Engineering Economic Decisions Lecture 1 Presentation based on the book Chan S. Park, Contemporary Engineering Economics Chapter 1, © Pearson Education International Edition Revised version with additional notes by János D. Pintér

Introduction and Motivation Discussion Topics Rational decision-making process The role of engineers in business What makes engineering economics decisions difficult? Strategic decisions The fundamental principles in engineering economics

Opening Story: Google

A Little Google History 1995 Developed in dorm room by Larry Page and Sergey Brin, graduate students at Stanford University Nicknamed BackRub (reflecting great taste… ) 1998 Raised $25 million to set up Google, Inc. Ran 100,000 queries a day out of a garage in Menlo Park 2005 Over 4,000 employees worldwide Over 8 billion pages indexed Estimated market value over $100 billion As of today, the value of Google is likely to be in the hundreds of billions range

Rational Decision-Making Process Recognize the decision problem Collect all needed (relevant) information Identify the set of feasible decision alternatives Define the key objectives and constraints Select the best possible and implementable decision alternative

A Simple Illustrative Example: Car to Lease – Saturn or Honda? Recognize the decision problem Collect all needed (relevant) information Identify the set of feasible decision alternatives Define the key objectives and constraints Select the best possible and implementable decision alternative Need to lease a car Gather technical and financial data Select cars to consider Wanted: small cash outlay, safety, good performance, aesthetics,… Choice between Saturn and Honda (or others) Select Honda

Engineering Economic Decisions Needed e.g. in the following (connected) areas: Profit! Then continue at the next stage… Manufacturing Design Financial planning Investment and loan Marketing

What Makes Engineering Economic Decisions Difficult What Makes Engineering Economic Decisions Difficult? Predicting the Future Estimating the required investments Estimating product manufacturing costs Forecasting the demand for a brand new product Estimating a “good” selling price Estimating product life and the profitability of continuing production

The Role of Engineers in Business Create & Design Engineering Projects Analyze Production Methods Engineering Safety Environmental Impacts Market Assessment Evaluate Expected Profitability Timing of Cash Flows Degree of Financial Risk Evaluate Impact on Financial Statements Firm’s Market Value Stock Price

Accounting vs. Engineering Economy Evaluating past performance Evaluating and predicting future events Accounting Engineering Economy Past Future Present

Key Factors in Selecting Good Engineering Economic Decisions Objectives, available resources, time and uncertainty are the key defining aspects of all engineering economic decisions

Large-Scale Engineering Projects These typically require a large sum of investment can be very risky take a long time to see the financial outcomes lead to revenue and cost streams that are difficult to predict All the above aspects (and some others not listed here) point towards the importance of EEA

Types of Strategic Engineering Economic Decisions in the Manufacturing Sector Service Improvement Equipment and Process Selection Equipment Replacement New Product and Product Expansion N.B. Cost reduction or profit maximization can be seen as generic (common, eventual) objectives In the most general sense, we have to make decisions under resource constraints, and in presence of uncertainty – not only in the EEA context

Example 1: Healthcare Service Improvement 1 Traditional Plan: Patients visit the service providers 2 New Strategy: Service providers visit the patients Which one of the two plans is more economical? The answer typically depends on the type of patients and the services offered. Examples? 1 2 service providers patients

Example 2: Equipment and Process Selection How do you choose between using alternative materials for an auto body panel? The choice of material will dictate the manufacturing process and the associated manufacturing costs

Example 3: Equipment Replacement Problem Key question: When is the right time to replace an old machine or equipment?

Example 4: New Product and Product Expansion Shall we build or acquire a new facility to meet the increased (increasing forecasted) demand? Is it worth spending money to market a new product?

Example 5: MACH 3 Project R&D investment: $750 million(!) Product promotion through advertising: $300 million(!) Priced to sell at 35% higher than the preceding Sensor Excel model (i.e., about $1.50 extra per razor) Question 1: Would consumers pay $1.50 extra for a shave with greater smoothness and less irritation? Question 2: What happens if the blade consumption drops more than 10% – due to the longer blade life of the new razor?...

Example 6: Cost Reduction Should a company buy new equipment to perform an operation that is now done manually? Should we spend money now, in order to save more money later? The answer obviously depends on a number of factors; can you name some of these?

