Simplex method of solving linear programming problem uses
only the corner points of the feasible region
The Simplex method is a widely used algorithm for solving Linear Programming Problems (LPPs). It is an iterative procedure that systematically searches for the optimal solution to an LPP.
A Linear Programming Problem involves optimizing (maximizing or minimizing) a linear objective function, subject to a set of linear equality and inequality constraints. The region defined by these constraints is called the feasible region.
The core principle behind the Simplex method is that if an optimal solution to a Linear Programming Problem exists, it will always be found at one of the corner points (also known as vertices or extreme points) of the feasible region. This is a fundamental theorem of linear programming.
The Simplex method begins at an initial corner point of the feasible region and then systematically moves to an adjacent corner point that improves the value of the objective function. This process continues until no further improvement is possible, indicating that the optimal solution has been reached at one of these corner points.
Let's analyze why other options do not accurately describe the Simplex method's approach:
In summary, the Simplex method is designed to efficiently navigate the corner points of the feasible region to find the optimal solution to a Linear Programming Problem.
In an Linear programming problem, the restrictions or limitations under which the objective function is to be optimised are called
For the linear programming problem:
Maximum Z = 3X1 + 2X2
Subject to
-2X1 + 3X2 ≤ 9
X1 – 5X2 ≥ - 20
X1, X2 ≥ 0
The above problem has
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Consider an LPP given as
Max Z = 2x 1 - x 2+ 2x 3
Subject to the constraints
2x 1+ x 2 ≤ 10
x 1+ 2x 2 - 2x 3 ≤ 20
x 1+ 2x 3 ≤ 5
x 1, x 2, x 3 ≥ 0
What shall be the solution of the LPP after applying first iteration of the Simplex Method?
Consider the following Linear programming problem (LPP):
Maximize z = x 1+ x 2
Subject to the constraints:
x 1+ 2x 2≤ 2000
x 1+ x 2≤ 1500
x 2≤ 600
and x 1, x 2≥ 0
The solution of the above LPP is: