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In the theory of linear programming, a basic feasible solution (BFS) is a solution with a minimal set of non-zero variables. Geometrically, each BFS corresponds to a vertex of the polyhedron of feasible solutions.
The space of all candidate solutions, before any feasible points have been excluded, is called the feasible region, feasible set, search space, or solution space. [2] This is the set of all possible solutions that satisfy the problem's constraints. Constraint satisfaction is the process of finding a point in the feasible set.
The satisfiability problem, also called the feasibility problem, is just the problem of finding any feasible solution at all without regard to objective value. This can be regarded as the special case of mathematical optimization where the objective value is the same for every solution, and thus any solution is optimal.
However, some problems have distinct optimal solutions; for example, the problem of finding a feasible solution to a system of linear inequalities is a linear programming problem in which the objective function is the zero function (i.e., the constant function taking the value zero everywhere).
The possible results of Phase I are either that a basic feasible solution is found or that the feasible region is empty. In the latter case the linear program is called infeasible. In the second step, Phase II, the simplex algorithm is applied using the basic feasible solution found in Phase I as a starting point.
A. The feasible set {b+L} ∩ K is bounded, and intersects the interior of the cone K. B. We are given in advance a strictly-feasible solution x^, that is, a feasible solution in the interior of K. C. We know in advance the optimal objective value, c*, of the problem. D. We are given an M-logarithmically-homogeneous self-concordant barrier F ...
Then row reductions are applied to gain a final solution. The value of M must be chosen sufficiently large so that the artificial variable would not be part of any feasible solution. For a sufficiently large M, the optimal solution contains any artificial variables in the basis (i.e. positive values) if and only if the problem is not feasible.
Otherwise, it has some solution x 0, and the set of all solutions can be presented as: Fz+x 0, where z is in R k, k=n-rank(A), and F is an n-by-k matrix. Substituting x = Fz + x 0 in the original problem gives: