isu ccee ce 203 rate of return analysis (eea chapter 7)

ISU CCEE

CE 203CE 203

Rate of Return Rate of Return AnalysisAnalysis(EEA Chapter 7)(EEA Chapter 7)

ISU CCEE

• “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

Rate of Return AnalysisRate of Return AnalysisRate of Return AnalysisRate of Return Analysis

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• P = F (P/F, i, n) or P = A(P/A, i, n) EEA 5• A = P (A/F, i, n) EEA 6

Rate of Return Analysis, Rate of Return Analysis, RoRRoRRate of Return Analysis, Rate of Return Analysis, RoRRoR

EEA 7: i?for benefits = costs

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• Internal RoR, i*, solve for i in :

– NPW = PWB – PWC = 0– EUAW = 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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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%

OR: Use SOLVER

In-class example 7-1In-class example 7-1In-class example 7-1In-class example 7-1

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Use plotting:

In-class example 7-1In-class example 7-1In-class example 7-1In-class example 7-1

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• 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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• NPW = $5000• EUAW = $800• RoR = 8%

What is easier to What is easier to understand?understand?What is easier to What is easier to understand?understand?

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• Investment: subsequent inflow >

initial amount

• Borrowing: subsequent outflow >

initial amount

• Usually (but not always) investigate initial cash flow– Investment if negative– Borrowing if positive

Investment vs. Borrowing Investment vs. Borrowing SituationSituationInvestment vs. Borrowing Investment vs. Borrowing SituationSituation

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Investment vs. Borrowing Investment vs. Borrowing Example Example Investment vs. Borrowing Investment vs. Borrowing Example Example

Year Cash Flow #1 Cash Flow #2

0 -$5,000 $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

Investment Borrowing

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• 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.)

Minimum Attractive Rate of Minimum Attractive Rate of ReturnReturnMinimum Attractive Rate of Minimum Attractive Rate of ReturnReturn

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• 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 (IRR) on the difference between the alternatives and compare to MARR If IRR > MARR, choose higher-cost alternative If IRR < MARR, choose lower-cost alternative


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In-class Example 7-2In-class Example 7-2In-class Example 7-2In-class Example 7-2

Year Alternative #1 Alternative #2

0 - $5,000 - $5,000

1 $4,500 $500

2 $1,400 $5,700

Payback alternatives for an initial investment of $5000 (sum of cash flows both > $0). MARR = 6%.

(RoR = 14.5%) (RoR = 11.9%)

Which is the better alternative? To answer, consider both RoR and MARR

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0 - $5,000 - $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 IRR

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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


Note: total or net cash flow for difference is positive

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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% = IRR

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0 - $5,000 - $5,000 $0

1 $4,500 $500 - $4000

2 $1,400 $5,700 +$4,300

Total C. F. +$900 + $1200 + $300


Since IRR = 7.5% > MARR = 6%, choose alternative #2

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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

Or, to look at it another way:Alternative #1

Year Action

0 Invest $5,000

1 Receive $500 and invest it at 6% (MARR)

2 Receive $5,700 + $500 (1 + .06) = $6,230

Alternative #2

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In-class Example 7-2 with MARR In-class Example 7-2 with MARR = 9%= 9%In-class Example 7-2 with MARR In-class Example 7-2 with MARR = 9%= 9%


0 - $5,000 - $5,000

1 $4,500 $500

2 $1,400 $5,700

Suppose MARR = 9% for payback alternatives for an initial investment of $5000:

Since DRoR = 7.5% < MARR = 9%, choose alternative #1

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In-class Example 7-2 with MARR In-class Example 7-2 with MARR = 9%= 9%In-class Example 7-2 with MARR In-class Example 7-2 with MARR = 9%= 9%

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 #1

Year Action

0 Invest $5,000

1 Receive $500 and invest it at 9% (MARR)

2 Receive $5,700 + $500 (1 + .09) = $6,245

Alternative #2

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In-class Example 7-2 but we are the In-class Example 7-2 but we are the borrowerborrowerIn-class Example 7-2 but we are the In-class Example 7-2 but we are the borrowerborrower


0 - $5,000 - $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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Year Cash flow

0 $1,020

1 - $2,000

2 $500

3 $500

What is the internal RoR (i*) for the cash flow shown in the table below?

0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F, i, 3). Solve for i.

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0 = 1020 – 2000(P/F, i, 1) + 500(P/F, i, 2) + 500(P/F, i, 3). Graphing solution (using EXCEL):

NPW vs. Interest Rate

-15

-10

-5

0

5

10

15

20

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

Interest Rate

Net

Pre

sen

t W

ort

h

NPW = 0 at i values of 5.24% and 27.4%

Two answers!

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Multiple values for ROR Multiple values for ROR possible!possible!Multiple values for ROR Multiple values for ROR possible!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)

isu ccee ce 203 rate of return analysis (eea chapter 7)

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