Crazy Picture. Maximum matching and simplex. z y x Maximum - - PowerPoint PPT Presentation

Crazy Picture.

Maximum matching and simplex.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0 x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints intersect.

x y z

Maximum matching and simplex.

y z x maxx +y +z x ≤ 1 x +z ≤ 1 z +y ≤ 1 y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints redundant.

x y z

Maximum matching and simplex.

y z x maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0 x y z

Maximum matching and simplex.

y z x maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0 x y z

Maximum matching and simplex.

maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

.3 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

.7 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X

Maximum matching and simplex.

.3 .7 .3 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path.

Maximum matching and simplex.

.7 .3 .7 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 maxx +y +z x +z ≤ 1 z +y ≤ 1 x ≥ 0 y ≥ 0 z ≥ 0

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 maxx +y +z x +z ≤ 1 a z +y ≤ 1 b x ≥ 0 y ≥ 0 z ≥ 0 c

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 1 1 maxx +y +z x +z ≤ 1 a = 1 z +y ≤ 1 b = 1 x ≥ 0 y ≥ 0 z ≥ 0 c = 1

Blue constraints tight. Sum: x +2z +y.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 1 1 maxx +y +z x +z ≤ 1 a = 1 z +y ≤ 1 b = 1 x ≥ 0 y ≥ 0 z ≥ 0 c = 1

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 1 1 maxx +y +z x +z ≤ 1 a = 1 z +y ≤ 1 b = 1 x ≥ 0 y ≥ 0 z ≥ 0 c = 1

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Maximum matching and simplex.

1 1 1 1 maxx +y +z x +z ≤ 1 a = 1 z +y ≤ 1 b = 1 x ≥ 0 y ≥ 0 z ≥ 0 c = 1

Blue constraints tight.

x y z X +1 −1 +1 Augmenting Path. Via Gaussian Elimination!

Convex Separator.

Convex Separator. Farkas

Convex Separator. Farkas Strong Duality!!!!!

Convex Separator. Farkas Strong Duality!!!!! Maybe.

Linear Equations.

Ax = b

Linear Equations.

Ax = b A is n ×n matrix...

Linear Equations.

Ax = b A is n ×n matrix... ..has a solution.

Linear Equations.

Ax = b A is n ×n matrix... ..has a solution. If rows of A are linearly independent.

Linear Equations.

Ax = b A is n ×n matrix... ..has a solution. If rows of A are linearly independent. yT A = 0 for any y

Linear Equations.

Ax = b A is n ×n matrix... ..has a solution. If rows of A are linearly independent. yT A = 0 for any y ..or if b in subspace of A.

Linear Equations.

Ax = b A is n ×n matrix... ..has a solution. If rows of A are linearly independent. yT A = 0 for any y ..or if b in subspace of A. x1 x2 x3

• k b

bad b

Strong Duality.

Strong Duality.

Later.

Strong Duality.

• Later. Actually. No.

Strong Duality.

• Later. Actually. No. Now

Strong Duality.

• Later. Actually. No. Now ...ish.

Special Cases:

Strong Duality.

• Later. Actually. No. Now ...ish.

Special Cases: min-max 2 person games and experts.

Strong Duality.

• Later. Actually. No. Now ...ish.

Special Cases: min-max 2 person games and experts. Max weight matching and algorithm.

Strong Duality.

• Later. Actually. No. Now ...ish.

Special Cases: min-max 2 person games and experts. Max weight matching and algorithm. Approximate: facility location primal dual.

Strong Duality.

• Later. Actually. No. Now ...ish.

Special Cases: min-max 2 person games and experts. Max weight matching and algorithm. Approximate: facility location primal dual. Today: Geometry!

Convex Body and point.

For a convex body P and a point b, b ∈ P or hyperplane separates P from b.

Convex Body and point.

For a convex body P and a point b, b ∈ P or hyperplane separates P from b. v,α, where v ·x ≤ α and v ·b > α.

Convex Body and point.

