Chapter 3
This presentation is the property of its rightful owner.
Sponsored Links
1 / 72

CHAPTER 3 PowerPoint PPT Presentation


  • 110 Views
  • Uploaded on
  • Presentation posted in: General

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.

Download Presentation

CHAPTER 3

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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.


Principles of money time relationships

Exchange money for shares of stock as proof of partial ownership


Time value of money

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 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

Determination of Interest Rate

Interest

Rate

Money Supply

MS1

ie

Money Demand

Quantity of Money


Simple and compound interest

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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

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


Plan 2 compound interest 8 yr

Plan 2 Compound Interest 8%/yr

  • Pay all at the End of 5 Years


Plan 3 simple interest paid annually

Plan 3: Simple Interest Paid Annually

  • Principal Paid at the End (balloon Note)


Plan 4 compound interest

Plan 4 Compound Interest

  • 20% of Principal Paid back annually


Plan 5 equal repayment plan

Plan 5 Equal Repayment Plan

  • Equal Annual Payments (Part Principal and Part Interest


Conclusion

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

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

Fn

………….

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

  • 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

  • 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 = ??

…………..

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

  • Write a Present worth expression

[1]

[2]


Uniform series present worth and capital recovery factors2

Uniform Series Present Worth and Capital Recovery Factors

  • Setting up the subtraction

[2]

-

[1]

=

[3]


Uniform series present worth and capital recovery factors3

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

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

$F

…………..

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

  • 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

  • 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

  • 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

  • Work with the following basic relationships


Estimating for i 7 3

Estimating for i = 7.3%

  • Form the following relationships


Interest rates that vary over time

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

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

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

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

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

(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

Multiply both sides by (1+i)


Subtracting 1 from 2

-

Subtracting [1] from [2]…..

2

1


The a g factor

The A/G Factor

  • 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

  • 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

  • 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)

(2)

Geometric Gradients

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


Geometric gradients2

Solve for Pg and simplify to yield….

Geometric Gradients

For the case i = g


Geometric gradient example

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

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

Fn = $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

Fn = $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. 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

  • 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

  • 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

  • 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

  • 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

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

  • 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

  • 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

  • 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

  • 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%


  • Login