Finicial Evaluation of Techologies - Speech or Presentation Example

Analysis of two energy proposals Two electri supply technologies have the following characteristics:   Capital cost Annual operating cost Lifetime (years)Salvage value (/cost) ($)Annual electricity supplied (kWh)Technology A (a sustainable energy option)20 00050020200025 000Technology B (a fossil fuel option)10 000300010-200025 0001. Calculate the simple payback period for technology A relative to technology B.Would a company that sets a two year maximum payback period for new investments spend the extra capital on technology A?

Project AProject BInitial Investment$20,000$10,000Annual electricity supplied$ 6,000$ 6,000Less: Maintenance cost 500 3,000Net annual benefit 5,500$ 3,000Payback period3.63 years3.3 years*Assume that the cost of electricity per kwh is 12cents –(In US)In simple calculations, Proj.B has a shorter time frame of return on investments which is 3.3 years as compared with 3.6 years of Proj. A. None of the projects fall within the required 2 years time frame.2. Calculate the internal rate of return for the additional investment in A compared to B over the assessment periods:a. 5 yearsb. 10 yearsc.

20 yearsProject AInitial investmentCash flowsPV 5%NPV-20,0005 yrs 0.78153000023,4453,44510 yrs.0.613960,00092,08572,08515 yrs. 0.481090,00043,29023,290Project BInitial InvestmentCash FlowsPV 5%NPV-10,0005 yrs 0.781530,00023,44513,44510 yrs 0.613960,00092,08582,08515 yrs. 0.481090,00043,29033,290 Both projects show positive PV and NPV at 5% rate of return.

However TSech B has a higher PV and positive NPV. Between two positsive proposals, one that gives a higher value is acceptable. In each case state whether the company would invest the extra capital in technology A if its minimum required real internal rate of return is 10%. (Be careful to take account of all replacements and salvage values during each assessment period.

But in the case of a technology that still has a useful lifetime remaining at the end of a period, do not seek to estimate its residual value; that is, only count the salvage value at the end of a lifetime.) Project A Cash flowsPV 10%0.6209NPVInitial cost20,0005 YEARS30,000Less: Depn. 5,000Net of cash flows25,00015,522.5-4477.50PROJECT BInitial Cost10,0005 YEARS30,000Less depreciation 5,000Net of cash flows25,000`15,522.505,552.50In both computation, Project B shows a higher positive NPV showing that “B” is acceptable than “A”.

In this case, company should invest its excess capital to “B” at 10% rate of return.3. Calculate the Present Worth (that is, the Net Present Value [NPV] of total costs) for each of the technologies for the real discount rates and periods of assessment as specified in the following tables (please present results in this format):Technology AAssessment periodDiscount rate   (years)5%10%15%20%5 0.7815 0.6209 0.497   0.40210 0.6139 0.3855 0.247 0.16215 0.4810 0.2394 0.123 0.06520 0.3769 0.1486 0.061 0.026Technology BAssessment periodDiscount rate   (years)5%10%15%20%5 0.7815 0.6209 0.497   0.40210 0.6139 0.3855 0.247 0.16215 0.4810 0.2394 0.123 0.06520 0.3769 0.1486 0.061 0.026Hence fill in the following table saying which technology would be selected for each of the cases on the basis of highest Present Worth (that is, NPV of lowest total cost):Project A Net Present ValueAssessmentPeriod5%10%15%20%-20,00053445-1,373-5090-509010179143130-5180-51801523,2901546-8980-89802025,228-2168-12680-12680Project B Net Present ValueAssessmentPeriod5%10%15%20%52244517627491011060102683413130482087201533290115451070-415020352287832-2680-7000At different discount rates and number of years, NPV of Proj.

B is higher than A.4. Using your answers to question 3, what are the lifecycle costs of both technologies over one lifecycle of technology A at (a) a 5% real discount rate; and (b) a 20% real discount rate. Which technology is preferred on this lifecycle cost basis in each caseABInitial investment20,00010,000PV of Operating cost3,769 4,860Residual value2,000 0Total life cycle cost25,76914,860Initial cost of project20,000 10,000,00LCC 5,7694,860Technology A has higher LCC than Tech. B and should be preferred.(a) Calculate the average unit cost of the power in present value terms (in cents/kWh) supplied by each technology over a period of 20 years at a discount rate of 5%.

Hint: use the answers from question 3 again to find the NPV of total costs for each technology over 20 years and then divide this amount by the total electricity supplied over this period. NPVkW/h suppliedAve. Cost per unitProject A25,228 25,000 1.10Project B35,228 25,000 1.415. Discuss briefly some of points emerging from this analysis of relevance to the financial comparison of sustainable energy supply options (in particular renewable) and current fossil-fuel technologies.

New technologies are being developed to replace use of fossil fuels used for power generation. Studies about renewable energies are being done by the authorities if its cost will compensate its use against cost of fossil fuels. Fossil fuels like coal and gas are the most commonly used around the world for power generation. This study looked at the cost of technology from fossil fuel and that of sustainable energy. A comparison of both presented important points in terms of capital and cost.1. In terms of technology, the average cost per kWh is much higher in Technology B which is 1.41 than 1.10 of A which means fossil energy will be more costly to produce and will be a higher price to consumer to borne.2. Technology B gives a higher NPV than Technology A for the same 20 year period.

A high positive value is an acceptable project proposition, and between two proposals, one having a higher value is considered.Annex1. Payback calculation using simple payback method.Payback APayback BInitial investment20,00010,000Annual electricity supplied6,0006,000Less maintenance cost5003000Net annual benefit5,50030003.63 yrs3.3 yrs.*Assume that the cost of electricity per kwh is 12cents –(In US)Formula used:Capital cost divided by net annual benefit2. Discount rate factors solving for PV and NPVTechnology AAssessment periodDiscount rate   (years)5%10%15%20%5 0.7815 0.6209 0.497   0.40210 0.6139 0.3855 0.247 0.16215 0.4810 0.2394 0.123 0.06520 0.3769 0.1486 0.061 0.026Assessment periodDiscount rate   (years)5%10%15%20%5 0.7815 0.6209 0.497   0.40210 0.6139 0.3855 0.247 0.16215 0.4810 0.2394 0.123 0.06520 0.3769 0.1486 0.061 0.0265%Tech ACash FlowPVFACTOR PVNPV 20,000    5 yrs 30,0000.

781523445344510 600000.6319379141791415 900000.4814329023,29020 120000.0.37694522825,228      10%2000030,000 0.620918627-1,3735 60000 0.385523130313010 90000 0.239421546154615 120000 0.148617832-216820           15%     520,00030,000 0.497 14910-509010 60000 0.24714820-518015 90000 0.12311070%-89801107020 120,0000.0617320%-126807320      20%           5 30000  0.4024860386010 60000 0.1629720872015 90000% 0.0655850485030 120000 0.0263000-7000TECHNOLOGY BTech BCASH FLOWFACTRPVNPV10,000 5%  530,000 0.

781523445224451060000 0.613936834268341590000 0.4810432903329020120,000 0.37694522835228       10%   30,000 0.62091862717627 60000 0.38552313013130 90000 0.23942154611545 120,000 0.1486178327832       15%   30,000 0.497 149104910 60000 0.247148204820 90000 0.123110701070 120,000 0.0617320-2680       20%   30,0000.4021206011060 60000 0.16297208720 90000 0.0655850-4150 120,000 0.0263000-7000