ISyE3025_Cycle1_compositehandout

ISyE3025_Cycle1_compositehandout - ISyE 3025 Engineering...

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Unformatted text preview: ISyE 3025 Engineering Economy Course Organization Overview Course Organization 1 ♦ Remote learning method ♦ How to follow this course ♦ Information sources Welcome ! ♦ Questions and feedback ♦ Text, calculators Copyright 1999. Georgia Tech Research Corporation. All rights reserved. Jack R. Lohmann and Gunter P. Sharp School of Industrial and Systems Engineering Georgia Institute of Technology Course Organization Remote learning Course Organization 2 ♦ Video lectures on demand On campus: T1 line Off campus: 56K modem Use RealPlayer 8.0 or G2 Typical lecture: 15 – 25 min. Temporary downloads, can’t save ♦ Course documents on web site Schedule, Homework Sample quiz questions Grades Web site: www.isye.gatech.edu/~engecon/ Latest announcements Course schedule items Administrative items Text material Bulletin Board Grades ISyE 3025 Fall 2002, Learning Cycle #1 How to follow course Read text 3 Download lecture material View lectures Work homework Take quiz Course Organization Information sources ♦ Quizzes Attend help sessions Course Organization 4 Contacting us 5 ♦ Web site Bulletin Board For questions of a general nature, that may be of interest to others ♦ E-mail engecon@isye.gatech.edu For questions that concern only you 1 Course Organization Text, calculators Course Organization 6 ♦ Text: PDF files on WebCT site that you may download and print for your own use in ISyE 3025 Quizzes 7 ♦ Brush up on spreadsheets ♦ Quizzes: Closed book, no note sheets ! Calculators allowed Every 3 weeks, approximately (see schedule) ♦ Sample quiz questions from previous semester on web site ♦ Attend help sessions to prepare for quizzes Course Organization Course Organization ♦ PowerPoint slide handouts of video streams available at GT Bookstore ♦ Calculator Recommend programmable Financial calculator OK but avoid cheap ones Grading policy 8 ♦ Take quiz, every 3 weeks, (4 total) ♦ General make-up quiz (Quiz 5), during final exam week ♦ Highest 3 grades count 30% each 4th highest grade counts 10% (Student may take 4 or 5 quizzes) ♦ Some opportunity for extra credit HW ♦ Individual make-up quizzes for illness or school-sponsored activity Course Organization Web site, e-mail www.isye.gatech.edu/~engecon/ 10 Course overview 9 ♦ Basic concepts ♦ Time value of money ♦ Economic decision criteria ♦ Multiple alternatives ♦ Income taxes ♦ Inflation ♦ Uncertainty ISyE 3025 Engineering Economy engecon@isye.gatech.edu Basic Concepts Jack R. Lohmann Copyright 1999. Georgia Tech Research Corporation. All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 School of Industrial and Systems Engineering Georgia Institute of Technology 2 Basic Concepts Basic Concepts Overview 1 Course focus Course Focus 2 This course is focused on the principles and procedures for making sound economic decisions 3 illustrative problems Course objectives 4 basic principles Basic Concepts Basic Concepts The Jackpot! 3 “1 million jackpot turns sour: Lotto winner kept from selling part of prize -and sues” (Newspaper headline) 4 An automobile insurance company offers customers the opportunity to either pay for 6 months insurance in full now or pay half now and half in 60 days. The second option requires a $5 service fee due now. If the premium was $150, would you take them up on their offer? Award: $50,000/yr for 20 yrs ($36,000/yr after taxes) Deal: Company pays $241,000 now (after taxes) for next 10 yrs payments ($500,000 before taxes) Basic Concepts Basic Concepts “Have we got a deal for you!” 5 The Cost of Paying Cash Investment $10,000.00 Money Mkt Rate 6.5% Investment Period 48 mo. Interest Earned $2,960.20 “You save $670.76 to finance your car than pay cash!” The Cost of Financing Amt Financed $10,000.00 A.P.R. 10.5% Term of Loan 48 mo. Interest Paid $2,289.44 ??? ISyE 3025 Fall 2002, Learning Cycle #1 Pay Now or Pay Later Course Objectives 6 For you to be able to make wise economic decisions . . . To be Proficient in comparing alternatives Able to see and assess tradeoffs Able to evaluate justifications prepared by others Comfortable using financial language Glad you took this course! 