In this method, we start with the North-West Corner (top left) and allocate the maximum amount allowable by the supply and demand to this variable i.e., X_{11}. The satisfied column/row is then crossed, meaning that remaining variables in this column/row equal zero. If a column and a row are satisfied simultaneously, only one of them is crossed out. After adjustment of the quantities of supply and that of demand for all left over (uncrossed-out) rows and columns, the minimum possible column row is allotted to the first uncrossed out element in the new column/row. The process gets completed when exactly one row or one column is left uncrossed out.

The procedure can be explained specifically in the following steps :-

**Step I**. Start the North West (top left) corner and compare the supply of source 1 (S_{1}) with the demand of destination center 1 (D_{1}). Three conditions are possible.

D_{1} ≤ S_{1} it means that the demand at destination center D_{1} is less than the supply at source S_{1}. In X_{11} (North West/top left corner) set X_{11} equal to D_{1} and proceed horizontally.

(a) D_{1} = S_{1} i.e., the demand is equal to supply, then set X_{11} equal to D_{1} and proceed diagonally.

(b) D_{1} > S_{1} i.e., the demand is more than supply, then set X_{11} = S_{1} and proceed vertically.

**Step ****II**. Proceed in this manner, step by step till a value in allotted to S-E/right bottom corner. The north-west corner rule can be best demonstrated by the example in hand.

- Set X
_{11}= 1000 i.e., the smaller of the amount available at S_{1}(1000) and that needed at D_{1}(2300). - Proceed to cell (BX) as per rule© above which demands that you should proceed vertically. 0
_{1}> S_{1}compare the quantity available at S_{2}(1500) with the amount required. Quantity available at 0_{1}(2300-1000 = 1300) and set X_{11}= 1300. - Proceed to cell BY (rule above) as now 0 < S. Here S
_{1}is 1500 and the demand is 1400. So set X_{12}= 1400. We are required to proceed horizontally to next cell. Since there is no other horizontal cell, the allocation ends here.

The transportation cost associated with the solution is

Z = 2000 × 1000 + 2500 × 1300 + 2700 × 1400.

= 20,00, 000 + 32,50,000 + 37,80,000

= 90,36,000.

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