Convex Cones

1
(Convex) Cones

• Def closed under nonnegative linear
  combinations, i.e. K is a cone provided
• a1, , ap ? K ? Rn, ?1, , ?p ? 0 ? ?i 1p
  ?i ai ? K
• ( Note Usually cones are defined as closed only
  under nonnegative scalar multiplication)
• Observations (Characteristics)
• Subspaces are cones
• For any family of cones Ki i ? I , ? i ? I
  Ki is a cone also.
• Any nonempty cone contains 0
• K1, K2 are cones, then so is K1K2 (xy) x ?
  K1, y ? K2
• Halfspaces are cones H x ? Rn ax ? 0

2

• Description of cones
• Any subset A ? Rn generates a cone K(A) (define
  K(?) 0 )
• K(A) ?1a1 ?pap p ? 1, ?i ? R , ai
  ? A,
• called conical span of A
• conical hull of A ? Ki ? A, Ki is cone Ki
  (outside description)
• They are the same.
• Finite basis result is false for cones (e.g. ice
  cream cones).
• Hence we will restrict our attention to the
  cones with finite conical basis.

3

• Conical dual
• For any A ? Rn, define the (conical) dual of A
  to be
• A x ? Rn Ax ? 0 , where Ax ? 0 means ax
  ? 0 for all a ? A.
• ( some people use Ax ? 0 )
• It is called a constrained cone since it is the
  solution set of some homogeneous inequalities.
• When A is a cone, A is defined as dual cone (or
  polar cone) of A.
• Note that A is always a cone a constrained
  cone.
• For A m ? n, A (with rows of A regarded as the
  vectors in the set A) is finitely constrained
  (polyhedron).

4
(conical) dual of A?Rn
Aa1, a2, a3
a1
a2
a3
5

• Prop Suppose A, B ? Rn. Then
• (1) B ? A ? A ? B
• (2) A ? A
• (3) A A
• (4) A A ? A is a constrained cone
• (5) If B ? A and B generates A conically, then
  A B.
• Pf) parallels the cases for subspaces.

6
A ? A
a1
a2
a3
Aa1, a2, a3
7
A A
a1
a2
a3
Aa1, a2, a3
8
A A ? A is a constrained cone
A constrained cone
9

• Thm (Weyl) Any nonempty finitely generated cone
  is polyhedral (finitely constrained).
• Pf) Use Fourier-Motzkin elimination (later). ?
• Cor 1 Among all subsets A ? Rn with finite
  conical basis
• A A ? A is a nonempty cone.
• Cor 2 Given A m?n, consider K yA y ? 0,
  L x Ax ? 0.
• Then K L, L K.

10
K L, L K
A
a1
a2
a3
Aa1, a2, a3
11

• Cor 3 (Farkas lemma)
• Given A m?n, c ? Rn, exactly one of the two
  holds
• (I) there exists y ? Rm s.t. yA c
• (II) there exists x ? Rm s.t. Ax ? 0, cx
  gt 0.
• Pf) Show (I) ? (II)
• (I) ? c ? K ? yA y ? 0
• ? c ? K (by Cor 1)
• ? ? x ? K (Ax ? 0) s.t. cx gt 0
• ? (II) holds ?

12
Farkas Lemma
A m ? n, c?Rn Case (1) c ?K (? y?0
such that yAc)
K
a1
c
a2
a3
13
Farkas Lemma
Case (2) c ?K (? x such that Ax?0, cxgt0)
K
a1
c
a2
a3
c
14

• Farkas lemma is core of LP duality theory
  (details later). There are many other forms of
  theorems of the alternative and they are
  important and powerful tools in optimization
  theory.
• ex)
• to verify that c (cx c for some x) is an
  optimal value of a LP (in minimization form) is
  the same as to verify that the following system
  has no solution.
• -cx c gt 0
• Ax b ? 0
• Truth of claim can be verified by giving a
  solution to the alternative system.
• question similar result possible for integer
  form?
• Finding projection of a polyhedron to a lower
  dimensional space (later)
• absence of arbitrage condition in finance theory.
• ...

