simplex method
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The document provides an overview of the simplex method for solving linear programming problems. It discusses: - The simplex method is an iterative algorithm that generates a series of solutions in tabular form called tableaus to find an optimal solution. - It involves writing the problem in standard form, introducing slack variables, and constructing an initial tableau. - The method then performs iterations involving selecting a pivot column and row, and applying row operations to generate new tableaus until an optimal solution is found. - It also discusses how artificial variables are introduced for problems with non-strict inequalities and provides an example solved using the simplex method.
simplex method
- 1. 6s-1 Linear Programming SIMPLEX METHOD Presented by:khushbu Kumari Deptt:Agricultural Economics M.Sc(Ag.) Previous
- 2. 6s-2 Linear Programming Simplex: A linear-programming algorithm that can solve problems having more than two decision variables. The simplex technique involves generating a series of solutions in tabular form, called tableaus. This process continues as long as a positive (negative) rate of profit (cost) exists. The simplex method is an iterative algorithm (a systematic solution procedure that keeps repeating a fixed series of steps, called, an iteration. WDDit is always advisable to use Simplex Method to avoid lengthy graphical procedure Simplex Method
- 3. 6s-3 Linear Programming SIMPLEX METHOD Step-1 Write the standard maximization problem in standard form, introduce slack variables to form the initial system, and write the initial tableau. Step-3 Select the pivot column Step-5 Select the pivot element and perform the pivot operation STOP The optimal solution has been found. STOP The linear programming problem has no optimal solution Step 2 Are there any negative indicators in the bottom row? Step 4 Are there any positive elements in the pivot column above the dashed line? Simplex algorithm for standard maximization problems
- 4. 6s-4 Linear Programming Slack Variables A mathematical representation of surplus resources.” In real life problems, it’s unlikely that all resources will be used completely, so there usually are unused resources. Slack variables represent the unused resources between the left-hand side and right-hand side of each inequality.
- 5. 6s-5 Linear Programming Simplex algorithm Initialization: setup to start iterations, including finding an initial solution Optimality test: is the current solution optimal? if no if yes stop Iteration: Perform an iteration to find a better solution
- 6. 6s-6 Linear Programming The simplex method in tabular form Steps: 1. Initialization: a. transform all the constraints to equality by introducing slack, surplus, and artificial variables as follows: Constraint type Variable to be added ≥ + slack (s) ≤ - Surplus (s) + artificial (A) = + Artificial (A)
- 7. 6s-7 Linear Programming Simplex method in tabular form b. Construct the initial simplex tableau Basic variable X1 … Xn S1 …... Sn A1 …. An RHS S Coefficient of the constraints b1 A bm Z Objective function coefficient In different signs Z value
- 8. 6s-8 Linear Programming 2. Test for optimality: Case 1: Maximization problem the current basic feasible solution is optimal if every coefficient in the objective function row is negative Case 2: Minimization problem the current BF solution is optimal if every coefficient in the objective function row is positive. Simplex method in tabular form
- 9. 6s-9 Linear Programming Simplex method in tabular form 3. Iteration Step 1: determine the entering basic variable by selecting the variable (automatically a nonbasic variable) with the most negative value (in case of maximization) or with the most positive (in case of minimization) in the last row (Z-row). Put a box around the column below this variable, and call it the “pivot column”
- 10. 6s-10 Linear Programming Simplex method in tabular form Step 2: Determine the leaving basic variable by applying the minimum ratio test as following: 1. Pick out each coefficient in the pivot column that is strictly positive (>0) 2. Divide each of these coefficients into the right hand side entry for the same row 3. Identify the row that has the smallest of these ratios 4. The basic variable for that row is the leaving variable, so replace that variable by the entering variable in the basic variable column of the next simplex tableau. Put a box around this row and call it the “pivot row”
- 11. 6s-11 Linear Programming Simplex method in tabular form Step 3: Solve for the new BF solution by using elementary row operations (multiply or divide a row by a nonzero constant; add or subtract a multiple of one row to another row) to construct a new simplex tableau, and then return to the optimality test. The specific elementary row operations are: 1. Divide the pivot row by the “pivot number” (the number in the intersection of the pivot row and pivot column) 2. For each other row that has a negative coefficient in the pivot column, add to this row the product of the absolute value of this coefficient and the new pivot row. 3. For each other row that has a positive coefficient in the pivot column, subtract from this row the product of the absolute value of this coefficient and the new pivot row.
