tutdemo.m

数字通信第四版原书的例程

echo on; clc
%TUTDEMO Tutorial for Optimization Toolbox.
%#########################################################################
%
%   This is a demonstration script-file for the OPTIMIZATION TOOLBOX 
%   It closely follows the Tutorial section of the users' guide
%
%   It consists of: 	(1) unconstrained optimization example
%			(2) constrained optimization example 
%			(3) constrained example using gradients
%			(4) bounded example
%			(5) equality constrained example
%			(6) unconstrained ex. using user-supplied settings
%
%   All the principles outlined in this demonstration apply to the other
%   routines: attgoal, minimax, leastsq, fsolve.
%
%   The routines differ from the Tutorial Section examples in the manual 
%   only in that M-files for the objective functions have not been coded 
%   up. Instead the expression form of the syntax has been
%   used where functions to be minimized are expressed as MATLAB string 
%   variables. 
%
%##########################################################################
%
pause % Strike any key to continue (Note: use Ctrl-C to abort)
clc

%###########################################################################
%  		 	DEMO 1: UNCONSTRAINED PROBLEM
%---------------------------------------------------------------------------
%
%	Consider initially the problem of finding a minimum to the function:
%
%			      2       2
%		f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%---------------------------------------------------------------------------
pause % Strike any key to continue


%   Take a guess at the solution

X0=[-1, 1];

pause % Strike any key to continue

options=[];    % Use default parameters

pause % Strike any key to start the optimization

[x,options]=fminu('exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)', X0, options);


%  The optimizer has found a solution at:
x
%  The function value at the solution is:
options(8)
pause % Strike any key to continue
%  The total number of function evaluations was: 
options(10)

% Note: by entering the expression explicitly in a string using the 
%        variable 'x' we avoid having to write an M-file.

pause % Strike any key for next demo
clc
%%###########################################################################
%  		 	DEMO 2: CONSTRAINED PROBLEM
%---------------------------------------------------------------------------
%
%	Consider the above problem with two additional constraints:
%
%			      2       2
%	minimize f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%	subject to: 1.5 + x(1).x(2) -x(1) -x(2) <=0
%		    -x(1).x(2) <= 10
%
%-------------------------------------------------------------------------
pause % Strike any key to continue

% Make a guess at the solution:

X0=[-1, 1];

options=[]; 	% Use default parameters

pause % Strike any key to start optimization:

[x,options]=constr(['f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1);',...
	'g(1)=1.5+x(1)*x(2)-x(1)-x(2);',...
	'g(2)=-x(1)*x(2)-10;'], X0, options);


% A solution to this problem has been found at:
x 
% The function value at the solution is: 
options(8)
pause % Strike any key to continue
% Both the constraints are active at the solution:
g   =[	1.5+x(1)*x(2)-x(1)-x(2)
	-x(1)*x(2)-10	]
% The total number of function evaluations was: 
options(10)

pause % Strike any key for next demo
clc

%############################################################################
%  		 	DEMO 3: BOUNDED EXAMPLE
%----------------------------------------------------------------------------
%
%	Consider the above problem with additional bound constraints:
%
%			      2       2
%	minimize f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%	subject to: 1.5 + x(1).x(2) -x(1) -x(2) <=0
%		    -x(1).x(2) <= 10
%
%        and  x(1) >=0
%             x(2) >=0
%
%----------------------------------------------------------------------------


pause % Strike any key to continue

% Again, make a guess at the solution:
  
X0=[-1, 1];

pause % Strike any key to continue

% This time we will use the bounded syntax:
%	X=CONSTR(`FUN', X0, OPTIONS, VLB, VUB) 

	VLB=zeros(1,2);	% Lower bounds X>0
	VUB=[]; 	% No upper bounds 

pause % Strike any key to start the optimization

[x,options]=constr(['f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1);',...
	'g(1)=1.5+x(1)*x(2)-x(1)-x(2);',...
	'g(2)=-x(1)*x(2)-10;'], X0, options, VLB, VUB);

% The solution to this problem has been found at:
x 
% The function value at the solution is: 
options(8)
% The constraint values at the solution are:
g   =[	1.5+x(1)*x(2)-x(1)-x(2)
	-x(1)*x(2)-10	]
% The total number of function evaluations was: 
options(10)

pause % Strike any key for next demo
clc
%%###########################################################################
%  		 	DEMO 4: USER-SUPPLIED GRADIENTS
%----------------------------------------------------------------------------
%  The above problem can be solved more efficiently and accurately if gradients
%  are supplied by the user. This demo shows how this may be performed.
%
%  Again the  problem is:
%
%			      2       2
%	minimize f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%	subject to: 1.5 + x(1).x(2) -x(1) -x(2) <=0
%		    -x(1).x(2) <= 10
%
%-------------------------------------------------------------------------

pause % Strike any key to continue

% Make a guess at the solution:
 
