Math Models of OR: Extreme Points and Basic Feasible...

Math Models of OR:

Extreme Points and Basic Feasible Solutions

John E. Mitchell

Department of Mathematical Sciences

RPI, Troy, NY 12180 USA

September 2018

Mitchell Extreme Points and Basic Feasible Solutions 1 / 26

Introduction

Outline

1 Introduction

2 Let x̄ be a basic feasible solution, show it is an extreme point

3 Let x̄ be an extreme point, show it is a basic feasible solutionExample

The general case


Introduction

Extreme points

Recall that if a standard form linear optimization problem

minx2IRn cT xsubject to Ax = b

x � 0

has an optimal solution then it has an optimal solution that is an

extreme point.

Definition

Let P be the feasible set of a linear program. A point x̄ 2 P is anextreme point of P if it is not on the line segment joining two otherpoints in P.


Q:extreme

Introduction

Basic feasible solutions

The simplex algorithm works with canonical form tableaus and moves

from basic feasible solution to adjacent basic feasible solution.

The basic feasible solution corresponding to a canonical form

tableau is obtained by setting the nonbasic variables equal to zero and

then finding the unique solution for the basic variables.


x , x - x3 4 4 X i " " Basic sequencet¥¥¥¥¥¥ ' s i n '115=4,4=2,x ,s 9

9X ,e x , 2×4=+620

Introduction

The relationship

We’ve seen graphically that basic feasible solutions correspond to

extreme points. We prove the following theorem to make this formal:

Theorem

Let x̄ be a feasible solution to a standard form linear optimizationproblem. The point x̄ is a basic feasible solution if and only if it is anextreme point of the feasible region of the linear optimization problem.

The theorem is an “if and only if” theorem, so we need to show two

directions.


Introduction

For example

x1

x2

0

1

2

2 4 6

x1 +

3x2 =

6, x3 =

0x1 = 3,x4 = 0

x2 = 2, x5 = 0

optimal contour

�x1 � 2x2 = �5

minx2IR2 �x1 � 2x2

subject to x1 + 3x2 6

x1 3

x2 2

x1, x2 � 0


t .EE:

Let x̄ be a basic feasible solution, show it is an extreme point

Outline

1 Introduction



The general case



Assume x̄ is not an extreme point and derive a

contradiction

Since x̄ is assumed to not be extreme, it must be the midpoint between

two other distinct feasible points x (1) and x (2), so x̄ = 1

2x (1) + 1

2x (2).

This means that for each component j = 1, . . . , n, we have

x̄j =1

2x (1)

j +1

2x (2)

j .

x (1) = (1, 4)

x (2) = (11, 2)

line segment

x̄ = (6, 3)


S h o u t i B F SAssume also I i s a B f f , g e t a contradiction


Nonbasic components of x̄We look in particular at nonbasic components of x̄ , so x̄j = 0.

For these components, we have

x̄j = 1

2x (1)

j + 1

2x (2)

j

= 0 for � 0 since � 0 since

nonbasic x (1) feasible x (2) feasible

components

It follows that we must have x (1)j = x (2)

j = 0 for all the nonbasic

components of x̄ .

But once the nonbasic components are fixed, the values of the basic

variables are then uniquely determined: they are equal to the

corresponding entries in b.

So we must have x (1) = x (2) = x̄ . So the points x (1) and x (2) are notdistinct, so x̄ is actually an extreme point.


I f i n o t extreme, couldhave

I = tax'" t # x " , for example.( i )

,X• ¥ N/ o ×

redefine

×")← {× '"+1×4)

Now I = ± (new +"1/+11×14)

E j =#f" + taxi"(8)i t (4)+ 'il't)

F¥tx÷÷¥l¥÷÷dSince x I", xs'"7 0 ,

m u s t have ×g'"ex ,"= O

Let x̄ be an extreme point, show it is a basic feasible solution

Outline

1 Introduction



The general case


Let x̄ be an extreme point, show it is a basic feasible solution

Introduction to proof

This direction is harder to prove, and the proof needs some linear

algebra.

We prove it by contradiction, so we assume x̄ is not a BFS and showit is not an extreme point.

To make the proof a little simpler, we assume x̄ has exactly m positivecomponents; the proof can be extended to handle the general case.


h o w tochartacterire?

A i s m e n

Let x̄ be an extreme point, show it is a basic feasible solution Example

Outline

1 Introduction



The general case



An example

We first work an example before abstracting the proof.

minx2IR6 3x1 + 2x2 + x3

subject to x1 + x2 + x3 + x4 = 6

2x1 + x3 + x5 = 4

�x1 + x2 + x6 = 2

x1, . . . , x6 � 0

The point x̄ = (1, 3, 2, 0, 0, 0) is feasible. Is it a BFS? The tableau is

x1 x2 x3 x4 x5 x6

0 3 2 1 0 0 0

6 1 1 1 1 0 0

4 2 0 1 0 1 0

2 �1 1� 0 0 0 1



Two pivots

x1 x2 x3 x4 x5 x6

0 3 2 1 0 0 0

6 1 1 1 1 0 0

4 2 0 1 0 1 0

2 �1 1� 0 0 0 1

R0 � 2R3,R1 � R3

�!

x1 x2 x3 x4 x5 x6

�4 5 0 1 0 0 �2

4 2 0 1� 1 0 �1

4 2 0 1 0 1 0

2 �1 1 0 0 0 1

R0 � R1,R2 � R1

�!

x1 x2 x3 x4 x5 x6

�6 3 0 0 �1 0 �3

4 2 0 1 1 0 �1

0 0 0 0 �1 1 �1

2 �1 1 0 0 0 1Mitchell Extreme Points and Basic Feasible Solutions 14 / 26

or:±÷i÷÷÷:c u l1 = - cu l 2

+ 2 (co13)

&


We don’t have a BFS

x1 x2 x3 x4 x5 x6

�6 3 0 0 �1 0 �3

4 2 0 1 1 0 �1

0 0 0 0 �1 1 �1

2 �1 1 0 0 0 1

The second constraint does not involve x1, x2, or x3.

