CHAPTER 3 PRINCIPLES OF MONEYTIME RELATIONSHIPS 1 Objectives

CHAPTER 3 PRINCIPLES OF MONEYTIME RELATIONSHIPS 1

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 2

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. 3

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

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 5

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

Determination of Interest Rate Money Supply MS 1 ie Money Demand Quantity of Money 7

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

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. . . 9

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” 10

Compound Interest: An Example • Investing $1000 for 3 year at 5% per year • P 0 = $1000, I 1 = $1, 000(0. 05) = $50. 00 • P 1 = $1, 000 + 50 = $1, 050 • New Principal sum at end of t = 1: = $1, 050. 00 • I 2 = $1, 050(0. 05) = $52. 50 • P 2=1050 + 52. 50 = $1102. 50 • I 3 = $1102. 50(0. 05) = $55. 125 = $55. 13 • At end of year 3 =1102. 50 + 55. 13 = $1157. 63 11

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 – 12 negative cash flows -- cash outflows

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. . 13

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 14

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 15

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 16

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 17

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 18

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. • Plan 1. 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 19

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

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

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

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

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

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. 25

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

Derivation by Recursion: F/P factor • • F 1 = P(1+i) F 2 = F 1(1+i)…. . but: F 2 = P(1+i)2 F 3 =F 2(1+i) =P(1+i)2 (1+i) = P(1+i)3 In general: F n P … …. N 0 FN = P(1+i)n FN = P(F/P, i%, n) 27

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 28

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

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

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

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

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 33

Section 3. 9 Lotto Example • If you win $5, 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, N = 19 years (1 st payment immediate), and i = 3% year • Solution: A = $5, 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 34

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

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 36

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 10 th deposit? 12. 5779 • Solution: • F= $2, 000 (F|A, 5%, 10) = $25, 156 • Again, due to compounding, F>Nx. A when i>0%. 37

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 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 (A | F, 10%, 40) = $ 38

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

Estimating for i = 7. 3% • Form the following relationships 40

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? 41

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? 42

Problem 3 -95 Solution • F 1 = -$1, 000(F/P, 15%, 1) - $1, 000 = -$2, 150 • F 2 = F 1 (F/P, 15%, 1) + $3, 000 = $527. 50 • F 4 = F 2 (F/P, 10%, 1)(F/P, 6%, 1) = $615. 07 43

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 44

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

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

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

Subtracting [1] from [2]…. . 2 1 48

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

Gradient Example $700 $600 $500 $400 $300 $200 $100 0 1 2 3 4 5 6 7 • PW(10%)Base Annuity = $379. 08 • PW(10%)Gradient Component= $686. 18 • Total PW(10%) = $379. 08 + $686. 18 • Equals $1065. 26 50

Geometric Gradients • An arithmetic (linear) gradient changes by a fixed dollar amount each time period. • A GEOMETRIC gradient changes by a fixed percentage each time period. • We define a UNIFORM RATE OF CHANGE (%) for each time period • Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next 51

Geometric Gradients: Increasing • Typical Geometric Gradient Profile • Let A 1 = the first cash flow in the series 0 1 A 1 2 3 4 ……. . n-1 n A 1(1+g)2 A 1(1+g)3 A 1(1+g)n-1 52

Geometric Gradients: Starting • Pg = The Aj’s time the respective (P/F, i, j) factor • Write a general present worth relationship to find Pg…. Now, factor out the A 1 value and rewrite as. . 53

Geometric Gradients (1) (2) Subtract (1) from (2) and the result is…. . 54

Geometric Gradients Solve for Pg and simplify to yield…. For the case i = g 55

Geometric Gradient: Example • Assume maintenance costs for a particular activity will be $1700 one year from now. • Assume an annual increase of 11% per year over a 6 -year time period. • If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0. • First, draw a cash flow diagram to represent the model. 56

Geometric Gradient Example (+g) • g = +11% period; A 1 = $1700; i = 8%/yr 0 1 2 3 $1700(1. 11)1 4 5 6 7 $1700(1. 11)2 $1700(1. 11)3 PW(8%) = ? ? $1700(1. 11)5 57

Example: i unknown • Assume on can invest $3000 now in a venture in anticipation of gaining $5, 000 in five (5) years. • If these amounts are accurate, what interest rate equates these two cash flows? $5, 000 0 1 2 3 4 5 • F = P(1+i)n $3, 000 • (1+i)5 = 5, 000/3000 = 1. 6667 • (1+i) = 1. 66670. 20 • i = 1. 1076 – 1 = 0. 1076 = 10. 76% 58

