Assumptions of Linear Programming



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Table1. Initial Solution




D1

D2

D3

Supply

S1

8 120

5

6

120

S2

15 30

10 50

12

80

S3

3

9 30

10 50

80

Demand

150

80

50

280

Table 2.




D1

D2

D3

Supply

ui

S1

8

120

5

+2

6

+2

120

U1=-7

S2

15

(-) 30

10

50 (+)

12

+1

80

U1=0

S3

3

(+) -11

9

30 (-)

10

50

80

U1=-1

Demand

150

80

50

280




vj

V1=15

V2= 10

V3=11




Since the unoccupied cell (S3, D1) has the largest negative opportunity cost of -11, therefore, ell (S3, D1) is entered into the new solution mix. The closed path for (S3, D1) is shown in Table 2. The maximum allocation to (S3, D1) is 30. However, when this amount is allocated to (S3, D1) both cells (S2, D1) and (S3, D2) become unoccupied because these tow have same allocations. Thus, the number of positive allocation became less than the required number, m + n - 1 = 3 + 3 - 1 = 5. Hence, this is a degenerate solution as shown in Table 3,

Table .3




D1

D2

D3

Supply

S1

8 120

5

6

120

S2

15

10 80

12

80

S3

3 30

9

10 50

80

Demand

150

80

50

280

To remove degeneracy a quantity is assigned to one of the cells that have become unoccupied so that there are m + n - 1 occupied cells. Assign to either (S3, D2) and proceed with the usual solution procedure. The optimal solution is given in Table 4.

Table 4.




D1

D2

D3

Supply

S1

8 70

5

6 50

120

S2

15

10 80

12

80

S3

3 80

9

10

80

Demand

150

80

50

280

12

  1. Formulate a linear Programming problem



  1. Which salesperson should be assigned to each region to minimize total time? Identify the optimal assignments and compute total minimum time. Use the Hungarian Method

Answer

Row reduction:

A B C D E

1 7 0 5 6 10

2 3 0 7 0 5

3 0 5 3 4 1

4 4 0 0 8 7

5 4 3 0 6 2

Coloumn reduction:

A B C D E

7 0 5 6 9

3 0 7 0 4

0 5 3 4 0

4 0 0 8 6

4 3 0 6 1

A B C D E

7 0 5 6 9

3 0 7 0 4

0 5 3 4 0

4 0 0 8 6

4 3 0 6 1

√ √


Final step is as follows:

A B C D E

6 0 5 5 8

3 1 8 0 4

0 6 4 4 0

4 0 0 7 5

3 3 0 5 0

B → 1


C → 4

A → 3


D → 2

E → 5


Total= 51 days
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