For a convex body P and a point b, b ∈ P or hyperplane separates P from b. v,α, where v ·x ≤ α and v ·b > α.

Convex Body and point.

For a convex body P and a point b, b ∈ P or hyperplane separates P from b. v,α, where v ·x ≤ α and v ·b > α. point p where (x −p)T (b −p) < 0 b p x

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P.

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0 b p x P (x −p)T (b −p) ≥ 0

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0 b p x P (x −p)T (b −p) ≥ 0 → ≤ 90◦ angle between − − − → x −p and − − − → b −p.

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0 b p x P (x −p)T (b −p) ≥ 0 → ≤ 90◦ angle between − − − → x −p and − − − → b −p. Must be closer point b on line from p to x.

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0 b p x P (x −p)T (b −p) ≥ 0 → ≤ 90◦ angle between − − − → x −p and − − − → b −p. Must be closer point b on line from p to x. All points on line to x are in polytope.

Proof.

For a convex body P and a point b, b ∈ A or there is point p where (x −p)T (b −p) < 0 b p x Proof: Choose p to be closest point to b in P. Done or ∃ x ∈ P with (x −p)T (b −p) ≥ 0 b p x P (x −p)T (b −p) ≥ 0 → ≤ 90◦ angle between − − − → x −p and − − − → b −p. Must be closer point b on line from p to x. All points on line to x are in polytope. Contradicts choice of p as closest point to b in polytope.

More formally.

b p x P Squared distance to b from p +(x −p)µ

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p.

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify:

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify: |p −b|2 −2µ|p −b||x −p|cosθ +(µ|x −p|)2.

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify: |p −b|2 −2µ|p −b||x −p|cosθ +(µ|x −p|)2. Derivative with respect to µ ...

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify: |p −b|2 −2µ|p −b||x −p|cosθ +(µ|x −p|)2. Derivative with respect to µ ... −2|p −b||x −p|cosθ +2(µ|x −p|2).

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify: |p −b|2 −2µ|p −b||x −p|cosθ +(µ|x −p|)2. Derivative with respect to µ ... −2|p −b||x −p|cosθ +2(µ|x −p|2). which is negative for a small enough value of µ

More formally.

b p x P Squared distance to b from p +(x −p)µ point between p and x (|p −b|− µ|x −p|cosθ)2 +(µ|x −p|sinθ)2 θ is the angle between x −p and b −p. b |p −b|−ℓcosθ Distance to new point. p θ p + µ(x −p) ℓ = µ|x −p| ℓsinθ ℓcosθ x Simplify: |p −b|2 −2µ|p −b||x −p|cosθ +(µ|x −p|)2. Derivative with respect to µ ... −2|p −b||x −p|cosθ +2(µ|x −p|2). which is negative for a small enough value of µ (for positive cosθ.)

Generalization: exercise.

Theorems of Alternatives.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A. y is normal.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A. y is normal. y in nullspace for column span.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A. y is normal. y in nullspace for column span. yT b = 0 = ⇒ b not in column span.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A. y is normal. y in nullspace for column span. yT b = 0 = ⇒ b not in column span. There is a separating hyperplane between any two convex bodies.

Generalization: exercise.

Theorems of Alternatives. Linear Equations: There is a separating hyperplane between a point and an affine subspace not containing it. From Ax = b use row reduction to get, e.g., 0 = 5. That is, find y where yT A = 0 and yT b = 0. Space is image of A. Affine subspace is columnspan of A. y is normal. y in nullspace for column span. yT b = 0 = ⇒ b not in column span. There is a separating hyperplane between any two convex bodies. Idea: Let closest pair of points in two bodies define direction.

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• 1

1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• 1

1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• 1

1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• 1

1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• 1

1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2 y where yT (b−Ax) < yT (0) < 0 for all x ≥ 0

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2 y where yT (b−Ax) < yT (0) < 0 for all x ≥ 0 → yT b < 0 and yT A ≥ 0.