- 12. 6s-12 Linear Programming Simplex method Example (All constraints are ) Solve the following problem using the simplex method Maximize Z = 3X1+ 5X2 Subject to X1 4 2 X2 12 3X1 +2X2 18 X1 , X2 0
- 13. 6s-13 Linear Programming Simplex method Solution Initialization 1. Standard form Maximize Z, Subject to Z - 3X1- 5X2 = 0 X1 + S1 = 4 2 X2 + S2 = 12 3X1 +2X2 + S3 = 18 X1 , X2, S1, S2, S3 0
- 14. 6s-14 Linear Programming Initial tableau 2. Initial tableau Basic variable X1 X2 S1 S2 S3 RHS S1 1 0 1 0 0 4 S2 0 2 0 1 0 12 S3 3 2 0 0 1 18 Z -3 -5 0 0 0 0 Pivot column Pivot row Pivot number Entering variable Leaving variable
- 15. 6s-15 Linear Programming Simplex tableau Notes: The basic feasible solution at the initial tableau is (0, 0, 4, 12, 18) where: X1 = 0, X2 = 0, S1 = 4, S2 = 12, S3 = 18, and Z = 0 Where S1, S2, and S3 are basic variables X1 and X2 are nonbasic variables The solution at the initial tableau is associated to the origin point at which all the decision variables are zero.
- 16. 6s-16 Linear Programming Optimality test By investigating the last row of the initial tableau, we find that there are some negative numbers. Therefore, the current solution is not optimal.
- 17. 6s-17 Linear Programming Iteration Step 1: Determine the entering variable by selecting the variable with the most negative in the last row. From the initial tableau, in the last row (Z row), the coefficient of X1 is -3 and the coefficient of X2 is -5; therefore, the most negative is -5. consequently, X2 is the entering variable. X2 is surrounded by a box and it is called the pivot column
- 18. 6s-18 Linear Programming Iteration Step 2: Determining the leaving variable by using the minimum ratio test as following: Basic variable Entering variable X2 (1) RHS (2) Ratio (2)(1) S1 0 4 None S2 Leaving 2 12 6 Smallest ratio S3 2 18 9
- 19. 6s-19 Linear Programming Iteration Step 3: solving for the new BF solution by using the eliminatory row operations as following: 1. New pivot row = old pivot row pivot number Basic variable X1 X2 S1 S2 S3 RHS S1 X2 0 1 0 1/2 0 6 S3 Z Note that X2 becomes in the basic variables list instead of S2
- 20. 6s-20 Linear Programming iteration 2. For the other row apply this rule: New row = old row – the coefficient of this row in the pivot column (new pivot row). For S1 1 0 1 0 0 4 - 0 (0 1 0 1/2 0 6) 1 0 1 0 0 4 For S3 3 2 0 0 1 18 - 2 (0 1 0 1/2 0 6) 3 0 0 -1 1 6 for Z -3 -5 0 0 0 0 - -5(0 1 0 1/2 0 6) -3 0 0 5/2 0 30
- 21. 6s-21 Linear Programming Iteration Basic variable X1 X2 S1 S2 S3 RHS S1 1 0 1 0 0 4 X2 0 1 0 1/2 0 6 S3 3 0 0 -1 1 6 Z -3 0 0 5/2 0 30 The most negative value; therefore, X1 is the entering variable The smallest ratio is 6/3 =2; therefore, S3 is the leaving variable This solution is not optimal, since there is a negative numbers in the last row
