EID170_2007FALL_EXAM3_STUDSOLN_[0]

# EID170_2007FALL_EXAM3_STUDSOLN_[0] - Final Project EID 170...

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Final Project EID 170 Prof. Barrett Dec 18, 2007

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2 Problems solved using Microsoft Excel Problem 1, Exam 1 Problem 2, Exam 1 Problem 3, Exam 1 Problem 4, Exam 1 Problem 1, Exam 2 Problem 2, Exam 2 Problem 3, Exam 2 Problem 4, Exam 2 Problems solved using Cash Flow Analyzer Problem 1, Exam 1 Problem 2, Exam 1 Problem 4, Exam 1 Problem 1, Exam 2 Problem 3, Exam 2 Problem 4, Exam 2
3 Exam 1, Problem 1. An engineer has proposed a design to improve the quality of an existing product. The design is expected to have a life of fifteen years. Its projected costs are \$250,000 to design and install, \$20,000 per year to operate and this cost is expected to increase by \$3,000 per year every year after the first. Additional maintenance is expected at years 5, 10 and 15. This cost will be \$20,000 the first time and will increase by \$10,000 each time thereafter. What must the value of the improvement be per year to recover all costs and economic growth of 10% per year in the first seven years of usage? Using Present-Worth analysis, we can determine the total costs of expenses for the lifetime of the machine (15 years), now (at time=0). The present value of this investment costs \$556,137.96. Then, to calculate the amount it will cost per year, we implement the Capital Recovery equation, which returns a value of \$114,223.80. This value represents the amount of money this investment will cost per year, at ac economic growth rate of 10% per year. Therefore, the investment must earn for the company a minimum of this value \$114,223.80 per year in order to pay for itself, any income greater than this value is profit (if the only expenses are those mentioned in the problem). The chart of values below shows the expenses per year and the Present Value equivalents at time = 0. Year Expenses PV(t=0) 0 \$250,000 \$250,000.00 1 \$20,000 \$18,181.82 2 \$23,000 \$19,008.26 3 \$26,000 \$19,534.18 4 \$29,000 \$19,807.39 5 \$52,000 \$32,287.91 6 \$35,000 \$19,756.59 7 \$38,000 \$19,500.01 8 \$41,000 \$19,126.80 9 \$44,000 \$18,660.30 10 \$77,000 \$29,686.83 11 \$50,000 \$17,524.69 12 \$53,000 \$16,887.43 13 \$56,000 \$16,221.21 14 \$59,000 \$15,536.54 15 \$102,000 \$24,417.99

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4 Solving the Problem by Cash Flow Analyzer Using the Capital Recovery equation (A/P, i, N) with the value calculated and displayed in the above picture (\$556,137.90), we determine that to recover all the costs of 15 years within 7 years, the value of the improvement must be \$114,223.80 per year.
5 Exam 1, Problem 2. A twenty five year old engineer is planning a five year sabbatical beginning twenty years from now. During the five years she plans to travel the world at an estimated cost of \$30,000 per year. To provide for these expenses she currently has \$10,000 that will earn 8%/yr-quarterly for the

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EID170_2007FALL_EXAM3_STUDSOLN_[0] - Final Project EID 170...

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