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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- 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 $-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=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% $20,000A 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 6
- Advanced 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 searchDexcon 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 yearsYour 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