Capital recovery cost= 10000 * (A/P, 6%,6) - 1500 * (A/F, 6%,6)
= 10000 * 0.203363 - 1500 * 0.143363
= 1818.58
A machine costs $10,000. The salvage value after 6 years will be $1.500. If the machine...
3. A machine tool costs $10,000 and does not have any salvage value. The manufacturer's warranty covers the maintenance and repair cost for the first year. The maintenance and repair costs for the second year will be $1,000 and will increase for $500 for the subsequent years. The operating costs for the first year will be $400 and will increase for $400 for the subsequent years. Consider an analysis period of 8 years and interest rate of 9%, and find...
15) A machine costs $50,000 today and has an estimated scrap (salvage) cash value of $10,000 after ten years. Inflation is 6% per year. The effective annual interest rate earned on money invested is 296. How much money needs to be set aside each year to replace the machine with an identical model ten years from now? (Hint #1: first determine the cost of the machine after inflation of the price for 10 years). (Hint #2: the salvage value is...
7. (10 points) A new production machine costs $120,000 and will have a $10,000 salvage value when disposed of in ten years. Annual repair costs are expected to be $0 in years 1, 2, 3 and 4, $5000 in year 5, $5500 in year 6, $6000 in year 7, $6500 in year 8, $7000 in year 9, and $7500 in year 10. If interest is 10%, what is the equivalent uniform annual cost of the new machine?
Machine A was purchased last year for $10,000 and had an estimated market value of $1,900 at the end of its 6-year life. Annual operating costs are $1,450. The machine will perform satisfactorily for the next five years. A salesman for another company is offering Machine B for $59,000 with an market value of $5,900 after 5 years. Annual operating costs will be $1,050. Machine A could be sold now for $16,000, and MARR is 50% per year. Using the...
A machine has an initial cost of $10,000 and a salvage value of $2,000 after a 5 year life. Annual benefits from using the machine are $5000 and the annual cost of operation is $1800. The tax rate is 50%. Assume straight line depreciation. Find the IRR? Show full steps, and equations, neatly, thanks
The cost of a machine is $10,000. The annual operation cost and maintenan machine is expected to save $1000 per year in labor costs. The salvage value a $3.000 Calculate machine's equivalent uniform annual worth (EUAW) at an interest rate of 50% The total equivalent uniform annual worth (EUAW) of an asset is given by: EUAW = EUAB (benefits) - EUAC (costs)
A piece of equipment costs $15,000 and is expected to have a salvage value of $1,000 in 6 years. If interest is at 15%, what is the cost of Capital Recovery plus Return? (Use MARR = 12% and tax rate = 40% in the problem unless otherwise stated.) - I don't think this applies to this problem.
3(a) A machine has a life of 20 years, costs $200,000 and has an estimated salvage value of $10,000. (i) For the Straight Line method of depreciation, what is the depreciation rate and what is the book value at the end of year 10? (ii) If the declining balance method is to be used, at what depreciation rate (i.e. the capital cost allowance CCA rate), the book value at the end of year 10 will be the same as for...
A company must decide whether to buy Machine A or Machine B. After 5 years Machine A will be replaced with another A. The initial cost for Machine A is $12,500, annual maintenance is $1,000, and the salvage value at 5 years is $10,000. Machine B has an initial cost of $20,000, 0 maintenance costs, and a salvage value of $10,000 at 10 years. Which machine should be purchased? Use a MARR of 10%.
10) You buy a machine now for $10,000. The machine can be depreciated using DDB depreciation with 5 years useful life and $2,000 salvage value. Three years later you sell the machine for $1,000. The annual net benefit is $8000. The inflation rate is 5% per year. The acceptable real after-tax rate of return after taking inflation into consideration is 10%. Combined incremental tax rate is 50%. Calculate PW of this investment a) $4102 b) $3654 c) $3181 d) $2706...