Question

--10/11□ -6/11 maths 8th assignment ​

Answer 1

Answer:

Add dummy row (M

5

)

A B C D E

M

1

9 11 15 10 11

M

2

12 9 - 10 9

M

3

- 11 14 11 7

M

4

14 8 12 7 8

M

5

0 0 0 0 0

Now apply row reduction

A B C D E

M

1

0 2 6 1 2

M

2

3 0 - 1 0

M

3

- 4 7 4 0

M

4

7 1 5 0 1

M

5

0 0 0 0 0

Number of crosed(bold rows and columns) line =4

order of matrix =5

4<5 (not optimal solution)

Now Modify matrix by subtracting minimum uncrossed(unbolded ) element from uncrossed element (unbolded ) element and adding same at T-point(element lying at junction of bolded rows and columns )

Modified element

A B C D E

M

1

0 1 5 1 2

M

2

4 0 - 2 1

M

3

- 3 6 4 0

M

4

7 0 4 0 1

M

5

1 0 0 1 0

Number of crossed line(bolded rows and column) =5

order of matrix =5

55 (optimal solution)

assignment

M

1

→A

M

2

→B

M

3

→E

M

4

→D

M

5

→C (C cannot be ssigned to any one)

Optical solution =9+9+7+7=32

Answer 2

$= \frac{ - 10}{11} \: \: \: () \: \frac{ - 6}{11}$

$= \frac{ - 10}{11} > \frac{ - 6}{11}$