Use Problem-Based Optimize solve the "Or Constraints" problem,which showed in the chapter of "Solver-Based Optimization Problem Setup"

Question

jin yong 2023-3-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1920910-use-problem-based-optimize-solve-the-or-constraints-problem-which-showed-in-the-chapter-of-sol

评论： jin yong 2023-3-3

How to solve the same problem with Problem-Based Optimize.

This example form https://ww2.mathworks.cn/help/releases/R2022b/optim/ug/or-instead-of-and-constraints.html shows with Solver-Based Optimization,how to How to solve the same problem with Problem-Based Optimize.

The problem suppose your feasible region is the L-shaped region: x is in the rectangle –1 ≤ x(1) ≤ 1, 0 ≤ x(2) ≤ 1 OR x is in the rectangle 0 ≤ x(1) ≤ 1, –1 ≤ x(2) ≤ 1.
Code for creating the figure

% Write the x and y coordinates of the figure, clockwise from (0,0)
x = [0,-1,-1,1,1,0,0];
y = [0,0,1,1,-1,-1,0];
plot(x,y)
xlim([-1.2 1.2])
ylim([-1.2 1.2])
axis equal

To represent a rectangle as a nonlinear constraint, instead of as bound constraints, construct a function that is negative inside the rectangle a ≤ x(1) ≤ b, c ≤ x(2) ≤ d:

function cout = rectconstr(x,a,b,c,d)
% Negative when  x is in the rectangle [a,b][c,d]
% First check that a,b,c,d are in the correct order
if (b <= a) || (d <= c)
    error('Give a rectangle a < b, c < d')
end
cout = max([(x(1)-b),(x(2)-d),(a-x(1)),(c-x(2))]);

Following the prescription of using the minimum of nonlinear constraint functions, for the L-shaped region, the nonlinear constraint function is:

function [c,ceq] = rectconstrfcn(x)
ceq = []; % no equality constraint
F(1) = rectconstr(x,-1,1,0,1); % one rectangle
F(2) = rectconstr(x,0,1,-1,1); % another rectangle
c = min(F); % for OR constraints

Suppose your objective function is

fun = @(x)exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);

Minimize fun over the L-shaped region:

opts = optimoptions(@fmincon,'Algorithm','interior-point','Display','off');
x0 = [-.5,.6]; % an arbitrary guess
[xsol,fval,eflag] = fmincon(fun,x0,[],[],[],[],[],[],@rectconstrfcn,opts)

xsol = 0.4998 -0.9996
fval = 2.4650e-07
eflag = 1
Clearly, the solution xsol is inside the L-shaped region. The exit flag is 1, indicating that xsol is a local minimum.

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Torsten 2023-3-2

编辑：Torsten 2023-3-2

Put the two functions as nested functions into a third function which you define as the function to be called by the optimizer.

jin yong 2023-3-2

在 MATLAB Online 中打开

clc,clear
prob=optimproblem
x=optimvar('x',2)
x0.x=[-0.5,0.6]
fun = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);
prob.Objective=fun
cons1=fcn2optimexpr(@optimer,x)
prob.Constraints=cons1
opts = optimoptions('fmincon','Algorithm','interior-point','Display','off');
[xsol,fval,eflag] = solve(prob,x0,'Options',opts)

nested function

function  con=optimer(x)
    function c= rectconstrfcn(x)
    
    % ceq = []; % no equality constraint
    F(1) = rectconstr(x,-1,1,0,1); % one rectangle
    F(2) = rectconstr(x,0,1,-1,1); % another rectangle
    c = min(F); % for OR constraints
    
    end
    
    function cout = rectconstr(x,a,b,c,d)
    % Negative when  x is in the rectangle [a,b][c,d]
    % First check that a,b,c,d are in the correct order
    
        if (b <= a) || (d <= c)
            error('Give a rectangle a < b, c < d')
        end
        
        cout = max([(x(1)-b),(x(2)-d),(a-x(1)),(c-x(2))]);
    end
end

Output argument "con" (and possibly others) not assigned a value in the execution with "untitled>optimer" function.

error: optim.problemdef.fcn2optimexpr

error optim.problemdef.fcn2optimexpr

error fcn2optimexpr

reason:

Function evaluation failed while attempting to determine output size. The function might contain an error, or might not be well-defined at the

automatically-chosen point. To specify output size without function evaluation, use 'OutputSize'.

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Answer 1

Torsten 2023-3-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1920910-use-problem-based-optimize-solve-the-or-constraints-problem-which-showed-in-the-chapter-of-sol#answer_1183880

编辑：Torsten 2023-3-2

在 MATLAB Online 中打开

clc,clear
prob=optimproblem;
x=optimvar('x',2);
x0.x=[-0.5,0.6];
fun = exp(x(1)) * (4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);
prob.Objective=fun;
cons1=fcn2optimexpr(@optimer,x);
prob.Constraints=cons1<=0;
opts = optimoptions('fmincon','Algorithm','interior-point','Display','off');
[xsol,fval,eflag] = solve(prob,x0,'Options',opts)
xsol = struct with fields:
    x: [2×1 double]
fval = 2.4649e-07
eflag = 
    OptimalSolution
xsol.x
ans = 2×1
    0.4998
   -0.9996

nested function

function  con=optimer(x)
    con = rectconstrfcn(x);
    function c= rectconstrfcn(x)
    
    % ceq = []; % no equality constraint
    F(1) = rectconstr(x,-1,1,0,1); % one rectangle
    F(2) = rectconstr(x,0,1,-1,1); % another rectangle
    c = min(F); % for OR constraints
    
    end
    
    function cout = rectconstr(x,a,b,c,d)
    % Negative when  x is in the rectangle [a,b][c,d]
    % First check that a,b,c,d are in the correct order
    
        if (b <= a) || (d <= c)
            error('Give a rectangle a < b, c < d')
        end
        
        cout = max([(x(1)-b),(x(2)-d),(a-x(1)),(c-x(2))]);
    end
end