Explain how to convert an unbalanced transportation problem to balanced one.
Answers
Minimise z =
Subjected to constraints
x11 for all i and j is said to be balanced transportation problem when total supply from all the sources is equal to the total demand in all destinations, otherwise, problem is said to be unbalanced transportation problem.
A transportation problem may have feasible solution only it is a balanced problem. An unbalanced problem can be made balanced by adding dummy supply centre (row) or dummy demand centre as per the requirement.
Unbalanced Transportation Problem:
If in a transportation problem, the sum of supply available from all sources is not equal to the sum of demands of all destinations, i.e. the problem is said to be unbalanced transportation problem.
But for a feasible solution to exist, total supply must be equal to the total demand thus it is necessary to convert these unbalanced T.P. into balanced one.
There are two possibilities:
Supply is in excess than the demand, we introduce a dummy demand centre (additional destination column) to the transportation problem to absorb the excess supply. The unit transportation cost for the cells of this dummy destination column are all set equal to zero, because these represents item that are not being made and sent.
Working Rule:
Whenever i.e. total demand exceeds total supply, we introduce a dummy supply centre (additional supply row) in the transportation problem to meet extra demand. The unit transportation cost for the cells of its dummy row are set equal to zero.
Example of Second Type:
Solve the transportation problem when the unit transportation costs, demand and supplies are as given below.
Solution:
Since the total demand ∑bj = 215 is greater than the total supply ∑ ai = 195 the problem is an unbalanced T.P.
We convert this into a balanced T.P. by introducing a dummy origin 04 with cost zero and giving supply equal to 215 – 195 = 20 units. Hence we have the converted problem as follows.
As this problem is balanced there exists a feasible solution to this problem. Using least cost method we get the following initial solution:
There are 7 independent non – negative allocation equals to m + n— 1. Hence, the solution is a non – degenerate one.
The total transportation cost
= (6 x 65) + (5 x 1)+(30 x 5)+(25 x 2)+(4 x 25) + (7 x 45) +(20 x 0)
= Rs. 1010
To find the optimal solution:
We apply the steps in MODI method to the above table.
First of all we calculate u1 vj & Δij.
using the relation cij = ui + vj for occupied cells
and Δij = cij – ( ui+ vj) for unoccupied cells.
Since all Δij is not greater or equal to zero then solution is not optimum.
... We introduce the cell (O3, D1) as this cell has most negative value of Δij, Hence this cell is selected to be brought into basis.
For cell O3 D1 we construct closed path as follows
O3D1→O3D3→O2D3→ O2D2→O1D2→ OxD1→O1D1→O3D3
We assign (+) and minus (-) alternately on the closed path starting (+) for the selected all O2D1, and adjusting relevant supplies/demand position, the revised allocation will be following table.
As the no of independent allocations are equal to m + n – 1, we check the optimality for this we again calculate ue uj & Δij.
Answer:
An unbalanced transportation problem is converted into a balanced transportation problem by introducing a dummy origin or a dummy destinations which will provide for the excess availability or the requirement the cost of transporting a unit from this dummy origin (or dummy destination) to any place is taken to be zero.