advanced24 min studyLesson 290 in Mathematics

Cutting planes and integer hulls

Derive valid inequalities that remove fractional relaxations while preserving every intended integer solution.

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What you’ll understand

  • Define the integer hull and distinguish validity from separation.
  • Derive a Chvátal–Gomory inequality using a justified rounding step.
  • Explain cover cuts, local validity, and qualified finite-convergence claims.

An LP relaxation may be weak because its polyhedron contains fractional corners that no integer decision can realize. A cutting plane improves the representation rather than splitting the search immediately. It adds an inequality satisfied by every relevant integer-feasible point but violated by the current fractional candidate. The central obligation is preservation: removing a fractional point is useful only if the cut does not remove a legitimate integer solution as well.

The ideal relaxation and a separation task

For a polyhedron PP, its integer hull is PI=conv⁡(P∩Zn)P_I=\operatorname{conv}(P\cap\mathbb Z^n). Linear optimization over the integer points and over this hull has the same supremum whenever the integer set is nonempty, because a linear objective on a convex combination is a convex combination of the constituent values. For rational polyhedra the integer hull is itself a rational polyhedron. This structural fact does not promise a small description: a useful formulation can require many inequalities, so explicitly listing the entire hull may be computationally unrealistic.

An inequality is valid if it holds at every integer-feasible point under consideration. It separates a candidate only if that candidate violates it. Both properties are required for an effective cut at the current LP solution. An inequality already implied by the relaxation is valid but cannot remove a feasible LP point. An arbitrary inequality that removes the candidate may be invalid because it excludes desirable integer points too. Separation algorithms search a specified family of valid inequalities; failure to find a violated member of that family does not prove the candidate belongs to the full integer hull.

PI=conv⁡(P∩Zn),aTx≤β valid on PI,aTxˉ>βP_I=\operatorname{conv}(P\cap\mathbb Z^n),\qquad a^Tx\le\beta\ \text{valid on }P_I,\quad a^T\bar x>\beta
A useful cut has two independently checked properties: validity for intended integer solutions and violation by the current fractional point.↔ Scroll wide equations horizontally.

Round only an integer-valued left side

Suppose P={x:Ax≤b}P=\{x:Ax\le b\} and all coordinates of the intended solution are integers. Choose nonnegative multipliers λ\lambda such that λTA\lambda^TA is an integer row. Combining the constraints gives λTAx≤λTb\lambda^TAx\le\lambda^Tb. Its left side is integer for every intended point, so it also satisfies λTAx≤⌊λTb⌋\lambda^TAx\le\lfloor\lambda^Tb\rfloor. This is a Chvátal–Gomory cut. The floor step is justified by the integrality of the combined left side, not by an instruction to round every coefficient or every inequality independently.

If some variables are continuous, an integer-looking coefficient row generally does not make the left side integer. Mixed-integer cut formulas therefore require different arguments that account for continuous terms. Likewise, a nonnegative combination is essential when adding inequalities in this direction; a negative multiplier could reverse an inequality. Rational scaling can help obtain integral combined coefficients, but multiplying an inequality by an arbitrary number and flooring its bound without checking the new left side is unsafe. A derivation should display the multipliers and explicitly name the integer quantity that permits rounding.

λ≥0,λTA∈Z1×n⟹(λTA)x≤⌊λTb⌋(x∈P∩Zn)\lambda\ge0,\quad \lambda^TA\in\mathbb Z^{1\times n}\quad\Longrightarrow\quad (\lambda^TA)x\le\lfloor\lambda^Tb\rfloor\quad(x\in P\cap\mathbb Z^n)
The weighted sum is a real valid inequality; integer-valued activity permits the strengthened rounded right-hand side.↔ Scroll wide equations horizontally.

