ie2140-09-soln-09

# Ie2140-09-soln-09 - IE 2140 Engineering Economy Solutions to Tutorial#9 1(Problem 3-8 I C2005 C2000 2005 I 2000 293 \$200,000 223 \$262,780.27

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IE2140 (2009) soln-9-1 IE 2140 Engineering Economy Solutions to Tutorial #9 1. (Problem 3-8) 27 . 780 , 262 \$ 223 293 000 , 200 \$ 2000 2005 2000 2005 I I C C 2. (Problem 3-11) Let C A = cost of new boiler, S A = 1.35 X C B = cost of old boiler, today S B = X Therefore 67 . 166 , 265 \$ 150 185 000 , 215 \$ B C 25 . 154 , 327 \$ 35 . 1 67 . 166 , 265 \$ 7 . 0 X X C A Total cost with options = \$327,155 + \$18,000 = \$ 345,154.25 3. (Problem 3-17) 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 hours 4 . 587 126 5 1 074 . 0 5 u u T C 5 = 587.4/5 = 117.5 hours

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IE2140 (2009) soln-9-2 4. (Problem 3-21) i x i y i x i 2 x i y i 1 26,500 75,000 702,250,000 1,987,500,000 2 29,000 84,500 841,000,000 2,450,500,000 3 31,400 82,500
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## This note was uploaded on 02/25/2011 for the course ECONS 2104 taught by Professor Prof during the Spring '08 term at National University of Singapore.

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Ie2140-09-soln-09 - IE 2140 Engineering Economy Solutions to Tutorial#9 1(Problem 3-8 I C2005 C2000 2005 I 2000 293 \$200,000 223 \$262,780.27

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