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
40
60
9
F3
8
20
18
Requirement
5
14
Answer
Penalty
19-10=9
40-30=10
20-8=12
40-19=21
30-8=22
50-40=10
20-10=10
(19)
(30)
(50)
(10)
(70)
(40)
(60)
8(8)
(20)
18/10
12
8/0
21
22
5(19)
7/2
5/0
X
10(20)
18/10/0
14/4
2(10)
7/2/0
14/4/2
7(40)
2(60)
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
I
II
III
IV
Warehouse
A
16
15
13
11
B
17
23
C
32
27
41
6
(21)
(16)
(15)
(13)
2
(17)
(18)
(14)
(23)
3
(32)
(27)
(41)
4
1
11(13)
11/0
15/4
4(23)
13/9
15/4/0
6(17)
13/9/3
6/0
3(18)
13/9/3/0
10/7
7(27)
12(18)
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
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