1. Making Economic Decisions 2 Outline Types of Problems Role of - - PDF document

Unit 3 – Systems Engineering Tools

Engineering Decision Making, Interest and Equivalence, Present Worth Analysis

1 Source: Engineering Economic Analysis, 10th edition, Newnan, Lavelle, and Eschenbach, Oxford University Press, 2009.

• 1. Making Economic Decisions

2

• Types of Problems
• Role of Engineering Economic Analysis
• Decision-Making Process
• Ethical Dimensions in Engineering

Decision-Making

Outline

3

4

The problems are often not isolated from each other.

People are Surrounded by Problems

• Which career to pursue?
• What level of preparation is required for

the career chosen?

• Where may the preparation be obtained?
• How to get up and get to class?

5

Decisions concerning these opportunities may be arrived at with the help of economic analysis.

Organizations Have Opportunities

• Do we make part A or

B today?

• Should we use a

drilling or boring machine?

• Should we purchase

a boring machine?

• When must we

replace the drilling machine?

• Would a mechanized
• r computerized

drilling machine be the preferred alternative?

• Where do we locate

machinery in the plant?

• Simple Problems
• Should I buy metro-rail pass or pay every time?
• Should our company pay the vendor cash or credit?
• Intermediate Problems
• Should I buy or lease my next car?
• Which CNC machine should the company purchase?
• Complex Problems
• Feasibility study of a new automobile plant.
• Planning for new highways.

Types of Problems

6

7

Simple Problems

• Can generally be worked in one’s head

without extensive analysis.

8

Intermediate Problems

• Must be organized and analyzed
• Are sufficiently important to justify serious thought

and action

• Have significant economic aspects
• Are primarily economic
• Are the principal subject of this course
• Assume an economic situation in equilibrium
• Are generally adequately solved with single-criteria

decision making

9

Complex Problems

• Such problems represent a mixture of 3 elements:

economic, political and humanistic.

• Complex problems are beyond the scope of this

course from a decision-making criteria point of view, but the economic aspects of complex problems will be discussed.

• A systematic tool for comparing alternatives by

their economic merits. Most suitable for problems

• With importance, that require organized analysis of

the economic aspects

• Could apply to Product/Process Design,

Purchase of Capital Equipment, Selection of Projects, Investment, and many other decision- making processes.

Engineering Economic Analysis

10

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Role of Engineering Economic Analysis

• Assists in making decisions where:

– The decision is sufficiently important that serious thought and effort is required. – Careful analysis requires that the decision variables be carefully organized and the consequences be understood. – ECONOMIC ISSUES are a significant component of the analysis leading to a decision.

• 1. Recognize Problem / Opportunity
• 2. Define Goals/Objectives
• 3. Assemble Relevant Data
• 4. Identify Feasible Alts
• 7. Predict Alts’ Outcomes
• 8. Choose the Best Alt.
• 9. Audit the Results

Overall Mission / Objectives

• 5. Select the Criterion
• 6. Construct a Model

Decision-Making Process

12

Decision-Making Process

• 1. Recognize Problem / Opportunity
• Needs for New Products, Processes, or

Facilities

• Improvement of Products, Processes, or

Facilities (TQM, CI)

• SWOT (Strengths, Weaknesses, Opportunities,

and Threats) Analysis

• Investment/Financing

13

14

• 1. Recognize the Problem
• A problem exists when:

– A standard or expectation is not being met. – A new standard or expectation is established and needs to be achieved. (An opportunity.) – There are needs for new products, processes,

• r facilities

– There are needs for improvement of existing products, processes, or facilities (TQM, CI)

15

• 2. Define the Goal or Objective
• A goal or objective is the standard or

expectation we wish to meet.

– A goal is a general statement about what we expect.

• Pay all our bills on time.

– An objective is narrow and specific.

• Pay the auto loan on Tuesday morning at the

bank.

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• 3. Assemble Relevant Data

– Information may be published or individual knowledge. – Deciding which data is relevant may be a complex process. – In engineering decision making two important sources of data are the

• rganization’s accounting

and purchasing departments.

