Question: Chapter 6 .
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Use the dual simplex method to save the following LPP:

Minimize, $z=2x_{1}+x_{2}$

subject to,

$3x_{1}+x_{2} \ge 3$

$4x_{1}+3x_{2} \ge 6$

$x_{1}+2x_{2} \le 3$

$x_{1}, \ x_{2} \ge 0$

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modified 6 weeks ago  • written 6 weeks ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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We have, $z=2x_{1}+x_{2}$

Maximise, $z'=-z=-2x_{1}-x_{2}-0s_{2}-0s_{3}-MA_{1}-MA_{2}$ -----------------(1)

Subject to, $3x_{1}+x_{2}+0s_{2}+0s_{3}+A_{1}+0A_{2}=3$ ---------------(2)

$4x_{1}+3x_{2}-s_{2}+0s_{3}+0A_{1}+A_{2}=6$ ---------------(3)

$x_{1}+2x_{2}+0s_{2}+0s_{3}+0A_{1}+0A_{2}=3$ ----------------(4)

Multiply (2) and (3) by M and to (1),

Maximise, $=z'=(-2+7M)X_{1}+(-1+4M)X_{2}-Ms_{2}+0s_{3}-A_{1}-0A_{2}-9M$

$=z'+(2-7M)X_{1}+(1-4M)X_{2}+Ms_{2}+0s_{3}+0A_{1}+0A_{2}=-9M$

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$X_{1}=\cfrac{3}{5}, \ X_{2}=\cfrac{6}{5}$

$z'_{max}=\cfrac{-12}{5}$

$z'_{min}=\cfrac{12}{5}$

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written 6 weeks ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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