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2 y where yT (b−Ax) < yT (0) < 0 for all x ≥ 0 → yT b < 0 and yT A ≥ 0. Farkas A: Solution for exactly one of:

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2 y where yT (b−Ax) < yT (0) < 0 for all x ≥ 0 → yT b < 0 and yT A ≥ 0. Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0

Ax = b, x ≥ 0

• 1

1 1 1

• x =
• −1

−1

• x1

x2 x3 Coordinates s = b −Ax. x ≥ 0 where s = 0? s1 s2 y where yT (b−Ax) < yT (0) < 0 for all x ≥ 0 → yT b < 0 and yT A ≥ 0. Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0 (2) yT A ≥ 0,yT b < 0.

Farkas 2

Farkas A: Solution for exactly one of:

Farkas 2

Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0

Farkas 2

Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0 (2) yT A ≥ 0,yT b < 0.

Farkas 2

Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0 (2) yT A ≥ 0,yT b < 0. Farkas B: Solution for exactly one of:

Farkas 2

Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0 (2) yT A ≥ 0,yT b < 0. Farkas B: Solution for exactly one of: (1) Ax ≤ b

Farkas 2

Farkas A: Solution for exactly one of: (1) Ax = b,x ≥ 0 (2) yT A ≥ 0,yT b < 0. Farkas B: Solution for exactly one of: (1) Ax ≤ b (2) yT A = 0,yT b < 0,y ≥ 0.

Strong Duality

(From Goemans notes.) Primal P z∗ = mincT x Ax = b x ≥ 0 Dual D :w∗ = maxbT y AT y ≤ c

Strong Duality

(From Goemans notes.) Primal P z∗ = mincT x Ax = b x ≥ 0 Dual D :w∗ = maxbT y AT y ≤ c Weak Duality: x,y- feasible P , D: xT c ≥ bT y.

Strong Duality

(From Goemans notes.) Primal P z∗ = mincT x Ax = b x ≥ 0 Dual D :w∗ = maxbT y AT y ≤ c Weak Duality: x,y- feasible P , D: xT c ≥ bT y. xT c −bT y = xT c −xT AT y = xT (c −AT y) ≥ 0

Strong duality If P or D is feasible and bounded then z∗ = w∗.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal. (a) ˜ x + µx ≥ 0 since ˜ x,x,µ ≥ 0.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal. (a) ˜ x + µx ≥ 0 since ˜ x,x,µ ≥ 0. (b) A(˜ x + µx) = A˜ x + µAx = b + µ ·0 = b.

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal. (a) ˜ x + µx ≥ 0 since ˜ x,x,µ ≥ 0. (b) A(˜ x + µx) = A˜ x + µAx = b + µ ·0 = b. Feasible

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal. (a) ˜ x + µx ≥ 0 since ˜ x,x,µ ≥ 0. (b) A(˜ x + µx) = A˜ x + µAx = b + µ ·0 = b. Feasible cT (˜ x + µx) = xT ˜ x + µcT x → −∞ as µ → ∞

Strong duality If P or D is feasible and bounded then z∗ = w∗. Primal feasible, bounded, value z∗. Claim: Exists a solution to dual of value at least z∗. ∃y,yT A ≤ c,bT y ≥ z∗. Want y.

• AT

−bT

• y ≤
• c

−z∗

• .

If none, then Farkas B says ∃x,λ ≥ 0.

• A

−b

• x

λ

• = 0
• cT

−z∗ x λ

• < 0

∃x,λ with Ax −bλ = 0 and ctx −z∗λ < 0 Case 1: λ > 0. A( x

λ ) = b, cT ( x λ ) < z∗. Better Primal!!

Case 2: λ = 0. Ax = 0,cT x < 0. Feasible ˜ x for Primal. (a) ˜ x + µx ≥ 0 since ˜ x,x,µ ≥ 0. (b) A(˜ x + µx) = A˜ x + µAx = b + µ ·0 = b. Feasible cT (˜ x + µx) = xT ˜ x + µcT x → −∞ as µ → ∞ Primal unbounded!

See you on Tuesday.