1.3 Lecture_1b

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Lecture 1 b:Linear Programming Problem (LPP):

Geometry of Linear Programming Problems

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong Kong

[email protected]

MAT581 SSMathematics for Logistics

Produced by Jeff Chak-Fu WONG 1

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TABLE OF C ONTENTS

Geometry of Linear Programming Problems

1. Geometry of a Constraint

2. Geometry of the Objection Function

3. Geometry of the Set of Feasible Solutions

BLE OF C ONTENTS 2

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G EOMETRY OF LINEAR PROGRAMMING PROBLEMS

EOMETRY OF LINEAR P ROGRAMMING P ROBLEMS 3

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Consider the geometry of linear programming problems by looking

at the geometric interpretation of a single constraint,

at a set of constraints, and

at the objection function.

EOMETRY OF LINEAR P ROGRAMMING P ROBLEMS 4

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If this inequality is reversed , the set of points

x = [x 1 x 2 xn ] R n

that satisfya T x bi

is also called a closed half-space .

EOMETRY OF A C ONSTRAINT 6

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Example 1 Consider the constraint2x + 3 y 6

and the closed half-space

H =x

y 2 3

x

y 6 ,

which consists of the points satisfying the constraint.

Note that

The points (3, 0) and (1, 1) satisfy the inequality and therefore are inH .

The points (3, 4) and ( 1, 3) do not satisfy the inequality andtherefore are not in H .

Every point on the line 2x + 3 y = 6 satises the constraint and thus

lies in H .


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Compute the x intercept to be x = 3 and the intercept to be y = 2 .These points have been plotted and the line connecting them hasbeen drawn in Figure 1(a).

y

x 2 4 6

2

4

(a)

y

x 2 6

2

4

4H

(b)

Figure 1: Closed half-space in two dimensions


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Since the origin does not lie on the line 2x + 3 x = 6 , we use the originas the test point.

y

x 2 4 6

2

4

(a)

y

x 2 6

2

4

4H

(b)

The coordinates of the origin satisfy the inequality, so that H lies

below the line and contains the origin as show in Figure 1(b).


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Example 2 The constraint in three variables,

4x + 2 y + 5 z 20,

denes the closed half-space in R 3 , where

H =

x

y

z

4 2 5

x

y

z

20 .

Let us graph H in R 3 by graphing the plane

4x + 2 y + 5 z = 20

and checking a test point.


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(5,0,0)

(0,10,0)

(0,0,4)

4x + 2y = 20

2x + 5z = 20

4x + 5z = 20

Figure 2: Closed half-space in three dimensions

The origin does not lie on the plane and thus can be used as a testpoint.

It satises the inequality so that the closed half-space contains theorigin as shown in Figure 2.


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In more than three dimensions, it is impossible to sketch a closedhalf-space.

However, one can

think about the geometry of closed half-spaces in anydimension and

use the lower dimension examples as models for our

computations.


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Example 3 The equation

4x + 2 y + 5 z = 20

denes a hyperplane H in R3

.The graph of this hyperplane, which is really a plane in this case, is shownin Figure 2.

(5,0,0)

(0,10,0)

(0,0,4)

4x + 2y = 20

2x + 5z = 20

4x + 5z = 20


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The hyperplane H is the boundary of the closed half-space H 1 dened bythe inequality

4x + 2 y + 5 z 20,

considered in Example 2.

(5,0,0)

(0,10,0)

(0,0,4)

4x + 2y = 20

2x + 5z = 20

4x + 5z = 20

The half-space H 1 extends below the hyperplane H and lies behind thepage.


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Also see that H is the boundary of the closed half-space H 2 dened by

4x + 2 y + 5 z 20.

The half-space H 2 extends above the hyperplane and reaches out of thepage.

(5,0,0)

(0,10,0)

(0,0,4)

4x + 2y = 20

2x + 5z = 20

4x + 5z = 20


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The hyperplane H dened by Eq. (1) divides R n into the two closedhalf-spaces

H 1 = {x R n |a T x b}

and

H 2 = {x Rn

|aT

x b}.

