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Given finitely encoded linear inequalities, a maximum feasible subsystem is a largest collection that can be simultaneously satisfied. Its decision version asks whether some $k$ inequalities admit a common solution. This is NP-complete, despite polynomial time feasibility testing for each fixed collection. For a graph, use nonnegative variables $x_v$, constraints $x_v\ge 1$ for vertices, and $M>|V|$ copies of $x_u+x_v\le 1$ for each edge. A subsystem of size $M|E|+k$ must retain a copy of every edge constraint and at least $k$ vertex constraints; those vertices form an independent set. Conversely an independent set of size $k$ supplies such a feasible subsystem by its indicator vector.

Introduce nonnegative slack variables $s_i$ and write $z=c^{\mathsf{T}}x$. A maximizing simplex dictionary is dual feasible when every nonbasic coefficient in the objective is nonpositive. After inserting a constraint, the dual simplex algorithm retains this property while repairing any negative basic value. For a leaving row $x_B=\beta +{\sum }_j{d}_j{x}_j$ with $\beta <0$, only $d_j>0$ can enter; choose a column minimizing the ratio of its objective penalty to $d_j$.

For the first linear program, $x_1$ has the best profit-to-resource ratio. Pivot it into the initial slack basis to obtainAll basic values are nonnegative and all reduced costs are nonpositive, so the optimal point is $(5,0,0,0)$, with value five. Adding the second constraint givesThe same basis remains feasible and optimal: the second constraint does not change the answer.

The third constraint gives the new rowThe current dictionary is dual feasible but primal infeasible. The eligible entering variables are $x_3,x_4$, with dual ratio-test values $2$ and $3$ respectively. Enter $x_3$ and leave $s_3$; solving that row and substituting into the old rows yieldsThis is a feasible optimal simplex dictionary. Thus the new optimal point is $(2,0,1,0)$, with value three. As an independent linear programming optimality certificate, dual multipliers $(3,0,2)$ have weighted objective bound $3\cdot 5+2(-6)=3$ and weighted coefficient row $(1,4,1,2)$, which dominates the profit row.

On inserting the fourth constraint the slack isIt has a negative constant and no eligible entering column. Since all its nonbasic variables are nonnegative, this row cannot be satisfied: the fourth linear program is infeasible, and adding the fifth constraint cannot restore feasibility. The requested optimal values are therefore

The arbitrary-data selection question is a maximum feasible subsystem problem. With rational input it is in NP: guess the selected row indices, and test their linear programming feasibility in polynomial time. Equivalently, a nonempty rational polyhedron has a feasible rational point of polynomially bounded encoding length, which serves as a certificate.

For an explicit NP-hardness reduction, take an independent set instance with graph $H$ having $r$ vertices and desired independent-set size $K$. Use variables $x_v\ge 0$, one inequality $-x_v\le -1$ for each vertex, and $M=r+1$ identical copies of $x_u+x_v\le 1$ for each edge. Ask for a feasible subsystem of size $M|E(H)|+K$. An independent set of size $K$ gives such a subsystem: include all edge rows and those $K$ vertex rows, and use its indicator vector. Conversely, omitting every copy of even one edge row leaves at most $M(|E(H)|-1)+r<M|E(H)|+K$ selectable rows. Hence any qualifying subsystem retains every edge inequality and at least $K$ vertex inequalities. Their vertices must be independent, since two adjacent selected vertices would force $x_u+x_v\ge 2$. The construction has polynomial size; $c$ is irrelevant to feasibility. Thus the rational-input decision problem is NP-complete, not merely difficult because there are many subsets to enumerate.