we explained north west corner and matrix minima concepts of transportation problem with examples.

1. DEPARTMENT OF MBA
A PRESENTATION ON
PROFIT MAXIMIZATION AND NORTH WEST CORNER
METHOD
SUBMITTED BY
19L31E0007 M JAYA CHANDRA

2. NORTH WEST CORNER METHOD
• the north-west corner rule is a method adopted to compute the initial feasible solution of the transportation problem. the name north-west
corner is given to this method because the basic variables are selected from the extreme left corner.
• the concept of north-west corner can be well understood through a transportation problem given below
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3. • in the table, three sources a, b and c with the production capacity of 50 units, 40 units, 60 units of product respectively is given. every day
the demand of three retailers d, e, f is to be furnished with at least 20 units, 95 units and 35 units of product respectively. the transportation
costs are also given in the matrix.
• in the table, three sources a, b and c with the production capacity of 50 units, 40 units, 60 units of product respectively is given. every day
the demand of three retailers d, e, f is to be furnished with at least 20 units, 95 units and 35 units of product respectively. the transportation
costs are also given in the matrix
• Similarly, in case the supply is more than the demand, then dummy source is created whose demand will be equivalent to the difference
between supply and demand. Again the cost associated with the dummy source will be zero.
• once the demand and supply are equal, the following procedure is
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4. • 1.select the north-west or extreme left corner of the matrix, assign as many units as possible to cell ad, within the supply
and demand constraints. such as 20 units are assigned to the first cell, that satisfies the demand of destination d while the
supply is in surplus.
• 2.now move horizontally and assign 30 units to the cell ae. since 30 units are available with the source a, the supply gets
fully saturated.
• 3.now move vertically in the matrix and assign 40 units to cell be. the supply of source b also gets fully saturated.
• 4.again move vertically, and assign 25 units to cell ce, the demand of destination e is fulfilled.
• 5.move horizontally in the matrix and assign 35 units to cell cf, both the demand and supply of origin and destination gets
saturated. now the total cost can be computed.
• the total cost can be computed by multiplying the units assigned to each cell with the concerned transportation cost.
therefore,
• total cost = 20*5+ 30*8+ 40*6+ 25*9+ 35*6 = rs 1015
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5. Find the maximum profit
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A B C D
1 3 6 2 4
2 7 1 4 4
3 3 8 5 8
4 6 4 3 7
5 5 2 4 3
6 5 7 6 4

6. • The given data is imbalanced.
To balance the matrix add dummy columns
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7. Dummy columns are added
A B C D
3 6 2 4 0 0
7 1 4 4 0 0
3 8 5 8 0 0
6 4 3 7 0 0
5 2 4 3 0 0
5 7 6 4 0 0
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8. Subtract all the elements from maximum element 8.
then minimize all the rows
5 2 6 4 8 8
1 7 4 4 8 8
5 0 3 0 8 8
2 4 5 1 8 8
3 6 4 5 8 8
3 1 2 4 8 8
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3 0 4 2 6 6
0 6 3 3 7 7
5 0 3 0 8 8
1 3 4 0 7 7
0 3 1 2 5 5
2 0 1 3 7 7

9. Minimize all the columns.
Then identify the single 0 cost in the rows
3 0 3 2 1 1
0 6 2 3 2 2
5 0 2 0 3 3
1 3 3 0 2 2
0 3 0 2 0 0
2 0 0 3 2 2
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3 0 3 2 1 1
0 6 2 3 2 2
5 0 2 0 3 3
1 3 3 0 2 2
0 3 0 2 0 0
2 0 0 3 2 2

10. after the assignment of rows and columns of the given matrix
the maximum profit is =6+7+8+0+0+6
= 27
3 0 3 3 1 1
0 6 2 4 2 2
4 0 1 0 2 2
0 2 2 0 1 1
0 3 0 3 0 0
2 0 0 4 2 2
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4 0 3 4 1 1
0 5 1 4 1 1
4 0 0 0 1 1
0 1 1 0 0 0
1 2 0 4 0 0
3 0 0 5 1 1

11. THANKING YOU FOR GIVING THIS
WONDERFUL OPPORTUNITY
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