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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.

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Chapter 3

CHAPTER 3

PRINCIPLES OF MONEY-TIME RELATIONSHIPS


Objectives of this chapter
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

  • 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.



Time value of money
Time Value of Money ownership

  • 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 Rate ownership

  • 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
Determination of Interest Rate ownership

Interest

Rate

Money Supply

MS1

ie

Money Demand

Quantity of Money


Simple and compound interest
Simple and Compound Interest ownership

  • 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 ownership

  • 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 ownership

  • 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
Compound Interest: An Example ownership

  • 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 and Cash Flows ownership

  • 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
Cash Flow Diagramming ownership

  • 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
Terminology and Symbols ownership

  • 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
The Cash Flow Diagram: CFD ownership

  • 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
END OF PERIOD Convention ownership

  • 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
EQUIVALENCE ownership

  • 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 ownership

  • 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
More on Economic Equivalence Concept ownership

  • 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
Plan 1 @ 8% Simple Interest ownership

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


Plan 2 compound interest 8 yr
Plan 2 Compound Interest 8%/yr ownership

  • Pay all at the End of 5 Years


Plan 3 simple interest paid annually
Plan 3: Simple Interest Paid Annually ownership

  • Principal Paid at the End (balloon Note)


Plan 4 compound interest
Plan 4 Compound Interest ownership

  • 20% of Principal Paid back annually


Plan 5 equal repayment plan
Plan 5 Equal Repayment Plan ownership

  • Equal Annual Payments (Part Principal and Part Interest


Conclusion
Conclusion ownership

  • 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: ownership

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

Finding Equivalent Values of Cash Flows- Six Scenarios


Derivation by recursion f p factor

F ownershipn

………….

N

P0

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)


Present worth factor from f p
Present Worth Factor from F/P ownership

  • 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
An Example ownership

  • 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

P = ?? ownership

…………..

1 2 3 .. .. n-1

n

0

Uniform Series Present Worth and Capital Recovery Factors

  • Annuity Cash Flow

$A per period


Uniform series present worth and capital recovery factors1
Uniform Series Present Worth and Capital Recovery Factors ownership

  • Write a Present worth expression

[1]

[2]


Uniform series present worth and capital recovery factors2
Uniform Series Present Worth and Capital Recovery Factors ownership

  • Setting up the subtraction

[2]

-

[1]

=

[3]


Uniform series present worth and capital recovery factors3
Uniform Series Present Worth and Capital Recovery Factors ownership

  • 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
Section 3.9 Lotto Example ownership

  • 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

$F ownership

…………..

N

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

Find $A given the Future amt. - $F

  • Annuity Cash Flow

$A per period

0


Example uniform series capital recovery factor
Example - Uniform Series Capital Recovery Factor ownership

  • 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
Example: Uniform Series Compound Amount Factor ownership

  • 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 example1
An Example ownership

  • 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
Basic Setup for Interpolation ownership

  • Work with the following basic relationships


Estimating for i 7 3
Estimating for i = 7.3% ownership

  • Form the following relationships


Interest rates that vary over time
Interest Rates that vary over time ownership

  • 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
Section 3.12 Multiple Interest Factors ownership

  • 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
Problem 3-95 Solution ownership

  • 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
Arithmetic Gradient Factors ownership

  • 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

A ownership1+n-1G

A1+n-2G

A1+2G

A1+G

0 1 2 3 n-1 N

Linear Gradient Example

  • Assume the following:

This represents a positive, increasing arithmetic gradient


Present worth gradient component

(n-1)G ownership

(n-2)G

3G

2G

1G

0G

0 1 2 3 4 ……….. n-1 n

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

Present Worth: Gradient Component

  • General CF Diagram – Gradient Part Only


To begin derivation of p g i n
To Begin- Derivation of P/G,i%,n ownership

Multiply both sides by (1+i)


Subtracting 1 from 2

- ownership

Subtracting [1] from [2]…..

2

1


The a g factor
The A/G Factor ownership

  • Convert G to an equivalent A

A/G,i,n =


Gradient example

0 1 2 3 4 5 6 7

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


Geometric gradients
Geometric Gradients 5 6 7

  • 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

0 1 2 3 4 …….. n-1 n

Geometric Gradients: Increasing

  • Typical Geometric Gradient Profile

  • Let A1 = the first cash flow in the series

A1

A1(1+g)

A1(1+g)2

A1(1+g)3

A1(1+g)n-1


Geometric gradients starting
Geometric Gradients: Starting …….. n-1 n

  • 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 gradients1

(1) …….. n-1 n

(2)

Geometric Gradients

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


Geometric gradients2

Solve for P …….. n-1 ng and simplify to yield….

Geometric Gradients

For the case i = g


Geometric gradient example
Geometric Gradient: Example …….. n-1 n

  • 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

0 1 2 3 4 5 6 7

$1700

$1700(1.11)1

$1700(1.11)2

$1700(1.11)3

PW(8%) = ??

$1700(1.11)5

Geometric Gradient Example (+g)

  • g = +11% per period; A1 = $1700; i = 8%/yr


Example i unknown

0 1 2 3 4 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

  • 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

F 5 n = $2000

0 1 2 . . . . . . ……. n

P = $1,000

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….

i = 5%/year; n is unknown!


Unknown number of years1

F 5 n = $2000

0 1 2 . . . . . . ……. n

P = $1,000

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


Section 3 16 nominal and effective interest rates
Section 3.16. 5 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
Relationship 5

  • 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
Nominal and Effective Interest Rates –Examples 5

  • 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
Nominal and Effective Interest Rates –Example 5

  • 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
Nominal and Effective Interest Rates 5

  • 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.


Example 12 nominal
Example: 12% Nominal 5

12% nominal for various compounding periods


Interest problems with compounding more often than once per year example a
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
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
Derivation of Continuous Compounding Year – Example B

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

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


Derivation of continuous compounding1
Derivation of Continuous Compounding Year – Example B

  • 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 compounding2
Derivation of Continuous Compounding Year – Example B

  • 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
Example Year – Example B

  • 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%


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