One of three machines must be purchased to meet an immediate need in company ACME. ACME uses a MARR of 10% in its analysis. Machine A costs 45,000 to install and will cost 9000 each year to run. Machine B has an operating cost of 9500 annually and costs 32,000 to install. Machine C can be bought for 51,000 and costs 7200 to run each year. All machines are expected to run for 8 years but only Machine C has a salvage value of 4,000. Using incremental ROR methods, identify the best economic choice.

We calcualte the present valeu of the maintenance cost like they were annuities. Then, we add the cost for the machine. in the case of machine C we will discount the salvage value

we will pick the lower of the cost.

Machine A

[tex]C \times \frac{1-(1+r)^{-time} }{rate} = PV\\[/tex]

C 9,000.00

time 8

rate 0.1

[tex]9000 \times \frac{1-(1+0.1)^{-8} }{0.1} = PV\\[/tex]

PV $48,014.3358

+ cost 45,000

net worth $ 93,014.34

machine B

[tex]C \times \frac{1-(1+r)^{-time} }{rate} = PV\\[/tex]

C 9,500.00

time 8

rate 0.1

[tex]9500 \times \frac{1-(1+0.1)^{-8} }{0.1} = PV\\[/tex]

PV $50,681.7989

+ 32,000 cost

net worth: 82,681.80

machine C

[tex]C \times \frac{1-(1+r)^{-time} }{rate} = PV\\[/tex]

C 7,200.00

time 8

rate 0.1

[tex]7200 \times \frac{1-(1+0.1)^{-8} }{0.1} = PV\\[/tex]

PV $38,411.4686

salvage value

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]

Maturity $4,000.0000

time 8.00

rate 0.10000

[tex]\frac{4000}{(1 + 0.1)^{8} } = PV[/tex]

PV 1,866.0295

cost: 51,000

net worth: 51,000 + 38,411.47 - 1,866.03 = 87.545,44‬