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FEASIBLE SOLUTION BY VAM

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