converte the anonymous constraint function using fcn2optimexpr

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jin yong
jin yong 2023-3-6
评论： Matt J 2023-3-7
Ran in:
N=20
N = 20
x=optimvar('x',N,'LowerBound',-1,'UpperBound',1);
y=optimvar('y',N,'LowerBound',-1,'UpperBound',1);
z=optimvar('z',N,'LowerBound',-2,'UpperBound',0);
elecprob=optimproblem;
cons1=x.^2+y.^2+(z+1).^2;
elecprob.Constraints.cons1=cons1<=1;
% elecprob.Constraints.plane1 = z <= -x-y;
% elecprob.Constraints.plane2 = z <= -x+y;
% elecprob.Constraints.plane3 = z <= x-y;
% elecprob.Constraints.plane4 = z <= x+y;
%% if use anonymous constraint function ra[lace the code upside,
%% the results are different,and there is something wrong with the
%% anonymous constraint function and fcn2optimexpr
con2=@(x,y,z)(z+abs(x)+abs(y));
cons2=fcn2optimexpr(con2,x,y,z);
elecprob.Constraints.cons2=cons2<=0;
expr=optimexpr(1);
for ii=1:(N-1)
for jj=(ii+1):N
exp(ii)=(x(ii)-x(jj))^2+(y(ii)-y(jj))^2+(z(ii)-z(jj))^2;
expr=expr+(exp(ii))^(-1/2);
end
end
elecprob.Objective=expr;
rng default % For reproducibility
x0 = randn(N,3);
for ii=1:N
x0(ii,:) = x0(ii,:)/norm(x0(ii,:))/2;
x0(ii,3) = x0(ii,3) - 1;
end
init.x = x0(:,1);
init.y = x0(:,2);
init.z = x0(:,3);
[sol,fval,flag]=solve(elecprob,init)
Solving problem using fmincon. Feasible point with lower objective function value found. Solver stopped prematurely. fmincon stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 3.000000e+03.
sol = struct with fields:
x: [20×1 double] y: [20×1 double] z: [20×1 double]
fval = 166.2480
flag =
SolverLimitExceeded
If use anonymous constraint function replace the code in the % noted , the fval results 166.2480 are different from right result :163.0099, and there is something wrong with the anonymous constraint function and fcn2optimexpr.
  1 个评论
Torsten
Torsten 2023-3-6
编辑：Torsten 2023-3-6
You shouldn't name a variable "exp" (see below) because exp is reserved for the exponential function in MATLAB. But the optimal value of the objective remains the same.
Are you sure about the optimal value being 163.0099 instead of 166.2480 ? Could you set the optimal x,y and z values as initial values and see what happens ?
N=20;
x=optimvar('x',N,'LowerBound',-1,'UpperBound',1);
y=optimvar('y',N,'LowerBound',-1,'UpperBound',1);
z=optimvar('z',N,'LowerBound',-2,'UpperBound',0);
elecprob=optimproblem;
cons1=x.^2+y.^2+(z+1).^2;
elecprob.Constraints.cons1=cons1<=1;
% elecprob.Constraints.plane1 = z <= -x-y;
% elecprob.Constraints.plane2 = z <= -x+y;
% elecprob.Constraints.plane3 = z <= x-y;
% elecprob.Constraints.plane4 = z <= x+y;
%% if use anonymous constraint function ra[lace the code upside,
%% the results are different,and there is something wrong with the
%% anonymous constraint function and fcn2optimexpr
con2=@(x,y,z)(z+abs(x)+abs(y));
cons2=fcn2optimexpr(con2,x,y,z);
elecprob.Constraints.cons2=cons2<=0;
expr=optimexpr;
for ii=1:(N-1)
for jj=(ii+1):N
val=(x(ii)-x(jj))^2+(y(ii)-y(jj))^2+(z(ii)-z(jj))^2;
expr=expr+val^(-1/2);
end
end
elecprob.Objective=expr;
rng default % For reproducibility
x0 = randn(N,3);
for ii=1:N
x0(ii,:) = x0(ii,:)/norm(x0(ii,:))/2;
x0(ii,3) = x0(ii,3) - 1;
end
init.x = x0(:,1);
init.y = x0(:,2);
init.z = x0(:,3);
opts = optimoptions(elecprob)
opts.MaxFunctionEvaluations = 10000
[sol,fval,flag]=solve(elecprob,init,Options=opts)
sol.x
sol.y
sol.z

