Vogel's Approximation Method - Part-2

Duration: 15 min

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This video is a lecture on solving a transportation problem in linear programming using Vogel's Approximation Method (VAM). The instructor, standing at a whiteboard, begins by presenting a transportation table with supply and demand values for four sources (Kolkata, Hyderabad, Delhi, Mumbai) and four destinations (Pune, Pune, Patna, Indore). The method involves calculating penalties for each row and column by finding the difference between the two smallest costs. The instructor then identifies the highest penalty, which is 5, and allocates the maximum possible quantity to the cell with the lowest cost in that row or column. This process is repeated, with the instructor marking allocations and recalculating penalties, until all supply and demand constraints are satisfied. The final step is to calculate the total transportation cost by summing the product of the allocated quantities and their respective costs, resulting in a total cost of 12,075.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a whiteboard displaying a transportation problem. The instructor, seen from the back, begins to explain the problem. The board shows a table with four sources (Kolkata, Hyderabad, Delhi, Mumbai) and four destinations (Pune, Pune, Patna, Indore). The supply for each source is 250, 250, 300, and 400 respectively, and the demand for each destination is 200, 225, 275, and 250. The instructor starts by calculating the penalties for each row and column, which are the differences between the two smallest costs in that row or column. He identifies the highest penalty, which is 5, and points to the corresponding row or column.

  2. 2:00 – 5:00 02:00-05:00

    The instructor proceeds with the Vogel's Approximation Method. He identifies the highest penalty, which is 5, and allocates the maximum possible quantity to the cell with the lowest cost in that row or column. He allocates 200 units to the cell (Kolkata, Pune) with a cost of 11, as this is the lowest cost in the row with the highest penalty. He then crosses out the column for Pune, as its demand is now satisfied. He recalculates the penalties for the remaining rows and columns, identifying the new highest penalty, which is 1, and allocates 25 units to the cell (Hyderabad, Patna) with a cost of 13. He crosses out the row for Hyderabad, as its supply is now satisfied.

  3. 5:00 – 10:00 05:00-10:00

    The instructor continues the process. He recalculates the penalties and identifies the highest penalty, which is 2, and allocates 25 units to the cell (Delhi, Indore) with a cost of 10. He crosses out the column for Indore, as its demand is now satisfied. He then recalculates the penalties again. The highest penalty is now 1, and he allocates 275 units to the cell (Mumbai, Patna) with a cost of 13. He crosses out the row for Mumbai, as its supply is now satisfied. He then allocates the remaining 25 units to the cell (Delhi, Pune) with a cost of 16, as this is the only remaining cell. He crosses out the row for Delhi, as its supply is now satisfied.

  4. 10:00 – 14:55 10:00-14:55

    The instructor now calculates the total transportation cost. He multiplies the allocated quantities by their respective costs and sums them up. The calculation is: (200 * 11) + (25 * 13) + (275 * 13) + (25 * 16) + (25 * 10) + (25 * 10). He writes this out on the board and calculates the sum, which is 2200 + 325 + 3575 + 400 + 250 + 250, resulting in a total cost of 12,075. He then writes the final answer on the board, indicating that the minimum transportation cost is 12,075.

The video provides a step-by-step demonstration of Vogel's Approximation Method for solving a transportation problem. The instructor systematically applies the method by calculating penalties, allocating quantities to the lowest-cost cells in the rows or columns with the highest penalties, and updating the supply and demand constraints. The process continues until all constraints are met, and the final total cost is calculated. The key learning point is the systematic approach to finding a good initial feasible solution for a transportation problem, which is a fundamental concept in operations research and linear programming.