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Now the problem is to find most economical shipment, i.e., to minimise the total transportation costs.įirst prepare a table showing the requirements and capacities of sites and plants respectively. There is no other condition any plant can transport the product to any site up to its requirement. The transportation costs in Rupees from plants to warehouses are shown in Table 2. These capacities and requirements are as per Table 1. Let there be three plants A, B and C and three sites 1, 2 and 3. To solve such problem ‘operation research’ helps, which involves a mathematical technique called ‘Transportation model’. When this happens, the question immediately arises as to which is the most economical shipment of product from different plants to the different sites. When a company have different manufacturing plants at different places, and have different sites for further distribution of products, we face a problem because if the production capacities of plants are different then it is not possible to ship all the different requirements from the nearest plant. On observing, we see that to get a basic solution, it is not necessary to work out the operation on the coefficients of A and B at each step, since we intend finally to equalise them to zero. It may be added that we can choose any number of basic variables and obtain the basic solution. Thus in this example, the basic solution is: If non- basic variables are zero, then the solution obtained is known as “Basic solution”. Thus we can rewrite the three equations as follows:įurther C, D and E are known as basic variables and A and B as non-basic variables.
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Now multiply the new third row by the coefficient of 2nd row against E (i.e. by -3/5) and add the result in the first row Now multiply the new third row by negative coefficient of first row against E (i.e. In the same way, eliminate the coefficient of E from all equations except the third by repeating the procedure Multiply new second row by 1 and add the resultant with third row we get Multiply new second row by - and add the resultant with 1st row In the same way, eliminate coefficient of D except second equation. In the same way, multiply the new first row with (+2) and add the resultant with third row. by -3) and add the result with the second row. Now multiply the new first row by negative coefficient of second row against C (i.e. Coefficient of C (i.e., 2) is called pivot and is circled in the first matrix, this causes the coefficient of C equal to unity. This can be done with the help of steps I, II and III described as under:ĭivide each entry in the first row by 2, coefficient of C in the first equation.