Answer 1

Step 1: Row minimum
Subtract the smallest element in each row from every element in its row.
The matrix obtained is given below:

	I	II	III	IV	V
A	30	0	55	60	80
B	15	0	10	40	55
C	30	0	45	60	75
D	0	0	30	30	60
E	20	0	35	45	70

Step 2: Column minimum
Subtract the smallest element in each column of assignment matrix obtained in step 1 from every element in its column.

	I	II	III	IV	V
A	30	0	45	30	25
B	15	0	0	10	0
C	30	0	35	30	20
D	0	0	20	0	5
E	20	0	25	15	15

Step 3:
Draw minimum number of vertical and horizontal lines to cover all zeros. F
First cover all rows and columns which have maximum number of zeros.

	I	II	III	IV	V
A	30	0	45	0	25
B	15	0	0	10	0
C	30	0	20	0	5
D	0	0	20	0	5
E	20	0	25	15	15

Step 4:
From step 3, minimum number of lines covering all the zeros are 3, which is less than order of matrix, i.e., 5.
∴  Select smallest element from all the uncovered elements, i.e., 15 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.

	I	II	III	IV	V
A	15	0	30	15	10
B	15	15	0	10	10
C	15	0	20	15	5
D	0	15	20	0	5
E	5	0	10	0	0

Step 5:
Draw minimum number of vertical and horizontal lines to cover all zeros.

	I	II	III	IV	V
A	15	0	30	15	10
B	15	15	0	10	0
C	15	0	20	15	5
D	0	15	20	0	5
E	5	0	10	0	0

Step 6:
From step 5, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.
∴ Select smallest element from all the uncovered elements, i.e., 5 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.

	I	II	III	IV	V
A	10	0	25	10	5
B	15	20	0	10	0
C	10	0	15	10	0
D	0	20	20	0	5
E	5	5	10	0	0

Step 7:
Draw minimum number of vertical and horizontal lines to cover all zeros.

	I	II	III	IV	V
A	10	0	25	10	5
B	15	20	0	10	0
C	10	0	15	10	0
D	0	20	20	0	5
E	5	5	10	0	0

Step 8:
From step 7, minimum number of lines covering all the zeros are 5, which is equal to order of the matrix, i.e., 5.
∴ Select a row with exactly one zero, enclose that zero in () and cross out all zeros in its respective column.
Similarly, examine each row and column and mark the assignment ().
∴ The matrix obtained is as follows:

	I	II	III	IV	V
A	10	0	25	10	5
B	15	20	0	10	0
C	10	0	15	10	0
D	0	20	20	0	5
E	5	5	10	0	0

Step 9:
The matrix obtained in step 8 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:

Routes	Dairy Plant	Distance (kms)
A	II	120
B	III	120
C	V	175
D	I	40
E	IV	70
		525

∴ Minimum distance travelled = 120 + 120 + 175 + 40 + 70 = 525 kms.