How to set lower bound lb as a function of optimization variable.

Question

Sol Elec 2024-5-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2117506-how-to-set-lower-bound-lb-as-a-function-of-optimization-variable

评论： Sol Elec 2024-5-12

在 MATLAB Online 中打开

I am solving one optimization using fmincon. Objectives function is a variable of x1,x2,x3.

B= 0; c =[ 0 0.1736 0.0693; 0 0 0.1736]

c = 2x3

0 0.1736 0.0693 0 0 0.1736

l_op = 6.3247;

rho = 1.225;

R=37.5; A = pi * R^2;

Z = 8;

n =3;

k= 0.1;

Th = sym('Th', [1, n]);

x = sym('x', [1, n]);

Lam = sym('Lam', [1, n]);

Pt = sym('P_opt', [1, n]);

Per = sym('Per', [1, n]);

z = sym('z', [1, n]);

z(1) = Z;

for i = 1:n

if i > 1

term_sum = 0;

for j = 1:i-1

term_sum = term_sum + ((1 - sqrt(1 - Th(j))) * c(j, i))^2;

end

z(i) = Z * (1 - sqrt(term_sum));

end

Lam(i) = (x(i) * R) / z(i);

Th(i) = (0.000086 * B - 0.0026) * Lam(i)^3 + (-0.0018 * B + 0.0481) * Lam(i)^2 + (0.008 * B - 0.165) * Lam(i) + (-0.0116 * B + 0.3);

Per(i) = 0.22 * (116 / (Lam(i) + 0.08 * B) - 4.06 / (B^3 + 1) - 0.4 * B - 5) * exp(-12.5 / (Lam(i) + 0.08 * B) + 0.4375 / (B^3 + 1));

Pt(i) = 0.5 * rho * A * Per(i) * z(i)^3;

end

objective = -sum(Pt)

objective =

Obj = matlabFunction(objective, 'Vars', {x});

% Lower and upper bounds

lb = 0.1687 * z(i) % z(i) is a function of x(i) *corrected

Unrecognized function or variable 'v'.

ub = 1.6867*ones(1,n);

% Initial guess

x0 = ones(1,n);

% Constraints

Aeq = [];

beq = [];

% Optimization using fmincon

options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'interior-point');

% % Optimize the objective function

[x_opt, fval_opt] = fmincon(Obj, x0, [], [], Aeq, beq, lb, ub, [], options);

In my optimization lower bound of the optimization is a function of optimaztion variables. How can I solve it? fmincon can be able to do it or another optimization tools can be chosen?

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John D'Errico 2024-5-10

That is not a "bound" constraint. It is an inequality constraint.

Sol Elec 2024-5-10

How can I modify it? Becuase z(i) is the function of x(i)? @John D'Errico

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Answer 1

Torsten 2024-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/2117506-how-to-set-lower-bound-lb-as-a-function-of-optimization-variable#answer_1455761

编辑：Torsten 2024-5-10

Use function "nonlcon" to define nonlinear equality and inequality constraints.

If v(i) depends linearly on the solution variable vector x, you could also use A and b in the call to "fmincon":

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)

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Sol Elec 2024-5-10

编辑：Sol Elec 2024-5-11

在 MATLAB Online 中打开

B= 0; cw =[ 0 0.1736 0.0693; 0 0 0.1736]

cw = 2x3

0 0.1736 0.0693 0 0 0.1736

l_op = 6.3247;

rho = 1.225;

R=37.5; A = pi * R^2;

Z = 8;

n =3;

k= 0.1;

Th = sym('Th', [1, n]);

x = sym('x', [1, n]);

Lam = sym('Lam', [1, n]);

Pt = sym('P_opt', [1, n]);

Per = sym('Per', [1, n]);

z = sym('z', [1, n]);

z(1) = Z;

for i = 1:n

if i > 1

term_sum = 0;

for j = 1:i-1

term_sum = term_sum + ((1 - sqrt(1 - Th(j))) * cw(j, i))^2;

end

z(i) = Z * (1 - sqrt(term_sum));

end

Lam(i) = (x(i) * R) / z(i);

Th(i) = (0.000086 * B - 0.0026) * Lam(i)^3 + (-0.0018 * B + 0.0481) * Lam(i)^2 + (0.008 * B - 0.165) * Lam(i) + (-0.0116 * B + 0.3);

Per(i) = 0.22 * (116 / (Lam(i) + 0.08 * B) - 4.06 / (B^3 + 1) - 0.4 * B - 5) * exp(-12.5 / (Lam(i) + 0.08 * B) + 0.4375 / (B^3 + 1));

Pt(i) = 0.5 * rho * A * Per(i) * z(i)^3;

end

objective = -sum(Pt)

objective =

Obj = matlabFunction(objective, 'Vars', {x});

% Define nonlinear constraint functions for each variable

c = cell(1, n);
for i = 1:n
    c{i} = @(x) 0.1687 * v(i) - x(i);
end

% Lower and upper bounds

lb = [];
ub = 1.6867 * ones(1, n);
% Initial guess
x0 = ones(1, n);
% Constraints
Aeq = [];
beq = [];
% Optimization using fmincon
options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'interior-point');
% Optimize the objective function
[x_opt, fval, exitflag, output] = fmincon(Obj, x0, [], [], Aeq, beq, [], ub, nonlcon, options);
Unrecognized function or variable 'x0'.

