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The Simplex Algorithm

The Simplex Algorithm. An Algorithm for solving Linear Programming Problems. We Start with a Linear Programming Problem. Maximise P = 4x +5y +3z Subject to the constraints 8x + 5y +2z  4 x +2y +3z  1. Setting up the Tableau.

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The Simplex Algorithm

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  1. The Simplex Algorithm An Algorithm for solving Linear Programming Problems

  2. We Start with a Linear Programming Problem • Maximise P = 4x +5y +3z • Subject to the constraints • 8x + 5y +2z  4 • x +2y +3z  1

  3. Setting up the Tableau • First rearrange the equation for P so that it is equal to zero : • P - 4x - 5y - 3z = 0

  4. Introduce Slack variables • 8x + 5y +2z  4 becomes • 8x + 5y + 2z + s = 4 • x +2y +3z  1 becomes • x +2y +3z + t = 1 • s and t are called the slack variables

  5. This is the Tableau (Fancy word for table)

  6. Finding a pivot • Chose any negative number in the first row • Consider the positive values in the column below it • Divide the value in the last column by the corresponding value in the chosen column and see which gives you the least • That tells you which is the pivot...it goes like this:

  7. 4/8 =1/2 1/1=11/2 is the least so 8 is the pivot

  8. The next step is to reduce the pivot to 1 by dividing equation by 8

  9. We now reduce the other elements in the column of the pivot to zero:

  10. We have now completed the first iteration of the algorithm • This tells us that • P = 2 when y = z = s = 0 and x = 1/2, t= 1/2 • P= 2 is not the optimal solution as we still have negative numbers in the first row.

  11. So is the pivot We now repeat the process: chose a negative number in the first row and find a new pivot:

  12. We now repeat the process: choose a negative number in the first row and find a new pivot:

  13. Reduce the pivot to 1 by dividing equation ‘by 13/8 to get equation ”

  14. Reduce the other elements in the column of the pivot to zero

  15. We have now completed the second iteration of the algorithm • This tells us that • P = 210/11 when z = s = t=0 and x = 3/11, y= 4/11 • P= 210/11 is the optimal solution as we have no negative numbers in the first row. • P= 210/11 is the maximum value; we have finished!

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