15
Absence of Arbitrage

• text Chapter 4, p167-169
• Text use the form
• (I) there exists some x ? 0 such that Ax b
• (II) there exists some vector p such that pA
  ? 0 and pb lt 0
• (here, columns of A are generators of a cone)
• Compare with
• (I) there exists y ? Rm s.t. yA c
• (II) there exists x ? Rm s.t. Ax ? 0, cx gt
  0.

16

• n different assets are traded in a market (single
  period)
• m possible states after the end of the period
• rsi return on investment of 1 dollar on asset
  i and the state is s at the end of the period
• payoff matrix R m?n

• xi amount held of asset i
• xi can be negative
• xi gt 0 has bought xi units of asset i, receive
  rsixi if state s occurs
• xi lt 0 short position, selling xi units of
  asset i at the beginning,
• with the promise to buy them back at the
  end.
• (sellers position in futures contract,
  payout rsixi, i.e. receiving a payoff
  of rsixi if state s occurs)

17

• Given a portfolio x, the resulting wealth when
  state s occurs is,
• ws ?i 1n rsixi ? w Rx
• Let pi be the price of asset i in the beginning,
  then cost of acquiring portfolio x is px. What
  are the fair prices for the assets?
• Absence of arbitrage condition asset prices
  should always be such that no investor can get a
  guaranteed nonnegative payoff out of a negative
  investment (no free lunch)
• Hence, if Rx ? 0, then we must have px ? 0,
  i.e. there exists no vector x such that xR ?
  0, xp lt 0.
• So, by Farkas lemma, there exists q ? 0 such
  that Rq p, i.e.
• pi ?s 1m qsrsi
• ( Here, R A, p b in Farkas lemma )

18
Fourier-Motzkin Elimination

• Solving system of inequalities (refer text
  section 2.8)
• Idea similar to Gaussian elimination.
• Eliminate one variable at a time with some
  mechanism reserved to recover the feasible values
  later.
• Given a system of inequalities and equalities

(I)
eliminate xn in (I) and obtain system (II)
which consists of linear equalities and
inequalities now in variables x1, x2, , xn-1 .
And we want to have (I) consistent ? (II)
consistent (have a solution), i.e. we want (x1,
x2, , xn ) satisfies (I) for some xn ? (x1, x2,
, xn-1 ) satisfies (II).
19

• Related concept projection of vectors to a lower
  dimensional space
• Def If x ( x1, x2, , xn) ? Rn and k ? n,
  the projection mapping ?k Rn ? Rk is defined
  as ?k(x) ?k(x1, x2, , xn) ( x1, , xk)
• For S ? Rn, ?k(S) ?k(x) x ? S
• Equivalently, ?k(S) (x1,, xk) ? xk1, , xn
  s.t. (x1, , xn)?S
• To determine whether a polyhedron P is nonempty,
  find ?n-1(P) ? ? ?1(P) and determine whether
  the one dimensional polyhedron is nonempty. (But
  it is inefficient)

20

• Elimination algorithm
• (0) If all coefficients of xn in (I) are 0,
  then take (II) same as (I).
• (1) ? some relation, say i-th, with ain ? 0,
  and this relation is . Then derive (II) from
  (I) by Gauss-Jordan elimination.
• ( ai1x1 ainxn bi ?
• xn 1/ain( bi - ai1x1 - - ainxn),
  substitute into (I).
• Clearly ( x1, , xn) solves (I) ? ( x1, ,
  xn-1) solves (II). )
• (continued)

21

• (continued)
• (2) Rewrite each constraint ?j1n aijxj ? bi
  as
• ainxn ? - ?j1n-1 aijxj bi, i 1, , m
• If ain?0, divide both sides by ain.
• By letting x (x1, , xn-1), we obtain
• xn ? di fix , if ain gt 0
• dj fjx ? xn , if ajn lt 0
• 0 ? dk fkx , if akn 0
• , where di, dj, dk ? R and fi, fj, fk ? Rn-1
• Let (II) be the system defined by
• dj fjx ? di fix , if ain gt 0 and ajn lt 0
• 0 ? dk fkx , if akn 0
• ( and remaining equations ) ?

