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    Pemrograman Non Linier

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    MIT and James Orlin 20032

    What is a non-linear program?

    maximize 3 sin x + xy + y 3 - 3z + log zSubject to x 2 + y2 = 1

    x + 4z 2z 0

    A non-linear program is permitted to havenon-linear constraints or objectives.

    A linear program is a special case of non-linearprogramming!

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    MIT and James Orlin 20033

    Nonlinear Programs (NLP)

    Nonlinear objective function f(x) and/or Nonlinear constraints gi(x).

    Today: we will present several types of non-linear programs.

    1 2 , , , ( )

    ( ) , 1, 2, ,

    n

    i i

    Let x x x x

    Max f x

    g x b i m

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    4

    Unimodal and Multimodal Functions A unimodal function f (x ) (in the range

    specified for x ) has a single extremum(minimum or maximum).

    A multimodal function f (x ) has two or moreextrema.

    There is a distinction between the globalextremum (the biggest or smallest between

    a set of extrema) and local extrema (anyextremum). Note : many numericalprocedures terminate at a local extremum.

    I f ( ) 0 f x at the extremum, the point iscalled a stationary point .

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    f (x )

    x

    local max (stationary)

    local min (stationary)

    global max (not stationary)

    stationary point(saddle point)

    global min (stationary)

    A multimodal function

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    6

    Multivariate Functions -Surface and Contour Plots

    We shall be concerned with basic properties of a scalar function f (x ) of n variables ( x 1,..., x n).

    If n = 1, f (x ) is a univariate functionIf n > 1, f (x ) is a multivariate function .

    n 1

    For any multivariate function, the equation

    z = f (x ) defines a surface in n +1 dimensionalspace .

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    In the case n = 2, the points z = f (x 1,x 2)represent a three dimensional surface.

    Let c be a particular value of f (x 1,x 2). Then f (x 1,x 2) = c defines a curve in x 1 and x 2 on theplane z = c.

    If we consider a selection of different valuesof c, we obtain a family of curves whichprovide a contour map of the function z =

    f (x 1,x 2).

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    1 2 21 2 1 2 2(4 2 4 2 1)

    x z e x x x x x contour map of

    - 3 -2 -1 0 1 2- 3

    - 2 . 5

    - 2

    - 1 . 5

    - 1

    - 0 . 5

    0

    0 .5

    1

    1 .5

    2

    x 2

    x 1

    0.20.40.7

    1.0

    1.7

    1.8 23

    4 56 z = 20

    1.8

    2

    1.7

    3

    4

    56

    local minimum

    saddle point

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    9

    Example: Surface and

    Contour Plots of PeaksFunction

    2 2 2

    1 1 23 5 2 2

    1 1 2 1 2

    2 21 2

    3(1 ) exp ( 1) 10 (0 .2 )exp( )

    1 3exp ( 1)

    z x x x x x x x x

    x x

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    10

    05

    1 01 5

    2 0

    0

    5

    1 0

    1 5

    2 0- 1 0

    - 5

    0

    5

    1 0

    x 2

    z

    x 1

    multimodal!

    1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0

    1 0

    2 0

    3 0

    4 0

    5 0

    6 0

    7 0

    8 0

    9 0

    1 0 0

    x 1

    x 2

    global max

    global min

    local min

    local max

    local max

    saddle

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    Gradient VectorThe slope of f (x ) at a point x x in the directionof the i th co-ordinate axis is

    The n -vector of these partial derivatives istermed the gradient vector of f , denoted by:

    NMMMMM

    f

    f

    x f

    xn

    ( )

    ( )

    ( )x

    x

    x

    1

    (a column vector )

    f

    xi

    ( )x

    x x

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    The gradient vector at a point x xis normal to the the contour through thatpoint in the direction of increasing f .

    x

    f ( )xincreasing f

    At a stationary point :

    f ( )x 0 (a null vector )

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    Example

    f x x x x

    f

    f x

    f

    x

    x x x

    x x x

    ( ) cos

    ( )

    ( )

    ( )sin

    cos

    x

    x

    x

    x

    L

    N

    MMMM

    O

    Q

    PPPP

    LNM

    1 22

    2 1

    1

    2

    22

    2 1

    1 2 12

    and the stationary point (points) are given bythe simultaneous solution(s) of:-

    x x x x x x

    22

    2 1

    1 2 1

    02 0

    sincos

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    e.g.

    f T

    ( )x c x x c f( )=

    Note : If f ( )x is a constant vector, f (x ) is then linear .

