TRANSPORTATION PROBLEMS

Introduction: A Transportation problem is Supply and Demand problem. There are some supply centres e. g. plants. And there are some demand centres e. g. warehouses or distribution centres.

Supply at each plant is given. Demand at each warehouse is given. And unit cost of transportation from each Supply centre to each Demand centre is given.

The solution of a Transportation problem is a two step process.

Step 1:                        To find Initial Feasible Solution (IFS). For this we can use any method, say – Vogel’s Approximation method (VAM), Least Cost method (LCM) or North West Corner Rule (NWCR).

But VAM is supposed to be the best method, because it gives IFS which is very near to the Optimal (Final) solution.

Step 2:                        From the IFS, to find the Optimal solution. The IFS may or may not be Optimal. If the IFS is not Optimal, then it can be improved to give a better result. This process of Testing & Improving the IFS is called Modified Distribution Method (MODI).

For IFS, there are many methods, but for Optimal solution there is only one method – MODI.

Note:   Sometimes, in examination problem, IFS is already given (some solution is given). And you are asked, whether it is an Optimal solution. In that case, do not find a new IFS by VAM or any other method. Directly apply MODI method. First, test the given solution. If it is Optimal, OK. If not Optimal, improve it and find actual Optimal solution.

Concepts:

1. The Transportation problem can be Balanced or Unbalanced problem.

A Balanced problem means total supply is equal to total demand, e. g. if total supply is 400 units and total demand is 400 units then it is balanced.

Where as, an Unbalanced problem means total supply is not equal to total demand. E. g.  if total supply is 400 units and total demand is 350 units, it is not balanced. Then first we need to balance the problem by taking a Dummy. (imaginary quantity). Hence, a Dummy will come on Demand side. Dummy quantity will be 50 (400 – 350 = 50).

2. Dummy: A Dummy is an imaginary quantity. The purpose of Dummy is to balance the problem. Since the Dummy is imaginary, all unit costs (or unit profits) for Dummy are always zero. Dummy can come as Supply or Demand, depending on problem. If Supply is less, Dummy will come in Supply side. If Demand is less, Dummy will come in Demand side.

3. The Transportation problem can be of Minimization type or Maximization type.

A Minimization Transportation problem involves cost data. The objective of solution is to minimize the total cost.

A Maximization Transportation problem involves sales, revenue or profit data. The objective of solution is maximization of total profit. We need to first convert the Maximization problem in Minimization problem. This conversion is called Regret matrix.  From the original profit values, we find out the highest profit value. From this highest profit, we subtract all profit values. The resulting matrix is Regret matrix. Then we solve it as a normal transportation problem.

Note:

1. But for calculating the final schedule, you must multiply quantity by original profit value. And not by Regret value.
1. In case of Unbalanced Maximization problem, first take Dummy. Dummy profit = 0.

Then convert in Regret matrix. Now, if highest profit in the problem is 21, then Regret for Dummy = 21 – 0 = 21.

4. Prohibited or Restricted problem: A Prohibited problem is the one in which there are one or more restrictions. E. g. say from Plant ‘A’ it is not possible to transport any quantity to Warehouse ‘Y’. This is a prohibited problem

Then we assign a very high or infinite cost (represented by M) to “Plant A – Warehouse Y” and proceed with solution. Throughout the solution steps, M does not change. Since M is infinity, no quantity or allocation is possible in M.

Even in case of Maximization problem, if there is Restriction, we take profit as M. When we convert to Regret matrix, M remains as M.

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Steps –

1. Take Dummy (if required) for balancing the problem & then convert in Regret matrix (if required) if it is a Maximization problem.
2. If there is any Restriction or Prohibition, convert it in M.
3. Find IFS. If the method for IFS is specified in question, use only that method. But if any method is not specified, VAM is preferable. (It will reduce your work when you test for Optimality using MODI).
4. After finding IFS, MODI begins. Find “U & V” values from allocations.
5. Then for empty cells, find Opportunity costs (Delta ∆). If any Delta is/are Positive, solution is not optimal. But if all Delta are Negative or zero, solution is optimal.
6. If any Delta is Positive, start a loop from maximum Positive delta. E. g. if there are two Positive delta (5 and 2), then start loop from 5. Not from 2.
7. Then write next table. Again find new “U, V, and Delta”. Repeat from Step no. 5.

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