5-73 Given the following data, use present worth analysis to find the best alternative, A, B, or C $10,000 15,000...
2. Given the following data, if the interest rate is 10%, use present worth analysis to find the best alternative, A, B, or C. А в Initial cost, $ 10,000 15,000 12.000 Annual benefit. S 6,000 10,000 5.000 Salvage value, S 1,000 -2.000 3,000 Useful life, years 2
4. Using prese sing present worth analysis, choose the best alternative from the information below, assuming interest is 12%, and a 15 year useful life. Alternative Initial Cost Uniform Net Annual Benefit $75,000 $35,000 $55,000 Salvage Value $7,500 $ 6,000 $10,000 $20,000 $15,000 $40,000 NPWar NPWb= NPWC Best Alter 2. ཆར་བར་ དམ་ Solve for EUAW (A) i = 8% 1000 1500 2,000 2500
I wonder how can I sovle this problem fast. if you know any fast
way pleas show me. I know LCD of 2,3 and 4 is 12 for alternative A,
I have $10,000 which keep repeat and I need to write 5 times
P(P/F,i,n) for calculating present value of $10,000. this takes
long time. any suggesiton for solving this question faster.in exam,
I have access to table and calculator.
5-73 Given the following data, use present worth analysis to find...
Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 3% per year, compounded annually. Alternative A 340 60 Alternative B 870 182 Initial Cost Annual Benefit Salvage Value Useful Life (yrs) 78 106 Alternative A, because it costs $65.43 less than Alternative B, in terms of present worth Alternative B, because it costs $65.43 more than Alternative A, in...
Question 1 10 pts Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 7% per year, compounded annually. Initial Cost Annual Benefit Alternative A 480 100 1100 Alternative B 1,310 260 |168 3 Salvage Value 116 Useful Life (yrs) O Alternative B, because it only incurs the initial cost once every three years instead of every two years O...
Problem (2): Consider the following three mutually exclusive alternatives. MARR is 10%. Alternative 1 10,000 Alternative 2 14,500 Alternative 3 20,000 $3,000 increasing by 500 each year thereafter negligible $5,000 Initial investment Annual yielded returns Salvage Value Service life $5,000 $5,000 negligible 6 a) Compute the payback (PB) period and discounted PB period of each alternative. Based on the PB period, which alternative do you recommend? b) Using Annual-worth analysis, which alternative do you recommend?
Given the following two alternatives, the present worth (PW) of E2 is closest to: Hint: Use positive sign for cash outflow and negative sign for cash inflow. Alt. E1 Alt. E2 Capital Investment, $ 6,000 12,000 Annual Expenses, $ 150 175 Useful life, years Salvage value, $ none 4,000 Less than $10,000 Between $10,000 - $10,500 Between $10,500 - $11,000 O Greater than $11,000
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 12% per year and a study period of 10 years. с $-44,000 $-12,000 $-34,000 $-7,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-1,500 $-1,200 $5,000 10 $1,200 5 The present worth of alternative C is $ -134497.32 and that of alternative D is $...
CALCULATE FOR B
PROBLEM The following costs are associated with three tomato-peeling machines being considered for use in a food canning plan Machine A S52,000 15,000 Machine B $67,000 12,000 Machine C $63,000 9,000 First cost Annual Maintenance & Operating costs Annual increase starting in year2 Annual benefit Salvage value Useful life, in years 38,000 13,000 4 37,000 22,000 12 250 31,000 19,000 If the canning company uses a MARR of 12%, which is the best alternative? Show your analysis...
show me through an equation please. thank you.
3. Using present worth analysis, decide which of the two options shown below is the best at 6% annual interest. First cost O&M Annual benefits Overhaul at the end of year 4 Salvage value Useful life, in years Machine A $15,000 $600 $5,000 $0 $1,000 Machine B $20,000 $400 $5,200 $1,000 $2,000 4