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Chap 2' The Geometry of LP

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(1- ) i 0 and sum up to 1 convex comb. of xi y (1- )z S. Linear Programming 2009 ... Def: For standard form problems, we say that two bases are adjacent if they ... – PowerPoint PPT presentation

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Title: Chap 2' The Geometry of LP


1
Chap 2. The Geometry of LP
  • In the text, polyhedron is defined as P x ?
    Rn Ax ? b . So some of our earlier results
    should be taken with modifications.
  • Thm 2.1
  • (a) The intersections of convex sets is convex.
  • (b) Every polyhedron is a convex set.
  • (c) Convex combination of a finite number of
    elements of a convex set also belongs to that
    set.
  • (recall that S closed for convex combination of
    2 points.
  • ? S closed for convex combination of a finite
    number of pts)
  • (d) Convex hull of a finite number of vectors
    (polytope) is convex.

2
  • Pf) (a) Let x, y ? ?i?I Si , Si convex ?
    x, y ? Si , ? i
  • ? ?x (1-?)y ? Si ? i since Si convex
  • ? ?x (1-?)y ? ?i?I Si , ? ? Si convex.
  • (b) Halfspace x ax ? b is convex.
  • P ? halfspaces ? From (a), P is convex
  • ( or we may directly show A( ?x (1-?)y ) ? b.
    )
  • (c) Use induction. True for k 2 by
    definition.
  • Suppose statement holds for k. Suppose ?k1 ?
    1.
  • Then ?i1k1 ?ixi ?k1xk1 (1-?k1) (
    ?i1k ( ?i / (1-?k1)) xi )
  • ?i / (1-?k1) ? 0 and sum up to 1, hence ?i 1k
    ( ?i / (1-?k1)) xi ? S. ? ?i 1k1 ?ixi ? S.
  • (d) Let S be the convex hull of vectors x1, … ,
    xk and y, z ? S
  • ? y ?i1k ?ixi , z ?i1k ?ixi for some
    ?i , ?i
  • ?y (1-?)z ? ? ?ixi (1-?) ? ?ixi ?i1k
    (??i (1-?)?i) xi
  • ??i (1-?)?i ? 0 and sum up to 1 ? convex
    comb. of xi
  • ? ?y (1-?)z ? S. ?

3
Extreme points, vertices, and b.f.ss
  • Def (a) Extreme point ( as we defined earlier)
  • (b) x ? P is a vertex if ? c ? Rn such that
    cx lt cy ? y ? P and y ? x. ( x is unique
    optimal solution of min cx, x ? P )
  • (c) Consider polyhedron P and x ? Rn. Then x
    is a basic solution if all equality constraints
    are active at x and ? n linearly independent
    active constraints among the constraints active
    at x.
  • ( basic feasible solution if x is basic
    solution and x ? P )
  • Note Earlier, we defined the extreme point same
    as in the text.
  • Vertex as 0-dimensional face ( dim (P) rank
    (A, b) n ) which is the same as basic
    feasible solution in the text.
  • We defined basic solution (and b.f.s) only for
    the standard LP. ( xB B-1b, xN 0 )
  • Definition (b) is new. It gives an equivalent
    characterization of extreme point. (b) can be
    extended to characterize a face F of P.

4
x3
x2
A
C
P
E
x1
D
B
  • Fig. 2.6 P (x1, x2, x3) x1x2x3 1, x1,
    x2, x3 ? 0
  • Three constraints active at A, B, C, D. Only
    two constraints active at E. Note that D is not
    a basic solution since it does not satisfy the
    equality constraint. However, if P is denoted as
    P (x1, x2, x3) x1x2x3 ? 1, x1x2x3 ? 1,
    x1, x2, x3 ? 0, D is a basic solution by the
    definition in the text, i.e. whether a solution
    is basic depends on the representation of P.

5
A
E
D
P
F
B
C
  • Fig. 2.7 A, B, C, D, E, F are all basic
    solutions. C, D, E, F are basic feasible
    solutions.

6
  • Comparison of definitions in the notes and the
    text

7
  • Thm 2.3 x ? P, then x vertex, extreme point,
    and b.f.s. are equivalent statements.
  • Pf) We follow the definitions given in the
    text. We already showed in the notes that
    extreme point and 0-dimensional face ( AIx bI,
    AI rank n, b.f.s. in the text) are equivalent.
  • To show all are equivalent, take the following
    steps
  • x vertex (1) ? x extreme point (2) ? x
    b.f.s. (3) ? x vertex
  • (1) x vertex ? x extreme point
  • Suppose x is vertex, i.e. ? c ? Rn such that
    x is unique min of min cx, x ? P.
  • If y, z ? P, y, z ? x, then cx lt cy and cx
    lt cz.
  • Hence cx lt c( ?y (1-?)z ), 0 ? ? ? 1 ?
    ?y (1-?)z ? x
  • Hence x cannot be expressed as convex
    combination of two other points in P
  • ? x extreme point.

