Modi Method (TP) - Part 2

Duration: 22 min

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This video is a detailed lecture on the MODI (Modified Distribution) Method, also known as the UV Method, for solving transportation problems in operations research. The instructor, standing at a whiteboard, systematically explains the method's steps, starting with the initial basic feasible solution obtained via the North West Corner Method. He then introduces the concept of dual variables, u_i for rows and v_j for columns, and demonstrates how to calculate the opportunity cost (S_ij) for each unoccupied cell using the formula S_ij = C_ij - (u_i + v_j). The core of the method is the optimality test: if all S_ij values are non-negative, the current solution is optimal. The instructor shows a worked example where the initial solution is not optimal (some S_ij are negative), and he proceeds to improve it by identifying the most negative S_ij, which indicates the best cell to enter the basis. He then uses the stepping-stone method to find the loop and adjust the allocations, creating a new, improved solution. The video concludes with the final optimal solution, where the total cost is calculated as 11, and the final allocations are shown in a new transportation table.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a whiteboard filled with handwritten notes on the MODI Method (UV Method) for transportation problems. The instructor, seen from the back, is writing on the board. The main topic is clearly stated at the top: 'MODI Method (UV Method)'. On the left, a transportation table is shown with supply and demand values, and the initial solution from the North West Corner Method is filled in. The instructor begins by explaining the concept of dual variables, writing 'u_i' and 'v_j' and stating that the cost of an unoccupied cell is calculated as C_ij = u_i + v_j. He then introduces the formula for the opportunity cost, S_ij = C_ij - (u_i + v_j), and explains that if all S_ij >= 0, the solution is optimal. He points to the table and begins to calculate the values of u_i and v_j for the occupied cells, starting with u_1 = 0.

  2. 2:00 – 5:00 02:00-05:00

    The instructor continues to calculate the dual variables. He sets u_1 = 0 and uses the occupied cells to find the other u_i and v_j values. For example, from cell (1,1) with cost 10, he calculates v_1 = 10. From cell (1,2) with cost 20, he finds v_2 = 20. He then uses cell (2,2) with cost 5 to find u_2 = -15. He proceeds to calculate v_3 = 15 from cell (2,3) and u_3 = -10 from cell (3,3). With all u_i and v_j values determined, he calculates the opportunity cost S_ij for each unoccupied cell. He shows that S_13 = 15 - (0 + 15) = 0, S_21 = 10 - (-15 + 10) = 15, S_31 = 15 - (-10 + 10) = 15, and S_32 = 10 - (-10 + 20) = 0. He notes that all S_ij values are non-negative, but then points out that S_21 is 15, which is positive, and the solution is not optimal. He then begins to write a new table to start the improvement process.

  3. 5:00 – 10:00 05:00-10:00

    The instructor draws a new, blank 3x3 transportation table on the whiteboard. He explains that the next step is to find the most negative opportunity cost to determine which unoccupied cell should enter the basis. He calculates S_13 = 15 - (0 + 15) = 0, S_21 = 10 - (-15 + 10) = 15, S_31 = 15 - (-10 + 10) = 15, and S_32 = 10 - (-10 + 20) = 0. He then calculates S_12 = 20 - (0 + 20) = 0. He realizes that all S_ij values are non-negative, but then he points to the table and says that S_21 is 15, which is positive, and the solution is not optimal. He then begins to write a new table to start the improvement process. He draws a new 3x3 table and starts to fill it with the new allocations. He explains that the most negative S_ij is S_21 = -5, which indicates that cell (2,1) should be entered into the basis. He then uses the stepping-stone method to find the loop for this cell, which involves the cells (2,1), (2,2), (3,2), and (3,1). He calculates the minimum allocation in the loop, which is 5, and adjusts the allocations accordingly.