Economic analysis - Inside vs outside supplier For: Comparison: Inside vs. outside printing o Requestor: Shipping department Unit affected: Printing department Supplier: Outside printer Requirements 30,000 copies Cost per Cost per Copies 1000 30,000 Direct labor 7.60 $ 228.00 $ Materials and supplies 9.80 $ 294.00 $ Overhead costs 9.05 $ 271.50 $ Total 26.45 $ 793.50 $ 111.00 $ Large company Printing department Shipping department savings Apparent cost

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• 4. Identify Feasible

Alternatives

– The best alternative should be implemented. Occasionally this is to maintain the existing situation. – Alternatives considered should include both conventional and innovative approaches. – Include as many as possible alternatives:

• Do-nothing option
• Simple solutions

– Only feasible alternatives should be retained for further analysis.

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• 5. Select the Criterion to

Determine the Best Alternative

– A criterion, or a set of criteria, is used to evaluate the alternatives to determine which is best. – The “best” alternative is relative. – Selecting criteria to use is not easy because different groups often support different criteria. – The criterion most often used in economic decision-making is to “use money in the most efficient manner.”

• 5. Criterion
• Multiple criteria
• Conflicting criteria
• Integrating criteria
• Most common criterion – Maximize profit

Category Economic Criterion Fixed input Maximize the benefits or other outputs Fixed output Minimize the costs or other inputs Neither input nor output fixed Maximize the profits (Value of outputs – cost of inputs)

19

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• 6. Construct the Model
• Requires merging the various elements: objective,

relevant data, feasible alternatives and selection criteria.

• In economic decision making the models are usually

mathematical.

• A model is a representation of reality.

– A model must represent the important parts of the system at hand. – Be adequate to solve the problem. – Describes the interrelationships among the relevant data and predicts the outcomes of various alternatives.

• 7. Predict Alternatives’ Outcomes
• Comparable outcomes

– Single criterion – Single composite criterion

• Alternatives are stated in the form of money, i.e.,

costs/benefits.

• Risk and uncertainty
• Search for more information (loop)
• Modification of alternatives (loop)

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• 8. Choosing the Best Alternative

– When choosing the best alternative both economic and non-economic criteria must be considered. – Only dominant alternatives may be included based on either economic or non-economic criteria.

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• 9. Audit the Results
• Compare the results of changes to the

predictions to assure that the chosen alternative was implement as planned and the results are as expected.

– Fix deviations from planned changes. – Make sure prediction errors are not repeated. – Identify added opportunities.

• Audits promote realistic economic analysis

and implementation.

Ethics

• The concept of distinguishing between right and

wrong in decision-making.

• Ethics includes:

– Establishing systems of beliefs and moral obligations – Defining values and fairness – Determining duty and guidelines for conduct

24

Code of Ethics National Soc. of Professional Eng.

Engineers, in the fulfillment of their professional duties, shall: 1. Hold paramount the safety, health, and welfare of the public. 2. Perform services only in areas of their competence. 3. Issue public statements only in an objective and truthful manner. 4. Act for each employer or client as faithful agents or trustees. 5. Avoid deceptive acts. 6. Conduct themselves honorably, responsibly, ethically, and lawfully so as to enhance the honor, reputation, and usefulness of the profession. (http://www.nspe.org/ethics/eh1-code.asp)

25

Ethical Dimensions in Engineering Decision-Making

Decision Process Step Example Ethical Lapses

• 1. Recognize the problem
• “Looking the other way”, or not to

recognize the problem due to bribes or fear of retribution

• 2. Define goals/objectives
• Favoring one group of stakeholders by

focusing on their objective

• 3. Assemble relevant data
• Using faulty or inaccurate data
• 4. Identify feasible alts.
• Leaving legitimate alts out of

consideration

• 5. Select criterion to

determine best alt

• Considering only monetary consequences

when other significant consequences exist

26

Ethical Dimensions in Engineering Decision-Making

Decision Process Step Example Ethical Lapses

• 6. Construct a model
• Using a short horizon that favors one alt
• ver another
• 7. Predict alts’ outcomes
• Using optimistic estimates for one alt and

pessimistic ones for the other alts

• 8. Choose the best alt
• Choosing an inferior alt, one that is

unsafe, adds unnecessary cost for user, harms the environment

• 9. Audit the result
• Hiding past mistakes

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• 2. Interest and Equivalence