3 Basic Concepts Four basic principles Basic Concepts 7 Alternatives: a choice among two or more things Some aspects certain, others uncertain Some choices obvious, some not Monetary & Non-monetary 1 possible choice: “do nothing” Monetary: amounts & timing Basic Concepts Basic Concepts 9 3 - An economic decision should be based on the differences among alternatives . . . Four basic principles 10 4 - An economic decision should be based on the objective of making the “best” use of limited resources All that is common is irrelevant to the decision Both monetary and nonmonetary (which also means that the past is irrelevant, except as a guide to predict future events, i.e., it is a “sunk cost”) Basic Concepts Summary 8 2 - An economic decision is no better than the forecasts describing each alternative Forecasting . . . not easy! 1 - An economic decision is no better than the alternatives considered . . . Four basic principles . . . Four basic principles 11 ISyE 3025 Engineering Economy Course focus The Concept of Equivalence (“Time Value of Money”) 3 illustrative problems Course objectives 4 basic principles Jack R. Lohmann Copyright 1999. Georgia Tech Research Corporation. All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 Copyright 1999. Georgia Tech Research Corporation. All rights reserved. School of Industrial and Systems Engineering Georgia Institute of Technology 4 Concept of Equivalence Overview Concept of Equivalence 1 Two viewpoints of “interest” 2 1 • Borrower’s viewpoint: Money paid for use of borrowed funds 2 • Investor’s viewpoint: Return, or capital growth, from the productive investment of capital Compound interest & capital growth “Time Value of Money” Interest rate: i = Amount accrued/unit time Concept of Equivalence Compound interest: An example Concept of Equivalence 3 What is the capital growth of $100.00 invested at i = 3% per year for 3 years? t 1 2 3 1 1 2 2 3 3 N N Concept of Equivalence What is the capital growth of $100.00 invested at i = 3% per year for 3 years? $109.27 = 100.00(1.03)3 ISyE 3025 Fall 2002, Learning Cycle #1 Capital at t 0 t Capital Interest from at t t to t+1 0 $100.00 0.03(100.00) = $3.00 $103.00 0.03(103.00) = $3.09 $106.09 0.03(106.09) = $3.18 $109.27 Our previous example . . . Compound interest: A formula 4 Interest from t to t+1 P P iP P+iP = P(1+i) P(1+i)+iP(1+i) = P(1+i)2 P(1+i)2+iP(1+i)2 = P(1+i)3 = P(1+i)N iP(1+i) iP(1+i)2 Concept of Equivalence 5 “Time Value of Money:” Example +$100 0 1 i = 0.05/yr 2 3 4 6 F=? 5 yrs F = P(1+i)N = $100(1.05)5 = $127.63 5 Concept of Equivalence Are these sums equivalent? Concept of Equivalence 7 Terminology: “Compounding” +$120 +$100 Calculating an equivalent amount money in the future amount given some amount of money in the present 0 1 2 3 4 5 yrs Equivalent at i = 0.03/year? F = 100(1.03)5 = $115.93 $100 at t = 0 is not equivalent to $120 at t = 5. Prefer $120 five years from today. Concept of Equivalence Another example . . . P=? Concept of Equivalence 9 Are these sums equivalent? +$500 +$300 i = 0.06/yr i = 0.06/yr 0 1 2 0 . . . 10 yrs F = P(1+i)N P = F(1+i)-N P = 500(1.06)-10 = $279.20 Calculating an equivalent amount money in the present amount given some amount of money in the future ISyE 3025 Fall 2002, Learning Cycle #1 1 2 10 +$500 . . . 