PRINCIPLES OF MONEY-TIME RELATIONSHIPS 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 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.

Exchange money for shares of stock as proof of partial ownership

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

RENTAL FEE PAID FOR THE USE OF SOMEONE ELSES MONEY 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

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

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

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...

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”

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

Parameters and Cash Flows 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

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 . . . .

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

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

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

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

Equality in terms of Economic Value 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

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

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

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

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

Plan 4 Compound Interest 20% of Principal Paid back annually

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

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.

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

Derivation by Recursion: F/P factor Fn N …………. 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)

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

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)=...

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

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

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

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

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

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

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

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%.

An Example Solution: A = $1,000,000 (A | F, 10%, 40) = $ 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) = $

Basic Setup for Interpolation Work with the following basic relationships

Estimating for i = 7.3% Form the following relationships

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?

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?

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

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

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

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

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

Subtracting [1] from [2]….. - 2 1

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

Gradient Example PW(10%)Base Annuity = $379.08 $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

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

Geometric Gradients: Increasing Typical Geometric Gradient Profile Let A1 = the first cash flow in the series 0 1 2 3 4 …….. n-1 n A1 A1(1+g) A1(1+g)2 A1(1+g)3 A1(1+g)n-1

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 A1 value and rewrite as..

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

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

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.

Geometric Gradient Example (+g) g = +11% per period; A1 = $1700; i = 8%/yr 0 1 2 3 4 5 6 7 PW(8%) = ?? $1700 $1700(1.11)1 $1700(1.11)2 $1700(1.11)3 $1700(1.11)5

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 (1+i)5 = 5,000/3000 = 1.6667 (1+i) = 1.66670.20 i = 1.1076 – 1 = 0.1076 = 10.76% $3,000

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? Draw the cash flow diagram as…. Fn = $2000 i = 5%/year; n is unknown! 0 1 2 . . . . . . ……. n P = $1,000

Unknown Number of Years P = $1,000 Fn = $2000 Solving we have….. (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

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.

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

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 ...

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

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.

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

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 = 15th 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

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

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

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

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 e0.18 – 1 = 1.1972 – 1 = 19.72%/year The 19.72% represents the MAXIMUM effective interest rate for 18% compounded anyway you choose!

Example Solution: er – 1 = 0.15 er = 1.15 ln(er) = ln(1.15) 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%