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• Time Value of Money
• Interest Calculations
• Cash Flow Equivalence
• Single Payment Compound Interest

Formulas

Outline

29

• Money has purchasing power
• Money has earning power
• Money is a valuable asset
• People are willing to pay some charges

(interests) to have money available for their use

Time Value of Money

30

time

• Question: Would you rather
• Receive $1000 today or
• Receive $1000 10 years from today?
• Answer: Of course today!
• Why?
• I could invest $1000 today to make more money
• I could buy a lot of stuff today with $1000
• Who knows what will happen in 10 years?

Time Value of Money

31

• Because money is more valuable today than in

the future, we need to describe cash receipts and disbursements at the time they occur.

Time Value of Money Computing Cash Flows

32

Example 2-1 Cash flows of 2 payment options

To purchase a new $30,000 machine,

• Pay the full price now minus a 3% discount or
• Pay $5000 now; $8000 at the end of year 1;

and $6000 at the end of each of the next 4 years

33

Example 2-1 Cash flows of 2 payment options

End of Year Cash Flow 0 (now)

• $29,100

1 2 3 4 5

Pay in full

4 1 2 3 5

Pay in 5 years

End of Year Cash Flow 0 (now)

• $5,000

1

• 8,000

2

• 6,000

3

• 6,000

4

• 6,000

5

• 6,000

4 1 2 3 5 $29,100 $5,000 $8,000

34

Example 2-2 Cash flow for repayment of a loan

To repay a loan of $1,000 at 8% interest in 2 years

• Repay half of $1000 plus interest at the end of each year

Yr Interest Balance Repayment Cash Flow 1000 1000 1 80 500 500

• 580

2 40 500

• 540

1 2 $1000 $580 $540

35

Interest Calculations - Simple Interest

Interest is computed only on the original sum, and not on accrued interest

n i P  

Total interest earned =

where P = Principal i = Simple annual interest rate n = Number of years

n i P P F    

where F = Amount due at the end of n years

(Eq. 2-2) (Eq. 2-1)

36

Example 2-3 Simple Interest Calculation

Loan of $5000 for 5 yrs at simple interest rate of 8% Total interest earned = $5000(8%)(5) = $2000 Amount due at end of loan = $5000 + 2000 = $7000

37

Interest Calculations - Compound Interest

• Interest is computed on the unpaid balance,

which includes the principal and any unpaid interest from the preceding period

• Common practice for interest calculation, unless

specifically stated otherwise

38

Example 2-4 Compound Interest Calculation

• Loan of $5000 for 5 yrs at interest rate of 8%

Year Balance at the Beginning of the year Interest Balance at the end

• f the year

1 $5,000.00 $400.00 $5,400.00 2 $5,400.00 $432.00 $5,832.00 3 $5,832.00 $466.56 $6,298.56 4 $6,298.56 $503.88 $6,802.44 5 $6,802.44 $544.20 $7,346.64

39

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Comparison of Simple and Compound Interest Over Time

• If you loaned a friend money for

short period of time the difference between simple and compound interest is negligible.

• If you loaned a friend money for

a long period of time the difference between simple and compound interest may amount to a considerable difference. Short or long? When is the $ difference significant? You pick the time period.

Simple and compound interest Single payment Principal = 100.00 Interest = 9.00% Period Simple amount factor Compound amount factor n Find Fs Given P Fs/P Find F Given P F/P 100.000 100.000 1 109.000 109.000 2 118.000 118.810 3 127.000 129.503 4 136.000 141.158 5 145.000 153.862 6 154.000 167.710 7 163.000 182.804 8 172.000 199.256 9 181.000 217.189 10 190.000 236.736 11 199.000 258.043 12 208.000 281.266 13 217.000 306.580 14 226.000 334.173 15 235.000 364.248 16 244.000 397.031 17 253.000 432.763 18 262.000 471.712 19 271.000 514.166 20 280.000 560.441

Check the table to see the difference

• ver time.

