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CHAPTER 3

CHAPTER 3. PRINCIPLES OF MONEY-TIME RELATIONSHIPS. Objectives Of This Chapter. Describe the return to capital in the form of interest Illustrate how basic equivalence calculation are made with respect to the time value of capital in Engineering Economy. Capital.

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CHAPTER 3

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  1. CHAPTER 3 PRINCIPLES OF MONEY-TIME RELATIONSHIPS

  2. Objectives Of This Chapter • Describe the return to capital in the form of interest • Illustrate how basic equivalence calculation are made with respect to the time value of capital in Engineering Economy

  3. Capital • Capital refers to wealth in the form of money or property that can be used to produce more wealth • Types of Capital • Equity capital is that owned by individuals who have invested their money or property in a business project or venture in the hope of receiving a profit. • Debt capital, often called borrowed capital, is obtained from lenders (e.g., through the sale of bonds) for investment.

  4. Exchange money for shares of stock as proof of partial ownership

  5. Time Value of Money • Time Value of Money • Money can “make” money if Invested • The change in the amount of money over a given time period is called the time value of money • The most important concept in engineering economy

  6. Interest Rate • INTEREST - THE AMOUNT PAID TO USE MONEY. • INVESTMENT • INTEREST = VALUE NOW - ORIGINAL AMOUNT • LOAN • INTEREST = TOTAL OWED NOW - ORIGINAL AMOUNT • INTEREST RATE- INTEREST PER TIME UNIT RENTAL FEE PAID FOR THE USE OF SOMEONE ELSES MONEY

  7. Determination of Interest Rate Interest Rate Money Supply MS1 ie Money Demand Quantity of Money

  8. Simple and Compound Interest • Two “types” of interest calculations • Simple Interest • Compound Interest • Compound Interest is more common worldwide and applies to most analysis situations

  9. Simple Interest • Simple Interest is calculated on the principal amount only • Easy (simple) to calculate • Simple Interest is: • (principal)(interest rate)(time); $I = (P)(i)(n) • Borrow $1000 for 3 years at 5% per year • Let “P” = the principal sum • i = the interest rate (5%/year) • Let N = number of years (3) • Total Interest over 3 Years...

  10. Compound Interest • Compound Interest is much different • Compound means to stop and compute • In this application, compounding means to compute the interest owed at the end of the period and then add it to the unpaid balance of the loan • Interest then “earns interest”

  11. Compound Interest: An Example • Investing $1000 for 3 year at 5% per year • P0 = $1000, I1 = $1,000(0.05) = $50.00 • P1 = $1,000 + 50 = $1,050 • New Principal sum at end of t = 1: = $1,050.00 • I2 = $1,050(0.05) = $52.50 • P2=1050 + 52.50 = $1102.50 • I3 = $1102.50(0.05) = $55.125 = $55.13 • At end of year 3 =1102.50 + 55.13 = $1157.63

  12. Parameters and Cash Flows • Parameters • First cost (investment amounts) • Estimates of useful or project life • Estimated future cash flows (revenues and expenses and salvage values) • Interest rate • Cash Flows • Estimate flows of money coming into the firm – revenues salvage values, etc. (magnitude and timing) – positive cash flows--cash inflows • Estimates of investment costs, operating costs, taxes paid – negative cash flows -- cash outflows

  13. Cash Flow Diagramming • Engineering Economy has developed a graphical technique for presenting a problem dealing with cash flows and their timing. • Called a CASH FLOW DIAGRAM • Similar to a free-body diagram in statics • First, some important TERMS . . . .

  14. Terminology and Symbols • P = value or amount of money at a time designated as the present or time 0. • F = value or amount of money at some future time. • A = series of consecutive, equal, end-of-period amounts of money. • n = number of interest periods; years • i = interest rate or rate of return per time period; percent per year, percent per month • t = time, stated in periods; years, months, days, etc

  15. The Cash Flow Diagram: CFD • Extremely valuable analysis tool • Graphical Representation on a time scale • Does not have to be drawn “to exact scale” • But, should be neat and properly labeled • Assume a 5-year problem

  16. END OF PERIOD Convention • A NET CASH FLOW is • Cash Inflows – Cash Outflows (for a given time period) • We normally assume that all cash flows occur: • At the END of a given time period • End-of-Period Assumption

  17. EQUIVALENCE • You travel at 68 miles per hour • Equivalent to 110 kilometers per hour • Thus: • 68 mph is equivalent to 110 kph • Using two measuring scales • Is “68” equal to “110”? • No, not in terms of absolute numbers • But they are “equivalent” in terms of the two measuring scales

  18. ECONOMIC EQUIVALENCE • Economic Equivalence • Two sums of money at two different points in time can be made economically equivalent if: • We consider an interest rate and, • No. of Time periods between the two sums Equality in terms of Economic Value

  19. More on Economic Equivalence Concept • Five plans are shown that will pay off a loan of $5,000 over 5 years with interest at 8% per year. • Plan1. Simple Interest, pay all at the end • Plan 2. Compound Interest, pay all at the end • Plan 3. Simple interest, pay interest at end of each year. Pay the principal at the end of N = 5 • Plan 4. Compound Interest and part of the principal each year (pay 20% of the Prin. Amt.) • Plan 5. Equal Payments of the compound interest and principal reduction over 5 years with end of year payments

  20. Plan 1 @ 8% Simple Interest • Simple Interest: Pay all at end on $5,000 Loan

  21. Plan 2 Compound Interest 8%/yr • Pay all at the End of 5 Years

  22. Plan 3: Simple Interest Paid Annually • Principal Paid at the End (balloon Note)

  23. Plan 4 Compound Interest • 20% of Principal Paid back annually

  24. Plan 5 Equal Repayment Plan • Equal Annual Payments (Part Principal and Part Interest

  25. Conclusion • The difference in the total amounts repaid can be explained (1) by the time value of money, (2) by simple or compound interest, and (3) by the partial repayment of principal prior to year 5.

