MMAN4400 Week 3 Introduction to Engineering Economics (Week 3 Annotation)

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School of Mechanical & Manufacturing Engineering Week 3 Lecture – Introduction to Engineering Economics
Week 3 Outline Engineering Economics 1 • Interest rate • Cost of capital (WACC) • Simple and compound interest • Cash flow diagram • Using standard notation
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High level outline of approaching a question 3
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Elizabeth’s Country Wares Case Study Elizabeth Hines, owner and operator of Elizabeth’s Country Wares (ECW), a decorative country housewares and ceramic pottery business based on Woodstock, Ontario, was trying to find a way to increase capacity. Because ECW was her primary occupation and source of income, Hines was eager to take advantage of the growing interest in one of her product lines from some of her primary customers – member stores of a national pharmacy chain. She was unsure whether she should purchase new equipment or outsource part or all of the production of these pieces. Hines was uncertain whether ECW could meet the forecasted demand of 3,000 units for the coming year, and she did no want to miss out on selling more of the line. However, she also believed there was even more earnings potential if she could focus her attention elsewhere in the business. Adopted from Richard Ivey School of Business – The University of Western Ontario
Opportunities??? Hire additional labour Purchase a new kiln Purchase a new decal equipment Outsource the product to China 7
Rate of Return Strategies Rate of Return Hire Additional Labour 55% Purchase a new kiln 60% Purchase new decal equipment 90% Outsource to China 195% 8 What else will you ask before making the choice for Elizabeth?
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Why Engineering Economy is Important to Engineers? • Engineers design and create. • Designing involves economic decisions. • Engineers must be able to incorporate economic analysis into their creative efforts. • Often engineers must select and implement from multiple alternatives. • Understanding and applying time value of money, economic equivalence, and cost estimation are vital for engineers. • A proper economic analysis for selection and execution is a fundamental task of engineering. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Time Value of Money (TVM) • TVM explains the change in the amount of money over time for funds owed by or owned by a corporation (or individual). • Corporate investments are expected to earn a return. • Investment involves money. • Money has a ‘time value’. The time value of money is the most important concept in engineering economy Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Engineering Economy • Engineering Economy involves: – Formulating, – Estimating, and – Evaluating expected economic outcomes of alternatives designed to accomplish a defined purpose • Easy-to-use math techniques simplify the evaluation. • Estimates of economic outcomes can be deterministic or stochastic in nature. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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General Steps for Decision Making Processes 1. Understand the problem – define objectives. 2. Collect relevant information. 3. Define the set of feasible alternatives. 4. Identify the criteria for decision making. 5. Evaluate the alternatives and apply sensitivity analysis. 6. Select the “best” alternative. 7. Implement the alternative and monitor results. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Steps in an Engineering Economy Study Problem description/ Objective statement Available data – Alternatives for solution Cash flows and other estimates Measure of worth criterion Engineering economic analysis Best alternative selection Implementation and monitoring One or more approaches to meet objective • Expected life • Revenue • Costs • Taxes • Project financing PW, FW, ROR, etc. Consider: • Non-economic factors • Risk analysis Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Interest and Interest Rate Interest – the manifestation of the time value of money. • Fee that one pays to use someone else’s money. • Difference between an ending amount of money and a beginning amount of money. 𝐼??????? = 𝐴????? ?𝑤?? ??𝑤 − ??𝑖??𝑖??? • Interest rate – Interest paid over a time period expressed as a percentage of principal. 𝐼??????? ???? % = 𝐼??????? ??????? ??? ?𝑖?? ??𝑖? ??𝑖??𝑖??? × 100% Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Rate of Return • Interest earned over a period of time is expressed as a percentage of the original amount (principal). 𝐼??????? ???? % = 𝐼??????? ??????? ??? ?𝑖?? ??𝑖? ??𝑖??𝑖??? × 100% • Borrower’s perspective – interest rate paid. • Lender’s or investor’s perspective – rate of return earned. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Commonly used Symbols • ? = time, usually in periods such as years or months. • ? = value or amount of money at a time ? designated as present or time zero. • ? = value or amount of money at some future time, such as at t = n periods in the future. • 𝐴 = series of consecutive, equal, end-of-period amounts of money. • ? = number of interest periods; years, months. • 𝑖 = interest rate or rate of return per time period; percent per year or month. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Cash Flows: Terms • Cash Inflows – Revenues 𝑅 , receipts, incomes, savings generated by projects and activities that flow in. Plus sign used. • Cash Outflows – Disbursements ? , costs, expenses, taxes caused by projects and activities that flow out. Minus sign used. • Net Cash Flow ??? for each time period: ??? = ???ℎ 𝑖????𝑤? − ???ℎ ??????𝑤? = 𝑅 − ? • End-of-period assumption: Funds flow at the end of a given interest period. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Cash Flows: Estimating • Point estimate – A single-value estimate of a cash flow element of an alternative. Cash inflow: Income = $150,000 per month • Range estimate – Min and max values that estimate the cash flow. Cash outflow: Cost is between $2.5 M and $3.2 M Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Cash Flow Diagrams What a typical cash flow diagram might look like: 0 1 2 ? ? − 1 𝑇𝑖?? … … … One time period Always assume end-of-period cash flow 0 1 2 ? ? − 1 𝑇𝑖?? … … … Show the cash flows (to approximate scale) Cash flows are shown as directed arrows: + (up) for inflow, – (down) for outflow Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
Rules in Establishing Cash Flow Diagrams • Define the perspective (lender or borrower) of the cash flow diagram. • Use ? = 0 to represent the present state. • Use ? = ? to represent the end of time period ? . • Use negative ? to represent time period before the present. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Cash Flow Diagram Example Plot observed cash flows over last 8 years and estimated sale next year for $150. Show present worth (P) arrow at present time, ? = 0. End of Year Income $ Cost $ Net Cash Flow – 7 0 2500 – 2500 – 6 750 100 650 – 5 750 125 625 – 4 750 150 600 – 3 750 175 575 – 2 750 200 550 – 1 750 225 525 0 750 250 500 1 900 275 625 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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−7 −6 𝑌??? −5 −4 −3 −2 −1 0 1 $2500 $650 $625 $600 $575 $550 $525 $500 $625 $? =? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Economic Equivalence • Definition: Combination of interest rate (rate of return) and time value of money to determine different amounts of money at different points in time that are economically equivalent. • How it works: Use rate 𝑖 and time ? in upcoming relations to move money (values of ? , ? and 𝐴 ) between time points ? = 0, 1, … , ? to make them equivalent (not equal) at the rate 𝑖 . Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Example of Equivalence Different sums of money at different times may be equal in economic value at a given rate. 0 𝑌??? 1 $ 110 $ 100 𝑅?𝑅 = 10% ?. ?. $100 now is economically equivalent to $110 one year from now, if the $100 is invested at a rate of 10% per year. . Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Simple Interest Simple Interest: Interest is calculated using principal only. 𝐼??????? = ??𝑖??𝑖??? × ?????? ?? ???𝑖??? × 𝐼??????? ???? 𝑖 = ??? Example: $100,000 lent for 3 years at simple 𝑖 = 10% per year. What is repayment after 3 years? 𝐼??????? = 100,000(3)(0.10) = $30,000 𝑇???? ??? = 100,000 + 30,000 = $130,000 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Compound Interest Compound Interest: Interest is based on principal plus all accrued interest. That is, interest compounds over time. 𝐼??????? = (??𝑖??𝑖??? + ??? ??????? 𝑖???????) (𝑖??????? ????) Interest for time period ? is: 𝑖 = ? 1 + ? 𝑛 − ? Where: ? is the principle 𝒓 is the interest rate (in decimal) per compound period 𝒏 is the number of compounding period Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Compound Interest Example Example: $100,000 lent for 3 years at 𝑖 = 10% per year compounded annually. What is repayment after 3 years? 𝑖 = ? 1 + ? 𝑛 − ? 𝑖 + ? = ? 1 + ? 𝑛 𝑖 + ? = 100,000 1 + 0.1 3 𝑅???𝑦???? = $133,100 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Minimum Attractive Rate of Return (MARR) • MARR is a reasonable rate of return (percent) established for evaluating and selecting alternatives. • An investment is justified economically if it is expected to return at least the MARR. • Also termed hurdle rate, benchmark rate and cut-off rate. Rate of return % Expected rate of return on a new proposal Range for the rate of return on accepted proposal, if other proposals were rejected. MARR: All proposals must offer at least MARR to be considered. Rate of return on “safe investment” Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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MARR Characteristics • MARR is established by the financial managers of the firm. • MARR is fundamentally connected to the cost of capital. • Both types of capital financing are used to determine the weighted average cost of capital (WACC) and the MARR. • MARR usually considers the risk inherent to a project. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Discount Rate • Discount rate ≠ Cash interest rate • Discount rate is used to determine the present value of future cash flows • Discount rate accounts for risk and uncertainty of the future • A high discount rate means greater uncertainty of the future cash flows • E.G. If you expect $1,000 in one year, and assuming a discount rate of 10%, then the $1,000 in a year’s time would be equivalent to $909.09 today. 30