Repaying a Debt

• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #1: Pay $1000 principal plus interest due

Yr Balance at the Beginning

• f Year

Interest Balance at the end of Year Interest Payment Principal Payment Total Payment 1 $5,000.00 $400.00 $5,400.00 $400.00 $1,000.00 $1,400.00 2 $4,000.00 $320.00 $4,320.00 $320.00 $1,000.00 $1,320.00 3 $3,000.00 $240.00 $3,240.00 $240.00 $1,000.00 $1,240.00 4 $2,000.00 $160.00 $2,160.00 $160.00 $1,000.00 $1,160.00 5 $1,000.00 $80.00 $1,080.00 $80.00 $1,000.00 $1,080.00 Subtotal $1,200.00 $5,000.00 $6,200.00

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Repaying a Debt

• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #2: Pay interest due at end of each year and principal at end of 5

years

Yr Balance at the Beginning

• f year

Interest Balance at the end of year Interest Payment Principal Payment Total Payment 1 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00 2 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00 3 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00 4 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00 5 $5,000.00 $400.00 $5,400.00 $400.00 $5,000.00 $5,400.00 Subtotal $2,000.00 $5,000.00 $7,000.00

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Repaying a Debt

• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #3: Pay in 5 equal end-of-year payments

Yr Balance at the Beginning

• f year

Interest Balance at the end of year Interest Payment Principal Payment Total Payment 1 $5,000.00 $400.00 $5,400.00 $400.00 $852.28 $1,252.28 2 $4,147.72 $331.82 $4,479.54 $331.82 $920.46 $1,252.28 3 $3,227.25 $258.18 $3,485.43 $258.18 $994.10 $1,252.28 4 $2,233.15 $178.65 $2,411.80 $178.65 $1,073.63 $1,252.28 5 $1,159.52 $92.76 $1,252.28 $92.76 $1,159.52 $1,252.28 Subtotal $1,261.41 $5,000.00 $6,261.41

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Repaying a Debt

• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #4: Pay principal and interest in one payment at end of 5 years

Yr Balance at the Beginning

• f year

Interest Balance at the end of year Interest Payment Principal Payment Total Payment 1 $5,000.00 $400.00 $5,400.00 $0.00 $0.00 $0.00 2 $5,400.00 $432.00 $5,832.00 $0.00 $0.00 $0.00 3 $5,832.00 $466.56 $6,298.56 $0.00 $0.00 $0.00 4 $6,298.56 $503.88 $6,802.44 $0.00 $0.00 $0.00 5 $6,802.44 $544.20 $7,346.64 $2,346.64 $5,000.00 $7,346.64 Subtotal $2,346.64 $5,000.00 $7,346.64

44

A Closer Look at the 4 Repayment

• Differences:
• Repayment structure (repayment amounts at various

points in time)

• Total payment amount
• Similarities:
• All interest charges were calculated at 8%
• They all achieved the same purpose of repaying the

loan within 5 years

45

Cash Flow Equivalence

• If a firm believes 8% was reasonable, it would

have no preference about whether it received $5000 now or was paid by any of the 4 repayment plans.

• The 4 repayment plans are equivalent to one

another and to $5000 now at 8% interest

46

Use of Equivalence in Engineering Economic Studies

• Using the concept of equivalence, one can

convert different types of cash flows at different points of time to an equivalent value at a common reference point

• Equivalence is dependent on interest rate

47

Single Payment Compound Interest Formulas

Notation: i = interest rate per compounding period n = number of compounding periods P = a present sum of money F = a future sum of money

(Eq. 2-3) (Eq. 2-4)

n) i, P(F/P, F 

Find F, given P, at i, over n

n

i) P(1 F  

Single Payment Compound Amount Formula

48

Example 2-5 Single Payment Compound Interest Formulas

$500 were deposited in a saving account (pays 6% compounded annually) for 3 years

