash Flow, MARR=15%

n C.F. Compounding Factor (MARR=15%)

Future Value

0 -300,000 (1+0.15)5=2.011 -603,407

1 90,000 (1+0.15)4=1.749 157,410

2 90,000 (1+0.15)3=1.521 136,879

3 90,000 (1+0.15)2=1.323 119,025

4 90,000 (1+0.15)1=1.150 103,500

5 90,000 (1+0.15)0=1.000 90,000

Total 3,407

Since NFV(15%)=3.407M won>0 , it is economically better for a firm to undertake the project. It is noted that NPV(15%)=NFV(15%)(1+MARR)-

N=3,407(1+0.15)-5=1,694

Calculate the NFV of the Equipment Project

41

(unit: 000 won)

Page 42: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

42

Def. 1: The term of ROR is defined as an interest rate earned on unpaid balance of an installment loan.

Def. 2: The term of ROR is the break-even interest rate, i*, which makes the sum of the present value of the cash inflows equal that of the cash outflows.

Mathematical Relationship:

Def. 3: The term of Return on invested capital is defined as an interest rate earned on the unrecovered balance of the initial investment cost by the end of the project life and makes the unrecovered balance zero at the end of the project life. This rate is generally coined as an internal rate of return , IRR.

* * *( ) ( ) ( )

0

PW i PW i PW i

C. I nfl ow C. Outfl ow

Rate of Return

Page 43: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

Ex 2.14] Make an investment decision based on the IRR criteria

Given] Cash Flows, MARR=15%

Sol.]

1) Calculate the IRR using MS-Excel

3*2***

)1(

000,90

)1(

000,90

)1(

000,90000,300)(

iiiiNPV

0)1(

000,90

)1(

000,905*4*

ii

Calculate the IRR

43

Page 44: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

2) Calculate the IRR using Mathematica 7.0

44

Calculate the IRR

Page 45: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

3) Calculate the IRR using the computer program

45

Calculate the IRR

Page 46: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

The Characteristics of the Real Investment Project(a) Uncertainty (b) irreversibility (c) the flexibility of the investment time

The reason for why uncertainty exists in the investment project : all the benefits and costs happen in the future time

What-if framework Wait-see framework

A Project Evaluation and Uncertainty

46

Page 47: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

1) Sensitivity Analysis

2) Scenario analysis

3) Break-even point analysis

4) Decision tree analysis

5) Probabilistic approach

6) ROV etc.

A Variety of the Techniques Dealing with Uncertainty

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Page 48: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

-Def.: It is the techniques which allows us to see what will happen at the final variable of interest(ex. NPV) when one of the variables changes.

Step 2

Indentify changes in the economic value by varying the value of the variable selected by a constant rate

Step 1

Evaluate the economic value using the base scenario

step 3

Take the same procedure for the remaining variable as in step 2

Step 4

Draw a graph for the varied economic values

Sensitivity Analysis

48

Page 49: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

The NPV profile for the project corresponding to the varying interest rate

A COST OF CAPITAL FOR THE PROJECT

n Cash flow

0 -300,000

1 90,000

2 90,000

3 90,000

4 90,000

5 90,000

total

IRR=15.0268%

49

(unit: 000 won)

Page 50: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

WEIGHTED AVERAGE COST OF CAPITAL (WACC)

Capital =equity + debt A cost of capital must reflect the cost of debt and equity WACC: The WACC is calculated as following

)()1)(( emd rT

Etr

T

DWACC

단 , T: total capital, D: total debt E: total equity tm: marginal tax rate, rd: cost of debt re: cost of equity

50

Page 51: CHAPTER 2: THE GENERAL CONCEPTS ON ENGINEERING ECONOMICS 1

CALCULATE THE WACC

Ex 2.14]

Given] Total capital(T)=300M won, total debt(D)=100M won,

Total equity(E)=200M won, Marginal tax rate(tm)=25%, cost of debt(rd)=6%, cost of equity(re)=12%

Sol.]

095.0

)12.0(3

2)25.01)(06.0(

3

1

)()1)((

emd rT

Etr

T

DWACC

51