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    Hessian Matrix (Curvature Matrix

    The second derivative of a n - variablefunction is defined by the n 2 partialderivatives:

    x f x i n j ni j

    ( ), , , ; , ,

    x

    HG KJ 1 1 written as:

    2 2

    2 f

    x xi j f

    xi j

    i j i

    ( ) , , ( ) , .x x

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    These n 2 second partial derivatives areusually represented by a square, symmetric

    matrix, termed the Hessian matrix , denotedby:

    H x x

    x x

    x x

    2

    2 2

    2 2( ) ( )

    ( ) ( )

    ( ) ( )

    NMMMMM

    f

    f

    x

    f

    x x

    f x x

    f x

    n

    n n

    12

    1

    12

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    Example : For the previous example:

    NMMMM Q

    PPPP LNM

    2

    2

    12

    2

    1 22

    1 2

    2

    22

    2 1 2 1

    2 1 1

    2

    2 2 f

    f x

    f x x

    f x x

    f x

    x x x x

    x x x( )

    cos sin

    sinx

    Note : If the Hessian matrix of f (x ) is aconstant matrix, f (x ) is then quadratic ,expressed as:

    f

    f f

    T T ( )

    ( ) , ( )

    x x Hx c x

    x Hx c x H

    12

    2

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    Convex and Concave Functions A function is called concave over a given

    region R if:

    f f f

    Ra b a b

    a b

    ( ( ) ) ( ) ( ) ( )

    , , .

    x x x x

    x x

    1 1

    0 1w here: an d

    The function is strictly concave if isreplaced by >.

    A function is called convex (strictly convex )if is replaced by (

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    concave function

    convex function

    x x a x b

    f (x )

    f x( ) 0

    x x a x b

    f (x )

    f x( ) 0

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    I f th en is c o n ca v e .

    I f th en is c o n v e x .

    f x f

    x f x

    f x f x f x

    ( ) ( )

    ( ) ( )

    2

    2

    2

    2

    0

    0

    For a multivariate function f (x ) theconditions are:-

    Strictly convex +ve def convex +ve semi def

    concave -ve semi def strictly concave -ve def

    f (x ) H (x ) Hessian matrix

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    Tests for Convexity and ConcavityH is +ve def (+ve semi def) iff

    x Hx x 0T 0 0( ), .H is -ve def (-ve semi def) iff

    x Hx x 0T 0 0( ), .Convenient tests : H (x ) is strictly convex(+ve def) ( convex) (+ve semi def)) if:

    1. all eigenvalues of H (x ) areor 2. all principal determinants of H (x ) are

    0 0( )

    0 0( )

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    H (x ) is strictly concave (-ve def)(concave (- ve semi def)) if:

    1. all eigenvalues of H (x ) areor 2. the principal determinants of H (x )

    are alternating in sign:

    0 0( )

    1 2 3

    1 2 3

    0 0 0

    0 0 0

    , , ,

    ( , , , )

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    Example f x x x x x( ) 2 3 212 1 2 222 2

    1 2 21 1 21

    2

    1 2 22 2

    ( ) ( ) ( )4 3 4 3

    ( ) ( )3 4 4

    f f f x x x x x x

    f f x x

    x x

    x x x

    x x

    1 24 3 4 3( ) , 4, 73 4 3 4

    H x

    22

    1 2

    4 3eigenvalues: | | 8 7 0

    3 41 Hence, ( ) is strictly conv, 7. e .x f

    I H

    x

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    Convex Regionx a

    x b

    non convex region

    A convex set of points exist if for any two points, x aand x b, in a region, all points:

    x x x a b( ) ,1 0 1

    on the straight line joining x a and x b are in the set.If a region is completely bounded by concavefunctions then the functions form a convex region.

    x a

    x b

    convex region

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    Necessary and Sufficient Conditionsfor an Extremum of an

    Unconstrained Function A condition N is necessary for a result R if R canbe true only if N is true.

