Two Phase Method - Part 2

This video lesson is available to enrolled students.

Enroll to watch — Coal India Management Trainee (CS) Recruitment 2026

AI Summary

An AI-generated summary of this video lecture.

The video is a lecture on solving a linear programming problem using the Two Phase Method, a technique for handling constraints that do not have an initial basic feasible solution. The instructor begins by presenting the problem, which includes an objective function to maximize and a set of constraints. He explains that the standard Simplex Method cannot be applied directly because the constraints are of the 'greater than or equal to' type, which prevents the immediate identification of a basic feasible solution. To overcome this, he introduces the Two Phase Method. In Phase I, an auxiliary objective function is created to minimize the sum of artificial variables (A1), with the goal of driving this sum to zero. The instructor sets up the initial simplex tableau for Phase I, showing the coefficients of the variables, including the artificial variable A1, and the right-hand side values. He then demonstrates the first iteration of the Simplex Method, identifying the entering variable (the one with the most negative coefficient in the objective row) and the leaving variable (determined by the minimum ratio test). The pivot element is identified, and the instructor proceeds to perform row operations to create a new tableau. The process continues, with the instructor updating the tableau, calculating new values, and checking for optimality. The video shows the step-by-step progression of the algorithm, including the calculation of ratios and the updating of the solution vector. The instructor's goal is to find a feasible solution where the artificial variable A1 is zero, indicating that a basic feasible solution has been found for the original problem. The video concludes with the final tableau of Phase I, setting the stage for Phase II, where the original objective function is optimized using the feasible basis found in Phase I.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The video opens with a whiteboard displaying a linear programming problem. The instructor, standing to the right, begins to explain the problem. The objective is to maximize Z = 5x1 - x2 + 3x3, subject to the constraints: 2x1 + x2 - x3 = 20, 6x1 + 5x2 + 10x3 = 76, and 8x1 - 3x2 + 6x3 = 50, with all variables being non-negative. The instructor notes that the constraints are of the 'greater than or equal to' type, which means the standard Simplex Method cannot be applied directly because there is no initial basic feasible solution. He introduces the Two Phase Method as a solution. The first step is to create an auxiliary problem to find a basic feasible solution. The objective for Phase I is to minimize Z' = A1, where A1 is an artificial variable. The instructor writes the initial simplex tableau for Phase I, with the objective row showing coefficients for the variables x1, x2, x3, A1, S1, S2, S3, and the solution column. The tableau shows the coefficients of the constraints and the right-hand side values. The instructor points to the tableau, explaining that the goal is to make the artificial variable A1 zero, which would mean a feasible solution has been found for the original problem.

2. 2:00 – 5:00 02:00-05:00

   The instructor continues to work on the Phase I simplex tableau. He identifies the entering variable by looking for the most negative coefficient in the objective row (Z' - A1). The coefficient for A1 is -1, which is the most negative, so A1 is the entering variable. He then performs the minimum ratio test to determine the leaving variable. The ratios are calculated by dividing the right-hand side (Soln) by the corresponding coefficient in the A1 column. The ratios are 20/2 = 10, 76/5 = 15.2, and 50/(-3) = -16.67. The minimum positive ratio is 10, which corresponds to the first row. Therefore, S1 is the leaving variable. The pivot element is the intersection of the A1 column and the S1 row, which is 2. The instructor circles this pivot element. He then begins to perform row operations to make the pivot element 1 and to eliminate the other entries in the A1 column. He divides the first row by 2 to make the pivot element 1. The new first row becomes: 1, 0.5, -0.5, 1, 0, 0, 0, 10. He then uses this new row to eliminate the A1 coefficient in the other rows. For the second row, he subtracts 5 times the new first row from the original second row. For the third row, he adds 3 times the new first row to the original third row. The instructor writes down the new values for the second and third rows, updating the tableau.

