Lecture10

Objective: Introducing the Internal Rate of Return (IRR) method for comparing projects and selecting the method to use in project comparisons.

Lecture 10

Topics to be Covered: - The Internal Rate of Return (IRR) - Internal Rate of Return Comparisons - IRR for Independent Projects - IRR for Mutually Exclusive Projects - Multiple IRRs - External Rate of Return (ERR) Methods - When to Use the ERR

Comparison Methods Part 2

- Rate of Return and Present/Annual Worth Methods Compared 1

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- Equivalence of Rate of Return and Present/Annual Worth Methods - Why Choose One Method Over the Other?

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Internal Rate of Return (IRR)

Internal Rate of Return (IRR), cont’d. How to calculate the IRR for complex cash flows:

The IRR is that interest rate at which a project just breaks even. Example 5.1 (p. 127): Suppose $100 is invested today in a project that returns $110 in one year.

PW(disbursements) = PW(receipts) FW(disbursements) = FW(receipts) AW(disbursements) = AW(receipts)

$110 1 year

- Solve equation for i* using trial-and-error and linear interpolation. $100 P = F(P/F,i*,1)

or

The value of i* could be positive or negative. A negative IRR means that the project is losing money rather than earning it.

100 = 110/(1+ i*)

i* = 0.1 = 10% 3

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Internal Rate of Return Comparisons 1- IRR for Independent Projects

Example 5.2 (p. 128): Clem is considering buying a tuxedo. It would cost $500, but would save him $160 per year in rental charges over its five-year life. What is the IRR for this investment.

- The project is accepted if its IRR > the MARR value

PW(disbursements) = $500

- The project is considered marginally accepted if its IRR = MARR

PW(receipts) = $160(P/A,i*,5) $500 = $160(P/A,i*,5)

or (P/A,i*,5) = 3.125

Using interest rate tables, trial -and-error, and linear interpolation i* = 18.14

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Example 5.3 (p. 130):

Example 5.3 (p. 130): MARR = 12% AW(receipts) - AW(disbursements) = 0 5000(A/F, i*, 10) + 15000 + 5000(A/G, i*,10) -120000(A/P, i*,10) - 10000 = 0 OR (A/F, i*, 10) + 1 + (A/G, i*,10) - 24(A/P , i*,10) = 0 i* (or IRR) can be obtained using: - Trial-and-error alone or with linear interpolation, or - Spreadsheet i* = 13.6% > MARR, the company should buy the new canner 7

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Example 5.4 (p. 132):

Internal Rate of Return Comparisons

Consider two investments. The first costs $1 today and returns $2 in one year. The second costs $1000 and returns $1900 in one year. Which is the preferred investment. Your MARR is 70%.

2- IRR for Mutually exclusive projects The projects must have equal lives if they are compared using the IRR method

The first project has i* = 100%

Algorithm: 1. Sort the projects from the lowest first cost to the highest. Start with the least first cost project. Call this the current best. 2. Challenge the current best with the next most expensive project. If the challenger is successful, i.e., if the incremental investment has an IRR ≥ MARR, then make the challenger the current best. 3. Repeat Step 2 until there are no further challengers.

The second project has i* = 90% - Do not choose the project with the highest IRR - Observe that the least cost investment provides a rate of return (IRR) that exceeds MARR. -Use the incremental investment analysis!

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Example 5.5 (p. 133): Monster Meats can buy a new meat slicer system for $50,000. They estimate it will save them $11,000 per year in labour and operating costs. The same system with an automatic loader is $68,000 and will save approximately $14,000 per year. The life of either system is thought to be eight years. Monster Meats has three feasible alternatives: Alternative First cost Annual Savings 1-DN $0 $0 2-Meat slicer alone $50,000 $11,000 3-Meat slicer with automatic loader $68,000 $14,000 Monster Meats uses a MARR of 12% for this type of project. Which alternative is better?

Incremental Investment: -(1000 - 1) + (1900 - 2)(P/F, i*,1) = 0 Or

(P/F, i*,1) = 0.52634 i* = 89.98% The incremental investment has IRR > MARR, Thus the second investment should be chosen

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Step1: Consider the Meat slicer alone

Example 5.6 (p. 135):

-50000+11000(P/A, i*, 8) = 0 solve for i*:

MARR = 15%. Which alternative should be chosen?

i* = 14.5% > MARR

Lathe

Meat slicer alone is better than the DN alternative

First Cost Annual Savings

1

2

$100 000

$150 000

$200 000

3

$255000

4

25 000

34 000

46 000

55 000

Step2: Consider the Meat slicer with automatic loader -68000+14000(P/A, i*, 8) = 0 solve for i*:

Solution:

i* = 12.5% > MARR

Lathe1: Step3: Consider the incremental investment

-100000+25000(P/A, i* ,10) = 0

-(68000-50000)+(14000-11000)(P/A, i*, 8) = 0 solve for i*: i* = 7% < MARR Monster meat should not buy the automatic loader

Solve for i*:

i* = 21.4% > MARR Lathe1 is considered current best

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Cont. of Example 5.6 Lathe2 is “challenger” of the “current best” -(150000-100000)+(34000-25000) (P/A, i* ,10) = 0

Cont. of Example 5.6

Solve for i*:

Lathe4 is “challenger” of the “current best”

i* = 12.4% < MARR

-(255000-200000)+(55000-46000) (P/A, i* ,10) = 0

Lathe2 fails the challenge. Lathe1 is still the current best

Solve for i*:

i* = 10.1% < MARR

Lathe4 fails the challenge. Lathe3 is still the current best

Lathe3 is “challenger” of the “current best” -(200000-100000)+(46000-25000) (P/A, i* ,10) = 0 Solve for i*:

The best alternative is Lathe3

i* = 16.4% > MARR

Lathe3 becomes the current best 15

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Flowchart for Comparing Mutually Exclusive Alternatives

Some Helpful Hints z

Is it necessary to check the IRR for each project? z

z

z

z

z

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Yes: pairwise comparison answers “which is better?”

What happens if we must do one project, but none of the projects have an IRR greater than the MARR? z

z

No, but it may be helpful

Can we use IRR for “cost only” problems?

e.g. several cost-only plans follow the algorithm: pairwise comparison answers “is Challenger better than Current Best?”

Must we calculate each incremental IRR accurately? z

Not if you want only to pick the best

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