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 2 Engineering Economics ID: 648274
Download Presentation The PPT/PDF document "Chapter 7 Rate of Return Analysis" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Chapter 7Rate of Return Analysis
1Slide2
Chapter ContentsInternal Rate of Return Rate of Return Calculations
Plot of NPW versus interest rate
i
Fees or DiscountsExamplesIncremental AnalysisUsing Spreadsheet
2
Engineering EconomicsSlide3
Rate of return analysis is the most frequently used exact analysis technique in industry.Major advantagesRate 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).Rate of Return Analysis
3
Engineering EconomicsSlide4
Internal Rate of ReturnWhat 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.
4
Engineering EconomicsSlide5
Internal Rate of ReturnInternal 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)
5
Engineering EconomicsSlide6
6 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
Internal Rate of Return (IRR)
Engineering EconomicsSlide7
ExampleA 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?
When the compound interest tables are visited the value of
i
where (F/A,
i
%, 26)=80 is found as 8%, so
i
=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
7
Engineering EconomicsSlide8
ExampleA 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?
When the compound interest tables are visited the value of
i
where (F/A,
i
%, 26)=80 is found as 8%, so
i
=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
8
Engineering EconomicsSlide9
Example – EXCEL solutionRATE(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
9
Engineering EconomicsSlide10
ExampleCash 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
10
Engineering EconomicsSlide11
EXAMPLE CONTINUES
-8.85
YEAR
CASH FLOW
0
($500)
1
$100
2
$150
3
$200
4
$250
11
Nov. 2, 2011
Engineering EconomicsSlide12
QUESTION CONTINUES
-8.85
12
Nov. 2, 2011
Engineering EconomicsSlide13
INTERPOLATION
5%
15%
X%
30.95
-8.85
5-X
30.95
10
39.80
0
13
Engineering EconomicsSlide14
INTERPOLATION
5%
15%
X%
30.95
-8.85
5-X
30.95
-10
39.80
0
14
Engineering EconomicsSlide15
INTERPOLATION
12%
15%
X%
3.350
-8.850
12-X
3.350
-3
12.200
0
15
Engineering EconomicsSlide16
INTERPOLATION
12%
15%
X%
3.350
-8.850
12-X
3.350
-3
12.200
0
16
Engineering EconomicsSlide17
EXCEL solution IRR(C1:C5) = 12.83%
C1 ~ C5 stores the stream of the 5 cash flows:
-500, 100, 150, 200, 250
17
Engineering EconomicsSlide18
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?
18
Engineering EconomicsSlide19
EXAMPLE CONTINUESYear
Cash Flow
0
10,000
1
10,000
2
10,000
3
10,000
4
0
5
(9240)
6
(9240)
7
(9240)
8
(9240)
9
(9240)
a)
b)
19
Engineering EconomicsSlide20
INTERPOLATION:
c)
Since the rate is low, the loan
looks
like a good
choice.
20
Engineering EconomicsSlide21
INTERPOLATION:
c)
Since the rate is low, the loan
looks
like a good choice
21
Engineering EconomicsSlide22
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.
22
Engineering Economics
EXCEL SolutionSlide23
Plot of NPW versus Interest Rate Borrowing Cases
Year
Cash Flow
0
200
1
-50
2
-50
3
-50
4
-50
5
-50
:
:
:
:
23
p. 218
Engineering EconomicsSlide24
Plot of NPW versus Interest RateInvestment Cases
Year
Cash Flow
0
-200
1
50
2
50
3
50
4
50
5
50
:
:
:
:
24
Engineering EconomicsSlide25
Fees or DiscountsQuestion: 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?
25
Engineering EconomicsSlide26
QUESTION CONTINUESYear
Pay Cash
Borrow from Seller
0
($180,000)
($20,000)
1
($47,484)
2
($47,484)
3
($47,484)
4
($47,484)
5
($47,484)
26
Engineering EconomicsSlide27
QUESTION CONTINUES
INTERPOLATION:
27
Engineering EconomicsSlide28
QUESTION CONTINUES
INTERPOLATION:
This is a relatively high rate of interest, so that borrowing from a bank and paying cash to the property owner is better.
28
Engineering EconomicsSlide29
EXCEL SolutionCombined cash flows (difference between options 1 & 2):At time 0: -$160,000EOY 1-5: $47,484IRR =
rate(5, 47484, -160000)
=
14.78% per yearIRR = irr(a1:a6) = 14.78%
per year
29
Engineering EconomicsSlide30
Loan and Investments are EverywhereQuestion: 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
30
Engineering EconomicsSlide31
QUESTION CONTINUESWeek
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)
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.
31
Effective annual interest
= (1+.15)^52 – 1 = 1432% !
Engineering EconomicsSlide32
EXCEL SolutionCombined cash flows (difference between the 2 options):At time 0: -$84EOM 1~11: $16
IRR =
rate(11, 16, -84)
= 14.92% per monthIRR = irr
(C1: C12) = 14.92% per month
32
Engineering EconomicsSlide33
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?
33
Engineering EconomicsSlide34
QUESTION CONTINUESMonthBuy Option
Lease Option
0
($30,000)
($2,000)
1-23
($2,000)
24
$3,000
0
To find IRR%,
set
cash flows of Buy and Lease options equal in PW terms
34
Engineering EconomicsSlide35
35
Engineering Economics
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 monthSlide36
Combined cash flows (difference of buy & lease):At time 0: -$28000EOM 1~23: $2000EOM 24: $3000
IRR =
irr
(e1:e25) = 4.97% per month
EXCEL Solution
36
Engineering EconomicsSlide37
When there are two alternatives, rate of return analysis is often performed by computing the incremental rate of return, Δ
IRR,
on the difference between
the two alternatives.
