Question

Precision Dyes is analyzing two machines to determine which one it should purchase. The company requires a rate of return of 14 percent and uses straight-line depreciation to a zero book value over the life of its equipment. Machine A has a cost of $512,000, annual operating costs of $34,200, and a four-year life. Machine B costs $798,000, has annual operating costs of $21,500, and has a six-year life. Whichever machine is purchased will be replaced at the end of its useful life. The firm should purchase Machine _____ because it lowers the firm's annual costs by approximately _______ as compared to the other machine.

Answer 1

Answer:

It would Pick Machine A

as it saves 16,790.73 as compared with machine B

Explanation:

We will calcualtethe present value of each machine.

But we have a problem, because one machine life is 4 years, and the other is 6 years. So to compare we need to match their useful life

4 year machine A and we have 6 years machine B

We will calculate the present worth for machine A

MACHINE A

$C \times \frac{1-(1+r)^{-time} }{rate} = PV$

C 34200

time 6

rate 0.14

$34200 \times \frac{1-(1+0.14)^{-4} }{0.14} = PV$

PV $99,648.96

Then we calculate the Net Worth

512,000 + 99,649 = 611,649

Now we do the same for Machine B

MACHINE B

$C \times \frac{1-(1+r)^{-time} }{rate} = PV$

C 21500

time 6

rate 0.14

$21500 \times \frac{1-(1+0.14)^{-6} }{0.14} = PV$

PV $83,606.35

Net Worth 798,000 + 83,606.35 = 881,606

Now how do we compare the 4 year machine with the 6 year machine ?

We will check which annuity payment equals the net worth for each project at our cost of capital rate

MACHINE A

$PV \div \frac{1-(1+r)^{-time} }{rate} = C$

PV $611,649

time 4

rate 0.14

$-226711.591116911 \times \frac{1-(1+0.14)^{-6} }{0.14} = PV$

C -209,920.86

MACHINE B

$PV \div \frac{1-(1+r)^{-time} }{rate} = C$

PV $881,606.00

time 6

rate 0.14

$-226711.591116911 \times \frac{1-(1+0.14)^{-6} }{0.14} = PV$

C -226,711.59

Now we compare each annuity:

-209,920.86 - (-226,711.59) = 16,790.73