lec 17 multivariable ot

8/9/2019 Lec 17 Multivariable OT

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Optimality Criteria• The definition of a local, a global or an inflection point

remains the same as for single variable functions;

• However optimality criteria for multi-variable functionsare different.• In multivariable functions the gradient of a function is

not a scalar quantity, instead it is a vector.• The optimality criteria can be derived by using the

definition of a local optimal point and a Taylor expansion

of the function.• We present these results here.


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Optimality Criteria

The unconstrained optimization problemconsidered in this section is stated as following:

Find a vector of optimization variables x = (x 1 , x 2 , x 3 , . . . , x n )T in order to minimize f (x)where f( x ) is termed as objective function


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The first order optimality condition for the minimum of f(x)can be derived by considering linear expansion of the functionaround the optimum point x* using Taylor Series:

Necessary Condition for Optimality:

*)((x*)*)()( T x x f x f x f −+≈

*)((x*)*)()( T x x f x f x f −=−

Where f(x*) is the gradient of function f(x) and x - x* is the distance.


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Unconstrained Problems:

• If the x* is a minimum point thenthis condition can only be ensuredif f(x)=0; The gradient of f(x)must vanish at the optimum.

• Thus the first order necessary

condition for the minimum of afunction is that its gradient is zeroat the optimum.

• This condition is true for amaximum point also and for anyother point where the slope is zero.

• Therefore, it is only a necessary

condition and is not sufficientcondition.

Conditions for Optimality

Graph of f(x) = x(-cos(1) – sin(1) + sin(x))

-6 -4 -2 0 2 4 6

-12

-8

-4

0

4

8

12 f(x)

x

Local max

Local min

Inflection

Local min


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Conditions for Optimality

-6 -4 -2 0 2 4 6

-12

-8

-4

0

4

8

12

df(x)/dx

x-6 -4 -2 0 2 4 6

-12

-8

-4

0

4

8

12

f(x)

x

Local

max

Local

min

Inflection

Local

min


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How to Plot Multivariable functions

% matlab function for 3D plotclear all[X,Y] = meshgrid(-8:.5:8);R = sqrt(X.^2 + Y.^2) + eps;

Z = sin(R)./R;mesh(X,Y,Z)%contour(X,Y,Z,20)

MATLAB PROGRAM:


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Contours for 3D plots

% matlab functionclear all[X,Y] = meshgrid(-8:.5:8);

R = sqrt(X.^2 + Y.^2) + eps;Z = sin(R)./R;contour(X,Y,Z,100)

MATLAB PROGRAM:

The direction of steepestascent (gradient) isgenerally perpendicular,

or orthogonal, to theelevation contour.


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Optimality Conditions – Unconstrained Case

• Let x* be the point that we think is the minimum for f(x)• Necessary condition (for optimality):

f(x*) = 0• A point that satisfies the necessary condition is a stationary

point

• It can be a minimum, maximum, or saddle point• How do we know that we have a minimum?• Answer: Sufficiency Condition:

The sufficient conditions for x* to be a strict local minimumare: f(x*) = 0

2 f(x*) is positive definite


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Definition of GradientThe gradient vector of a function f, denotedas f , tells us that from an arbitrary point

• Which direction is the steepest ascend/descend?

• That is the Direction that will yield the greatestchange in f.

• How much we will gain by taking that step?

• Indicate by the magnitude of f = || f || 2∂∂

∂∂∂∂

=

n x f

x f x f

f

Μ2

1


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Gradient – ExampleProblem : Employ gradient to evaluate the steepest ascent directionfor the function f ( x , y ) = xy 2 at point (2, 2).

Solution:

8)2)(2(2y2

4)2(y 22

===∂

∂

===∂∂

x y

f x f

4 unit

8 unit

944.884of magnitudeascentof Magnitude22

=+== f

ly.respectivedirectionsyandxinrsunit vectotheare, where

84or8

4directionascentSteepest

ji

ji f +==


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Gradient – ExampleThe direction of steepest ascent (gradient) is generallyperpendicular, or orthogonal, to the elevation contour.

01

23

45

02

4

60

50

100

150

f (x, y) = xy2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

clear all[x,y] = meshgrid([0:.2:5]);z = x.*y.^2;

% mesh(x,y,z)contour(x,y,z,20)


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Testing Optimum Point for one-D• For 1-D problems:

If f' ( x' ) = 0

andIf f" ( x' ) < 0 , then x' is a maximum pointIf f" ( x' ) > 0 , then x' is a minimum pointIf f" ( x' ) = 0 , then x' is a saddle point

• What about for multi-dimensional problems?


