Feasible Solution¶

Definition¶

A feasible solution assigns values to all decision variables and satisfies every constraint, including their domain restrictions. A solution violating at least one constraint is infeasible.

Interpretation¶

Feasibility asks whether a decision is allowed, independently of how good it is. A decision is feasible when it satisfies every constraint of the model, including the restrictions on the values of the variables.

The feasible pentagon with two feasible points, one inside and one on the boundary, an infeasible point outside, and a point with a negative coordinate.

The green points $(15,40)$ and $(40,20)$ are feasible; $(40,20)$ lies on two boundary lines, which is allowed because the constraints are $\le$ . The red point $(30,50)$ fails a single constraint, finishing time, and is therefore infeasible, as is any point with $x_s<0$ $0$ .

All constraints must hold together: one violated inequality is enough to reject a decision, and a large surplus in one constraint does not make up for a violation in another. In the figure, $(30,50)$ would give Giapetto a profit of $3\cdot30+2\cdot50=190$ $190$ , more than the best feasible profit $180$ , but it needs $110$ finishing hours where only $100$ are available.

The feasible solutions together form the feasible region, and an optimal solution is a feasible solution with the best objective value. A feasible solution without a proof of optimality, for example one found by a heuristic, can still be carried out, and its objective value bounds the optimum.

A common slip is to treat a boundary point as infeasible. Constraints with $\le$ or $\ge$ include equality, so a plan that uses a resource exactly is feasible.

Key Results¶

An attractive objective value does not compensate for a violated constraint.
An LP-feasible fractional solution may be infeasible after integer restrictions are imposed.
Every optimal solution is feasible.

Examples¶

In the figure, $(15,40)$ satisfies $70\le100$ $100$ , $55\le80$ and $15\le40$ $40$ ; $(40,20)$ is also feasible. By contrast, $(30,50)$ requires $2\cdot30+50=110>100$ finishing hours, and any $x_s<0$ violates nonnegativity.

For $x_1\le4$ $4$ , $x_2\le5$ $5$ , $2x_1+x_2\le9$ $9$ , $x_1,x_2\ge0$ $0$ , $(2,5)$ is feasible.
In the same model, $(4,5)$ is infeasible because $2(4)+5=13>9$ $9$ .

Prerequisites¶

Constraint

Sources¶

OR Modelling · Week 1

Aliases: feasible point

Last updated: 20 October 2026 03:02