Find Solution of Assignment problem using Hungarian method (MIN case)

Work\Job	I	II	III
A	6	3	5
B	5	9	2
C	5	7	8

Solution:
The number of rows = 3 and columns = 3

	`I`	`II`	`III`
`A`	6	3	5
`B`	5	9	2
`C`	5	7	8

Here given problem is balanced.

Step-1: Find out the each row minimum element and subtract it from that row

	`I`	`II`	`III`
`A`	3 `3=6-3`	0 `0=3-3`	2 `2=5-3`	(-3) Minimum element of `1^(st)` row
`B`	3 `3=5-2`	7 `7=9-2`	0 `0=2-2`	(-2) Minimum element of `2^(nd)` row
`C`	0 `0=5-5`	2 `2=7-5`	3 `3=8-5`	(-5) Minimum element of `3^(rd)` row

Step-2: Find out the each column minimum element and subtract it from that column.

	`I`	`II`	`III`
`A`	3 `3=3-0`	0 `0=0-0`	2 `2=2-0`
`B`	3 `3=3-0`	7 `7=7-0`	0 `0=0-0`
`C`	0 `0=0-0`	2 `2=2-0`	3 `3=3-0`
	(-0) Minimum element of `1^(st)` column	(-0) Minimum element of `2^(nd)` column	(-0) Minimum element of `3^(rd)` column

Step-3: Make assignment in the opporunity cost table
(1) Rowwise cell `(A,II)` is assigned

(2) Rowwise cell `(B,III)` is assigned

(3) Rowwise cell `(C,I)` is assigned


Rowwise & columnwise assignment shown in table

	`I`	`II`	`III`
`A`	3	[0] (1) Rowwise cell `(A,II)` is assigned	2
`B`	3	7	[0] (2) Rowwise cell `(B,III)` is assigned
`C`	[0] (3) Rowwise cell `(C,I)` is assigned	2	3

Step-4: Number of assignments = 3, number of rows = 3
Which is equal, so solution is optimal

Optimal assignments are

	`I`	`II`	`III`
`A`	3 Original cost 6	[0] Original cost 3	2 Original cost 5
`B`	3 Original cost 5	7 Original cost 9	[0] Original cost 2
`C`	[0] Original cost 5	2 Original cost 7	3 Original cost 8

Optimal solution is

Work	Job	Cost
`A`	`II`	3
`B`	`III`	2
`C`	`I`	5
	Total	10

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