Compute the present value for the alternative below if the analysis period is 8 years: Alternative: • First cost: 5000 • Uniform annual benefit: 1800 • Useful life in years: 4 Answer:
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- In AW OF MACHINE A FIRST COST - 27056 ANNUAL COST - 12.786 Salvage value - 4172 Life year - 3 In AWA MACHINE B first cost - 102616 Salvage value - 9501 Life - Infine ( all values in $) Sorry may b picture is not so clear please tell me how to calculateAs an engineer, you are responsible to find the best alternative that is economically best. The cash flows shown in the table and the MARR is 11 % per year. Then please answer the following questions. Cash Flow Project 01 Project 02 Project 03 First Cost, $ -90000 220000 230000 Annual Cost, $ Per Year -60000 -21000 -17000 Overhaul Every 10 Years, $ _ _ -60000 Salvage Value, $ 8000 29000 11000 Life, Years 3 10 a) Find the Aw for Project 01. b) calculate the AW for Project 02 c) Calculate Aw for Project 03. What is your decision?For the cash flows shown, determine the incremental cash flow between machines B and A for (a) year 0, (b) in year 3, and (c) in year 6. Machine First Cost, $ АОС, $/year Salvage value, $ 3,000 Life, years B -15,000 -25,000 -1,600 -400 6,000 3 4 O (a) = -10000 (b) = -1800 (c) = 1200 %3D !3! O (a) = -10000 (b) = 13200 (c) = 4200 !3! %3! O (a) = -10000 (b) = 13200 (c) =6000 %3! %3D O (a) = -10000 (b) = 13200 (c) = 13200 %3D
- Required information The TT Racing and Performance Motor Corporation wish to evaluate two alternative machines for NASCAR motor tune- ups. Machine First cost, $ Annual operating cost, $ per year Life, years Salvage value, $ -254,000 -40,000 The better alternative is machine S 3 20,400 Use the AW method at 9% per year to select the better alternative. The annual worth of machine R is $- S -360,500 -50,000 5 19,600 47694, and the annual worth of machine S is $- 10970! Required information A company that manufactures magnetic flow meters expects to undertake a project that will have the cash flows estimated. First cost, $ Equipment replacement cost in year 2, $ Annual operating cost, $/year Salvage value, $ Life, years -740,000 -300,000 -790,000 250,000 4 At an interest rate of 10% per year, what is the equivalent annual cost of the project? Find the AW value using tabulated factors. The equivalent annual cost of the project is $- 1,007,097.68.A solid-waste recycling plant is considering two types of storage bins. Use ROR evaluation and an MARR of 46% per year to determine which should be selected. Storage Bin First Cost, $ AOC, $ per Year Salvage Value, $ Life, Years P -24,000 -4000 1800 3 Q -35,000 -2000 2600 6 Since A/* is (Click to select) MARR of 46%, (Click to select) should be selected.
- Buying Equipment 1 from XYZ company and company ABC will give the production similar productivity input of 400,000.00 per year. The equipment from company XYZ has a purchase price of 200,000.00, annual maintenance of 5,000.00, and production life of 10 years, while the equipment from company ABC has a purchase price of 100,000.00, annual maintenance of 2,000.00, and production life of 12. Using present worth method, what is the present values of their purchase profit? Use MARR of 20%Buying Equipment 1 from XYZ company and company ABC will give the production similar productivity input of 400,000.00 per year. The equipment from company XYZ has a purchase price of 200,000.00, annual maintenance of 5,000.00, and production life of 10 years, while the equipment from company ABC has a purchase price of 100,000.00, annual maintenance of 2,000.00, and production life of 12. Using present worth method, what is the total present worth of each alternatives? Use MARR of 20%In comparing the machines on a present worth basis, the present worth of machine X, the cost of money is 10% per year, is closest to Machine X Machine Y Initial cost, $ -66,000 -46,000 Annual cost, $/year Salvage value, $ Life, years -10,000 -15,000 10,000 24,000 3. $-103,908 $-87,840 None of the answers S-65,270
- Two alternatives are being considered for the upgrades to a detector at an X-ray synchrotron facility. Which alternative would be selected based the present worth, if the interest rate is 10% per year? Please note: The alternatives are mutually exclusive! Alternative A B First Cost, $ Operational & Maintenance Costs, $ -25,000 ||-10,000 Salvage Value, $ Life, years |-100,000|-140,000 20,000 30,000 4 12 Alternative A, PW= -$250,349 а. Alternative A, PW= -$355,931 Ob. Alternative B, PW = -$198,578 C. Alternative A, PW= -$120,968 Od.At i = 10% per year, a machine has a first cost of $8,000 and estimated operating costs and year-end salvage values as shown. The annual equivalent cost if you operate the machine for 3 years is closest to which value? Year 1 234N 5 Operating Cost, $ -1000 7000 -1200 5000 -1500 4200 -2000 3000 -3000 2000 Salvage Value, $Question 5: A metallurgical engineer is considering two materials for use in a space vehicle. All estimates are made. (a) Which should be selected on the basis of a present worth comparison at an interest rate of 12% per year? Vishes First cost, $ Maintenance cost, $ per year Salvage value, $ Life, years Material X Material Y -15,000 -35,000 -9,000 2,000 -7,000 20,000 5 5