View PoliciesCurrent Attempt in ProgressA project that cost $63000 has a useful life of 5 years and a salvage value of $3000. The internal rate of return is 12% and the annualrate of return is 18%. The annual net income isO $3960.○ $5940.O $5400.O $3600.Save for LaterAttempts: 0 of 1 usedSubmit Answer

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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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For the following alternatives compute the Delta B/C ratio of Alternative C minus Alternative A. Use 9% as MARR. (Remember for our convention, salvage value is a minus cost.) A B C D Initial Investment Annual Benefit Salvage Value Useful Life (years) $1,000 $350 $300 $2,500 $600 $800 $4,000 $650 $2,000 $2,000 $500 $600 7 O1.73 O1.57 O1.30 O1.29 O1.46 O1.59
A machine cost is 3000$ and salvage value 850$, if the rate of interest is 9% and number of years are 5 then what willI be the present worth for fixed output condition? Select one: a. 2448$ b. 2418$ c. 2427$
Solve step by step in digital format, explain How would you set up the function in Excel to get the IRR of an investment of $22,500 over 6 years, saving $1,600 per year, with a final payment of $7,800? Option A Option B Initial cost -16,700 -18,500 Annual operating cost -990 -1270 Rescue value 8600 9500 Useful life 10 10
Given the following data of equipment A and B. A First Cost P 50 000 P 150 000 2000 Salvage Value Annual Maintenance 6000 Economic life 6000 3000 5 15 The MARR is 12%. Use sinking fund depreciation. What is the rate of return for the additional initial investment on equipment B?
A project has a first cost of $14,000, uniform annual benefits of $2,400, and a salvage value of $3,000 at the end of its 10-year useful life. What is its net present worth at an interest rate of 12%? Select one: O a. $300 O b. $526 O c. $450 O d. $600 Next page R03-07.06.2021 Time:14.30 Jump to. Rabiu (Log out) _204 prime Video AK tu O 4x O ENG 14:38 24/06/2021
Using the following two relationships: AW = CR + A of AOC CR = -P(A/P,i,n) + S(A/F,i,n) %3D Calculate the Annual Worth (AW) based on the data for the following project: Corporate MARR = 10% Initial Investment Cost $1,000,000 Anticipated Project Life = 10 years Salvage Value at the end of 10 years $100,000 Annual Cost of Operation %3D $50,000 10% Compound Interest Factors
Two machines X and Y below are being considered for purchase. Machine X Machine Y 1200 300 200 Initial Cost Uniform annual benefit Salvage value Life, year 100 150 160 3 6 Find the exact incremental Rate of Retum? O 7.7%
Finance Machine X has an intial cost of $10,000. It is expected to last 12 years, to cost $200 per year to maintain and to have a salvage value of $1,000 at the end of its useful life. The equivalent uniform annual cost of the machine at 8% interest is most nearly ___________. A. $1,160 B. $1,507 C. $1,580 D. $1,475 E. None of the above Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism. Do not provide Excel Screet shot rather use tool table Answer completely.
Question 3. Choose the best piping systems as shown in the table below using Annual Worth method (MARR: 6%). Concrete piping UPVC piping Initial cost (OR) Ca Initial cost (OR) Annual 0&M (OR/y)" 10000 Annual O&M (OR/yr) 5000 Rehabilitation cost at the 120000 5th year Useful life (years)¤ 10 Useful life (years)= 20 Ca 1720102 1136000 1828000:
Check whether the project is accepted or rejected and recommend the best project, MARR=6% and n=8 years, using Payback period, ROI, NPV, Aw, IRR methods. ID Project I Project II increase Annual Initial Annual cost revenues initial Annual Salvage value annual salvage value running in cost revenues maintenance revnues cost 1910095 280000 60000 2000 28000 400000 72000 2000 40000 12000
a) Use Repeatability Assumption to compare between three alternatives X, Y and Z at (MARR=18% per year). Project Investment cost Production rate per year 20,000 Price per unit Annual expenses Salvage value Useful life (years) A В C $90,000 $50,000 $35,000 24,000 18,000 $6 $2.8 $4 $6,000 $5,000 $4,000 $10,000 $18,000 3
5. The data below are estimated for a project study. į = 10% Plan A Initial Investment Annual Operating Cost Life Salvage Value Annual Revenue Plan B Initial Investment Annual Revenue Annual Disbursement Life Salvage Value P 35,000 P 6,450 4 years none 19,000 P 50,000 P 25,000 P 13830 8 years none Which plan would you recommend? Use Present Worth Method and 8 years of study period. Profit for Plan A = 9047.85 Profit for Plan B =9, 591.13

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