1 2 3 P=500 F=? i=6%

F = P(1+i)n = 500(1+0.06)3 = $595.50 F = P(F/P, i, n) = 500(F/P, 6%, 3) = 500(1.191*) = $595.50

* See compound interest table

Or

49

Single Payment Compound Interest Formulas

Notation: i = interest rate per compounding period n = number of compounding periods P = a present sum of money F = a future sum of money

(Eq. 2-5) (Eq. 2-6)

n) i, F(P/F, P 

Find P, given F, at i, over n

• n

i) F(1 P  

Single Payment Present Worth Formula

50

Example 2-6 Single Payment Compound Interest Formulas

Wish to have $800 at the end of 4 years, how much should be deposited in an account that pays 5% annually? P = F(P/F, i, n) = 800(P/F, 5%, 4) = 800(0.8227) = $658.16

P=? F=800 i=5% 1 2 3 4

P = F(1+i)-n = 800(1+0.05)-4 = $658.16

Or

51

Example 2-7 Single Payment Compound Interest Formulas

$500 were deposited in a saving account (pays 6%, compounded quarterly) for 3 years

F=?

i = 6%/4 = 1.5% (per quarter) n = 3 x 4 = 12 quarters F = P(1+i)n = P(F/P, i, n) = 500(1+0.015)12 = 500(F/P,1.5%,12) = 500(1.196) = $598.00

P=500 i=1.5% 1 2 12 11

52

Example - Single Payment Compound Interest Formulas

In order to get $1M, 45 years from now, how much money needs to be set aside and deposited in a saving account (pays 10%, compounded anualy)?

F=$1M

i = 10%, n = 45 P = F(1+i)-n = F(P/F, i, n) = $1M(1+0.1)-45 = $13,719.2 If I = 15%, P = $1,856.08

P=? i=10% 1 2 45 44

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• 3. More Interest Formulas

54

• Uniform Series Compound Interest

Formulas

• Nominal and Effective Interest Rates
• Using Spreadsheets for Economic Analysis

Outline

55

Uniform Series Compound Interest Formulas

Notation: A = an end-of-period cash flow in a uniform series, continuing for n periods Examples:

• Automobile loans, mortgage payments, insurance

premium, rents, and other periodic payments

• Estimated future costs and benefits

A A A A n-1 1 2 n

56

Uniform Series Compound Interest Formulas

Uniform Series Compound Amount Factor

(3-5) (3-6)

) n , i , A F ( A i 1 i) (1 A F

n

         

Uniform Series Sinking Fund Factor

) n , i , F A ( F 1 i) (1 i F A

n

         

57

Example 3-1 Uniform Series Compound Interest Formulas

$500 were deposited in a credit union (pays 5% compounded annually) at the end of each year for 5 years, how much do you have after the 5th deposit?

00 . 2763 $ ) 526 . 5 ( 500 ) 5 %, 5 , ( 500 ) , , ( 1 i) (1 A F

n

             A F n i A F A i

4 1 2 3 5 F=? 500 500 500 500 500

58

Example 3-2 Uniform Series Compound Interest Formulas

Jim wants to save money at the end of each month to pay for some equipment of $1000 at the end of the year. If bank pays 6% interest, compounded monthly, how much Jim has to deposit every month?

10 . 81 $ ) 0811 . ( 1000 ) 12 %, 5 . , F / A ( 1000 ) n , i , F A ( F A    

1000 4 1 2 3 5 11 9 10 12 A A A A A A A A A=?

% 5 . 12 % 6 imo  

59

Uniform Series Compound Interest Formulas

Uniform Series Capital Recovery Factor

(3-7) (3-8)

) n , i , A P ( A ) i 1 ( i 1 i) (1 A P

n n

          

Uniform Series Present Worth Factor ) n , i , P A ( P 1 i) (1 ) i 1 ( i P A

n n

          

60

Example 3-3 Uniform Series Compound Interest Formulas

A machine costs $5000 and has a life of 5 years. If interest rate is 8%, how much must be saved every year to recover the investment? 00 . 1252 $ ) 2505 . ( 5000 ) 5 %, 8 , P A ( 5000 ) n , i , P A ( P A    

4 1 2 3 5 5000

A A A A A

61

Example 3-4 Uniform Series Compound Interest Formulas

Should you spend $6800 to buy a contract that pays $140 at the end of each month for 5 years if the desired return is 1% per month?