    R

    A condition S is sufficient for a result R if R istrue if S is true.

    S R

    A condition T is necessary and sufficient for aresult R iff T is true.T R

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    There are two necessary and a single sufficientconditions to guarantee that x * is an extremum of afunction f (x ) at x = x *:

    1. f (x) is twice continuously differentiable at x*.

    2. , i.e. a stationary point exists at x*.

    3. is +ve def for a minimum to existat x *, or -ve def for a maximum to exist at x *

    f ( )*x 0

    2

    f ( ) ( )* *x H x

    1 and 2 are necessary conditions; 3 is asufficient condition.Note : an extremum may exist at x* eventhough it is not possible to demonstrate thefact using the three conditions.

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    Example: Consider:

    The gradient vector is: NM f x x x x x

    x x x( )

    .x

    45 2 2 4 4

    4 4 2 21 2 1

    31 2

    2 1 12

    yielding three stationary points located by settingand solving numerically: f ( )x 0

    x *=( x 1,x 2) f (x *) eigenvaluesof

    2 f ( )xclassification

    A.(-1.05,1.03) -0.51 10.5 3.5 global min

    B.(1.94,3.85) 0.98 37.0 0.97 local min

    C.(0.61,1.49) 2.83 7.0 -2.56 saddle

    where:

    NM2 1

    22 1

    1

    2 12 4 2 4

    2 4 4 f

    x x x

    x( )x

    f x x x x x x x x x( ) .x 4 4 5 4 2 2 21 2 12 22 1 2 14 12 2

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    Interpretation of the Objective Functionin Terms of its Quadratic Approximation

    If a function of two variables can be approximatedwithin a region of a stationary point by a quadratic

    function :

    f x x x xh h

    h h

    x

    xc c

    x

    x

    h x h x h x x c x c x

    ( , )1 212 1 2

    11 12

    12 22

    1

    21 2

    1

    2

    12 11 1

    2 12 22 2

    212 1 2 1 1 2 2N

    MQPNM

    QP

    NM

    QP

    then the eigenvalues and eigenvectors of:

    H ( , ) ( , )* * * * x x f x xh h

    h h1 2

    21 2

    11 12

    12 22

    NMcan be used to interpret the nature of f (x 1,x 2) at: x x x x1 1 2 2 * *,

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    They provide information on the shape of

    f (x 1,x 2) at x x x x1 1 2 2 * *, If H( , )* * x x1 2 is +ve def, theeigenvectors are at right angles ( orthogonal ) andcorrespond to the principal axes of ellipticalcontours of f (x 1,x 2).

    A valley or ridge lies in the direction of theeigenvector associated with a relative smalleigenvalue.

    These interpretations can be generalized to the

    multivariate quadratic approximation :

    f T T ( )x x Hx c x 12

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    Topics Convex sets and convex programming First-order optimality conditions Examples Problem classes

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    Minimize f (x )s.t. g i(x ) ( , , =) bi, i = 1,, m

    x = ( x1,, xn)T is the n-dimensional vector of

    decision variables f (x ) is the objective function

    g i(x ) are the constraint functions

    bi are fixed known constants

    General NLP

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    Convex Sets

    Definition : A set S n

    is convex if every point on the linesegment connecting any two points x 1, x 2 S is also in S .

    Mathematically, this is equivalent to

    x 0 = x 1 + (1 )x 2 S for all such 0 1.

    x 1x 2

    x 1x 1x

    2

    x 2

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    S = {( x1, x2) : (0.5 x1 0.6) x2 1

    2( x1)2 + 3( x2)2 27; x1, x2 0}

    (Nonconvex) Feasible Region

    x 1

    x 2

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    Convex Sets and Optimization

    Let S = { xn

    : g i(x ) bi, i = 1,, m }

    Fact : If g i(x ) is a convex function for each i = 1,, m then S is aconvex set.

