Step 1. In column P, the difference between the two lowest cost elements is 100 which is entered as, (100) below column P, similarly the two smallest elements in row Q are 0 and 300 and the difference is. 300, we write their difference as (300) below column Q. Under column R. we write (500) and under column S we write (600). Similarly, against row A, we write (100), against row B (100) and against row C (300):
Step 2. We choose column S (having largest difference 600). In this column, cell BS has the lowest cost i.e., 100 and we allot 1 as maximum possible allocation of only 1 is possible.
Step 3. Cross out row B as the supply 1 is completely satisfied by the allocation made, 1, in step II.
Step 4. Write down the shrunken matrix after crossing out row B as follows.

P 
Q 

R 
S 
Supply 
A 
200 
300 
5 
1100 
700 
6 (100) 
C 
500 
800 

1500 
900 
10 (300) 
Demand 
7 
5 

3 
2 

We repeat step I and write the difference in rows and columns as shown above. In column Q least cost AQ, we make allocation of 5. Since this satisfies the condition of column Q completely, we cross out column Q and shrunken matrix is written as follows:
Once again step I is repeated and the difference in rows and columns are written as shown above. We now make allocations in cell AP as this is the least cost cell. Only I can be allotted in this cell since this satisfies row A, it is crossed off and the shrunken matrix is rewritten as follows:
In this, as cell CP has the lowest cost, maximum possible allocation of 6 is made here. Next lowest least cost cell is CS and I is allotted here and 3 in the cell CR.
The allocations made above are shown ill the allocation matrix given below:
Z = Rs. (200 × 1 + 300 × 5 + 100 × 1 + 500 × 900 × 1= 6 + 1500 × 3)
= Rs. 10,200
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