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lecture 8

lecture 8. Simplex method. 3.1- Solving of problem1. The mathematical model of problem is: Max Z= 3 x 1 + 5 x 2 S.t. x 1  4 (Plant 1 constraint) 2 x 2  12 (Plant 2 constraint) 3 x 1 + 2 x 2  18 (Plant 3 constraint) x 1 , x 2  0 .

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lecture 8

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  1. lecture 8 Simplex method

  2. 3.1-Solving of problem1 The mathematical model of problem is: Max Z= 3x1 + 5x2 S.t. x1  4 (Plant 1 constraint) 2x2 12 (Plant 2 constraint) 3x1+ 2x2 18 (Plant 3 constraint) x1, x2 0

  3. 1st Step: InitialSimplex tableau Non-Basic variables Basic variables We start by a basic solution Z=0.

  4. 2nd Step: Entering variable The largest negativecoefficient in the objective function is -5 then the entering variable will be X2. 3rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 12/2=6 then the leaving variable will be S2 .

  5. 4th Step: Pivoting Pivot item Pivot row Pivot column

  6. 4th Step: Pivoting All coefficients of objective function aren’t negatives or equal to zero, the optimal solution is not found. Then go to step2.

  7. 2nd Step: Entering variable The new largest negativecoefficient in the objective function is -3 then the entering variable will be X1. 3rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 6/3=2 then the leaving variable will be S3 .

  8. 4th Step: Pivoting Pivot item Pivot row Pivot column

  9. 4th Step: Pivoting All coefficients of objective function are now equal to zero. then the optimal solution is found. X1=2, X2=6, Z=36.

  10. 3.2- Solving of problem2 The mathematical model of problem is: Max Z= 10x1 + 15x2 S.t. 2x1 + 4x2  100 ( aluminum constraint) 3x1 + 2x2  80 ( steel constraint) x1 , x2  0

  11. 1st Step:InitialSimplex tableau Non-Basic variables Basic variables We start by a basic solution Z=0.

  12. 2nd Step: Entering variable The largest negativecoefficient in the objective function is -15 then the entering variable will be X2. 3rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 100/4=25 then the leaving variable will be S1 .

  13. 4th Step: Pivoting Pivot item Pivot row Pivot column

  14. 4th Step: Pivoting

  15. 2nd Step: Entering variable The new largest positive coefficient in the objective function is 5/2 then the entering variable will be X1. 3rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 30/2=15 then the leaving variable will be S2.

  16. 4th Step: Pivoting Pivot item Pivot row Pivot column

  17. 4th Step: Pivoting All coefficients of objective function are now equal to zero, the optimal solution is found: X1=15, X2=17.5, Z= 412.5.

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