Valid Inequalities for Integer Valid Inequalities for Integer Programs John E. Mitchell Department...

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  • Valid Inequalities for Integer Programs

    John E. Mitchell

    Department of Mathematical Sciences RPI, Troy, NY 12180 USA

    February 2015, 2019

    Mitchell Valid Inequalities 1 / 20

  • Polarity

    Outline

    1 Polarity

    2 Comparing Inequalities for Polyhedra

    3 Chvatal-Gomory Rounding Procedure

    Mitchell Valid Inequalities 2 / 20

  • Polarity

    Polar cone Definition The cone

    Π := {(π, π0) ∈ Rn+1 : πT x − π0 ≤ 0 for all x ∈ Q}

    is the polar of the polyhedron Q = {x ∈ Rn+ : Ax ≤ b}.

    Proposition

    Let {xk}k∈K and {r j}j∈J be the extreme points and rays of Q. Then Π is the polyhedral cone defined by

    πT xk − π0 ≤ 0 ∀k ∈ K πT r j ≤ 0 ∀j ∈ J.

    Follows since Q = {x =

    ∑ K λkx

    k + ∑

    J µj r j : λ ≥ 0, µ ≥ 0,

    ∑ K λk = 1}.

    Mitchell Valid Inequalities 3 / 20

  • Polarity

    Polar cone Definition The cone

    Π := {(π, π0) ∈ Rn+1 : πT x − π0 ≤ 0 for all x ∈ Q}

    is the polar of the polyhedron Q = {x ∈ Rn+ : Ax ≤ b}.

    Proposition

    Let {xk}k∈K and {r j}j∈J be the extreme points and rays of Q. Then Π is the polyhedral cone defined by

    πT xk − π0 ≤ 0 ∀k ∈ K πT r j ≤ 0 ∀j ∈ J.

    Follows since Q = {x =

    ∑ K λkx

    k + ∑

    J µj r j : λ ≥ 0, µ ≥ 0,

    ∑ K λk = 1}.

    Mitchell Valid Inequalities 3 / 20

  • Polarity

    The polar of Q and facets of Q

    Proposition If dim(Q) = n and π̄ 6= 0 then (π̄, π̄0) is an extreme ray of Π if and only if π̄T x ≤ π̄0 defines a facet of Q.

    Recall:

    Definition The cone

    Π := {(π, π0) ∈ Rn+1 : πT x − π0 ≤ 0 for all x ∈ Q}

    is the polar of the polyhedron Q = {x ∈ Rn+ : Ax ≤ b}.

    Mitchell Valid Inequalities 4 / 20

  • Polarity

    The polar of Q and facets of Q

    Proposition If dim(Q) = n and π̄ 6= 0 then (π̄, π̄0) is an extreme ray of Π if and only if π̄T x ≤ π̄0 defines a facet of Q.

    Recall:

    Definition The cone

    Π := {(π, π0) ∈ Rn+1 : πT x − π0 ≤ 0 for all x ∈ Q}

    is the polar of the polyhedron Q = {x ∈ Rn+ : Ax ≤ b}.

    Mitchell Valid Inequalities 4 / 20

  • Comparing Inequalities for Polyhedra

    Outline

    1 Polarity

    2 Comparing Inequalities for Polyhedra

    3 Chvatal-Gomory Rounding Procedure

    Mitchell Valid Inequalities 5 / 20

  • Comparing Inequalities for Polyhedra

    Comparing inequalities

    Assume we have two valid inequalities for x ∈ Q, namely

    πT x ≤ π0 (1) γT x ≤ γ0. (2)

    Definition The inequalities are equivalent if (γ, γ0) = µ(π, π0) for some µ > 0.

    Mitchell Valid Inequalities 6 / 20

  • Comparing Inequalities for Polyhedra

    Maximal valid inequality

    Definition If the inequalities are not equivalent and if there exists µ > 0 such that γ ≥ µπ and γ0 ≤ µπ0 then

    {x ∈ Rn+ : γT x ≤ γ0} ⊆ {x ∈ Rn+ : πT x ≤ π0}

    and (2) dominates or is stronger than (1). A maximal valid inequality is one that is not dominated by any other.

