Chapter 7x

8/14/2019 Chapter 7x

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CHAPTER 7

Rate of ReturnAnalysis: Single

AlternativeMcGrawHill

ENGINEERING ECONOMY,SixthEdition

by Blank and

Tarquin


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02/19/10 Authored by Don Smith, Texas A&M University 2004 2

Chapter 7 Learning Objectives

1. Definition of Rate of Return (ROR)

2. ROR using PW and AW

3. Calculations about ROR

4. Multiple RORs

5. Composite ROR

6. ROR of Bonds


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Section 7.1INTERPRETATION OF A RATE OF

RETURN VALUE

The IRR method is one of the popular

time-discounted measures of investment

worth that is related to the NPV approach.

DEFINITION follows


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7.1 INTERPRETATION OF A RATEOF RETURN VALUE

DEFINITION

ROR is either the interest rate paid onthe unpaid balance of a loan, or theinterest rate earned on theunrecovered investment balance of aninvestment such that the final

payment or receipt brings theterminal value to equal 0.


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7.1 INTERPRETATION OF A RATEOF RETURN VALUE

In rate of return problems you seekan unknown interest rate (i*) thatsatisfies the following:

PWi*(+ cash flows) PWi*( - cash

flows) = 0

This means that the interest rate i*, is an unknown parameter and mustbe solved or approximated.


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7.1 Unrecovered InvestmentBalance

ROR is the interest rateearned/charged on the unrecoveredinvestment balance of a loan or

investment project

ROR is not the interest rate earnedon the original loan amount or

investment amount


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7.1 The Loan ScheduleYear BOY Bal Payment Interest

Amount

Prin. Red.

Amount

UnPaidBalance

0 $1,000 0 -- --- $1,000

1 1,000 315.47 100.00 215.47 784.53

2 784.53 315.47 78.45 237.02 547.51

3 547.51 315.47 54.75 260.72 286.79

4 286.79 315.47 28.68 286.79 0


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For this loan the unpaid loanbalances at the end of each year are:

$1,000

784.53

547.51

286.79

0

0

1

2

3

4

Unpaid loan

balance is now0 at the end of

the life of theloan


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The URBt =4 is exactly 0 at a 10%

rate$1,000

784.53

547.51

286.79

0

0

1

2

3

4


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7.1 Reconsider the following

Assume you invest $1000 over 4years

The investment generates$315.47/year

Draw the cash flow diagram0 1 2 3

4

P=$-1,000

A =+315.47


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7.1 Investment Problem

What interest rate equates thefuture positive cash flows to theinitial investment?

We can state:

-$1000= 315.47(P/A, i*,4)

Where i* is the unknown interestrate that makes the PW(+) = PW(-)


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$1000= 315.47(P/A, i*,4)

Solve the above for the i* rate

(P/A,i*,4) = 1000/315.47 = 3.16987

Given n = 4 what value of i* yields aP/A factor value = 3.16987?

Interest Table search yields i*=10%


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i*=10% per year

Just like the loan problem, we cancalculate the unrecovered investmentbalances that are similar to theunpaid loan balances

See the next slide.


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7.1 Unrecovered InvestmentBalances (UIB)

We set up the following table

t C.F(t) Future Value for 1 year UIBt

0 -1,000 ---- -1,000

1 +315.47 -1000(1.10)+315.47= -784.53

2 +315.47 -784.53(1.10)+315.47= -547.51

3 +315.47 -547.51(1.10)+315.47= -286.79

4 +315.47 -286.79(1.10)+315.47= 0


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7.1 UIBs for the Example

See Figure 7.1t C.F(t) UIBt

0 -1,000 -1,000

1 +315.47 -784.53

2 +315.47 -547.51

3 +315.47 -286.79

4 +315.47 0

The unrecoveredinvestmentbalances have beencalculated at a 10%interest rate.

Note, the

investment is fullyrecovered at theend of year 4

The UIB = 0 at a

10% interest rate


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0 -1,000 -1,000

1 +315.47 -784.53

2 +315.47 -547.51

3 +315.47 -286.79

4 +315.47 0

The 10% rate is theonly interest rate

that will cause theUIB at the end ofthe projects life toequal exactly 0


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0 -1,000 -1,000

1 +315.47 -784.53

2 +315.47 -547.51

3 +315.47 -286.79

4 +315.47 0

Note, all of UIBsare negative at the

10% rate.

