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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.

- 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. ?

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.

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.

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.

- Comparison of definitions in the notes and the

text

- 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.

- (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. ?

- 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).

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 )

- 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.

- 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 ?

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.

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.

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.

- 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.

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.

- 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).

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.

- (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.

- ( 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) ?