Question

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,x_2,x₃≥0

Answer 1

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 = $\begin{pmatrix} 3 & 4 &3\\ \end{pmatrix}$

B= $\begin{pmatrix} 12 \\ 11\\ 8 \end{pmatrix}$

D =$\begin{pmatrix} 2& 4&0\\ 5 & 0&3\\ 6&1&0 \end{pmatrix}$

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 = $\begin{pmatrix} 12 & 11 &8\\ \end{pmatrix}$

B _dual = $\begin{pmatrix} 3 \\ 4\\ 3 \end{pmatrix}$

C _dual = $\begin{pmatrix} 2& 5&6\\ 4& 0&1\\ 0&3&0 \end{pmatrix}$

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