22

• Ex) x1 x2 ? 1
• x1 x2 2x3 ? 2
• 2x1 3x3 ? 3
• x1 - 4x3 ? 4
• -2x1 x2 - x3 ? 5
• ? 0 ? 1 - x1 - x2
• x3 ? 1 (x1/2) (x2/2)
• x3 ? 1 (2x1/3)
• -1 (x1/4) ? x3
• -5 2x1 x2 ? x3
• ? 0 ? 1 - x1 - x2
• -1 (x1/4) ? 1 (x1/2) (x2/2)
• -1 (x1/4) ? 1 (2x1/3)
• -5 2x1 x2 ? 1 (x1/2) (x2/2)
• -5 2x1 x2 ? 1 (2x1/3)

23

• Thm 2.10 The polyhedron Q (defined by system
  (II)) constructed by the elimination algorithm is
  equal to ?n-1(P) of P.
• Pf) If x ? ?n-1(P), ? xn such that (x, xn) ? P.
  In particular, x (x, xn) satisfies system (I),
  hence also satisfies system (II). It shows that
  ?n-1(P) ? Q.
• Let x ? Q. Then x satisfies
• min j ajn lt 0 (dj fjx) ? max i ain gt
  0 (di fix).
• Let xn be a number between the two sides of the
  above inequality. Then (x, xn) ? P, which shows
  Q ? ?n-1(P). ?
• Observe that for x (x1, , xn), we have
  ?n-2(?n-1(x)) ?n-2(x).
• Also ?n-2(?n-1(P)) ?n-2(P). Hence obtain
  ?1(P) recursively to determine if P is empty or
  to find a solution.
• A solution in P can be recovered recursively
  starting from ?1(P) and finding xi that lies in
  min j ajn lt 0 (dj fjx), max i ain gt 0
  (di fix) .

24

• Cor 2.4 Let P ? Rnk be a polyhedron. Then,
  the set x ? Rn there exists y ? Rk such that
  (x, y) ? P is also a polyhedron.
• ( Will be used to prove the Weyls Theorem.
  Other proof technique is not apparent.)
• Cor 2.5 Let P ? Rn be a polyhedron and A be an
  m?n matrix. Then the set Q Ax x ? P is
  also a polyhedron.
• Pf) Q y ? Rm y Ax, x ? P. Hence Q is
  the projection of the polyhedron (x, y) ? Rnm
  y Ax, x ? P onto the y coordinates. ?
• Cor 2.6 The convex hull of a finite number of
  vectors (called polytope) is a polyhedron.
• Pf) The convex hull ?i1k ?ixi ?i ?i 1, ?i
  ? 0 is the image of the polyhedron ( ?1, ,
  ?k) ?i ?i 1, ?i ? 0 under the mapping that
  maps ( ?1, , ?k) to ?i ?ixi. (The mapping can
  be expressed as A?, where the columns of the
  matrix A are xi vectors. We will see a different
  proof later.) ?

25
Remarks

• FM elimination not efficient as an algorithm.
  Number of inequalities grows exponentially as we
  eliminate variables.
• Can also handle strict inequalities.
• Can solve LP problem max cx Ax ? b ?
  Consider Ax ? b, z cx and eliminate x and
  find z as large as possible in the one
  dimensional polyhedron. Solution can be
  recovered by backtracking.
• FM gives an algorithm to find the projection of P
  (x, y) ? Rnp Ax Gy ? b onto the x space
  Prx(P) x ? Rn (x, y) ? P for some y ? Rp.
• But how can we characterize Prx(P) for arbitrary
  P?

26

• Concept of projection becomes important in recent
  optimization theory (especially in integer
  programming) as new techniques using projections
  have been developed.
• (e.g. RLT (reformulation and linearization
  technique))
• Formulation in a higher dimensional space and use
  the projection to lower dimensional space may
  give stronger formulation in integer programming.
• (e.g. Node edge variable formulation stronger
  than edge formulation for weighted maximal
  b-clique problem)
• Given a complete undirected graph G (V, E),
  weight ce, e ?E. Find a clique (complete
  subgraph) of size (number of nodes in the clique)
  at most b and sum of edge weights in the clique
  is maximum.

