1 / 22

A Polyhedral Approach to Cardinality Constrained Optimization

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.

mareo
Download Presentation

A Polyhedral Approach to Cardinality Constrained Optimization

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A Polyhedral Approach to Cardinality Constrained Optimization Ismael Regis de Farias Jr. and Ming Zhao University at Buffalo, SUNY

  2. Summary • Problem definition • Relation to previous work • Simple bound inequalities • Further research

  3. Problem Definition • Given c1n , Amn ,bm1, and ln1, un1≥ 0,find xn1that: • maximizes • cx • subject to • Ax  b, • −l≤x ≤u, • and at most k variables are nonzero

  4. Motivation • Portfolio selection • Feature selection in data mining

  5. Polyhedral approach Derive within a branch-and-cut scheme strong inequalities valid for: Pi= conv {x  Rn : jN 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

  6. 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 …

  7. 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

  8. To take advantage of previous work … first, we scale and translate the variables, i.e. P = conv {x  [0, 1]n : jN aj xjb and xj • βj , j N, for at most k variables}, and second, we consider the pieces of P

  9. 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 : jN aj xjb and xj βj , j N, for at most k variables}. 

  10. 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

  11. Example P = conv {x  [0, 1]2: 6 x1 + 4 x27, and x1 = ½ or x2= ½}. x2 1 ½ 6 x1 + 4 x2=7 x1 ½ 1

  12. Example P = conv {x  [0, 1]2: 6 x1 + 4 x27, and x1 = ½ or x2= ½}. x2 1 P{2} PN ½ P P{1} x1 ½ 1

  13. Example P = conv {x  [0, 1]2: 6 x1 + 4 x27, and x1 = ½ or x2= ½}. x2 1 −3 y1 + 2 y22 3 y1 + 2 y22 at most 1 nonzero at most 1 nonzero ½ −3 y1 − 2 y22 3 y1 − 2 y22 at most 1 nonzero at most 1 nonzero x1 ½ 1

  14. When aj 0 and b > 0 … • Proposition The inequality jN xj k is facet-defining iff an−k+ …+ an−1 b and a1+ an−k+2+ …+ an b.  • Proposition Whenan−k+ …+ an−1 b and a1+ an−k+2+ …+ an> b, the inequality a1x1+2≤j≤n−k−1max {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). 

  15. It then follows that … • Proposition The inequality: • jW (xj–βj)/(1 – βj)–jN−W (xj–βj)/βj k is valid  W  N, and it is facet-defining “under certain conditions”. 

  16. In the same way … • Proposition The inequality: • a1(x1–β1)/(1 – β1)+2≤j≤n−k−1, jWmax {aj ,Δ}(xj–βj)/(1 – βj)+2≤j≤n−k−1, jN−Wmax {aj,Δ} (xj–βj)/βj + Δn−k≤j≤n , jW(xj–βj)/(1 – βj) + Δn−k≤j≤n, jN−W(xj–βj)/βj ≤k Δ defines a facet of P “under certain conditions”. 

  17. Example P = conv {x  [0, 1]2: 6 x1 + 4 x27, and x1 = ½ or x2= ½}. x2 1 −x1 + x21/2 x1 + x23/2 (y1 + y21 and −3 y1 + 2 y22) (y1 + y21 and 3 y1 + 2 y22) x1 − x21/2 x1 + x2≥1/2 (y1 + y21 and 3 y1 − 2 y22) (y1 + y21 and −3 y1 − 2 y22) x1 ½ 1

  18. 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

  19. 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}.

  20. 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+ αx4≤ 1, which gives α= ⅓. Lifting now with respect to x3, we obtain 3x1+αx3+x4≤ 3, which gives α= 2, and so 3x1+2x3+x4≤ 3.

  21. 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

  22. Further Research • Separation routines and computational testing • Inequalities derived from intersection of knapsacks • Special results for feature selection in data mining

More Related