fmincon gradient of nonlinear inequality constraints must have size???

Greeting guys! I run my code and I get the following error:

"Gradient of nonlinear inequality constraints must have size 4-by-12."

My code includes two objective functions and two nonlinear inequality constraints and I'm using fmincon to solve my problem and I added gradient of my nonlinear constraints and got error.

clc;
clear;
close all;
%% Problem Definition
nVar=4;
VarSize=[1 nVar];
VarMin=10^(-3);
VarMax=10^2;
%% Weighted-Sum Approach
%N=60;
%w1=linspace(0,1,N);
%w2=1-w1;
FWS=@(x) MyCost(x);
x0=unifrnd(VarMin,VarMax,VarSize);
x0=x0';
A=[];
b=[];
LB=[10^(-3);10^(-3);10^(-3);10^(-3)];
UB=[10^2;10^2;10^2;10^2];
nonlcon = @NLC2;
options=optimoptions('fmincon','Display','iter','Algorithm','interior-point','SpecifyConstraintGradient',true,'TolFun',10^(-12),'TolCon',10^(-10),'MaxFunEvals',4*10^3);
[x,fval,exitflag,output,lambda,grad,hessian]=fmincon(FWS,x0,A,b,[],[],LB,UB,nonlcon,options)

function [c,ceq,gradc,gradceq]=NLC2(x)
n=2;
c(1,:)=[Obj1(x)-x(n:end);-x(n:end)];
c(2,:)=[Obj2(x)-x(n:end);-x(n:end)];
ceq=[];
if nargout>1
    gradc(1,:)=[gradient(Obj1(x)-x(n:end),x);gradient(-x(n:end),x)];
    gradc(2,:)=[gradient(Obj2(x)-x(n:end),x);gradient(-x(n:end),x)];
    gradceq=[];
end
end

function z = MyCost(x)
K=3;
s=zeros(1,K);
for i=1:K
s(i) = (10^(K-i))*2; 
end
S = diag(s);
M=60;
w1=linspace(0,1,M);
w2=1-w1;
for j=1:M
z1 = w1(j)*Obj1(x);
z2 = w2(j)*Obj2(x);
end
n=2;
z = z1+z2+x(n:end)'*S*x(n:end);
end

function z = Obj2(x)
yref = 0;
p1 = -0.0053913;
p2 = 0.011457;
p3 = 0.039991;
p4 = -0.1433;
p5 = 0.070996;
p6 = 0.31844;
p7 = -0.64456;
p8 = 0.33539;
p9 = 0.30767;
p10 = -0.37571;
p11 = 0.071788;
mu = 50.5;
sigma = 29.011;
n=2;
g = (x(1:n-1)-mu)/sigma;
z = sum(((yref-(p1*g.^10+p2*g.^9+p3*g.^8+p4*g.^7+p5*g.^6+p6*g.^5+p7*g.^4+p8*g.^3+p9*g.^2+p10*g+p11)).^2)); 
end

function z = Obj1(x)
yref = 0;
p1 = -6.2039e-06;
p2 = 0.012729;
p3 = -0.040506;
p4 = -0.035217;
p5 = 0.28562;
p6 = -0.3329;
p7 = -0.19587;
p8 = 0.75971;
p9 = -0.54118;
p10 = -0.021845;
p11 = 0.098187;
mu = 50.5;
sigma = 29.011;
n=2;
g = (x(1:n-1)-mu)/sigma;
z = sum(((yref-(p1*g.^10+p2*g.^9+p3*g.^8+p4*g.^7+p5*g.^6+p6*g.^5+p7*g.^4+p8*g.^3+p9*g.^2+p10*g+p11)).^2));
end

Answer 1

[Matt J]

What is mysterious about the error message? It has told you that your gradc output has to be 4x12 and in your NLC2 you clearly only give it 2 rows.

if nargout>2
    gradc(1,:)=[gradient(Obj1(x)-x(n:end),x);gradient(-x(n:end),x)];
    gradc(2,:)=[gradient(Obj2(x)-x(n:end),x);gradient(-x(n:end),x)];
    gradceq=[];
end

Why are gradc(3,:) and gradc(4,:) omitted?

[Ali Esmaeilpour]

there are two objective functions hence there are two gradients.

[Bruno Luong]

Nope you have *one* objective function that is the sum of two terms (three if one count the weighted regularization term).

[Matt J]

No, the dimensions of gradc are required to be [M,N] where M is the number of unknown variables and N is the number of constraints. You have 4 unknown variables and you have created 12 constraints (see below), so gradc needs to be 4x12 as the error message has told you.

K>> whos c
  Name      Size            Bytes  Class     Attributes

  c         2x6                96  double       

[Ali Esmaeilpour]

eh... ok, but how to obtain those gradc(3,:) or gradc(4,:)???

the only thing I know is to define syms x and then put those Obj1 and Obj2 in gradient(Obj1,x) to have gradient of Obj1 with respect to x.

[Bruno Luong]

You constraints doesn't look right, you use Obj1 and Obj2 in the score then in the constraint by removing the variable x?

I don't know what you are trying to do but IMO the whole thing needs to be reformulated, not just the gradient calculation.

[Matt J]

If you don't know the formula for your gradients, then you should probably just set SpecifyConstraintGradient=false.

But similar to Bruno, your constraints look crazy to me. Before you do anything, I would start a new thread describing your problem in words and equations (not code).

[Ali Esmaeilpour]

why crazy? x(n:end) are slack variables. I mean I've devided my decision variable x to have main variables and slack variables hence x(n:end) are slack variables and x(1:n-1) are main decision variables. it's common in multi-objective optimization, but I still wonder why there must be gradc(3,:) and gradc(4,:) because when you say gradc(3,:) the only thing I can imagine is gradient of gradient of Obj1 etc.

function [c,ceq,gradc,gradceq]=NLC2(x)
n=2;
c=[Obj1(x)-x(n:end);-x(n:end);
   Obj2(x)-x(n:end);-x(n:end)];
ceq=[];
end

[Ali Esmaeilpour]

my optimization problem and constraints are as follows in the peicture:

[Matt J]

In the definition of the F_i(x), there is no x on the right hand side? Also what is ? Is this a known constant?

[Ali Esmaeilpour]

no we got delta'*S*delta in main problem where S is weighing matrix and delta is vector of slack variables which is x(n:end) in code and Fstar is constant. it is utopia value of F and it is constant.

[Matt J]

In the equation defining the F_i(x), there is no x on the right hand side. What is the dependence of F_i on x supposed to be?

[Ali Esmaeilpour]

F(x) is yref-y and y is closed loop response of your system. when you got your closed loop response you find the equation of it through basic fitting. I got 2 inputs and 2 outputs hence two F(x) etc.

[Matt J]

OK, well the first thing you need to do is write down a function that computes from a given parameter vector x=[x1,x2,x3,x4]

function y=closedloopresponse(x,i,k)

  %%% What goes in here????

end

Once that is specified, it will be easy to build the rest.

[Ali Esmaeilpour]

yeah done already. Obj1(x) and Obj2(x) and MyCost(x) are for that purpose.

[Matt J]

No, the function I asked you to write depends on x, i, and k. Obj1 and Obj2 are not in that form.

In any case, I don't want to refer to your original code. There are too many mistakes in it to know for certain what is intended and how things there are supposed to match up with your equations. For example, I don't see a variable called y in your code like there is in your equations. I don't see a variable called F. I don't see i, k, or delta,...

So let's start from scratch, starting with the functional form of y_ik(x):

function y=closedloopresponse(x,i,k)

  %%% What goes in here????
  
end