RelativeResourceManager

# RelativeResourceManager - 1 ECOR3800 ENGINEERING ECONOMICS...

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Unformatted text preview: 1 ECOR3800 ENGINEERING ECONOMICS Assignment 2 (2010) Solutions Question 1 (a) : NPW(A) = -10,000+2,500(P/A,8%,12) = \$8,840.20 NPW(B) = -19,000+3,500(P/A,8%,15) = \$10,958.18 NPW(C) = -5000+1000(P/A,8%,10) = \$1,710.08 NPW(D) = -20,000+4000(P/A,8%,10) = \$10,195.37 ROI(A) = (NPW/P)x100= (8,840.20)/10,000)x100= 88.4% ROI(B) = (10,958.18/19,000)x100 = 57.7% ROI(C) = 34.2% ROI(D) = (10,195.37/20,000)x100 = 51.0% Ranking: A,B,D,C (b) Capital required = \$54,000 Question 2: Capital available = \$44,000 Try combinations: PW of A+B+C+\$10,000 (left over) = [10,000+8,840.20]+[19,000+10,958.18] +[5,000+1,710.08]+10,000 = \$65,508.46 PW of B+C+D = [19,000+10,958.18] +[5,000+1,710.08]+[20,000+10,195.37] =\$66,863.80 (Best) PW of B+D +\$5,000 (left over) = [19,000+10,958.18]+[20,000+10,195.37]+5,000 =\$65,153.55 PW of A+D+C+9,000 =[[10,000+*,8840.20] + [20,000+10,195.37] + [5,000+1,710.08] +9,000 = \$64,745.65 PW of A+D+\$14,000 (left over) = [10,000+8,840.20]+[20,000+10,195.37]+14,000 = \$63,035.57 Select Projects B,C,D. ______________________________________________________________________________ Question 3: PW of cost of existing Testing Machine called A = 9,000+ 9,0000(P/A,10%,9)=\$60,831.21 (cheapest) PW of cost of New Model Machine called B = 40,000+40,000(P/F,10%,3)= \$70,052.59 2 (a) NPW Incremental: NPW for A B = -PW of cost of (B)+PW of cost of (A) = -\$70,052.59 + \$60,831.21 = - \$9,221.38. Do not go to B. A is the choice. (b) IRR Incremental: NPW Incremental: NPW for A B = 0 = - [PW of cost of (B)+PW of cost of (A) = - [40,000+40,000(P/F,i%,3)] + [9,000+ 9,0000(P/A,i%,9)] . Solve by trial and error. IRR is 4.2% <MARR Do not go to B. A is the choice....
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## This note was uploaded on 02/10/2011 for the course ECON 3800 taught by Professor Asnerob during the Spring '09 term at Spring Arbor University.

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RelativeResourceManager - 1 ECOR3800 ENGINEERING ECONOMICS...

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