11 yrs P = 500(1.06)-11 = $263.39 $300 at t = 0 is not equivalent to $500 at t = 11 when i = 0.06/yr. Prefer $300 today. Concept of Equivalence Terminology: “Discounting” 8 Concept of Equivalence 11 One more example . . . Find F i = 0.1/yr 50 0 50 50 1 2 3 12 F=? 50 4 yrs F = 50(1.1)3 + 50(1.1)2 + 50(1.1)1 + 50 = $232.05 6 Concept of Equivalence Concept of Equivalence One more example . . . Find P 13 P= 1 2 50(1.1)-1 3 14 232.05 158.49 i = 0.1/yr 50 50 50 50 P = ? i = 0.1/yr 50 50 50 50 0 F and P equivalent? 0 4 yrs 1 2 3 4 yrs Is $158.49 at t = 0 equivalent to $232.05 at t = 4? 50(1.1)-2 + + 50(1.1)-3 + 50(1.1)-4 = $158.49 Yes! 158.49(1.1)4 = $232.05 Concept of Equivalence Concept of Equivalence Summary 15 Summary 16 “Time Value of Money” Compound interest & capital growth Two views of “interest” F = P(1+i)N ISyE 3025 Engineering Economy Equivalence depends on: 1 - amount of money; 2 - their timing; and 3 - an interest rate Equivalence Formulas Equivalence Formulas (sometimes called Equivalence Factors) Jack R. Lohmann Copyright 1999. Georgia Tech Research Corporation. All rights reserved. 3 examples: compounding, discounting, both School of Industrial and Systems Engineering Georgia Institute of Technology ISyE 3025 Fall 2002, Learning Cycle #1 Overview 1 Purpose 4 classic equivalence models Single cash flow Uniform cash flow series Arithmetic gradient cash flow series Geometric gradient cash flow series 7 Equivalence Formulas Purpose 2 Equivalence Formulas A cash flow convention 3 Several cash flow conventions exist: this course will use end-of-period models To facilitate performing equivalence calculations $10,000 01 2 N-1 N Yr, Qtr Yr, Mo, Wk, Day, etc “End-of-Year Convention” Equivalence Formulas Equivalence Formulas Four classic equivalence models $ Time 4 Single cash flow $ $ T0 T1 T2 T3 . . . TN-1 TN T0 = past, present, or future TN = N periods after T0 F = P(1+i)TN-T0 = P(1+i)N = P(F/P,i,N) Appendix Equivalence Formulas Equivalence Formulas Single cash flow: An example 100 3 4 F=? 56 5 F=? P Uniform series Arithmetic gradient series Geometric gradient series $ Single cash flow: The model 78 What future amount F at time t = 8 is equivalent to 100 at t = 3 if i = 0.05/year? F = 100(F/P,0.05,5) = 100(1.276) = 127.60 ISyE 3025 Fall 2002, Learning Cycle #1 6 Single cash flow: The model 7 F P=? T0 T1 T2 T3 . . . TN-1 TN P = F(1+i)T0-TN = F(1+i)-N = F(P/F,i,N) Appendix 8 Equivalence Formulas Equivalence Formulas Single cash flow: Another example P=? 8 Uniform cash flow series: The model AAA 500 -3 -2 -1 0 1 . . . 6 7 9 A A F=? T0 T1 T2 T3 . . . TN-1 TN What amount P at t = - 3 is equivalent to 500 at t = 7 if i = 0.06/year? F = A(1+i)N-1+A(1+i)N-2+ . . . +A(1+i)+A Algebra [(1+i)-1] F = A[(1+i)N-1+(1+i)N-2+ . . . +(1+i)+1] [(1+i)-1] F[(1+i)-1]= A[(1+i)N+(1+i)N-1 +(1+i)N-2 + . . . +(1+i) -(1+i)N-1 -(1+i)N-2 - . . . -(1+i)-1] P = 500(P/F,0.06,10) = 500(0.5584) = 279.20 i Equivalence Formulas Uniform cash flow series: The model AAA Equivalence Formulas 10 50 A = F[i/{(1+i)N-1}] = F(A/F,i,N) Equivalence Formulas AAA 50 Arithmetic gradient series: The model A = P[i(1+i)N/{(1+i)N-1}] = P(A/P,i,N) ISyE 3025 Fall 2002, Learning Cycle #1 13 (N-1)G (N-2)G P P = A(1+i)-1+A(1+i)-2+ . . . +A(1+i)-N P = A[{(1+i)N-1}/i(1+i)N] Algebra = A(P/A,i,N) 50 Equivalence Formulas 12 AA T0 T1 T2 T3 . . . TN-1 TN 50 01 2 3 4 yrs What future amount F at t = 4 is equivalent to $50 each year for the next four years if i = 0.1/year? F = 50(F/A,0.1,4) = 50(4.641) = 232.05 F = A[{(1+i)N-1}/i ] = A(F/A,i,N) Uniform cash flow series: The model 11 F=? AAF T0 T1 T2 T3 . . . TN-1 TN P Uniform cash flow series: Example G 2G A T0 T1 T2 T3 . . . TN-1 TN P = G[{(1+i)N-iN-1]/[i2(1+i)N] = G(P/G,i,N) A = G[{(1+i)N-iN-1]/[i(1+i)N-1] = G(A/G,i,N) 9 Equivalence Formulas Equivalence Formulas An example Maintenance costs 01 2 34 14 Service contract 01 2 34 - 3000 A=? - 4000 - 5000 - 6000 P=? An example: Another way Maintenance costs 01 2 34 Service contract 01 2 34 - 3000 A=? - 4000 - 5000 - 6000 i = 0.08/yr P = - 3000(P/A,.08,4) - 1000(P/G,.08,4) = - 14,586 15 i = 0.08/yr A = - 3000 -1000(A/G,.08,4) = - 4404 A = -14,586(A/P,.08,4) = - 4404 Equivalence Formulas Equivalence Formulas Geometric gradient series: The model P A1(1+g)N-1 A1(1+g)2 A1(1+g)1 A1 T0 T1 16 A1 P g>0 (1+g)N-1 A1(1+g)2 A1(1+g)1 A1 g=0 g<0 T0 T1 T2 T3 TN-1 TN P = [A1/(1+i) + A1(1+g)/(1+i) … + A1(1+g)N-1/(1+i)N] 17 g>0 g=0 g<0 T2 T3 TN-1 TN A1[{1-(1+g)N(1+i)-N}/{i-g}] for i ≠ g P= A1N(1+i)-1 for i = g = A1(P/A1,i,g,N) Equivalence Formulas An example Geometric gradient series: The model Equivalence Formulas 18 An example 19 F=? Assume you plan to save 10% of your salary each year and invest it in an account at 8%/year, how much would you accumulate in the account in 10 years? 6000(1.06)9 6000(1.06)2 6000(1.06) 6000 Assume your current salary is $60,000, and raises are expected at the rate of 6%/year. P = 6000(P/A1,.08,.06,10) = 6000(8.5246) = 51,148 ISyE 3025 Fall 2002, Learning Cycle #1 P 0 1 2 3 … 10 F = 51,148(F/P,.08,10) = 132,662 10 Equivalence Formulas Equivalence Formulas Summary 20 Purpose 4 classic equivalence models Single cash flow (F/P,i,n) & (P/F,i,n) Uniform cash flow series (F/A,i,n), (P/A,i,n), (A/F,i,n), (A/P,i,n) Summary 21 4 classic equivalence models Arithmetic gradient cash flow series (P/G,i,n) Geometric gradient cash flow series (P/A1,i,g,n) Interest Rates ISyE 3025 Engineering Economy Overview Interest Rates 1 Time scale conversions “Nominal” versus “Effective” rates Jack R. Lohmann Copyright 1999. Georgia Tech Research Corporation. All rights reserved. School of Industrial and Systems Engineering Georgia Institute of Technology Interest Rates Interest Rates Time scale conversion: Example If you were offered i4 = 0.04 per quarter year, what semiannual rate, i2, would make you in different between i4 = 0.04 and i2? F P 0 0 1 23 1 1 Year 4 qtr yrs 2 half yrs ISyE 3025 Fall 2002, Learning Cycle #1 2 Time scale conversion: Example P 0 0 3 F 1 23 1 1 Year 4 qtr yrs 2 half yrs F = P(1+i4)4 = P(1+i2)2 i2 = (1+i4)4/2-1 = (1.04)2-1 = 0.0816 11 Interest Rates Interest Rates Time scale conversions: Formula F P 0 0 4 1 2 ... 1 ... 1 Year Nominal vs. Effective Rates 5 Nominal Annual Interest Rate, r =MiM Credit cards often charge 1.5%/mo for unpaid balances and report it as “18% compounded monthly,” i.e., 12(0.015) = 0.18. M2 M1 iM1 = (1+iM2)M2/M1-1 Interest Rates Interest Rates Nominal vs. Effective Rates 6 Another view of “effective” interest 7 A bank offers loans of $2500 for 30 months at an advertised interest rate of 2%/mo. The monthly payments are computed as follows: Interest = 0.02(2500)30 = $1500; Processing Fee = $50; Total Owed = 2500+1500+50 = $4050; Monthly Payment = 4050/30 = $135. What is the effective annual interest rate? Effective Annual Interest Rate: i = (1+iM)M-1 = (1.015)12 - 1 = 0.196 Interest Rates Interest Rates Solution +2500 30 mos ... -135 2500 = 135(P/A,i12,30) i12 = 0.036 per month i = (1.036)12 - 1 = 0.529/year 8 Summary 9 Time scale conversions iM1 = (1+iM2)M2/M1-1 Nominal vs effective interest rates r =MiM i = (1+iM)M-1 i should include all costs Copyright 1999. Georgia Tech Research Corporation. All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 12 ...
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This note was uploaded on 09/07/2009 for the course ISYE 3025 taught by Professor Lee during the Spring '09 term at Georgia Institute of Technology.

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