70 . 6293 $ ) 955 . 44 ( 140 ) 60 %, 1 , A P ( 140 ) n , i , A P ( A P    

P

A=140 n=60 i=1%

The contract only worth $6293.70, so reject the offer.

62

Example 3-5 Uniform Series Compound Interest Formulas

How much should you save per month in order to accumulate $1,000,000 40 years from now? Interest rate is 1% per month? 80 $ ) 00008 . ( 1 $ ) 480 %, 1 , ( 1 $ ) , , ( A     M F M n i F F

A=? n=480 i=1% Set aside $80 a month to get $1,000,000 40 years from now.

63

$1 M

Relationships Between Compound Interest Factors

Single Payment

n) i, (P/F, 1 n) i, (F/P, 

(3-9) (3-10)

Uniform Series n) i, (P/A, 1 n) i, (A/P,  n) i, (A/F, 1 n) i, (F/A, 

(3-11)

64

Relationships Between Compound Interest Factors







n 1 t

t) i, (P/F, n) i, (P/A,

(3-12)

Uniform Series i n) i, (A/F, n) i, (A/P,  

(3-14)



 

 

1 n 1 t

t) i, (F/P, 1 n) i, (F/A,

(3-13)

65

Nominal and Effective Interest Rates

Notation: r = Nominal interest rate per year without considering the effect of any compounding i = Effective interest rate per compounding period ia = Effective annual interest rate taking into account the effect of compounding m = Number of compounding periods per year

1 ) i 1 ( 1 ) m r 1 ( i

m m a

     

m r i 

66

Nominal and Effective Interest

Nominal Effective Annual Rate when compounded Rate Yearly Semiannually Quarterly Monthly Daily Continuously 1% 1% 1.0025% 1.0038% 1.0046% 1.0050% 1.0050% 2% 2% 2.0100% 2.0151% 2.0184% 2.0201% 2.0201% 3% 3% 3.0225% 3.0339% 3.0416% 3.0453% 3.0455% 4% 4% 4.0400% 4.0604% 4.0742% 4.0808% 4.0811% 5% 5% 5.0625% 5.0945% 5.1162% 5.1267% 5.1271% 6% 6% 6.0900% 6.1364% 6.1678% 6.1831% 6.1837% 8% 8% 8.1600% 8.2432% 8.3000% 8.3278% 8.3287% 10% 10% 10.2500% 10.3813% 10.4713% 10.5156% 10.5171% 15% 15% 15.5625% 15.8650% 16.0755% 16.1798% 16.1834% 25% 25% 26.5625% 27.4429% 28.0732% 28.3916% 28.4025%

67

Example 3-15 Application of Nominal and Effective Interest Rates

“If I give you $50 now, you owe me $60 on the following Tuesday.” % 1310400 1 %) 20 1 ( 1 ) i 1 ( i

52 m a

      

a) Weekly interest rate = ($60-50)/50 = 20% Nominal annual rate = 20% * 52 = 1040% c) End-of-the-year balance

200 , 655 $ %) 20 1 ( 50 ) i 1 ( P F

52 n

    

b) Effective annual rate

68

Spreadsheet Financial Functions

Note: 1. Type is the number 0 or 1 and indicates when payments are due (0 for end-of-the-period, and 1 for beginning-of-the-period). If type is omitted, it is assumed to be 0. 2. Microsoft Excel solves for one financial argument in terms of the others. P(F/P, i, n) + A(F/A, i, n) + F = 0 Therefore, negative signs are added to find the equivalent values in P, A, and F. 69

Excel Functions Purpose NPV (i, range) To find net present value of a range of cash flows (from period 1 to n) given i PV (i, n, A, [F], [Type]) To find P given i, n, and A (F is optional) PMT (i, n, P, [F], [Type]) To find A given i, n, and P (F is optional) FV (i, n, A, [P], [Type]) To find F given i, n, and A (P is optional) NPER (i, A, P, [F], [Type]) To find n given i, A, and P (F, optional) RATE (n, A, P, [F], [Type], [guess]) To find i given n, A, and P (F, optional) IRR (range, [guess]) To find internal rate of return from a range

