Hi! Please check the attached picture for the solution.
Alternative J = $23.50 million
Alternative K = $30.51 million
Alternative J should be selected.
Thank you.
Use the imputed market value technique to determine the better alternative below. The MARR is 12%...
Use the imputed market value technique to determine the better alternative below. The MARR is 13% per year and the study period is six years. Capital Investment, millions Annual Expenses, millions Useful Life, years Market Value (End of useful life) Alternative J 41 11 6 0 Alternative K 66 18 10 0 Click the icon to view the interest and annuity table for discrete compounding when the MARR is 13% per year. The present worth of Alternative J over six...
Use the imputed market value technique to determine the better alternative below. The MARR is 10% per year and the study period is six years. 0 Alternative Alternative K Capital Investment, millions 47 62 Annual Expenses, millions 11 15 Useful Life, years 6 9 Market Value (End of useful life) Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. The present worth of Alternative J over six years is...
Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
A city is spending $19.5 million on a new sewage system. The expected life of the system is 50 years, and it will have no market value at the end of its life Operating and maintenance expenses for the system are projected to average $0.4 million per year. If the city's MARR IS 11% per year, what is the capitalized worth of the system? The study period is 100 years. Click the icon to view the interest and annuity table...
engineering economy
The AW of Alternative A is?
The AW of Alternative B is?
Two mutually exclusive alternatives are being considered. The MARR is 15% per year. General inflation is 4.5% / year Based on the data below, perform an appropriate analysis to select the most economical alternative. Assume that the market value grows at the general inflation rate. Alternative A Alternative B 51700D5240,000 Initial investment Annual revenue (actual $) $43,000 $48,000 $3,000 in year 1 increasing by $300 each...
Evaluate a combined cycle power plant on the basis of the PW method when the MARR is 15 % per year . Pertinent cost data are as follows. Power Plant (S) Investment cost $11,000 Useful life 12 years $4,000 S1,000 Market value (EOY 12) Annual operating expenses Overhaul cost-end of 5th year Overhaul cost-end of 10th year $200 $650 Click the icon to view the interest and annuity table for discrete compounding when the MARR is 15% per year. PW(15%)...
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...
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...
Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? Investment cost Expected life Market (salvage) value Annual receipts Annual expenses Proposal A $9,500 6 years -$1,100 $7,000 $4,000 A negative market value means that there is a net cost to dispose of an asset. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 17% per year. The FW of the following engineering...
(a) Assuming an infinite planning horizon, which project is a
better choice at MARR = 12%\
The present worth for project B1 is $ thousand
The present worth for project B2 is $ thousand
(b) With a 10 year planning horizon, which project is a better
choice at MARR = 12%
Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the...