An asset costs $100,000 and has a useful life of 10 years. The salvage value at the end of 10 years is estimated to be $10,000. Using the Modified Accelerated Cost Recovery System (ACRS), find the book value of the asset at the end of year 3.

An asset is purchased for $90,000. It is expected to have a useful life of six years and a salvage value of $18,000 at the end of its useful life. Find the BV at the end of the second year using the SL method.

An automated assembly robot that cost $352,000 has a depreciable life of 5 years with a $85,000 salvage value. The MACRS (Modified Accelerated Cost Recovery System) depreciation rates for years 1, 2, 3, and 6 are 20.00%, 32.00%, 19.20%, and 5.76%, respectively. What is the book value at the end of year 3? Year 5? Year 6? The book value at the end of year 3 is $ .   The book value at the end of year 5 is $ .   The book value at the end of year 6 is $ .

An automated assembly robot that cost $300,000 has a recovery period of five years with an expected $50,000 salvage value. If the MACRS depreciation rates for years 1, 2, and 3 are 20.0%, 32.0%, and 19.2%, respectively, what is the depreciation recapture, capital gain, or capital loss, provided the robot was sold after 3 years for $80,000?

2) The initial cost of a new m/c is $10,000. The annual operating cost is $1,000/yr for first 3 years, and then becomes $3,000/yr after that. The m/c needs a major repair at the end of 5th year, which costs $5,000. The m/c has 10 years useful life with salvage value of $4,000. Calculate EUAC for keeping the m/c for 10 years. (i=10%/yr)

An oven at a bakery is being considered for replacement. Its salvage value and maintenance costs for several years are given in the table below. The first cost of a new oven is $80,000, including installation, with a MARR of 10%. What is the economic service life for this oven?  Year Salvage value at end of year Maintenance costs 0 $20,000 $ 1 75,000 500 2 70,000 500 3 66,000 1000 4 62,000 3000

Two techniques can be used to produce expansion anchors. Technique A costs $90,000 initially and will have a $12,000 salvage value after 3 years. The operating cost with this method will be $33,000 in year 1, increasing by $2600 each year. Technique B will have a first cost of $113,000, an operating cost of $7000 in year 1, increasing by $7000 each year,and a $43,000 salvage value after its 3-year life. At an interest rate of 13% per year, which technique should be used on the basis of a present worth analysis? Notice that there are no revenues.   Please work out and do not use excel, however if you use excel please show how to input everything needed down to the formula, thank you!

A machine costs $210,000, has a $14,000 salvage value, is expected to last ten years, and will generate an after-tax income of $43,000 per year after straight-line depreciation. Compute the payback period.