MATLAB优化工具箱教程

gooobooo · 发表于 2018-9-17 09:17:13

echo off
%优化工具箱简明教程
%TUTDEMO Tutorial for Optimization Toolbox.
% $Revision: 1.12 $ $Date: 1997/11/29 01:23:27 $
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 example 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) + 2x(2) + 1)
%
%--------------------------------------------------------------------------
pause % Strike any key to continue
% The function to be minimized is
fun = 'exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1)';
pause % Strike any key to continue
% Take a guess at the solution
x0 = [-1 1];
pause % Strike any key to continue
% Use default parameters
options = [];
pause % Strike any key to start the optimization
[x, options] = fminu(fun, x0, options);
% The optimizer has found a solution at:
x
pause % Strike any key to continue
% 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) + 2x(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
% Objective Function and constraints
funf = 'f = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);';
fung = 'g = [1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10];';
fun = [funf fung];
pause % Strike any key to continue
% Take a guess at the solution
x0 = [-1 1];
pause % Strike any key to continue
% Use default parameters
options = [];
pause % Strike any key to start the optimization
[x, options] = constr(fun, x0, options);
% A solution to this problem has been found at:
x
pause % Strike any key to continue
% 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]
pause % Strike any key to continue
% 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) + 2x(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
% Objective Function and constraints
funf = 'f = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);';
fung = 'g = [1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10];';
fun = [funf fung];
pause % Strike any key to continue
% Again, make a guess at the solution
x0 = [-1 1];
pause % Strike any key to continue
% Use default parameters
options = [];
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(fun, x0, options, vlb, vub);
% The solution to this problem has been found at:
x
pause % Strike any key to continue
% The function value at the solution is:
options(8)
pause % Strike any key to continue
% The constraint values at the solution are:
g = [1.5 + x(1)*x(2) - x(1) - x(2); - x(1)*x(2) - 10]
pause % Strike any key to continue
% 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) + 2x(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
% Objective function and constraints
funf = 'f = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);';
fung = 'g = [1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10];';
fun = [funf fung];
pause % Strike any key to continue
% Make a guess at the solution:
x0 = [-1 1];
pause % Strike any key to continue
% Use default parameters
options = [];
pause % Strike any key to continue
% Partial derivatives
dfdx1 = 'exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1)+4*exp(x(1))*(2*x(1)+x(2));';
dfdx2 = '4*exp(x(1))*(x(1)+x(2)+0.5);';
dfdg = '[x(2)-1, -x(2); x(1)-1, -x(1)];';
grad = ['df=[',dfdx1, dfdx2,']; dg=', dfdg];
pause % Strike any key to start the optimization
[x, options] = constr(fun, x0, options, [], [], grad);
% As before the solution to this problem has been found at:
x
pause % Strike any key to continue
% 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]
pause % Strike any key to continue
% 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) + 2x(2) + 1)
%
% subject to x(1) + x(2) = 0
% 1.5 + x(1).x(2) - x(1) - x(2) <= 0
% - x(1).x(2) <= 10
%
%--------------------------------------------------------------------------
pause % Strike any key to continue
% Objective function and constraints
funf = 'f = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);';
fung = 'g = [x(1) + x(2); 1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10];';
fun = [funf fung];
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 indicate the number of equality constraints
% by setting options(13) to the number of constraints. The equality
% constraints must be contained the in first few elements of g.
% One equality constraint
clear options
options(13) = 1;
pause % Strike any key to start the optimization
[x, options] = constr(fun, x0, options);
% The solution to this problem has been found at:
x
pause % Strike any key to continue
% 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]
pause % Strike any key to continue
% 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) + 2x(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
% The function to be minimized is
fun = 'exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1)';
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:
clear options
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(fun, x0, options);
% The optimizer has found a solution at:
x
pause % Strike any key to continue
% 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
[x, options] = fminu(fun, x0, options);
% The optimizer has found a solution at:
x
pause % Strike any key to continue
% 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
%
%------------------------------Demo End------------------------------------
pause % Strike any key for main menu
echo off
disp('End of demo')

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