8
  • (continued)
  • (2) x extreme point ? x b.f.s.
  • Suppose x is not a b.f.s.. Let I i aix
    bi
  • Since x is not a b.f.s., the number of linearly
    independent vectors ai in I lt n.
  • Hence ? nonzero d ? Rn such that aid 0, ? i ?
    I.
  • Consider y x ?d, z x - ?d. But, y, z ?
    P for sufficiently small positive ?, and x
    (yz)/2, which implies x is not an extreme
    point.
  • (3) x b.f.s. ? x vertex
  • Let x be a b.f.s. and let I i aix bi
  • Let c ?i?I ai . Then cx ?i?I aix ?i?I
    bi
  • ? x ? P, we have cx ?i?I aix ? ?i?I bi
    cx, hence x optimal.
  • For uniqueness, equality holds ? aix bi , i
    ? I.
  • Since x is b.f.s., it is the unique solution of
    aix bi , i ? I
  • Hence x is a vertex. ?

9
  • Note Whether x is a basic solution depends on
    the representation of P. However, x is b.f.s.
    iff x extreme point and x being extreme point
    is independent of the representation of P. Hence
    the property of being a b.f.s. is also
    independent of the representation of P.
  • Cor 2.1 For polyhedron P ? ?, there can be
    finite number of basic or basic feasible
    solutions.
  • Def Two distinct basic solutions are said to be
    adjacent if we can find n-1 linearly independent
    constraints that are active at both of them. (
    In Fig 2.7, D and E are adjacent to B A and C
    are adjacent to D.)
  • If two adjacent basic solutions are also
    feasible, then the line segment that joins them
    is called an edge of the feasible set ( one
    dimensional face).

10
2.3 Polyhedra in standard form
  • Thm 2.4 P x Ax b, x ? 0 , A m ? n,
    full row rank.
  • Then x is a basic solution ? x satisfies Ax
    b and ? indices B(1), … , B(m) such that AB(1),
    … , AB(m) are linearly independent and xi 0, i
    ? B(1), … , B(m).
  • Pf) see text.
  • ( To find a basic solution, choose m linearly
    independent columns AB(1), … , AB(m). Set xi 0
    for all i ? B(1), … , B(m), then solve Ax b
    for xB(1), … , xB(m). )
  • Def basic variable, nonbasic variable, basis,
    basic columns, basis matrix B. (see text)
  • ( BxB b ? xB B-1b )

11
  • Def For standard form problems, we say that two
    bases are adjacent if they share all but one
    basic column.
  • Note A basis uniquely determines a basic
    solution.
  • Hence if have two different basic solutions ?
    have different basis.
  • But two different bases may correspond to the
    same basic solution. (e.g. when b 0 )
  • Similarly, two adjacent basic solutions ? two
    adjacent bases
  • Two adjacent bases with different basic
    solutions ? two adjacent basic solutions.
  • However, two adjacent bases only not necessarily
    imply two adjacent basic solutions. The two
    solutions may be the same solution.

12
  • Check that full row rank assumption on A results
    in no loss of generality.
  • Thm 2.5 P x Ax b, x ? 0 ? ?, A m ?
    n, rank is k lt m.
  • Q x AIx bI, x ? 0 , I i1, … , ik
    with linearly indep. rows.
  • Then P Q.
  • Pf) Suppose first k rows of A are linearly
    independent.
  • P ? Q is clear. Show Q ? P.
  • Every row ai of A can be expressed as ai
    ?j1k ?ijaj.
  • Hence, for x ? P, bi aix ?j1k ?ij ajx
    ?j1k ?ij bj , i 1, … , m
  • i.e. bi is also linear combination of bj , j ?
    I.
  • Suppose y ? Q, then ? i 1, … , m,
  • aiy ?j1k ?ij ajy ?j1k ?ij bj bi
  • Hence, y ? P ? Q ? P ?

13
2.4 Degeneracy
  • Def 2.10 A basic solution x ? Rn is said to be
    degenerate if more than n of the constraints are
    active at x.
  • Def 2.11 P x ? Rn Ax b, x ? 0 , A m ?
    n, full row rank.
  • Then x is a degenerate basic solution if more
    than n m of the components of x are 0 ( i.e.
    some basic variables have 0 value)
  • For standard LP, if we have more than n m
    variables at 0 for a basic feasible solution x,
    it means that more than n m of the
    nonnegativity constraints are active at x in
    addition to the m constraints in Ax b.
  • The solution can be identified by defining n-m
    nonbasic variables ( value 0). Hence,
    depending on the choice of nonbasic variables, we
    have different bases, but the solution is the
    same.