Further Areas of Strategic Engineering Economic Decisions in the Service Sector Commercial Transportation Logistics and Distribution Healthcare Industry Electronic Markets and Auctions Financial Engineering and Banking Retail Hospitality and Entertainment Customer Service and Maintenance

U.S. Gross Domestic Product (GDP) Distribution by Sector Manufacturing 14% Healthcare 14% Agriculture 2% Total 30% Service sector 70%

The Four Fundamental Principles of Engineering Economics 1: An instant dollar is worth more than a distant dollar… 2: Only the relative (pair-wise) difference among the considered alternatives counts… 3: Marginal revenue must exceed marginal cost, in order to carry out a profitable increase of operations 4: Additional risk is not taken without an expected additional return of suitable magnitude Background and explanatory notes will follow later on

Principle 1 An instant dollar is worth more than a distant dollar… Today 6 months later

Principle 2 Only the cost (resource) difference among alternatives counts Option Monthly Fuel Cost Monthly Maintenance Cash paid at signing (cash outlay ) Monthly payment Salvage Value at end of year 3 Buy $960 $550 $6,500 $350 $9,000 Lease $2,400 The data shown in the green fields are irrelevant items for decision making, since their financial impact is identical in both cases

Principle 3 Marginal (unit) revenue has to exceed marginal cost, in order to increase production Manufacturing cost 1 unit Marginal revenue Sales revenue 1 unit

Principle 4 Additional risk is not taken without a suitable expected additional return Investment Class Potential Risk Expected Return Savings account (cash) Lowest 1.5% Bond (debt) Moderate 4.8% Stock (equity) Highest 11.5% A simple illustrative example. Note that all investments imply some risk: portfolio management is a key issue in finances

Summary The term engineering economic decision refers to any investment or other decision related to an engineering project The five main types of engineering economic decisions are (1) service improvement, (2) equipment and process selection, (3) equipment replacement, (4) new product and product expansion, and (5) cost reduction The factors of time, resource limitations and uncertainty are key defining aspects of any investment project Notice that all listed decision types can be seen and modeled as a constrained decision (optimization) problem Question: are you familiar with the basic optimization concepts? If not, then a brief introduction will be presented (soon)

Engineering Economic Decisions Lecture No.2 Fundamentals of Engineering Economics © 2004 by Chan S. Park

Chapter 1 Engineering Economic Decisions Rational Decision- making Process The Engineer’s Role in Business Types of Strategic Engineering Economic Decisions Fundamental Principles in Engineering Economics Bose Corporation Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Bose Corporation Dr. Amar Bose, a graduate of electrical engineering, an MIT professor, and Chairman of Bose Corporation. He invented a directional home speaker system that reproduces the concert experience. He formed Bose corporation in 1964 and became the world’s No.1 speaker maker. He became the 288th wealthiest American in 2002 by Forbes magazine. Fundamentals of Engineering Economics © 2004 by Chan S. Park

Engineering Economics Overview Rational Decision-Making Process Economic Decisions Predicting Future Role of Engineers in Business Large-scale engineering projects Types of strategic engineering economic decisions Fundamentals of Engineering Economics © 2004 by Chan S. Park

Rational Decision-Making Process Recognize a decision problem Define the goals or objectives Collect all the relevant information Identify a set of feasible decision alternatives Select the decision criterion to use Select the best alternative Fundamentals of Engineering Economics © 2004 by Chan S. Park

Which Car to Lease? Saturn vs. Honda Recognize a decision problem Define the goals or objectives Collect all the relevant information Identify a set of feasible decision alternatives Select the decision criterion to use Select the best alternative Need a car Want mechanical security Gather technical as well as financial data Choose between Saturn and Honda Want minimum total cash outlay Select Honda Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park

Engineering Economic Decisions Manufacturing Profit Planning Investment Marketing Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Predicting the Future Required investment Forecasting product demand Estimating selling price Estimating manufacturing cost Estimating product life Fundamentals of Engineering Economics © 2004 by Chan S. Park