This means thatthe investment isunrecovered

throughout the life.


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7.1 Pure Investment

The basic definition of ROR is theinterest rate that will cause theinvestment balance at the end of the

project to exactly equal 0

If there is only one such interestrate that will cause this, the

investment is said to be a PUREinvestment or, Conventionalinvestment


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Unrecovered means that the

investment balance for a given year is

negative.

If a projects UIBs are all negative

(under-recovered) then that

investment will possess one unique

interest rate to cause the UIBt=n to

equal 0


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7.1 ROR - Explained

ROR is the interest rate earned onthe unrecovered investment balancesthroughout the life of the investment.

ROR is not the interest rate earnedon the original investment

ROR (i*) rate will also cause the

NPV(i*) of the cash flow to = 0.


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Section 7.2ROR using Present Worth

PW definition of ROR

PW(-CFs) = PW(+CFs)

PW(-CFs) - PW(+CFs) = 0

AW definition of ROR

AW(-CFs) = AW(+CFs)

AW(-CFs) - AW(+CFs) = 0


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7.2 ROR using Present Worth

Finding the ROR for most cash flows is a

trial and error effort.

The interest rate, i*, is the unknown

Solution is generally an approximation

effort

May require numerical analysis

approaches


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See Figure 7.2

0 1 2 3 4

5

-

$1,000

+$500

+$1,500

Assume you invest $1,000 at t = 0: Receive$500 @ t=3 and $1500 at t = 5. What is theROR of this project?


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Write a present worth expression, set equal to 0

and solve for the interest rate that satisfies the

formulation.

1000 = 500(P/F, i*,3) +1500(P/F, i*,5)

Can you solve this directly for the value ofi*?

NO!

Resort to trial and error approaches


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1000 = 500(P/F, i*,3) +1500(P/F, i*,5)

Guess at a rate and try it

Adjust accordingly

Bracket

Interpolate

i* approximately 16.9% per year onthe unrecovered investmentbalances.


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7.2 Trial and Error Approach

Iterative procedures require an initial

guess for i*

One makes an educated first guess and

calculates the resultant PV at the guess

rate.


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7.2 Trial and Error Approach

If the NPV is not = 0 then another i* value is

evaluated. Until NPV close to 0

The objective is to obtain a negative PV foran i* guess value then.

Adjust the i* value to obtain a positive PV

given the adjusted i* value

Then interpolate between the two i* values


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7.2 Example 7.3 In Excel

Excel Setup for ROR

MARR= 10.00%

Life 10 Year Cash Flow

0 -$500,000

1 $10,000

2 $10,000

3 $10,0004 $10,000

5 $10,000

6 $10,000

7 $10,000

8 $10,000

9 $10,000

10 $710,000$300,000

ROR-Guess 0%

ROR 5.16%

NPV= -$168,674.03

=IRR(D6:D16,D19)

D19 = Guess Value


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7.2 Example 7.2 InvestmentBalances

Investment balances

at the i* rate

The time t = 10balance = 0 at i*

As it should!


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CHAPTER VII

McGra

wHill

ENGINEERING ECONOMYFifth Edition

Blank andTarquin

Section 7.3Cautions when using the

ROR Method


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Section 7.3Cautions when using the

ROR Method

Important Cautions toremember when using the ROR

method


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7.3 Cautions when using theROR Method No.1

Many real-world cash flows may possess

multiple i* values

More than one i* value that will satisfy thedefinitions of ROR

If multiple i*s exist, which one, if any, is the

correct i*???

au ons w en us ng e


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. au ons w en us ng eROR Method. 2. ReinvestmentAssumptions

Reinvestment assumption for the ROR

method is not the same as the reinvestment

assumption for PW and AW

PW and AW assume reinvestment at the

MARR rate

ROR assumes reinvestment at the i* rate

Can get conflicting rankings with ROR vs.

