we calculate the Net Present Value of the project as = Present value of Inflow- Present Value of Outflow.
Year | Net Inflow ($) (a) |
PV Factor @ 17% (b) |
Present Value ($) (a*b) |
1 | 3000 | 0.8547 | 2564.1 |
2 | 3000 | 0.7305 | 2191.5 |
3 | 3000 | 0.6244 | 1873.2 |
4 | 3000 | 0.5337 | 1601.1 |
5 | 3000 | 0.4561 | 1368.3 |
6 | 1900 | 0.3898 | 740.62 |
10,339 | |||
Investment | 9,500 | ||
Net Present Value | 839 |
Future Worth of Project is $839*2.5652 = $2152
Determine the FW of the following engineering project when the MARR is 17% per year. Is...
Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? a 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 project is $. (Round to the nearest dollar.)
Determine the FW of the following engineering project when the MARR is 15% per year. Is the project acceptable? Project Investment Cost $10,000 Expected life 5 years Market (Salvage) Value* -$1,000 Annual Receipts $8,000 Annual Expenses $4,000 * A negative market value means that there is a cost to dispose of an asset.
Question (1) Use FW method, MARR 15% Project Investment Cost 10,000 Expected Life 5 years Market (Salvage) Value* -1,000 Annual Receipts 8,000 Annual Expenses 4,000 Answer: FW(15%) = - 10,000 (2.0114) + 4000(6.7424) - 1,000 = 5,855.60 Is this project acceptable? Yes, but I would like to see more profit.
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%)...
0.8. When MARR = € 15% per year, determine whether the project with the following cash- flow diagram is acceptable. Use ERR method to make your decision. $7,000 $7,000 $7,000 $7,000 | End of Year $6,000 $14,000
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...
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...
Solve for A and B, Engineering Economy please solve it right! Question Help %) Problem 6-52 (algorithmic) Compare alternatives A and B with the present worth method if the MARR is 15% per year. Which one would you recommend? Assume repeatability and a study period of 20 years. $40,000 $7,000 at end of year 1 and increasing by $700 per year thereafter $7,000 every 5 years $15,000 $14,000 at end of year 1 and increasing by $1,400 per year thereafter...
(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...
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...