Role of Engineers in Business Create & Design Engineering Projects Evaluate Expected Profitability Timing of Cash Flows Degree of Financial Risk Evaluate Impact on Financial Statements Firm’s Market Value Stock Price Analyze Production Methods Engineering Safety Environmental Impacts Market Assessment Fundamentals of Engineering Economics © 2004 by Chan S. Park

Accounting vs. Engineering Economy Evaluating past performance Evaluating and predicting future events Accounting Engineering Economy Past Future Present Fundamentals of Engineering Economics © 2004 by Chan S. Park

Two Factors in Engineering Economic Decisions The factors of time and uncertainty are the defining aspects of any engineering economic decisions Fundamentals of Engineering Economics © 2004 by Chan S. Park

A Large-Scale Engineering Project Requires a large sum of investment Takes a long time to see the financial outcomes Difficult to predict the revenue and cost streams Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Types of Strategic Engineering Economic Decisions in Manufacturing Sector Service Improvement Equipment and Process Selection Equipment Replacement New Product and Product Expansion Cost Reduction Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Service Improvement How many more jeans would Levi need to sell to justify the cost of additional robotic tailors? Fundamentals of Engineering Economics © 2004 by Chan S. Park

Equipment & Process Selection How do you choose between Plastic SMC and Steel sheet stock for the auto body panel? The choice of material will dictate the manufacturing process for the body panel as well as manufacturing costs. Fundamentals of Engineering Economics © 2004 by Chan S. Park

Which Material to Choose? Fundamentals of Engineering Economics © 2004 by Chan S. Park

Equipment Replacement Problem Now is the time to replace the old machine? If not, when is the right time to replace the old equipment? Fundamentals of Engineering Economics © 2004 by Chan S. Park

New Product and Product Expansion Shall we build or acquire a new facility to meet the increased demand? Is it worth spending money to market a new product? Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Example - MACH 3 Project R&D investment: $750 million Product promotion through advertising: $300 million Priced to sell at 35% higher than Sensor Excel (about $1.50 extra per shave). Question 1: Would consumers pay $1.50 extra for a shave with greater smoothness and less irritation? Question 2: What would happen if blade consumption dropped more than 10% due to the longer blade life of the new razor? Gillette’s MACH3 Project Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Cost Reduction Should a company buy equipment to perform an operation now done manually? Should spend money now in order to save more money later? Fundamentals of Engineering Economics © 2004 by Chan S. Park

Types of Strategic Engineering Economic Decisions in Service Sector Commercial Transportation Logistics and Distribution Healthcare Industry Electronic Markets and Auctions Financial Engineering Retails Hospitality and Entertainment Customer Service and Maintenance Fundamentals of Engineering Economics © 2004 by Chan S. Park

U.S. Gross Domestic Products (GDP) Manufacturing (14%) Service sector (80%) Healthcare (14%) Agriculture (2%) Fundamentals of Engineering Economics © 2004 by Chan S. Park

Industrial Employment Industry 1993 Employment distribution 1983-94 National Average 1994-2005 Projected Change Manufacturing 12.6% -0.70% -7.2% Services 30.5% 60.0% 39.0% Retail trade 16.7% 31.1% 13.0% Financial 8.0% 26.8% 6.3% Source: Bureau of Economic Analysis/Bureau of Labor Statistics Fundamentals of Engineering Economics © 2004 by Chan S. Park

Example - Healthcare Delivery Which plan is more economically viable? Traditional Plan: Patients visit each service provider. New Plan: Each service provider visits patients : patient : service provider Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamental Principles of Engineering Economics Principle 1: A nearby dollar is worth more than a distant dollar Principle 2: All it counts is the differences among alternatives Principle 3: Marginal revenue must exceed marginal cost Principle 4: Additional risk is not taken without the expected additional return Fundamentals of Engineering Economics © 2004 by Chan S. Park

Principle 1: A nearby dollar is worth more than a distant dollar Today 6-month later Fundamentals of Engineering Economics © 2004 by Chan S. Park

Principle 2: All it counts is the differences among alternatives Option Monthly Fuel Cost Monthly Maintenance Cash outlay at signing Monthly payment Salvage Value at end of year 3 Buy $960 $550 $6,500 $350 $9,000 Lease $2,400 Irrelevant items in decision making Fundamentals of Engineering Economics © 2004 by Chan S. Park