14
A
D
C
P
B
E
  • Fig 2.9 A and C are degenerate basic feasible
    solutions. B and E are nondegenerate. D is a
    degenerate basic solution.

15
A
x40
x30
B
x50
P
x10
x60
x20
  • Fig 2.11 (n-m)-dimensional illustration of
    degeneracy. Here, n6, m4. A is nondegenerate
    and basic variables are x1, x2, x3, x6. B is
    degenerate. We can choose x1, x6 as the nonbasic
    variables. Other possibilities are to choose x1,
    x5, or to choose x5, x6.

16
  • Degeneracy is not purely geometric property, it
    may depend on representation of the polyhedrom
  • ex) P x Ax b, x ? 0 , A m ? n
  • P x Ax ? b, -Ax ? -b, x ? 0
  • We know that P P, but representation is
    different.
  • Suppose x is a nondegenerate basic feasible
    solution of P.
  • Then exactly n m of the variables xi are
    equal to 0.
  • For P, at the basic feasible solution x, we
    have n m variables set to 0 and additional 2m
    constraints are satisfied with equality. Hence,
    we have n m active constraints and x is
    degenerate.

17
2.5 Existence of extreme points
  • Def 2.12 Polyhedron P ? Rn contains a line if
    ? a vector x ? P and a nonzero d ? Rn such that x
    ?d ? P for all ? ? R.
  • Note that if d is a line in P, then A(x ?d) ?
    b for all ? ? R
  • ? Ad 0
  • Hence d is a vector in the lineality space S. (
    in P SKQ )
  • Thm 2.6 P x ? Rn aix ? bi, i 1, … , m
    ? ?, then the following are equivalent.
  • (a) P has at least one extreme point.
  • (b) P does not contain a line.
  • (c) ? n vectors out of a1, … , am, which are
    linearly independent.
  • Pf) see proof in the text.

18
  • Note that the conditions given in Thm 2.6 means
    that the lineality space S 0
  • Cor 2.2 Every nonempty bounded polyhedron
    (polytope) and every nonempty polyhedron in
    standard form has at least one basic feasible
    solution (extreme point).

19
2.6 Optimality of extreme points
  • Thm 2.7 Consider the LP of minimizing cx over
    a polyhedron P. Suppose P has at least one
    extreme point and there exists an optimal
    solution.
  • Then there exists an optimal solution which is
    an extreme point of P.
  • Pf) see text.
  • Thm 2.8 Consider the LP of minimizing cx over
    a polyhedron P. Suppose P has at least one
    extreme point.
  • Then, either the optimal cost is - ?, or there
    exists an extreme point which is optimal.

20
  • (continued)
  • Idea of proof in the text)
  • Consider any x ? P. Let I i aix bi
  • Then we move to y x ?d, where aid 0, i ?
    I and cd ? 0.
  • Then either the optimal cost is - ? ( if the
    half line d is in P and cd lt 0 ) or we meet a
    new inequality which becomes active ( cost does
    not increase).
  • By repeating the process, we eventually arrive
    at an extreme point which has value not inferior
    to x.
  • Therefore, for any x in P, there exists an
    extreme point y such that cy ? cx. Then we
    choose the extreme point which gives the smallest
    objective value with respect to c.

21
  • ( alternative proof of Thm 2.8)
  • P S K Q. Pointedness of P implies S
    0.
  • Hence x ? P ? x ?i qidi ?j rjuj , where
    dis are extreme rays of K and ujs are extreme
    points of P and qi ? 0, rj ? 0, ?j rj 1.
  • Suppose ? i such that cdi lt 0, then LP is
    unbounded.
  • ( For x?P, x ?di ? P for ? ? 0. Then c(
    x?di ) ? - ? as ? ? ?)
  • Otherwise, cdi ? 0 for all i , take u such
    that cu minj cuj
  • Then ? x ? P,
  • cx ?i qi(cdi) ?j rj(cuj) ? ?j rj(cuj) ?
    (cu) ?j rj cu.
  • Hence LP is either unbounded or ? an extreme
    point of P which is an optimal solution.
  • Proof here shows that the existence of an
    extreme ray di of the pointed recession cone Ax ?
    0 ( if have min problem and polyhedron is Ax ?
    b) such that cdi lt 0 is the necessary and
    sufficient condition for unboundedness of the LP.
  • ( If P has at least one extreme point, then LP
    unbounded
  • ? ? an extreme ray di in recession cone K such
    that cdi lt 0) ?
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