Title: PRINCIPLES OF MONEYTIME RELATIONSHIPS
1CHAPTER 3
 PRINCIPLES OF MONEYTIME RELATIONSHIPS
2Objectives 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
3Capital
 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.
4Exchange money for shares of stock as proof of
partial ownership
5Time 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
6Interest 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
7Determination of Interest Rate
Interest Rate
Money Supply
MS1
ie
Money Demand
Quantity of Money
8Simple and Compound Interest
 Two types of interest calculations
 Simple Interest
 Compound Interest
 Compound Interest is more common worldwide and
applies to most analysis situations
9Simple 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...
10Compound 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
11Compound 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
 P21050 52.50 1102.50
 I3 1102.50(0.05) 55.125 55.13
 At end of year 3 1102.50 55.13 1157.63
12Parameters 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 flowscash inflows  Estimates of investment costs, operating costs,
taxes paid negative cash flows  cash outflows
13Cash 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 freebody diagram in statics
 First, some important TERMS . . . .
14Terminology 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, endofperiod
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
15The 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 5year problem
16END 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
 EndofPeriod Assumption
17EQUIVALENCE
 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
18ECONOMIC 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
19More 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
20Plan 1 _at_ 8 Simple Interest
 Simple Interest Pay all at end on 5,000 Loan
21Plan 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)
23Plan 4 Compound Interest
 20 of Principal Paid back annually
24Plan 5 Equal Repayment Plan
 Equal Annual Payments (Part Principal and Part
Interest
25Conclusion
 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.
26Finding Equivalent Values of Cash Flows Six
Scenarios
 Given a
 Present sum of money
 Future sum of money
 Uniform endofperiod series
 Present sum of money
 Uniform endofperiod series
 Future sum of money
 Find its
 Equivalent future value
 Equivalent present value
 Equivalent present value
 Equivalent uniform endofperiod series
 Equivalent future value
 Equivalent uniform endofperiod series
27Derivation by Recursion F/P factor
 F1 P(1i)
 F2 F1(1i)..but
 F2 P(1i)(1i) P(1i)2
 F3 F2(1i) P(1i)2 (1i)
 P(1i)3
 In general
 FN P(1i)n
 FN P(F/P,i,n)
28Present Worth Factor from F/P
 Since FN P(1i)n
 We solve for P in terms of FN
 P F1/ (1i)n F(1i)n
 Thus
 P F(P/F,i,n) where
 (P/F,i,n) (1i)n
29An 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)...
30Uniform Series Present Worth and Capital Recovery
Factors
A per period
31Uniform Series Present Worth and Capital Recovery
Factors
 Write a Present worth expression
1
2
32Uniform Series Present Worth and Capital Recovery
Factors
 Setting up the subtraction
2

1
3
33Uniform 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
34Section 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 lotterys
calculation of A  P 250,000 250,000(P A, 3, 19)
 P 250,000 3,580,950 3,830,950
35Sinking Fund and Series Compound amount factors
(A/F and F/A)
Find A given the Future amt.  F
A per period
0
36Example  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
37Example 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 (FA, 5, 10) 25,156
 Again, due to compounding, FgtNxA when igt0.
38An 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)
39Basic Setup for Interpolation
 Work with the following basic relationships
40Estimating for i 7.3
 Form the following relationships
41Interest 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?
42Section 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 reintegrated. See page 93.  Example Problem 395
 What is the value of the following CFD?
43Problem 395 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
44Arithmetic 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
45Linear Gradient Example
This represents a positive, increasing arithmetic
gradient
46Present Worth Gradient Component
 General CF Diagram Gradient Part Only
47To Begin Derivation of P/G,i,n
Multiply both sides by (1i)
48Subtracting 1 from 2..
2
1
49The A/G Factor
 Convert G to an equivalent A
A/G,i,n
50Gradient 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
51Geometric 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
52Geometric Gradients Increasing
 Typical Geometric Gradient Profile
 Let A1 the first cash flow in the series
A1
A1(1g)
A1(1g)2
A1(1g)3
A1(1g)n1
53Geometric Gradients Starting
 Pg The Ajs 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..
54Geometric Gradients
Subtract (1) from (2) and the result is..
55Geometric Gradients
For the case i g
56Geometric 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
6year 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.
57Geometric Gradient Example (g)
 g 11 per period A1 1700 i 8/yr
58Example 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
 F P(1i)n
 (1i)5 5,000/3000 1.6667
 (1i) 1.66670.20
 i 1.1076 1 0.1076 10.76
3,000
59Unknown 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.
i 5/year n is unknown!
60Unknown Number of Years
 (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
61Section 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.
62Relationship
 Relation between nominal interest and effective
interest i(1r/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
63Nominal 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(10.18/4)4
10.1925 or 19.25  (2) Semiannual compounding i(10.18/2)2
10.1881 or 18.81  (3) Monthly compounding ...
64Nominal 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 (10.169/365)365 10.184 or 18.4 per year
65Nominal and Effective Interest Rates
 Two situations well deal with in Chapter 3
 (1) Cash flows are annual. Were given r per year
and M. Procedure find i/yr (1r/M)M1 and
discount/compound annual cash flows at i/yr.  (2) Cash flows occur M times per year. Were
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.
66Example 12 Nominal
12 nominal for various compounding periods
67Interest 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 _at_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
68Interest 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 (10.12/2)121 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
halfyear  F 1000(F/P, 6, 20) 3000(F/P, 6, 12)
1500(F/P, 6, 8)  11,634.50
69Derivation of Continuous Compounding
 We can state, in general terms for the EAIR
Now, examine the impact of letting m approach
infinity.
70Derivation of Continuous Compounding
 We redefine the general form as
 From the calculus of limits there is an important
limit that is quite useful.
ieff. er 1
71Derivation 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!
72Example
 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