Principle 3: Marginal revenue must exceed marginal cost 1 unit Manufacturing cost Marginal revenue Sales revenue 1 unit Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Principle 4: Additional risk is not taken without the expected additional return Investment Class Potential Risk Expected Return Savings account (cash) Low/None 1.5% Bond (debt) Moderate 4.8% Stock (equity) High 11.5% Fundamentals of Engineering Economics © 2004 by Chan S. Park

Fundamentals of Engineering Economics © 2004 by Chan S. Park Summary The term engineering economic decision refers to all investment decisions relating to engineering projects. The five main types of engineering economic decisions are (1) service improvement, (2) equipment and process selection, (3) equipment replacement, (4) new product and product expansion, and (5) cost reduction. The factors of time and uncertainty are the defining aspects of any investment project. Fundamentals of Engineering Economics © 2004 by Chan S. Park

Time Value of Money Lecture No.3

Chapter 2 Time Value of Money Interest: The Cost of Money Economic Equivalence Interest Formulas – Single Cash Flows Equal-Payment Series Dealing with Gradient Series Composite Cash Flows. Power-Ball Lottery

What Do We Need to Know? To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time. To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis. Fundamentals of Engineering Economics, copyright © 2004 by Professor C. S. Park

Time Value of Money Money has a time value because it can earn more money over time (earning power). Money has a time value because its purchasing power changes over time (inflation). Time value of money is measured in terms of interest rate. Interest is the cost of money—a cost to the borrower and an earning to the lender

Delaying Consumption

Which Repayment Plan? End of Year Receipts Payments Plan 1 Plan 2 $20,000.00 $200.00 Year 1 5,141.85 Year 2 Year 3 Year 4 Year 5 30,772.48 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate) Fundamentals of Engineering Economics, copyright © 2004 by Professor C. S. Park

Cash Flow Diagram

End-of-Period Convention Interest Period 1 End of interest period Beginning of Interest period 1

Methods of Calculating Interest Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

Simple Interest P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 8% N = 3 years End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $80 $1,080 2 $1,160 3 $1,240

Simple Interest Formula

Compound Interest Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

Compound Interest P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 8% N = 3 years End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $80 $1,080 2 $86.40 $1,166.40 3 $93.31 $1,259.71

Compounding Process $1,080 $1,166.40 $1,259.71 1 $1,000 2 3 $1,080 $1,259.71 1 $1,000 2 3 $1,080 $1,166.40

$1,259.71 1 2 3 $1,000

Compound Interest Formula

Some Fundamental Laws The Fundamental Law of Engineering Economy

Practice Problem: Warren Buffett’s Berkshire Hathaway Went public in 1965: $18 per share Worth today (August 22, 2003): $76,200 Annual compound growth: 24.58% Current market value: $100.36 Billion If he lives till 100 (current age: 73 years as of 2003), his company’s total market value will be ?

Market Value Assume that the company’s stock will continue to appreciate at an annual rate of 24.58% for the next 27 years.

EXCEL Template In 1626 the Indians sold Manhattan Island to Peter Minuit Of the Dutch West Company for $24. If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now? As of Year 2003, the total US population would be close to 275 millions. If the total sum would be distributed equally among the population, how much would each person receive?

Excel Solution =FV(8%,377,0,1) = $3,988,006,142,690

Practice Problem Problem Statement If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?

Solution F 0 1 2 3 4 5 6 7 8 9 10 $100 $200

? Practice problem Problem Statement Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn 10% interest, what would be the balance at the end of 4 years? ? $1,210 1 4 2 3 $1,500 $1,000 $1,000

? $1,210 1 3 2 4 $1,000 $1,000 $1,500 $1,100 $1,000 $1,210 $2,981 $2,100 $2,310 + $1,500 -$1,210 $1,100 $2,710

Solution n = 0 n = 1 n = 2 n = 3 n = 4 End of Period Beginning balance Deposit made Withdraw Ending n = 0 $1,000 n = 1 $1,000(1 + 0.10) =$1,100 $2,100 n = 2 $2,100(1 + 0.10) =$2,310 $1,210 $1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,500 $2,710 n = 4 $2,710(1 + 0.10) =$2,981 $2,981

