Strong Formulations for Convex Functions over Non-Convex Domains

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1 Strong Formulations for Convex Functions over Non-Convex Domains Daniel Bienstock and Alexander Michalka University JMM 2013

2 Introduction Generic Problem: min Q(x), s.t. x F,

3 Introduction Generic Problem: min Q(x), s.t. x F, Q(x) convex, especially: convex quadratic

4 Introduction Generic Problem: min Q(x), s.t. x F, Q(x) convex, especially: convex quadratic F nonconvex

5 Introduction Generic Problem: min Q(x), s.t. x F, Q(x) convex, especially: convex quadratic F nonconvex Examples: F is a mixed-integer set

6 Introduction Generic Problem: min Q(x), s.t. x F, Q(x) convex, especially: convex quadratic F nonconvex Examples: F is a mixed-integer set F is constrained in a nasty way, e.g. x 1 x 2 3 x 3 sin(x 4 ) + 2 cos(x 5 ) = 4

7 Or, Convex constraint: Q(x) q, and x F, Q(x) convex, especially: convex quadratic F nonconvex Examples: F is a mixed-integer set F is constrained in a nasty way, e.g. x 1 x 2 3 x 3 sin(x 4 ) + 2 cos(x 5 ) = 4

8 Exclude-and-cut min z, s.t. z Q(x), x F 0. Let ˆF be a convex relaxation of F

9 Exclude-and-cut min z, s.t. z Q(x), x F 0. Let ˆF be a convex relaxation of F 1. Let (x, z ) = argmin{ z : z Q(x), x ˆF }

10 Exclude-and-cut min z, s.t. z Q(x), x F 0. Let ˆF be a convex relaxation of F 1. Let (x, z ) = argmin{ z : z Q(x), x ˆF } 2. Find an open set S s.t. x S and S F =. Examples: lattice-free sets, geometry

11 Exclude-and-cut min z, s.t. z Q(x), x F 0. Let ˆF be a convex relaxation of F 1. Let (x, z ) = argmin{ z : z Q(x), x ˆF } 2. Find an open set S s.t. x S and S F =. Examples: lattice-free sets, geometry 3. Add to the formulation an inequality az + α T x α 0 valid for { (x, z) : x R n S, z Q(x) } but violated by (x, z ).

12 = feasible region

13 = feasible region convex relaxation

14 = feasible region convex relaxation = optimum over relaxation

15 ALLOWED EXCLUDED ALLOWED = feasible region convex relaxation = optimum over relaxation

16 Initial Research Goals Given an exclusion zone, can we obtain an improved lower bound for the optimization problem?

17 Initial Research Goals Given an exclusion zone, can we obtain an improved lower bound for the optimization problem? Answer: use S-Lemma.

18 Initial Research Goals Given an exclusion zone, can we obtain an improved lower bound for the optimization problem? Answer: use S-Lemma. Can we incorporate the improvement into the formulation by adding linear inequalities?

19 The SUV problem given full-dimensional polyhedra P 1,..., P K in R d, find a point closest to the origin not contained inside any of the P h.

20 The SUV problem given full-dimensional polyhedra P 1,..., P K in R d, find a point closest to the origin not contained inside any of the P h. min x 2 s.t. x R d K h=1 int(p h ),

21 The SUV problem given full-dimensional polyhedra P 1,..., P K in R d, find a point closest to the origin not contained inside any of the P h. min x 2 s.t. x R d K h=1 int(p h ), (application: X-ray lythography; see Ahmadia (2010)) (yes, it is NP-hard)

23 Typical values for d (dimension): less than 20; usually even smaller Typical values for K (number of polyhedra): possibly hundreds, but often less than 50 Very hard problem

24 First problem setting Let Q(x) be a strongly convex function on R d, Let P R d be closed and such that R d P.

25 First problem setting Let Q(x) be a strongly convex function on R d, Let P R d be closed and such that R d P. Want to produce a linear inequality description for: { (x, q) R d+1 : Q(x) q, } x R d int(p).

26 First-order cut: Given y R d, q Q(y) + T Q(y)(x y)

27 First-order cut: Given y R d, is valid for all x R d, q Q(y) + T Q(y)(x y)

28 First-order cut: Given y R d, is valid for all x R d, does not cut anything off q Q(y) + T Q(y)(x y)

29 First-order cut: Given y R d, q Q(y) + T Q(y)(x y) is valid for all x R d, does not cut anything off How can we use the structure of P to strengthen the inequality?

