Wolfe’s Method - UC Davis Mathematics jhaddock/Slides/Wolfe/UC... Wolfe’s Method Jamie...

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Transcript of Wolfe’s Method - UC Davis Mathematics jhaddock/Slides/Wolfe/UC... Wolfe’s Method Jamie...

  • Wolfe’s Method

    Jamie Haddock

    April 6, 2017

    Graduate Group in Applied Mathematics

    UC Davis

  • Minimum Norm Point in Polytope

    We are interested in solving the problem (MNP(P)):

    min x∈P ||x ||2

    where P is a polytope.

    Reminder: A polytope, P, is the convex hull of points p1, p2, ..., pm,

    P =

    { m∑ i=1

    λipi : m∑ i=1

    λi = 1, λi ≥ 0 for all i = 1, 2, ...,m } .

    1

  • Minimum Norm Point in Polytope

    We are interested in solving the problem (MNP(P)):

    min x∈P ||x ||2

    where P is a polytope.

    Reminder: A polytope, P, is the convex hull of points p1, p2, ..., pm,

    P =

    { m∑ i=1

    λipi : m∑ i=1

    λi = 1, λi ≥ 0 for all i = 1, 2, ...,m } .

    1

  • Minimum Norm Point in Polytope

    p1 p2

    p3

    p4

    p5

    P

    O

    2

  • Minimum Norm Point in Polytope

    p1 p2

    p3

    p4

    p5

    P

    O

    2

  • Formulation as QP

    This is a quadratic programming problem (QP).

    Given p1, p2, ..., pm ∈ Rn, define P = [p1p2...pm] ∈ Rn×m. Then our problem is which can be solved using interior-point methods.

    3

  • Formulation as QP

    This is a quadratic programming problem (QP).

    Given p1, p2, ..., pm ∈ Rn, define P = [p1p2...pm] ∈ Rn×m.

    Then our problem is which can be solved using interior-point methods.

    3

  • Formulation as QP

    This is a quadratic programming problem (QP).

    Given p1, p2, ..., pm ∈ Rn, define P = [p1p2...pm] ∈ Rn×m. Then our problem is

    min ||Py ||22 s.t. 1T y = 1

    y ≥ 0

    which can be solved using interior-point methods.

    3

  • Optimality Condition

    Theorem (Wolfe)

    Let P = {p1, p2, ..., pm}. Then x ∈ conv(P) is MNP(P) if and only if

    xTpj ≥ ||x ||22 for all j = 1, 2, ...,m.

    4

  • Optimality Condition

    Theorem (Wolfe)

    Let P = {p1, p2, ..., pm}. Then x ∈ conv(P) is MNP(P) if and only if

    xTpj ≥ ||x ||22 for all j = 1, 2, ...,m.

    p1 p2

    p3

    p4

    p5

    P

    O

    4

  • Optimality Condition

    Theorem (Wolfe)

    Let P = {p1, p2, ..., pm}. Then x ∈ conv(P) is MNP(P) if and only if

    xTpj ≥ ||x ||22 for all j = 1, 2, ...,m.

    p1 p2

    p3

    p4

    p5

    P

    O

    4

  • Optimality Condition

    Theorem (Wolfe)

    Let P = {p1, p2, ..., pm}. Then x ∈ conv(P) is MNP(P) if and only if

    xTpj ≥ ||x ||22 for all j = 1, 2, ...,m.

    p1 p2

    p3

    p4

    p5

    P

    O

    {x : pT4 x = ||p4||22}

    4

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P).

    5

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P).

    5

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P).

    5

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P).

    5

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P).

    5

  • Intuition and Definitions

    Idea: Exploit linear information about the problem in order to progress

    towards the nonlinear (quadratic) solution.

    Def: An affinely independent set of points Q = {q1, q2, ..., qk} is a corral if MNP(conv(Q)) ∈ relint(conv(Q)).

    q1 q2

    O

    q1 q2

    q3

    O

    Note: Singletons are corrals.

    Note: There is a corral of points in P whose convex hull contains

    MNP(P). 5

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C ∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1

    p3

    p2

    O

    P

    6

  • Sketch of Method

    x ∈ P = {p1,p2, ...,pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C ∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P

    6

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C ∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P

    C = {p1}

    6

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C ∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P

    C = {p1}

    6

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x||22} C = C ∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P

    C = {p1}

    6

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C∪ {pj} y = MNP(aff(C ))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P

    C = {p1,p2}

    6

  • Sketch of Method

    x ∈ P = {p1, p2, ..., pm} C = {x} while x is not MNP(P)

    pj ∈ {p ∈ P : xTp < ||x ||22} C = C ∪ {pj} y = MNP(aff(C))

    while y 6∈ conv(C ) z = argmin

    z∈conv(C)∩xy ||z − y ||2

    C = C − {pi} where pi , z are on different faces of

    conv(C )

    x = z

    y = MNP(aff(C ))

    x = y

    return x

    p1 = (0, 2)

    p2 = (3, 0)

    p3 = (−2, 1)

    p1 = x

    p3

    p2

    O

    P y

    C = {p1, p2}

    6

  • Sketch of Method