Contemporary Engineering Economics, 4th edition ©2007 Time Value of Money Lecture No.4 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition ©2007

Chapter Opening Story —Take a Lump Sum or Annual Installments Mrs. Louise Outing won a lottery worth $5.6 million. Before playing the lottery, she was offered to choose between a single lump sum $2.912 million, or $5.6 million paid out over 20 years (or $280,000 per year). She ended up taking the annual installment option, as she forgot to mark the “Cash Value box”, by default. What basis do we compare these two options? Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Year Option A (Lump Sum) Option B (Installment Plan) 1 2 3 19 $2.912M $280,000 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 What Do We Need to Know? To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time. To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis. Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Time Value of Money Money has a time value because it can earn more money over time (earning power). Money has a time value because its purchasing power changes over time (inflation). Time value of money is measured in terms of interest rate. Interest is the cost of money—a cost to the borrower and an earning to the lender This a two-edged sword whereby earning grows, but purchasing power decreases (due to inflation), as time goes by. Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 The Interest Rate Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Cash Flow Transactions for Two Types of Loan Repayment End of Year Receipts Payments Plan 1 Plan 2 Year 0 $20,000.00 $200.00 Year 1 5,141.85 Year 2 Year 3 Year 4 Year 5 30,772.48 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate) Contemporary Engineering Economics, 4th edition © 2007

Cash Flow Diagram for Plan 2 Contemporary Engineering Economics, 4th edition © 2007

End-of-Period Convention Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Methods of Calculating Interest Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Simple Interest P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 10% N = 3 years End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $100 $1,100 2 $1,200 3 $1,300 Contemporary Engineering Economics, 4th edition © 2007

Simple Interest Formula Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Compound Interest P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 10% N = 3 years End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $100 $1,100 2 $110 $1,210 3 $121 $1,331 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Compounding Process $1,100 $1,210 $1,331 1 $1,000 2 3 $1,100 $1,210 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Cash Flow Diagram $1,331 1 2 3 $1,000 Contemporary Engineering Economics, 4th edition © 2007

Relationship Between Simple Interest and Compound Interest Contemporary Engineering Economics, 4th edition © 2007

Compound Interest Formula Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Some Fundamental Laws The Fundamental Law of Engineering Economy Contemporary Engineering Economics, 4th edition © 2007

“The greatest mathematical discovery of all time,” Albert Einstein Compound Interest “The greatest mathematical discovery of all time,” Albert Einstein Contemporary Engineering Economics, 4th edition ©2007

Practice Problem: Warren Buffett’s Berkshire Hathaway Went public in 1965: $18 per share Worth today (June 22, 2006): $91,980 Annual compound growth: 23.15% Current market value: $115.802 Billion If his company continues to grow at the current pace, what will be his company’s total market value when reaches 100? ( lives till 100 (76 years as of 2006) Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Market Value Assume that the company’s stock will continue to appreciate at an annual rate of 23.15% for the next 24 years. Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 EXCEL Template In 1626 the Indians sold Manhattan Island to Peter Minuit of the Dutch West Company for $24. If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now? As of Year 2006, the total US population would be close to 300 millions. If the total sum would be distributed equally among the population, how much would each person receive? Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Excel Solution =FV(8%,380,0,1) = $5,023,739,194,020 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Practice Problem Problem Statement If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10? Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Solution F 0 1 2 3 4 5 6 7 8 9 10 $100 $200 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Practice problem Problem Statement Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn a 10% interest, what would be the balance at the end of 4 years? ? $1,210 1 4 2 3 $1,500 $1,000 $1,000 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 ? $1,210 1 3 2 4 $1,000 $1,000 $1,500 $1,100 $1,000 $1,210 $2,981 $2,100 $2,310 + $1,500 -$1,210 $1,100 $2,710 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Solution End of Period Beginning balance Deposit made Withdraw Ending n = 0 $1,000 n = 1 $1,000(1 + 0.10) =$1,100 $2,100 n = 2 $2,100(1 + 0.10) =$2,310 $1,210 $1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,500 $2,710 n = 4 $2,710(1 + 0.10) =$2,981 $2,981 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Economic Equivalence Lecture No.5 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Economic Equivalence What do we mean by “economic equivalence?” Why do we need to establish an economic equivalence? How do we establish an economic equivalence? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Economic Equivalence Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal. Contemporary Engineering Economics, 4th edition, © 2007