30 Definition: Given y P, say P is locally flat at y if B(µ, ρ) P with µ y 2 = ρ and ρ > 0. ρ y µ P R d P

31 Suppose P is locally flat at y. Write: a = µ y.

32 Suppose P is locally flat at y. Write: a = µ y. Lemma: For real α > 0 small enough the inequality q Q(y) + T Q(y)(x y) + αa T (x y) is valid,

33 Suppose P is locally flat at y. Write: a = µ y. Lemma: For real α > 0 small enough the inequality q Q(y) + T Q(y)(x y) + αa T (x y) is valid, i.e., for all (x, q) with x R d P and q Q(x).

34 Suppose P is locally flat at y. Write: a = µ y. Lemma: For real α > 0 small enough the inequality q Q(y) + T Q(y)(x y) + αa T (x y) is valid, i.e., for all (x, q) with x R d P and q Q(x). Inequality is tight at (y, Q(y)), and cuts-off points (x, Q(x)) and x int(p).

35 Suppose P is locally flat at y. Write: a = µ y. Lemma: For real α > 0 small enough the inequality q Q(y) + T Q(y)(x y) + αa T (x y) is valid, i.e., for all (x, q) with x R d P and q Q(x). Inequality is tight at (y, Q(y)), and cuts-off points (x, Q(x)) and x int(p). Largest possible α: lifted first-order inequality.

36 { (x, q) R d+1 : Q(x) q, } x R d int(p)

37 { (x, q) R d+1 : Q(x) q, } x R d int(p) Theorem. Any linear inequality valid for S is dominated by a lifted first-order inequality. More precisely, conv { (x, q) R d+1 : Q(x) q, x R d int(p) } = {(x, q) R d+1 : LFO + FO }

38 { (x, q) R d+1 : Q(x) q, } x R d int(p) Theorem. Any linear inequality valid for S is dominated by a lifted first-order inequality. More precisely, conv { (x, q) R d+1 : Q(x) q, x R d int(p) } = {(x, q) R d+1 : LFO + FO } How do we make this computationally practicable?

39 First problem setting Let Q(x) is a positive definite quadratic on R d, P = {x R d : a T i x b i, 1 i m} full dimensional

40 First problem setting Let Q(x) is a positive definite quadratic on R d, P = {x R d : a T i x b i, 1 i m} full dimensional Want to produce a linear inequality description for: { (x, q) R d+1 : Q(x) q, } x R d int(p).

41 First problem setting Let Q(x) is a positive definite quadratic on R d, P = {x R d : a T i x b i, 1 i m} full dimensional Want to produce a linear inequality description for: { (x, q) R d+1 : Q(x) q, } x R d int(p). change in coordinates S =. (x, q) Rd+1 : d xj 2 q, x R d int(p). j=1

42 {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} Here P = {x R d : a T i x b i, 1 i m} P is locally flat at any y iff a T i y = b i but a T j y < b j j i for some i.

43 {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} Here P = {x R d : a T i x b i, 1 i m} P is locally flat at any y iff a T i y = b i but a T j y < b j j i for some i. Lifted first-order inequality: q 2y T x y 2 α (a T i x b i ) for α > 0 appropriately chosen.

44 {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} Here P = {x R d : a T i x b i, 1 i m} P is locally flat at any y iff a T i y = b i but a T j y < b j j i for some i. Lifted first-order inequality: q 2y T x y 2 α (a T i x b i ) for α > 0 appropriately chosen. Theorem: Let (ˆx, ˆq) R d+1 with ˆv int(p). We can compute a lifted first-order inequality maximally violated by (ˆx, ˆq), by solving m linearly constrained convex quadratic programs on O(d) variables.

45 When does a point ˆx, d ˆx j 2 j=1 violate a lifted first-order inequality q 2y T x y 2 α (a T i x b i )?

46 When does a point ˆx, d ˆx j 2 j=1 violate a lifted first-order inequality q 2y T x y 2 α (ai T x b i )? It does, if and only if: d ˆx j 2 2y T ˆx + αai T ˆx < y 2 + αb j=1

47 When does a point ˆx, d ˆx j 2 j=1 violate a lifted first-order inequality q 2y T x y 2 α (ai T x b i )? It does, if and only if: d ˆx j 2 2y T ˆx + αai T ˆx < y 2 + αb j=1 This describes the interior of a ball,

