Required information The TT Racing and Performance Motor Corporation wish to evaluate two alternative machines for NASCAR motor tune- ups. Machine R S First cost, $ -290,000 -350,500 k Annual operating cost, $ per year -40,000 -50,000 Life, years 5 Salvage value, $ 21,900 16,000 t ht Use the AW method at 9% per year to select the better alternative. The annual worth of machine R is $- 67885, and the annual worth of machine S is $- 37437 ences The better alternative is machine S
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- The AW method is to be used to select the better alternative of the two machines listed below, at 10% per year. Answer the below questions: Machine R Machine S First cost, $ 250,000 497,732 Annual operating cost, $ per year 40,000 50,000 Salvage value, $ Life, years 20,000 30,000 3. 5. The AW of Machine S=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, $ R 8 -261,000 -40,000 14 21,100 Use the AW method at 9% per year to select the better alternative. The annual worth of machine R is $-[ The better alternative is machine R -325,500 -50,000 5 18,900 and the annual worth of machine S is $-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.
- Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis? Machine X Machine Y First cost, $ -200,000 -400,000 Annual operating cost, $ per year -40.000 -50,000 Salvage value, $ 80,000 90,000 Life, years 3 Solution: PW Machine x =5 (Use the right sign) PW Machine Y = 5 (Use the right sign) So, the best alternative is MachineReference: Case Study S Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below. Initial cost Benefits/year Machine X $120,000 $20,000 for the first 10 years and $9,000 for the next 10 years Life Salvage value $40,000 MARR 5.2. The NPW of machine X is A) $35,158 B) $48,192 C) $50,752 Machine Y $96,000 $12,000 per year for 20 years. 20 years 8% $20,000Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (i.e., low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 9% per year, which alternative should be selected? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-100,000 $-30,000 $2,000 3 years LS $-300,000 $-5,000 $15,000 6 years
- A manufacturing company is trying to decide between the two machines shown below. Determine which machine should be selected on the basis of rate of return. Assume the MARR is 20% per year. Machine A Machine B Initial Cost, $ -18,000 -35,000 Annual operating cost, $/year -4,000 -3,600 Salvage value, $ 1,000 2,700 Life, years 3 6Advanced Technologies, Inc. is evaluating two alternatives to produce its new plastic filament with tribological (low friction)properties for creating custom bearing for 3-D printers. The estimates associateu with its alternatives are shown below. Use LCM way and a MARR of 20% per year. Method DDM LS First Cost $ -165,000 -375,000 M&O cost, $/year -50,000 -25,000 30,000 Salvage value $ Life, years 2 26°C Mostly clear P Type here to searchYour boss has told you to evaluate the cost of two machines.After some questioning, you are assured that they have thecosts shown at the right. Assume:a) The life of each machine is 3 years.b) The company thinks it knows how to make 14% oninvestments no riskier than this one.Determine via the present value method which machine topurchase. MACHINE A MACHINE BOriginal cost $13,000 $20,000Labor cost per year 2,000 3,000Floor space per year 500 600Energy (electricity) per year 1,000 900Maintenance per year 2,500 500Total annual cost $ 6,000 $ 5,000Salvage value $ 2,000 $ 7,000
- 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 comparisonA firm is trying to decide which of two machines to purchase as described in in the table below. Using the interest rate 9% or 0.09, use present worth analysis to determine which machine, if either, should be purchased. Show all your work. Machine: A - B First Cost: $800 - $600 Annual Net Benefit: $130 - $230 Salvage Value: $40 - $20 Usefil Life (years): 9 - 3Emerson Electric manufactures compressors for air conditioners. It needs replacement equipment to improve one of its manufacturing lines. Select between two options using the MARR of 14% per year and a future worth analysis for the expected use period. What are the future values of each option? Option First cost, S A B -64,000-76,000 -16,000-22,000 AOC, $ per year Expected salvage value 8,000 11,000 Expected use, years 3 6