I am getting error. What should I do here? @Torsten

Walter Roberson 2024-5-10

编辑：Walter Roberson 2024-5-10

在 MATLAB Online 中打开

You really need to look further into your lb being empty. lb being empty means the same as -infinity as the lower bound. However, you have division by several of your variables, so you would pass through infinity at 0.

B= 0; cw =[ 0 0.1736 0.0693; 0 0 0.1736]

cw = 2x3

0 0.1736 0.0693 0 0 0.1736

l_op = 6.3247;

rho = 1.225;

R=37.5; A = pi * R^2;

Z = 8;

n =3;

k= 0.1;

Th = sym('Th', [1, n]);

x = sym('x', [1, n]);

Lam = sym('Lam', [1, n]);

Pt = sym('P_opt', [1, n]);

Per = sym('Per', [1, n]);

z = sym('z', [1, n]);

z(1) = Z;

for i = 1:n

if i > 1

term_sum = 0;

for j = 1:i-1

term_sum = term_sum + ((1 - sqrt(1 - Th(j))) * cw(j, i))^2;

end

z(i) = Z * (1 - sqrt(term_sum));

end

Lam(i) = (x(i) * R) / z(i);

Th(i) = (0.000086 * B - 0.0026) * Lam(i)^3 + (-0.0018 * B + 0.0481) * Lam(i)^2 + (0.008 * B - 0.165) * Lam(i) + (-0.0116 * B + 0.3);

Per(i) = 0.22 * (116 / (Lam(i) + 0.08 * B) - 4.06 / (B^3 + 1) - 0.4 * B - 5) * exp(-12.5 / (Lam(i) + 0.08 * B) + 0.4375 / (B^3 + 1));

Pt(i) = 0.5 * rho * A * Per(i) * z(i)^3;

end

objective = -sum(Pt)

objective =

Obj = matlabFunction(objective, 'Vars', {x});

c = cell(1, n);

for i = 1:n

c{i} = @(x) 0.1687 * v(i) - x(i);

end

lb = [];

ub = 1.6867 * ones(1, n);

% Initial guess

w0 = ones(1, n);

% Constraints

Aeq = [];

beq = [];

%added

x0 = [.1, .2, .3];

nonlcon = [];

% Optimization using fmincon

options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'interior-point');

% Optimize the objective function

[x_opt, fval, exitflag, output] = fmincon(Obj, x0, [], [], Aeq, beq, [], ub, nonlcon, options);

First-order Norm of Iter F-count f(x) Feasibility optimality step 0 4 -5.332679e+03 0.000e+00 8.704e+04 Objective function returned complex; trying a new point... 1 12 -5.118628e+05 0.000e+00 3.377e+05 1.381e+00 2 16 -1.393827e+06 0.000e+00 4.918e+05 2.165e+00 Objective function returned complex; trying a new point... 3 22 -1.404491e+06 0.000e+00 6.320e+05 5.980e-01 4 27 -1.410640e+06 0.000e+00 7.113e+05 5.000e-01 5 32 -1.597991e+06 0.000e+00 2.592e+05 7.208e-01 6 38 -1.609659e+06 0.000e+00 2.175e+05 1.637e-01 7 45 -1.630059e+06 0.000e+00 1.653e+05 3.450e-01 8 49 -1.641995e+06 0.000e+00 2.363e+04 1.519e-01 9 53 -1.642236e+06 0.000e+00 2.077e+03 1.828e-02 10 57 -1.642238e+06 0.000e+00 3.969e+01 1.745e-03 11 61 -1.642238e+06 0.000e+00 9.937e-01 4.049e-05 12 65 -1.642238e+06 0.000e+00 1.398e-01 8.400e-07 13 69 -1.642238e+06 0.000e+00 1.000e-01 9.049e-08 14 73 -1.642238e+06 0.000e+00 4.100e-02 2.187e-07 15 77 -1.642238e+06 0.000e+00 2.894e-02 2.661e-08 16 81 -1.642238e+06 0.000e+00 7.093e-02 1.317e-08 Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.

x_opt

x_opt = 1x3

1.2516 1.1989 1.2815

fval

fval = -1.6422e+06

Walter Roberson 2024-5-11

在 MATLAB Online 中打开

You have the problem that v is not defined.

