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be used to show what the PW becomes 5. (30 points An area can be irrigated by pumping water from a nearby river. Two competing

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be used to show what the PW becomes 5. (30 points An area can be irrigated by pumping water from a nearby river. Two competing installations are being considered. The MARR is 10% per year and electric power for the pumps costs $0.04 per kWh. 1 hp=0.746 kilowatts. a. At what level of operation (hours per year) would you be indifferent between the Two pumping systems? If the pumping system is expected to operate 2.500 hours per year, which system should be recommended? b. Perform a sensitivity analysis on the efficiency of Pump A. Over what range of pumping efficiency is Pump A preferred to Pump B? Assume 2,500 hours of operation per year. Pump A 6 In. System Pump B 8 In. System 15 hp 0,60 Operating load on motor Efficiency of pump motor Cost of installation Market value Useful life $2,410 $80 10 hp 0.75 $4,820 $0 8 years 8 years unknown hrs. ual operating cost ishox 0.746W 0.04# IP lishe Xhre 10.04kW A lkyon lkyph cetharency II efberenay | xhre 80.04kW | (ishet [ hrs I hrs. - 0.7466 248 18hpx 0.746 HW I The Wh 0.75 2-0.7466 (P/A, 8.1080p-108 Extra Problem 6. [10 points) The heating, ventilating, and air-conditioning (HVAC) system in a commercial building has become unreliable and inefficient. Data of new HVAC system is given below. The useful life of the rebuilt system, however, is quite uncertain. The estimated probabilities of various useful lives are provided. Assume that MARR = 10% per year and the estimated market value of the rebuilt system at the end of its useful life is zero. Based on this information, what are E(PW), V(PW), and SD(PW) of the project's cash flows? Also, what is the probability of PW 20? What decision would you make regarding the project, and how would you justify your decision using the available information? Estimate Economic Factor Capital investment Annual savings Increased annual revenue $521,000 48,600 31,000 Useful Life, Year (N) P (N) 0.1 0.2 0.3 15 0.2 0.1 0.05 0.05 = 1.00 16

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