Least Cost Method

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This video presents a step-by-step solution to a balanced transportation problem using the Least Cost Method, a technique for minimizing total transportation cost. The instructor begins by setting up a transportation table with supply and demand values, identifying the total supply and demand as 950, confirming the problem is balanced. He then explains the method's core principle: allocating units to the cell with the lowest cost first, and then moving to the next lowest cost cell, while respecting supply and demand constraints. The process is demonstrated by systematically filling the table, starting with the lowest cost of 11 (Kolkata to Hyderabad), then 13 (Kolkata to Pune), and so on, until all supply and demand are satisfied. The final solution is a feasible transportation plan, and the instructor calculates the total cost by multiplying the allocated quantities by their respective unit costs and summing the results, arriving at a total cost of 12,220. The video concludes by showing the final allocation table and the total cost calculation.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The video opens with a whiteboard displaying a transportation problem titled 'Linear Programming Problem'. The instructor, seen from the back, begins by writing the problem's data. A transportation table is shown with three supply points (Hyderabad, Pune, Mumbai) and four demand points (Kolkata, Delhi, Indore, Puna). The supply values are 250, 300, and 400, and the demand values are 250, 300, 150, and 250. The instructor writes the total supply and demand as 250 + 300 + 400 = 950 and 250 + 300 + 150 + 250 = 950, confirming the problem is balanced. He then introduces the 'Least Cost Method' as the solution technique, writing 'Least Cost Method' and 'D.B.F.S' (likely referring to the method's name) on the board.

2. 2:00 – 5:00 02:00-05:00

   The instructor begins the allocation process using the Least Cost Method. He identifies the cell with the lowest cost, which is 11 in the Kolkata-Hyderabad cell. He allocates the minimum of the supply (250) and demand (250), which is 250, to this cell. He then crosses out the first row (Hyderabad) as its supply is exhausted. Next, he identifies the next lowest cost, which is 13 in the Kolkata-Pune cell. He allocates 250 to this cell, but since the demand for Kolkata is already met, he allocates the remaining 250 to the Kolkata-Pune cell. He then crosses out the first column (Kolkata) as its demand is exhausted. The instructor then moves to the next lowest cost, which is 16 in the Delhi-Hyderabad cell, but since Hyderabad's supply is exhausted, he moves to the next lowest cost, which is 16 in the Delhi-Pune cell. He allocates 300 to this cell, exhausting the demand for Delhi. He then crosses out the second column (Delhi) as its demand is met.

3. 5:00 – 10:00 05:00-10:00

   The instructor continues the allocation process. He identifies the next lowest cost, which is 17 in the Kolkata-Indore cell, but Kolkata's demand is already met. He then looks at the remaining cells and finds the next lowest cost, which is 18 in the Mumbai-Indore cell. He allocates 150 to this cell, exhausting the demand for Indore. He then crosses out the third column (Indore) as its demand is met. The next lowest cost is 19 in the Mumbai-Pune cell. He allocates 250 to this cell, exhausting the supply for Mumbai. He then crosses out the third row (Mumbai) as its supply is exhausted. The final allocation is 250 to the Pune-Mumbai cell, which is the only remaining cell. The instructor then calculates the total cost by multiplying the allocated quantities by their respective unit costs and summing the results.

4. 10:00 – 15:00 10:00-15:00

   The instructor calculates the total cost of the transportation plan. He writes the formula for total cost: Total Cost = (250 x 11) + (250 x 13) + (300 x 16) + (150 x 18) + (250 x 19). He then calculates each term: 250 x 11 = 2750, 250 x 13 = 3250, 300 x 16 = 4800, 150 x 18 = 2700, and 250 x 19 = 4750. He adds these values together: 2750 + 3250 + 4800 + 2700 + 4750 = 12,220. He writes the final answer as Total Cost = 12,220. He then reviews the final allocation table, which shows the quantities allocated to each cell.

5. 15:00 – 16:03 15:00-16:03

   The instructor concludes the lesson by summarizing the final solution. He points to the final allocation table on the whiteboard, which shows the quantities allocated to each cell: 250 to Kolkata-Hyderabad, 250 to Kolkata-Pune, 300 to Delhi-Pune, 150 to Mumbai-Indore, and 250 to Pune-Mumbai. He reiterates that the total cost of this transportation plan is 12,220. He emphasizes that this is the minimum cost solution using the Least Cost Method. The video ends with the instructor standing in front of the completed whiteboard, which shows the problem, the method, the allocation table, and the final cost calculation.

The video provides a comprehensive, step-by-step demonstration of solving a balanced transportation problem using the Least Cost Method. The core concept is to minimize total transportation cost by systematically allocating units to the cell with the lowest cost first, while respecting supply and demand constraints. The method progresses by identifying the lowest cost cell, allocating the maximum possible quantity, and then eliminating the exhausted row or column. This process is repeated until all supply and demand are satisfied. The final solution is a feasible transportation plan, and the total cost is calculated by summing the products of the allocated quantities and their respective unit costs. The video effectively illustrates the entire process, from problem setup to final cost calculation, providing a clear example of this fundamental optimization technique.