Equivalence from Personal Financing Point of View If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N. Contemporary Engineering Economics, 4th edition, © 2007

Alternate Way of Defining Equivalence P F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i. N = F N Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem At an 8% interest, what is the equivalent worth of $2,042 now in 5 years? $2,042 If you deposit $2,042 today in a savings account that pays an 8% interest annually. how much would you have at the end of 5 years? 1 2 3 4 5 3 F = 5 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution Contemporary Engineering Economics, 4th edition, © 2007

i = ? Example 3.3 At what interest rate would these two amounts be equivalent? i = ? $3,000 $2,042 5 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equivalence Between Two Cash Flows Step 1: Determine the base period, say, year 5. Step 2: Identify the interest rate to use. Step 3: Calculate equivalence value. $3,000 $2,042 5 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Example - Equivalence Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%. 0 1 2 3 4 5 P F $2,042 $3,000 $2,205 $2,382 $2,778 $2,572 Contemporary Engineering Economics, 4th edition, © 2007

Equivalent Cash Flows Are Equivalent at Any Common Point in Time Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem V 1 2 3 4 5 $100 $80 $120 $150 $200 = 1 2 3 4 5 Compute the equivalent lump-sum amount at n = 3 at 10% annual interest. Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Approach 1 2 3 4 5 $100 $80 $120 $150 $200 V Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 1 2 3 4 5 $100 $80 $120 $150 $200 V Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem $500 $1,000 0 1 2 3 0 1 2 3 A B C Given: i = 10%, Find: C that makes the two cash flow streams to be indifferent Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Approach $500 $1,000 0 1 2 3 0 1 2 3 A B C Step 1: Select a base period to use, say n = 2. Step 2: Find the equivalent lump sum value at n = 2 for both A and B. Step 3: Equate both equivalent values and solve for unknown C. Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution $500 $1,000 0 1 2 3 0 1 2 3 A B C For A: For B: To Find C: Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem $500 $1,000 0 1 2 3 0 1 2 3 $502 $502 $502 A B At what interest rate would you be indifferent between the two cash flows? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Approach Step 1: Select a base period to compute the equivalent value (say, n = 3) Step 2: Find the equivalent worth of each at n = 3. $1,000 $500 A 0 1 2 3 $502 $502 $502 B 0 1 2 3 Contemporary Engineering Economics, 4th edition, © 2007

Establish Equivalence at n = 3 Find the solution by trial and error, say i = 8% Contemporary Engineering Economics, 4th edition, © 2007

Interest Formulas for Single Cash Flows Lecture No.6 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, ©2007

Types of Common Cash Flows in Engineering Economics Contemporary Engineering Economics, 4th edition, © 2007

Equivalence Relationship Between P and F Contemporary Engineering Economics, 4th edition, © 2007

Single Cash Flow Formula- Compound-Amount Factor Single payment compound amount factor (growth factor) Given: Find: F F N P Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years? F = ? i = 10% 8 $2,000 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Single Cash Flow Formula – Present-Worth Factor Single payment present worth factor (discount factor) Given: Find: F N P Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Practice Problem You want to set aside a lump sum amount today in a savings account that earns 7% annual interest to meet a future expense in the amount of $10,000 to be incurred in 6 years. How much do you need to deposit today? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution $10,000 6 P Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solving for i Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution EXCEL Solution using RATE Function =RATE(5,0,-10,20)=14.87% Contemporary Engineering Economics, 4th edition, © 2007

Rule of 72 – Number of Years Required to Double Your Investment Contemporary Engineering Economics, 4th edition, © 2007

