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1. An engineer analyzed four independent alterna- tives by the present worth method. On the basis of her results, the alternative(s) she should select are: Alternative D Present worth. S-5000-200-3000-1000 (a) Only D (b) Can't tell from this information (c) All of them (d) None of them 5.50 The capitalized cost of an initial investment of $200,000 and annual investments of $30,000 for- ever at an interest rate of 10% per year is closest to (a) S-230,000 (b) S-300,000 (c) S-500.000 (d) S-2,300,000 5.51 An upgraded version of a CNC machine has a first cost of $200.cod. an annual operating cost of 560.000 and a salvage value of $50.000 after its 8-year life. At an interest rate of 10% per year, the capitalized cost is closest to (a) S-93,116 (b) S-100,060 CS-931.160 (d) S-1,000,600 5.52 The cost of maintaining a permanent monument is expected to be $70,000 now and $70.000 every 10 years forever. At an interest rate of 10% per year, the capitalized cost is nearest: (a) S-11,393 (b) S-58.930 (c) S-84,360 d) S-113.930 5 Problems 5.53 through 5.55 are based on the following cash flows for alternatives X and Y at an interest rate of 10% per year. Machine Initial cost, - 146,000 -220,000 AOC, S/year -15.000 - 10,000 Annual revenue, S/year 80,000 75,000 Salvage values 10,000 25.000 Life, years 3 6 5.53 In comparing the alternatives on a present worth basis, the PW of machine X is closest to: (a) $23.160 (5) S40,560 (e) $58,950 (d) $72,432 5.54 In comparing the alternatives on a future worth basis, the FW of machine X is closest to: (a) $23.160 () $40,560 (c) S58,950 (d) $71,860 5.55 The capitalized cost of machine Y is closest to (a) $17.726 (b) $86,590 (c) $177,260 (d) $207,720 Problems 5.57 through 5.61 are based on the following cash flows for alternatives A and B at an interest rate of 10% per year. Alternative First costs -90,000 -750,000 AOC. Syear -50,000 - 10,000 Salvage values 8.000 2,000,000 Life, years 5.57 You have been asked to compare the alternatives on the basis of a present worth comparison. The PW of alternative A is closest to (a) S-724,320 (b) S-530,520 (c) S-388.950 (d) S-72,432 5.58 The present worth of alternative B is nearest: (a) S-85.000 (b) S-750,000 (c) S-850,000 (d) S-950,000 5.59 The capitalized cost of alternative B is nearest: (a) S-590,000 (1) S-625,000 (c) S-734,000 (d) 5-850,000 5.60 All of the following equations for calculating the capitalized cost of alternative B are correct, except: (a) CC, -750,000 - 10,000/0.10 (b) CC-750,000 - 10,000/0.10 + 2,000,000(0.10) () CC= (-750,000(0.10) - 10,0000.10 (d) CC, = -750,000 - 10,000/0.10 + 2,000,000/(1+0.10) 5.61 The future worth of alternative B is closest to: (a) S-85,000 (b) S-750,000 (c) S-850,000 (d) -00

2. An automaton asset with a high first cost of (c) select alternative C $10 million has required capital recovery (CR) of (d) select the Do Nothing alternative $1,985,000 per year. The correct interpretation of this CR value is that: 6.46 If you have the annual worth of an alternative with (a) the owner must pay an additional S1.985,000 a 5-year life, you can calculate its perpetual annual each year to retain the asset. worth by: (b) cach year of its expected life, a net revenue of (a) no calculation needed. The perpetual annual S1.985,000 must be realized to recover the worth is equal to the annual worth S10 million first cost and the required rate of (b) multiplying the annual worth by (A/P.1.5) return on this investment (c) dividing the annual worth by i (c) each year of its expected life, a net revenue of (d) multiplying the annual worth by i S1.985,000 must be realized to recover the SIO million first cost 6,47 The annual worth for years 1 through infinity of (d) the services provided by the asset will stop if $50,000 now. S10,000 per year in years through less than $1,985,000 in net revenue is re- 15, and $20,000 per year in years 16 through infin- ported in any year ity at 10% per year is closest to (a) less than $16,900 6.42 If you have the capitalized cost of a certain alterna- (b) S16,958 tive that has a 5-year life, you can get its annual (c) SI7.395 worth by: (d) $19.575 (a) multiplying the CC by (b) multiplying the CC by (A/F.1.5) 6.48 An alumnus of West Virginia University wishes to (c) multiplying the CC by (P/1.1.5) start an endowment that will provide scholarship (d) multiplying the CC by (A/P.1.5) money of S40,000 per year beginning in year 5 and continuing indefinitely. The donor plans to give 6.43 The AW values of three cost alternatives are money now and for each of the next 2 years. If the $-23,000 for alternative A, S-21,600 for B, and size of each donation is exactly the same, the amount that must be donated each year at i = 8% $-27,300 for C. On the basis of these results, the decision is to per year is closest to: (a) select alternative A (a) S190,820 (b) select alternative B (b) S122.280 (c) select alternative C (c) S127,460 (d) S132,040 (d) select the Do Nothing alternative 6,44 The initial cost of a packed-bed degassing reactor 6.49 You will make equal deposits into your retirement for removing tribalomethanes from potable water account each year for 10 years starting now (i.e.. years 0-9). If you expect to withdraw $50,000 per is $84,000. The annual operating cost for power, year forever beginning 30 years from now, and the site maintenance, etc. is $13,000. If the salvage funds will earn interest at 10% per year, the size of value of the pumps, blowers, and control systems the 10 deposits is nearest: is expected to be $9000 at the end of 10 years, the (a) S4239 (b) S4662 AW of the packed-bed reactor, at an interest rate of (c) S4974 (d) S5471 8% per year, is closest to (a) S-26.140 6,50 To get the AW of a cash flow of $10,000 that oc- (b) S-25.518 curs every 10 years forever, with the first one oc- (c) S-24.60 curring now, it is correct to (d) S-13.140 (a) multiply the $10,000 by (A/P.1,10) (b) multiply the $10,000 by (A/F.1,10) 6,45 The AW amounts of three revenue alternatives are (c) multiply the $10,000 by i $-23,000 for alternative A. $-21,600 for B. and (d) multiply the $10,000 by (A/Fin) and then $-27,300 for C. On the basis of these AW values, multiply by i the correct decision is to: (a) select alternative A 6.51 The estimates for two alternatives are to be com- (b) select alternative B pared on the basis of their perpetual equivalent

3.The Chilliwack Inn had an average dinner check of $9.50, when 4,000 customers were served on weekly basis. Patrick Lee, the owner, learned from statistics report that the local economy has improved and on average people has an income increase of 4.50%. Thus, he decided to increase the dinner menu prices. The result was that 3,200 guests were served weekly and the average check was $11.40. Compute the price elasticity of demand and the income elasticity of demand. Explain your results in detail (keep two decimals).