Simplex Method

1. Simplex Method Maximization by Dr. Arshadzaheer

2. Steps of Simplex Method Express the problem in equation form Write the inequalities in form of equalities Determine an initial feasible solution Write initial tableau / iteration Determine the criteria for optimality Select an entering variable using the optimality condition Select a leaving variable using the feasibility condition Write the optimal solution

3. Steps of Simplex Method (Details) There are following steps related to solving a Maximization problem by simplex method • Express the problem in equation form • In the case the given problem is in the form of statements we need to translate it in equations • Write the inequalities in form of equalities • In this step we encounter normally three kinds of problem

4. Steps of Simplex Method (Details)

5. Steps of Simplex Method (Details)

6. Steps of Simplex Method (Details)

7. Steps of Simplex Method (Details) 3. Determine an initial feasible solution To determine an initial feasible solution see how much equations and variables are there. This will help in deciding how much variables would be given arbitrary values Arbitrary values = # of Variables -- # of Equations Suppose we have 6 variables and 4 equations; Arbitrary values = 6 – 4 = 2 So the two variables will be given arbitrary values equal to zero All the variables and the RHSs of equations should be non negative

8. Steps of Simplex Method (Details) 4.Write initial tableau / Iteration The values from the initial solution will be put into the initial tableau; Non basic Variables having zero values called non basic variable All the non zero variables from initial solution • The coefficient of constraints will be written here The RHS of every equation will came here The coefficient of objective function

9. Steps of Simplex Method (Details) 5. Determine the criteria for optimality Solution of current tableau is said to be optimal if all the objective function co-efficient becomes Non Negative / Positive

10. Steps of Simplex Method (Details) 6.Select an entering variable using the optimality condition The entering variable in a maximization probe is non basic variable with most negative co efficient in the f- row. The optimum is reached at the iteration where all the f- row coefficients are non negative

11. Steps of Simplex Method (Details) 7.Select a leaving variable using the feasibility condition In case of maximization the leaving variable is the basic variable associated with the smallest non negative ratio with the strictly positive denominator. There is no need to take ratios for negative and zero because they can not be entered into basic variable column.

12. Steps of Simplex Method (Details) 8. Write the optimal solution When the solution criteria is satisfied, the values of the decision variables will be taken from tableau from corresponding RHS. Put these values in the original maximization function to calculate optimal point.

13. Illustration No 1

14. Illustration Maximize: f=5x1+ 4x2 (Objective Function) Subject to: (Constraints) 6x1+ 4x2≤ 24 (1) x1+ 2x2≤ 6 (2) -x1+ x2≤ 1 (3) x2≤ 2 (4) (Non-Negative Constraints) x1, x2 ≥ 0

15. Rewriting Constraints equations in homogeneous order Step1.Rewriting Constraints equations in homogeneous order 6x1+ 4x2 ≤ 24 (1) x1+ 2x2 ≤ 6 (2) -x1+ x2≤ 1 (3) 0x1+ x2 ≤ 2 (4) x1, x2 ≥ 0

16. Inequalities Constraints in Equation Form Write inequalities constraints in equation form by adding slack variable “S”. e.g. 6x1+ 4x2+ S = 24 (S is slack variable) Let S1, S2, S3 and S4 be the slack variables for first, second, third and fourth constraints respectively. 6x1+ 4x2+ S1 = 24 …………. (5) x1+ 2x2+ S2 = 6 …………. (6) -x1+ x2 + S3 = 1 …………. (7) 0x1+ x2+ S4 = 2 …………. (8) x1, x2,S1, S2,S3, S4 ≥ 0 f-5X1 -4X2 =0

17. Initial feasible solution How many variable will be given Arbitrary values = # of Variables -- # of Equations 6 - 4 = 2 Let x1= 0, x2 = 0 Putting above values in objective function (f=5x1+ 4x2) and equation 5-8, f = 0 S1 = 24 S2 = 6 S3 = 1 S4 = 2 x1, x2,S1, S2,S3, S4 ≥ 0

18. Pivot Column Pivot Row Initial Tableau / Iteration Pivot No

19. Some cross checking measures The basic variables are equal to 1 under their respective columns and all the other values in columns will be zero, so if you don’t have zero or one on these position it means you have done some thing wrong.

20. Calculation for next Iteration = [1 2/3 1/6 0 0 0 4] New Row = Old Row – Pivot Column Coefficient x New Pivot Row New S2 Row = [1 2 0 1 0 0 6] - (1)[1 2/3 1/6 0 0 0 4] = [0 4/3 -1/6 1 0 0 2]

21. Calculation New S3 Row =[-1 1 0 0 1 0 1] -(-1) [1 2/3 1/6 0 0 0 4] = [0 5/3 1/6 0 1 0 5] New S4 Row = [0 1 0 0 0 1 2] - (0) [1 2/3 1/6 0 0 0 4] = [0 1 0 0 0 1 2] New f Row = [-5 -4 0 0 0 0 0] -(-5) [1 2/3 1/6 0 0 0 4] = [0 -2/3 5/6 0 0 0 20]

22. Calculation Still the solution is not optimal as there is -2/3 as negative coefficient of objective function so we again repeat the whole process

23. Calculation New Pivot Row= 3/4[0 4/3 -1/6 1 0 0 2] [0 1 -1/8 3/4 0 0 3/2]

24. Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New x1 Row = [1 2/3 1/6 0 0 0 4] - (2/3)[0 1 -1/8 3/4 0 0 3/2] = [1 0 1/4 -1/2 0 0 3] New S3 Row =[0 5/3 1/6 0 1 0 5] - 5/3[0 1 -1/8 3/4 0 0 3/2] = [0 0 3/8 -5/4 1 0 5/2] New S4 Row = [0 1 0 0 0 1 2] - (1) [0 1 -1/8 3/4 0 0 3/2] = [0 0 1/8 -3/4 0 1 1/2] New f Row = [0 -2/3 5/6 0 0 0 20] - (-2/3) [0 1 -1/8 3/4 0 0 3/2] = [0 0 3/4 1/2 0 0 21]

25. Criteria for optimality Solution of current tableau is said to be optimal if all the objective function co-efficient becomes Non Negative / Positive

26. Criteria for optimality satisfied Criteria for Optimum Solution because all coefficients (of objective function) are non-negative or zero

27. Optimal Solution X1= 3 X2= 3/2 f= 21 Cross checking of maximization point put values of X1 and X2 from above solution into original objective function f=5x1+ 4x2 =5 (3) + 4(3/2) =21

28. Stay Blessed