Q 62,paper 3,J 12. The optimal solution of the following assignment problem using Hungarian method is

	I	II	III	IV
A	8	26	17	11
B	13	28	4	26
C	38	19	18	15
D	19	26	24	10

Options

----(A) (B) (C) (D)

(A) (I) (II) (III) (IV)

(B) (I) (III) (II) (IV)

(C) (I) (III) (IV) (II)

(D) (I) (IV) (II) (III)

Answer(B)

Explanation.

1.    Substract the minimum entry from each row.ex 8 from row 1, 4 from row 2, 15 from row 3, 10 from row 4.

	I	II	III	IV
A	0	18	9	3
B	9	24	0	22
C	23	4	3	0
D	9	16	14	0

2.    Substract minimum of column from all entries of table.example 0 from column 1. 4 from column 2,0 from column 3 and 4.

	I	II	III	IV
A	0	14	9	3
B	9	20	0	22
C	23	0	3	0
D	9	12	14	0

3.    Cover all zero with minimum number of horizontal or vertical line.

	I	II	III	IV
A	0	14	9	3
B	9	20	0	22
C	23	0	3	0
D	9	12	14	0

4.Since the minimal number of lines is 4, an

optimal assignment of zeros is possible and we are

finished.

4.  Replace all 0 values with original values.

	I	II	III	IV
A	8	14	9	3
B	9	20	4	22
C	23	19	3	15
D	9	12	14	10

Solution is A1(8),b3(4),C2(19), Div(10).