0 1 0 0 minimize subject to 0 0 1 0 3 4 2 0 3y1 4y2

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Unformatted text preview: 1 0 ≤ y1 , y2 , y3 P maximize subject to D −4x1 − 2x2 − x3 −x1 − x2 + 2x3 ≤ −3 −4x1 − 2x2 + x3 ≤ −4 x1 + x2 − 4x3 ≤ 2 0 ≤ x1 , x2 , x3 -1 -4 1 -4 -1 -2 1 -2 2 1 -4 -1 1 0 0 0 Dual feasible! 0 1 0 0 minimize subject to 0 0 1 0 -3 -4 2 0 −3y1 − 4y2 + 2y3 −y1 − 4y2 + y3 ≥ −4 −y1 − 2y2 + y3 ≥ −2 2y1 + y2 − 4y3 ≥ −1 0 ≤ y1 , y2 , y3 Not primal feasible. The tableau below is said to be dual feasible because the objective row coefficients are all non-positive, but it is not primal feasible. -1 -4 1 -4 -1 -2 1 -2 2 1 -4 -1 1 0 0 0 0 1 0 0 0 0 1 0 -3 -4 2 0 The tableau below is said to be dual feasible because the objective row coefficients are all non-positive, but it is not primal feasible. -1 -4 1 -4 -1 -2 1 -2 2 1 -4 -1 1 0 0 0 0 1 0 0 0 0 1 0 -3 -4 2 0 A tableau is optimal if and only if it is both primal feasible and dual feasible. The tableau below is said to be dual feasib...
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