Question

Boris is considering buying a new lawnmower. He has a choice between X mower and a Y mower. The salvage value of each mower at the end of its service life is zero.

	X	Y
First Cost	$350	$120
Life	10 years	4 years
Annual Gas	$60	$40
Annual Maintenance	$30	$60

Find which alternative is preferable using the IRR method and a MARR of 5 percent. Answer the following question by using present worth computations to find the IRRs. Use the least common multiple of service lives. Which of the two alternatives should be chosen? Use the ERR method if necessary.

(It is important to use least common multiple of service lives in this question.)

Answer 1

The two alternatives have different useful life in order to have common period of analysis take LCM of 4 and 10 we get 20. Thus, 20 years would be the common analysis period.

Alternative X has a higher initial investment in comparison to alternative Y. Therefore, alternative X is the challenger and Alternative Y is defender.

We can determine the incremental cash flow by subtracting the cash flow of alternative Y (defender) from the cash flow of alternative X (challenger).

Refer the attached picture

The net present worth of the incremental cash flow can be expressed as

We can determine the Incremental IRR using the trial and error method. Assume rate = 5% (MARR)

Now assume rate = 4%

On solving above equation we get

Still the value of incremental cash flow is negative thus we are required to reduce the value further. Assume rate =3%

From above calculation we can see the NPW at a rate of 3% is a positive number and at a rate of 4% the amount is negative. Hence the value of $ 0 less in between 3% to 4%. We can determine the value using linear interpolation technique.

Incremental IRR = 3.93%

The incremental IRR is less than the MARR. Hence, reject alternative X and accept alternative Y.

Solved it using excel too got the same value. Refer the attached screenshot of excel.

ERR method is not required as we can select the alternative using ∆IRR method.

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