- 22. 6s-22 Linear Programming Iteration Apply the same rules we will obtain this solution: Basic variable X1 X2 S1 S2 S3 RHS S1 0 0 1 1/3 -1/3 2 X2 0 1 0 1/2 0 6 X1 1 0 0 -1/3 1/3 2 Z 0 0 0 3/2 1 36 This solution is optimal; since there is no negative solution in the last row: basic variables are X1 = 2, X2 = 6 and S1 = 2; the nonbasic variables are S2 = S3 = 0 Z = 36
- 23. 6s-23 Linear Programming Simplex method in case of Artificial variables Solve the following linear programming problem by using the simplex method: Min Z =2 X1 + 3 X2 S.t. ½ X1 + ¼ X2 ≥ 4 X1 + 3X2 20 X1 + X2 = 10 X1, X2 0
- 24. 6s-24 Linear Programming . Solution Step 1: standard form Min Z, s.t. Z – 2 X1 – 3 X2 - M A1 -M A2 = 0 ½ X1 + ¼ X2 + S1 = 4 X1 + 3X2 - S2 + A1 = 20 X1 + X2 + A2 = 10 X1, X2 ,S1, S2, A1, A2 0
- 25. 6s-25 Linear Programming . Step 2: Initial tableau Basic variables X1 2 X2 3 S1 0 S2 0 A1 M A2 M RHS S1 ½ ¼ 1 0 0 0 4 A1 1 3 0 -1 1 0 20 A2 1 1 0 0 0 1 10 Z -2 -3 0 0 -M -M 0
- 26. 6s-26 Linear Programming . To correct this violation before starting the simplex algorithm, the elementary row operations are used as follows: New (Z row) = old (z row) ± M (A1 row) ± M (A2 row) In our case, it will be positive since M is negative in the Z row, as following: Old (Z row): -2 -3 0 0 -M -M 0 M (A1 row): M 3M 0 -M M o 20M M (A2 row): M M 0 0 0 M 10M New (Z row):2M-2 4M-3 0 -M 0 0 30M It becomes zero
- 27. 6s-27 Linear Programming The initial tableau will be: Basic variables X1 2 X2 3 S1 0 S2 0 A1 M A2 M RHS S1 1/2 1/4 1 0 0 0 4 A1 1 3 0 -1 1 0 20 A2 1 1 0 0 0 1 10 Z 2M-2 4M-3 0 -M 0 0 30M • Since there is a positive value in the last row, this solution is not optimal • The entering variable is X2 (it has the most positive value in the last row) • The leaving variable is A1 (it has the smallest ratio)
- 28. 6s-28 Linear Programming . First iteration Basic variables X1 2 X2 3 S1 0 S2 0 A1 M A2 M RHS S1 5/12 0 1 1/12 -1/12 0 7/3 X2 1/3 1 0 -1/3 1/3 0 20/3 A2 2/3 0 0 1/3 -1/3 1 10/3 Z 2/3M-1 0 0 1/3M-1 1-4/3M 0 20+10/3M • Since there is a positive value in the last row, this solution is not optimal • The entering variable is X1 (it has the most positive value in the last row) • The leaving variable is A2 (it has the smallest ratio)
- 29. 6s-29 Linear Programming . Second iteration Basic variables X1 X2 S1 S2 A1 A2 RHS S1 0 0 1 -1/8 1/8 -5/8 1/4 X2 0 1 0 -1/2 1/2 -1/2 5 X1 1 0 0 1/2 -1/2 3/2 5 Z 0 0 0 -1/2 ½-M 3/2-M 25 This solution is optimal, since there is no positive value in the last row. The optimal solution is: X1 = 5, X2 = 5, S1 = ¼ A1 = A2 = 0 and Z = 25
- 30. 6s-30 Linear Programming . In the final tableau, if one or more artificial variables (A1, A2, …) still basic and has a nonzero value, then the problem has an infeasible solution. All other notes are still valid in the Big M method.
- 31. 6s-31 Linear Programming THANK YOU