show the formula Homework Answers
| Petroleum Based Feedstock | |||||
| Year | First Cost | Annual cost | Salvage | Annual Revenue | Cash Flow |
| 0 | -350000 | -350000 | |||
| 1 | -130000 | 300000 | 170000 | ||
| 2 | -130000 | 300000 | 170000 | ||
| 3 | -130000 | 300000 | 170000 | ||
| 4 | -130000 | 300000 | 170000 | ||
| 5 | -130000 | 300000 | 170000 | ||
| 6 | -350000 | -130000 | 45000 | 300000 | -135000 |
| 7 | -130000 | 300000 | 170000 | ||
| 8 | -130000 | 300000 | 170000 | ||
| 9 | -130000 | 300000 | 170000 | ||
| 10 | -130000 | 300000 | 170000 | ||
| 11 | -130000 | 300000 | 170000 | ||
| 12 | -130000 | 45000 | 300000 | 215000 | |
| NPW | 882604.13 | ||||
| Inorganic Based Feedstock | |||||
| Year | First Cost | Annual cost | Salvage | Annual Revenue | Cash Flow |
| 0 | -120000 | -120000 | |||
| 1 | -60000 | 290000 | 230000 | ||
| 2 | -60000 | 290000 | 230000 | ||
| 3 | -60000 | 290000 | 230000 | ||
| 4 | -120000 | -60000 | 34000 | 290000 | 144000 |
| 5 | -60000 | 290000 | 230000 | ||
| 6 | -60000 | 290000 | 230000 | ||
| 7 | -60000 | 290000 | 230000 | ||
| 8 | -120000 | -60000 | 34000 | 290000 | 144000 |
| 9 | -60000 | 290000 | 230000 | ||
| 10 | -60000 | 290000 | 230000 | ||
| 11 | -60000 | 290000 | 230000 | ||
| 12 | -60000 | 34000 | 290000 | 264000 | |
| NPW | 1703103.55 | ||||
As NPW of Inorganic Based Feedstock is higher it should be selected
Showing formula in Excel
| Petroleum Based Feedstock | |||||
| Year | First Cost | Annual cost | Salvage | Annual Revenue | Cash Flow |
| 0 | -350000 | =I8+J8+K8+L8 | |||
| 1 | -130000 | 300000 | =I9+J9+K9+L9 | ||
| 2 | -130000 | 300000 | =I10+J10+K10+L10 | ||
| 3 | -130000 | 300000 | =I11+J11+K11+L11 | ||
| 4 | -130000 | 300000 | =I12+J12+K12+L12 | ||
| 5 | -130000 | 300000 | =I13+J13+K13+L13 | ||
| 6 | -350000 | -130000 | 45000 | 300000 | =I14+J14+K14+L14 |
| 7 | -130000 | 300000 | =I15+J15+K15+L15 | ||
| 8 | -130000 | 300000 | =I16+J16+K16+L16 | ||
| 9 | -130000 | 300000 | =I17+J17+K17+L17 | ||
| 10 | -130000 | 300000 | =I18+J18+K18+L18 | ||
| 11 | -130000 | 300000 | =I19+J19+K19+L19 | ||
| 12 | -130000 | 45000 | 300000 | =I20+J20+K20+L20 | |
| NPW | =NPV(6%,M9:M20)+M8 | ||||
| =M39-M21 | |||||
| Inorganic Based Feedstock | |||||
| Year | First Cost | Annual cost | Salvage | Annual Revenue | Cash Flow |
| 0 | -120000 | =I26+J26+K26+L26 | |||
| 1 | -60000 | 290000 | =I27+J27+K27+L27 | ||
| 2 | -60000 | 290000 | =I28+J28+K28+L28 | ||
| 3 | -60000 | 290000 | =I29+J29+K29+L29 | ||
| 4 | -120000 | -60000 | 34000 | 290000 | =I30+J30+K30+L30 |
| 5 | -60000 | 290000 | =I31+J31+K31+L31 | ||
| 6 | -60000 | 290000 | =I32+J32+K32+L32 | ||
| 7 | -60000 | 290000 | =I33+J33+K33+L33 | ||
| 8 | -120000 | -60000 | 34000 | 290000 | =I34+J34+K34+L34 |
| 9 | -60000 | 290000 | =I35+J35+K35+L35 | ||
| 10 | -60000 | 290000 | =I36+J36+K36+L36 | ||
| 11 | -60000 | 290000 | =I37+J37+K37+L37 | ||
| 12 | -60000 | 34000 | 290000 | =I38+J38+K38+L38 | |
| NPW | =NPV(6%,M27:M38)+M26 | ||||
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show the formula
Compare the following alternatives on the basis of Present worth Analysis at an...
Compare 10 years the alternatives C and D on the basis of a present worth analysis...
Compare 10 years the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
An incremental ROR analysis was made with two alternatives, G and H, as shown in Table...
An incremental ROR analysis was made with two alternatives, G and H, as shown in Table 1. The incremental cash flows are given in Table 2. When the LCM method is used, what is the cash flow at Year 4 (Cash flow for (1) in the table). Table 1. Two Alternatives -250,000 -150,000 80,000 50,000 First Cost ($) Annual Revenue ($/year) Salvage ($) Life (years) 40,000 30,000 Table 2. Incremental Cash flow Incremental Cash flow ($) Year O PITTITITY OOO...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /=...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis...
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 $...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year....
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Please dont use excel,show me the formula used 7. Compare the alternatives shown below on the...
Please dont use excel,show me the formula used
7. Compare the alternatives shown below on the basis of a future worth analysis, using an interest rate of 8% per year. Р First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years -23.000 -4,000 3,000 -30,000 -2.500 1.000
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%
QUESTION 3For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $26,5383000010000Annual cost, $/year8,0606,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:A- AW for machine A=QUESTION 4For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $1500021,66710000Annual cost, $/year8,8706,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:B- AW for machine B=
For the below Me alternatives, which machine should be selected based on the future worth analysis....
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different...
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at i= 11% per year on the basis of an annual worth evaluation. 1 T First Cost AOC per Year Salvage Value Life Land $-150,000 $-97,000 $29,000 4 years Incineration -720,000 $-62,000 $370,000 6 years Contract...
Compare 10 years the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
An incremental ROR analysis was made with two alternatives, G and H, as shown in Table 1. The incremental cash flows are given in Table 2. When the LCM method is used, what is the cash flow at Year 4 (Cash flow for (1) in the table). Table 1. Two Alternatives -250,000 -150,000 80,000 50,000 First Cost ($) Annual Revenue ($/year) Salvage ($) Life (years) 40,000 30,000 Table 2. Incremental Cash flow Incremental Cash flow ($) Year O PITTITITY OOO...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
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 $...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Please dont use excel,show me the formula used
7. Compare the alternatives shown below on the basis of a future worth analysis, using an interest rate of 8% per year. Р First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years -23.000 -4,000 3,000 -30,000 -2.500 1.000
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
Problem 06.022 Evaluating Alternatives by Annual Worth Analysis An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at i= 11% per year on the basis of an annual worth evaluation. 1 T First Cost AOC per Year Salvage Value Life Land $-150,000 $-97,000 $29,000 4 years Incineration -720,000 $-62,000 $370,000 6 years Contract...
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