Incremental Analysis
37
Engineering EconomicsSlide38
Incremental AnalysisThe 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 (ΔIRR) on difference between alternatives:
Two -Alternative
Situations
Decision
Δ
IRR≥MARR
Choose
the higher-cost alternative
Δ
IRR<MARR
Choose
the lower-cost alternative
38
Engineering EconomicsSlide39
Incremental AnalysisQuestion: If any money not invested here may be invested elsewhere at the MARR of 6% which alternative (1 or 2) given below should you select using the internal rate of return (IRR) analysis?
Year
Alternative 1
Alternative 2
0
($20)
($30)
1
$33
$45
39
Engineering EconomicsSlide40
Question ContinuesYear
Alternative 1
Alternative 2
Alt2-Alt1
0
($20)
($30)
($30)-($20)=($10)
1
$33
$45
$45-$33=$12
40
Effective annual interest
(P/F,i,1) = 1/(1+i) = 0.833
1+i =1/0.833 = 1.2000
i
= 20%
Engineering EconomicsSlide41
Incremental AnalysisQuestion: When an equipment is installed it will save $800 per year in costs. The equipment has an estimated useful life of 5 years and no salvage value. Two suppliers are willing to install the equipment
Supplier
1
will provide the equipment in return for three beginning-of-year annual payments of $500 each.
Supplier
2
will provide the equipment for $1345.
If the MARR is 15%, which supplier should be selected?
41
Engineering EconomicsSlide42
QUESTION CONTINUESYear
0
1
2
3
4
5
Supplier 1
($500)
($500)
($500)
$800
$800
$800
$800
$800
Supplier 2
($1345)
$800
$800
$800
$800
$800
Supp2-Supp1
($845)
$500
$500
0
0
0
42
Engineering EconomicsSlide43
QUESTION CONTINUES
43
Engineering Economics
EXCEL Solution:
IRR =
rate(2, 500, -845)
=
12.00%
per yearSlide44
Think-Pair-ShareThe company uses a MARR of 14%. Based on the rate of return, which is the most desirable alternative for a 10-yr analysis period? (Use incremental ROR analysis to choose a better alternative)
Try
i= 25%
44
Engineering EconomicsSlide45
SolutionRequire an incremental ROR analysis to choose a better alternative between the two given alternatives.Year Alt B- Alt A0 -$165,000
1-10 $78,000
10 $20,000
PW of incremental costs – PW of incremental benefits
= 165,000 – (78,000(P/A, i, 10) + 20,000(P/F,i
, 10)) = 0
You want it to be higher than 14% which is MARR
Try 15%
165,000 – (78,000(5.019) + 20,000(0.2472)) = – 231,426
Try 50%
165,000 – [78,000(1.965) + 20,000(0.0173)] = 11,384
MARR < 15% <
Δ
IRR < 50%.
Choose Alternative B
.
45
Engineering EconomicsSlide46
IRR = rate(10,78000,-165000,20000, ,0.4) = 46.35%/yrIRR =
irr
(a1:a11)
= 46.35% per year
EXCEL Solution
46
Engineering EconomicsSlide47
Fancy Gadgets, Inc. has developed a new Thing-A-May-Jig at a cost of $2,000,000. Projected profits from the sale of Thing-A-May-Jigs for the next five years are: $300,000, $400,000, $500,000, $600,000, and $250,000. At the end of the fifth year, the production equipment associated with the Thing-A-May-Jig project will be disposed of for $750,000. Determine the rate of return for the Thing-A-May-Jig project. Try i
= 10 and
i
=12Think-Pair- Share
47
Engineering EconomicsSlide48
48
Engineering Economics
EXCEL Solution: IRR =
irr
(a1:a6)
= 10.16% per yearSlide49
Think-Pair- ShareAssuming that alternatives are replaced at the end of their useful life, determine the better alternative using annual cash flow analysis at an interest rate of 10%.
49
Engineering EconomicsSlide50
50
Engineering EconomicsSlide51
Analysis PeriodQuestion: If the MARR is 15% which machine shown below should be bought using the ΔIRR analysis comparison?
Machine 1
Machine 2
Initial
Cost
($300)
($700)
Uniform
Annual Benefits
$175
$200
End-of-useful-life
salvage value
$150
$250
Useful life
(years)
4
8
51
Engineering EconomicsSlide52
QUESTION CONTINUESYear
Machine 1
Machine 2
Mach2- Mach1
0
($300)
($700)
($400)
1
$175
$200
$25
2
$175
$200
$25
3
$175
$200
$25
4
$175
$200
$175
$150
($300)
5
$175
$200
$25
6
$175
$200
$25
7
$175
$200
$25
8
$175
$200
$25
$150
$250
$100
52
Machine 1
Machine 2
Initial
Cost
($300)
($700)
Uniform
Annual Benefits
$175
$200
End-of-useful-life
salvage value
$150
$250
Useful life
(years)
4
8
Engineering EconomicsSlide53
QUESTION CONTINUES
INTERPOLATION:
53
Engineering EconomicsSlide54
QUESTION CONTINUES
INTERPOLATION:
54
Engineering EconomicsSlide55
USING SPREADSHEETIRR = irr(b1:b9) = 2.36% per year
55
Engineering EconomicsSlide56
PW is a (polynomial) function of i, PW(i).Example
PW(
i
) = –400 + 25(P/A, i, 8) + 150(P/F, i, 4) +100(P/F, i, 8)
= –400 + 25i–1[1–(1+i)
–8
] + 150(1+i)
–4
+ 100(1+i)
–8
PW(
i
) = 0 may have multiple solutions for
i
.
56
Engineering Economics
Appendix 7ASlide57
End of Chapter 757Engineering Economics