Vogel’s Approximation Method

Vogel's Approximation Method, Unit Cost Penalty Method

Step1

For all row of the table, determine the smallest and the next to smallest cost. Also determine the difference between them for every row. These are known as penalties. Place them aside by enclosing them in the parenthesis against the respective rows. At the same time, compute penalties for each column.

Step 2

Find the row or column with the highest penalty. If a tie takes place then use an arbitrary choice. Assume the largest penalty related to the ith row have the cost cij. Allot the largest possible amount xij = min (ai, bj) in the cell (i, j) and cross either ith row or jth column in the usual manner.

Step 3

Repeat again the calculation of the row and column penalties for the reduced table and then go for step 2. Repeat the process until all the requirements are satisfied or fulfilled.

Determine the initial basic feasible solution with the use of vogel's approximation method

1.

 

W1

W2

W3

W4

Availability

F1

19

30

50

10

7

F2

70

30

40

60

9

F3

40

8

70

20

18

Requirement

5

8

7

14

 

 

 

 

 

 

 

Answer

 

W1

W2

W3

W4

Availability

Penalty

F1

19

30

50

10

7

19-10=9

F2

70

30

40

60

9

40-30=10

F3

40

8

70

20

18

20-8=12

Requirement

5

8

7

14

 

 

Penalty

40-19=21

30-8=22

50-40=10

20-10=10

 

 

 

 

 

 

W1

W2

W3

W4

Availability

Penalty

F1

(19)

(30)

(50)

(10)

7

9

F2

(70)

(30)

(40)

(60)

9

10

F3

(40)

8(8)

(70)

(20)

18/10

12

Requirement

5

8/0

7

14

 

 

Penalty

21

22

10

10

 

 

 

 

W1

W2

W3

W4

Availability

Penalty

F1

5(19)

(30)

(50)

(10)

7/2

9

F2

(70)

(30)

(40)

(60)

9

20

F3

(40)

8(8)

(70)

(20)

18/10

20

Requirement

5/0

X

7

14

 

 

Penalty

21

X

10

10

 

 

 

 

W1

W2

W3

W4

Availability

Penalty

F1

5(19)

(30)

(50)

(10)

7/2

40

F2

(70)

(30)

(40)

(60)

9

20

F3

(40)

8(8)

(70)

10(20)

18/10/0

50

Requirement

X

X

7

14/4

 

 

Penalty

X

X

10

10

 

 

 

 

 

W1

W2

W3

W4

Availability

Penalty

F1

5(19)

(30)

(50)

2(10)

7/2/0

40

F2

(70)

(30)

(40)

(60)

9

20

F3

(40)

8(8)

(70)

10(20)

X

X

Requirement

X

X

7

14/4/2

 

 

Penalty

X

X

10

50

 

 

 

 

 

W1

W2

W3

W4

Availability

Penalty

F1

5(19)

(30)

(50)

2(10)

X

X

F2

(70)

(30)

7(40)

2(60)

X

X

F3

(40)

8(8)

(70)

10(20)

X

X

Requirement

X

X

X

X

 

 

Penalty

X

X

X

X

 

 

 

Initial Basic Feasible Solution

x11 = 5, x14 = 2, x23 = 7, x24 = 2, x32 = 8, x34 = 10

The transportation cost comes out to be 5 (19) + 2 (10) + 7 (40) + 2 (60) + 8 (8) + 10 (20) = Rs. 779

 

2.

 

 

 

Stores

Availability

 

 

I

II

III

IV

 

Warehouse

A

21

16

15

13

11

B

17

18

14

23

13

C

32

27

18

41

19

Requirement

 

6

10

12

15

 

 

 

 

 

 

 

 

 

Answer

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

(13)

11

2

B

(17)

(18)

(14)

(23)

13

3

C

(32)

(27)

(18)

(41)

19

9

Requirement

 

6

10

12

15

 

 

Penalty

 

4

2

1

10

 

 

 

 

 

 

 

 

 

 

 

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

11(13)

11/0

2

B

(17)

(18)

(14)

(23)

13

3

C

(32)

(27)

(18)

(41)

19

9

Requirement

 

6

10

12

15/4

 

 

Penalty

 

4

2

1

10

 

 

 

 

 

 

 

 

 

 

 

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

11(13)

X

X

B

(17)

(18)

(14)

4(23)

13/9

3

C

(32)

(27)

(18)

(41)

19

9

Requirement

 

6

10

12

15/4/0

 

 

Penalty

 

15

9

4

18

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

11(13)

X

X

B

6(17)

(18)

(14)

4(23)

13/9/3

3

C

(32)

(27)

(18)

(41)

19

9

Requirement

 

6/0

10

12

X

 

 

Penalty

 

15

9

4

X

 

 

 

 

 

 

 

 

 

 

 

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

11(13)

X

X

B

6(17)

3(18)

(14)

4(23)

13/9/3/0

4

C

(32)

(27)

(18)

(41)

19

9

Requirement

 

X

10/7

12

X

 

 

Penalty

 

X

9

4

X

 

 

 

 

 

 

 

 

 

 

 

 

 

Stores

Availability

Penalty

 

 

I

II

III

IV

 

 

Warehouse

A

(21)

(16)

(15)

11(13)

X

X

 

B

6(17)

3(18)

(14)

4(23)

X

X

 

C

(32)

7(27)

12(18)

(41)

X

X

 

Requirement

 

X

X

X

X

 

 

Penalty

 

X

X

X

X

 

 

 

 

 

 

 

 

 

 

 

                               

Initial Basic Feasible Solution

x14 = 11, x21 = 6, x22 = 3, x24 = 4, x32 = 7, x33 = 12

The transportation cost comes out to be 11 (13) + 6 (17) + 3 (18) + 4 (23) + 7 (27) + 12 (18) = Rs. 796