Solution – Analytical Approach N = ? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution - Rule of 72 Approximating how long it will take for a sum of money to double Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Number of Years Required to Double an Initial Investment at Various Interest Rates Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Uneven Payment Series How much do you need to deposit today (P) to withdraw $25,000 at n =1, $3,000 at n = 2, and $5,000 at n =4, if your account earns 10% annual interest? 1 2 3 4 $25,000 $3,000 $5,000 P Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 $25,000 Uneven Payment Series $3,000 $5,000 1 2 3 4 P $25,000 $3,000 $5,000 + + 1 2 3 4 1 2 3 4 1 2 3 4 P2 P4 P1 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Check 1 2 3 4 Beginning Balance 28,622 6,484.20 4,132.62 4,545.88 Interest Earned (10%) 2,862 648.42 413.26 454.59 Payment +28,622 -25,000 -3,000 -5,000 Ending Balance $28,622 0.47 Rounding error It should be “0.” Contemporary Engineering Economics, 4th edition, © 2007

Interest Formulas – Equal Payment Series Lecture No.7 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007

Equal Payment Series Equivalent Future Worth F P 1 2 N A A A N 1 2 N A A A P N 1 2 N Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equal Payment Series – Compound Amount Factor F A A A 1 2 N N 1 2 F = 1 2 N A A A Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Process of Finding the Equivalent Future Worth, F F A A(1+i)N-2 A A A A(1+i)N-1 N 1 2 1 2 N Contemporary Engineering Economics, 4th edition, © 2007

Another Way to Look at the Compound Amount Factor Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equal Payment Series Compound Amount Factor (Future Value of an Annuity) F 0 1 2 3 N A Example: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A,6%,5) = $28,185.46 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Validation Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Finding an Annuity Value F 0 1 2 3 N A = ? Example: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F,7%,5) = $869.50 Contemporary Engineering Economics, 4th edition, © 2007

Handling Time Shifts in a Uniform Series First deposit occurs at n = 0 i = 6% 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Annuity Due Excel Solution Beginning period =FV(6%,5,5000,0,1) Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Sinking Fund Factor F 0 1 2 3 N A Example: College Savings Plan Given: F = $100,000, N = 8 years, and i = 7% Find: A Solution: A = $100,000(A/F,7%,8) = $9,746.78 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Excel Solution Given: F = $100,000 i = 7% N = 8 years Find: =PMT(i,N,pv,fv,type) =PMT(7%,8,0,100000,0) =$9,746.78 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Example 3.15 Combination of a Uniform Series and a Single Present and Future Amount Contemporary Engineering Economics, 4th edition, © 2007

Solution: A Two-Step Approach Step 1: Find the required savings at n = 5. Step 2: Find the required annual contribution (A) over 5 years. Contemporary Engineering Economics, 4th edition, © 2007

Comparison of Three Different Investment Plans – Example 3.16 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Investor A: Investor B: Investor C: Contemporary Engineering Economics, 4th edition, © 2007

How Long Would It Take to Save $1 Million? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Loan Cash Flows Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Capital Recovery Factor P 1 2 3 N A = ? Example: Paying Off an Educational Loan Given: P = $21,061.82, N = 5 years, and i = 6% Find: A Solution: A = $21,061.82(A/P,6%,5) = $5,000 Contemporary Engineering Economics, 4th edition, © 2007

Example 3.17 Loan Repayment Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Using Interest Factor: Using Excel: Contemporary Engineering Economics, 4th edition, © 2007

Example 3.18 – Deferred Loan Repayment Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 A Two-Step Procedure Contemporary Engineering Economics, 4th edition, © 2007

Present Worth factor – Find P, Given A, i, and N 1 2 3 N A Contemporary Engineering Economics, 4th edition, © 2007

Example 3.19 Louise Outing’s Lottery Problem Given: A = $280,000 i = 8% N = 19 Find: P Using interest factor: P =$280,000(P/A,8%,19) = $2,689,008 Using Excel: =PV(8%,19,-280000) Contemporary Engineering Economics, 4th edition, © 2007

Interest Formulas (Gradient Series) Lecture No.8 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007

Linear Gradient Series P Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Gradient Series as a Composite Series of a Uniform Series of N Payments of A1 and the Gradient Series of Increments of Constant Amount G. Contemporary Engineering Economics, 4th edition, © 2007

Example – Present value calculation for a gradient series $2,000 $1,750 $1,500 $1,250 $1,000 1 2 3 4 5 How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? P =? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Method 1: Using the (P/F, i, N) Factor $2,000 $1,750 $1,500 $1,250 $1,000 1 2 3 4 5 $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03 P =? Contemporary Engineering Economics, 4th edition, © 2007

