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Simplex Algorithm
- 1. Linear programming simplex method This presentation will help you to solve linear programming problems using the Simplex tableau Steve Bishop 2004, 2007, 2012 1
- 2. This is a typical linear programming problem Keep left clicking the mouse to reveal the next part Maximise P = 2x + 3y+ 4z The objective function Subject to 3x +2y + z ≤ 10 The constraints 2x + 5y+ 3z ≤15 x, y ≥ 0 2
- 3. The first step is to rewrite objective formula so it is equal to zero We then need to rewrite P = 2x + 3y + 4z as: P – 2x – 3y – 4z = 0 This is called the objective equation 3
- 4. Next we need to remove the inequalities from the constraints This is done by adding slack variables Adding slack variables s and t to the inequalities, we can write them as equalities: 3x +2y + z + s = 10 2x + 5y+ 3z + t = 15 These are called the constraint equations 4
- 5. Next, these equations are placed in a tableau P x y z s t values 5
- 6. First the Objective equation P – 2x – 3y – 4z =0 P x y z s t values 1 -2 -3 -4 0 0 0 6
- 7. Next, the constraint equations 3x +2y +z +s = 10 2x + 5y + 3z + t = 15 P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 7
- 8. We now have to identify the pivot First, we find the pivot column The pivot column is the column with the greatest negative value in the objective equation Mark this column with an arrow P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 8
- 9. The next step in identifying the pivot is to find the pivot row This is done by dividing the values of the constraints by the value in the pivot column P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 9
- 10. 10 divided by 1 = 10 15 divided by 3 = 5 The lowest value gives the pivot row This is marked with an arrow P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 10
- 11. The pivot is where the pivot column and pivot row intersect This is normally marked with a circle P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 11
- 12. We now need to make the pivot equal to zero To do this divide the pivot row by the pivot value P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 12
- 13. P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2/3 5/3 3/3 0 1/3 15/3 13
- 14. It is best to keep the values as fractions P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2/3 5/3 1 0 1/3 5 14
- 15. Next we need to make the pivot column values into zeros To do this we add or subtract multiples of the pivot column P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2/3 5/3 1 0 1/3 5 15
- 16. The pivot column in the objective function will go to zero if we add four times the pivot row The pivot column value in the other constraint will go to zero if we subtract the pivot row P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2/3 5/3 1 0 1/3 5 16
- 17. Subtracting 4 X the pivot row from the objective function gives: P x y z s t values 1+ -2+ -3+ -4+ 0+ 0+ 0+ 4(0) 4(2/3) 4(5/3) 4(1) 4(0) 4(1/3) 4(5) 0 3 2 1 1 0 10 0 2/3 5/3 1 0 1/3 5 17
- 18. Subtracting 4 X the pivot row from the objective function gives: P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 3 2 1 1 0 10 0 2/3 5/3 1 0 1/3 5 18
- 19. Subtracting 1 X the pivot row from the other constraint gives: P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 3- 2- 1- 1- 0- 10- 2/3 5/3 1 0 1/3 5 0 2/3 5/3 1 0 1/3 5 19
- 20. Subtracting 1 times the pivot row from the other constraint gives: P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 0 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 20
- 21. This then is the first iteration of the Simplex algorithm P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 0 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 21
- 22. If any values in the objective equation are negative we have to repeat the whole process Here, there are no negative values. This is then the optimal solution. P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 0 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 22
- 23. We now have to determine the values of the variables that give the optimum solution P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 0 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 23
- 24. Reading the objective function: P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 1 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 24
- 25. Reading the objective function: The maximum value for P is then 20 For this to be the case x, y and t must be zero In general any variable that has a value in the objective function is zero. 25
- 26. From the objective function x, y and t are zero – substituting these values in to the constraint rows gives: s = 5 and z = 5 P x y z s t values 1 2/3 11/3 0 0 4/3 20 0 7/3 1/3 0 1 -1/3 5 0 2/3 5/3 1 0 1/3 5 26
- 27. The solution to the linear programming problem: maximise: P = 2x + 3y+ 4z subject to: 3x +2y + z ≤ 10 2x + 5y+ 3z ≤15 x, y ≥ 0 is: P = 20 x = 0, y = 0 and z = 5 27
- 28. There are 9 steps in the Simplex algorithm 2. Rewrite objective formula so it is equal to zero 3. Add slack variables to remove inequalities 4. Place in a tableau 5. Look at the negative values to determine pivot column 6. Divide end column by the corresponding pivot value in that column 7. The least value is the pivot row 8. Divide pivotal row by pivot value – so pivot becomes 1 9. Add/subtract multiples of pivot row to other rows to zero 10. When top element are ≥ 0 then optimised solution 28
- 29. Go to this web site to solve any linear programme problem using the simplex method: For on-line exercises go here: 29
Editor's Notes
- #5: Steve Bishop 2004