The Vogel’s approximation method or VAM gives a good start up to the solution than any other method. The solution of VAM provides us with an optimum result or very close to optimum result to the starting solution. This process involves not only lest cost Cij but also the cost that exceeds Cij. So the steps are as follows:
Step I – look at the cost matrix given below
Distribution centres supply
A X11 Rs. 2000 X12 Rs. 5380
1000 1000 (3380)
Factories B X21 Rs. 2500 X22 Rs. 2700
C X23 Rs. 2550 X32 Rs.1700
Demand 2300 1400
Now the difference between the smallest and the second smallest cost elements have to be identified in each column and then note it down in the column within brackets, i.e., in column X there is a difference of 500 and 100in the second column.
Now the difference between the smallest and the second smallest elements in the row are written at the right side of each bracket i.e. ill row A 3380, in row B 200 andRow C 850.
The difference is finally noted down because the elements written within the brackets below the columns or rows are the penalty incurred due to the failing in the least cost cell in the row or column.
Step II – the row or column with the maximum difference is allocated to the least cost cell is thus selected. If you find the same values, then you are free to choose anyone value. Things will be clear if you look at this example, here 3380 is the greatest difference, so row A is selectedas the least cost cell containing 1000 in the cell AX is thus allocated.
Step III– now the rows or columns that satisfy the conditions are crossed out. Here in this example row, A is crossed out. Below is the matrix without row A.
B Rs. 2500 Rs. 2700
X21 X22 1500(200)
C Rs. 2550 Rs.1700
X23 X32 = 1200 1200 (850)
Step I and Step II is thus repeated till all the allocations are done. After row A, column Y shows the highest difference as CY has avalue of 1200. There the conditions are not fulfilled. Not only this, but row C also carries a huge difference (850) compared to the other rows. This is the reason we allocate 1200 to the cell CY which is the least cost cell in row C.
So here is the final form of the matrix given below:
B Rs. 2500 Rs. 2700
X21= 1300 X22= 200
As the cell, BX has the least cost, so the maximum allocations of 1300 are made here. In the cell BY, 200 could be allocated.
Below is the list of all the allocations made in one matrix.
Rs. 200 Rs.5380
Rs.2500 Rs. 2700
Therefore the cost of transportation is related to the following calculation:
Z = Rs. (2000 x 1000 + 2500 x 1300 + 1700 x 1200 + 2700 x 200) = Rs. 78, 30, 000
Links of Previous Main Topic:-
- Concept of capital expenditure
- Learning objectives and chapter outline
- Limitations of operations research
- Linear programming learning objectives and outline of chapter
- Introduction learning objectives
- Duality in linear programming
- Learning objectives
Links of Next Finance Topics:-
- Performing optimality test
- The stepping stone method
- The modified distribution modi method or uv method
- Learning objectives and chapter outline in assignment model
- Minimization problems
- Learning objectives the transportation problems
- Special case of traveling sales man problem
- Replacement theory learning objectives and chapter outline
- Learning objectives and chapter outline for waiting line queuing theory