Optimization Problem with Error using optim.problemdef.OptimizationProblem/solve

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Jason Qian 2021-2-1

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评论： Matt J 2021-2-1

Hi, I'm currrently working on creating an optimization problem that finds the maximum of an equation that has 3 constraints. My code is:

prob = optimproblem('ObjectiveSense','max');
x=optimvar('x',2,1);
prob.Objective = 1+(x(1))^2 * ((x(2)-1)^3) * (exp(-x(1)-x(2)));
cons1 = x(2) >= log(x(1));
cons2 = x(1) + x(2) <= 6;
prob.Constraints.cons1 = cons1;
prob.Constraints.cons2 = cons2;
show(prob);
sol=solve(prob);
sol.x; 

I keep getting the same errors:

Error using optim.problemdef.OptimizationProblem/solve

SOLVE requires a non-empty initial point structure to solve a nonlinear problem.

Please let me know if you have any advice. Thanks!

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Answer 1

Matt J 2021-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/732708-optimization-problem-with-error-using-optim-problemdef-optimizationproblem-solve#answer_611593

Since your problem only has two unknowns, you can do a surface plot of the objective to find a decent initial guess.

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Matt J 2021-2-1

编辑：Matt J 2021-2-1

在 MATLAB Online 中打开

[X,Y] = meshgrid(0:0.005:6,0:0.005:6);

Z = 1+(X).^2 .* ((Y-1).^3) .* (exp(-X-Y));

Z(Y<log(X))=nan;

Z(X+Y>6)=nan;

[zmax,i]=max(Z(:));

sol0.x=[X(i),Y(i)], zmax %Initial guess

sol0 = struct with fields:

x: [2 4]

zmax = 1.2677

h=surf(X,Y,Z,'EdgeColor','none','FaceAlpha',0.5);

xlabel 'X'; ylabel 'Y';

prob = optimproblem('ObjectiveSense','max');

x=optimvar('x',2,1,'LowerBound',0,'UpperBound',6);

prob.Objective = 1+(x(1))^2 * ((x(2)-1)^3) * (exp(-x(1)-x(2)));

cons1 = x(2) >= log(x(1));

cons2 = x(1) + x(2) <= 6;

prob.Constraints.cons1 = cons1;

prob.Constraints.cons2 = cons2;

[sol,zopt]=solve(prob,sol0);

Solving problem using fmincon. Initial point is a local minimum that satisfies the constraints. Optimization completed because at the initial point, the objective function is non-decreasing in feasible directions to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

sol.x, zopt

ans = 2×1

2 4

zopt = 1.2677

Jason Qian 2021-2-1

Thank you so much. That makes a lot more sense. I really appreciate all the help!

Matt J 2021-2-1

You're quite welcome, but please Accept-click the answer to indicate that it resolved your question.

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