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A Polyhedral Approach to Cardinality Constrained Optimization

Explore the polyhedral approach to optimize cardinality-constrained problems, with cutting-plane techniques, strong inequalities, and facet definitions. Discover its application in portfolio and feature selection.

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A Polyhedral Approach to Cardinality Constrained Optimization

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  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

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