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采纳的回答

Matt J
Matt J 2023-3-6
Ran in:
N=20;
x=optimvar('x',N,'LowerBound',-1,'UpperBound',1);
y=optimvar('y',N,'LowerBound',-1,'UpperBound',1);
z=optimvar('z',N,'LowerBound',-2,'UpperBound',0);
elecprob=optimproblem;
cons1=x.^2+y.^2+(z+1).^2;
elecprob.Constraints.cons1=cons1<=1;
%% if use anonymous constraint function ra[lace the code upside,
%% the results are different,and there is something wrong with the
%% anonymous constraint function and fcn2optimexpr
con2=@(x,y,z)(z+abs(x)+abs(y));
cons2=fcn2optimexpr(con2,x,y,z);
elecprob.Constraints.cons2=cons2<=0;
[I,J]=find(triu(ones(N),1));
elecprob.Objective=sum(1./ sqrt( (x(I)-x(J)).^2 + (y(I)-y(J)).^2 + (z(I)-z(J)).^2) );
rng default % For reproducibility
x0 = randn(N,3);
for ii=1:N
x0(ii,:) = x0(ii,:)/norm(x0(ii,:))/2;
x0(ii,3) = x0(ii,3) - 1;
end
init.x = x0(:,1);
init.y = x0(:,2);
init.z = x0(:,3);
opts = optimoptions(elecprob);
opts.MaxIterations = 50000;
opts.OptimalityTolerance=1e-16;
opts.StepTolerance=1e-16;
opts.FunctionTolerance=1e-16;
opts.Algorithm='active-set';
[sol,fval,flag]=solve(elecprob,init,Options=opts);
Solving problem using fmincon. Feasible point with lower objective function value found. Solver stopped prematurely. fmincon stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 6.000000e+03.
fval
fval = 162.9497
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更多回答（1 个）

