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At any stage if there is a tie in the minimum cost so that two or more routes have the same least cost of transportation then conceptually either of them may be selected. However a better initial solution is obtained if the route chosen is the one where largest quantity can be assigned. Thus if there are three cells for which the least cost value is equal then consider each one of these one by one and determine the quantity ( by ) reference to the demand and supply quantities given which can be despatched and choose the cell with the largest quantity. If there is still a tie then either of them may be selected. To begin with the lowest of all cost elements is 1 for the by with corresponding supply and demand being 24 & 12. According assign 12 to this cell delete the Colum and reduce the quantity available to 12. The next cost is Rs. 2 in cell BX, DX, AY, AZ. But we cannot allocate to cell BX as the demand for B is exhausted. Now we see that demand of cell AY i ,e, 13 is more than available supply of 24-12 = 12 units. For cell DX demand 19 is greater than supply 15 and for cell AZ 13 is less than supply 21. Out of these maximum amount of 15 units can be allocated to cell DX only. So we choose cell DX and allocate 15 units to it. Delete the row headed X as well and adjust the available demand at D to 4.
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