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Testing Optimum Point for Two-D

• For 2-D problems, if a point is an optimum point, then

0and0 =∂∂

=∂∂

y f

x f

• In addition, if the point is a maximum point, then

0and0 22

2

2<

∂∂<

∂∂

y f

x f

• Question : If both of these conditions aresatisfied for a point, can we conclude that thepoint is a maximum point?


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• For 2-D functions, we also have to take intoconsideration of

• That is, whether a maximum or a minimum occursinvolves both partial derivatives w.r.t. x and y and thesecond partials w.r.t. x and y.

Testing Optimum Point for two-D systems

):note(222

x y f

y x f

y x f

∂∂∂=

∂∂∂

∂∂∂


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• Also known as the matrix of second partial derivatives.• It provides a way to discern if a function has reached an

optimum or not.

Hessian Matrix (or Hessian of f )

∂

∂

∂∂

∂∂∂

∂∂∂

=

2

22

2

2

2

y

f

x y

f y x f

x f

H

∂∂

∂∂∂

∂∂∂

∂∂∂

∂∂

∂∂∂

∂∂∂

∂∂∂

∂∂

=

2

2

2

2

1

2

2

2

22

2

12

2

1

2

1

2

21

2

nnn

n

nn

x f

x x f

x x f

x x f

x f

x x f

x x f

x x f

x f

Λ

ΜΜ

Λ

Λ

H

n=2


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• Suppose gradient f and Hessian H is evaluated at

x* = ( x *1, x* 2, …, x* n).

• If f = 0, the point x* is a stationary point.

• Further if H is positive definite, then x* is a minimum .

• If - H is positive definite (or H is negative definite) , then x* is amaximum point .

• If H is indefinite (neither positive nor negative definite), then x*is a saddle point.

• If H is singular, no conclusion (need further investigation)

Testing Optimum Point (General Case)


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Assuming that the partial derivatives are continuous at and near the pointbeing evaluated. For function with two variables (i.e. N = 2 ),

pointsaddleahas),(then,0If

maximumlocalahas),(then,0and0If

minimumlocalahas),(then,0and0If

2

2

2

2

22

2

2

2

2

y x f

y x f x

f y x f x

f

y x f

y f

x f

<

<∂∂>

>∂

∂>

∂∂∂−

∂∂

∂∂=

H

H

H

H

The quantity | H| is equal to the determinant of the Hessian matrix of f .

Testing Optimum Point (Special case – function with two variables)


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Principal Minors Test• This test usually requires less computational effort than the eigen-value

test. If all principal minors, A i for i = 1, 2, ..., n, of n х n matrix A in thequadratic form f (x) = 0.5 x T Ax are known, then the sign of the

quadratic form is determined as follows:

1. Positive definite if A i > 0 for all i = 1, ...., n.

2. Positive semi-definite if A i ≥ 0 for all i = 1,..., n.

3. Negative definite { if А i < 0 for all i = l, 3, 5,... (odd indices) ;

or А i > 0 for all i = 2, 4, 6,... (even indices) ;

2. Negative semi-definite { if А i ≤ 0 for all i = l, 3, 5,... (oddindices);

or А i ≥ 0 for all i = 2, 4, 6,... (even indices) ;

2. Indefinite if none of the above cases applies.


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Example 4.1:

+−

−+−=

∂∂

∂∂

y x

y x

y f

x f

4

22

−

−

=

∂∂∂∂∂

∂∂∂∂∂41

12

//

//222

222

y f x y f

y x f x f

Find all stationary points for the following function. Using Optimalityconditions, classify them as minimum, maximum or inflection points.

The objective function is : -2x + x 2 –xy +2y 2

The gradient vector :

The Hessian Matrix :


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Example 4.1:The function f(x, y) at the point x* = 1.14286, y* = 0.285714 is

f = -1.14286 ;

The point is minimum point.Since Hessian matrix is positive define, we know the function is convex.

Therefore any minimum is a global minimum.

% Matlab program to draw contour of function[X,Y] = meshgrid(-1:.1:2);Z = -2.*X + X.*X - X.*Y + 2.*Y.*Y;contour(X,Y,Z,100)


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Example 4.1:• Contour graph using MATLAB

Graphical

presentation offunction andminimum at point(x*, y*)


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Example 1:Important observations: • The minimum point does not change if we add a

constant to the objective function.• The minimum point does not change if we multiply theobjective function by a positive constant.

• The problem changes from minimization to maximizationproblem if we multiply the objective function by anegative sign.

• The unconstrained problem is a convex problem if theobject function is convex . For convex problems any localminimum is also a global minimum .

lec 17 multivariable ot

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