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2. Simplex method example ( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm
  3. Maximization example-1
  4. Maximization example-2
  5. Maximization example-3
  6. BigM method Algorithm
  7. Minimization example-1
  8. Minimization example-2
  9. Minimization example-3
  10. Degeneracy example-1 (Tie for leaving basic variable)
  11. Degeneracy example-2 (Tie first Artificial variable removed)
  12. Unrestricted variable example
  13. Multiple optimal solution example
  14. Infeasible solution example
  15. Unbounded solution example
Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method

5. Maximization example-3
(Previous example)
7. Minimization example-1
(Next example)

6. BigM method Algorithm





BigM Method Steps (Rule)
Step-1: Formulate the Problem

a. Formulate the mathematical model of the given linear programming problem.

b. All the `X_(B_i) > 0`. So if any `X_(B_i) < 0` then multiply the corresponding constraint by -1 to make `X_(B_i) > 0`. So sign `<=` changed to `>=` and vice varsa

c. Convert LP problem into standard form by adding slack, surplus and artificial variables for each constraint.

c. In the objective function, assign a 0 coefficient to slack and surplus variables. Then add -M (for Maximization case) or +M (for Minimization case) to artificial variables. (M is very large positive value)
Step-2: Find out the Initial basic solution

Find the initial basic feasible solution by setting zero value to the decision variables. That shown in the example-1
Step-3: Test for Optimality

a. Calculate the values of `c_j - z_j` in the last row of simplex table.

b. For Maximization case

1. If all `c_j - z_j <= 0` , the current basic feasible solution is the optimal solution.

2. In `c_j - z_j > 0`, then select the variable that has largest `c_j - z_j` and enter this variable into the new table. This column is called key column (pivot column).

c. For Minimization case

1. If all `c_j - z_j >= 0` , the current basic feasible solution is the optimal solution.

2. In `c_j - z_j < 0`, then select the variable that has smallest `c_j - z_j` and enter this variable into the new table. This column is called key column (pivot column).
Step-4: Test for Feasibility (variable to leave the basis)

a. Find the ratio by dividing the values of `X_B` column by the positive values of key column (say `a_(ij)>0`)

b. Find the minimum ratio and this row is called key row (pivot row) and corresponding variable will leave the solution.

c. The intersection element of key row and key column is called key element (pivot element).
Step-5: Determine the new solution

a. The new values of key row can be obtained by dividing the key row elements by the pivot element.

eg. `R_1`(new)`=``R_1`(old) `-:3`

b. The numbers in the remaining rows can be computed by utilizing the following formula:

Row(new) = Row(old) - (value of key column and Row(old)) `xx` KeyRow(new)

eg. `R_2`(new)`=``R_2`(old) `- 2` `R_1`(new)
Step-6: Repeat the procedure

Goto step 3 and repeat the procedure until all the values of `c_j - z_j<=0` (For Maximization case) or `c_j - z_j>=0` (For Minimization case).



This material is intended as a summary. Use your textbook for detail explanation.
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5. Maximization example-3
(Previous example)
7. Minimization example-1
(Next example)





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