MANUAL SOLUTIONS NOT EXCELDOWNVOTE IF EXCELDetermine the difference between the capitalized cost of the timber and steel penstock for a hydroelectricplant with interest of 10%:Timber SteelPhp50,00010 yearsPhp2,000Php80,00030 yearsNoneFirst CostEstimated LifeScrap ValueAnnual Maintenance Php1,200 Php200Show calculations for the capitalized cost of each item separately.

Capitalized cost refers to the present value of cash outflow related to the asset. we discount the…

MANUAL SOLUTIONS NOT EXCEL DOWNVOTE IF EXCEL Determine the difference between the capitalized cost of the timber and steel penstock for a hydroelectri plant with interest of 10%: Timber Steel First Cost Php50,000 Php80,000 Estimated Life 10 years 30 years None Scrap Value Php2,000 Annual Maintenance Php1,200 Php200 Show calculations for the capitalized cost of each item separately.

MANUAL SOLUTIONS NOT EXCEL DOWNVOTE IF EXCEL Determine the difference between the capitalized cost of the timber and steel penstock for a hydroelectri plant with interest of 10%: Timber Steel First Cost Php50,000 Php80,000 Estimated Life 10 years 30 years None Scrap Value Php2,000 Annual Maintenance Php1,200 Php200 Show calculations for the capitalized cost of each item separately.

FINANCIAL ACCOUNTING

10th Edition

ISBN:9781259964947

Author:Libby

Publisher:Libby

Chapter1: Financial Statements And Business Decisions

Section: Chapter Questions

Problem 1Q

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MANUAL SOLUTIONS NOT EXCELDOWNVOTE IF EXCEL Determine the difference between the capitalized cost of the timber and steel penstock for a hydroelectric plant with interest of 10%: Timber Steel First Cost Php50,000 Php80,000 Estimated Life 10 years 30 years Scrap Value Php2,000 None Annual Maintenance Php1,200 Php200 Show calculations for the capitalized cost of each item separately.
Its required to select one of the two machines, if you know that the firms MARR-12% and if the costs are shown below: Project A Project B Initial cost, $ 7650 12900 Maintenance cost, S/year 1200 900 Salvage value, $ 2000 Economic life, year 4 Compare the two-alternative using: 1-Equavlent Annual worth comparison. 2- Present worth comparison
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR-10% Machine B 27,724 6,000 Machine A Machine C First cost, $ 15000 10000 Annual cost, S/year Salvage value, $ Life, years 8,293 4,000 4,000 5,000 1,000 Answer the below questions: B- AW for machine B= For the below ME alternatives, which machine should be selected based on the AW analysis. MARR 10%. Machine A 15000 Machine B Machine C 12,409 30000 First cost, $ Annual cost, $/year Salvage value, $ Life, years 17,181 6,000 4,000 4,000 5.000 1,000 Answer the below questions: C- AW for machine C=
RLC Manufacturing is planning to purchase a cutting equipment. Information are as follows: Equipment 1 Equipment 2 First Cost P 12,000 P 18,000 Salvage Value P 600 P 2,000 Annual Operation P 3,200 P 2,500 Annual Maintenance P 1,200 P 1,000 Taxes & Insurance 3% 3% Life, years 10 15 Money is worth at least 16%. Which equipment should be selected? Use: a. Rate of Return Method Rate of Return Method Annual Cost Method NOTE: Show cashflow diagram.
Assignment 1 a) LIFE CYCLE COSTS FOR PROPOSED BRIDGE REPLACEMENT STEEL PANEL TRUSS DESIGN Year Activity Cost 0 Bridge replacement 595,000 15 Replace asphalt paving 15,000 30 Replace asphalt paving 15,000 40 Replace deck, regalvanize trusses, 306,000 55 Replace asphalt paving 15,000 70 Replace asphalt paving 15,000 Calculate the present value of the costs for the above project cash flow with a discounting rate of 5% b)THREE-SPAN PRE-STRESSED GIRDER DESIGN Year Activity Cost 0 Bridge replacement…
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A 15000 B- AW for machine B= 8,342 4,000 Answer the below questions : Machine B 6 20,447 6,000 5,000 Machine C 10000 4,000 1,000
Given the two machines' data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value Annual operation 2,000.00 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B: ANSWER for ALTERNATIVE A: Blank 1 ANSWER for ALTERNATIVE B: Blank 2
Given the two machines’ data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value 0 2,000.00 Annual operation 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B:
Three alternatives have the following cost data associated with them: Data Alt. 1 Alt. 2 Alt. 3 Useful Life, Years 10 10 10 First Cost $1,125,000 $1,980,000 $1,650,000 Annual Benefit 265,000 589,000 435,000 Annual M&O Costs 95,000 97,000 91,000 Salvage Value 145,000 205,000 178,000 Determine whích machine is better to purchase, if i = 10%. 1. Use a B/C 2. Use a Payback Period
[The following information applies to the questions displayed belad Project Y requires a $321,000 investment for new machinery wityy vifour-year life and no salvage value. The project yield owing annual results. Cash flows occur evenly withit veach year. ( Annual Asounts Sales of new product Expenses Materials, labor, and overhead (except depreciation) Depreciation-Machinery Selling, general, and administrative expenses Incone 2. Determine Project Y's payback period Project Y Project Y Numerator: Initial investment $ 3. Compute Project Y's accounting rate of return Not present value Payback Period T Denominator: 7 Annual net cash flow Numerator 321,000 $ 180,000 Accounting Rate of Return Denominator Net Cash Flows Present Value of Annuity at a Project Y $ 375,000 80,250 4. Determine Project Y's net present value using 7% as the discount rate. (Do not round intermediate calculations. Round your present value factor to 4 decimals and final answers to the nearest whole dollar) S $ 99,750…
A piece of production equipment is to be replaced immediately because it no longer meets quality requirements for the end product. The two best alternatives are a used piece of equipment (E1) and a new automated model (E2). The economic estimates for Alt E1 Alt E2 Capital Investment PhP 14,000 PhP 65,000 Aппual Expenses 14,000 9,000 each are shown in the accompanying table. Useful Life (years) 5 20 Market Value (at the end of useful life) 8,000 13,000 The MARR is 15% per year. Which alternative is preferred, based on the (a) coterminated assumption with a five-year study period and an imputed market value for Alternative b? Use FW. ANSWER: FW (E1) = PhP Blank 1 FW (E2) = PhP Blank 2 The best alternative is EBlank 3
Calculate the NPV of an asset replacement given the following information: Investment = $180,000; salvage recovered now = $70,000; operating cash flows = $180,000; PV of salvage forgone = $5,500; PV of salvage recovered = $14,000; CCATS = $35,000. Multiple Choice $119,500 $121,500 $115,500 $113,500 $117,500