Write the dual of the following LPP after reducing it to canonical form.
Min Z = 3x1 + 4x2 + 3x3
Subject to
2x1+4x2 =12
5x1+3x3 ≥11
6x1+ x2 ≥ 8
x1,x2,x3≥0
The LPP is given as;
3x1 + 4x2 + 3x3 = Min Z
2x1 + 4x2 + 0x3 =12
5x1 + 0x2 + 3x3 ≥ 11
6x1+ x2 + 0x3 ≥ 8
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
We construct the matrix of coefficients for the free problem, direct problem, and forward problem and name them A, B, C respectively, where;
A = "\\begin{pmatrix}\n 3 & 4 &3\\\\\n \n\\end{pmatrix}"
B= "\\begin{pmatrix}\n 12 \\\\\n \n 11\\\\\n8\n\\end{pmatrix}"
D ="\\begin{pmatrix}\n 2& 4&0\\\\\n 5 & 0&3\\\\\n6&1&0\n\n\\end{pmatrix}"
To find the dual of A, we compute the transpose matrix of B and to compute the dual of B, we find the tranpose matrix of A. The dual of C is the transpose matrix of C
A dual = "\\begin{pmatrix}\n 12 & 11 &8\\\\\n \n\\end{pmatrix}"
B dual = "\\begin{pmatrix}\n 3 \\\\\n \n 4\\\\\n3\n\\end{pmatrix}"
C dual = "\\begin{pmatrix}\n 2& 5&6\\\\\n 4& 0&1\\\\\n0&3&0\n\n\\end{pmatrix}"
To find dual of the problem, we maximize the objective function. Also, we have the same number of constraints in the dual problem as that of the direct problem.
A maximized objective function will contain either the inequality sign ""\\le" " or " ="
Therefore, the dual LPP is given as follows;
12y1 +11y2 + 8y3 = Max Z
2y1 +5y2 + 6y3 "\\le" 3
4y1 + y3 "\\le" 4
3y2 "\\le" 3
y1, y2, y3 ≥ 0
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