chapter research

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 writers 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 writers view in ONE paragraph; give the reference location for this writer. 10 points3. State your views on this topicagain 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
Engineering Economics
2
Rate of Return Analysis
? Rate of return analysis is the most frequently used
exact analysis technique in industry.
 Rate of return is a single figure of merit that is
 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
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
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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