Proof of Validity for Gomory Fractional Cuts

Q: What happens if $ f_i = 0 $?

If $ f_i = 0 $, the basic variable $ x_i $ is already an integer in the current optimal tableau row, providing no cut. A cut is only generated when $ \bar{b}_i $ is fractional.

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The Formal Theorem

Consider an optimal simplex tableau row for an integer program represented as

x_i + \sum_{j \in N} \bar{a}_{ij} x_j = \bar{b}_i

, where

N

is the index set of non-basic variables and

x_i

is a basic integer variable. Let

\bar{a}_{ij} = \lfloor \bar{a}_{ij} \rfloor + f_{ij}

and

\bar{b}_i = \lfloor \bar{b}_i \rfloor + f_i

where

0 \leq f_{ij}, f_i < 1

. The Gomory Fractional Cut is defined as:

\sum_{j \in N} f_{ij} x_j \geq f_i

Analytical Intuition.

Imagine the simplex method as a traveler seeking the highest peak in a landscape, only to find the summit resides in a forbidden zone of non-integers. The Gomory cut is a mathematical 'fence' we build precisely at the boundary of this forbidden zone. By examining a single row of the optimal tableau where a basic variable

x_i

is forced to be non-integer

\bar{b}_i

, we extract the 'fractional fingerprint' of that constraint. We observe that

x_i + \sum \lfloor \bar{a}_{ij} \rfloor x_j + \sum f_{ij} x_j = \lfloor \bar{b}_i \rfloor + f_i

. Because

x_i

and the integer

\sum \lfloor \bar{a}_{ij} \rfloor x_j

must be integers, the remaining fractional parts must balance the equation. Since

x_j \geq 0

, the term

\sum f_{ij} x_j

must be at least

f_i

to account for the fractional remainder

f_i

. We have effectively carved away the non-integer vertex without sacrificing a single feasible integer point, forcing the optimizer to backtrack and find a valid integer solution.

CAUTION

Institutional Warning.

Students often struggle to see why $f_{ij} x_j$ doesn't exclude integer points. Crucially, the cut relies on $x_j \geq 0$ and the fact that $f_i > 0$ ; any integer point that satisfies the original constraints will automatically satisfy the inequality, leaving the feasible set intact while pruning the non-integer vertex.

Academic Inquiries.

Why is this cut valid?

It is valid because any integer solution to the original system must satisfy the cut equation by construction; the cut is a linear combination of existing constraints that forces a fractional remainder to be non-negative.

What happens if $f_i = 0$ ?

If $f_i = 0$ , the basic variable $x_i$ is already an integer in the current optimal tableau row, providing no cut. A cut is only generated when $\bar{b}_i$ is fractional.

Standardized References.

Definitive Institutional SourceWolsey, L. A., Integer Programming.

Foundational

The Convexity of the Feasible Region of a Linear Program

Exploring the cinematic intuition of The Convexity of the Feasible Region of a Linear Program.

Intermediate

The Fundamental Theorem of Linear Programming: Existence of an Optimal Extreme Point Solution

Exploring the cinematic intuition of The Fundamental Theorem of Linear Programming: Existence of an Optimal Extreme Point Solution.

Intermediate

Equivalence of Basic Feasible Solutions and Extreme Points

Exploring the cinematic intuition of Equivalence of Basic Feasible Solutions and Extreme Points.

Advanced

Characterization of Unboundedness in Linear Programming

Exploring the cinematic intuition of Characterization of Unboundedness in Linear Programming.

Institutional Citation

Reference this proof in your academic research or publications.

NICEFA Visual Mathematics. (2026). Proof of Validity for Gomory Fractional Cuts: Visual Proof & Intuition. Retrieved from https://nicefa.org/library/linear-and-integer-programming/proof-of-validity-for-gomory-fractional-cuts

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