Chapter 7 - Help-A-Bull

Report
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
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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
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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.
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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)
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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
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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
When
the10%,
compound
interest tables are visited the value of i
26 yrs @
F/A = 109.182
where
26)=80
is found as 8%, so i=8%
26 yrs (F/A,
@ 8%,i%,
F/A
= 79.954
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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%
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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
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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
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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
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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
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INTERPOLATION
5%
30.95
30.95
5-X
10
5 X
5  15

X%
0
15%
-8.85
30.95  0
30 . 95  (  8 . 85 )
39.80

30.95
39.80
X  1 2 . 78  13 %
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INTERPOLATION
5%
30.95
30.95
5-X
-10
5 X
5  15

X%
0
15%
-8.85
30.95  0
30 . 95  (  8 . 85 )
39.80

30.95
39.80
X  1 2 . 78  13 %
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INTERPOLATION
12%
3.350
3.350
12-X
-3
12  X
12  15

X%
0
15%
-8.850
3 . 350  0
3 . 350  (  8 . 850 )
12.200

3 . 350
12 . 200
X  1 2 . 82  13 %
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INTERPOLATION
12%
3.350
3.350
12-X
-3
12  X
12  15

X%
0
15%
-8.850
3 . 350  0
3 . 350  (  8 . 850 )
12.200

3 . 350
12 . 200
X  1 2 . 82  13 %
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EXCEL solution
IRR(C1:C5) = 12.83%
C1 ~ C5 stores the stream of the 5 cash flows:
-500, 100, 150, 200, 250
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Example
A student, who will graduate after 4 years, borrows
$10,000 per year at 5% interest rate at the beginning of
each year. No interest is charged till graduation. If the
student makes five equal annual payments after the
graduation (end-of-period payments).
a) What is each payment after the graduation?
b) Calculate IRR of loan?
(hint: use cash flow from when the student started
borrowing the money to when it is all paid back)
c) Is the loan attractive to the student?
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EXAMPLE CONTINUES
Year
Cash
Flow
0
10,000
1
10,000
2
10,000
3
10,000
4
0
5
(9240)
Net Present Wo rth  0
6
(9240)
10,000  10,000 ( P/A, i%, 3)
7
(9240)
8
(9240)
 [$ 9 , 2 40 ( P/A, i%, 5)](P/F, i%,4)  0
9
(9240)
a) Loan Payment 
$40,000(A/ P, 5%, 5)  $ 40,000(0.2 310)
b) To find IRR%
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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 , 2 40 ( 4.580)](0. 8885)  690
Try i  2%
10,000  10,000 ( 2 . 884 )  [$ 9 , 2 40 ( 4.713)](0. 9238)   1, 389
INTERPOLATION:
3  IRR
32

690  0
690  1389

690
2079
IRR  %2.67
c) Since the rate is low, the loan looks like a good choice.
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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 , 2 40 ( 4.580)](0. 8885)  690
Try i  2%
10,000  10,000 ( 2 . 884 )  [$ 9 , 2 40 ( 4.713)](0. 9238)   1, 389
INTERPOLATION:
3  IRR
32

690  0
690  1389

690
2079
IRR  %2.67
c) Since the rate is low, the loan looks like a good choice
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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.
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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
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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)
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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?
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QUESTION CONTINUES
Year
0
Pay Cash
($180,000)
Borrow
from Seller
($20,000)
1
($47,484)
2
($47,484)
3
($47,484)
4
($47,484)
5
($47,484)
Engineering Economics
D ow n P aym ent
 200,000  10%
 20,000
A nnual P aym ents
 (200, 000  20, 000)(A /P ,10% ,5)
 180,000(0.2638)
 47,484
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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
between
12%(3.605)
INTERPOLATION:
be
and 15%(3.352) .
12  IRR
12  15

3 . 605  3 . 37
3 . 605  3 . 352

0 . 235
 0 . 928
0 . 253
IRR  14 . 79 %
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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
between
12%(3.605)
INTERPOLATION:
be
and 15%(3.352) .
12  IRR
12  15

