Transportation Problem in Operational Research/ Quantitative Techniques for Management

 

Transportation Problem in Operational Research/ Quantitative Techniques for Management 

The transportation problem in operational research is concerned with finding the minimum cost of transporting a single commodity from a given number of sources (e.g. factories) to a given number of destinations (e.g. warehouses). These types of problems can be solved by general network methods, but here we use a specific transportation algorithm.

The data of the model include

1. The level of supply at each source and the amount of demand at each destination.

2. The unit transportation cost of the commodity from each source to each destination.

Since there is only one commodity, a destination can receive its demand from more than one source. The objective is to determine how much should be shipped from each source to each destination so as to minimize the total transportation cost.

Types of Transportation Problems in Operational Research

1. Balanced Transportation Problem
2. Unbalanced Transportation Problem

The solution to the transportation problem

Stage I: Finding an initial basic feasible solution.

Stage II: Checking for optimality

Existence of Feasible Solution: A necessary and sufficient condition for the existence of a feasible solution to the general transportation problem is that

Total supply = Total demand

Existence of Basic Feasible Solution: The number of basic variables of the general transportation problem at any stage of the feasible solution must be (m + n – 1). Now degenerate basic feasible solution (a feasible solution) involving exactly (m + n – 1) positive variables is known as a non-degenerate basic feasible solution otherwise it is said to be degenerate basic feasible. These allocations should be independent positions in case of non-degenerate basic feasible solutions.

• Optimum Solution: A feasible solution is said to be optimal if it minimizes the total transportation cost.
• Unbalance TP If total supply is not equal to total demand, then it balances with the dummy source or destination.
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Finding an Initial Basic Feasible Solutions

1. North- west corner method [NWCM]
2. Least cost entry method [LCEM]
3. Vogel’s approximation method (or Penalty method) [VAM]
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Steps for North-West Corner Method

1. Allocate the maximum amount allowable by the supply and demand constraints to the variable x11 (i.e. the cell in the top left corner of the transportation tableau).
2. If a column (or row) is satisfied, cross it out. The remaining decision variables in that column (or row) are non-basic and are set equal to zero. If a row and column are satisfied simultaneously, cross only one out (it does not matter which).
3. Adjust supply and demand for the non-crossed out rows and columns.
4. Allocate the maximum feasible amount to the first available non-crossed out element in the next column (or row).
5. When exactly one row or column is left, all the remaining variables are basic and are assigned the only feasible allocation.

Steps for Least Cost Method

1. Assign as much as possible to the cell with the smallest unit cost in the entire tableau. If there is a tie then choose arbitrarily.
2. Cross out the row or column which has satisfied supply or demand. If a row and column are both satisfied then cross out only one of them.
3. Adjust the supply and demand for those rows and columns which are not crossed out.
4. When exactly one row or column is left, all the remaining variables are basic and are assigned the only feasible allocation.

Steps for Vogel’s Approximation Method

1. 1. Determine a penalty cost for each row (column) by subtracting the lowest unit cell cost in the row (column) from the next lowest unit cell cost in the same row (column).
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3. 2. Identify the row or column with the greatest penalty cost. Break the ties arbitrarily (if there are any). Allocate as much as possible to the variable with the lowest unit cost in the selected row or column. Adjust the supply and demand and cross out the row or column that is already satisfied. If a row and column are satisfied simultaneously, only cross out one of the two and allocate a supply or demand of zero to the one that remains.
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• If there is exactly one row or column left with a supply or demand of zero, stop.
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• -If there is one row (column) left with a positive supply (demand), determine the basic variables in the row (column) using the Minimum Cell Cost Method. Stop.
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• -If all of the rows and columns that were not crossed out have zero supply and demand (remaining), determine the basic zero variables using the Minimum Cell Cost Method. Stop.
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