Transcript Linear Programming (Optimization)

II.2
Strong Valid Inequalities and Facets for
Structured Integer Programs
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1. Introduction
 Strength of the valid inequalities is important (want facet or high dimensional
face). Use the structure of the problem to obtain strong valid inequalities.
 Node packing problem: 𝑆 = 𝑥 ∈ 𝐵𝑛 : 𝑥𝑖 + 𝑥𝑗 ≤ 1, ∀ 𝑖, 𝑗 ∈ 𝐸
dim(conv(𝑆)) = 𝑛 (𝑛 unit vector + 0 vector)
If 𝐶 ⊆ 𝑁 is a clique,
𝑗∈𝐶 𝑥𝑗
If 𝐶 is a maximal clique,
≤ 1 is a valid inequality.
𝑗∈𝐶 𝑥𝑗
≤ 1 defines a facet of conv(𝑆).
Pf) enough to show that there exist 𝑛 affinely independent feasible points
satisfying the clique constraint at equality.
Let 𝐶 = {1, … , 𝑘} be a maximal clique. For each node 𝑗 ∉ 𝐶, ∃ 𝑙(𝑗) ∈ 𝐶
which is not connected to 𝑗 since 𝐶 is a maximal clique.
Consider the characteristic vectors of the packings, {1}, … , {𝑘}, {𝑘 + 1, 𝑙(𝑘 +
1)}, … , {𝑛, 𝑙(𝑛)}.
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2
𝐺:
2
𝐻:
6
3
6
3
1
5
4
5
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 𝑥 1 = 12(0, 1, 1, 1, 1, 1) is an extreme point of the initial polytope (clique
constraints and nonnegativities).
To cut off 𝑥 1 , consider odd hole (chordless cycle, odd cycle with no crossing
edges). Let 𝐻 ⊆ 𝑉, H induces an odd hole, then
𝑗∈𝐻 𝑥𝑗
≤ ( 𝐻 − 1)/2 is a valid inequality. (𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 ≤ 2)
Moreover, it gives a facet of the convex hull of node packings for the
subgraph with node set 𝐻 (five linearly independent vectors satisfying the
inequality at equality). But, it does not give a facet of conv(𝑆) for 𝐺.
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 Consider α𝑥1 + (𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 ) ≤ 2.
Can we make  > 0 and large while the inequality remains valid?
When 𝑥1 = 0, it is valid for any  > 0.
When 𝑥1 = 1, 𝛼 ≤ 2 - (𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 ). But 𝑥1 = 1 implies 𝑥2 =
⋯ = 𝑥6 = 0, so 𝛼 ≤ 2.
Therefore, 2𝑥1 + (𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 ) ≤ 2 is a valid inequality.
Moreover, (1, 0, 0, 0, 0, 0) vector in addition to the five vectors we already
have are linearly independent and satisfy the constraint at equality ⟹ facet.
 Lifting procedure: Systematic method to obtain facet (high dimensional face)
from a facet for the low dimensional restricted problem
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 Prop 1.1: Suppose 𝑆 ⊆ 𝐵𝑛 , 𝑆 𝛿 = 𝑆⋂{𝑥 ∈ 𝐵𝑛 : 𝑥1 = 𝛿}, 𝛿 = 0,1 , and
𝑛
𝑗=2 𝜋𝑗 𝑥𝑗
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≤ 𝜋0 (1.7) is valid for 𝑆 0 . If 𝑆1 = ∅, then 𝑥1 ≤ 0 is valid for S. If
𝑆 ≠ ∅, then 𝛼1 𝑥1 + 𝑛𝑗=2 𝜋𝑗 𝑥𝑗 ≤ 𝜋0 (1.8) is valid for 𝑆 for any 𝛼1 ≤ 𝜋0 − 𝜁,
where 𝜁 = max{ 𝑛𝑗=2 𝜋𝑗 𝑥𝑗 : 𝑥 ∈ 𝑆1 }.
Moreover, if 𝛼1 = 𝜋0 − 𝜁 and (1.7) gives a face of dimension 𝑘 of conv(𝑆0),
then (1.8) gives a face of dimension 𝑘 + 1 of conv(𝑆). [If (1.7) facet of
conv(𝑆0), (1.8) facet of conv(𝑆).]
Pf) If 𝑥 ∈ 𝑆 0 ⟹ 𝛼1 𝑥1 +
If 𝑥 ∈ 𝑆1 ⟹ 𝛼1 𝑥1 +
𝑛
𝑗=2 𝜋𝑗 𝑥𝑗
𝑛
𝑗=2 𝜋𝑗 𝑥𝑗
≤ 𝜋0 valid.
= 𝛼1 +
𝑛
𝑗=2 𝜋𝑗 𝑥𝑗
≤ 𝛼1 + 𝜁 ≤ 𝜋0 valid.
(1.7) gives a 𝑘 −dimensional face of conv(𝑆0) ⟹ ∃ 𝑥 𝑖 ∈ 𝑆 0 such that 𝑥1𝑖 =
0, 𝑖 = 1, … , 𝑘 + 1. Also 𝑥 𝑖 satisfies 𝛼1 𝑥1 + 𝑛𝑗=2 𝜋𝑗 𝑥𝑗 ≤ 𝜋0 at equality.
Let 𝜁 =
𝑛
∗
𝑗=2 𝜋𝑗 𝑥𝑗 ,
𝑥 ∗ ∈ 𝑆1
𝛼1 = 𝜋0 − 𝜁 ⟹ 𝑥 ∗ satisfies 𝛼1 𝑥1 +
𝑆1
𝑛
𝑗=2 𝜋𝑗 𝑥𝑗
= 𝜋0 and 𝑥1∗ = 1 since 𝑥 ∗ ∈
⟹ 𝑥 ∗ cannot be written as affine combination of {𝑥 1 , … , 𝑥 𝑘+1 }, so the 𝑘 + 2
vectors {𝑥 ∗ , 𝑥 1 , … , 𝑥 𝑘+1 } are affinely independent.