48 When does a point ˆx, d ˆx j 2 j=1 violate a lifted first-order inequality It does, if and only if: q 2y T x y 2 α (a T i x b i )? d ˆx j 2 2y T ˆx + αai T ˆx < y 2 + αb j=1 This describes the interior of a ball, which must be contained in int(p)

49 Geometrical characterization int(p) cut off region y

50 Geometrical characterization int(p) cut off region Characterization: x R d int(p) iff, for each ball B(µ, R) P, y

51 Geometrical characterization int(p) cut off region Characterization: x R d int(p) iff, for each ball B(µ, R) P, y x µ 2 R 2

52 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m}

53 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m} where Q i = conv(s) = conv (Q 1 Q 2... Q m ), (x, q) Rd+1 : d j=1 x 2 j q, ai T x b i.

54 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m} where Q i = conv(s) = conv (Q 1 Q 2... Q m ), (x, q) Rd+1 : d j=1 x 2 j q, ai T x b i. Ceria and Soares (1999) min { q : (x, q) S} is an SOCP

55 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m} where Q i = conv(s) = conv (Q 1 Q 2... Q m ), (x, q) Rd+1 : d j=1 x 2 j q, ai T x b i. Ceria and Soares (1999) min { q : (x, q) S} is an SOCP with O(md) variables and m conic constraints

56 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m} where Q i = conv(s) = conv (Q 1 Q 2... Q m ), (x, q) Rd+1 : d j=1 x 2 j q, ai T x b i. Ceria and Soares (1999) min { q : (x, q) S} is an SOCP with O(md) variables and m conic constraints Separation: solve SOCP, use SOCP Farkas Lemma, get linear cut

57 S = {(x, q) R d+1 : d j=1 x 2 j q, x R d int(p)} P = {x R d : a T i x b i, 1 i m} where Q i = conv(s) = conv (Q 1 Q 2... Q m ), (x, q) Rd+1 : d j=1 x 2 j q, ai T x b i. Ceria and Soares (1999) min { q : (x, q) S} is an SOCP with O(md) variables and m conic constraints Separation: solve SOCP, use SOCP Farkas Lemma, get linear cut = LFO cut

58 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0. j=1 P = {x R d : x T Ax 2c T x + b 0}

59 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0 j=1 P = {x R d : x T Ax 2c T x + b 0}. Ceria and Soares (1999) infinite SOCP

60 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0 j=1. P = {x R d : x T Ax 2c T x + b 0}

61 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0 j=1. P = {x R d : x T Ax 2c T x + b 0} P µ ρ

62 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0 j=1. P = {x R d : x T Ax 2c T x + b 0} P µ ρ Separation problem: given(ˆx, ˆq) R d+1, find B(µ, ρ) P to max ρ (ˆq 2µ Tˆx + µ T µ)

63 Second setting: separating across a quadratic set For A 0, polynomially separable linear inequality description for: d (x, q) Rd+1 : xj 2 q, x T Ax 2c T x + b 0 j=1. P = {x R d : x T Ax 2c T x + b 0} P µ ρ Separation problem: given(ˆx, ˆq) R d+1, find B(µ, ρ) P to max ρ (ˆq 2µ Tˆx + µ T µ) = ρ ˆx µ 2 ˆq + ˆx 2

64 Separation problem: min µ,ρ Subject to: µ 2 ρ 2 x T µ {x : x µ 2 ρ} P

65 Separation problem: min µ,ρ Subject to: µ 2 ρ 2 x T µ {x : x µ 2 ρ} P Theorem: Optimal choices for µ and ρ are given by: ˆµ = ˆθb + (I ˆθA) x and ˆρ = ˆµ 2 2 x T ˆµ + x 2 ˆθ( x T A x 2b T x + c). Here, ˆθ = 1 λ maxa.

66 Separating accross general quadratics Π. = { (x, w, z) R d R R : z x T Qx + q T x, w x T Ax } (A 0, Q 0).

67 Separating accross general quadratics Π. = { (x, w, z) R d R R : z x T Qx + q T x, w x T Ax } (A 0, Q 0). Linear transformation Π is the set of points R d R R s.t. z x 2 + q T x, w x T Λx (Λ > 0). Write P. = {(x, w) R d R : x T Λx w 0}, and for µ R d, ν R, Then M(µ, ν). = {(x, w) R d R : λ max x µ 2 + (ν w) 0}. x R d int(p) iff x R d int(m(µ, ν)), for all µ, ν with M(µ, ν) P.

