I am assigned to write a research paper on the following topic and the following are instructions on how to do the paper I also attached a powerpoint slides that have chapter one in it and also attached a sample of how the paper should look likeIn a single Word doc (or similar editable application), respond to the next four items:1. Restate one point/idea (from your assigned chapter) in your own words. It could be the same as you posted on the Discussions (above). In any case, the idea must have two or more possible viewpoints.You must do this to get any credit for parts 2, 3, and 4.1. Research to find one writer (public, that is, someone who has been published in print, or other medium) who agrees with the point. State that writers view in ONE paragraph; give the reference location for this writer. 10 points2. Research to find one writer (as above) who disagrees with the point. State that writers view in ONE paragraph; give the reference location for this writer. 10 points3. State your views on this topicagain in one paragraph. 16 pointsPost your document to this Assignment (in Canvas) regardless of which chapter you have to cover. That is, everyone posts their work to the same Canvas Assignment. You are to do this research ONLY for your assigned chapter.Everyone does this for their specific chapter.

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Chapter 7

Rate of Return Analysis

1

Chapter Contents

? Internal Rate of Return

? Rate of Return Calculations

? Plot of NPW versus interest rate i

? Fees or Discounts

? Examples

? Incremental Analysis

? Using Spreadsheet

Engineering Economics

2

Rate of Return Analysis

? Rate of return analysis is the most frequently used

exact analysis technique in industry.

? Major advantages

Rate of return is a single figure of merit that is

readily understood.

Calculation of rate of return is independent from

the minimum attractive rate of return (MARR).

Engineering Economics

3

Internal Rate of Return

What is the internal rate of return (IRR)?

IRR is the interest rate at which present worth or

equivalent uniform annual worth is equal to 0.

In other words, the internal rate of return is the

interest rate at which the benefits are equivalent

to the costs.

Engineering Economics

4

Internal Rate of Return

? Internal rate of return is commonly used to evaluate

the desirability of investments or projects.

? IRR can be used to rank multiple prospective projects.

? Because the internal rate of return is a rate quantity,

it is an indicator of the efficiency, quality, or yield of

an investment.

? To decide how to proceed, IRR will be compared to

preselected minimum attractive rate of return

(Chapter 8)

Engineering Economics

5

Internal Rate of Return (IRR)

Given a cash flow stream, IRR is the interest rate i which yields

a zero NPW (i.e., the benefits are equivalent to the costs), or a

zero worth at any point in time. This can be expressed in 5

different ways as follows.

?

?

?

?

?

NPW = 0

PW of benefits PW of costs = 0

PW of benefits = PW of costs

PW of benefits/PW of costs = 1

EUAB EUAC = 0

Engineering Economics

6

Example

A person invests $1000 at the end of each year. If the

person would like to have $80,000 in savings at EOY 26

what interest rate should he select?

Net Present Wo rth = 0 = Net Future Worth

– $1,000 (F/A, i%,26) + $80,000 = 0

(F/A, i%,26) = $80,000/$1 ,000 = 80

Checking the Tables

26 yrs @ 6%, F/A = 59.156

26 yrs @ 10%, F/A = 109.182

26 yrs @ 8%, F/A = 79.954

Engineering Economics

7

Example

A person invests $1000 at the end of each year. If the

person would like to have $80,000 in savings at EOY 26

what interest rate should he select?

Net Present Wo rth = 0 = Net Future Worth

– $1,000 (F/A, i%,26) + $80,000 = 0

(F/A, i%,26) = $80,000/$1 ,000 = 80

When the compound interest tables are visited the value of i

where (F/A, i%, 26)=80 is found as 8%, so i=8%

Engineering Economics

8

Checking the Tables

26 yrs @ 6%, F/A = 59.156

26 yrs @ 10%, F/A = 109.182

26 yrs @ 8%, F/A = 79.954

Example

A person invests $1000 at the end of each year. If the

person would like to have $80,000 in savings at EOY 26

what interest rate should he select?

