IRR is the minimum rate of return acceptable from a project. It can be calculated with the use of IRR function/formula of EXCEL/Financial Calculator. The basic function/formula for calculating IRR is given as below:
NPV = 0 = Cash Flow Year 0 + Cash Flow Year 1/(1+IRR)^1 + Cash Flow Year 2/(1+IRR)^2 + Cash Flow Year 3/(1+IRR)^3 + Cash Flow Year 4/(1+IRR)^4 + Cash Flow Year 5/(1+IRR)^5 + Cash Flow Year 6/(1+IRR)^6 + Cash Flow Year 7/(1+IRR)^7 + Cash Flow Year 8/(1+IRR)^8 + Cash Flow Year 9/(1+IRR)^9
IRR is calculated with the use of EXCEL as below:
where
IRR (Alt. A) = IRR(B3:B14) = 26.29%
IRR (Alt. B) = IRR(C3:C15) = 17.04%
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Based on the above calculations, it can be concluded that the increment in cost from A to B is not justified. It is because, even though both the alternatives provide IRR greater than MARR, IRR of alternative A is higher than that of alternative B.
5. Compare the following two alternatives by the IRR method, given MARR of 8%/year. Is the...
Please don't provide the answer by excel calculation. I need to understand the formula and step by step.Thank you. 5. Compare the following two alternatives by the IRR method, given MARR of 8%/year. Is the incement in cost form A to B justified? Construction cost $ Benefits $/yr Salvage $ Service Life (yrs) 510,000 145,000 10,000 11 775,000 155,000 20,000 12
Question 1 The cash flows given in table below are for two different alternatives. MARR =10% Data IN Initial Cost Annual Benefits Salvage Value Useful Life in years M $20,000 $6,000 $5,000 $80,000 $10,000 $20,000 a) Determine the annual worth of alternative M b) Determine the annual worth of alternative N
Compare the following 2 alternatives using the Net Present Worth (NPS) method – rate is 3% per year. Draw all cash flow diagrams. Please show work using a formula. Alt. Construction Cost ($) Benefit ($/yr.) Service Life (yrs.) A 380,000 200,000 7 B 450,000 220,000 7
4. After checking the feasibility of each alternative below, compare them using the Incremental IRR method to decide which one is best. MARR=10%. Suggestion: Use the NEUA-0 defintion of the IRR. Alt А Cosntruction cost s 50000 75000 95000 Annual benefits S/yr | Life yrs 15000 18000 20000
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability assumption is acceptable. Aternative I: Initial investment of $45,000, net revenue the first year of $8,000, increasing $4,000 per year for the six year useful life. Salvage value is estimated to be $6500. Alternative II: Initial investment of $60,000, uniform annual revenue of $12,000 for the five year useful life. Slavage value is estimated to be $9,000.
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate of 3% per year. Use first the B/C method to determine feasibility and then the incremental B/C method to determine the oprimum level of investment. Alt Construction Cost $ Annual Benefits $/yr Life yrs A 105,000 40,000 5 B 230,000 52,000 6 C 350,000 64,000 7 D 600,000 100,000 8
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate of 3% per year. Use first the B/C method to determine feasibility and then the incremental B/C method to determine the oprimum level of investment. Alt Construction Cost $ Annual Benefits $/yr Life yrs A 105,000 40,000 5 B 230,000 52,000 6 C 350,000 64,000 7 D 600,000 100,000 8
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...
4. Compare the following altemtives given a market rate of 3.45% per year and an inflation rate of 3% per year. Use first the B/C method to determine feasibility and then the incremental B/C method to determine the oprimum level of investment. Life yrs Alt A Construction Costs Annual Benefits S/yr 105,000 40,000 230,000 52.000 350.000 64,000 600,000 100,000 6
QUESTION 6 Data for two mutually exclusive alternatives are given below. Alternatives B $4,000 $800 А Initial Cost $5,000 Annual Benefits (beginning at end of $1,500 year 1) Annual Costs (beginning at end of year $500 1) Salvage Value $500 Useful Life (years) 5 $200 $0 10 Compute the net present worth for each alternative and choose the better alternative. MARR = 8%