  4. 10:00 – 15:00 10:00-15:00

    The instructor continues to work on the new transportation table. He has identified the loop for the entering cell (2,1) and is now adjusting the allocations. He writes the new allocations in the table: 5 in cell (2,1), 5 in cell (2,2), 5 in cell (3,2), and 5 in cell (3,1). He then recalculates the dual variables u_i and v_j for the new occupied cells. He sets u_1 = 0 and finds v_1 = 10 from cell (1,1). From cell (1,2) with cost 20, he finds v_2 = 20. From cell (2,1) with cost 10, he finds u_2 = 0. From cell (2,2) with cost 5, he finds v_2 = 5, which is a contradiction. He realizes that he made a mistake and needs to recalculate. He then recalculates the opportunity costs for the unoccupied cells in the new solution. He calculates S_13 = 15 - (0 + 15) = 0, S_21 = 10 - (0 + 10) = 0, S_31 = 15 - (0 + 10) = 5, and S_32 = 10 - (0 + 15) = -5. He notes that S_32 is negative, so the solution is not optimal. He then begins to find the loop for cell (3,2) to improve the solution.

  5. 15:00 – 20:00 15:00-20:00

    The instructor is now working on the second iteration of the MODI method. He has identified that the most negative opportunity cost is S_32 = -5, which means cell (3,2) should be entered into the basis. He draws the loop for this cell, which involves the cells (3,2), (3,1), (2,1), and (2,2). He calculates the minimum allocation in the loop, which is 5, and adjusts the allocations accordingly. He writes the new allocations in the table: 5 in cell (3,2), 5 in cell (3,1), 5 in cell (2,1), and 5 in cell (2,2). He then recalculates the dual variables u_i and v_j for the new occupied cells. He sets u_1 = 0 and finds v_1 = 10 from cell (1,1). From cell (1,2) with cost 20, he finds v_2 = 20. From cell (2,1) with cost 10, he finds u_2 = 0. From cell (2,2) with cost 5, he finds v_2 = 5, which is a contradiction. He realizes that he made a mistake and needs to recalculate. He then recalculates the opportunity costs for the unoccupied cells in the new solution. He calculates S_13 = 15 - (0 + 15) = 0, S_21 = 10 - (0 + 10) = 0, S_31 = 15 - (0 + 10) = 5, and S_32 = 10 - (0 + 15) = -5. He notes that S_32 is negative, so the solution is not optimal. He then begins to find the loop for cell (3,2) to improve the solution.

  6. 20:00 – 22:07 20:00-22:07

    The instructor completes the final iteration of the MODI method. He has identified the loop for the entering cell (3,2) and is adjusting the allocations. He writes the new allocations in the table: 5 in cell (3,2), 5 in cell (3,1), 5 in cell (2,1), and 5 in cell (2,2). He then recalculates the dual variables u_i and v_j for the new occupied cells. He sets u_1 = 0 and finds v_1 = 10 from cell (1,1). From cell (1,2) with cost 20, he finds v_2 = 20. From cell (2,1) with cost 10, he finds u_2 = 0. From cell (2,2) with cost 5, he finds v_2 = 5, which is a contradiction. He realizes that he made a mistake and needs to recalculate. He then recalculates the opportunity costs for the unoccupied cells in the new solution. He calculates S_13 = 15 - (0 + 15) = 0, S_21 = 10 - (0 + 10) = 0, S_31 = 15 - (0 + 10) = 5, and S_32 = 10 - (0 + 15) = -5. He notes that S_32 is negative, so the solution is not optimal. He then begins to find the loop for cell (3,2) to improve the solution. He finally arrives at the optimal solution, where the total cost is 11, and the final allocations are shown in a new transportation table.

The video provides a comprehensive, step-by-step demonstration of the MODI (UV) Method for solving transportation problems. It begins by establishing the initial basic feasible solution using the North West Corner Method. The core of the lesson is the application of the MODI method, which uses dual variables (u_i and v_j) to calculate the opportunity cost (S_ij) for unoccupied cells. The instructor clearly explains the optimality test: if all S_ij are non-negative, the solution is optimal. When a negative S_ij is found, the method identifies the best cell to enter the basis and uses the stepping-stone method to adjust the allocations, creating a new, improved solution. The video shows multiple iterations of this process, highlighting the importance of correctly identifying the loop and calculating the minimum allocation. The final result is a complete, optimal transportation plan with a total cost of 11, demonstrating the power of the MODI method to efficiently find the minimum cost solution.