Balanced and Unbalanced Transportation Problems

The two categories of transportation problems are balanced and unbalanced transportation problems. As we all know, a transportation problem is a type of Linear Programming Problem (LPP) in which items are carried from a set of sources to a set of destinations based on the supply and demand of the sources and destinations, with the goal of minimizing the total transportation cost. It is also known as the Hitchcock problem.

Introduction to Balanced and Unbalanced Transportation Problems

Balanced Transportation Problem

The problem is considered to be a balanced transportation problem when both supplies and demands are equal.

Unbalanced Transportation Problem

Unbalanced transportation problem is defined as a situation in which supply and demand are not equal. A dummy row or a dummy column is added to this type of problem, depending on the necessity, to make it a balanced problem. The problem can then be addressed in the same way as the balanced problem.

Methods of Solving Transportation Problems

There are three ways for determining the initial basic feasible solution. They are

1. NorthWest Corner Cell Method.

2. Vogel’s Approximation Method (VAM).

3. Least Call Cell Method.

Balanced Transportation Problem

The following is the basic framework of the balanced transportation problem:

The destinations D1, D2, D3, and D4 in the above table are where the products/goods will be transported from various sources O1, O2, O3, and O4. The supply from the source Oi is represented by S_i. The demand for the destination Dj is d_j. If a product is delivered from source Si to destination Dj, then the cost is called C_ij.

Let us now explore the process of solving the balanced transportation problem using one of the ways known as the NorthWest Corner Method in this article.

Solving Balanced Transportation problem by Northwest Corner Method

Consider this scenario:

With three sources (O1, O2, and O3) and four destinations (D1, D2, D3, and D4), what is the best way to solve this problem? The supply for the sources O1, O2, and O3 are 300, 400, and 500, respectively. Demands for the destination D1, D2, D3, and D4 are 250, 350, 400, and 200, respectively.

Solution:

The starting point for the North West Corner technique is (O1, D1), which is the table’s northwest corner. The cost of transportation is calculated for each value in the cell. As indicated in the diagram, compare the demand for column D1 with the supply from source O1 and assign a minimum of two to the cell (O1, D1).

Column D1’s demand has been met, hence the entire column will be canceled. The supply from the source O1 is still 300 – 250 = 50.

Analyze the northwest corner, i.e. (O1, D2), of the remaining table, excluding column D1, and assign the lowest among the supply for the appropriate column and rows. Because the supply from O1 is 50 and the demand for D2 is 350, allocate 50 to the cell (O1, D2).

Now, row O1 is canceled because the supply from row O1 has been completed. Hence, the demand for Column D2 has become 350 – 50 = 50.

The northwest corner cell in the remaining table is (O2, D2). The shortest supply from source O2 (400) and the demand for column D2 (300) is 300, thus putting 300 in the cell (O2, D2). Because the demand for column D2 has been met, the column can be deleted, and the remaining supply from source O2 is 400 – 300 = 100.

Again, find the northwest corner of the table, i.e. (O2, D3), and compare the O2 supply (i.e. 100) to the D2 demand (i.e. 400) and assign the smaller (i.e. 100) to the cell (O2, D2). Row O2 has been canceled because the supply from O2 has been completed. Column D3 has a leftover demand of 400 – 100 = 300.

Continuing in the same manner, the final cell values will be:

It should be observed that the demand for the relevant columns and rows is equal in the last remaining cell, which was cell (O3, D4). In this situation, the supply from O3 was 200, and the demand for D4 was 200, therefore this cell was assigned to it. Nothing was left for any row or column at the end.

To achieve the basic solution, multiply the allotted value by the respective cell value (i.e. the cost) and add them all together.

I.e., (250 × 3) + (50 × 1) + (300 × 6) + (100 × 5) + (300 × 3) + (200 × 2) = 4400.

Solving Unbalanced Transportation Problem

An unbalanced transportation problem is provided below. Because the sum of all the supplies, O1, O2, O3, and O4, does not equal the sum of all the demands, D1, D2, D3, D4, and D5, the situation is unbalanced.

Solution:

The idea of a dummy row or dummy column will be applied in this type of scenario. Because the supply is more than the demand in this situation, a fake demand column will be inserted, with a demand of (total supply – total demand), i.e. 117 – 95 = 22, as seen in the image below. A fake supply row would have been introduced if demand was greater than supply.

Now this problem has been changed to a balanced transportation problem, and it can be addressed using any of the ways listed below to solve a balanced transportation problem, such as the northwest corner method mentioned earlier.