a polyhedral approach to cardinality constrained optimization
DESCRIPTION
A Polyhedral Approach to Cardinality Constrained Optimization. Ismael Regis de Farias Jr. and Ming Zhao University at Buffalo, SUNY. Summary. Problem definition Relation to previous work Simple bound inequalities Further research. Problem Definition. - PowerPoint PPT PresentationTRANSCRIPT
A Polyhedral Approach to Cardinality Constrained Optimization
Ismael Regis de Farias Jr. and Ming ZhaoUniversity at Buffalo, SUNY
Summary
• Problem definition
• Relation to previous work
• Simple bound inequalities
• Further research
Problem Definition
Given c1n , Amn , bm1 , and ln1 , un1 ≥ 0, find xn1 that:
maximizes
cx
subject to
Ax b,
−l ≤ x ≤ u,
and at most k variables are nonzero
Motivation
• Portfolio selection
• Feature selection in data mining
Polyhedral approach
Derive within a branch-and-cut scheme strong
inequalities valid for:
Pi = conv {x Rn : jN aij xj bi , −l ≤ x ≤ u,
and at most k variables are nonzero},
i {1, …, m}, to use as cutting planes in the
branch-and-cut
Previous work
• Bienstock (1996): critical set inequalities
• de Farias and Nemhauser (2003): cover inequalities.
However, the present case is more general and
the polyhedral structure is much richer …
Example
Let P = conv {x [−1, 1]6 : 6 x1 + 4 x2 + 3 x3 + 2 x4 + x5 + x6 6 and at most 3 variables are nonzero}. The following inequalities define facets of P:• 6 x1 + 4 x2 + 3 x3 + 2 x4 + x5 + x6 6• 4 x2 + 3 x3 + 2 x4 + x5 + x6 6• 4 x2 + 3 x3 + 2 x4 + x5 6• 4 x2 + 3 x3 + 2 x4 + x6 6• 4 x2 + 2 x4 + x5 + x6 6• 4 x2 + 2 x4 + x6 6• 4 x2 + 2 x4 + x5 6
To take advantage of previous work
… first, we scale and translate the variables,
i.e. P = conv {x [0, 1]n : jN aj xj b and xj
βj , j N, for at most k variables}, and
second, we consider the pieces of P
The pieces are defined as follows …
Proposition Let W N, XW = {x Rn : xj
βj j W and xj ≤ βj j N − W}, and PW = P ∩ XW . Then, PW = conv (S ∩ XW), where S = {x [0, 1]n : jN aj xj b and xj βj , j N, for at most k variables}. �
For each piece …
i.e. for a given W, we change the variables
as:
• yj ← (xj – βj) / (1 – βj), j W
• yj ← (βj – xj) / βj , j N − W
Example
P = conv {x [0, 1]2 : 6 x1 + 4 x2 7, and x1
=
½ or x2 = ½}.
½
½
1
1x1
x2
6 x1 + 4 x2 = 7
Example
P = conv {x [0, 1]2 : 6 x1 + 4 x2 7, and x1 =
½ or x2 = ½}.
½
½
1
1x1
x2
PN
P{1}
P{2}
P
Example
P = conv {x [0, 1]2 : 6 x1 + 4 x2 7, and x1 =
½ or x2 = ½}.
½
½
1
1x1
x2
3 y1 + 2 y2 2−3 y1 + 2 y2 2
3 y1 − 2 y2 2−3 y1 − 2 y2 2
at most 1 nonzero
at most 1 nonzeroat most 1 nonzero
at most 1 nonzero
When aj 0 and b > 0 …
Proposition The inequality jN xj k is facet-defining iff an−k + …+ an−1 b and a1 + an−k+2 + …+ an b. �
Proposition When an−k + …+ an−1 b and a1 + an−k+2 + …+ an > b, the inequality a1x1 +2≤j≤n−k−1 max {aj , Δ} xj +Δ n−k≤j≤n xj ≤ k Δ defines a facet of P, where Δ = (b − n−k−2≤i≤n
ai). �
It then follows that …
Proposition The inequality:
jW (xj – βj)/(1 – βj)–jN−W (xj – βj)/βj k is valid W N, and it is facet-defining “under certain conditions”. �
In the same way …
Proposition The inequality:
a1(x1 – β1)/(1 – β1) +2≤j≤n−k−1, jW max {aj ,
Δ}(xj – βj)/(1 – βj)+2≤j≤n−k−1, jN−W max {aj,
Δ} (xj – βj)/βj + Δ n−k≤j≤n , jW (xj – βj)/(1 – βj) + Δ n−k≤j≤n, jN−W (xj – βj)/βj ≤k Δ defines a facet of P “under certain conditions”. �
Example
P = conv {x [0, 1]2 : 6 x1 + 4 x2 7, and x1 =
½ or x2 = ½}.
½
1
1x1
x2
x1 + x2 3/2
x1 − x2 1/2x1 + x2 ≥ 1/2
(y1 + y2 1 and 3 y1 + 2 y2 2)
(y1 + y2 1 and 3 y1 − 2 y2 2)
−x1 + x2 1/2
(y1 + y2 1 and −3 y1 − 2 y2 2)
(y1 + y2 1 and −3 y1 + 2 y2 2)
Critical sets and covers
• By fixing, at 0 or 1, variables with positive or negative coefficients, we can obtain implied critical sets or cover inequalities that define facets in the projected polytope.
• Then, by lifting the fixed variables, we obtain strong inequalities valid for P
Example
Let P = conv {x [0,1]5: 6x1 + 4x2 − 3x3 − 2x4
+ x5 6 and at most 2 variables are positive}.
Fix x3 = 1 and x4 = 0. The inequality:
6x1 + 4x2 + 3x5 9
defines a facet of P ∩ {x [0,1]5: x3 = 1 and
x4 = 0}.
Simple bound inequalities
Let P = conv {x [0,1]4: 6x1 − 4x2 + 3x3 − x4
3 and at most 2 variables are positive}. Fix x3 = x4
= 0. Then, x1 1 defines a facet of P ∩ {x [0,1]4:
x3 = x4 = 0}. Lifting with respect to x4, we obtain x1 +
α x4 ≤ 1, which gives α = ⅓. Lifting now with
respect to x3, we obtain 3x1 + α x3 + x4 ≤ 3, which
gives α = 2, and so 3x1 + 2 x3 + x4 ≤ 3.
Additional results
• Two families of lifted cover inequalities
• Two families of inequalities derived from simple bounds
• Necessary and sufficient condition for “pieces of a facet” to be a facet
Further Research
• Separation routines and computational testing
• Inequalities derived from intersection of knapsacks
• Special results for feature selection in data mining