11.8 Vogel’s Approximation Method

Vogel’s Approximation Method (VAM) is a heuristic technique used to generate an initial feasible solution for a balanced transportation problem. It is widely regarded as one of the most effective initial allocation methods, as it incorporates both cost efficiency and opportunity cost into the decision-making process.

This method typically results in a lower total transportation cost than both the North-West Corner and Least Cost methods, making it a preferred choice for generating high-quality initial solutions.

Determining the Total Transportation Cost Using Vogel’s Approximation Method (VAM)

Step 1: Calculate Row and Column Penalties

The first step in Vogel’s Approximation Method is to compute the penalty values for each row and column. These penalties represent the opportunity cost of not choosing the lowest-cost allocation in a given row or column. The penalty is calculated as the difference between the two lowest unit transportation costs in each row or column.

Row Penalties:

• S1: [latex]6 − 2 = 4[/latex]
• S2: [latex]10 – 4 = 6[/latex]
• S3: [latex]6 − 4 = 2[/latex]

Column Penalties:

• D1: [latex]6 – 4 = 2[/latex]
• D2: [latex]6 − 2 = 4[/latex]
• D3: [latex]10 – 6 = 4[/latex]
• D4: [latex]8 − 4 = 4[/latex]

Step 2: Select the Row or Column with the Highest Penalty

Among all the calculated penalties, the highest is 6, which corresponds to row S2. This indicates that failing to allocate to the lowest-cost cell in this row would result in the greatest cost inefficiency.

• The lowest-cost cell in row S2 is (S2/D1), with a unit transportation cost of $4.

Allocation:

• Allocate the minimum of the available supply and demand:
  • 50 units to satisfy the demand D1

Update:

• S2 supply is reduced to 30 units, [latex]80-50=30[/latex]
• D1 demand is fully satisfied, and the column is eliminated from further consideration

Step 3: Recalculate Penalties (Excluding Column D1)

With D1 now fully satisfied and removed from the tableau, we recalculate the penalty values for the remaining rows and columns.

Updated Row Penalties:

• S1: [latex]8 − 2 = 6[/latex]
• S2: [latex]12 – 10 = 2[/latex]
• S3: [latex]6 − 4 = 2[/latex]

Updated Column Penalties:

• D2: [latex]6 – 2 = 4[/latex]
• D3: [latex]10 − 6 = 4[/latex]
• D4: [latex]8 – 4 = 4[/latex]

Step 4: Select the Row or Column with the Highest Penalty

The highest penalty is 6, corresponding to row S1. This indicates that failing to allocate to the lowest-cost cell in this row would result in the greatest cost inefficiency.

• The lowest-cost cell in row S1 is (S1/D2), with a unit transportation cost of $2.

Allocation:

• Allocate the minimum of the available supply and demand:
  • The minimum is the supply of 60 units

Update:

• S1 supply is now exhausted: [latex]60 – 60 = 0[/latex]
• D2 demand is partially fulfilled: [latex]70 − 60 = 10[/latex]

Row S1 is now eliminated from further consideration

Step 5: Recalculate Penalties (Excluding Row S1 and Column D1)

With S1 and D1 now eliminated from the tableau, we recalculate the penalty values for the remaining rows and columns.

Updated Row Penalties:

• S2: [latex]12−10 = 2[/latex]
• S3: [latex]6 − 4 = 2[/latex]

Updated Column Penalties:

• D2: [latex]12 – 6 = 6[/latex]
• D3: [latex]10 − 6 = 4[/latex]
• D4: [latex]18 – 4 = 14[/latex]

Step 6: Select the Row or Column with the Highest Penalty

The highest penalty is 14, corresponding to column D4. This indicates that failing to allocate to the lowest-cost cell in this column would result in the greatest cost inefficiency.

• The lowest-cost cell in column D4 is (S3/D4), with a unit transportation cost of $4.

Allocation:

• Allocate the minimum of the available supply and demand:
  • The minimum is 40 units from D4

Update:

• S3 supply is reduced to 100 – 40 = 60 units
• D4 demand is fully satisfied, and the column is removed from further consideration

Step 7: Recalculate Penalties (Excluding S1, D1, and D4)

With row S1 and columns D1 and D4 now eliminated from the tableau, we recalculate the penalty values for the remaining rows and columns.

Updated Row Penalties:

• S2: [latex]12−10=2[/latex]
• S3: [latex]6−6=0[/latex]

Updated Column Penalties:

• D2: [latex]12−6=6[/latex]
• D3: [latex]10−6=4[/latex]

Step 8: Select the Row or Column with the Highest Penalty

The highest penalty is 6, corresponding to column D2. This indicates that failing to allocate to the lowest-cost cell in this column would result in the greatest cost inefficiency.

• The lowest-cost cell in column D2 is (S3/D2), with a unit transportation cost of $6.

Allocation:

• Allocate the minimum of the remaining supply and demand:
  • The minimum is 10 units to satisfy D2

Update:

• S3 supply is reduced to: [latex]60−10=50[/latex] units
• D2 demand is now fully satisfied, and the column is removed from further consideration

Final Allocation: Completing the Tableau

At this stage, only two unallocated cells remain—one in each of the remaining rows (S2 and S3) and one column (D3). Since no further penalty calculations are meaningful with only one option per row and column, we proceed directly with the final allocations.

Final Allocations:

• Allocate 50 units to (S3/D3)
  • Remaining supply at S3: [latex]50 − 50 = 0[/latex]
    Remaining demand at D3: [latex]80 − 50 = 30[/latex]
• Allocate  30 units to (S2/D3)
  • Remaining supply at S2: [latex]30 − 30 = 0[/latex]
    Remaining demand at D3: [latex]30 − 30 = 0[/latex]

Result:

• All supplies are exhausted
• All market demands are fully satisfied

This completes the initial feasible solution using Vogel’s Approximation Method.

After completing all allocations using Vogel’s Approximation Method, the total transportation cost is calculated as follows:

[latex]\begin{align*} \text{Total Cost}&= (\small\text{S2/D1} \times \small{4}) + (\small\text{S1/D2} \times 2) + (\small\text{S3/D2} \times 6)+ (\small\text{S2/D3} \times 10) + (\small\text{S3/D3} \times 6) + (\small\text{S3/D4} \times 4) \\ \text{Total Cost}&= (50 \times 4) + (60 \times 2) + (10 \times 6) + (30 \times 10) + (50 \times 6) + (40 \times 4)\\ \text{Total Cost}&= 200 + 120 + 60 + 300 + 300 + 160 \\ \text{Total Cost}&= 1140 \end{align*}[/latex]

Thus, the initial basic feasible solution obtained using VAM results in a total transportation cost of $1140, which is lower than the $1160 obtained using the Least Cost Method and significantly better than the $1760 from the North-West Corner Method.

This outcome highlights one of the key advantages of VAM: it often produces a more cost-effective initial solution by incorporating both transportation costs and opportunity costs into the allocation process.

However, it is important to note that even though VAM typically yields a near-optimal solution, it does not guarantee optimality. In the following sections, we will explore optimization techniques