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 Lifting is also possible from 𝑆1 to S.
 Prop 1.2: Suppose (1.7) is valid for 𝑆1 . If 𝑆 0 = ∅, then 𝑥1 ≥1 is valid for 𝑆.
If 𝑆 0 ≠ ∅, then 𝛾1 𝑥1 + 𝑛𝑗=2 𝜋𝑗 𝑥𝑗 ≤ 𝜋0 + 𝛾1 (1.9) is valid for 𝑆 for any 𝛾1 ≥
𝜁 − 𝜋0 , where 𝜁 = max{ 𝑛𝑗=2 𝜋𝑗 𝑥𝑗 : 𝑥 ∈ 𝑆 0 }.
Moreover, if 𝛾1 = 𝜁 − 𝜋0 and (1.7) gives a face of dimension 𝑘 of conv(𝑆1),
then (1.9) gives a face of dimension 𝑘 + 1 of conv(𝑆). [If (1.7) facet of
conv(𝑆0), (1.8) facet of conv(𝑆).]
 When 𝛼1 = 𝜋0 − 𝜁 in Prop 1.1 or when 𝛾1 = 𝜁 − 𝜋0 in Prop 1.2, we say that
the lifting is maximum.
 Lifting is done sequentially one variable at a time and the strength of the lifted
inequality depends on the sequence of the variables lifted.

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 Prop 1.3: Let 𝑁\𝑁1 = {1, 2, … , 𝑡} and suppose when Prop 1.1 is applied
sequentially using maximum lifting in the order (𝑖1 , 𝑖2 , … , 𝑖𝑘−1 , 𝑖𝑘 , … , 𝑖𝑡 ), the
inequality 𝑗∈𝑁\𝑁1 𝛼𝑗 𝑥𝑗 + 𝑗∈𝑁1 𝜋𝑗 𝑥𝑗 ≤ 𝜋0 is obtained.
′
Then for any order (𝑖1′ , 𝑖2′ , … , 𝑖𝑘−1
, 𝑖𝑘′ , … , 𝑖𝑡′ ) with 𝑖𝑗′ = 𝑖𝑗 for 𝑗 = 1, … , 𝑘 − 1, the
resulting inequality 𝑗∈𝑁\𝑁1 𝛼𝑗′ 𝑥𝑗 + 𝑗∈𝑁1 𝜋𝑗 𝑥𝑗 ≤ 𝜋0 obtained by maximum
lifting has 𝛼𝑖′𝑘 ≤ 𝛼𝑖𝑘 .
Pf) 𝑖𝑘 = 𝑖𝑠′ for some 𝑠 > 𝑘. Then
𝛼𝑖𝑘 = 𝜋0 − max{
0 for 𝑗 > 𝑘}
= 𝜋0 − max{
𝑘−1
𝑗=1 𝛼𝑖𝑗 𝑥𝑖𝑗
𝑘−1
𝑗=1 𝛼𝑖𝑗 𝑥𝑖𝑗
+
𝑛 : 𝑥 = 1 and 𝑥 =
𝜋
𝑥
∶
𝑥
∈
𝑆
∩
{𝑥
∈
𝐵
𝑗
𝑗
𝑖𝑘
𝑖𝑗
𝑗∈𝑁1
+
𝑗∈𝑁1 𝜋𝑗 𝑥𝑗
: 𝑥 ∈ 𝑆 ∩ {𝑥 ∈ 𝐵𝑛 : 𝑥𝑖𝑠′ = 1 𝑎𝑛𝑑 𝑥𝑖 ′ =
+
𝑗∈𝑁1 𝜋𝑗 𝑥𝑗
+
𝑗
0 for 𝑘 ≤ 𝑗 ≤ 𝑡, 𝑗 ≠ 𝑠}
≥ 𝜋0 − max{
𝑘−1
𝑗=1 𝛼𝑖𝑗 𝑥𝑖𝑗
𝑠−1 ′
𝑗=𝑘 𝛼𝑖𝑗′ 𝑥𝑖𝑗′
: 𝑥 ∈ 𝑆 ∩ {𝑥 ∈ 𝐵𝑛 : 𝑥𝑖𝑠′ =
1 𝑎𝑛𝑑 𝑥𝑖 ′ = 0 for 𝑗 > 𝑠}
𝑗
= 𝛼𝑖′𝑘

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