Question

Google is evaluating three companies to buy a motor from. The motor will be used for the next 5 years. If only one machine can be chosen, do nothing is an alternative and the minimum attractive rate of return is 16.5% which motor should be chosen, if any? (use incremental rate of return analysis.)

	Alternative
	Large	Medium	Small
First Cost	225,000	127,000	74,000
Annual benefit	78,000	44,000	29,000
Maintenance and Operating Cost	39,000	12,000	8,500
Salvage value	16,500	6,800	3,100

Answer 1

ANSWER:

I = 16.5% AND N = 5 YEARS

We will find incremental ror by putting pw to zero and we will subtract large from medium as first cost of large is more.

pw(large - medium) = first cost(large - medium) + annual benefit (large - medium) (p/a,i,n) + moc (large - medium) (p/a,i,n) + salvage value (large - medium) (p/f,i,n)

0 = (-225,000 - (-127,000) ) + (78,000 - 44,000) (p/a,i,5) + (-39,000 - (-12,000) ) (p/a,i,5) + (16,500 - 6,800) (p/f,i,5)

0 = (-225,000 + 127,000) + 34,000(p/a,i,5) + (-39,000 + 12,000) (p/a,i,5) + 9,700(p/f,i,5)

0 = -98,000 + 34,000(p/a,i,5) - 27,000(p/a,i,5) + 9,700(p/f,i,5)

0 = -98,000 + 7,000(p/a,i,5) + 9,700(p/f,i,5)

98,000 = 7,000(p/a,i,5) + 9,700(p/f,i,5)

solving via trial and error we get that i is between -19% and -20% and solving further we get that i is -19.29% and therefore we reject large and select medium and now compare it with small as marr is greater then the ror.

pw(medium - small) = first cost(medium - small) + annual benefit (medium - small) (p/a,i,n) + moc (medium - small) (p/a,i,n) + salvage value (medium - small) (p/f,i,n)

0 = (-127,000 - (-74,000) ) + (44,000 - 29,000) (p/a,i,5) + (-12,000 - (-8,500) ) (p/a,i,5) + (6,800 - 3,100) (p/f,i,5)

0 = (-127,000 + 74,000) + 15,000(p/a,i,5) + (-12,000 + 8,500) (p/a,i,5) + 3,700(p/f,i,5)

0 = -53,000 + 15,000(p/a,i,5) - 4,500(p/a,i,5) + 3,700(p/f,i,5)

0 = -53,000 + 10,500(p/a,i,5) + 3,700(p/f,i,5)

53,000 = 10,500(p/a,i,5) + 3,700(p/f,i,5)

solving via trial and error we get that i is between 1% and 2% and solving further we get that i is 1.9% and so we reject medium and select small and compare it with do nothing as marr is greater then the ror.

pw(small - do nothing) = first cost(small - do nothing) + annual benefit (small - do nothing) (p/a,i,n) + moc (small - do nothing) (p/a,i,n) + salvage value (small - do nothing) (p/f,i,n)

0 = (-74,000 - 0) + (29,000 - 0) (p/a,i,5) + (-8,500 - 0) (p/a,i,5) + (3,100- 0) (p/f,i,5)

0 = -74,000 + 29,000(p/a,i,5) + (- 8,500) (p/a,i,5) + 3,100(p/f,i,5)

0 = -74,,000 + 29,000(p/a,i,5) - 8,500(p/a,i,5) + 3,100(p/f,i,5)

0 = -74,000 + 20,500(p/a,i,5) + 3,100(p/f,i,5)

74,000 = 20,500(p/a,i,5) + 3,100(p/f,i,5)

solving via trial and error we get that i is between 12% and 13% and solving further we get that i is 12.88% and so we reject small and select do nothing as the marr is greater then ror.

so do nothing is the right answer and no motor will be chosen.