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Solutions to Homework _2

# Solutions to Homework _2 - Solutions to HW#2 IE 1040 Fall...

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IE 1040 Fall, 2005 3-6 (a) I R = + + + = 013 131 0 20 150 0 06 140 . ( ) . ( ) . ( ) 154.9 I C = + + + = 013 176 0 20 210 0 06 172 . ( ) . ( ) . ( ) 203.4 (b) C C = \$314,300 203 4 154 9 . . = \$412,710 3-10 C B (8 years ago) = \$2,500,000 C B (now) = \$2,500,000(1.12) 8 = \$6,189,908 S A = 1,500,000 lbs/yr S B = 500,000 lbs/yr X = 0.65 C A = C B S S A B X = \$6,189,908 1,500,000 500,000 0 65 . = \$12,641,919 3.14 1 mile = 5,280 feet; 25 miles = 132,000 feet; 50 yards = 150 feet A light pole is to be installed every 50 yards or every 150 feet. Thus, 132,000 feet/150 feet = 880 light poles need to be installed. Cost of installing light poles = 880(\$2,500) = \$2,200,000 Total cost = \$15,000/mile(25 miles) + \$2,200,000 = \$2,575,000 3-16 K = 126 hours; s = 0.95 (95% learning curve); n = (log 0.95)/(log 2) = -0.074 (a) Z 8 = 126(8) -0.074 = 108 hours ; Z 50 = 126(50) -0.074 = 94.3 hours (b) C 5 = T 5 /5;

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Solutions to Homework _2 - Solutions to HW#2 IE 1040 Fall...

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