Net Present Wo rth = 0 = Net Future Worth

– $1,000 (F/A, i%,26) + $80,000 = 0

(F/A, i%,26) = $80,000/$1 ,000 = 80

When the compound interest tables are visited the value of i

where (F/A, i%, 26)=80 is found as 8%, so i=8%

Engineering Economics

9

Example EXCEL solution

RATE(n, A, P, F, type, guess)

rate(26, 1000, 0, -80000) = 8%

rate (26, -1000, 0, 80000) = 8%

A, P, F must have different signs (+ or )!

IRR(value range, guess)

value range = the cash flow stream

Engineering Economics

10

Example

Cash flows for an investment are shown in the following

figure. What is the IRR to obtain these cash flows?

YEAR

CASH FLOW

0

($500)

1

$100

2

$150

3

$200

4

$250

Engineering Economics

11

YEAR

CASH

FLOW

0

($500)

1

$100

2

$150

3

$200

4

$250

EXAMPLE CONTINUES

EUAW = EUAB – EUAC = 0

100 + 50( A / G, i %, 4) – 500( A / P, i %, 4) = 0

Try i = 5%

100 + 50( A / G,5%, 4) – 500( A / P,5%,4)

100 + 50(1.439) – 500(0.2820) = 30.95

Try i = 15%

100 + 50( A / G,15%, 4) – 500( A / P,15%, 4)

100 + 50(1.326) – 500(0.3503) = -8.85

-10.20

Nov. 2, 2011

Engineering Economics

12

QUESTION CONTINUES

EUAW = EUAB – EUAC = 0

100 + 50( A / G, i %, 4) – 500( A / P, i %, 4) = 0

Try i = 5%

100 + 50( A / G,5%, 4) – 500( A / P,5%,4)

100 + 50(1.439) – 500(0.2820) = 30.95

Try i = 15%

100 + 50( A / G,15%, 4) – 500( A / P,15%, 4)

100 + 50(1.326) – 500(0.3503) = -8.85

-10.20

Nov. 2, 2011

Engineering Economics

13

INTERPOLATION

5%

30.95

30.95

5-X

10

X%

0

15%

-8.85

39.80

5- X

30.95 – 0

30.95

=

=

5 – 15 30.95 – (-8.85) 39.80

X = 12.78 ? 13%

Engineering Economics

14

INTERPOLATION

5%

30.95

30.95

5-X

-10

X%

0

15%

-8.85

39.80

5- X

30.95 – 0

30.95

=

=

5 – 15 30.95 – (-8.85) 39.80

X = 12.78 ? 13%

Engineering Economics

15

INTERPOLATION

12%

3.350

3.350

12-X

-3

X%

0

15%

-8.850

12.200

12 – X

3.350 – 0

3.350

=

=

12 – 15 3.350 – (-8.850) 12.200

X = 12.82 ? 13%

Engineering Economics

16

INTERPOLATION

12%

3.350

3.350

12-X

-3

X%

0

15%

-8.850

12.200

12 – X

3.350 – 0

3.350

=

=

12 – 15 3.350 – (-8.850) 12.200

X = 12.82 ? 13%

Engineering Economics

17

EXCEL solution

IRR(C1:C5) = 12.83%

C1 ~ C5 stores the stream of the 5 cash flows:

-500, 100, 150, 200, 250

Engineering Economics

18

Example

A student, who will graduate after 4 years, borrows

$10,000 per year at 5% interest rate at the beginning of

each year. No interest is charged till graduation. If the

student makes five equal annual payments after the

graduation (end-of-period payments).

a) What is each payment after the graduation?

b) Calculate IRR of loan?

(hint: use cash flow from when the student started

borrowing the money to when it is all paid back)

c) Is the loan attractive to the student?

