METHODOLOGY RECOMMENDATION FOR ONE‐CRITERION TRANSPORTATION PROBLEMS: CAKMAK METHOD

Tanyel Çakmak,Filiz Ersöz

doi:10.3846/16484142.2007.9638128

Abstract

Transportation problems (TP) are one of the most prominent fields of application of the mathematical disciplines to optimization and operations research. In general, there are three starting basic feasible solution methods: Northwest Corner, Least Cost Method, VAM – Vogel's Approximation Method. The three methods differ in the quality of the starting basic solution. In this study, we actually show a new method for starting basic feasible solution to one‐criterion‐transportation problems: Çakmak Method. This method can be used for balanced or unbalanced one-criterion transportation problems, and gives the basic feasible optimum solution accordingly.

Highlights

Transportation problems(TP) are historically one of the most prominent fields of application of the mathematical disciplines to optimization and operations research [1, 2]
A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points
While trying to find the best way, generally, a variable cost of transporting the product from source to destination point or a similar constraint should be taken into consideration

Summary

Introduction

Transportation problems(TP) are historically one of the most prominent fields of application of the mathematical disciplines to optimization and operations research [1, 2]. A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. While trying to find the best way, generally, a variable cost of transporting the product from source (supply) to destination (demand) point or a similar constraint should be taken into consideration. Finding Basic Feasible Solution to Transportation Problem is unlike other Linear Programming problems, a balanced TP with m supply points and n demand points is easier to solve, it has m + n equality constraints. The reason for that is, if a set of decision variables (xij’s) satisfy all but one constraint, the values for xij’s will satisfy that remaining constraint automatically

Transportation models

D2 D3 D4 D5 Supplies

Conclusions