3 . 605  3 . 37
3 . 605  3 . 352

0 . 235
 0 . 928
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.
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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
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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
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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.
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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
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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?
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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 )`
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0  28, 000  3,000(P /F,i% ,24)  2,000(P /A , i% , 23 )`
T ry 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
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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
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Incremental Analysis
When there are two alternatives, rate of return
analysis is often performed by computing the
incremental rate of return, ΔIRR, on the
difference between the two alternatives.
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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 (ΔIRR) on difference between alternatives:
Two -Alternative
Situations
Decision
ΔIRR≥MARR
Choose the higher-cost alternative
ΔIRR<MARR
Choose the lower-cost alternative
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Incremental Analysis
 Question: 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
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Effective annual interest
(P/F,i,1) = 1/(1+i) = 0.833
Question Continues
1+i =1/0.833 = 1.2000
i = 20%
Year Alternative 1 Alternative 2
Alt2-Alt1
0
($20)
($30)
($30)-($20)=($10)
1
$33
$45
$45-$33=$12
PW Alt.2-Alt.1  0   10  12(P/F,i% ,1)
(P /F,i% ,1)  10/12  0.833
From tables, the above value is read at 20%
S ince 20%  6% (M A R R ),
H igher-cost alternative(A lt.2) w ill be s elected.
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Incremental Analysis
 Question: 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
Supplier1 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?
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QUESTION CONTINUES
Year
PW
0
1
2
3
4
5
($500)
($500)
$800
$800
$800
$800
$800
Supplier 1
($500)
Supplier 2
($1345)
$800
$800
$800
$800
$800
Supp2-Supp1
($845)
$500
$500
0
0
0
Suppl.2 - Suppl.1
 0   845  500(P/F, i%,1)  500(P/F, i%,2)
 845  500( 0.8929 )  500( 0.7972 )
 845  4 46 . 45  398.60  0
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QUESTION CONTINUES
PW
Suppl.2 - Suppl.1
 0   845  500(P/F, i%,1)  500(P/F, i%,2)
 845  500( 0.8929 )  500( 0.7972 )
 845  4 46 . 45  398.60  0
The underlined
values refers to 12% interest t able
Since 12%  15%(MARR),
Lower - cost alternativ e(Supplier 1) will be selected.
EXCEL Solution:
IRR = rate(2, 500, -845) = 12.00% per year
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Think-Pair-Share
The 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%
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44
Solution
Require an incremental ROR analysis to choose a better alternative between
the two given alternatives.
Year
Alt B- Alt A
0
-$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% < ΔIRR < 50%. Choose Alternative B.
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EXCEL Solution
IRR = rate(10,78000,-165000,20000, ,0.4) = 46.35%/yr
IRR = irr(a1:a11) = 46.35% per year
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Think-Pair- Share
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-AMay-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=12
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EXCEL Solution: IRR = irr(a1:a6) = 10.16% per year
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Think-Pair- Share
Assuming 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%.
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Analysis Period
 Question: 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
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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
QUESTION CONTINUES
Year
Machine 1
Machine 2
Mach2- Mach1
0
($300)
($700)
($400)
1
$175
$200
$25
2
$175
$200
$25
3
$175
$200
$25
$175
$200
4
$150
$175
($300)
5
$175
$200
$25
6
$175
$200
$25
7
$175
$200
$25
$175
$200
$25
$150
$250
$100
8
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QUESTION CONTINUES
PW
Machl.2 - Mach.1
0
 4 00  25(P/A, i%,8)  150(P/F, i%,4)  100(P/F, i%,8)  0
Try i  1%
 4 00  25(7.652)  150(0.9610 )  100(0.9235 )  27 . 80
Try i  5%
 4 00  25(6.463)  150(0.8227 )  100(0.6768 )   47 . 34
INTERPOLATION: 1  IRR
1 5

27 . 80  0
27 . 80  (  47 . 34 )
IRR  %2.48
Since 2.48%  15%(MARR),
Lower - cost alternativ
e(Machine1 ) will be selected.
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QUESTION CONTINUES
PW
Machl.2 - Mach.1
0
 4 00  25(P/A, i%,8)  150(P/F, i%,4)  100(P/F, i%,8)  0
Try i  1%
 4 00  25(7.652)  150(0.9610 )  100(0.9235 )  27 . 80
Try i  5%
 4 00  25(6.463)  150(0.8227 )  100(0.6768 )   47 . 34
INTERPOLATION: 1  IRR
1 5

27 . 80  0
27 . 80  (  47 . 34 )
IRR  %2.48
Since 2.48%  15%(MARR),
Lower - cost alternativ
e(Machine1 ) will be selected.
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USING SPREADSHEET
IRR = irr(b1:b9) = 2.36% per year
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55
Appendix 7A
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.
Engineering Economics
56
End of Chapter 7
Engineering Economics
57

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