Minimizing the Cost of Transportation

$\text { Minimise } z=\sum_{i=1}^{m} \sum_{j=1}^{n} C_{i j} X_{i j}$

$\sum_{j=1}^{n} X_{i} \leq a_{i}, i=1,2,3 \ldots m$

(Demand constraint)

$\sum_{j=1}^{n} X_{i j} \geq b_{j}(j=1,2,3 \ldots n)$

(Supply constraint)

$X_{i j} \geq 0(i=1,2,3 \ldots m, j=1,2,3 \ldots n)$

This is a linear program with m, n decision variables, m+n functional constraints, and m, n non-negative constraints.

m=Number of sources, n= Number of destinations, a_i= Capacity of i^th source (in tons, pounds, litres, etc.), b_j =Demand of j^th destination (in tons, pounds, litres, etc.)

c_ij = cost coefficients of material shipping (unit shipping cost) between i^th source

and j^th destination (in $ or as a distance in kilometres, miles, etc.), x_ij= amount of material shipped between i^th source and j^th destination (in tons, pounds, litres etc.)

A necessary and sufficient condition for the existence of a feasible

$\sum_{i=1}^{m} a_{i}=\sum_{j=1}^{n} b_{j}$

Remark. The set of constraints

represents m+n equations in non-negative variables. Each variable appears in exactly two constraints, one is associated with the origin and the other is associated with the destination.

Unbalanced Transportation Problem

The transportation problem is known as an unbalanced transportation problem. There are two cases:
Case (1)

$\sum_{i=1}^{m} a_{i}>\sum_{j=1}^{n} b_{j}$

Case (2)

Introduce a dummy origin in the transportation table; the cost associated with this origin is set equal to zero. The availability at this origin is:

$\sum_{i=1}^{m} a_{i}$