Compute an extreme point of this set in the given direction.
s = S.extreme(x)
s = extreme(S, x)
S |
A convex set described as YSet object. Class: YSet |
x |
A point given as vector. Note that for YSet with symmetric matrix variable,
the point x must be given as vector with symmetric terms. Class: double |
s |
The output structure with the information about the extreme point and the exit status from the optimization. Class: struct |
s.exitflag |
Exit status from the optimization, i.e. an integer value that informs
if the result was feasible (1), or otherwise (different from 1). Class: double |
s.how |
A string that informs if the result was feasible ('ok'), or if any problem appeared through optimization. Class: char |
s.x |
Computed extreme point that lies on the boundary of the set S. Class: double |
s.supp |
The support of this set in the direction x which represents the optimal value of the objective
function in the optimization problem ![]() Class: double |
x = sdpvar(2,1);
F = [ [-3 0.3;0.1 -1;-0.1 2]*x<=[0.8;2.1;1.5] ];
F = [F; 0.3*x'*x-4*x(1)+2*x(2)<=0.1];
S = YSet(x,F);Compute the extreme point in the direction of the point v=[0;2].
v = [6;0];
s = S.extreme(v);The computed extreme point is lying the edge of the set. We can plot the set and the point s.x.
S.plot; hold on; text(s.x(1),s.x(2),'\bf x');
Plotting... 21 of 40
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© 2010-2013 Colin Neil Jones: EPF Lausanne, colin.jones@epfl.ch
© 2010-2013 Martin Herceg: ETH Zurich, herceg@control.ee.ethz.ch