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

• 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