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Weighted Average Cost of Capital WACC • WACC is one approach to determine the discount rate for a project or company. • WACC is the average cost of raising money. 𝑊𝐴?? = ? ? + ? 𝑅 ? + ? ? + ? (𝑅 𝐷 )(1 − 𝑇 ? ) Where: ? is the market value of the firm’s equity ? is the market value of the firm’s debt 𝑅 ? is the cost of equity 𝑅 ? is the cost of debt 𝑇 ? is the corporate tax rate 31
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WACC Example A company wants to rise $1,000,000 in capital for plant expansion. The company sells 8,000 shares of stock at $100 each to rise $800,000. The shareholders expect 6% of return on their investment, so the cost of equity is 6%. The remaining $200,000 is acquired by financing, the loan is 5%, so the cost of debt is 5%. Assume a corporate tax rate of 30%. The WACC is: 𝑊𝐴?? = 800,000 1,000,000 0.06 + 200,000 1,000,000 0.05 1 − 0.3 = 0.055 This means for every $1 the company rises, they must pay $0.055 to their investor 32
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Types of Financing • Equity Financing – Funds either from retained earnings, new stock issues, or owner’s infusion of money. • Debt Financing – Borrowed funds from outside sources – loans, bonds, mortgages, venture capital pools, etc. Interest is paid to the lender on these funds. For an economically justified project: 𝑹?𝑹 ≥ 𝑴𝑨𝑹𝑹 > 𝑾𝑨𝑪𝑪 Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Single Payment Factors (F/P and P/F) Single payment factors involve only P and F Cash flow diagrams are as follows ? = ? 1 + 𝑖 𝑛 ? = ? 1 1 + 𝑖 𝑛 0 1 2 ? ? − 1 … 𝑖 = given Terms in parentheses or brackets are called factors. Values are in tables for 𝒊 and 𝒏 values. Factors are represented in standard factor notation such as (?/?, 𝒊, 𝒏) , where letter to left of slash is what is sought; letter to right represents what is given. ? is given ? =? 0 1 2 ? ? − 1 … ? =? ? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example A company is examining its cash flow requirements for the next 7 years. The company is expects to spend $7,000 two years from now, $9,000 three years from now, and $5,000 five years from now. What is the future worth of the expenditures assuming 𝑖 = 10% p.a. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Uniform Series Involving P/A and A/P ? = 𝐴(?/𝐴, 𝑖, ?) 𝐴 = ?(𝐴/?, 𝑖, ?) 0 1 2 ? ? − 1 … ? =? 𝐴 is given The uniform series factors that involve ? and 𝑨 are derived as follows: • Cash flow occurs in consecutive interest periods. • Cash flow amount is same in each interest period. 0 1 2 ? ? − 1 … ? is given 𝐴 =? Note: P is one period Ahead of first A value Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example A company will make the move to expand to a new facility when it has a total value of $1.2mil. If the fund currently has $400,000 and the company adds $50,000 per year, how many years will it take for the account to reach the desire value, if 𝑖 = 10 % p.a. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Arithmetic Gradients Arithmetic gradients change by the same amount each period. The cash flow diagram for the ? ? of an arithmetic gradient is: 0 1 2 3 4 … ? − 1 ? ? 𝐺 =? $? $2? $3? $(? − 2) G $(? − 1) G Standard factor notation is: ? ? = ?(?/?, 𝒊, 𝒏) Note that P G is located Two Periods Ahead of the first change that is equal to G. Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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Typical Arithmetic Gradient Cash Flow 0 1 2 3 4 5 ? 𝑇 =? $400 $450 $500 $550 $600 0 1 2 3 4 5 ? 𝐴 =? $400 0 1 2 3 4 5 ? 𝐺 =? $50 $100 $150 $200 ? 𝑻 = ? 𝑨 + ? ? = ???(?/𝑨, ??%, ?) + ??(?/?, ??%, ? ) Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example The future worth in year 10 of a decreasing geometric gradient series of cash flows was found to be $80,000. If the interest rate was 10% p.a. and the annual rate of decrease was 8% p.a., what was the cash flow amount in year 1? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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In-class Example An engineer invests $10,000 now and then increases his deposit by $1000 each year through year 20, how much will be in the account immediately after the last deposit, if the account grows by 12% per year? Source: Blank, L., Tarquin, A., Engineering Economy 7 th Ed, McGraw-Hill, New York, N.Y
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