| Machine A | |||||
| Discount rate | 0.1 | ||||
| Year | 0 | 1 | 2 | 3 | 4 |
| Cash flow stream | -30000 | -5000 | -5000 | -5000 | 0 |
| Discounting factor | 1 | 1.1 | 1.21 | 1.331 | 1.4641 |
| Discounted cash flows project | -30000 | -4545.45 | -4132.23 | -3756.57 | 0 |
| NPV = Sum of discounted cash flows | |||||
| NPV Machine A = | -42434.26 | ||||
| Where | |||||
| Discounting factor = | (1 + discount rate)^(Corresponding period in years) | ||||
| Discounted Cashflow= | Cash flow stream/discounting factor | ||||
| Equvalent annuity(EAA)= | -13386.77 | ||||
| Required rate = | 0.1 | ||||
| Year | 0 | 1 | 2 | 3 | 4 |
| Cash flow stream | 0 | -13386.8 | -13386.8 | -13386.8 | -13386.77 |
| Discounting factor | 1 | 1.1 | 1.21 | 1.331 | 1.4641 |
| Discounted cash flows project | 0 | -12169.8 | -11063.4 | -10057.7 | -9143.344 |
| Sum of discounted future cashflows = | -42434.26 | ||||
| Discounting factor = | (1 + discount rate)^(Corresponding period in years) | ||||
| Discounted Cashflow= | Cash flow stream/discounting factor | ||||
| Machine B | |||||
| Discount rate | 0.1 | ||||
| Year | 0 | 1 | 2 | 3 | |
| Cash flow stream | -41000 | -4000 | -4500 | 3000 | |
| Discounting factor | 1 | 1.1 | 1.21 | 1.331 | |
| Discounted cash flows project | -41000 | -3636.36 | -3719.01 | 2253.944 | |
| NPV = Sum of discounted cash flows | |||||
| NPV Machine B = | -4610143.00% | ||||
| Where | |||||
| Discounting factor = | (1 + discount rate)^(Corresponding period in years) | ||||
| Discounted Cashflow= | Cash flow stream/discounting factor | ||||
| Equvalent annuity(EAA)= | -18538.07 | ||||
| Required rate = | 0.1 | ||||
| Year | 0 | 1 | 2 | 3 | |
| Cash flow stream | 0 | -18538.1 | -18538.1 | -18538.1 | |
| Discounting factor | 1 | 1.1 | 1.21 | 1.331 | |
| Discounted cash flows project | 0 | -16852.8 | -15320.7 | -13927.9 | |
| Sum of discounted future cashflows = | -46101.43 | ||||
| Discounting factor = | (1 + discount rate)^(Corresponding period in years) | ||||
| Discounted Cashflow= | Cash flow stream/discounting factor | ||||
16 MARR value of 10 ro% per year, antes shown below on the basis of an annual worth analysis, Use year Machine B Ma...
Use FW analysis to compare between machines A and B if the MARR is 10% Machine A Machine B First Cost, $ -20,000 -30,000 Annual Cost, $ year -9000 -7000 Salvage Value, $ 4000 6000 Life 3 6 The Future Worth of machine A The Future Worth of machine B Which machine should be Selected A. $-68,960 B. Machine B C. $-203,272 D. $-101,156 E. Machine A F. $-122,168 G. $-103,245
Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2 Machine 1 20,000 9000 4000 First cost,S Annual cost,S/year Salvage value, S Life, years 100,000 -7000 Infinite A.-20000(A/P 1096,3)-9000+4000(A/F,1096,3) B. $-15832.40 C. $-170,000 Equation for AWI= (Answer 2 decimals) $ | CC2- $ Selection= E. $-180,000 F 2 G. $-166,540
Question 2 Compare the machines shown below on the basis of their capitalized cost. Use 10% per year Machine 2...
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
perform a present worth analysis per the following project using a MARR of 8% The initial coast of the project is $100000, 60% of the initial cost will borrowed from a bonk using a loan that charges 5% interest componded annually. The loan will repaid with the equal annual payment .The project has life of 8 years. Revenne from the project will be $ 5000 in the first year and increasing by $3500 every year after that (ie: $5000 at...
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
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-
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=
Q2. Evaluate an electronic fabrication machine on the basis of the annual worth method when the MARR is 10% per year. Relevant cost data are as follows: (7 Marks) Investment cost Useful life Market (salvage) value at end of useful life Annual operating expenses Overhead cost-end of 8th year Overhead cost-end of 12th year Electronic Fabrication Machine $18,000 15 years $6,000 $450 $1000 $1500 Using Aw(i) method with short explai Please don't use excel use AW(i) factor
Either of the cost alternatives shown below can be used in a chemical refining process. If the company’s MARR is 15% per year, determine which should be selected on the basis of an incremental ROR analysis. A B First cost ,$ − 40,000 − 61,000 Annual cost, $/year − 25,000 − 19,000 Salvage value, $ 8,000 11,000 Life, years 5 5 5 - A. B. C. D. E. F. The incremental rate of return computed using a present worth analysis...
For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10%.Machine AMachine BMachine CFirst cost, $ 15000 30000 10,360Annual cost, $/year 8,320 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :C- PW for machine C =