Management Science Lecture Notes

15.053 z

February 13, 2007

The Geometry of Linear Programs \ u 2 0 1 3

the geometry of LPs illustrated

1

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Quotes of the day You don't understand anything

Marvin Minsky

One finds limits by pushing \ue000 them. \ue000 Herbert Simon

2

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Goal of this Lecture\ue000 z\

u e 0 0 1 Present the Geometry of Linear Programs \u2013 \ u e 0 0 1 A key way of looking at LPs

\ u e 0 0 1 Others are algebraic and economic 0 0 1 Some basic concepts 0 0 1 2-dimensional (2 variable) linear programs) 0 0 1 3-dimensional (3 variable) linear programs 0 0 1 Properties of the set of feasible solutions and of optimal

\u2022

\u \u2013 \u \u2013 \u \u2013 \u \u2013

e e e e

solutions

\ u e 0 0 1 generalizable to all linear programs

\u2022

3

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A Two Variable Linear Program (a variant of the DTC example)

z = 3x + 5y

objective

2x +

3y

≤ 10

x +

2y

≤

6

y

≤

5

≤

4

y ≤

3

x + x

(1)

x, y ≥ 0 4

We could have used the original variable names of K and S, but it is simpler to use x and y since we usually think of the two axes as the x and y axis.

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Finding an optimal solution z

z

Introduce yourself to your partner Try to find an optimal solution to the linear program, without looking ahead.

5

Finding an optimal solution to a 2-variable LP can be challenging until you have

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Some Basic Concepts

y A point is represented as a pair (x, y). For example, (2, 3).

5 4

Sometimes, we will call (x, y)

3 2 1

1

2

3

4

5

We go through this review pretty quickly

6

x

6

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Lines y

Every pair of (distinct) points determines a unique line. p1 = (1, 5) p1

5 4 3

p2

2

Alternative representation of the line: (1-λ)p1 + λp2 for -∞ ≤ λ ≤ ∞. L = (1, 5) + λ(3, -3) for -∞ ≤ λ ≤ ∞.

1

1

2

3

4

5

6

x

7

The alternative representation is really important, as we shall see on the next few slides.

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Rays y

Every pair of (distinct) points determines a unique ray beginning at the first point. p1 = (1, 5)

p1

5

p2 = (4, 2)

4

Ray: (1-λ)p1 + λp2 for 0 ≤ λ ≤ ∞.

3

p2

2

= (1, 5) + λ(3, -3) for 0 ≤ λ ≤ ∞.

1

1

2

3

4

5

6

x

8

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Line segments y

Every pair of points determines a unique line segment. p1 = (1, 5)

p1

5

p2 = (4, 2)

4

Segment: (1-λ)p1 + λp2 for 0 ≤ λ ≤ 1.

3

p2

2

= (1, 5) + λ(3, -3) for 0 ≤ λ ≤ 1.

1

1

2

3

4

5

6

x

9

We keep seeing (1-λαμβδα)p1 + lambda p2 as the formula. But the representation of the line segment is the most useful for our purposes.

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Inequalities An inequality with two variables determines a unique half-plane

y 5

x+ 2y ≤ 6

4 3 2 1

1

2

3

4

5

6

x

10

A half plane contains the line as well as all points on one side of the line.

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Graphing the Feasible Region y

We will construct

5 4 3 2 1

1

2

3

4

5

6

x

11

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Graph the Constraints: 2x+ 3y ≤ 10 (1) x ≥ 0 , y ≥ 0. (6)

y 5 4 3

2x + 3y = 10

2 1

1

2

3

4

5

6

x

12

OK. This is really three constraints despite what was said on the last slide.

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Add the Constraint:

x + 2y ≤ 6

(2)

y 5 4 3 2

x + 2y = 6

1

1

2

3

4

5

6

x

13

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Add the Constraint:

x+ y ≤ 5

A constraint is

y 5

x + y = 5

4 3

“x + y = 5” is redundant

2 1

1

2

3

4

5

6

x

14

We don’t concern ourselves much with redundant constraints 15.053. In principle,

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Add the Constraints:

x ≤ 4; y ≤ 3

y 5

We have now

4 3 2 1

1

2

3

4

5

6

x

15

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How do we maximize z = 3x + 5y ? Let’s avoid adding a 3rd dimension.

x

Find a feasible solution such that 3x + 5y = p for different values of p.

5

Choose p as large as possible.

4 3 2 1

1

2

3

4

5

6

y

16

If you can do so, try to maximize the objective function before looking ahead.

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y 3

How do we maximize 3x + 5y ? Is there a feasible solution such that 3x + 5y = 8? Is there a feasible solution such that 3x + 5y = 11?

2

1

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y 3

Find the maximum value p such that there is a feasible solution with 3x + 5y = p. Move the line with profit p parallel as much as possible. This is called the geometric method for optimizing in 2D

2

3x + 5y = 16 The optimal

1

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Corner Points z

A corner point of the feasible region is a point that is not the midpoint of two other points of the feasible region.

Where are the

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Solving for the Corner Point

y 3 2

In two dimensions, a corner point lies at the intersection of two lines. x + 2y = 6 2x + 3y = 10 x +2y = 6

2x +3y = 10

2x + 4y = 12 2x + 3y = 10

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Solving for Corner Points In three dimensions, a corner point is the intersection of three constraints. (3 planes) 0≤x≤2 0≤y≤2 z

0≤z≤2 x - y + z ≤3

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An important difference between the geometry and the algebra

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Remainder of the lecture: Corner Points. Does every linear program with a feasible solution

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Theorem. If there is a feasible solution, and if there is no feasible line, then there is a corner point.

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If there is an optimal solution, and if there is a

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satisfies and p2 are each both ofinthe thelinear set and equalities if p3 is onand the line inequalities. segment joining The above p1 and is an p2,outline then p3of the must also proof why be this inisthe true set. with two variables. Indeed, the proof idea can be generalized to show that it is true regardless of the number of variables.

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z z z

infeasibility and unboundedness Corner Points and their significance Convexity of the feasible region