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EXERCISE 8 Analyze and solve the following problems. 1. Four mutually exclusive alternatives are being evaluated, and their costs and revenues are itemized in the table below: Mutually Exclusive Alternative I II III IV Capital Investment (Php) 1,000,000 1,520,000 1,840,000 2,200,000 Annual revenues less expenses 152,000 319,000 359,000 415,000 (Php) Market value (Php) 100,000 0 150,000 200,000 Useful life (years 12 12 12 12 a. If the MARR is 15% per year and the analysis period is 12 years, use the PW method and the conventional payback period method to determine which alternative should be selected. b. If the total capital investment budget available is Php2,000,000, which alternative should be selected? c. Which rule applies? Why? SOLUTION:2. Two possible routes for a power line are under study. Data on the routes are as follows: Around the Lake Under the Lake Length (km) 15 5 First cost (Php) 200,000/km 1,000,000/km Maintenance (Php) 8,000/km/yr 16,000/km/yr Useful life (years) 15 15 Salvage value (Php) 120,000/km 200,000/km Yearly power loss (Php) 20,000/km 20,000/km Annual property taxes 2% of first cost 2% of first cost (Php) If 7% interest is used, should the power line be routed around the lake or under the lake? Use AW method in your analysis. SOLUTION:NAME: SCORE: COURSE/YEAR: DATE: 3. An investor is considering buying some land for Php5 million and constructing an office building on it. Three buildings are being analysed: 2-storey 5-storey 10-storey Cost of building 20 M 40 M 105 M (excluding cost of land) Resale value* of land & 10 M 15 M 20 M building after 20-year horizon Annual net rental income 3.5 M 5.25 M 12.8 M *Resale value to be considered a reduction in cost, rather than a benefit. a. Using benefit-cost ratio analysis and and 8% MARR, determine which alternative, if any should be selected? b. Compute the IRR for each alternative. SOLUTION:4. Consider the following EDY cash ows for two mutually exclusive alternatives one must be chosen: Lithium Inn Caital investment 'h- 6D 14D, [I Annual .mESEE Ph- 25, 24, Useful life are E Market value ' . The MARE is 5% per year. a. Determine which alternative should be selected if the repeatabith assumption applies. b. Determine which alternative should be selected it" the analysis period is 13 years, the repeatability assumption does not apply, and a battery system can be leased for thB per year after the useful life of either battery is over. ss,ooo er SoLUTIoN: If this is not the case, we obviously would not invest more than the minimum amount of capital required, and we may even do nothing at all. Stated simply, our rule will keep as much capital as possible invested at a rate of return equal to or greater than the MARE. Two rules were given in Chapter 2 for facilitating the correct analysis and comparison of mutually exclusive alternatives when the time value of money is not a factor [present economy studies}. For convenience, these rules are repeated here and extended to account for the time value of money: when revenues and other economic benets are present and vary RULE 1; among alternatives, choose the alternative that Minimizes overall protability. That is, select the alternative that has the greatest positive equivalent worth at 1' = MERE and satises all project requirements. when revenues and other economic benets are not present or are RULE 2: constant among all alternatives, consider only the costs and select the alternative that minimizes the total cost. That is, select the alternative that has the least negative equivalent worth at 1' = HARE and satisfies all project requirements. THE. STUDY [Annuals] Panto]: The study {analysis} period, sometintes called the planning horizon, is the selected time period over which mutually exclusive alternatives are compared. The determination of the study period for a decision situation may be inuenced by several factors such as the service period required, the useful life of the shorter lived alternative, the useful life of the longerlived alternative, and company policy. The key point is that the selected study period must be appropn'ate 1th the decision situation under hwesh'gatt'ott Example 5.2 Consider the four plastic molding presses of Example 5.1. Suppose that each press is still capable of producing 120,000 total units per year, but the estimated reject rate is different for each alternative. This means that the expected revenue will differ among the alternatives since only nondefective units can be sold. The additional data for the four presses are summarized below. The life of each press (and the study period) is five years. Press PI P2 P3 P4 Reject rate 8. 4% 0.3% 2.6% 5.6% If all nondefective units can be sold for Php3.75 per unit, which press should be chosen? Use AW method in your analysis. SOLUTION: AW (10%) = Php3.75(120,000) (1-0.084)-Php240,000( A/ P,10%,5)-Php312,000 AW (10%) = Php36,888.60 AW (10%)p2 = Php3.75(120,000) (1-0.003) -Php304,000( 4/ P,10%,5)-Php291,280 AW (10%), Php77,175.57 AW (10%) ; = Php3.75(120,000)(1-0.026)-Php496,000( 4/ P,10%,5)-Php251,920 AW (10%) ; = Php55,536.45 AW (10%) , = Php3.75(120,000) (1-0.056) -Php520,000( A/ P,10%,5)-Php228,800 AW (10%)PA = Php58,825.31 Alternative P2 maximizes the annual worth of overall profitability and is preferred (versus P4 in Example 5.1).THE COTERMINATED ASSUMPTION Example 5.4 Suppose that Example 5.3 is modified such that an analysis period of six years is used (coterminated assumption) instead of 12 years, which was based on repeatability and the least common multiple of the useful lives. Perhaps the responsible manager did not agree with the repeatability assumption and wanted a six-year analysis period because it is the planning horizon used in the company for small investment projects. SOLUTION: An assumption used for an investment alternative (when useful life is less than the study period) is that all cash flows will be reinvested by the firm at the MARR until the end of the study period. This assumption applies to Alternative A, which has a four-year useful life (two years less than the study period), Assumed reinvestment of cash flow at the MARR for 2 years B Using FW to analyze the situation: FW (10%) =[-Php35,000(F / P,10%,4) + Php12,550(F / A,10%,4)](F / P, 10%, 2) FW (10%) = Php8,471.27 FW (10%), =-Php50,000( F / P,10%, 6)+Php14,800(F / A,10%,6) FW (10%) = Php25,612.98 Based on the FW of each alternative at the end of the six-year study period, we would select Alternative B because it has the larger value