All constraints can be converted to ≤ by multiplying by -1. So we have;
Max z=5x1−2x2subject to −3x1−2x2≤−16 x1− x2≤4 −x1 ≤−5and x1≤0;x2is unrestricted.
Since the primal has two variables and three constraints, then the dual will have three variables and two constraints. Also, the x2 variable is unrestricted in the primal, therefore the second constraint in the dual shall be equality.
Dual program is;
Min z=−16y1+4y2−5y3subject to −3y1+y2−y3≥5 −2y1−y2 =−2and y1,y2≥0.
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