2) The initial cost of a new m/c is $10,0000. The annual maintenance cost is $3,000 for first 2 years, and then increases $1,000 every year after that ($3000 for 1st year, $3000 for 2nd year, $4000 for 3rd year, $5000 for 4th year, ...). The m/c has 10 years useful life with salvage value of $4,000. Calculate EUAC for keeping the m/c. (i=10%/yr)

2) The initial cost of a new m/c is $10,0000. The annual maintenance cost is $3,000 for first 2 years, and then increases $1,000 every year after that ($3000 for 1st year, $3000 for 2nd year, $4000 for 3rd year, $5000 for 4th year, ...). The m/c has 10 years useful life with salvage value of $4,000. Calculate EUAC for keeping the m/c. (i=10%/yr)

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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onsider the following financial informationabout a retooling project at a computer manufacturing company:• The project costs $2.5 million and has a five-yearservice life.• The retooling project can be classified as sevenyear property under the MACRS rule.• At the end of the fifth year, any assets held for theproject will be sold. The expected salvage valuewill be about 10% of the initial project cost.• The firm will finance 40% of the project moneyfrom an outside financial institution at an interestrate of 10%. The firm is required to repay the loanwith five equal annual payments.• The firm’s incremental (marginal) tax rate on theinvestment is 35%.• The firm’s MARR is 18%.With the preceding financial information,(a) Determine the after-tax cash flows.(b) Compute the annual equivalent worth for thisproject.
A "standard" model of a dozer costs $20,000 and has an annual operating expense of $450. The dozer will be replaced in 6 years when the salvage value is expected to be $2,000. A "super" model can be purchased for $25,000, but will have a salvage value of $7,000 when retired in 6 years. Its operating expenses are also $450 a year. The purchaser's other investment opportunities are 5%. Compare these alternatives by using the annual equivalent method.
Margaret has a project with a $30560 first cost that returns $4320 per year over its 10-year life. It has a salvage value of $3000 at the end of 10 years. If the MARR is 5 percent, what is the payback period of this project if cash flow is continuous?
Mr Dela Cruz is going to buy a new machine for manufacturing its product: Machine A has a first cost of P50,000; annual maintenance P6,000; life 12 years and salvage value of P2,000. Machine B has a first cost of P140,000; annual maintenance P2,500, life 36 years and salvage value P10,000. Money is worth 9%.Compute the present worth cost of B P156,060 P166,080 P186,070 P154,292
Margaret has a project with a $28 000 first cost that returns $5000 per year over its 10-year life. It has a salvage value of $3000 at the end of 10 years. If the MARR Is 15 percent, what is the annual worth of this project? (Ctrl) -
A building cost P8.5 million and the salvage value is P50,000 after 23 years. The annual maintenance cost is P85,000, costs of repair is P350,000 every 3 years. Find the capitalized cost if money worth 14% per annum.What is the renewal cost?What is the capitalized cost?
Annual maintenance costs of a tunnel considered to have an eternal life are $ 500.is worth. The interest rate is considered to be an average of 5% per year. CapitalizedSince the cost is estimated to be $ 15,000, how much is the initial investment cost of the tunnel?can it? (Answer: $ 5,000)
An injection molding system has a first cost of $200,000 and an annual operating cost of $69,000 in years 1 and 2, increasing by $5,500 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within its maximum useful life of 5 years. Using a MARR of 11% per year, determine the ESL and the respective AW value of the system. The ESL is ]year(s) and AW value of the system is $|
please as soon as possible . thank you
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!
Rework the given problem, assuming the following additional information:• The machine is classified as a seven-year MACRS recovery period. Thetax rate is 40%. and the after-tax MARR is 10%.

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