Rate of Return Analysis CE 203 ISU

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CE 203
Rate of Return
Analysis
(EEA Chapter 7)
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Rate of Return Analysis
• “Equivalent” cash flows: same value at
some given time for a given interest rate
• Internal rate of return (definitions):
– interest rate such that, for given payment
schedule, loan is paid off with final payment
– interest rate such that, for given payment
schedule, unrecovered investment = 0
at final payment
– interest rate such that benefits = costs
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Rate of Return Analysis, RoR
• P = F (P/F, i, n) or P = A(P/A, i, n)
• A = P (A/F, i, n)
EEA 7: i?
for benefits = costs
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EEA 5
EEA 6
Rate of Return Analysis
• Internal RoR, i*, solve for i in :
– NPW
– EUAW
=
=
PWB – PWC
= 0
EUAB – EUAC = 0
• To solve for i* :
–
–
–
–
Iterative solution (get close, interpolate)
Use “solver”
Plot NPW or EUAW, “read” i* at NPW = 0
Spreadsheet (Excel or ???)
» RATE (N, A, P, F, Type, guess)
» IRR (value, guess)
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In-class example 7-1
If you invest $10,000 now and are paid
$5,200 at the end of each of the next two
years, what is the internal rate of return?
Use iteration, then interpolation to find i
NPW = $ 5,200(1+i)-1 + $ 5,200(1+i)-2 - $10k = 0
Try 2%
= $ 5,098 + $ 4,998 - $ 10,000 = $ 96
Try 3%
= $ 5,045 + $ 4,902 - $ 10,000 = -$ 53
Interest rate, from linear interpolation
2% + 96/(96+53)(3-2) = 2.64%
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OR: Use SOLVER
In-class example 7-1
Use plotting:
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Rate of Return Analysis
• Chapter 7: compare two alternatives
• Chapter 8: compare three+ alternatives
• Advantages of RoR analysis:
– More widely understood
– Single value of “merit”
– Most widely used (but maybe not in CE?)
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What is easier to understand?
• NPW = $5000
• EUAW = $800
• RoR = 8%
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Investment vs. Borrowing Situation
• Investment: subsequent inflow >
initial amount
• Borrowing: subsequent outflow >
initial amount
• Usually (but not always) investigate initial
cash flow
– Investment if negative
– Borrowing if positive
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Investment vs. Borrowing Example
Investment
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Borrowing
Year
0
Cash Flow #1
-$5,000
Cash Flow #2
$5,000
1
$1,000
-$1,000
2
$3,000
-$3,000
3
$2,000
-$2,000
Sum = $1,000
Investment
Sum = - $1,000
Borrowing
Minimum Attractive Rate of Return
• Minimum Attractive Rate of Return (MARR)
Rate of return (RoR) below which we will not invest
(because we can invest elsewhere at MARR or
simply decide not to invest if RoR is < MARR)
• MARR is the highest of
– Interest rate for borrowing money
– Average interest rate for the cost of capital
(loans, bonds, stock, etc.)
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Rate of Return Analysis
• RoR criterion: If internal rate of return (i*) P >
MARR, the investment is considered acceptable
(but not necessarily the best)
• RoR analysis for two alternatives
– Determine the cash flow for the difference between
alternatives (highest total cash flow alternative minus
lower total cash flow alternative)
– Determine the incremental rate of return (DIRR) on the
difference between the alternatives and compare to MARR
 If DIRR > MARR, choose higher-cost alternative
 If DIRR < MARR, choose lower-cost alternative
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In-class Example 7-2
Payback alternatives for an initial investment of
$5000 (sum of cash flows both > $0). MARR = 6%.
Year
0
Alternative #1
- $5,000
Alternative #2
- $5,000
1
$4,500
$500
2
$1,400
$5,700
(RoR = 14.5%)
(RoR = 11.9%)
Which is the better alternative?
To answer, consider both RoR and MARR
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In-class Example 7-2
Payback alternatives for an initial investment of
$5000 (sum of cash flows both > $0). MARR = 6%.