Method 2: Using the Gradient Factor Contemporary Engineering Economics, 4th edition, © 2007

Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N) Contemporary Engineering Economics, 4th edition, © 2007

Example 3.21 – Find the Equivalent Uniform Deposit Plan Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Contemporary Engineering Economics, 4th edition, © 2007

Example 3.22 Declining Linear Gradient Series Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Contemporary Engineering Economics, 4th edition, © 2007

Types of Geometric Gradient Series Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Present Worth Factor Contemporary Engineering Economics, 4th edition, © 2007

Example 3.23 Annual Power Cost if Repair is Not Performed Contemporary Engineering Economics, 4th edition, © 2007

Solution – Adopt the new compressed-air system Contemporary Engineering Economics, 4th edition, © 2007

Example 3.24 Jimmy Carpenter’s Retirement Plan – Save $1 Million Contemporary Engineering Economics, 4th edition, © 2007

What Should be the Size of his first Deposit (A1)? Contemporary Engineering Economics, 4th edition, © 2007

Unconventional Equivalence Calculations Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equivalent Present Worth Calculation – Brute Force Approach using Only P/F Factors Contemporary Engineering Economics, 4th edition, © 2007

Equivalent Present Worth Calculation – Grouping Approach $50 $100 $150 $150 $150 $150 $200 1 2 3 4 5 6 7 8 9 Contemporary Engineering Economics, 4th edition, © 2007

Unconventional Equivalence Calculations – A Personal Savings Problem Situation 1: If you make 4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount (A) can be withdrawn over 4 subsequent years? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Unconventional Equivalence Calculations – An Economic Equivalence Problem Situation 2: What value of A would make the two cash flow transactions equivalent if i = 10%? Contemporary Engineering Economics, 4th edition, © 2007

Method 1: Establish the Economic Equivalence at n = 0 Contemporary Engineering Economics, 4th edition, © 2007

Method 2: Establish the Economic Equivalence at n = 4 Contemporary Engineering Economics, 4th edition, © 2007

Multiple Interest Rates F = ? Find the balance at the end of year 5. 6% 4% 4% 6% 5% 2 4 5 1 3 $400 $300 $500 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution Contemporary Engineering Economics, 4th edition, © 2007

Cash Flows with Missing Payments 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Missing payment Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution P = ? Add $100 to offset the change $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 Pretend that we have the 10th Payment in the amount of $100 i = 10% Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Approach P = ? $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Equivalent Cash Inflow = Equivalent Cash Outflow Contemporary Engineering Economics, 4th edition, © 2007

Equivalence Relationship Contemporary Engineering Economics, 4th edition, © 2007

Unconventional Regularity in Cash Flow Pattern $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C Payments are made every other year Contemporary Engineering Economics, 4th edition, © 2007

Approach 1: Modify the Original Cash Flows $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A A A A A A A A A A A A A A Contemporary Engineering Economics, 4th edition, © 2007

Relationship Between A and C $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A A A A A A A A A A A A A A Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution i = 10% C A A A =$1,357.46 Contemporary Engineering Economics, 4th edition, © 2007

Approach 2: Modify the Interest Rate Idea: Since cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period. How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%. (1+0.10)(1+0.10) = 1.21 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution $10,000 i = 21% 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Example 3.25 – At What Value of C would Make the Two Cash Flows Equivalent? Figure: 03-38 Contemporary Engineering Economics, 4th edition, © 2007

Example 3.26 Establishing a College Fund Figure: 03-39 Contemporary Engineering Economics, 4th edition, © 2007

Solution: Establish the Economic Equivalence at n = 18 Figure: 03-40 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Example 3.27 Calculating an Unknown Interest Rate with Multiple Factors Figure: 03-41 Contemporary Engineering Economics, 4th edition, © 2007

Establish an economic Equivalence at n =7 Figure: 03-42 Contemporary Engineering Economics, 4th edition, © 2007

Linear Interpolation to Find an Unknown Interest Rate Figure: 03-43 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Linear Interpolation Contemporary Engineering Economics, 4th edition, © 2007

Using the Goal Seek Function to Find the Unknown Interest rate Figure: 03-44EXM Contemporary Engineering Economics, 4th edition, © 2007