Observe that H 1 H 2 = H , the original hyperplane.

In other words, a hyperplane is the intersection of two closed

half-spaces .


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Recall that

A feasible solution to a LPP is a point in R n that satises all theconstraints of the problem and the non-negativity restrictions.

It then follows that this set of feasible solutions is the intersection ofall the closed half-spaces determined by the constraints.

Specically, the set of solutions to an inequality ( or )constraints is a single closed half-space, whereas the set of

solutions to an equality constraint is the intersection of twoclosed half-spaces.


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Recall that






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Recall that






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Example 4 Sketch the set of all feasible solutions satisfying the set of inequalities

2x + 3 y 6

x + 2 y 4x 0

y 0.


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Solution

The set of solutions to the rst inequality, 2x + 3 y 6, is shown as theshaded region in Figure 3(a), and the set solutions to the secondinequality, x + 2 y 4, form the shaded region in Figure 3(b).

y

x 2 6

2

4

4

2x + 3y = 6

(a)

y

2

4

-4 x

-x + 2y = 4

(b)

Figure 3: Set of all feasible solutions (two dimensions)

In determining these regions, we have used the origin as a test

point.


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The regions satisfying the third and fourth constraints as shown inFigures 4(a) and (b), respectively.

y

x y = 0

(a)

y

x

x = 0

(b)



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The point (1, 1) was used as a test point to determine these regions.The intersection of the regions in Figures 3 and 4 is shown in Figure 5;it is the set of all feasible solutions to the given set of constraints.

y

-4x

-x + 2y = 4

2

3

2x + 3y = 6



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G EOMETRY OF THE O BJECTION FUNCTION

The objective function of any LPP can be written as

c T x .

If k is a constant, then the graph of the equation

cT

x = k

is a hyperplane.

Assume that we have a LPP that asks for a maximum value of the

objective function.In solving this problem, we are searching for points x in the set offeasible solutions for which the value of k as large as possible.

EOMETRY OF THE O BJECTION FUNCTION 26

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Geometrically we are looking for a hyperplane that intersects the

set of feasible solutions and for which k is a maximum.

The value of k measures the distance from the origin to thehyperplane.

One can think of starting with very large values of k and thendecreasing them until we nd a hyperplane that just touches the setof feasible solutions.


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Example 5 Consider the LPP

Maximise z = 4 x + 3 y

subject to

x + y 4

5x + 3 x 2 15

x, y 0


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The set of feasible solutions (the shaded region) and thehyperplanes

z = 9 , z = 12 , z = 27

2 , and z = 15

are shown in Figure 6.

z = 9

Figure 6: Objective function hyperplanes (two dimensions)


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z = 9 , z = 12 , z = 27

2 , and z = 15


z = 9z = 12



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z = 9 , z = 12 , z = 27

2 , and z = 15


z = 9z = 12

z = 27/2



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z = 9 , z = 12 , z = 27

2 , and z = 15


z = 9z = 12

z = 27/2

z = 15



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Note that it appears that the maximum value of the objectivefunction is 272 , which is obtained when x =

32 , y =

52 .

This conjecture will be veried in a later lecture.


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A linear programming problem may not have a solution if the set of

feasible solutions is unbounded.


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Example 6 Consider the LPP

Maximise z = 2 x + 5 y

subject to

3x + 2 y 6

x + 2 x 2 2

x, y 0


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The graph of the set of feasible solutions and the graphs ofhyperplanes are shown as the shaded region in Figure 10. E.g.,

z = 6 , z = 14 , and z = 20 .

z = 6z = 14

z = 20

x + 2y = 2

- 3x + 2y = 6

Figure 10: Unbounded objective function hyperplanes (two dimen-sions)

Observe that in each case there are points that lie to the right ofthe hyperplane and that are still in the set of feasible solutions.

Evidently the value of the objective function can be made


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arbitrarily large.