B= 0; cw =[ 0 0.1736 0.0693; 0 0 0.1736]

cw = 2x3

0 0.1736 0.0693 0 0 0.1736

l_op = 6.3247;

rho = 1.225;

R=37.5; A = pi * R^2;

Z = 8;

n =3;

k= 0.1;

Th = sym('Th', [1, n]);

x = sym('x', [1, n]);

Lam = sym('Lam', [1, n]);

Pt = sym('P_opt', [1, n]);

Per = sym('Per', [1, n]);

z = sym('z', [1, n]);

z(1) = Z;

for i = 1:n

if i > 1

term_sum = 0;

for j = 1:i-1

term_sum = term_sum + ((1 - sqrt(1 - Th(j))) * cw(j, i))^2;

end

z(i) = Z * (1 - sqrt(term_sum));

end

Lam(i) = (x(i) * R) / z(i);

Th(i) = (0.000086 * B - 0.0026) * Lam(i)^3 + (-0.0018 * B + 0.0481) * Lam(i)^2 + (0.008 * B - 0.165) * Lam(i) + (-0.0116 * B + 0.3);

Per(i) = 0.22 * (116 / (Lam(i) + 0.08 * B) - 4.06 / (B^3 + 1) - 0.4 * B - 5) * exp(-12.5 / (Lam(i) + 0.08 * B) + 0.4375 / (B^3 + 1));

Pt(i) = 0.5 * rho * A * Per(i) * z(i)^3;

end

objective = -sum(Pt)

objective =

Obj = matlabFunction(objective, 'Vars', {x});

c = cell(1, n);

for i = 1:n

c{i} = @(x) 0.1687 * v(i) - x(i);

end

lb = [];

ub = 1.6867 * ones(1, n);

% Initial guess

w0 = ones(1, n);

% Constraints

Aeq = [];

beq = [];

%added

x0 = [.1, .2, .3];

nonlcon = @(x) cellfun(@(C) C(x), c);

% Optimization using fmincon

options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'interior-point');

% Optimize the objective function

[x_opt, fval, exitflag, output] = fmincon(Obj, x0, [], [], Aeq, beq, [], ub, nonlcon, options);

Unrecognized function or variable 'v'.

Error in solution>@(x)0.1687*v(i)-x(i) (line 33)
c{i} = @(x) 0.1687 * v(i) - x(i);

Error in solution>@(C)C(x) (line 46)
nonlcon = @(x) cellfun(@(C) C(x), c);

Error in solution>@(x)cellfun(@(C)C(x),c) (line 46)
nonlcon = @(x) cellfun(@(C) C(x), c);

Error in constrfunEvaluator (line 5)
[cin, ceq] = feval(Nonlcon, x, self.FunArgs.AdditionalParameters{:});

Error in OptimFunctions/constraints (line 415)
[cin_,ceq_,cingrad_,ceqgrad_] = self.ConstraintFunAndGrad(self,self.ConFcn{3},...

Error in OptimFunctions/constraint_first_eval (line 662)
[cin,ceq,self] = self.constraints(X0);

Error in fmincon (line 516)
funObj.constraint_first_eval(X);

x_opt

fval

Sol Elec 2024-5-11

编辑：Sol Elec 2024-5-11

在 MATLAB Online 中打开

Thans for the reply @Walter Roberson. v(i) is a typo error. v(i) should be replaced to z(i), and z(i) is a function of x(i).

I think the error is here.

c = cell(1, n);
for i = 1:n
    c{i} = @(x) 0.1687 * z(i) - x(i);
end

But I am getting

c =

1×3 cell array

{@(x)0.1687*z(i)-x(i)} {@(x)0.1687*z(i)-x(i)} {@(x)0.1687*z(i)-x(i)}

but my constraint is

{@(x1)0.1687*z(1)-x(1)} {@(x2)0.1687*z(2)-x(2)} {@(x3)0.1687*z(3)-x(3)}

here z(1) is constant, z(2) is a function of x(1) and z(3) is a function of x(1) and x(2).

z(1) =8;

z(2) =8 - 8*(((217*((8775*x1^3)/32768 - (4329*x1^2)/4096 + (99*x1)/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) ;

z(3) = 8 - 8*(((217*(7/10 - (4329*x2^2)/(64*(8*(((217*((8775*x1^3)/32768 - (4329*x1^2)/4096 + (99*x1)/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^2) - (8775*x2^3)/(64*(8*(((217*((8775*x1^3)/32768 - (4329*x1^2)/4096 + (99*x1)/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^3) - (99*x2)/(16*(8*(((217*((8775*x1^3)/32768 - (4329*x1^2)/4096 + (99*x1)/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)))^(1/2))/1250 - 217/1250)^2 + ((693*((8775*x1^3)/32768 - (4329*x1^2)/4096 + (99*x1)/128 + 7/10)^(1/2))/10000 - 693/10000)^2)^(1/2) ;

