An automobile leasing company has a contract with a new car dealer to do major repairs for $720 per car. The leasing company estimates that for $400,000, it could buy equipment to service their own cars at a cost of $300 per car.

Required:

a) If the equipment will have a salvage value of 10% of its first cost after 15 years, the minimum number of cars that must require major servicing each year to justify the equipment at a MARR of 10% per year is closest to _________.

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TomShelby TomShelby · Answer 1 · 2020-05-16T04:19:51+02:00

Answer:

a) 123 cars will breakeven the project.

Explanation:

We need to solve for the equivalent annual cost of the equipment and then, solve for the car to achieve a finnancial break-even:

Salvage value present value

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

Maturity $40,000.00

time 15.00

rate 0.10000

[tex]\frac{40000}{(1 + 0.1)^{15} } = PV[/tex]

PV 9,575.6820

Present value of the Equipment

400,000 - 9,576 = 390,424

Equivalent annual cost:

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

PV 390,424.00

time 15

rate 0.1

[tex]390424 \div \frac{1-(1+0.1)^{-15} }{0.1} = C\\[/tex]

C $ 51,330.518

Each car generates 720 of reveneu with a cost of 300 dollar the contribution is 420 per car

51,330 equivalent annual equipment cost

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420 contribution margin per car

break even = 122.12 = 123 cars