27

• Weyls Theorem
• Any nonempty finitely generated cone is
  polyhedral (i.e. finitely constrained).
• Pf) K yA y ? 0 for A m ? n
• x x yA 0, y ? 0 is a consistent
  system in (x, y)
• Use FM elimination to get rid of ys
• ( x some linear homog. system in x is
  consistent
• Write these relation as Bx ? 0
• Then K x Bx ? 0 -- polyhedral ?
• Note that we get homogeneous system Bx ? 0 if we
  apply FM.

28

• Minkowskis Theorem
• Any polyhedral cone is nonempty and finitely
  generated.
• Pf) Let L be a polyhedral cone. Clearly L ? ?.
• We know that L L from earlier Prop.
• Part 2 of Cor. 2 says L is finitely generated (
  L K )
• By Weyls Thm, L itself is polyhedral.
• By part 2 of Cor. 2 (L) is finitely generated.
• ? L L from above ? L is finitely
  generated. ?
• FM elimination leads to Weyl-Minkowski cone
  representation.
• (nonempty finitely generated cone ? finitely
  constrained
• (polyhedral))
• Extend this result to affine version
• Def The set of all convex combinations of a
  finite point set is called a polytope.

29

• Affine Weyl Theorem (i.e. finitely generated
  are finitely constrained)
• Suppose P x ? Rn x yB zC, y ? 0, z
  ? 0, ?i zi 1 ,
• B p ? n, C q ? n.
• Then ? matrix A m ? n and b ? Rm s.t. P
  x ? Rn Ax ? b.
• (special case where B is vacuous is that
  polytope is polyhed.)
• Pf) (use technique called homogenization)
• If P ? (i.e. B, C vacuous, i.e. p q 0),
  take A 0,,0, b -1.
• If P ? ?, consider P ? Rn1 defined as

Observe that x?P ? (x, 1)?P
30

• and P is finitely generated nonempty cone in
  Rn1.
• Apply Weyls Thm to P in Rn1 to get
• P (x, xn1) A(x, xn1) ? 0 for some
  A m ? (n1)
• i.e. A A d with d last column of A
• Define b -d, then A A -b
• Observe that x ? P ? (x, 1) ? P ?
  A(x, 1) ? 0
• ? (A -b)(x, 1) ? 0
• ? Ax ? b ?
• Note that we changed the problem as the problem
  involving a cone in Rn1, for which we know more
  properties, and used the results for cones to
  prove the theorem.

31

• Affine Minkowski Theorem (i.e. finitely
  constrained are finitely generated)
• Suppose P x ? Rn Ax ? b, A m ? n, b ?
  Rm.
• Then ? matrices, B p ? n, C q ? n such that
• P x ? Rn x yB zC, y, z ? 0, ?i
  zi 1
• Pf) For P ?, take p q 0, i.e. B, C
  vacuous.
• Otherwise, again homogenize and consider

Then x ? P ? (x, 1) ? P P is a
polyhedral cone, so Minkowskis Thm
applies. Hence ? matrices, B l ? (n1) such
that P (x, xn1) ? Rn1 (x, xn1)
yB , y ? Rl
32

• (continued)
• Break B into 2 parts so that all rows of B
  with 0 last component come as top rows and rows
  with nonzero last component come as bottom rows.
  Note that all nonzero values in the last
  column of B must be gt 0.

It doesnt change P. Then we have x ? P ?
(x, 1) ? P
i.e. x ? P ? x yB zC, y ? 0, z ? 0, ?i
zi 1 ?
33

• Geometric view of homogenization in Affine
  Minkowski Thm

R
P?Rn1
1
0
Rn
Px Ax?b
34

• Think about similar picture for affine Weyl.
• Affine Weyl, Minkowski Thm together provides
  Double Description Thm
• We can describe polyhedron as
• (finite) intersection of halfspaces
• ?
• ?
• (finite) conical combination of points convex
  combination of points ( i.e. P C Q, where C
  is a cone and Q is a polytope).
• Existence of different representations has been
  shown. Next question is how to identify the
  representation.