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Unformatted text preview: CEE 291 Problem Solving Using Computer Tools Homework 3 soln : Excel and Matlab Intro Assigned: 2/2/2010 Due: 2/8/2010 by 5:00 pm in the ECOW2 drop box and in the hall box outside 1261 EH. EXCEL 1. ENGINEERING ECONOMICS Engineering economics involves a set of principles for comparing alternative, physically feasible plans/solutions to a problem in order to determine the economically optimal one. Such a comparison requires that all relevant costs and benefits of alternative plans are comparable; thus, benefits and cost must be expressed in the same monetary units, and time for realization of benefits and costs are expressed with discounting techniques using compound interest expressions. Benefits and costs associated with projects occur at various times (e.g., initial investment occurs at the beginning, operation and maintenance occur through the project life, major replacement/rehabilitation occur periodically, benefits typically accrue over time). Three common discounting formulas are: ( 29 n i 1 P F + =- future amount, F , from single, initial payment, P ( 29 ( 29 - + + = 1 i 1 i 1 i P A n n- annual payment amount , A , to recover initial capital investment, P ( 29 - + = 1 i 1 i F A n- annual investment amount, A , to accumulate amount F with n = period of time in years of economic life of project (time beyond when operational cost exceeds incremental benefits) and i = discount (interest) rate, assumed constant over period (for public works involves a value judgment based upon maximizing social welfare and may be based upon a market rate for risk free investment and allows only slow change in the rate). Applications a) To pay for one year of school, a student borrows $12,000 from the government. The money must be repaid in 20 years with annual payments at an interest rate of 3%. Determine the students required annual payment. A $ P $ I n years $ 806.59 $ 12,000.00 0.03 20 b) Two alternatives, A and B, are being considered for a new section of an aqueduct that transports water from a reservoir to a community. Both alternatives have the same flow capacity. Plan A uses a tunnel, while plan B uses a section of lined canal and a section of steel flume. The initial and maintenance costs and economic life for each plan and its components are shown in the table. Assuming a constant interest rate of 4% over the 100 year project life, compare both alternatives on the basis of annual costs and on the basis of present worth of costs. Plan Facility type, Initial Cost, life expectancy, yrs Annual Maintenance Cost A Tunnel, $450,000, 100 yrs $4,000 B Canal, $120,000, 100 yrs Canal lining, $50,000, 20 yrs Flume, $90,000, 50 yrs $10,500 Tunnel PWC Canal PWC $ 551,940.80 $ 580,414.76 The tunnel option has a lower present worth of costs....
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This note was uploaded on 05/04/2010 for the course CEE 291 taught by Professor Hoopes during the Spring '10 term at Wisconsin.
- Spring '10