Comparing Alternatives with Rate of Return

Use Incremental Analysis

Topics

In this section, we • recall the definition of ROR,• discuss decision situations for multiple

alternatives, and• discuss the appropriate decision

methodology for each situation

Rate of Return

• Recall that the ROR of an investment is the interest rate that makes

• NPW = 0• NAW = 0• PW Benefits = PW Costs• AW Benefits = AW Costs• For a single project, accept if ROR ≥

MARR.

Decisions Involving Multiple Projects

Given a set of possible projects,• any subset of the projects may be

selected.• only one may be selected, but one must

be chosen.• only one may be selected, but it is OK to

choose none.

Non-Mutually Exclusive Alternatives

• Suppose we can select any subset of the projects.

• Solution Methodology:– Compute the ROR of each alternative– Select each alternative for which ROR ≥ MARR

Example 1:

• MARR = 13%

Alternative Investment AnnualIncome

AnnualCost

Life

A $100,000 $50,000 $30,000 9

B $85,000 $44,000 $20,000 5

C $60,000 $30,000 $12,000 5

Analysis

– ROR of A = 13.7%– ROR of B = 12.7%– ROR of C = 15.3%

• Both A and C have ROR larger than our MARR. Therefore, select both A and C.

Mutually Exclusive Alternatives

• Suppose we must select one, only one, but at least one, alternative.

• Solution Methodology:– Each increment of investment must yield the

MARR.– Perform Incremental Analysis:

Recall Incremental Analysis1. Rank the alternatives in increasing order of

investment.2. Select as the defender the alternative with

the smallest investment.3. Let the challenger be the alternative with

the next higher investment.4. Accept or reject the challenger on the basis

of the return on the extra investment. The winner becomes the defender.

5. If the highest level of investment has been reached, stop. Otherwise return to step 3.

Example 2: Comparing cost alternativesAlternative Investment Operating Cost

A $5000 $240

B $3000 $875

C $4000 $500

D $2500 $1000

Ranked Alternatives

• MARR = 15%• Alternatives: All have five-year life and no

salvage• Ranked Alternatives : Rank by order of

increasing investment: D, B, C, A.Alternative Investment Operating Cost

D $2500 $1000

B $3000 $875

C $4000 $500

A $5000 $240

Example 2: Incremental Analysis

• Defender: Project D.• Next greater investment (Challenger):

Project B. • Is the extra investment in B over D

justified?– Incremental Investment: -$500– Incremental Benefit: $125– NAW= -500 (A/P, i, 5) + 125 => ROR = 8% <

MARR– Decision: Reject Challenger, Project B.

Example 2: Incremental Analysis (cont’d)

• Defender: Project D.• Next greater investment (Challenger):

Project C. • Is the extra investment in C over D

justified?– Incremental Investment: -$1500– Incremental Benefit: $500– NAW= -1500 (A/P, i, 5) + 500 => ROR = 20%

> MARR– Decision: Accept Challenger, Project C.

Example 2: Incremental Analysis (cont’d)

• Defender: Project C.• Next greater investment (Challenger):

Project A. • Is the extra investment in A over C justified?

– Incremental Investment: -$1000– Incremental Benefit: $260– NAW= -1000 (A/P, i, 5) + 260 => ROR = 9% <

MARR– Decision: Reject Challenger, – Keep Defender, Project C

Example 3: Projects with Benefits

Alternative Investment Net AnnualBenefits

Life

A $100,000 $20,000 9

B $85,000 $24,000 5

C $60,000 $18,000 5

• Choose one of the three

Example 3 (cont’d)

• MARR = 13% (select one and only one)• Ranked Projects: C, B, A

Alternative Investment Net AnnualBenefits

Life

C $60,000 $18,000 5

B $85,000 $24,000 5

A $100,000 $20,000 9

Example 3: Incremental Analysis

• Defender: Project C• Challenger: Project B• Is the extra investment in B over C

justified?– Incremental Investment: -$25,000– Incremental Benefit: $6000– NAW= -25 (A/P, i, 5) + 6 => ROR = 6.4% <

MARR– Decision: Reject Challenger, Keep Project C

Example 3: Incremental Analysis (cont’d)

• Defender: Project C• Challenger: Project A• Is the extra investment in A over C

justified?– Incremental Investment: -$40,000– Incremental Benefit: $2000– Useful Life: Project lives are different!!

Example 3: Incremental Analysis (cont’d)

• Select a common study period, say 45 years.

A

…

10 20 30 40

C

100,000

20,000

18,000

60,000

Example 3: Incremental Analysis (cont’d)

• Compute the incremental cash flows.

• The cash flows of A-C represent a non-simple investment.

A-C 10 20 30 40

40,000

2,000

60,000

60,000

Example 3: Incremental Analysis (cont’d)

• We must verify graphically (or with another method) that the interest found correspond to a rate of return. Luckily, it does. The rate of return of the incremental cash flows is 11.6%.

• Therefore reject the challenger, project A and accept C.

An Easier way for Different Lives

• Find ROR of A - C• NAW(A - C) = NAW(A) - NAW(C) = 0• -100(A/P, i, 9) + 20 - [-60(A/P, i, 5) + 18]• -100(A/P, i, 9) + 60(A/P, i, 5) + 2 = 0• i NAW(A - C)• 0% 2889• 5% 1789• 10% 463• 15% -1058• 12% -123• ROR of A - C is 11.6%. Reject A - C and choose C

Mutually Exclusive Alternatives with a Do-Nothing Alternative

• Suppose we can select one alternative or no alternative

• Solution Methodology: – Compute ROR of each alternative– Reject any alternatives that do not yield

the MARR– If only one remains, choose that

alternative– If more than one remains, do incremental

analysis to select the best

Example 4

• MARR = 13% (select one or none)

Alternative Investment NetIncome

Life ROR Life

A $100,000 $20,000 9 13.7% 9

B $85,000 $24,000 5 12.7% 5

C $60,000 $18,000 5 15.3% 5

Example 4 (cont’d)

• Since B does not return 13% it can be discarded.

• We most compare A and C, by computing the rate of return of the extra investment of A over C.

• The rate of return of A over C is 11.6%. Since the return on the extra investment is less than 13% (the MARR) reject the extra investment.

• We should choose option C.

Conclusion:

• Measure of merit is the ROR (a percentage)

• For a single alternative, accept if ROR ≥ MARR

• For multiple alternatives: Accept an increment if the ROR for the increment≥ MARR