Unknown Number of Years • Some problems require knowing the number of time periods required given the other parameters • Example: • How long will it take for $1, 000 to double in value if the discount rate is 5% per year? Fn = $2000 • Draw the cash flow diagram as…. i = 5%/year; n is unknown! 0 1 2 . . . ……. n P = $1, 000 59

Unknown Number of Years Fn = $2000 • Solving we have…. . 0 1 2 . . . ……. n P = $1, 000 • (1. 05)x = 2000/1000 • Xln(1. 05) =ln(2. 000) • X = ln(1. 05)/ln(2. 000) • X = 0. 6931/0. 0488 = 14. 2057 yrs • With discrete compounding it will take 15 years 60

Section 3. 16. Nominal and Effective Interest Rates • Nominal interest (r) = interest compounded more than one interest period per year but quoted on an annual basis. • Example: 16%, compounded quarterly • Effective interest (i) = actual interest rate earned or charged for a specific time period. • Example: 16%/4 = 4% effective interest for each of the four quarters during the year. 61

Relationship • Relation between nominal interest and effective interest: i=(1+r/M)M -1, where • i = effective annual interest rate • r = nominal interest rate per year • M = number of compounding periods per year • r/M = interest rate per interest period 62

Nominal and Effective Interest Rates –Examples • Find the effective interest rate per year at a nominal rate of 18% compounded (1) quarterly, (2) semiannually, and (3) monthly. • (1) Quarterly compounding; i=(1+0. 18/4)4 -1=0. 1925 or 19. 25% • (2) Semiannual compounding; i=(1+0. 18/2)2 -1=0. 1881 or 18. 81% • (3) Monthly compounding. . . 63

Nominal and Effective Interest Rates –Example • A credit card company advertises an A. P. R. of 16. 9% compounded daily on unpaid balances. What is the effective interest rate per year being charged? r = 16. 9% M = 365 • Solution: ieff = (1+0. 169/365)365 -1=0. 184 or 18. 4% per year 64

Nominal and Effective Interest Rates • Two situations we’ll deal with in Chapter 3: • (1) Cash flows are annual. We’re given r per year and M. Procedure: find i/yr = (1+r/M)M-1 and discount/compound annual cash flows at i/yr. • (2) Cash flows occur M times per year. We’re given r per year and M. Find the interest rate that corresponds to M, which is r/M per time period (e. g. , quarter, month). Then discount/compound the M cash flows per year at r/M for the time period given. 65

Example: 12% Nominal 12% nominal for various compounding periods 66

Interest Problems with Compounding more often than once per Year – Example A • If you deposit $1, 000 now, $3, 000 four years from now followed by five quarterly deposits decreasing by $500 per quarter at an interest rate of 12% per year compounded quarterly, how much money will you have in you account 10 years from now? r/M = 3% per quarter and year 3. 75 = 15 th Quarter P @yr. 3. 75 = P qtr. 15 = 3000(P/A, 3%, 6) - 500(P/G, 3%, 6) = $9713. 60 F yr. 10 = F qtr. 40 = 9713. 60(F/P, 3%, 25) + 1000(F/P, 3%, 40) = = $23, 600. 34 67

Interest Problems with Compounding more often than once per Year – Example B • If you deposit $1, 000 now, $3, 000 four years from now, and $1, 500 six years from now at an interest rate of 12% per year compounded semiannually, how much money will you have in your account 10 years from now? • i per year = (1+0. 12/2)12 -1 = 0. 1236 • F = $1, 000(F/P, 12. 36%, 10) + $3, 000(F/P, 12. 36%, 6) +$1, 500(F/P, 12. 36%, 4) or r/M = 6% per half-year • F = 1000(F/P, 6%, 20) + 3000(F/P, 6%, 12)+ 1500(F/P, 6%, 8) • = $11, 634. 50 68

Derivation of Continuous Compounding • We can state, in general terms for the EAIR: Now, examine the impact of letting “m” approach infinity. 69

Derivation of Continuous Compounding • We re-define the general form as: • From the calculus of limits there is an important limit that is quite useful. ieff. = er – 1 70

Derivation of Continuous Compounding • Example: • What is the true, effective annual interest rate if the nominal rate is given as: – r = 18%, compounded continuously Solve e 0. 18 – 1 = 1. 1972 – 1 = 19. 72%/year The 19. 72% represents the MAXIMUM effective interest rate for 18% compounded anyway you choose! 71

Example • An investor requires an effective return of at least 15% per year. What is the minimum annual nominal rate that is acceptable if interest on his investment is compounded continuously? • Solution: er – 1 = 0. 15 er = 1. 15 ln(er) = ln(1. 15) r = ln(1. 15) = 0. 1398 = 13. 98% 72