  26. Given a: Present sum of money Future sum of money Uniform end-of-period series Present sum of money Uniform end-of-period series Future sum of money Find its: Equivalent future value Equivalent present value Equivalent present value Equivalent uniform end-of-period series Equivalent future value Equivalent uniform end-of-period series Finding Equivalent Values of Cash Flows- Six Scenarios

  27. Fn …………. N P0 Derivation by Recursion: F/P factor • F1 = P(1+i) • F2 = F1(1+i)…..but: • F2 = P(1+i)(1+i) = P(1+i)2 • F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3 In general: FN = P(1+i)n FN = P(F/P,i%,n)

  28. Present Worth Factor from F/P • Since FN = P(1+i)n • We solve for P in terms of FN • P = F{1/ (1+i)n} = F(1+i)-n • Thus: P = F(P/F,i%,n) where (P/F,i%,n) = (1+i)-n

  29. An Example • How much would you have to deposit now into an account paying 10% interest per year in order to have $1,000,000 in 40 years? • Assumptions: constant interest rate; no additional deposits or withdrawals Solution: P= 1000,000 (P/F, 10%, 40)=...

  30. P = ?? ………….. 1 2 3 .. .. n-1 n 0 Uniform Series Present Worth and Capital Recovery Factors • Annuity Cash Flow $A per period

  31. Uniform Series Present Worth and Capital Recovery Factors • Write a Present worth expression [1] [2]

  32. Uniform Series Present Worth and Capital Recovery Factors • Setting up the subtraction [2] - [1] = [3]

  33. Uniform Series Present Worth and Capital Recovery Factors • Simplifying Eq. [3] further The present worth point of an annuity cash flow is always one period to the left of the first A amount A/P,i%,n factor

  34. Section 3.9 Lotto Example • If you win $5,000,000 in the California lottery, how much will you be paid each year? How much money must the lottery commission have on hand at the time of the award? Assume interest = 3%/year. • Given: Jackpot = $5,000,000, N = 19 years (1st payment immediate), and i = 3% year • Solution: A = $5,000,000/20 payments = $250,000/payment (This is the lottery’s calculation of A P = $250,000 + $250,000(P | A, 3%, 19) P = $250,000 + $3,580,950 = $3,830,950

  35. $F ………….. N Sinking Fund and Series Compound amount factors (A/F and F/A) Find $A given the Future amt. - $F • Annuity Cash Flow $A per period 0

  36. Example - Uniform Series Capital Recovery Factor • Suppose you finance a $10,000 car over 60 months at an interest rate of 1% per month. How much is your monthly car payment? • Solution: A = $10,000 (A | P, 1%, 60) = $222 per month

  37. Example: Uniform Series Compound Amount Factor • Assume you make 10 equal annual deposits of $2,000 into an account paying 5% per year. How much is in the account just after the 10th deposit? 12.5779 • Solution: • F= $2,000 (F|A, 5%, 10) = $25,156 • Again, due to compounding, F>NxA when i>0%.

  38. An Example • Recall that you would need to deposit $22,100 today into an account paying 10% per year in order to have $1,000,000 40 years from now. Instead of the single deposit, what uniform annual deposit for 40 years would also make you a millionaire? • Solution: A = $1,000,000 (A | F, 10%, 40) = $

  39. Basic Setup for Interpolation • Work with the following basic relationships

  40. Estimating for i = 7.3% • Form the following relationships

  41. Interest Rates that vary over time • In practice – interest rates do not stay the same over time unless by contractual obligation. • There can exist “variation” of interest rates over time – quite normal! • If required, how do you handle that situation?

  42. Section 3.12 Multiple Interest Factors • Some situations include multiple unrelated sums or series, requiring the problem be broken into components that can be individually solved and then re-integrated. See page 93. • Example: Problem 3-95 • What is the value of the following CFD?

  43. Problem 3-95 Solution • F1 = -$1,000(F/P,15%,1) - $1,000 = -$2,150 • F2 = F 1 (F/P,15%,1) + $3,000 = $527.50 • F4 = F 2 (F/P,10%,1)(F/P,6%,1) = $615.07

  44. Arithmetic Gradient Factors • An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a contestant amount over n time periods. • A linear gradient is always comprised of TWO components: • The Gradient component • The base annuity component • The objective is to find a closed form expression for the Present Worth of an arithmetic gradient

  45. A1+n-1G A1+n-2G A1+2G A1+G 0 1 2 3 n-1 N Linear Gradient Example • Assume the following: This represents a positive, increasing arithmetic gradient

  46. (n-1)G (n-2)G 3G 2G 1G 0G 0 1 2 3 4 ……….. n-1 n We want the PW at time t = 0 (2 periods to the left of 1G) Present Worth: Gradient Component • General CF Diagram – Gradient Part Only

  47. To Begin- Derivation of P/G,i%,n Multiply both sides by (1+i)

  48. - Subtracting [1] from [2]….. 2 1

  49. The A/G Factor • Convert G to an equivalent A A/G,i,n =

  50. 0 1 2 3 4 5 6 7 Gradient Example $700 $600 $500 $400 $300 $200 $100 • PW(10%)Base Annuity = $379.08 • PW(10%)Gradient Component= $686.18 • Total PW(10%) = $379.08 + $686.18 • Equals $1065.26

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