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Unformatted text preview: z 4 x 1 + x 2 = 012 x 1 + 6 x 2 + x 3 = 6 4 x 1 4 x 2 + x 4 = 4 1 4 1 0 0 0 12 6 1 0 6 0 4 4 0 1 4 Initial Simplex Table Sec 2.7 Friday, February 05, 2010 8:07 PM Section2dot7 Page 1 z 4 x 1 + x 2 = 012 x 1 + 6 x 2 + x 3 = 6 4 x 1 4 x 2 + x 4 = 4 Initial Simplex Table pc 1 4 1 0 0 0 12 6 1 0 6 0 4 4 0 1 4 pivot entry is 4 because it is the only positive entry in pc R' 3 = 1/4 R 3 , R' 1 = R 1 + 4R' 3 , R' 2 = R 2 + 12R' 3 2.7.1 Friday, February 05, 2010 8:07 PM Section2dot7 Page 2 z 4 x 1 + x 2 = 012 x 1 + 6 x 2 + x 3 = 6 4 x 1 4 x 2 + x 4 = 4 Initial Simplex Table pc 1 4 1 0 0 0 12 6 1 0 6 0 4 4 0 1 4 pivot entry is 4 because it is the only positive entry in pc R' 3 = 1/4 R 3 , R' 1 = R 1 + 4R' 3 , R' 2 = R 2 + 12R' 3 pc 1 0 3 0 1 4 0 0 6 1 3 18 0 1 1 0 1/4 1 there are no positive entries in pc, and so no pivot entries 2.7.2 Friday, February 05, 2010 8:07 PM Section2dot7 Page 3 z 4 x 1 + x 2 = 012 x 1 + 6 x 2 + x 3 = 6 4 x 1 4 x 2 + x 4 = 4 Initial Simplex Table pc 1 4 1 0 0 0 12 6 1 0 6 0 4 4 0 1 4 pivot entry is 4 because it is the only positive entry in pc R' 3 = 1/4 R 3 , R' 1 = R 1 + 4R' 3 , R' 2 = R 2 + 12R' 3 pc 1 0 3 0 1 4 0 0 6 1 3 18 0 1 1 0 1/4 1 there are no positive entries in pc, and so no pivot entries The algorithm is stuck. There is still a negative entry in the top row, so we have not reached the maximum yet, but we can't pivot any more. What happened ?!? 2.7.3 Friday, February 05, 2010 8:07 PM Section2dot7 Page 4 Let's plot the feasibility region:12 x 1 + 6 x 2 ≤ 6 line 2 x 1 + x 2 ≤ 1 (0, 1), (1/2, 0) test pt (0, 0): 12(0) + 6(0) = 0 ≤ 6 Yes! 1 2 3 1 2 3 x 1 ≥ 0, x 2 ≥ 0 2.7.4 Friday, February 05, 2010 8:08 PM Section2dot7 Page 5 Let's plot the feasibility region:12 x 1 + 6 x 2 ≤ 6 line 2 x 1 + x 2 ≤ 1 (0, 1), (1/2, 0) test pt (0, 0): 12(0) + 6(0) = 0 ≤ 6 Yes!...
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This note was uploaded on 06/07/2011 for the course MATH 125 taught by Professor Tom during the Fall '07 term at University of Illinois at Urbana–Champaign.
 Fall '07
 Tom
 Math, Linear Algebra, Algebra

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