    Convex Programming Theorem : Let xn

    and let f (x ) be aconvex function defined over a convex constraint set S . If a

    finite solution exists to the problem

    Minimize { f (x ) : x S }

    then all local optima are global optima. If f (x ) is strictlyconvex, the optimum is unique.

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    Max f ( x 1,, x n)

    s.t. gi ( x 1,, x n) b i i = 1,,m

    x 1 0,, x n 0

    is a convex program if f isconcave and each g i isconvex .

    Convex Programming

    Min f ( x 1,, x n)

    s.t. gi ( x 1,, x n) b i i = 1,,m

    x 1 0,, x n 0

    is a convex program if f isconvex and each g i isconvex .

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    x11 2 3 4 5

    1

    2

    3

    4

    5

    x2

    Maximize f (x ) = ( x1 2) 2 + ( x2 2) 2

    subject to 3 x1 2 x2 6

    x1 + x 2 3 x1 + x2 7

    2 x1 3 x2 4

    Linearly Constrained Convex Functionwith Unique Global Maximum

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    (Nonconvex) Optimization Problem

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    First-Order Optimality Conditions

    Minimize { f (x) : g i(x ) bi, i = 1,, m }

    Lagrangian: L(x ,) f (x ) i g i (x ) bi i 1

    m

    L(x ,) f (x ) i g i (x )i 1

    m

    0

    Optimality conditions

    Stationarity:

    Complementarity: ig i(x) = 0, i = 1,, m

    Feasibility: g i(x ) bi, i = 1,, m

    Nonnegativity: i 0, i = 1,, m

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    Commercial optimization software cannot guarantee that a

    solution is globally optimal to a nonconvex program.

    Importance of Convex Programs

    NLP algorithms try to find a point where the gradient of theLagrangian function is zero a stationary point andcomplementary slackness holds.

    Given L(x, ) = f (x) + (g(x) b)

    we want L(x, ) = f (x) + g(x) = 0 g(x) b) = 0

    g(x) b 0, 0For a convex program, all local solutions are global optima.

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    NLP Problem Classes Constrained vs. unconstrained Convex programming problem Quadratic programming problem

    f (x) = a + cT

    x + xT

    Qx , Q 0 Separable programming problem

    f (x) = j =1 ,n f j ( x j )

    Geometric programming problem

    g(x) = t =1 ,T ct Pt (x), Pt (x) = ( x 1a t 1) . . . ( x n

    a tn ), x j > 0

    Equality constrained problems

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    What You Should Know AboutNonlinear Programming

    How to identify a convex program. How to write out the first-order optimality

    conditions. The difference between a local and global

    solution. How to classify problems.

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    Outline of Part 1

    Equality-Constrained Problems

    Lagrange Multipliers

    Kuhn-Tucker Conditions

    Kuhn-Tucker Theorem

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    Equality-Constrained Problems

    solving the problem as an unconstrainedproblem by explicitly eliminating K independent variables using the equalityconstraints

    GOAL

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    Example 5.1

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    Optimisation with EqualityConstraints

    min ( );( ) ;x x x

    h x 0

    n

    f subject to: m constraints (m n)

    Elimination of variables:

    example: min ( ), x x

    f x x

    x x1 2

    4 5

    2 3 6

    12

    22

    1 2

    x

    (a)

    s. t. (b)

    and substituting into (a) :- f x x x( ) ( )2 22

    226 3 5

    Using (b) to eliminate x 1 gives: x x1 26 32 (c)

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    At a stationary point

    f x x

    x x

    x x

    ( )( )

    .*

    2

    22 2

    2 2

    0 6 6 3 10 0

    28 36 1286

    Then using (c): x x

    126 3

    210 71*

    *

    .

    Hence, the stationary point (min) is: (1.071, 1.286)

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    What if?

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    Lagrange Multipliers

    Converting constrained problem to an unconstrained problem with help of

    certain unspecified parameters known as LagrangeMultipliers

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    Lagrange Multipliers

    Lagrange

    function

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    Lagrange Multipliers

    Lagrange

    multiplier

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    If:

    f x

    f x

    h x

    h x

    1 2

    1 2

    0 nontrivial nonunique solutionsfor dx 1 and dx 2 will exist.