jin yong
jin yong 2023-3-6
编辑：Torsten 2023-3-6
Ran in:
the optimal value fval will be 163.0099.
replace the con2=@(x,y,z)(z+abs(x)+abs(y))
cons2=fcn2optimexpr(con2,x,y,z)
elecprob.Constraints.cons2=cons2<=0
with :elecprob.Constraints.plane1 = z <= -x-y;
elecprob.Constraints.plane2 = z <= -x+y;
elecprob.Constraints.plane3 = z <= x-y;
elecprob.Constraints.plane4 = z <= x+y;
the result will be: fval=163.0099
clear
N=20;
x=optimvar('x',N,'LowerBound',-1,'UpperBound',1);
y=optimvar('y',N,'LowerBound',-1,'UpperBound',1);
z=optimvar('z',N,'LowerBound',-2,'UpperBound',0);
elecprob=optimproblem;
cons1=x.^2+y.^2+(z+1).^2;
elecprob.Constraints.cons1=cons1<=1;
elecprob.Constraints.plane1 = z <= -x-y;
elecprob.Constraints.plane2 = z <= -x+y;
elecprob.Constraints.plane3 = z <= x-y;
elecprob.Constraints.plane4 = z <= x+y;
%% if use anonymous constraint function ra[lace the code upside,
%% the results are different,and there is something wrong with the
%% anonymous constraint function and fcn2optimexpr
% con2=@(x,y,z)(z+abs(x)+abs(y));
% cons2=fcn2optimexpr(con2,x,y,z);
% elecprob.Constraints.cons2=cons2<=0;
expr=optimexpr;
for ii=1:(N-1)
for jj=(ii+1):N
val=(x(ii)-x(jj))^2+(y(ii)-y(jj))^2+(z(ii)-z(jj))^2;
expr=expr+val^(-1/2);
end
end
elecprob.Objective=expr;
rng default % For reproducibility
x0 = randn(N,3);
for ii=1:N
x0(ii,:) = x0(ii,:)/norm(x0(ii,:))/2;
x0(ii,3) = x0(ii,3) - 1;
end
init.x = x0(:,1);
init.y = x0(:,2);
init.z = x0(:,3);
opts = optimoptions(elecprob)
opts =
fmincon options: Options used by current Algorithm ('interior-point'): (Other available algorithms: 'active-set', 'sqp', 'sqp-legacy', 'trust-region-reflective') Set properties: No options set. Default properties: Algorithm: 'interior-point' BarrierParamUpdate: 'monotone' CheckGradients: 0 ConstraintTolerance: 1.0000e-06 Display: 'final' EnableFeasibilityMode: 0 FiniteDifferenceStepSize: 'sqrt(eps)' FiniteDifferenceType: 'forward' HessianApproximation: 'bfgs' HessianFcn: [] HessianMultiplyFcn: [] HonorBounds: 1 MaxFunctionEvaluations: 3000 MaxIterations: 1000 ObjectiveLimit: -1.0000e+20 OptimalityTolerance: 1.0000e-06 OutputFcn: [] PlotFcn: [] ScaleProblem: 0 SpecifyConstraintGradient: 0 SpecifyObjectiveGradient: 0 StepTolerance: 1.0000e-10 SubproblemAlgorithm: 'factorization' TypicalX: 'ones(numberOfVariables,1)' UseParallel: 0 Show options not used by current Algorithm ('interior-point')
opts.MaxFunctionEvaluations = 10000
opts =
fmincon options: Options used by current Algorithm ('interior-point'): (Other available algorithms: 'active-set', 'sqp', 'sqp-legacy', 'trust-region-reflective') Set properties: MaxFunctionEvaluations: 10000 Default properties: Algorithm: 'interior-point' BarrierParamUpdate: 'monotone' CheckGradients: 0 ConstraintTolerance: 1.0000e-06 Display: 'final' EnableFeasibilityMode: 0 FiniteDifferenceStepSize: 'sqrt(eps)' FiniteDifferenceType: 'forward' HessianApproximation: 'bfgs' HessianFcn: [] HessianMultiplyFcn: [] HonorBounds: 1 MaxIterations: 1000 ObjectiveLimit: -1.0000e+20 OptimalityTolerance: 1.0000e-06 OutputFcn: [] PlotFcn: [] ScaleProblem: 0 SpecifyConstraintGradient: 0 SpecifyObjectiveGradient: 0 StepTolerance: 1.0000e-10 SubproblemAlgorithm: 'factorization' TypicalX: 'ones(numberOfVariables,1)' UseParallel: 0 Show options not used by current Algorithm ('interior-point')
[sol,fval,flag]=solve(elecprob,init,Options=opts)
Solving problem using fmincon. Local minimum found that satisfies the constraints. Optimization completed because 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 = struct with fields:
x: [20×1 double] y: [20×1 double] z: [20×1 double]
fval = 163.0099
flag =
OptimalSolution
sol.x
ans = 20×1
-0.5374 0.0000 -0.4736 -0.2989 0.5374 -1.0000 -0.7883 0.0000 0.7883 0.4736
sol.y
ans = 20×1
0.0305 -0.4877 0.0000 0.9271 0.0305 0.0000 0.5278 0.0000 0.5278 0.0000
sol.z
ans = 20×1
-1.8428 -1.8730 -0.4736 -1.2260 -1.8428 -1.0000 -1.3162 -0.0000 -1.3162 -0.4736

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