Obtain an initial and optimum solution to the following transportation problem
using Vogel’s Approximation method.
To Available
7 3 2 2
2
1 3 3
3 4 6 5
Demand 4 1 5
To obtain an initial solution using Vogel’s Approximation method, follow these steps:
Step 1: Set up the transportation problem matrix with the given supplies and demands:
To Available
7 3 2 2
2 - 1 3 3
3 4 6 5
Demand: 4 1 5
Step 2: Calculate the penalties for each row and column. The penalty is determined by finding the difference between the smallest and second smallest cost in each row or column.
For the rows:
Row 1 penalty = 3 - 2 = 1
Row 2 penalty = 3 - 1 = 2
Row 3 penalty = 6 - 5 = 1
For the columns:
Column 1 penalty = 7 - 3 = 4
Column 2 penalty = 4 - 2 = 2
Column 3 penalty = 6 - 2 = 4
Step 3: Choose the row or column with the highest penalty, and allocate as much as possible in the cell with the smallest cost for that row or column. In the case of a tie, prioritize the row.
In this case, the highest penalty is 4 for columns 1 and 3. Column 2 has a penalty of 2. Since column 1 and 3 have the same penalty, we prioritize column 1.
Allocate as much as possible in the cell with the smallest cost in column 1:
Allocate 2 units from row 2 in the cell (2, 1) with cost 1. After allocation, update the available supply and demand:
To Available
7 3 2 0
- - 1 3 3
3 4 6 5
Demand: 4 1 5
Step 4: Recalculate the penalties for the remaining rows and columns:
For the rows:
Row 1 penalty = 3 - 2 = 1
Row 3 penalty = 6 - 5 = 1
For the columns:
Column 2 penalty = 4 - 3 = 1
Column 3 penalty = 6 - 2 = 4
Step 5: Choose the row or column with the highest penalty and allocate as much as possible in the cell with the smallest cost for that row or column.
In this case, the highest penalty is 4 for column 3. We allocate as much as possible in the cell with the smallest cost in column 3:
Allocate 4 units from row 3 in the cell (3, 3) with cost 5. After allocation, update the available supply and demand:
To Available
7 3 - 0
- - 1 3 3
3 4 - 1
Demand: 4 1 1
Step 6: Recalculate the penalties for the remaining rows and columns:
For the rows:
Row 1 penalty = 3 - 2 = 1
For the columns:
Column 2 penalty = 3 - 3 = 0
Step 7: Choose the row or column with the highest penalty and allocate as much as possible in the cell with the smallest cost for that row or column.
In this case, the highest penalty is 1 for row 1. We allocate as much as possible in the cell with the smallest cost in row 1:
Allocate 2 units from column 2 in the cell (1, 2) with cost 3. After allocation, update the available supply and demand:
To Available
7 - - 0
- - 1 - 1
3 4 - 1
Demand: 2 1 1
Step 8: Recalculate the penalties for the remaining rows and columns:
For the rows:
No remaining rows.
For the columns:
Column 1 penalty = 7 - 7 = 0
Step 9: Allocate the remaining demand to the remaining cells.
In this case, allocate 2 units from column 1 in the cell (1, 1) with cost 7, and allocate 1 unit from row 3 in the cell (3, 2) with cost 4. After allocation, the transportation table is fully balanced:
To Available
- - - 0
- - 1 - 0
1 - - 0
Demand: - - -
This is the initial solution using Vogel's Approximation method.