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.

4

Financing Definition Instrument Description

• Debt financing

• Equity financing

• Borrow money

• Sell partial ownership of company;

• Bond

• Stock

• Promise to pay principle & interest;

• Exchange shares of stock for ownership of company;

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

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

AMOUNT ORIGINAL

UNITTIME PER INTEREST RATE INTEREST

InterestRate

Quantity of Money

ieMoney Demand

Money SupplyMS1

Determination of Interest Rate

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•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

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

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.

26

Finding Equivalent Values of Cash Flows- Six Scenarios

• 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

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)

P0

F

n

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

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)=...

Uniform Series Present Worth and Capital Recovery Factors

• Annuity Cash Flow

$A per period

P = ??

0

………….. n 1 2 3 .. ..

n-1

Uniform Series Present Worth and Capital Recovery Factors

• Write a Present worth expression

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

[1]

2 3 1

1 1 1 1..

1 (1 ) (1 ) (1 ) (1 )n n

PA

i i i i i

[2]

Uniform Series Present Worth and Capital Recovery Factors

• Setting up the subtraction

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

[1]

[2]

-

1

1 1

1 (1 ) (1 )n

iP Ai i i

= [3]

2 3 1

1 1 1 1..

1 (1 ) (1 ) (1 ) (1 )n n

PA

i i i i i

Uniform Series Present Worth and Capital Recovery Factors

• Simplifying Eq. [3] further

1

1 1

1 (1 ) (1 )n

iP Ai i i

(1 )

(1 ) 1

n

n

i iA P

i

(1 ) 1 0

(1 )

n

n

iP A for i

i i

/ %, P A i n factor

A/P,i%,n factor

The present worth point of an annuity cash flow is always one period to the left of the first A amount

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

Sinking Fund and Series Compound amount factors (A/F and F/A)

• Annuity Cash Flow

0

………….. N

$A per period

$F

Find $A given the Future amt. - $F

1

(1 )nP F

i

(1 )

(1 ) 1

n

n

i iA P

i

(1 ) 1n

Ai

iF

)=A

(1 1F

ni

i

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) = $

Basic Setup for Interpolation

•Work with the following basic relationships

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

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

• Assume the following:

0 1 2 3 n-1 N

A1+G

A1+2G

A1+n-2G

A1+n-1G

This represents a positive, increasing arithmetic gradient

Present Worth: Gradient Component

• General CF Diagram – Gradient Part Only

0 1 2 3 4 ……….. n-1 n

1G2G

3G(n-2)G

(n-1)G

0G

We want the PW at time t = 0 (2 periods to the left of 1G)

To Begin- Derivation of P/G,i%,n

( / , %,2) 2 ( / , %,2) ...

...+ [(n-2)G](P/F,i,n-1)+[(n-1)G])P/F,i,n)

P G P F i G P F i

( / , %,2) 2( / , %,2) ...

...+ [(n-2)](P/F,i,n-1)+[(n-1)])P/F,i,n)

{}

P G P F i P F i

2 3 n-1 n

1 2 n-2 n-1P=G ...

(1+i) (1+i) (1+i) (1+i)

Multiply both sides by (1+i)

2 3 n-1 n

1 2 n-2 n-1P=G ...

(1+i) (1+i) (1+i) (1+i)

Subtracting [1] from [2]…..

11 2 n-2 n-1

1 2 n-2 n-1P(1+i) =G ...

(1+i) (1+i) (1+i) (1+i)

-

G (1 ) 1P=

i (1 ) (1 )

N

N N

i N

i i i

2

1

( / , %, )P G i N factor

The A/G Factor

• Convert G to an equivalent A

( / , , )( / , , )A G P G i n A P i n

G (1 ) 1P=

i (1 ) (1 )

N

N N

i N

i i i

(1 )

(1 ) 1

N

N

i i

i

1

(1 ) 1N

nGi i

A/G,i,n =

50

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

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…. 2 11 1 1 1

1 2 3

(1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

n

g n

A A g A g A gP

i i i i

Now, factor out the A1 value and rewrite as..

Geometric Gradients

1 2 1

1 2 3

1 (1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

n

g n

g g gP A

i i i i

(1)

(1+g)Multuply both sides by to create another equation

(1+i)1 2 1

1 2 3

(1+g) (1+g) 1 (1 ) (1 ) (1 )...

(1+i) (1+i) (1 ) (1 ) (1 ) (1 )

n

g n

g g gP A

i i i i

(2)

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

Geometric Gradients

1 1

1+g (1 ) 11

1+i (1 ) 1

n

g n

gP A

i i

Solve for Pg and simplify to yield….

1

11

1 g i

n

g

gi

P Ai g

1

(1 )g

nAP

i

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

$1700 $1700(1.11)1

$1700(1.11)2

$1700(1.11)3

$1700(1.11)5

PW(8%) = ??

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?

0 1 2 3 4 5

$3,000

$5,000

•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%

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….

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

i = 5%/year; n is unknown!

Unknown Number of Years

• Solving we have…..

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

•(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

61

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.

62

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

63

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 ...

64

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

65

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-1and 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.

66

Example: 12% NominalNo. of EAIR EAIR

Comp. Per. (Decimal) (per cent)Annual 1 0.1200000 12.00000%semi-annual 2 0.1236000 12.36000%Quartertly 4 0.1255088 12.55088%Bi-monthly 6 0.1261624 12.61624%Monthly 12 0.1268250 12.68250%Weekly 52 0.1273410 12.73410%Daily 365 0.1274746 12.74746%Hourly 8760 0.1274959 12.74959%Minutes 525600 0.1274968 12.74968%seconds 31536000 0.1274969 12.74969%

12% nominal for various compounding periods

67

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 QuarterP @yr. 3.75 = P qtr. 15= 3000(P/A, 3%, 6) - 500(P/G, 3%, 6) = $9713.60F yr. 10 = F qtr. 40= 9713.60(F/P, 3%, 25) + 1000(F/P, 3%, 40) == $23,600.34

68

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

69

Derivation of Continuous Compounding

• We can state, in general terms for the EAIR:

(1 ) 1mrim

Now, examine the impact of letting “m” approach infinity.

70

Derivation of Continuous Compounding• We re-define the general form as:

(1 ) 1 1 1

rm

rmr r

m m

•From the calculus of limits there is an important limit that is quite useful.

1lim 1 2.71828

h

he

h

lim 1 ,

m

r

m

re

m

ieff.= er – 1

71

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!

72

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%