Date Table 2 (MARR-10%) First cost, S Annual cost, S per year Salvage value, S Life, years -40,00...
Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2 Machine 1 20,000 9000 4000 First cost,S Annual cost,S/year Salvage value, S Life, years 100,000 -7000 Infinite A.-20000(A/P 1096,3)-9000+4000(A/F,1096,3) B. $-15832.40 C. $-170,000 Equation for AWI= (Answer 2 decimals) $ | CC2- $ Selection= E. $-180,000 F 2 G. $-166,540 Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2...
please show work that is easy to UNDERSTAND 3. A machine is purchased for S70,000. Life is 10 years with a S 10,000 salvage. MARR is 10%, and the tax rate is 50%. Cash operating costs are $4500 per year. Five year MACRS (20, 32, 19.2. 11.52, 11.52,5.76) depreciation will be used. Calculate the annual equivalent revenue requirements for this machine. 3. A machine is purchased for S70,000. Life is 10 years with a S 10,000 salvage. MARR is 10%,...
Purchase $320.000 Lease Alternative Initial Cost Lease Annual Operating costs Salvage Value Life, years $40 000 $7.000 $8500 $ 80.000 5 5 c. PW of Purchase Option a. PW of the lease Option 0. Salvage value of purchase Option Decision to Purchase or Lease 567.942 9-193,337 C5-309 5515 5-49,072 5-319.327 * Lease You are given the following is about two machine with 10% Com First costs 30 000 Antal maintenance con 111000 Persodi me cost every years Salvare values 6000...
NOT IN EXCEL 7.For the following table assume a MARR of 6% per year, and a useful life for each alternative of six years. Which equals the study period. The rank order of alternatives from least capital to greatest capital investment is A, C, B. Complete the PW-Incremental Analysis by selecting the preferred alternative. (5ptos) V A Capital Investment A Annual Revenues A Annual Cost A Market Value APW A $15,000 4,000 1,000 6,000 $3,982 A(C-A) A(B-C) $2,000 $3,000 900...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Date: 19 March 2019 No: Name: EGN 3212 Engineering Economy LO3- PW & AW Analysis ,000, and Annual 1. For a project following information is known: Life 20 years, Initial cost $ 20 (10 marks] Maintenance-S 1500 per Calculate the pw of this project for-12%. 2. Refer to cash flow diagram for 1-12%. Calculate Aw for 1 life cycle. (10 marks] 0 4 5 Al-5 = $3,000 $5,000 Date: 19 March 2019 No: Name: EGN 3212 Engineering Economy LO3- PW...
11.14 A piece of onboard equipment has a first cost of $600,000, an annual cost of $92,000, and a salvage value that decreases to zero by $150,000 each year of the equipment's maximum useful life of 5 years. Assume the company's MARR is 10% per year. (a) Determine the ESL by hand. (b) Use a spreadsheet with a graph indicating the capital recovery, AOC, and total AW per year to determine the ESL
For the following table, assume a MARR of 12% per year and a useful life for each alternative of eight years which equals the study period. The rank-order of alternatives from least capital investment to greatest capital investment is Z ·Y? W? X Complete the incremental analysis by selecting the prefer ed altemative. Do nothing" is not an option. $250 $400 $100 Capital investment ? Annual cost savings ? Market value ? PW (12%) 70 100 138 90 50 67...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...