Year
0
Alternative #1
- $5,000
Alternative #2
- $5,000
1
$4,500
$500
2
$1,400
$5,700
Total C. F.
+$900
+ $1200
Both total cash flows are positive, so both are “investments”;
Alternative 2 has larger total, so use Alt. 2 – Alt. 1 for DIRR
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In-class Example 7-2
Payback alternatives for an initial investment of
$5000 (sum of cash flows both > $0). MARR = 6%.
Alt. #2 – Alt. #1
$0
Year
0
Alt. #1
- $5,000
Alt. #2
- $5,000
1
$4,500
$500
- $4000
2
$1,400
$5,700
+$4,300
Total C. F.
+$900
+ $1200
+ $300
Note: total or net cash flow for difference is positive
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In-class Example 7-2
Payback alternatives for an initial investment of
$5000 (sum of cash flows both > $0). MARR = 6%.
Year
Alt. #1
Alt. #2
Alt. #2 – Alt. #1
0
- $5,000
- $5,000
$0
1
$4,500
$500
- $4000
2
$1,400
$5,700
+$4,300
Total C. F.
+$900
+ $1200
+ $300
Need i such that NPW = 0 = - 4000 (1 + i) -1 + 4300 (1 + i) -2
For this simple case, i = 300/4000 = .075 = 7.5% = DIRR
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In-class Example 7-2
Payback alternatives for an initial investment of
$5000 (sum of cash flows both > $0). MARR = 6%.
Year
0
Alt. #1
- $5,000
Alt. #2
- $5,000
Alt. #2 – Alt. #1
$0
1
$4,500
$500
- $4000
2
$1,400
$5,700
+$4,300
Total C. F.
+$900
+ $1200
+ $300
Since DIRR = 7.5% > MARR = 6%, choose alternative #2
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In-class Example 7-2
Or, to look at it another way:
Alternative #1
Year Action
0
Invest $5,000
1
Receive $4,500 and invest it at 6% (MARR)
2
Receive $1,400 + $4,500 (1 + .06) = $6,170
Alternative #2
Year Action
0
Invest $5,000
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1
Receive $500 and invest it at 6% (MARR)
2
Receive $5,700 + $500 (1 + .06) = $6,230
In-class Example 7-2 with MARR = 9%
Suppose MARR = 9% for payback alternatives for an
initial investment of $5000:
Year
0
Alternative #1
- $5,000
Alternative #2
- $5,000
1
$4,500
$500
2
$1,400
$5,700
Since DRoR = 7.5% < MARR = 9%, choose alternative #1
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In-class Example 7-2 with MARR = 9%
Alternative #1
Year Action
0
Invest $5,000
1
Receive $4,500 and invest it at 9% (MARR)
2
Receive $1,400 + $4,500 (1 + .09) = $6,305
Alternative #2
Year Action
0
Invest $5,000
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1
Receive $500 and invest it at 9% (MARR)
2
Receive $5,700 + $500 (1 + .09) = $6,245
In-class Example 7-2 but we are the borrower
Year
0
Alternative #1
- $5,000
Alternative #2
- $5,000
1
$4,500
$500
2
$1,400
$5,700
First, if MARR = 6%, we would choose neither
alternative and go to bank to get $$$.
If forced to choose, selection criterion is
reversed: we would choose Alternative #1.
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In-class Example 7-3
What is the internal RoR (i*) for the cash flow
shown in the table below?
Year
Cash flow
0
$1,020
1
- $2,000
2
$500
3
$500
0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) +
500(P/F, i, 3). Solve for i.
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In-class Example 7-3
0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F,
i, 3). Graphing solution (using EXCEL):
NPW = 0 at
i values of
5.24% and
27.4%
NPW vs. Interest Rate
20
Net Present Worth
15
10
5
0
1
3
5
7
9
11
13
15
17
19
-5
-10
-15
Interest Rate
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21
23
25
27
29
Two
answers!
Multiple values for ROR possible!
…there may be as many positive values for i*
as there are sign changes in cash flow table
(in example, +1020 to -2000 to +500)
…try the modified internal rate of return (p.
238 of the textbook)
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