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G EOMETRY OF THE SET OF FEASIBLE SOLUTIONSLet is explore the question of where in the set of feasible solutions weare likely to nd a point at which the objective function takes on itsoptimal value.

First show that

if x 1 and x 2 are two feasible solutions, then any point on the line segment joining these two points is also a feasible solution.

EOMETRY OF THE SET OF FEASIBLE SOLUTIONS 38

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The line segment joining x 1 and x 2 is dened as

{ x R n | x = x 1 + (1 )x 2 , 0 1}.

Observe that, if = 0 , we get x 2 , and if = 1 , we get x 1 .

The points of the line segment at which 0 < < 1 are called theinterior points of the line segment.

x 1 and x 2 and called its end points .


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Suppose that x 1 and x 2 are feasible solutions of a LPP.

Ifa T x bi

is a constraint of the problem, then we have

a T x 1 bi and a T x 2 bi .


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For any pointx = x 1 + (1 )x 2 , 0 1,

on the line segment joining x 1 and x 2 , we have

a T x = a T ( x 1 + (1 )x 2 )

= a T x 1 + (1 )a T x 2

b i + (1 )bi

= bi .

Hence, x also satises the constraint.

This result also holds if the inequality in the constraint is reversedor if the constraint is an equality.

Thus, the line segment joining any two feasible solutions to a LPPis contained in the set of feasible solutions.


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Consider now two feasible solutions x 1 and x 2 to a LPP in generalform with objective function

c T x .

If the objective function has the same value k at x 1 and x 2 , then

proceeding as above we can easily show thatit has the value of k at any point on the line segment joining x 1 and x 2 .

Suppose that the value of the objective function is different at x 1and x 2 , say

c T x 1 < cT x 2 .


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If x = x 1 + (1 )x 2 , 0 < < 1, is any interior point of the linesegment joining x 1 and x 2 , then

c T x = c T ( x 1 + (1 )x 2 )

= c T x 1 + (1 )c T x 2

< cT x 1 + (1 )c T x 2 (since c T x 1 < c T x 2 )

= c T x 2 .

That is, the value of the objective function at any interior point ofthe line segment is less than its value at one end point .

Likewise, we may show that the value of the objective function atany interior point of the line segment is greater than its value at oneend point .


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Summarizing, we conclude that

on a given line segment joining two feasible solutions to a LPP,the objective function

either

is a constant

or

attains a maximum at one end point and a minimum at theother.

Thus, the property that a set contains the line segment joining

any two points in it has strong implications for linearprogramming.

The following denition gives a name to this property.


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Denition 1 A subset S of R n is called convex if any two distinct pointsx 1 and x 2 in S the line segment joining x 1 and x 2 lies in S .

That is, S is convex if, whenever x 1 and x 2 S , so does

x = x 1 + (1 )x 2 , for 0 1.


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Example 7 The sets in R 2 in Figures 11 and 12 and are convex.

B

C

D

A

A

B

C

A

D C

B

x

x

x

x 2

x

x

2

1

2

x 1

1

2x

Figure 11:


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x

x

x

x

x x

1

2

2

12

1

x

x

1

2

O

O

y

x

A

y

x

z

y

x

x

y

Figure 12:


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The sets in R 2 in Figure 13 are not convex.

x

x

x

x

x x

x

x 22

2

21

1

1

1

Figure 13:


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The following results help to identify convex sets.


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Theorem 1 A closed half-space is a convex set.

Theorem 2 A hyperplane is a convex set.

Theorem 3 The intersection of a nite collection of convex sets is convex.

Theorem 4 Let A be an m n matrix, and let b be a vector in R m . The setof solutions to the system of linear equations A x = b , if it is not empty, is aconvex set.


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Convex sets are of two types: bounded and unbounded.

To dene a bounded convex set, we rst need the concept of arectangle.

A rectangle in R n is a set,

R = { x R n | a i x i bi }

where a i < b i , i = 1 , 2, , n, are real numbers.

A bounded convex set is one that can be enclosed in a rectanglein R n . An unbounded convex set cannot be so enclosed.


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