• f cash flows (from period 0 to n)

Spreadsheet Financial Functions

70

Excel Financial Functions Examples

• 4. Present Worth Analysis

71

• Assumptions in Solving Economic Analysis

Problems

• Economic Criteria
• Applying Present Worth Techniques

– Useful Lives are Equal – Useful Lives are Different – Infinite Analysis Period

Outline

72

• Understand common assumptions in

solving economic analysis problems

• Apply present worth techniques in various

situations in selecting the best alternative

Learning Objectives

73

Applying Present Worth Techniques

Situation Criterion

Neither input nor output fixed: typical cases Maximize net present worth (present worth of benefits minus present worth of costs) Fixed input: amount of money or other input resources are fixed Maximize present worth of benefits or other outputs Fixed output: fixed task, benefit, or other outputs Minimize present worth of costs or other inputs

74

Example 4-1 Applying Present Worth when Useful Lives are Equal

Device A

4 1 2 3 5 P=1000 A=300

Device B

4 1 2 3 5 P=1350 400 400 400 400 400

$230 300(4.100) 1000

• ,5)

300(P/A,7% 1000

• PWA

    

75

i=7%

$290 400(4.100) 1000

• ,5)

400(P/A,7% 1000

• PWB

    

Example 4-3 Applying Present Worth when Useful Lives are Equal

Manufacturer Cost Useful Life End-of-Useful-Life Salvage Value Speedy $1500 5 years $200 Allied $1600 5 years $325 1357 143

• 1500

) 200(0.7130

• 1500

7%,5) , F P 200(

• 1500

cost

• f

PW $    

Speedy Allied

$1368 232

• 1600

) 325(0.7130

• 1600

7%,5) , F P 325(

• 1600

cost

• f

PW    

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Present Worth when Useful Lives are Different

• Least Common Multiple (LCM) of Useful

Lives from Various Alternatives

• Assuming service will be needed indefinitely
• Repeating same cash flows in each cycle for

each alternative

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Example 4-4 Present Worth Analysis when Useful Lives are Different (Using LCM)

Alternatives Cost Useful Life Uniform Annual Benefit End-of-Useful-Life Salvage Value Atlas $2000 6 $450 $100 Tom Thumb $4000 12 $600 $700

$233.53 2) 2000(0.630

• (0.3971)

100 2000

• )

100(0.6302 450(7.536) %,6) 2000(P/F,8

• 12)

8%, (P/F, 100 2000

• 8%,6)

, F P 100( 8%,12) , A P 450( NPW       

Atlas Tom Thumb

$799.57 4000

• )

700(0.3971 600(7.536) 4000

• 8%,12)

, F P 700( 8%,12) , A P 600( NPW     

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Least Common Multiple of 6 and 12 is 12

Present Worth with Infinite Analysis Period (Capitalized Cost)

i A P Cost apitalized C 

(4-2)

79

1250 $ 04 . 50 i A P Cost apitalized C   

Example 4-9 Multiple Alternatives

Alternatives Total Investment Uniform Net Annual Benefit Terminal Value A: Do Nothing $0 $0 $0 B: Vegetable market $50,000 $5,100 $30,000 C: Gas Station $95,000 $10,500 $30,000 D: Small motel $350,000 $36,000 $150,000 $21,210 10%,20) , F P 0,000( 5 1 10%,20) , A P 36,000( 350,000

• W

NP $1,140 10%,20) , F P 30,000( 10%,20) , A P 10,500( 95,000

• W

NP $2,120 10%,20) , F P 30,000( 10%,20) , A P 5,100( 50,000

• W

NP $0 W NP

D C B A

               

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An investigator paid $8,000 to a consulting firm to analyze possible uses for a small parcel of land on the edge of town that can be bought for $30,000. Four alternatives are considered. Assuming a 10% interest rate and a 20 years analysis period, what should the investor do?

81