    This is achieved by setting

    h x

    h x

    h x

    h x

    f x

    h x

    f x

    h x

    1 2

    1 2

    1 1 2 2

    with

    where is known as a Lagrange multiplier .

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    If an augmented objective function , called theLagrangian is defined as:

    L x x f x x h x x( , , ) ( , ) ( , )1 2 1 2 1 2 we can solve the constrained optimisation problem bysolving:

    L x

    f x

    h x

    L x

    f x

    h x

    L h x x

    1 1 1

    2 2 2

    1 2

    0

    0

    0

    V||W||

    provides equations (a) and (b)

    re-statement of equality constraint( , )

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    Generalizing : To solve the problem:

    min ( );

    subject to: ( ) ; constraints ( )

    f

    m m n

    n

    xx x

    h x 0

    define the Lagrangian :

    ( , ) ( ) ( ),T m L f x x h x and the stationary point (points) is obtained from:-

    ( )( , ) ( )

    ( , ) ( )

    T

    L f

    L

    x x

    h xx x 0

    x

    x h x 0

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    Example Consider the previous exampleagain. The Lagrangian is:-

    L x x x x

    L x

    x

    L x

    x

    L x x

    4 5 2 3 6

    8 2 0

    10 3 0

    2 3 6 0

    12

    22

    1 2

    11

    22

    1 2

    ( )

    (a)

    (b)

    (c)

    Substituting (a) and (b) into (c) gives:

    3 9 30

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    which agrees with the previous result.

    x x

    x x

    1 2

    1 2

    4310 2

    910

    6 0 30

    74281

    15

    14

    10 71 90

    70

    1286

    , .

    . , .Hence,

    4 -3 -2 -1 0 1 2 3 4-4

    -3

    -2

    -1

    0

    1

    2

    3

    4

    x 1

    x 2

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    Example 5.2

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    Test whether the stationary pointcorresponds to a minimum

    positive definite

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    Example 5.3

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    positivedefinite

    negativedefinite

    ax

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    Limitations of Analytical Methods

    The computations needed to evaluate the aboveconditions can be extensive and intractable.Furthermore, the resulting simultaneousequations required for solving x *, * and * areoften nonlinear and cannot be solved without

    resorting to numerical methods.

    The results may be inconclusive.

    For these reasons, we often have to resort tonumerical methods for solving optimisationproblems, using computer codes (e.g.. MATLAB )

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    Kuhn-Tucker Conditions

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    NLP problem

    Kuhn Tucker conditions

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    Kuhn-Tucker conditions(aka Kuhn-Tucker Problem)

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    Example 5.4

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    Example 5.4

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    Example 5.4

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    Kuhn-Tucker Theorems

    1. Kuhn Tucker Necessity Theorem

    2. Kuhn Tucker Sufficient Theorem

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    Kuhn-Tucker Necessity Theorem

    For certain special NLP problems, theconstraint qualification is satisfied:1. When all the inequality and equality

    constraints are linear2. When all the inequality constraints are

    concave functions and equalityconstraints are linear

    ! When the constraint qualification isnot met at the optimum, there may notexist a solution to the KTP

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    Example 5.5

    x* = (1, 0)

    and for k=1,.,K are linearly independent at the optimum

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    Example 5.5

    x* = (1, 0)

    No Kuhn-Tucker point at theoptimum

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    Example 5.6

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    Kuhn-Tucker Sufficiency Theorem

    Let f(x) be convex

    the inequality constraints g j(x) for j=1,,J beall concave function the equality constraints hk (x) for k=1,,K belinear

    If there exists a solution (x*,u*,v*) thatsatisfies KTCs, then x* is an optimal solution

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    t d t i t

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    82

    -2 0 2 4 6 8-2

    -1

    0

    1

    2

    3

    5

    6

    7

    8

    x1

    2

    (1.0 0 ,4.90 )

    x2=0

    h1(x )=0

    g1(x

    )=0105

    0-5

    -10

    -20-25

    -30-40-50-60

    contours and constraints

    We test each Kuhn Tucker condition in turn:

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    83

    We test each Kuhn-Tucker condition in turn:(a) All functions are seen by inspection to be twice

    differentiable.

