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Solve the below LP using SIMPLEX method.\newlineMaximize −x1−x2+4x3,\text{Maximize } -x_{1}-x_{2}+4x_{3}, subject to x1+x2+2x3≤9,\text{ subject to } x_{1}+x_{2}+2x_{3} \leq 9,x1+x2−x3≤2,x_{1}+x_{2}-x_{3} \leq 2,−x1+x2+x3≤4,-x_{1}+x_{2}+x_{3} \leq 4,x1,x2,x3≥0.x_{1}, x_{2}, x_{3} \geq 0.

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Q. Solve the below LP using SIMPLEX method.\newlineMaximize −x1−x2+4x3,\text{Maximize } -x_{1}-x_{2}+4x_{3}, subject to x1+x2+2x3≤9,\text{ subject to } x_{1}+x_{2}+2x_{3} \leq 9,x1+x2−x3≤2,x_{1}+x_{2}-x_{3} \leq 2,−x1+x2+x3≤4,-x_{1}+x_{2}+x_{3} \leq 4,x1,x2,x3≥0.x_{1}, x_{2}, x_{3} \geq 0.
  1. Introduce slack variables: Introduce slack variables to convert inequalities into equations.\newlineLet x4x_{4}, x5x_{5}, and x6x_{6} be the slack variables for the respective constraints.\newlinex1+x2+2x3+x4=9x_{1} + x_{2} + 2x_{3} + x_{4} = 9\newlinex1+x2−x3+x5=2x_{1} + x_{2} - x_{3} + x_{5} = 2\newline−x1+x2+x3+x6=4-x_{1} + x_{2} + x_{3} + x_{6} = 4
  2. Set up initial tableau: Set up the initial simplex tableau.\newline\begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Basis} & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & x_{6} & \text{RHS} \ \hline x_{4} & 1 & 1 & 2 & 1 & 0 & 0 & 9 \ x_{5} & 1 & 1 & -1 & 0 & 1 & 0 & 2 \ x_{6} & -1 & 1 & 1 & 0 & 0 & 1 & 4 \ Z & -1 & -1 & 4 & 0 & 0 & 0 & 0 \ \hline \end{array}
  3. Identify entering variable: Identify the entering variable.\newlineThe entering variable is the one with the highest coefficient in the objective function row, which is x3x_{3}.
  4. Identify leaving variable: Identify the leaving variable using the minimum ratio test.\newlineRatios for x3x_{3}: 92\frac{9}{2}, cannot use 2−1\frac{2}{-1} (negative), 41\frac{4}{1}.\newlineThe smallest positive ratio is 44, so x6x_{6} will leave.
  5. Perform row operations: Perform row operations to make x3x_{3} the basic variable in place of x6x_{6}.\newlineRow x6x_{6} becomes: (Row x6+Row x1)/1(\text{Row } x_{6} + \text{Row } x_{1}) / 1\newlineNew Row x6x_{6}: 0x1+2x2+2x3+0x4+0x5+1x6=40x_{1} + 2x_{2} + 2x_{3} + 0x_{4} + 0x_{5} + 1x_{6} = 4

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