Introduction to Unconstrained Optimization - Scilab

3 Answers to exercises
3.1 Answers for section 1.10
Answer of Exercise 1.1 (Contours of quadratic functions) By differientiating the equation 25
with respect to x, we get
g(x) = b + 1 2 Hx + 1 2 xT H. (62)
We have x T H = H T x. By hypothesis, the matrix H is symetric, which implies that H T = H.
Hence, we have x T H = Hx. This allows to simplify the equation 62 into
g(x) = b + Hx. (63)
We can differentiate the gradient, so that the Hessian matrix is H.
The following function quadratic defines the required function and returns its function value
f, its gradient g and its Hessian matrix H.
function [f,g,H] = quadratic ( x )
H = [
2 1
1 4
]
b = [
1
2
]
f = b’ * x + 0.5 * x’ * H * x;
g = b + H * x
endfunction
The following script creates the contour plot which is presented in figure 27.
function f = quadraticC ( x1 , x2 )
f = quadratic ( [x1 x2]’ )
endfunction
xdata = linspace ( -2 ,2 ,100);
ydata = linspace ( -2.5 ,1.5 ,100);
contour ( xdata , ydata , quadratic , [0.5 2 4 8 12] )
In the following session, we define the point x and compute the function value, the gradient
and the Hessian matrix at this point.
-->x = [
--> 1
--> 1
--> ]
x =
1.
1.
-->[ f , g , H ] = quadratic ( x )
H =
2. 1.
1. 4.
g =
4.
7.
f =
7.
35