PV and AW

. au ons w en us ng e


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. au ons w en us ng eROR Method: 3. ComputationalDifficulties

ROR method is computationally more

difficult than PW/AW

Can become a numerical analysis problem

and the result is an approximation

Conceptually more difficult to understand

7 3 Cautions when using the


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7.3 Cautions when using theROR Method: 4. Special Procedurefor Multiple Alternatives

For analysis of two or more alternatives

using ROR one must resort to a different

analysis approach as opposed to the PW/AW

methods

For ROR analysis of multiple alternatives

must apply an incremental analysisapproach


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7.3 Cautions when using theROR Method: ROR is more difficult!

ROR is computationally more difficult

But is a popular method with financial

managersROR is used internally by a substantial

number of firms

Suggest using PW/AW methods where

possible


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7.3 Valid Ranges for usable i*rates

Mathematically, i* rates must be:

*100% i < +

If an i*


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Section 7.4Multiple Rates of Return

A class of ROR problems exist that will

possess multiple i* values

Capability to predict the potential for

multiple i* values

Two tests can be applied


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7.4 Tests for Multiple i* values

1. Cash Flow Rule of Signs

2. Cumulative Cash Flow Rule of Signs

test

Example follows


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7.4 Cash Flow Rule of Signs Test

The total number of real values i*s is

always less than or equal to the number of

sign changes in the original cash flow series.

Follows from the analysis of a n-th degree

polynomial

A 0 value does not count as a sign change

Example follows


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7.4 C.F. Rule of Signs example

Consider Example 7.4

Year Cash Flow

0 $2,0001 -$5002 -$8,1003 $6,800

+--

+

Result: 2 sign changes in the CashFlow

l C l f Si


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7.4 Results: CF Rule of SignsTest

Two sign changes in this example

Means we can have a maximum of 2

real potential i* values for this problem

Beware: This test is fairly weak and

the second test must also be performed


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7.4 Norstroms Test

Norstroms Test1972 works with the

accumulated cash flow

For the example problem theaccumulated cash flow (ACF) is.

7 4 A l t d CF Si T t


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7.4 Accumulated CF Sign Test(ACF)

For the problem form the accumulated cash

flow from the original cash flow.

Year Cash Flow Accum.

C.F0 $2,000 $2,0001 -$500 $1,500

2 -$8,100 -$6,600

3 $6,800 $200

Count sign changeshere

7 4 A l t d CF i


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7.4 Accumulated CF signs -Example

Year Cash Flow Accum.C.F

0 $2,000 $2,0001 -$500 $1,500

2 -$8,100 -$6,6003 $6,800 $200

+

+-+

2 sign changes in the ACF.

A.C.F


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7.4 ACF Sign Test States:

A sufficient but not necessary condition

for a single positive i* value is:

The ACF value at year N is > 0

There is exactly one sign change in

the ACF


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7.4 ACF Test - continued

If the value of the ACF for year N is

0 then an i* of 0% exists

If the value of ACF for year N is > 0,this suggests an i* > 0

If ACF for year N is < 0 there mayexist one or more negative i* values

but not always


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Alternatively:

If the ACF in year 0 < 0

And.

One sign change in the ACF series then

Have a unique i* value!


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If the number of sign changes in the

ACF is 2 or greater this strongly suggest

that multiple rates of return exist.


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7.4 Example 7.4 ACF Sign Test

Year Cash Flow Accum.C.F

0 $2,000 $2,0001 -$500 $1,500

2 -$8,100 -$6,6003 $6,800 $200

2 Sign Changeshere

Strong evidence that we have multiple i*

values

ACF(t=3) = $200 > 0 suggests positive i* (s)


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7.4 Excel Analysis of Ex. 7.4

We find two i* values:

{7.47%,41.35%}

Year Cash Flow

0 $2,000

1 -$500

2 -$8,100

3 $6,800Sum $200

ROR-Guess 0%

ROR 7.47%

NPV= $200.00

ROR-Guess 30%

ROR 41.35%

First i* using aguess value of 0%

Second i* valueusing guess value

of 30%


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7.4 Investment (Project) Balance

Examine the Investment or Project

balances at each i* rateInv Bal

i-rate 7.47%

Year Project Balances@ i* Rate

0 $2,000.00

1 $1,649.362 -$6,327.473 $0.00

Inv Bal

i-rate 41.35%


0 $2,000.00

1 $2,327.042 -$4,810.693 $0.00


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7 4 Investment Balances at both


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7.4 Investment Balances at bothi*s

Important Observations

The IBs for the terminal year (3) bothequal 0

Means that the two i* values are validRORs for this problem

Note: The IB amounts are not all the

same for the two i* values.IB amounts are a function of theinterest rate used to calculate theinvestment balances.


