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- 5-27image If produced by Method A, a product’s initial capital cost will be $100,000, its annual operating cost will be $20,000, and its salvage value after 3 years will be $20,000. With Method B there is a first cost of $150,000, an annual operating cost of $10,000, and a $50,000 salvage value after its 3-year life. Based on a present worth analysis at a 15% interest rate, which method should be used?Engineering economy - ENGR 3322 The International Parcel Service has installed a new radio frequency identification system to help reduce the number of packages that are incorrectly delivered. The capital investment in the system is $65,000, and the projected annual savings are tabled below. The system’s market value at the EOY five is negligible, and the MARR is 18% per year. Calculate the present worth of the project. a. $ 35,730 b. $ 36,730 c. $ 37,730 d. None of the choicesData for two alternatives are as follows: A B Investment ₱ 35,000 ₱ 50,000 Annual benefits ₱ 20,000 ₱ 25,000 Annual O and M ₱ 6,450 ₱ 13,830 Estimated life, years 4 8 Net Salvage value ₱ 3,500 0 Using an interest rate of 20%, which alternative should be chosen?
- 29. Solve Problem using annual worth method. CLC Publishing is considering the purchase of an offset printing machine. A study of three available alternatives shows the following estimates: A B C Acquisition Cost P700 000.00 P850 000.00 P1 000 000.00 Shipping Cost 5% 5% 8% Net Annual Benefit P155 000.00 P145 000.00 P170 000.00 Useful Life 8 years 16 years 16 years Salvage Value P35 000.00 P70 000.00 P80 000.00 (Ans. ATA= — P799.02; ATB = P3916.33; ATC = — P815.40)Don Garlis is a landscaper. He is considering the purchase of a new commercial lawn mower, either the Atlas or the Zippy. Construct a choice table for the interest rates of 0-100%. Atlas Zippy Initial cost $6,700 $16,900 Annual maintenance cost $1,500 $1,200 Annual benefit $4,000 $4,500 Salvage value $1,000 $3,500 Useful live, in years 3 6Municipal Engineer wants to evaluate three alternatives for supplementing the water supply. 1st alternative – continue deep well pumping at an annual cost of $10,500 2nd alternative – install a 10” pipeline from a surface reservoir. First cost is $25,000 and annual pumping cost is $7,000 3rd alternative – install a 20” pipeline from the reservoir. First cost of $34,000 and annual pumping cost of $5,000. Life of all alternatives is 20 years. For the second and third alternatives, salvage value is 10% of first cost. With interest at 8%, which alternative should the engineer recommend? Use present worth analysis PW (deepwell) = ? PW (10”pipeline) = ? PW (20”pipeline) = ?
- A project is being considered that has a first cost of $12,500, creates $5000 in annual cost savings, requires $3000 in annual operating costs, and has a salvage value of $2000 after a project life of 3 years. If interest is 10% per year, which formula calculates the project’s present worth? (a) PW = 12,500(P/F, 10%, 1) + (− 5000 + 3000) (P/A, 10%, 3) − 2000(F/P, 10%, 3) (b) PW = − 12,500 + (5000 − 3000) (P/A, 10%, 3 ) − 2000(P/F, 10%, 3) (c) PW = 12,500(F/P, 10%, 3) + (5000 − 3000) (F/A, 10%, 3) + 2000 (d) PW = − 12, 500 + 5000(P/A, 10%, 3) − 3000 (P/A, 10%, 3) + 2000(P/F, 10%, 3)Consider the three mutually exclusive alternatives below. A B C Initial Cost $350 $125 $200 Uniform Annual Benefit $90 $27.7 $80 Assume each alternative has a 5-year useful life and no salvage value. Based on a 5% interest rate and the benefit-cost ratio, which alternative should be selected? Group of answer choices A B No answer text provided. They are all equalPlease no written by hand and no emage Solve in excel Carp, Inc. wants to evaluate two machines for packaging their products.Machine A:Initial cost is $700,001st year O&M cost is 18,000; this cost increases $900 each year.The annual benefits are $154,000It can be sold at the end of 10 years useful life for $145,000 Machine B:Initial cost is $1,600,001st year O&M cost is 28,000; this cost increases $650 each year.The annual benefits are $300,000It can be sold at the end of 20 years useful life for $210,000The companies uses an interest rate of 15% Use annual cash flow analysis to decide which is the most desirable alternative.
- A series of alternative projects have the following series of discount annualized costs and benefits. Which project is your best choice? What is your second-best choice? Please list your calculation details. Project Alternative A B C D E Present Value of Benefits 100 160 90 70 180 Present Value of Costs 60 90 40 30 1204. A government is planning to implement a new traffic control and surveillance system. The maintenance cost of the system is $1,200,000 per year. It is believed that the new system can help 3,000 driver saving 8 hours per year. The average wage of the people in this country is $150 per hour. It is expected that the system can be used for 5 years, i.e., the first benefit and cost are B1 and C1 respectively, while the last are B5 and C5. (a) What is the value of time saved by the new system per year? (b) Suppose the market interest rate is 5%. Calculate the net present value of this system. Should the system be launched? (c) Suggest two reasons to explain why the market interest rate in part (b) may be too high to evaluate the project. Should the system be launched if the interest rate is adjusted according to your suggestions?SUBJECT: ENGINEERING ECONOMICS (a) Identify the Given and the Unknown or what is being asked in the problem (b)Provide the formula to be used (c)Show the complete solution. The final answer is already provided.. At 5% annual interest, what is the difference in the present and future value of P100 paid at the endof each year for 10 years and P100 paid at the beginning of each year? Answer: P value difference = P38.61 F value difference = P62.89