Question 2 Compute the Equivalent Annual Worth of this alternative. The annual interest rate is 8%. The life of this alternative is 4 years. First cost: $93,000 Annual cost: $7,000 Annual income: $19,000 Salvage value: $11,000 O a. EAW = $617 O b. EAW = -$13,636 OC. EAW = $3,148 O d. EAW = $6,158
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- E2 A steel bridge on Louisiana state highway near the Gulf of Mexico is costing $450.000 yearlyin maintenance large chipping, priming, and painting. It originaly cost $1.600.000 when it wasbuilt 15 years ago. The Louisiana bridge engineers estimate that its remaining life is 10 years,then it will need to be replaced because of increased traffic. Its salvage value at any point intime is zero, because the cost of demolition will most like equal its value as scrap steel.A concrete bridge is considered to be the best challenger. It will cost $3.000.000 to build and$100.000 annually in maintenance costs. Its estimated life is 50 years. Its resale value may becounted as zero at any time during its life.No taxes of any kind will be considered for this government project. All costs are in constantdollars of year 0. Inflation may be ignored. Assume that annual benefits for either structure areexactly the same. A discount rate of 10 percent is to be used in analysis.(a) What is the economic life…Please 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 $210,000The companies uses an interest rate of 15% Use annual cash flow analysis to decide which is the most desirable alternative.Solar plant investment or cost establish 40000$, Maintenance Cost Expected customer Demand (KWH) Year 0 0 0 350 1500 1 300 1200 2 200 1000 3 200 800 4 150 700 5 What is the cost Consumption Kilowatt Per Hour? please tell me how can I solve it , as you know we have Present worth, annual worth and future worth! please i need correct answer because i have post it before and they give me wrong answer , do not search it because these number is randomly , i am looking for the way of answer in case of the question has change that i can solve it by your clarification . rate 10%
- A factory with capacity of 700,000 units per year operates at 62% capacity. The annual income is $430,000.00. Annual fixed costs are $190,000.00 and the variable costs are $0.348 per unit. a) What is the current loss or profit? b) What is the breakeven point?.Two lathes are being considered in the manufacture of certain machine parts. Data is given below, all cost in peso: LATHE A LATHE B First Cost 40,000 56,000 Salvage Value 5,000 7,000 Annual Maintenance 2,000 2,800 Operation, Cost/hour 4 3.5 Life, in years 10 12 Time per part (hours) 0.40 0.25 REQUIRED: Determine the number of machine parts/year that could be produced so that 2 lathes will be equally economical if the MARR is 18%. Use AWM If the number of parts is 10,000 units, which lathe will you recommend? Use ROR If the number of parts is 10,000 units, which lathe will you recommend? Use PWM If the number of parts is 10,000 units, which lathe will you recommend? Use EUACIf the MARR=10%, compute the value of X that makes the alternatives equally desirable. Do not use spreadsheets. Alternatives Machine A Machine B First cost $12,000 $20,000 Annual Operating cost $1,400/year X Salvage value $2,000 $3,000 Life 4 years 8 years
- Motor A and B is providing 150hp and it is being used for 10 hours per day, 250 days in a year if the motor is using the electricity. Additional cost for both motors is the annual maintenance cost worth 1200 and other manufacturing overhead costing 550 per month. Motor A has an original price of 50,000 while 45000 for Motor B, and both have the salvage value of 3000 after 5 years. Machine efficiency is 92% and 90% respectively.A. How much is the electrical cost if the money is worth 15%? Electrical cost = pesos ( 2 decimal places ) B. With the same initial cost, salvage value, expected life, number of working days per year, annual maintenance cost and manufacturing overhead. How much is the fuel cost if the interest rate is 10% compounded annually if motor A needs to be refilled by 2L/ output while motor B requires a 3L/output, and the target output per day is 10 outputs?Given the two machines’ data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value 0 2,000.00 Annual operation 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B: note:round off final answer to 2 decimal ANSWER for ALTERNATIVE A: ANSWER for ALTERNATIVE B:Alternative 3 is incorrect EUAC=( Equivalent Annaul COst of Initial Investment)+ (Expected Moderate annaual flood damage cost )+ (expected severe annaual flood cost )
- Given the two machines’ data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value 0 2,000.00 Annual operation 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B: ANSWER for ALTERNATIVE A: Blank 1 ANSWER for ALTERNATIVE B: Blank 2The First cost of a passenger bus is ₽800,000. Its estimated life is 4 years with no salvage value. The operating cost per year of 300 days of operation are as follows: Tires, ₽24,000; gasoline, 60 liters per day at ₽15 per liter; Oil,₽80 per day; maintenance and repair, ₽40,000; labor ₽800 per day; miscellaneous expense, ₽18,000; depreciation, sinking fund at 10%. If the average passenger fare is ₽3.50 for each passenger one way and expected profit or return is 15% determine the minimum average number of passengers that should be transported each day. Use the AW method and Future worth method.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 choices