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7.4 PV Plot of 7.4

-150.000

-100.000

-50.000

0.000

50.000

100.000

150.000

200.000

250.000

0.00 0.20 0.40 0.60InterestRate

PV-

$$

i* = 7.47%

i*=41.35%


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7.4PV Plot - continued

-150.000

-100.000

-50.000

0.000

50.000

100.000

150.000

200.000

250.000

0.00 0.20 0.40 0.60

PV < 0If the

MARR isbetween

the two i*values thisinvestment would berejected!


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7.4 Comments on ROR

Multiple i* values lead tointerpretation problems

If multiple i*s which one, if any is

the correct one to use in ananalysis?

Serves to illustrate the

computational difficulties associatedwith ROR analysis

Section 7.5 provides an alternativeROR approach Composite ROR (C-ROR

Section 7 5


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Section 7.5Composite ROR Approach

Consider the following investment

-$10,000

0 1 2 3 45

+$8,000

+$9,000

Determine the ROR as.


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7 5 Composite ROR Approach:


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7.5 Composite ROR Approach:IBs

The Investment Balances ar i* are: Year Cash Flow Project Balances

@ i* Rate

0 -$10,000 -$10,000.00

1 $0 -$11,681.592 $8,000 -$5,645.95

3 $0 -$6,595.36

4 $0 -$7,704.435 $9,000 $0.00

Sum $7,000

ROR-Guess 0%

ROR 16.82%

All IBs are

negative for t= 0 4: IB(5) =0

Conventional

(pure)investment


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7.5 Composite ROR Approach

i* = 16.82%

-$10,000

0 1 2 3 45

+$8,000

+$9,000

Question: Is it reasonable toassume that the +8,000 can beinvested forward at 16.82%?


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Remember .

ROR assumes reinvestment at

the calculated i* rate

What if it is not practical for the+$8,000 to be reinvested forward

one year at 16.82%?


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7.5 Reinvestment Rates

Most firms can reinvest surplusfunds at some conservative marketrate of interest in effect at the time

the surplus funds become available.

Often, the current market rate is lessthan a calculated ROR value

What then is the firm to do with the+$8,000 when it comes in to the firm?


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7.5 Reinvesting

Surplus funds must be put to gooduse by the firm

These funds belong to the owners not to the firm!

Owners expect such funds to be putto work for benefit of future wealth ofthe owners


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Consider the followingrepresentation.

TheFirm

Project

Invested Funds

Returns

back


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Or, put in another context.

TheFirm

Project

Project borrows from the

firm

Project Lends back to the

firm


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Or, put in another context.

TheFirm

Project

Project borrows from thefirm

At the i* rate (16.82%)

Project Lends back to the

firm

But can the firm reinvestthese funds at 16.82%?Probably not!


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So, we may have to consider areinvestment rate that is closer to thecurrent market rate for reinvestment

of the $8,000 for the next timeperiod(s)

Assume a reasonable market rate is

say, 8% per year.

Call this rate an external rate - c


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Finding i is a much more involvedprocess

Prior to digital computers, only verysmall (N


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7.5 Investment Balances are:

IB(7.468%)

Inv Bal

i-rate 7.468%


0 $2,000.00

1 $1,649.36

2 -$6,327.473 $0.00

Inv Bal

i-rate 41.352%


0 $2,000.00

1 $2,327.04

2 -$4,810.693 $0.00

IB(41.352%)


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7.5 Project Balances @ 41.35%

Project Balances

-$6,000

-$5,000

-$4,000

-$3,000

-$2,000

-$1,000

$0$1,000

$2,000

$3,000

0 1 2 3

Years

PB's

-$

Over-recovered

Under-recovered


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Project Balances

-$8,000

-$6,000

-$4,000

-$2,000

$0

$2,000

$4,000

0 1 2 3

Years

PB's

-$

7.5 Project Balances @ 7.468%%

Over-recovered

Under-recovered


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7.5 Assume you Reinvest at 20%

c = 20%

Positive IBs are reinvested at 20% -

Not at the computed i* rate

Now, what is the modified ROR i?

