PRINCIPLES OF MONEY-TIME RELATIONSHIPS

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CHAPTER 3

• PRINCIPLES OF MONEY-TIME RELATIONSHIPS

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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

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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.

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Exchange money for shares of stock as proof of
partial ownership
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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

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

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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...

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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

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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
• 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

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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

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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 . . . .

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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

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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

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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

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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

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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
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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

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Plan 1 _at_ 8 Simple Interest

• Simple Interest Pay all at end on 5,000 Loan

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Plan 2 Compound Interest 8/yr

• Pay all at the End of 5 Years

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Plan 3 Simple Interest Paid Annually

• Principal Paid at the End (balloon Note)

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Plan 4 Compound Interest

• 20 of Principal Paid back annually

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Plan 5 Equal Repayment Plan

• Equal Annual Payments (Part Principal and Part
  Interest

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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.

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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

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Derivation 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)

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Present 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

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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)...

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Uniform Series Present Worth and Capital Recovery
Factors

• Annuity Cash Flow

A per period
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Uniform Series Present Worth and Capital Recovery
Factors

• Write a Present worth expression

1
2
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Uniform Series Present Worth and Capital Recovery
Factors

• Setting up the subtraction

2
-
1

3
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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
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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 lotterys
  calculation of A
• P 250,000 250,000(P A, 3, 19)
• P 250,000 3,580,950 3,830,950

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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
0
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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

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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 (FA, 5, 10) 25,156
• Again, due to compounding, FgtNxA when igt0.

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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)

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Basic Setup for Interpolation

• Work with the following basic relationships

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Estimating for i 7.3

• Form the following relationships

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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?

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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?

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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

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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

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Linear Gradient Example

• Assume the following

This represents a positive, increasing arithmetic
gradient
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Present Worth Gradient Component

• General CF Diagram Gradient Part Only

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To Begin- Derivation of P/G,i,n
Multiply both sides by (1i)
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Subtracting 1 from 2..
2
1
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The A/G Factor

• Convert G to an equivalent A

A/G,i,n
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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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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

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Geometric 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)n-1
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Geometric 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..
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Geometric Gradients
Subtract (1) from (2) and the result is..
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Geometric Gradients
For the case i g
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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.

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Geometric Gradient Example (g)

• g 11 per period A1 1700 i 8/yr

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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

• 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
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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.

i 5/year n is unknown!
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Unknown Number of Years

• 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

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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.

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Relationship

• 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

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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(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 ...

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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 (10.169/365)365 -10.184 or 18.4 per year

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Nominal 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)M-1 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.

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Example 12 Nominal
12 nominal for various compounding periods
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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 _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
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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 (10.12/2)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

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Derivation of Continuous Compounding

• We can state, in general terms for the EAIR

Now, examine the impact of letting m approach
infinity.
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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
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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!
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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