@Walter Roberson @Torsten

Torsten 2024-5-11

编辑：Torsten 2024-5-11

在 MATLAB Online 中打开

% Lower and upper bounds
lb = [];   
ub = 1.6867*ones(1,n);
% Initial guess
x0 = ones(1,n);
% Constraints
Aeq = [];
beq = [];
% Optimization using fmincon
options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'interior-point');
% % Optimize the objective function
[x_opt, fval, exitflag, output] = fmincon(Obj, x0, [], [], Aeq, beq, [], ub, @nonlcon, options)
function [c,ceq] = nonlcon(x)
  z(1) = 8; 
  z(2) = 8 - 8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) ;
  z(3) = 8 - 8*(((217*(7/10 - (4329*x(2)^2)/(64*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^2) - (8775*x(2)^3)/(64*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^3) - (99*x(2))/(16*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)))^(1/2))/1250 - 217/1250)^2 + ((693*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/10000 - 693/10000)^2)^(1/2) ;
  c(1) = 0.1687*z(1)-x(1);
  c(2) = 0.1687*z(2)-x(2);
  c(3) = 0.1687*z(3)-x(3);
  ceq = [];
end

Torsten 2024-5-11

编辑：Torsten 2024-5-11

在 MATLAB Online 中打开

Then do everything in the script part:

z = @(x)[8;...
         8 - 8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) ;...
         8 - 8*(((217*(7/10 - (4329*x(2)^2)/(64*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^2) - (8775*x(2)^3)/(64*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)^3) - (99*x(2))/(16*(8*(((217*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/1250 - 217/1250)^2)^(1/2) - 8)))^(1/2))/1250 - 217/1250)^2 + ((693*((8775*x(1)^3)/32768 - (4329*x(1)^2)/4096 + (99*x(1))/128 + 7/10)^(1/2))/10000 - 693/10000)^2)^(1/2) ];
nonlcon = @(x)deal(0.1687*z(x)-x,[])
[x_opt, fval, exitflag, output] = fmincon(Obj, x0, [], [], Aeq, beq, [], ub, nonlcon, options)

I'm not sure if the input x-vector to "nonlcon" is a column vector. If it's a row vector, you will have to transpose either x or z in the above code fragment.

Walter Roberson 2024-5-11

在 MATLAB Online 中打开

but my constraint is

{@(x1)0.1687*z(1)-x(1)} {@(x2)0.1687*z(2)-x(2)} {@(x3)0.1687*z(3)-x(3)}

The first of those accepts a parameter that for the purposes of the function will be referred to ax x1 . It proceeds to ignore that parameter and to try to calculate 0.1687*z(1)-x(1) . It happens that x(1) is defined because of the

x = sym('x', [1, n]);

so you are defining an anonymous function that refers to unresolved symbolic variables. That will fail at evaluation time, as the result of the nonlinear constraints must be numeric and not contain any unresolved symbolic functions.

I suspect that what you are trying to say is that you expect

{@(x)0.1687*z(1)-x(1)} {@(x)0.1687*z(2)-x(2)} {@(x)0.1687*z(3)-x(3)}

but instead you are seeing

{@(x)0.1687*z(i)-x(i)} {@(x)0.1687*z(i)-x(i)} {@(x)0.1687*z(i)-x(i)}

What you have to realize is that at that point, i is a captured variable. The first of those expressions evaluates to approximately

> functions(c{1})

struct with fields:

function: '@(x)0.1687*z(i)-x(i)'

type: 'anonymous'

file: ''

workspace: {[1×1 struct]}

within_file_path: ''

>> ans.workspace{1}

ans =

struct with fields:

z: [0.814723686393179 0.905791937075619 0.126986816293506]

i: 1

When a @() expression is constructed, the code after the @() expression is saved literally (except with extra spaces taken out and with commas inserted into lists), along with copies of the current values of all variables that have been mentioned by name; when the anonymous function is executed, the saved workspace is inserted and the code is executed.

To be explicit, when a @() expression is constructed, MATLAB does not evaluate any of the terms -- MATLAB would not construct

@(x) 0.1687*0.814723686393179-x(1)

Torsten 2024-5-12

编辑：Torsten 2024-5-12

在 MATLAB Online 中打开

I just saw that z is defined in your loop. In this case, simply use

zfun = matlabFunction(z,'Vars',{x})
nonlcon = @(x)deal(0.1687*zfun(x)-x,[])

Here it is assumed that the x passed from "fmincon" to "nonlcon" is a row vector.

Sol Elec 2024-5-12

Thanks a lot for your valuable input regarding this. @Torsten @Walter Roberson.

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How to set lower bound lb as a function of optimization variable.

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