Engineering Economics

19

EXAMPLE CONTINUES

Year

Cash

Flow

0

10,000

1

10,000

2

10,000

3

10,000

4

0

5

(9240)

Net Present Worth = 0

6

(9240)

10,000 + 10,000(P/A, i%, 3)

7

(9240)

8

(9240)

9

(9240)

a) Loan Payment =

$40,000(A/P, 5%, 5) = $40,000(0.2310)

b) To find IRR%

– [$9,240(P/A, i%, 5)](P/F, i%,4) = 0

Engineering Economics

20

10,000 + 10,000(P/A, i%, 3) – [$9,240(P/A, i%, 5)](P/F,i%,4) = 0

Try i = 3%

10,000 + 10,000(2.829) – [$9,240(4.580)](0.8885) ? 690

Try i = 2%

10,000 + 10,000(2.884) – [$9,240(4.713)](0.9238) ? -1,389

INTERPOLATION:

3 – IRR

690 – 0

690

=

=

3- 2

690 + 1389 2079

IRR = %2.67

c) Since the rate is low, the loan looks like a good choice.

Engineering Economics

21

10,000 + 10,000(P/A, i%, 3) – [$9,240(P/A, i%, 5)](P/F,i%,4) = 0

Try i = 3%

10,000 + 10,000(2.829) – [$9,240(4.580)](0.8885) ? 690

Try i = 2%

10,000 + 10,000(2.884) – [$9,240(4.713)](0.9238) ? -1,389

INTERPOLATION:

3 – IRR

690 – 0

690

=

=

3- 2

690 + 1389 2079

IRR = %2.67

c) Since the rate is low, the loan looks like a good choice

Engineering Economics

22

EXCEL Solution

a) pmt(5%, 5, -40000) = $9,238.99 per month.

b) irr(g1:g10) = 2.66%.

c) Since the rate is low, the loan looks like a good choice.

Engineering Economics

23

Plot of NPW versus Interest Rate

Borrowing Cases

NPW

Year Cash Flow

$50.00

0

1

2

3

4

5

200

-50

-50

-50

-50

-50

($25.00)

:

:

:

:

($50.00)

$25.00

$0.00

0%

5%

10%

15%

20%

p. 218

Engineering Economics

24

Plot of NPW versus Interest Rate

Investment Cases

NPW

Year Cash Flow

0

1

2

3

4

5

:

:

-200

50

50

50

50

50

:

:

$50.00

$25.00

$0.00

0%

5%

10%

15%

20%

($25.00)

($50.00)

Engineering Economics

25

Fees or Discounts

? Question:

Option 1: If a property is financed through a loan

provided by a seller, its price is $200,000 with 10%

down payment and five annual payments at 10%.

Option 2: If a property is financed through the same

seller in cash, the seller will accept 10% less. However,

the buyer does not have $180,000 in cash.

What is the IRR for the loan offered by seller?

Engineering Economics

26

QUESTION CONTINUES

Year

0

Pay Cash

($180,000)

Borrow

from Seller

Down Payment

($20,000)

= 20,000

1

($47,484)

2

($47,484)

3

($47,484)

4

($47,484)

5

($47,484)

Engineering Economics

= 200,000 ?10%

Annual Payments

= (200, 000 – 20, 000)(A/P,10%,5)

= 180,000(0.2638)

= 47,484

27

QUESTION CONTINUES

To find IRR%, setting cash flows equal in PW terms

– 180,000 = -20,000 – 47,484(P/A , i%, 5)

(P/A, i%, 5) = (180,000 – 20,000)/47 ,484 = 3.37

Looking in the tables for the above value, IRR% should be

between 12%(3.605) and 15%(3.352) .

INTERPOLATION:

12 – IRR 3.605 – 3.37 0.235

=

=

= 0.928

12 – 15

3.605 – 3.352 0.253

IRR = 14.79%

Engineering Economics

28

QUESTION CONTINUES

To find IRR%, setting cash flows equal in PW terms

– 180,000 = -20,000 – 47,484(P/A , i%, 5)

(P/A, i%, 5) = (180,000 – 20,000)/47 ,484 = 3.37

Looking in the tables for the above value, IRR% should be

between 12%(3.605) and 15%(3.352) .