    (b) We assume the Lagrange multipliers exist

    (c) Are the constraints satisfied?

    h yes

    g yesbinding

    g yes not active

    g yes not active

    g yes not active

    12 2

    12 2

    22

    3

    4

    25 10 0 490 0 0 1 0

    10 10 0 10 0 10 490 490 34 0 0 1 0

    490 1 1921 0

    1

    490 0

    : ( . ) ( . ) .

    : ( . ) ( . ) ( . ) ( . ) .

    : ( ( . ) .

    :

    : .

    1.00-3).00 -2

    2

    h f h di i d

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    84

    To test the rest of the conditions we need todetermine the Lagrange multipliers using thestationarity conditions. First we note that fromcondition (e) we require:

    j jg j* *( ) , , , ,x 0 1 2 3 4

    1 1

    2 2

    3 3

    4 4

    0

    0

    0

    0

    * *

    * *

    * *

    * *

    ( )

    ( )

    ( )

    ( )

    can have any value because

    must be zero because

    must be zero because

    must be zero because

    g

    g

    g

    g

    x

    x

    x

    x

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    85

    Hence:

    L

    NM O

    QPL

    NMO

    QP L

    NM O

    QP

    x

    x

    L x x

    L x x x1

    2

    4 2 10 2 4 2 8

    2 2 10 2 98 98 0 2

    2 8

    98 0 2

    4

    9810 15

    1 1 1 1 1 1

    2 1 2 1 2 1 1

    1

    1

    1 1

    * * * * * *

    * * * * * * *

    *

    ** *

    ( )

    ( ) . . .

    . . .. , =0.754

    Now consider the stationarity condition:* *2 3

    2 2 2 2 21 2 1 1 2 1 1 1 2 2

    ( , , ) 0 where, since 0

    =4 12 (25 ) (10 10 34)

    L

    L x x x x x x x x

    x x

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    86

    Now we can check the remaining conditions:

    (d) Are j j* , , , , ? 0 1 2 3 4

    1 2 3 40 754 0* * * *. , Hence, the answer is yes

    (e) Are j jg j* *( ) , , , , ?x 0 12 3 4

    Yes , because we have already used this above.

    (f) Is the Lagrangian function at a stationary point?

    Yes , because we have already used this above.

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    87

    Hence, all the Kuhn-Tucker conditions aresatisfied and we can have confidence in the

    solution :

    x* = (1.00,4.90)

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    Example 5.4 f(x) be convex the inequality constraints g j (x) for j=1,,J be all concave function the equality constraints hk (x) for k=1,,K be linear

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    Example 5.4 f(x) be convex

    semi-definite

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    Example 5.4 f(x) be convex the inequality constraints g j (x) for j=1,,J be all concave function

    v

    g1(x) linear, hence both convex andconcave

    negative definite

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    Example 5.4 f(x) be convex the inequality constraints g j (x) for j=1,,J be all concave function the equality constraints hk (x) for k=1,,K be linear

    v

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    Remarks

    For practical problems, the constraintqualificationwill generally hold. If the functions are

    differentiable, a KuhnTucker point is a possiblecandidate for the optimum. Hence, many of theNLP methods attempt to converge to a KuhnTucker point.

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    Remarks

    When the sufficiency conditions of Theorem 5.2hold, a KuhnTucker point automaticallybecomes the global minimum. Unfortunately,the sufficiency conditions are difficult to verify,and often practical problems may not possessthese nice properties. Note that the presenceof one nonlinear equality constraint is enough

    to violate the assumptions of Theorem 5.2

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    Remarks

    The sufficiency conditions of Theorem 5.2 havebeen generalized further to nonconvexinequality constraints, nonconvex objectives,and nonlinear equality constraints. These usegeneralizations of convex functions such asquasi-convex and pseudoconvex functions