Right Maximum Minimum Budget Ultimate Procedure for Protected and Crazy Rectangular Network

: In this Research, proposed algorithm namely Right Maximum Minimum Allotment method is applied to find the optimal feasible solution to reduce the cost from the basic feasible solution for transportation problems. The proposed algorithm is a unique way to reach the feasible (or) may be optimal (for some extant) solution without disturbance of degeneracy condition


Introduction
The transportation problem is a special type of linear programming problem where the objective consists in minimizing transportation cost of a given commodity from a number of sources or origins (e.g.factory, manufacturing facility) to a number of destinations (e.g.warehouse, store) [1] [2].Each source has a limited supply (i.e.maximum number of products that can be sent from it) while each destination has a demand to be satisfied (i.e.minimum number of products that need to be shipped to it) [3][4].The cost of shipping from a source to a destination is directly proportional to the number of units shipped [5] [6].
Transportation problems have been widely studied in Medical Science and Operations Research.It is one of the fundamental problems of network flow problem which is usually use to minimize the transportation cost for Bio medical Engineering with number of sources and number of destination (Hospitals) while satisfying the supply limit and demand requirement for the medicines [7].
Transportation models play an important role in medicine distribution management along with cost minimizing and improving service in effective manner [8].Some early approaches have devised solution procedure for the transportation problem with precise supply and demand parameters with respect to the medicine supply management in quick delivery with minimum cost [9][10].

Algorithm
Right Maximum Minimum Allotment (RMxMiA) STEP 1: Construct the TT for the given pay off matrix (POM).

Note:
If the given pay off is not balanced then balance the POM by introducing Zero column and Zero row along with the equated demand and supply cost in the very last iteration.
Example 1 : Consider the following Balanced POM to achieve minimum cost.

Turkish Journal of Computer and Mathematics Education
Vol.12 No.10 (2021), 6861 -6866 6862 Research Article By Applying the above said Procedure, We get Step 1: Here the Maximum cost is 12 in (3, 2) is a Pivot Element in the table 1.1; it is highlighted with Violet Colour.Select the minimum cost 1 in the Pivot element column and allot the maximum possible demand 35 units.Blue colour marked row is not considered for the next iteration.

Table .1.1
Step 2: Here the Maximum cost is 11 in (2, 1) is a Pivot Element in the table 1.2; it is highlighted with Violet Colour.Select the minimum cost 2 from the right side of the Maximum Pivot element and allot the maximum possible demand 50 units.Blue colour marked column is not considered for the next iteration.

Table .1.2
Step 3: Here the Maximum cost is 11 in (2, 1) is a Pivot Element in the table 1.3; it is highlighted with Violet Colour.Select the minimum cost 3 from the right side of the Maximum Pivot element and allot the maximum possible demand 35 units.Blue colour marked row is not considered for the next iteration.

Table .1.3
Step 4: Here the Maximum cost is 11 in (2, 1) is a Pivot Element in the table 1.4; it is highlighted with Violet Colour.Select the minimum cost 8 from the right side of the Maximum Pivot element and allot the maximum possible demand 5 units.Blue colour marked row is not considered for the next iteration.

Table .1.4
Step 5: Here the Maximum cost is 10 in (3, 1) is a Pivot Element in the table 1.5; it is highlighted with Violet Colour.Supply the maximum possible supply 85 units in (3, 1) and 5 units in (3,4) which leads to the solution satisfying all the conditions.

Table .1.5
The resulting Optimum feasible solution is given below.

RESULT:
Example 2 : Consider the following Balanced POM to achieve minimum cost.

Table .2
By Applying the above said Procedure, We get the resulting Optimum feasible solution for Example 2.

STEP 2 :STEP 3 :STEP 4 :
Find the Maximum cost from the given POM and find the minimum cost from the right side of the maximum cost as a Pivot element occurred in the column.Allot the demand from the corresponding supply for the Pivot element which follows Step 2. Repeat the procedure until the degeneracy conditions prevails.