3. 5:00 – 10:00 05:00-10:00

   The instructor has updated the tableau after the first iteration of Phase I. He now checks for optimality. The objective row (Z' - A1) has coefficients: 0, 0.5, -0.5, 0, 0, 0, 0, 10. Since there is a negative coefficient (-0.5) in the x3 column, the solution is not optimal, and another iteration is needed. The entering variable is x3. He performs the minimum ratio test for x3. The ratios are 10 / (-0.5) = -20, 76 / 10 = 7.6, and 50 / 6 = 8.33. The minimum positive ratio is 7.6, which corresponds to the second row. Therefore, S2 is the leaving variable. The pivot element is the intersection of the x3 column and the S2 row, which is 10. The instructor circles this pivot element. He then performs row operations. He divides the second row by 10 to make the pivot element 1. The new second row becomes: 0, 0.5, 1, 0, 0.1, 0, 0, 7.6. He then uses this new row to eliminate the x3 coefficient in the other rows. For the first row, he adds 0.5 times the new second row to the original first row. For the objective row, he adds 0.5 times the new second row to the original objective row. The instructor writes down the new values for the first row and the objective row, updating the tableau. He then checks for optimality again. The objective row now has coefficients: 0, 0.75, 0, 0, 0.05, 0, 0, 13.8. Since all coefficients are non-negative, the solution is optimal for Phase I. The value of the artificial variable A1 is 0, which means a basic feasible solution has been found for the original problem.

4. 10:00 – 14:02 10:00-14:02

   The instructor now transitions to Phase II of the Two Phase Method. He removes the artificial variable A1 and its column from the tableau, as it is no longer needed. He also removes the Phase I objective row (Z' - A1) and replaces it with the original objective function, Z = 5x1 - x2 + 3x3. The new objective row is: 0, 0, 0, 0, 0, 0, 0, 0, but with the coefficients of the original objective function. The instructor writes the new objective row as: 5, -1, 3, 0, 0, 0, 0, 0. He then checks for optimality. The coefficients in the objective row are: 5, -1, 3, 0, 0, 0, 0. Since there are negative coefficients (-1 and 3), the solution is not optimal. The entering variable is x2 (the one with the most negative coefficient). He performs the minimum ratio test for x2. The ratios are 10 / 0.5 = 20, 7.6 / 0.5 = 15.2, and 50 / 1 = 50. The minimum positive ratio is 15.2, which corresponds to the second row. Therefore, S2 is the leaving variable. The pivot element is the intersection of the x2 column and the S2 row, which is 0.5. The instructor circles this pivot element. He then performs row operations. He divides the second row by 0.5 to make the pivot element 1. The new second row becomes: 0, 1, 2, 0, 0.2, 0, 0, 15.2. He then uses this new row to eliminate the x2 coefficient in the other rows. For the first row, he subtracts 0.5 times the new second row from the original first row. For the objective row, he adds 1 times the new second row to the original objective row. The instructor writes down the new values for the first row and the objective row, updating the tableau. He then checks for optimality again. The objective row now has coefficients: 5, 0, 5, 0, 0.2, 0, 0, 15.2. Since all coefficients are non-negative, the solution is optimal. The optimal solution is x1 = 0, x2 = 15.2, x3 = 0, and the maximum value of Z is 15.2.

The video provides a comprehensive, step-by-step demonstration of the Two Phase Method for solving a linear programming problem. It begins by identifying the challenge of an initial infeasible solution due to 'greater than or equal to' constraints. The method is broken down into two distinct phases. Phase I is dedicated to finding a basic feasible solution by minimizing the sum of artificial variables. The instructor meticulously walks through the first iteration of the Simplex Method, identifying the pivot element, performing row operations, and checking for optimality. The process is repeated until the artificial variable is driven to zero, confirming a feasible solution. The video then transitions to Phase II, where the original objective function is optimized using the feasible basis found in Phase I. The instructor demonstrates the final iteration, showing how the optimal solution is reached. The entire process is illustrated on a whiteboard, with clear labeling of the tableau, variables, and calculations, making it an excellent educational resource for understanding this advanced optimization technique.