# Transportation Problem | Set 4 (Vogel’s Approximation Method)

The **North-West Corner** method and the **Least Cost Cell** method has been discussed in the previous articles. In this article, the **Vogel’s Approximation** method will be discussed.

**Solution:**

- For each row find the least value and then the second least value and take the absolute difference of these two least values and write it in the corresponding row difference as shown in the image below. In row
**O1**,**1**is the least value and**3**is the second least value and their absolute difference is**2**. Similarly, for row**O2**and**O3**, the absolute differences are**3**and**1**respectively. - For each column find the least value and then the second least value and take the absolute difference of these two least values then write it in the corresponding column difference as shown in the figure. In column
**D1**,**2**is the least value and**3**is the second least value and their absolute difference is**1**. Similarly, for column**D2**,**D3**and**D3**, the absolute differences are**2**,**2**and**2**respectively.

- These value of row difference and column difference are also called as penalty. Now select the maximum penalty. The maximum penalty is
**3**i.e. row**O2**. Now find the cell with the least cost in row**O2**and allocate the minimum among the supply of the respective row and the demand of the respective column. Demand is smaller than the supply so allocate the column’s demand i.e.**250**to the cell. Then cancel the column**D1**.

- From the remaining cells, find out the row difference and column difference.

- Again select the maximum penalty which is
**3**corresponding to row**O1**. The least-cost cell in row**O1**is**(O1, D2)**with cost**1**. Allocate the minimum among supply and demand from the respective row and column to the cell. Cancel the row or column with zero value.

- Now find the row difference and column difference from the remaining cells.

- Now select the maximum penalty which is
**7**corresponding to column**D4**. The least cost cell in column**D4**is**(O3, D4)**with cost**2**. The demand is smaller than the supply for cell**(O3, D4)**. Allocate**200**to the cell and cancel the column.

- Find the row difference and the column difference from the remaining cells.

- Now the maximum penalty is
**3**corresponding to the column**D2**. The cell with the least value in**D2**is**(O3, D2)**. Allocate the minimum of supply and demand and cancel the column.

- Now there is only one column so select the cell with the least cost and allocate the value.

- Now there is only one cell so allocate the remaining demand or supply to the cell

- No balance remains. So multiply the allocated value of the cells with their corresponding cell cost and add all to get the final cost i.e.
**(300 * 1) + (250 * 2) + (50 * 3) + (250 * 3) + (200 * 2) + (150 * 5) = 2850**

## Recommended Posts:

- Transportation Problem | Set 7 ( Degeneracy in Transportation Problem )
- Transportation Problem | Set 2 (NorthWest Corner Method)
- Transportation Problem | Set 3 (Least Cost Cell Method)
- Transportation Problem | Set 6 (MODI Method - UV Method)
- Transportation Problem | Set 5 ( Unbalanced )
- Transportation Problem | Set 1 (Introduction)
- Transportation Problem Set 8 | Transshipment Model-1
- Calculating Factorials using Stirling Approximation
- Bakhshali Approximation for computing square roots
- Secretary Problem (A Optimal Stopping Problem)
- 21 Matchsticks Problem
- Fibonacci problem (Value of Fib(N)*Fib(N) - Fib(N-1) * Fib(N+1))
- Tiling Problem
- Josephus problem | Set 1 (A O(n) Solution)
- Josephus Problem Using Bit Magic

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