Cover inequalities and algorithmic limits

For a binary knapsack ∑iaixi≤B\sum_i a_ix_i\le B with nonnegative weights, a cover is an index set CC whose weights sum to more than BB. All its items cannot be selected simultaneously, giving ∑i∈Cxi≤∣C∣−1\sum_{i\in C}x_i\le|C|-1. This proof uses both the overweight condition and binary variables. Minimal covers can be strengthened by lifting additional variables, but any lifted coefficient needs its own validity argument. Cover cuts expose combinatorial exclusions absent from the original average resource inequality and often improve a node relaxation before further branching.

A globally valid cut preserves every integer-feasible point of the original model; a locally valid cut may rely on branch-specific restrictions and can only be used where those restrictions hold. Keeping that scope is essential when sharing cuts between nodes. For rational pure-integer polyhedra, repeated Chvátal–Gomory closure reaches the integer hull after finitely many rounds, but that theorem neither says one cut suffices nor grants a small worst-case algorithm. Arbitrary choices of valid cuts need not reproduce the full closure or inherit a finite-step guarantee. Branch-and-cut combines validated cuts with exhaustive branching, retaining the bound and coverage logic of the previous lesson.

A weighted cut closes the gap

For nonnegative integer x,yx,y with 2x+y≤42x+y\le4 and x+2y≤4x+2y\le4, derive a cut removing the fractional LP point (4/3,4/3)(4/3,4/3).

  1. Multiply each resource inequality by 1/31/3 and add. The combined coefficient row is (1,1)(1,1), which is integral, and the combined right-hand side is 8/38/3.

  2. Since x+yx+y is integer at every intended point, strengthen x+y≤8/3x+y\le8/3 to x+y≤2x+y\le2. The fractional candidate has sum 8/38/3, so it violates the cut.

  3. For objective 3x+2y3x+2y, the cut and the original bound x≤2x\le2 imply 3x+2y=2(x+y)+x≤63x+2y=2(x+y)+x\le6. The integer point (2,0)(2,0) attains six, closing the objective gap without enumerating all feasible integer points.

The result

The valid cut x + y ≤ 2 removes the fractional root solution and yields a certified integer optimum of six with an independently checked objective bound.

Explore the model

Interactive lab

Interactive lab · vectors

A cut has a normal and a separate bound

Pause and Reset; set angle to 45 degrees and magnitude to 1. Predict equal projections 1/21/\sqrt2, then inspect them. This unit normal is parallel to the cut x+y≤2x+y\le2. Reset, set angle to 45 and magnitude to 2, and predict doubled components: the same halfspace requires doubling its right-hand side too.

A cut has a normal and a separate boundPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA normal direction alone does not specify a valid inequality: its bound and domain matter. The instrument draws one normal vector, not a feasible region or an automatic cutting-plane proof.

Check your understanding

For integer x and y, a valid weighted combination gives x + y ≤ 8/3. Which strengthened inequality is justified solely by integrality?

Try it yourself

For binary x1,x2,x3x_1,x_2,x_3 with 4x1+3x2+2x3≤54x_1+3x_2+2x_3\le5, derive the cover inequality from C={1,2}C=\{1,2\} and show that it removes a feasible fractional point of the LP relaxation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The two cover weights exceed capacity; seek a fractional pair with sum greater than one while keeping its weighted load at most five.

Compare with the solution

Since 4+3>54+3>5, both x1x_1 and x2x_2 cannot equal one, so x1+x2≤1x_1+x_2\le1 is valid for all binary-feasible points. The relaxed point (x1,x2,x3)=(1/2,1,0)(x_1,x_2,x_3)=(1/2,1,0) has load 4(1/2)+3=54(1/2)+3=5 and satisfies all box bounds, but its cover sum is 3/2>13/2>1. Thus the valid cover cut separates it without excluding any feasible binary selection.

Take it with you

  • A cut needs both a validity proof and an actually violated candidate.
  • Chvátal–Gomory rounding requires an integer-valued combined left side.
  • Local cut assumptions and finite-closure theorems must not be confused with arbitrary-cut convergence guarantees.

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