X0=[-1, 1];


options=[]; 	% Use default parameters

pause % Strike any key to enter function and constraints into a string:


fun = [
% Objective Function and constraints
	'f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1);', ...
	'g(1)=1.5+x(1)*x(2)-x(1)-x(2);',...  %Constraints
	'g(2)=-x(1)*x(2)-10;'];


pause % Strike any key to load the partial derivatives into a string: 

jacob = ['t= exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1);',...
	'gf=[t+4*exp(x(1))*(2*x(1)+x(2));',...
		'4*exp(x(1))*(x(1)+x(2)+0.5)];',...
	'gg = [ x(2)-1,   -x(2);',... 
		'x(1)-1,  -x(1)];'];


pause % Strike any key to start the optimization

[x,options]=constr(fun, X0, options, [], [], jacob);

% As before the solution to this problem has been found at:
x 
% The function value at the solution is: 
options(8)
pause % Strike any key to continue
% Both the constraints are active at the solution:
g   =[	1.5+x(1)*x(2)-x(1)-x(2)
	-x(1)*x(2)-10	]
% The total number of function evaluations was: 
options(10)

pause % Strike any key for next demo
clc
%############################################################################
%  		 	DEMO 5: EQUALITY CONSTRAINED EXAMPLE
%---------------------------------------------------------------------------
%
%	Consider the above problem with an additional equality constraint
%
%			      2       2
%	minimize f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%	subject to: x(1)+ x(2) = 0
%                   1.5 + x(1).x(2) -x(1) -x(2) <=0
%		    -x(1).x(2) <= 0
%
%-------------------------------------------------------------------------


 % Again, make a guess at the solution: 

X0=[-1, 1];

pause % Strike any key to continue

% This time we will indicate the number of equality constraints
% by setting options(13) to the number of constraints. 
% The equality constraints must be contained the first few elements of g

options(13)=1; % One equality constraint

pause % Strike any key to start the optimization

[x,options]=constr(['f = exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1);',...
	'g(1)=x(1)+x(2);',...
	'g(2)=1.5+x(1)*x(2)-x(1)-x(2);',...
	'g(3)=-x(1)*x(2)-10;'], X0, options);

% The solution to this problem has been found at:
x 
% The function value at the solution is: 
options(8)
pause % Strike any key to continue
% The constraint values at the solution are:
g   =[	x(1)+x(2)
	1.5+x(1)*x(2)-x(1)-x(2)
	-x(1)*x(2)-10	]
% The total number of function evaluations was: 
options(10)

pause % Strike any key  for next demo
clc


%############################################################################
%  		 	DEMO 6: CHANGING THE DEFAULT SETTINGS
%---------------------------------------------------------------------------
%
%	Consider the original unconstrained problem
%
%			      2       2
%	minimize f(x)=exp(x(1)) .(4x(1)  + 2x(2) + 4x(1)x(2) + 1
%
%	this time we will solve it more accurately by overriding the 
%       default termination criteria (OPTIONS(2) and OPTIONS(3)). 
%-------------------------------------------------------------------------


pause % Strike any key to continue

% Again, make a guess at the solution:

X0=[-1, 1];

pause % Strike any key to continue

% Override the default termination criteria:
options(2)=1e-8;   	% Termination tolerance on X
options(3)=1e-8;	% Termination tolerance on F

pause % Strike any key to start the optimization

[x,options]=fminu('exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)', X0, options);

%  The optimizer has found a solution at:
x
%  The function value at the solution is:
options(8)
pause % Strike any key to continue
%  The total number of function evaluations was: 
options(10)


% 	Note: warning messages may be displayed because of problems
% 	in the calculation of finite difference gradients at the 
%	solution point. They are an indication that tolerances are overly 
%	stringent, however, the optimizer always continues to make steps 
%	towards the solution. 

pause % Strike any key to continue

% If we want a tabular display of each iteration we can set options(1)=1
% as follows:

options=1;  % Set display parameter to on, the others to default. 

pause % Strike any key to start the optimization

X0=[-1, 1];
[x,options]=fminu('exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)', X0, options);

%  The optimizer has found a solution at:
x
%  The function value at the solution is:
options(8)
pause % Strike any key to continue
%  The total number of function evaluations was: 
options(10)

%  At each major iteration the table displayed consists of: 
%			- 	number of function values
%			-	function value
%			-	step length used in the line search
%			-	gradient in the direction of search
%			-	line search procedures
%	

%------------------------------Demo End-----------------------------
pause % Strike any key for main menu
echo off