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## Note: Important point to remember for VAM - Case of

Tie

If the "smallest cost" in a row or column are repeating, then difference for that row or column is "0".

In VAM, we have to select the row or column which is having higher difference

- But if there is a tie in selection, then we have to select the row or column which contains minimum cost.
- In case there's a tie in minimum cost too, select the cell in which maximum allocation can be done.

The given problem is already balanced.

→ Select row/column with highest difference. In the same row/column, select the cell with minimum cost, then allocate smallest value of demand or supply in that cell.

→ Here, we have [3] as the highest difference. Selecting column with [3] as column difference and finding cell with minimum cost.

→ As we can see here "1" in the second row is the minimum cost in this last column with highest difference of [3].

→ So, allocating 10 to "1"(min. cost) with highest column difference.

→ Remove the row/column whose supply or demand is fulfilled and prepare new matrix as shown below.

→ Check that here, we have multiple highest difference as "[2]".

→ * [Tie]* We have to select the row/column which has minimum cost included. (check note above for this step).

→ So, selecting first column with highest column difference as [2] and minimum cost as "1" and allocating same as we have done in

→ Repeat the procedure until all allocations are done.

→ You may get * [Tie]* again in this and further steps. Just repeat

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