INTERPOLATION:

12 – IRR 3.605 – 3.37 0.235

=

=

= 0.928

12 – 15

3.605 – 3.352 0.253

IRR = 14.79%

This is a relatively high rate of interest, so that borrowing from a bank and paying

cash to the property owner

is better.

29

Engineering Economics

EXCEL Solution

Combined cash flows (difference between options 1 & 2):

At time 0:

EOY 1-5:

-$160,000

$47,484

IRR = rate(5, 47484, -160000) = 14.78% per year

IRR = irr(a1:a6) = 14.78% per year

Engineering Economics

30

Loan and Investments are Everywhere

? Question: A student will decide whether to buy weekly

parking permit or summer semester parking permit from

USF. The former costs $16 weekly; the latter costs $100 due

May 17th 2010; in both cases the duration is 12 weeks.

Assuming that the student pay the weekly fee on every

Monday:

a) What is the rate of return for buying the weekly permit?

b) Is weekly parking attractive to student?

*Total 12 weeks

Engineering Economics

31

Effective annual interest

= (1+.15)^52 1 = 1432% !

QUESTION CONTINUES

Week

Weekly Semester

May 17

0

($16)

($100)

May 24

1

($16)

May 31

2

($16)

June 7

3

($16)

June 14

4

($16)

June 21

5

($16)

June 28

6

($16)

July 5

7

($16)

July 12

8

($16)

July 19

9

($16)

July 26

10

($16)

August 2

11

($16)

Engineering Economics

a) To find IRR%, set cash flows

equal in PW terms

100 = 16 16 (P/A,i%,11)

(P/A,i%,11) = (100 – 16) / 16

(P/A,i%,11) = 5.25

Looking in the table for the

above value:

IRR = 15%

b) Nominal interest rate for 52

weeks

IRR 15%/week or 15*52

= 780%/yr

Since the rate is high, paying

the semester fee looks like a

good choice.

32

EXCEL Solution

Combined cash flows (difference between the 2 options):

At time 0:

-$84

EOM 1~11: $16

IRR = rate(11, 16, -84) = 14.92% per month

IRR = irr(C1: C12) = 14.92% per month

Engineering Economics

33

Rate Of Return Calculations

Question: There are two options for an equipment: Buy

or Lease for 24 months. The equipment might be

either leased for $2000 per month or bought for

$30,000. If the plan is to buy the equipment, the

salvage value of the equipment at EOM 24 is $3,000.

What is the IRR or cost of the lease?

Engineering Economics

34

QUESTION CONTINUES

Month

Buy Option

Lease Option

0

($30,000)

($2,000)

1-23

24

($2,000)

$3,000

0

To find IRR%,

set cash flows of Buy and Lease options equal in PW terms

-30,000 + 3,000(P/F,i%,24) = -2,000 – 2,000(P/A, i%, 23)

0 = 28, 000 – 3,000(P/F,i%,24) – 2,000(P/A, i%, 23)`

Engineering Economics

35

0 = 28, 000 – 3,000(P/F,i%,24) – 2,000(P/A, i%, 23)`

Try i = 6%

28, 000 – 3,000(0.2470) – 2000(12.303) = 2653

Try i = 5%

28000 3000(0.3101) 20000(13.489) = 91.7

Try i = 4%

28000 3000(0.3477) 2000(14.148) = -1339.1

i 5% per month

Engineering Economics

36

EXCEL Solution

Combined cash flows (difference of buy & lease):

At time 0: -$28000

EOM 1~23: $2000

EOM 24:

$3000

IRR = irr(e1:e25) = 4.97% per month

Engineering Economics

37

Incremental Analysis

When there are two alternatives, rate of return

analysis is often performed by computing the

incremental rate of return, ?IRR, on the

difference between the two alternatives.

Engineering Economics

38

Incremental Analysis

? The cash flow for the difference between alternatives is

calculated by taking the higher initial-cost alternative

minus the lower initial-cost alternative.

? The following decision path is made for incremental rate

of return (?IRR) on difference between alternatives:

Two -Alternative

Situations

Decision

?IRR=MARR

Choose the higher-cost alternative

?IRR

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