68 Π = { (x, w, z) R d R R : z x 2 +q T x, w x T Λ x}, P = {(x, w) R d R : x T Λx w}, M(µ, ν) = {(x, w) R d R : λ max x µ 2 + ν w}.

69 Π = { (x, w, z) R d R R : z x 2 +q T x, w x T Λ x}, P = {(x, w) R d R : x T Λx w}, M(µ, ν) = {(x, w) R d R : λ max x µ 2 + ν w}. Geometric characterization: x R d int(p) iff x R d int(m(µ, ν)), µ, ν with M(µ, ν) P.

70 Π = { (x, w, z) R d R R : z x 2 +q T x, w x T Λ x}, P = {(x, w) R d R : x T Λx w}, M(µ, ν) = {(x, w) R d R : λ max x µ 2 + ν w}. Geometric characterization: x R d int(p) iff x R d int(m(µ, ν)), µ, ν with M(µ, ν) P. So, valid inequality for any µ, ν with M(µ, ν) P: λ max µ 2 λ max (2µ + q) T x + (ν w) + λ max z 0 Separation problem, given ( x, w) int(p) subject to: w ν λ max µ 2 + 2λ max x T µ + 2λ max q T x ν + min x { λ max x µ 2 x T Λx } 0.

71 Separation problem, given ( x, w) int(p) subject to: w ν λ max µ 2 + 2λ max x T µ + 2λ max q T x ν + min x { λ max x µ 2 x T Λx } 0. Lemma: This problem can be explicitly solved in polynomial time.

72 Separation problem, given ( x, w) int(p) subject to: w ν λ max µ 2 + 2λ max x T µ + 2λ max q T x ν + min x { λ max x µ 2 x T Λx } 0. Lemma: This problem can be explicitly solved in polynomial time. But we started with Π. = { (x, w, z) R d R R : z x T Qx + q T x, w x T Ax } and transformed it into: Π = { (x, w, z) R d R R : z x 2 + q T x, w x T Λ x},

73 Separation problem, given ( x, w) int(p) subject to: w ν λ max µ 2 + 2λ max x T µ + 2λ max q T x ν + min x { λ max x µ 2 x T Λx } 0. Lemma: This problem can be explicitly solved in polynomial time. But we started with Π. = { (x, w, z) R d R R : z x T Qx + q T x, w x T Ax } and transformed it into: Π = { (x, w, z) R d R R : z x 2 + q T x, w x T Λ x}, Theorem. Eigenspace not necessary for poly-time separation (only max eigenvalue of A).

74 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T.

75 Example: f (x) =. 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)?

76 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)? f (x) = U(x) L(x), where U(x) L(x). = (x 1 + x 2 ) 2 + (x 1 + x 3 ) 2 + (x 2 + x 3 ) 2,. = 2(x1 2 + x2 2 + x3 2 ).

77 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)? f (x) = U(x) L(x), where U(x) L(x). = (x 1 + x 2 ) 2 + (x 1 + x 3 ) 2 + (x 2 + x 3 ) 2,. = 2(x1 2 + x2 2 + x3 2 ). Linear transformation: x p (in R 3 )

78 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)? f (x) = U(x) L(x), where U(x) L(x). = (x 1 + x 2 ) 2 + (x 1 + x 3 ) 2 + (x 2 + x 3 ) 2,. = 2(x1 2 + x2 2 + x3 2 ). Linear transformation: x p (in R 3 ). f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ).

79 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)? f (x) = U(x) L(x), where U(x) L(x). = (x 1 + x 2 ) 2 + (x 1 + x 3 ) 2 + (x 2 + x 3 ) 2,. = 2(x1 2 + x2 2 + x3 2 ). Linear transformation: x p (in R 3 ). f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ).

80 Example: f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ( x), where x = (1/2, 1/2, 1/2) T. Can we prove positive lower bound on f ( x)? f (x) = U(x) L(x), where U(x) L(x). = (x 1 + x 2 ) 2 + (x 1 + x 3 ) 2 + (x 2 + x 3 ) 2,. = 2(x1 2 + x2 2 + x3 2 ). Linear transformation: x p (in R 3 ). f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ). x (0, 0, 2 3 (x 1 + x 2 + x 3 )) T

81 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ((1/2, 1/2, 1/2) T ).. f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ). x (0, 0, 2 3 (x 1 + x 2 + x 3 )) T Paraboloid cut: p p (p 3 2α 3) 2 + ɛ 2p p p2 3.