So our given feasible solution is not a bfs.



A little linear algebra

Equivalently, we can say that this means the original first threecolumns of A are linearly dependent.

In fact, notice that

2

41 1 1

2 0 1

�1 1 0

3

5

2

41

1

�2

3

5 =

2

40

0

0

3

5 .

Because of this equality, we have a direction we can move in and still

maintain feasibility.


F td , I - d both feasible

I = ÷ ( I td)+ t h - d )

7- D =Eli'?'""}= (1,312,...) ✓i e - e d@D = (1,1,-2,490)feqifl.ee#-xee%Ad=0


The direction

Let d = (1, 1,�2, 0, 0, 0)T 2 IR6. Then

x̂ := x̄ + td

satisfies Ax̂ = b for any t , positive or negative, since Ax̄ = b and

Ad = 0.

Further, provided �1 t 1, we get x̂ � 0.



Two new feasible points

Want to show x̄ is not extreme.

Let’s take the two new feasible points x (1) and x (2) with t = 1 and

t = �1, respectively (the x4, x5, and x6 components of d , x (1), x (2) and

x̄ are all zero, so we don’t write them out explicitly):

x (1) = x̄ + d =

2

41

3

2

3

5 +

2

41

1

�2

3

5 =

2

42

4

0

3

5

x (2) = x̄ � d =

2

41

3

2

3

5 �

2

41

1

�2

3

5 =

2

40

2

4

3

5



x̄ is not extreme

Notice that

1

2x (1) +

1

2x (2) =

1

2

2

42

4

0

3

5 +1

2

2

40

2

4

3

5 =

2

41

3

2

3

5 = x̄ .

So x̄ is not an extreme point.



Graph this example

x1

x2

x3

0

4

2 62

6

2

6 x1 + x2 + x3 = 6, x4 = 0



Graph this example

x1

x2

x3

0

4

2 62

6

2

6 x1 + x2 + x3 = 6, x4 = 0

�x1 + x2 = 2, x6 = 0



Graph this example

x1

x2

x3

0

4

2 62

6

2

6 x1 + x2 + x3 = 6, x4 = 0

�x1 + x2 = 2, x6 = 0

2x1 + x3 = 4, x5 = 0



Graph this example

x1

x2

x3

0

4

2 62

6

2

6 x1 + x2 + x3 = 6, x4 = 0

�x1 + x2 = 2, x6 = 0

2x1 + x3 = 4, x5 = 0

portion offeasible region with

x4 = x5 = x6 = 0

(2, 4, 0) = x (1)

(0, 2, 4) = x (2)

(1, 3, 2) = x̄



Linear algebra characterization of BFS

The example illustrates the following result:

Let x̄ be a feasible solution to a standard form linear program with

m ⇥ n constraint matrix A with rank equal to m.

The point x̄ is a basic feasible solution if and only if the columns of

A corresponding to the positive components of x̄ are linearly

independent.


Let x̄ be an extreme point, show it is a basic feasible solution The general case

Outline

1 Introduction



The general case



Return to the general case

Recall, we assume x̄ is not a basic feasible solution and try to derive a

contradiction.

Without loss of generality, we can assume the m positive components

of x̄ are the first m components, by renumbering the components if

necessary. So we have

x̄j

⇢> 0 for j = 1, . . . ,m= 0 for j = m + 1, . . . , n


- - -


Row reducing A

We can write the constraint matrix A as

A =

2

64 B|{z}m columns

N|{z}(n�m) columns

3

75

If x̄ was a basic feasible solution, we would be able to use

elementary row operations to turn this into a canonical form, so

A = [B N] �!hI N̂

i

for some m ⇥ (n � m) matrix N̂, where I is the m ⇥ m identity matrix.


positive component,

✓o l ' x

✓' {Imponenti

a f I


But we don’t have a BFS

Since x̄ is not a basic feasible solution, this row reduction cannot be

possible. That means the columns of B are linearly dependent, which

is equivalent to stating that

there exists a vector dB 2 IRm with BdB = 0,where dB has at least one nonzero component.

We can extend dB out to a vector d 2 IRn by appending zeroes:

let dj =

⇢dB

j for j = 1, . . . ,m0 for j = m + 1, . . . , n

so d =

dB

0

�m components

(n � m) components



Moving in our direction

Notice that

Ad = [B N]

dB

0

�= BdB = 0,

so for any scalar t we have

A(x̄ + td) = Ax̄ + tAd = b + 0 = b.

Further, since dj = 0 if x̄j = 0, we have x̄ + td � 0 for t sufficiently

close to zero (positive or negative).

Thus, choose t̄ > 0 so that both x̄ + t̄d and x̄ � t̄d are feasible.

Then notice that x̄ is the midpoint of these two feasible points:

x̄ =1

2

�x̄ + t̄d

�+

1

2

�x̄ � t̄d

�.

Thus, x̄ is not an extreme point. Thus, we’ve proved the

contrapositive, so we do indeed have the result that if x̄ is an extreme

point then it is a basic feasible solution.


Math Models of OR: Extreme Points and Basic Feasible...

Documents

Transcript of Math Models of OR: Extreme Points and Basic Feasible...