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

 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

 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