Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Basics of Engineering Economy Chapter 1 Foundations of Engineering Economy

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Chapter 1 - Foundations PURPOSE Understand the fundamental concepts of engineering economy TOPICS  Definition and study approach  Interest rate, ROR, and MARR  Equivalence  Interest – simple and compound  Cash flow diagrams  Rules of 72 and 100  Spreadsheet introduction

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved 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 factors – safety, environmental, political, public acceptance, etc.  Fundamental terminology:  Alternative -- stand-alone solution  Cash flows -- estimated inflows (revenues) and outflows (costs) 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 Eng. Econ.)

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec Interest Rate, ROR, MARR  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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec Interest Rate and ROR Interpretations Borrower’s perspective 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 Invest (or lend) $5,000 for one year; receive $5,350 Interest earned = $350 Rate of return = 350/5,000 = 7% RATE OF RETURN

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Cost of capital (COC) – interest rate paid for funds to finance projects MARR – Minimum ROR needed for an alternative to be justified and economically acceptable. MARR ≥ COC. If COC = 5% and 6% must be realized, MARR = 11% Always, for acceptable projects ROR ≥ MARR > COC Sec ROR and MARR ROR

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec 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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved 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)(n periods)(interest rate) Example: Invest $250,000 in a 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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved 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 Example: Invest $250,000 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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sct 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 t; $ 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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sct 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 = $5,000Find: A = ? per year i = 5% per year n = 10 years t = year 1, 2, …, 10 (F not used here)

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec 1.7 – Cash Flow Estimates Cash inflow – receipt, revenue, income, saving Cash outflows – cost, expense, disbursement, loss Net cash flow (NCF) = inflow – outflow End-of-period convention: all cash flows and NCF occur at the end of an interest period

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec 1.7 – Cash Flow Diagrams Year 1Year 5 Time, t Cash flow - Cash flow Find P in year 0, given 3 cash flows P = ? Typical time scale or 5 years

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved 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

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec 1.8 – Rule of 72 ( and 100) Approximate n = 72 / i  Estimates # of years (n) for an amount to double (2X) at a stated compound interest rate  e.g., at i = 10%, $1,000 doubles to $2,000 in ~7.2 years  Solution for i estimates compound rate to double in n years Approximate i = 72 / n  For simple interest, doubling time is exact, using rule of 100 n = 100/i or i = 100/n  $1,000 doubles in 10 years at 10% simple interest

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 by McGraw-Hill All Rights Reserved Sec 1.9 – Introduction to Spreadsheet Functions To display Excel Function Present value, P= PV( i %, n,A,F) Future value, F= FV( i %, n,A,P) Annual amount, A = PMT( i %, n,P,F) # of periods, n= NPER( i %,A,P,F) Compound rate, i= RATE( 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