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Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.1 Engineering Economic Analysis 9th.

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Presentation on theme: "Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.1 Engineering Economic Analysis 9th."— Presentation transcript:

1 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.1 Engineering Economic Analysis 9th Edition Chapter 3 INTEREST AND EQUIVALENCE

2 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.2 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

3 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.3 Computing Cash Flows Cash flows have: Costs (disbursements) > a negative number Benefits (receipts) > a positive number Example 3-1

4 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.4 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

5 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.5 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

6 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.6 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 ( x 2) = $118 Would you accept payment with simple interest terms? Would a bank?

7 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.7 Compound Interest Interest that is computed on the original unpaid debt and the unpaid interest Total interest earned = I n = P (1+i) n - P Where P – present sum of money i – interest rate n – number of periods (years) I 2 = $100 x (1+.09) 2 - $100 = $18.81

8 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.8 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 = $ Would you be more likely to accept payment with compound interest terms? Would a bank?

9 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.9 Comparison of Simple and Compound Interest Over Time 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. Short or long? When is the $ difference significant? You pick the time period. Check the table to see the difference over time.

10 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.10 Four Ways to Repay a Debt Plan Repay Principal Repay InterestInterest Earned 1Equal annual installments Interest on unpaid balance Declines 2End of loanInterest on unpaid balance Constant 3Equal annual installmentsDeclines at increasing rate 4End of loanCompound and pay at end of loan Compounds at increasing rate until end of loan

11 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.11 Loan Repayment – Four Options This calculator is partially complete. If you complete the calculator you can earn 10 bonus points for your team.

12 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.12 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.

13 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.13 Given the choice of these two plans which would you choose? YearPlan 1Plan 2 1$1400$ Total$6200$7000 To make a choice the cash flows must be altered so a comparison may be made.

14 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.14 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.

15 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.15 Repayment Plans Establish the Interest Rate 1.Principal outstanding over time 2.Amount repaid over time As an example: If F = P (1 + i) n Then i=(F/P) 1/n -1

16 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.16 Application of Equivalence Calculations Pick an alternative. Which would you choose? Change the interest rate. What happens at 8%,15%,3%?

17 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.17 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)

18 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.18 Single Payment Compound Interest Year Beginning balance Interest for period Ending balance 1PiPP(1+i) 2 iP(1+i)P(1+i) 2 3 iP(1+i) 2 P(1+i) 3 nP(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

19 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.19 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.

20 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.20 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)

21 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.21 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

22 Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.22 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


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