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Engineering Economic Analysis 9th Edition Chapter 3 INTEREST AND EQUIVALENCE Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 Economic Decision Components • Where economic decisions are immediate we need to consider: • amount of expenditure • taxes • Where economic decisions occur over a considerable period of time we also need to consider: • interest • inflation Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 2 Computing Cash Flows • Cash flows have: • Costs (disbursements) > a negative number • Benefits (receipts) > a positive number Example 3-1 End of Year 0 1 2 Cash flow $ (1,000.00) $ 580.00 $ 580.00 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 3 Time Value of Money • Money has value • Money can be leased or rented • The payment is called interest • If you put $100 in a bank at 9% interest for one time period you will receive back your original $100 plus $9 Original amount to be returned = $100 Interest to be returned = $100 x .09 = $9 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 4 Simple Interest • Interest that is computed only on the original sum or principal • Total interest earned = I = P x i x n • Where • P – present sum of money • i – interest rate • n – number of periods (years) I = $100 x .09/period x 2 periods = $18 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 5 Future Value of a Loan with Simple Interest • Amount of money due at the end of a loan • F = P + P i n or F = P (1 + i n ) • Where • F = future value F = $100 (1 + .09 x 2) = $118 • Would you accept payment with simple interest terms? • Would a bank? Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 6 Compound Interest • Interest that is computed on the original unpaid debt and the unpaid interest • Total interest earned = In = P (1+i)n - P • Where • P – present sum of money • i – interest rate • n – number of periods (years) I2 = $100 x (1+.09)2 - $100 = $18.81 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 7 Future Value of a Loan with Compound Interest • Amount of money due at the end of a loan • F = P(1+i)1(1+i)2…..(1+i)n or F = P (1 + i)n • Where • F = future value F = $100 (1 + .09)2 = $118.81 • Would you be more likely to accept payment with compound interest terms? • Would a bank? Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 8 Comparison of Simple and Compound Interest Over Time Simple and compound interest Single payment •If you loaned a friend money for short period of time the difference between simple and compound interest is negligible. •If you loaned a friend money for a long period of time the difference between simple and compound interest may amount to a considerable difference. Principal = Interest = Check the table to see the difference over time. Short or long? When is the $ difference significant? You pick the time period. 100.00 9.00% Simple Compound Period amount factor amount factor Find Fs Given P Find F Given P n Fs/P F/P 0 100.000 100.000 1 109.000 109.000 2 118.000 118.810 3 127.000 129.503 4 136.000 141.158 5 145.000 153.862 6 154.000 167.710 7 163.000 182.804 8 172.000 199.256 9 181.000 217.189 10 190.000 236.736 11 199.000 258.043 12 208.000 281.266 13 217.000 306.580 14 226.000 334.173 15 235.000 364.248 16 244.000 397.031 17 253.000 432.763 18 262.000 471.712 19 271.000 514.166 20 280.000 560.441 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 9 Four Ways to Repay a Debt Plan Repay Principal 1 3 Equal annual Interest on installments unpaid balance End of loan Interest on unpaid balance Equal annual installments 4 End of loan 2 Repay Interest Compound and pay at end of loan Interest Earned Declines Constant Declines at increasing rate Compounds at increasing rate until end of loan Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 0 Loan Repayment – Four Options Loan Repayment Option Calculator This calculator is partially complete. If you complete the calculator you can earn 10 bonus points for your team. $5,000 Principal 10.00% Interest rate (enter as .1 for 10%) 10 Years Plan 1 Enter 1 through 4 Principal payment Equal annual installments Interest payment EOY on unpaid principal Amount owed at the beginning of the year Years Interest owed for that year Total owed at the end of year Principal payment Total end of year payment 1 2 3 4 5 6 7 8 9 10 5,000 4,500 4,000 3,500 3,000 2,500 2,000 1,500 1,000 500 500 450 400 350 300 250 200 150 100 50 2,750 5,500 4,950 4,400 3,850 3,300 2,750 2,200 1,650 1,100 550 500 500 500 500 500 500 500 500 500 500 5,000 1,000 950 900 850 800 750 700 650 600 550 7,750 1 2 3 4 5 5,000 5,000 5,000 5,000 5,000 500 500 500 500 500 2,500 5,500 5,500 5,500 5,500 5,500 0 0 0 0 5,000 5,000 500 500 500 500 5,500 7,500 1 2 3 4 5 5,000 4,500 4,000 3,500 3,000 500 450 400 350 300 2,000 5,500 4,950 4,400 3,850 3,300 314 345 380 418 459 1,915 814 814 814 814 814 4,069 1 2 3 4 5 5,000 5,500 6,050 6,655 7,321 500 550 605 666 732 3,053 5,500 6,050 6,655 7,321 8,053 -500 -550 -605 -666 7,321 5,000 0 0 0 0 8,053 8,053 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 1 Equivalence • When an organization is indifferent as to whether it has a present sum of money now or the assurance of some other sum of money (or series of sums of money) in the future, we say that the present sum of money is equivalent to the future sum or series of sums. Each of the plans on the previous slide is equivalent because each repays $5000 at the same 10% interest rate. Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 2 Given the choice of these two plans which would you choose? Year Plan 1 Plan 2 1 2 3 $1400 1320 1240 $400 400 400 4 5 Total 1160 1080 $6200 400 5400 $7000 To make a choice the cash flows must be altered so a comparison may be made. Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 3 Technique of Equivalence • Determine a single equivalent value at a point in time for plan 1. • Determine a single equivalent value at a point in time for plan 2. Both at the same interest rate. •Judge the relative attractiveness of the two alternatives from the comparable equivalent values. Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 4 Repayment Plans Establish the Interest Rate Equivalence Calculator 1. Principal outstanding over time 2. Amount repaid over time $5,000 8.00% 5 Plan 1 Principal payment Equal annual installments Interest payment EOY on unpaid principal Year s Interest owed for that year Amount owed at the beginning of the year 1 2 3 4 5 Totals 5,000 4,000 3,000 2,000 1,000 Interest paid over time Total owed over time As an example: 400 320 240 160 80 1,200 1,200 15,000 5,400 4,320 3,240 2,160 1,080 = 8.00% $4,876.63 9.00% 5 Plan 1 Principal payment Equal annual installments Interest payment EOY on unpaid principal i)n If F = P (1 + Then i=(F/P)1/n-1 Total owed at the end of year Year s 1 2 3 4 5 Totals Interest owed for that year Amount owed at the beginning of the year 4,877 3,901 2,926 1,951 975 Total owed at the end of year 439 351 263 176 88 1,317 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 5,316 4,252 3,189 2,126 1,063 1 5 Application of Equivalence Calculations Pick an alternative. Which would you choose? Change the interest rate. What happens at 8%,15%,3%? Comparing alternatives Interest rate Year 0 1 2 3 4 5 6 7 8 9 10 A $600 $115 $115 $115 $115 $115 $115 $115 $115 $115 $115 P $1,306.63 A $212.65 F $3,389.05 10.00% Alternative B C -$600 -$850 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 -$115 -$80 ($1,306.63) ($1,341.57) ($212.65) D $850 $80 $80 $80 $80 $80 $80 $80 $80 $80 $80 $1,341.57 ($218.33) $218.33 ($3,389.05) ($3,479.68) $3,479.68 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. Present worth Annual worth Future worth 1 6 Interest Formulas • To understand equivalence, the underlying interest formulas must be analyzed. • Notation: I = Interest rate per interest period n = Number of interest periods P = Present sum of money (Present worth) F = Future sum of money (Future worth) Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 7 Single Payment Compound Interest Year Beginning balance Interest for period Ending balance 1 P iP P(1+i) 2 P(1+i) iP(1+i) P(1+i)2 3 P(1+i)2 iP(1+i)2 P(1+i)3 n P(1+i)n-1 iP(1+i)n-1 P(1+i)n P at time 0 increases to P(1+i)n at the end of time n. Or a Future sum = present sum (1+i)n Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 8 Notation for Calculating a Future Value • Formula: F=P(1+i)n is the single payment compound amount factor. • Functional notation: F=P(F/P,i,n) F=5000(F/P,6%,10) • F =P(F/P) which is dimensionally correct. Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1 9 Notation for Calculating a Present Value • P=F(1/1+i)n=F(1+i)-n is the single payment present worth factor. • Functional notation: P=F(P/F,i,n) P=5000(P/F,6%,10) Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 2 0 Compound Interest Factors Examples F=P(F,i,n) P=F(F,i,n) F=$5000 i=0.10 n=5 P=? F=P(1+i)–n=$5000(1+0.10)–5 =$5000(1.611)=$8055 F=P(F/P,10,5)=$5000(1.611) =$8055 P=F(P/F,10,5)=$8055(.62092) =$5000 10.00% n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 40 50 60 72 100 Single Amount Factor Compound Amount Factor Present Worth Factor F/P $1.00 $5,000.00 1.100 $5,500.00 1.210 $6,050.00 1.331 $6,655.00 1.464 $7,320.50 1.611 $8,052.55 1.772 $8,857.81 1.949 $9,743.59 2.144 $10,717.94 2.358 $11,789.74 2.594 $12,968.71 2.853 $14,265.58 3.138 $15,692.14 3.452 $17,261.36 3.797 $18,987.49 4.177 $20,886.24 4.595 $22,974.86 5.054 $25,272.35 5.560 $27,799.59 6.116 $30,579.55 6.727 $33,637.50 10.835 $54,173.53 17.449 $87,247.01 45.259 $226,296.28 117.391 $586,954.26 304.482 $1,522,408.20 955.594 $4,777,969.09 13,780.612 $68,903,061.70 P/F $1.00 $8,052.55 0.90909 $7,320.50 0.82645 $6,655.00 0.75131 $6,050.00 0.68301 $5,500.00 0.62092 $5,000.00 0.56447 $4,545.45 0.51316 $4,132.23 0.46651 $3,756.57 0.42410 $3,415.07 0.38554 $3,104.61 0.35049 $2,822.37 0.31863 $2,565.79 0.28966 $2,332.54 0.26333 $2,120.49 0.23939 $1,927.72 0.21763 $1,752.47 0.19784 $1,593.15 0.17986 $1,448.32 0.16351 $1,316.66 0.14864 $1,196.96 0.09230 $743.22 0.05731 $461.48 0.02209 $177.92 0.00852 $68.60 0.00328 $26.45 0.00105 $8.43 0.00007 $0.58 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 2 1 18% Compounded Monthly • 18% interest: Assume a yearly rate if not stated • Compounded monthly: Indicates 12 periods/year • [18%/year] / [12months/year] = 1.5% / month Effective vs Nominal Interest Comparator Nominal Interest rate Effective Interest rate Number of years 9.00% 9.42% 1.00 @ 365 per year Periods/year Single Amount Factor Compound Amount Factor i 9.00% 0.02% n 1.00 365 F/P $1.00 $500.00 1.090 $545.00 1.094 $547.08 Present Worth Factor P/F $1.00 $547.08 0.99975 $501.91 0.91394 $500.00 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 2 2