    Proposition

    Let πT x ≤ π0 be a valid inequality for Q = {x ∈ Rn+ : Ax ≤ b}. If Q 6= ∅ then πT x ≤ π0 is either equivalent to or dominated by an inequality of the form uT Ax ≤ uT b for some u ∈ Rm+.

    These are nonnegative combinations of the constraints Ax ≤ b.

    Mitchell Valid Inequalities 7 / 20

  • Comparing Inequalities for Polyhedra

    Maximal valid inequality

    Definition If the inequalities are not equivalent and if there exists µ > 0 such that γ ≥ µπ and γ0 ≤ µπ0 then

    {x ∈ Rn+ : γT x ≤ γ0} ⊆ {x ∈ Rn+ : πT x ≤ π0}

    and (2) dominates or is stronger than (1). A maximal valid inequality is one that is not dominated by any other.

    Proposition

    Let πT x ≤ π0 be a valid inequality for Q = {x ∈ Rn+ : Ax ≤ b}. If Q 6= ∅ then πT x ≤ π0 is either equivalent to or dominated by an inequality of the form uT Ax ≤ uT b for some u ∈ Rm+.

    These are nonnegative combinations of the constraints Ax ≤ b.

    Mitchell Valid Inequalities 7 / 20

  • Comparing Inequalities for Polyhedra

    Example S = {x ∈ Z2+ : −x1 + 2x2 ≤ 4, 5x1 + x2 ≤ 20, −2x1 − 2x2 ≤ −7}

    x1

    x2

    0 2 4

    2

    4

    Mitchell Valid Inequalities 8 / 20

  • Comparing Inequalities for Polyhedra

    Example S = {x ∈ Z2+ : −x1 + 2x2 ≤ 4, 5x1 + x2 ≤ 20, −2x1 − 2x2 ≤ −7}

    x1

    x2

    0 2 4

    2

    4 Extreme points of S:(

    2 2

    ) ,

    ( 2 3

    ) ,

    ( 3 3

    ) ,

    ( 4 0

    )

    Mitchell Valid Inequalities 8 / 20

  • Comparing Inequalities for Polyhedra

    Polar for the example

    x1

    x2

    0 2 4

    2

    4 Extreme points of S:(

    2 2

    ) ,

    ( 2 3

    ) ,

    ( 3 3

    ) ,

    ( 4 0

    )

    (π, π0) satisfy: 2π1 + 2π2 − π0 ≤ 0 2π1 + 3π2 − π0 ≤ 0 3π1 + 3π2 − π0 ≤ 0 4π1 − π0 ≤ 0

    Mitchell Valid Inequalities 9 / 20

  • Comparing Inequalities for Polyhedra

    Extreme rays of polar cone

    Facets of conv(S) give extreme rays of polar cone.

    Polar cone satisfies

    2π1 + 2π2 − π0 ≤ 0 2π1 + 3π2 − π0 ≤ 0 3π1 + 3π2 − π0 ≤ 0 4π1 − π0 ≤ 0

    Facet: x2 ≤ 3. Corresponds to π1 = 0, π2 = 1, π0 = 3. Facet: 3x1 + x2 ≤ 12. Corresponds to π1 = 3, π2 = 1, π0 = 12. Facet: x1 + x2 ≥ 4. Corresponds to π1 = −1, π2 = −1, π0 = −4. Facet: x1 ≥ 2. Corresponds to π1 = −1, π2 = 0, π0 = −2.

    Each extreme ray satisfies two of the polar constraints at equality.

