Example 3 The new dual dictionary � 7 18 7 6 1 ˆ z 3 ˆ z 1 ˆ z 4 ˆ z 6 ˆ z 5 ˆ

Example 3 the new dual dictionary ? 7 18 7 6 1 ˆ z 3

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Example 3 The new dual dictionary: - ζ = 7 + - 18 - 7 - 6 1 ˆ z 3 ˆ z 1 ˆ z 4 ˆ z 6 ˆ z 5 ˆ z 2 = 1 4 - 2 1 2 - 1 7 3 2 - 4 ˆ z 3 ˆ z 1 ˆ z 4 ˆ z 6 . The new dual dictionary is unbounded, thus the new primal LP is infeasible with the new constraint. Lizhi Wang ([email protected]) IE 534 Linear Programming October 30, 2017 12 / 15
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Case 4: x x n +1 . . . x n + k , c c n +1 . . . c n + k , A A (: ,n +1) ... A (: ,n + k ) Adding new variables could possibly affect dual feasibility, but not primal feasibility. If y * satisfies the new dual constraints, then it is still optimal. Otherwise, apply Simplex on new primal dictionary. If we add all the k new primal variables into the non-basis, then the new primal dictionary is: ζ = c > B A - 1 B b + ( c 0 > N 0 - c > B A - 1 B A 0 N 0 ) x 0 N 0 x B = A - 1 B b - A - 1 B A 0 N 0 x 0 N 0 . Lizhi Wang ([email protected]) IE 534 Linear Programming October 30, 2017 13 / 15
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Example 4 Suppose we add a new variable x 3 : x 0 = x 1 x 2 x 3 , c 0 = - 1 - 1 1 , A 0 = - 2 - 1 3 - 2 4 - 2 - 1 3 1 . The old optimal solution: ( x * 1 = 7 , x * 2 = 0) and ( y * 1 = 0 , y * 2 = 0 , y * 3 = 1) . y * satisfies the constraint 3 - 2 1 y * 1 . The new optimal solution is ( x 0 1 = 7 , x 0 2 = 0 , x 0 3 = 0) . Lizhi Wang ([email protected]) IE 534 Linear Programming October 30, 2017 14 / 15
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Example 4 A = [-2 -1 3; -2 4 -2; -1 3 1]; b = [4; -8; -7]; c = [-1; -1; 1]; dictionary(A,b,c,[4 1 5],[6 2 3]); Output is zeta = -7.0000 - 1.0000 w6 - 4.0000 x2 - 0.0000 x3 w4 = 18.0000 + 2.0000 w6 + 7.0000 x2 - 1.0000 x3 x1 = 7.0000 + 1.0000 w6 + 3.0000 x2 + 1.0000 x3 w5 = 6.0000 + 2.0000 w6 + 2.0000 x2 + 4.0000 x3 Lizhi Wang ([email protected]) IE 534 Linear Programming October 30, 2017 15 / 15
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