EMT320W1Lecture
- 1. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 1 Basics of Engineering Economy by Leland Blank and Anthony Tarquin Slides to accompany Chapter 1 Foundations of Engineering Economy Slides to accompany Blank and Tarquin Basics of Engineering Economy, 2nd ed.
- 2. Chapter 1 - Foundations PURPOSE Understand and apply the fundamental concepts of engineering economy TOPICS Definition and role study approach and terms Interest rate, ROR, and MARR Equivalence Interest – simple and compound Symbols Cash flows and diagrams
- 3. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 3 Sec 1.1 – Definition of Engineering Economy Sec 1.2 – Elements of a Study DEFINITION: Techniques that simplify comparison of alternatives on an economic basis Most project decisions consider additional (noneconomic; intangible) factors – safety, environmental, political, legal, public acceptance, ethical, etc. Fundamental terminology: Alternative -- stand-alone solution Cash flows -- estimated inflows (revenues, savings) and outflows (costs, expenses) for an alternative Evaluation criteria -- basis used to select ‘best’ alternative; usually money (currency of the country) Time value of money -- change in amount of money over time (Most important concept in Engineering Economy)
- 4. 1 - 4 Sec 1.3 - Interest Rate, ROR, MARR interest accrued per time unit x 100% original amount Interest is a manifestation of time value of money Calculated as difference between an ending amount and a beginning amount of money Interest = end amount – original amount Interest rate is interest over specified time period based on original amount Interest rate (%) = Interest rate and rate of return (ROR) have same numeric value, but different interpretations © 2014 by McGraw-Hill Education All Rights Reserved
- 5. 1 - 5 Sec 1.3 - Interest Rate and ROR Interpretations Borrower’s perspective (money paid to lender) Take loan of $5,000 for one year; repay $5,350 Interest paid = $350 Interest rate = 350/5,000 = 7% INTEREST RATE Investor’s perspective (money earned by lender) Invest (or lend) $5,000 for one year; receive $5,350 Interest earned = $350 Rate of return = 350/5,000 = 7% RATE OF RETURN (ROI has same interpretation) © 2014 by McGraw-Hill Education All Rights Reserved
- 6. 1 - 7 Sec 1.4 - Equivalence Different sums of money at different times may be equal in economic value $100 now $106 one year from now Interest rate = 6% per year Interpretation: $94.34 last year, $100 now, and $106 one year from now are equivalent only at an interest rate of 6% per year $94.34 last year -1 0 1 © 2014 by McGraw-Hill Education All Rights Reserved
- 7. 1 - 8 Sec 1.5 – Simple and Compound Interest Simple interest is always based on the original amount, which is also called the principal Interest per period = (principal)(interest rate) Total interest = (principal)(number of periods)(interest rate) Example: Invest $250,000 in a 3-year bond at 5% per year simple Interest each year = 250,000(0.05) = $12,500 Interest over 3 years = 250,000(3)(0.05) = $37,500 © 2014 by McGraw-Hill Education All Rights Reserved
- 8. 1 - 9 Sec 1.5 – Simple and Compound Interest Compound interest is based on the principal plus all accrued interest Interest per period = (principal + accrued interest)(interest rate) Total interest = (principal)(1+interest rate) n periods – principal Total due = principal(1+interet rate)n periods Example: Invest $250,000 for 3 years at 5% per year compounded Interest, year 1 = 250,000(0.05) = $12,500 Interest, year 2 = 262,500(0.05) = $13,125 Interest, year 3 = 275,625(0.05) = $13,781 Interest over 3 years = 250,000(1.05)3 – 250,000 = $39,406 Total due after 3 years = 250,000(1.05)3 = $289,406 © 2014 by McGraw-Hill Education All Rights Reserved
- 9. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 10 Sec 1.6 - Terminology and Symbols t = time index in periods; years, months, etc. P = present sum of money at time t = 0; $ F = sum of money at a future time; $ A = series of equal, end-of-period cash flows; currency per period, $ per year n = total number of periods; years, months i = compound interest rate or rate of return; % per year
- 10. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 11 Sec 1.6 - Terminology and Symbols Example: Borrow $5,000 today and repay annually for 10 years starting next year at 5% per year compounded. Identify all symbols. Given: P = $5,000 Find: A = ? per year i = 5% per year n = 10 years t = year 1, 2, …, 10 (F not used here)
- 11. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 12 Sec 1.7 – Cash Flow Estimates Cash inflows – receipts, revenue, income, savings Cash outflows – cost, expenses, disbursements, losses Net cash flow (NCF) = inflows – outflows End-of-period convention: all cash flows and NCF occur at the end of an interest period
- 12. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 13 Sec 1.7 – Cash Flow Diagrams 0 1 2 3 4 5 Year 1 Year 5 Time, t 0 1 2 3 4 5 + Cash flow - Cash flow Find P in year 0, given 3 cash flows P = ? Typical time scale for 5 years
- 13. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 14 Sec 1.7 – Cash Flow Diagrams Example: Find an amount to deposit 2 years from now so that $4,000 per year can be available for 5 years starting 3 years from now. Assume i = 15.5% per year
- 14. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 15 Sec 1.8 – Introduction to Spreadsheet and Financial Calculator Functions To display Spreadsheet Function Calculator Function Present value, P = PV(i%,n,A,F) PV(i,n,A,F) Future value, F = FV(i%,n,A,P) FV(i,n,A,P) Annual amount, A = PMT(i%,n,P,F) PMT(i,n,P,F) # of periods, n = NPER(i%,A,P,F) n(i,A,P,F) Compound rate, i = RATE(n,A,P,F) i(n,A,P,F) i for input series = IRR(first_cell:last_cell) P for input series = NPV(i%,second_cell:last_cell) + first_cell Items to remember Spreadsheets: Omitted parameters are assumed to be zero Interior parameters omitted require entry of comma Calculators: Procedures and notation varying between manufacturers
- 15. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 16 Sec 1.9 – Ethics and Economics Decisions Most bad ethical decisions have a fundamentally common base of money Economic decisions made at the start of a project or during its lifetime always involve money Engineering economic analysis and decisions are based primarily on money and often involve ethical challenges Universal (common) morals – Beliefs held by most individuals. Wrong to murder, do physically harm, cheat, lie, etc. Individual (personal) morals – Beliefs of a person. How universal moral is interpreted Usually parallels universal morals; however, can conflict
- 16. © 2014 by McGraw-Hill Education All Rights Reserved 1 - 17 Sec 1.9 – Ethics and Economics Decisions This where economic decisions can conflict with ethical practices due to temptations of money (usually profit or personal financial gain), position, power, favors, etc. Professional ethics – Practices used by an individual in professional activities that reflect their personal morals. Standards and Codes of Ethics guide people in their activities to demonstrate honesty and integrity ? An engineer’s ethical actions and judgments are guided by this national code