    Mitchell Valid Inequalities 10 / 20

  • Comparing Inequalities for Polyhedra

    Comparing valid inequalities for conv(S) The inequalities 3x1 + 4x2 ≤ 24 and x1 + x2 ≤ 6 are both valid for conv(S).

    x1

    x2

    0 2 4

    2

    4

    x 1 +

    x 2 =

    6

    3x1 + 4x2 =

    24

    x1 + x2 ≤ 6 dominates 3x1 + 4x2 ≤ 24: for x ≥ 0, if x1 + x2 ≤ 6 then 3x1 + 4x2 ≤ 24.

    Mitchell Valid Inequalities 11 / 20

  • Comparing Inequalities for Polyhedra

    Dominating inequalities

    x1 + x2 ≤ 6 dominates 3x1 + 4x2 ≤ 24: for x ≥ 0, if x1 + x2 ≤ 6 then 3x1 + 4x2 ≤ 24.

    How can we see this algebraically?

    x1 + x2 ≤ 6 is equivalent to 4x1 + 4x2 ≤ 24.

    Then the coefficients on the left hand side are at least as large as the coefficients on the left for 3x1 + 4x2 ≤ 24, and the right hand side is no larger.

    So it is harder to satisfy the constraint x1 + x2 ≤ 6.

    Mitchell Valid Inequalities 12 / 20

  • Comparing Inequalities for Polyhedra

    Getting constraints for conv(S) from Q Take linear combinations of the constraints of Q.

    Take 411(−x1 + 2x2 ≤ 4) + 3

    11(5x1 + x2 ≤ 20): Get valid constraint for Q: x1 + x2 ≤ 7611 = 6

    10 11 .

    Exploit integrality to round down RHS, get valid constraint for conv(S):

    x1 + x2 ≤ 6

    Take 12(−2x112x2 ≤ −7): Get valid constraint for Q: −x1 − x2 ≤ −3.5. Exploit integrality to round down RHS, get valid constraint for conv(S):

    −x1 − x2 ≤ 4

    Mitchell Valid Inequalities 13 / 20

  • Comparing Inequalities for Polyhedra

    Getting constraints for conv(S) from Q Take linear combinations of the constraints of Q.

    Take 411(−x1 + 2x2 ≤ 4) + 3

    11(5x1 + x2 ≤ 20): Get valid constraint for Q: x1 + x2 ≤ 7611 = 6

    10 11 .

    Exploit integrality to round down RHS, get valid constraint for conv(S):

    x1 + x2 ≤ 6

    Take 12(−2x112x2 ≤ −7): Get valid constraint for Q: −x1 − x2 ≤ −3.5. Exploit integrality to round down RHS, get valid constraint for conv(S):

    −x1 − x2 ≤ 4

    Mitchell Valid Inequalities 13 / 20

  • Comparing Inequalities for Polyhedra

    The two valid inequalities for conv(S)

    x1

    x2

    0 2 4

    2

    4 x 1 +

    x 2 =

    4

    x 1 +

    x 2 =

    6

    One of the inequalities gives a facet of conv(S), the other does not.

    Mitchell Valid Inequalities 14 / 20

  • Chvatal-Gomory Rounding Procedure

    Outline

    1 Polarity

    2 Comparing Inequalities for Polyhedra

    3 Chvatal-Gomory Rounding Procedure

    Mitchell Valid Inequalities 15 / 20

  • Chvatal-Gomory Rounding Procedure

    Chvatal-Gomory Rounding Procedure

    Let P = {x ∈ Rn+ : Ax ≤ b} and S = P ∩ Zn. Assume A is m × n. Denote the entries of A by aij . The Chvatal-Gomory rounding procedure generates valid linear inequalities for S from valid linear inequalities for P as follows:

    1 Choose multipliers u ∈ Rm+ and construct the following inequality that is valid for P:

    uT Ax ≤ uT b,

    or equivalently

    n∑ j=1

    ( m∑

    i=1

    uiaij

    ) xj ≤

    m∑ i=1

    uibi . (3)

    Mitchell Valid Inequalities 16 / 20

  • Chvatal-Gomory Rounding Procedure

    Step 2

    2 Round down the left hand side