82 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ((1/2, 1/2, 1/2) T ).. f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ). x (0, 0, 2 3 (x 1 + x 2 + x 3 )) T Paraboloid cut: or, p p (p 3 2α 3) 2 + ɛ 2p p p2 3. f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ

83 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. McCormick relaxation gives zero lower bound on f ((1/2, 1/2, 1/2) T ).. f (x) F (p) = (p1 2 + p2 2 + p2 3 ) (2p p p2 3 ). x (0, 0, 2 3 (x 1 + x 2 + x 3 )) T Paraboloid cut: or, When valid? p p (p 3 2α 3) 2 + ɛ 2p p p2 3. f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ

84 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ

85 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2.

86 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4 5 (x 1 + x 2 + x 3 )

87 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4 5 (x 1 + x 2 + x 3 ) At (.5,.5,.5)T get.98

88 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4 5 (x 1 + x 2 + x 3 ) At (.5,.5,.5)T get.98 α = 1/2, ɛ = 2, valid when x 1 + x 2 + x 3 3/2.

89 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4 5 (x 1 + x 2 + x 3 ) At (.5,.5,.5)T get.98 α = 1/2, ɛ = 2, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4(x 1 + x 2 + x 3 ) 5.

90 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. Paraboloid cut: f (x) 8α(x 1 + x 2 + x 3 ) 12α 2 ɛ α = ɛ = 1/10, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4 5 (x 1 + x 2 + x 3 ) At (.5,.5,.5)T get.98 α = 1/2, ɛ = 2, valid when x 1 + x 2 + x 3 3/2. get: f (x) 4(x 1 + x 2 + x 3 ) 5. At (.5,.5,.5) T get 1.

91 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. f (x) 4 5 (x 1 + x 2 + x 3 ) valid for x 1 + x 2 + x 3 3/2 f (x) 4(x 1 + x 2 + x 3 ) 5 valid for x 1 + x 2 + x 3 3/2

92 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. f (x) 4 5 (x 1 + x 2 + x 3 ) valid for x 1 + x 2 + x 3 3/2 f (x) 4(x 1 + x 2 + x 3 ) 5 valid for x 1 + x 2 + x 3 3/2 Disjunction: either { (x, f ) (x, f ) : x [0, 1] 3, j x j 3/2, f 4 5 (x 1 + x 2 + x 3 ) or { (x, f ) (x, f ) : x [0, 1] 3, } j x j 3/2, f 4(x 1 + x 2 + x 3 ) 5. },

93 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. f (x) 4 5 (x 1 + x 2 + x 3 ) valid for x 1 + x 2 + x 3 3/2 f (x) 4(x 1 + x 2 + x 3 ) 5 valid for x 1 + x 2 + x 3 3/2 Disjunction: either { (x, f ) (x, f ) : x [0, 1] 3, j x j 3/2, f 4 5 (x 1 + x 2 + x 3 ) or { (x, f ) (x, f ) : x [0, 1] 3, } j x j 3/2, f 4(x 1 + x 2 + x 3 ) 5. }, Proves: f (x) (x 1 + x 2 + x 3 ) valid throughout,

94 f (x). = 2(x 1 x 2 + x 1 x 3 + x 2 x 3 ) over [0, 1] 3. f (x) 4 5 (x 1 + x 2 + x 3 ) valid for x 1 + x 2 + x 3 3/2 f (x) 4(x 1 + x 2 + x 3 ) 5 valid for x 1 + x 2 + x 3 3/2 Disjunction: either { (x, f ) (x, f ) : x [0, 1] 3, j x j 3/2, f 4 5 (x 1 + x 2 + x 3 ) or { (x, f ) (x, f ) : x [0, 1] 3, } j x j 3/2, f 4(x 1 + x 2 + x 3 ) 5. }, Proves: f (x) (x 1 + x 2 + x 3 ) valid throughout, at (.5,.5.,.5) T get

95 Ongoing work Complement of union of two polyhedra Complement of union of two ellipsoids Complement of union of ellipsoid + half-plane

Optimizing Convex Functions over Non-Convex Domains

Optimizing Convex Functions over Non-Convex Domains Daniel Bienstock and Alexander Michalka University Berlin 2012 Introduction Generic Problem: min Q(x), s.t. x F, Introduction Generic Problem: min Q(x),