# 4 - Carleton University Department of Civil and...

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Carleton University Department of Civil and Environmental Engineering ECOR3800A Engineering Economics ASSIGNMENT 4 (SOLUTIONS) Due Date: April 1, 2010 1. These are mutually exclusive alternatives with unequal lives. The use of Annual Worth Method (AW) is appropriate. AW(Alt.1) = -\$16,000(A/P,8%,4) - \$3,000 + \$11,000 + \$0(A/F,8%,4) = \$3,169.27 AW(Alt.2) = -\$20,000(A/P,8%,8) - \$5,000 + \$11,500 + \$4,000(A/F,8%,8) = \$3,395.76. Alternative 2 is the choice. 2. Let S be the Salvage of Alternative 1 Set 16,000(A/P,8%,4) + 8,500 + S(A/F,8%,4) = AW(2) found in question 1 Solve for S. S = \$1,021 Therefore S has to vary by \$1,021 - \$0 =\$1,021 in order to reverse the choice in question 1. 3. Profit = revenue costs Therefore, Z=[(3n/4) (n 3 8n 2 +25n + 30)/25] = - [(4n 3 + 32n 2 - 25n -120)/100] (a) To find the break-even points, the profit equation is set equal to zero. The positive roots of this equation yield break-even points of n = 3.26, and n = 6.23. This can be shown graphically. (b) To find the maximum profit, the profit equation is differentiated and set equal to zero. dz/dn = - [(1/100)(12n 2 64n + 25)] = 0 The quadratic formula shows a minimum profit at n = 0.425 and the desired maximum profit at an output level of n = 4.91.

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