Consider each of the alter-tax cash ows shown in the table below. Suppose that projects B and C are mutually exclusive. Suppose also that the required service period is eight years and that the company is considering leasing comparable equipment with an annual lease expense of $3,000. payable at the end of each year for the remaining years of the required service period. Which project is a better choice at 15%? a Click the icon to view the cash flows for the projects. a Click the icon to View the interest factors for discrete compounding when i=15 per year. The present worth of project B is 3; thousand. (Round to one decimal place.) o More Info An electrical utility is experiencing a sharp power demand that continues to grow at a high rate in a certain local area. Two alternatives are under consideration. Each is designed to provide enough capacity during the next 25 years, and both will consume the same amount of fuel, so fuel cost is not considered in the analysis. . Alternative A. Increase the generating capacity now so that the ultimate demand can be met without additional expenditures later. An investment of $24 million would be required, and it is estimated that this plant facility would be in service for 25 years and have a salvage value of $0.9 million. The annual operating and maintenance costs (including income taxes) would be $0.6 million. Alternative B. Spend $11 million now and follow this expenditure with future additions during the 10th year and the 15th year. These additions would cost $14 million and $7 million, respectively. The facility would be sold 25 years from now with a salvage value of $1.25 million. The annual operating and maintenance costs (including income taxes) will be $350,000 initially and will increase to $0.45 million after the second addition (from the 11th year to the 15th year) and to $0.55 million during the final 10 years. (Assume that these costs begin one year subsequent to the actual addition.) On the basis of the present-worth criterion, if the firm uses 10% as a MARR, which alternative should be undertaken? Note: Adopt incremental cost approach. Click the icon to view the interest factors for discrete compounding when i = 10% per year. i More Info - X The present worth of Alternative A is $ |million. (Round to one decimal place.) Single Payment Equal Payment Series Compound Present Compound Sinking Present Capital Amount Worth Amount Fund Worth Recovery Factor Factor Factor Factor Factor Factor (F/P, i, N) (P/F, i, N) F/A, i, N (A/F, i, N) (P/A, i, N) (A/P, i, N) 1.1000 0.9091 1.0000 1.0000 0.9091 1.1000 1.2100 0.8264 2. 1000 0.4762 1.7355 0.576 A W N - Z 1.3310 0.7513 3.3100 0.3021 2.4869 0.4021 1.4641 0.6830 4.6410 0.2155 3.169 0.315 1.6105 0.6209 6.1051 0.1638 3.7908 0.2638 1.7716 0.5645 7.7156 0. 1296 4.355 0.2296 1.9487 0.5132 9.487 0. 1054 4.8684 0.205 2.1436 .466 11.435 0.0874 5.3349 0.1874 2.3579 0.4241 13.5795 0.0736 5.7590 0.173 2.5937 0.3855 15.9374 0.0627 6.1446 0. 1627 2.8531 .350 18.5312 0.0540 6.4951 0. 1540 3.1384 0.3186 21.3843 0.0468 5.8137 0. 1468 .452 0.2897 24.5227 0.0408 .103 0.140 3.7975 .2633 27.975 0.0357 7.3667 0.135 4.1772 0.2394 31.7725 0.0315 7.6061 .1315 16 4.5950 0.2176 35.9497 0.0278 7.8237 0.1278 17 5.0545 0. 1978 10.5447 0.0247 8.0216 .124 5.5599 .179 45.5992 0.0219 8.2014 .1219 6.1159 0.1635 51.1591 0.0195 3.364 0.1195 20 6.7275 0. 1486 57.2750 0.0175 8.5136 0.1175 7.4002 0. 1351 64.0025 0.0156 8.6487 0.1156 8.1403 0.122 71.4027 0.0140 8.7715 0.1140 8.9543 .1117 79.5430 0.0126 8.8832 0.1126 9.8497 0. 1015 88.4973 0.0113 8.9847 0.1113 10.8347 0.0923 98.3471 0.0102 0.1102 Print DoneYou are considering two types of machines for a manufacturing process. . Machine A has a rst cost of $75,200, and its salvage value at the end of six years of estimated service life is $21,000. The operating costs of this machine are estimated to be $6,800 per year. Extra income taxes are estimated at $2,400 per year. . Machine B has a rst cost of $44,000, and its salvage value at the end of six years' service is estimated to be negligible. The annual operating costs will be $11,500. Compare these two mutually exclusive alternatives by the present-worth method at i=13". a Click the icon to view the interest factors for discrete compounding when i = 13% per year. The present worth for machine A is $ thousand. (Round to the nearest whole number.)