Must perform a recursive analysis


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7.5 Recursive IBs are.

Let Ibj = Fj

F0 = +2000 (> 0; invest at 20%)

F1 = 2000(1.20) 500 = +1900

+1900 > 0; invest at 20%

F2 = 1900(1.20) 8100 = -5820

-5820 < 0; invest at i rate

Under-recovered


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F2 = 1900(1.20) 8100 = -5820

-5820 < 0; invest at i rate

F3 =-5820(1+i) + 6800

Since N = 3, F3

= 0

Solve; -5820(1+i) + 6800 = 0

Repeated from the

previous slide forclarity


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7.5 Single conditional ROR value

-5820(1+i) + 6800 = 0

i = [6800/5820] 1

i = 16.84% given c = 20%

If MARR = 20% and i = 16.84% thisinvestment would be rejected

Reduced the problem to a singleconditional ROR value.


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7.5 What ifc = i*1 (7.47%)?

Here, we examine what happens IFwe assume the reinvestment rate, c,equals one of the computed i* values.

This approach assumes that thereinvestment rate, c, is one of the i*rates

Recursive calculations follow.



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(@7.47%)

Let IBj = Fj

F0 = +2000 (> 0; invest at 7.47%)

F1 = 2000(1.0747) 500 = +1649.40

>0: Over-recovered balance

F2 = +1649.40(1.074) 8100 =

-6327.39

Now, under-recovered balance


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F2 = +1649.40(1.074) 8100 =

-6327.39

F3 =-6327(1+i) + 6800

Since N = 3, F3must = 0 (terminal IB

condition)

Solve; -6327(1+i) + 6800 = 0


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7.5 Single conditional ROR value

Must solve:

-6327.39(1+i) + 6800 = 0

(1+i) = 6800/6327.39

i = 0.0747 = 7.47%

Given c = one of the i*s, the i willequal the i* used as the

reinvestment rate!

Section 7 6


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Section 7.6Rate of Return on a Bond

Investment

Review Chapter 5 Section on Bonds

Bond problems represent a classicalROR type problem

Bond problems represent aconventional investment with a

unique ROR

Example Follows:


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7.6 Example 7.8

Purchase Price: $800/bond

4%, $1000 bond

Life = 20 years

Interest paid semiannually

Question: If you pay $800 per bond

what is the ROR (yield) on thisinvestment?


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7.6 Ex. 7.8: Cash Flow Diagram

. ..

0 1 2 3 439 40

$800

F40 =

$1000

$I/6 mos = $1000(0.04/2) = +$20.00 (every 6months)

A = +$20/6 months


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7.6 Ex. 7-8: Closed Form Setup

Setup is:

0 = -$800 +20(P/A,i*,40 + $1000(P/F,i*,40)

Solve for i*

Manual or computer solution yields:i*=2.87%/6 months

Nominal ROR/year = (2.87%)(2) = 5.74%/yr,

C.S.A.

Effective ROR/year = (1.0287)2 1 =5.82%/yr


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Chapter 7 Summary

ROR is a popular analysis method

Understood in part by practitioners

Presents computational difficulties

No ROR may exist

Multiple RORs may exist


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Chapter 7 Summary cont.

For cash flows where multiple i*values exist, one must decide to:

Go with one of the i* rates or,

Calculate the composite rate, i

If an exact ROR is not required, it is

suggested to go with PW or AW


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For cash flows where multiple i*values exist, one must decide to:

Go with one of the i* rates or,

Calculate the composite rate, i

If an exact ROR is not required, it is

suggested to go with PW or AW

h


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To have a possible i* value with agiven cash flow, there must be atleast one negative CF amount in the

series

IROR can be demanding and is bestaccomplished using a computer

algorithmIROR is often misunderstood bypractitioners but is a popular methodof analysis

Ch 7 S


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In real world applications one mayencounter a cash flow that possess nousable i* value!

These types of problems are difficultto explain to uninformed decisionmakers.

ENGINEERING ECONOMYSixthEditi


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McGrawHill

EditionBlank andTarquin

END OF SLIDE SET

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Chapter 7x