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and we want to know the sensibility of this optimum with respect to a. The following derivative can be calculated, relating the change of the objective function with respect to a change in the constrain:

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell fraction numerator straight partial differential calligraphic L over denominator straight partial differential a end fraction end cell cell equals left parenthesis fraction numerator straight partial differential f over denominator straight partial differential x end fraction right parenthesis left parenthesis fraction numerator straight partial differential x over denominator straight partial differential a end fraction right parenthesis plus left parenthesis fraction numerator straight partial differential lambda over denominator straight partial differential a end fraction right parenthesis left parenthesis a minus g left parenthesis x right parenthesis right parenthesis plus lambda left parenthesis 1 minus fraction numerator straight partial differential g over denominator straight partial differential x end fraction right parenthesis left parenthesis fraction numerator straight partial differential x over denominator straight partial differential a end fraction right parenthesis end cell end table (2.43)

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row blank cell equals left parenthesis fraction numerator straight partial differential f over denominator straight partial differential x end fraction minus lambda fraction numerator straight partial differential g over denominator straight partial differential x end fraction right parenthesis left parenthesis fraction numerator straight partial differential x over denominator straight partial differential a end fraction right parenthesis plus left parenthesis a minus g left parenthesis x right parenthesis right parenthesis left parenthesis fraction numerator straight partial differential lambda over denominator straight partial differential a end fraction right parenthesis plus lambda end cell end table (2.44)

       table attributes columnalign right left columnspacing 0em 2em end attributes row blank cell equals lambda end cell end table (2.45)

       lambda equals fraction numerator straight partial differential calligraphic L over denominator straight partial differential a end fraction (2.46)

      this means that λ is the variation of the lagrangian (and hence the objective function), for a small variation on the parameter a (see Chapter 3 for more details about dual variables).

      Example 2.8

      Consider the following optimization problem

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell min blank end cell cell x squared plus y squared end cell row blank cell x plus y equals 1 end cell end table (2.47)

      If the problem were unconstrained, the solution would be x = 0, y = 0; however, the solution must fulfill the constraint x + y = 1. Therefore, a new function is defined as follows:

       calligraphic L left parenthesis x comma y comma lambda right parenthesis equals x squared plus y squared plus lambda left parenthesis 1 minus x minus y right parenthesis (2.48)

      with the following optimal conditions

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 2 x minus lambda equals 0 end cell end table (2.49)

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 2 y minus lambda equals 0 end cell end table (2.50)

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 1 minus x minus y equals 0 end cell end table (2.51)

      This is a linear system of equation with solution x = 1/2, y = 1/2, and λ = 1 that constitutes the optimum of the problem.

      2.7 The Newton’s method

      Consider a set of algebraic equations S(x) = 0 where S :

n
n is continuous and differentiable. Thus, the following approximation is made:

       S left parenthesis x right parenthesis equals S left parenthesis x subscript 0 right parenthesis plus calligraphic J left parenthesis x subscript 0 right parenthesis straight capital delta x (2.52)

       table row cell capital delta x end cell cell leftwards arrow left square bracket calligraphic J left parenthesis x right parenthesis right square bracket to the power of negative 1 end exponent S left parenthesis x right parenthesis end cell end table (2.53)

       table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row x cell not stretchy leftwards arrow x minus straight capital delta x end cell end table (2.54)

      Example 2.9

      Consider

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