For the alternatives in Example 8–6, conduct a pairwise incremental comparison. Which alternative is the best at an interest rate of 10%? Each of the five alternatives has no salvage value at the end of a 20-year useful life.

Question

Alternatives
E	D	C	B	A
$9000	$1000	$6000	$2000	$4000	Cost
785	117	761	410	639	Annual benefit

Accepted Answer

1. The first step is to reorder the alternatives in order of increasing cost. This ensures that for each pairwise comparison the increment is an investment. In this example the order becomes D, B, A, C, E.

Alternatives
E	C	A	B	D
$9000	$6000	$4000	$2000	$1000	Cost
785	761	639	410	117	Annual benefit

2. Calculate the IRR of the least expensive alternative to see if it is better than doing nothing at a MARR of 10%.

[latex]PW_{D}=0=-\$1000+\$117(P/A,IRR_{D},20)[/latex]

Solving this without a spreadsheet may take several tries and interpolation, but using the Excel function = RATE(life, annual benefit, –first cost) makes it easy and the answer exact:

[latex]IRR_{D}=9.94\%[/latex]

Because investment D earns less than 10%, doing nothing is preferred to doing D.

3. Doing nothing is still our defender, and the next Alternative, B, becomes the challenger. Calculate the IRR of Alternative B to see if it is better than doing nothing at a MARR of 10%.

[latex]PW_{B}=0=-\$2000+\$410(P/A,IRR_{B},20)[/latex]

Using the Excel function = RATE(life, annual benefit, –first cost), we find

[latex]IRR_{B}=19.96\%[/latex]

Because investment B earns more than 10%, Alternative B is preferred to doing nothing.

4. Alternative B is now our defender, and the next Alternative, A, becomes the challenger. This comparison must be made incrementally.

[latex]PW_{A-B}=0=-(\$4000-\$2000)+(\$639-\$410)(P/A,IRR_{A-B},20)[/latex]

Using the Excel function = RATE(life, annual benefit, −first cost), we have

[latex]IRR_{A-B}=9.63\%[/latex]

Because the incremental investment A − B earns less than 10%, Alternative B is preferred to doing Alternative A.

5. Alternative B is still our defender, and the next Alternative, C, becomes the challenger. This comparison must be made incrementally.

[latex]PW_{C-B}=0=-(\$6000-\$2000)+(\$761-\$410)(P/A,IRR_{C-B},20)[/latex]

Using the Excel function = RATE(life, annual benefit, −first cost), we have

[latex]IRR_{C-B}=6.08\%[/latex]

Because the incremental investment C − B earns less than 10%, Alternative B is preferred to doing Alternative C.

6. Alternative B is still our defender, and the final Alternative, E, becomes the challenger. This comparison must be made incrementally.

[latex]PW_{E-B}=0=-(\$9000-\$2000)+(\$785-\$410)(P/A,IRR_{E-B},20)[/latex]

Using the Excel function = RATE(life, annual benefit, –first cost)

[latex]IRR_{E-B}=0.67\%[/latex]

Because the incremental investment E − B earns less than 10%, Alternative B is preferred to doing Alternative E.

At an interest rate of 10%, Alternative B is the best choice. But only by referring to Figure 8–6 can we see that Alternative B is the best choice for all interest rates between 9.63% (where Alternative A is equally attractive) and 19.96% (where doing nothing is equally attractive). With other data, the curve for Alternative B might intersect at an incremental comparison we hadn’t done. Also this pairwise comparison has not even compared Alternatives A and C so we cannot complete the choice table.

Question 8.7: For the alternatives in Example 8–6, conduct a pairwise incr...

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