Insufficient number of outputs from right hand side of equal sign to satisfy assignment

I've seen other questions and answers on this topic, but could not find the answer to my problem.

function [c,ceq] = nonlcon(r)
s = [1 2   0.18     0.0997;
    2	3   0.3     0.102;
    3	4   0.5     0.1045;
    4	5   0.5     0.1064];
vv = 6;
ceq(1) = (s(1,1)-r(1)).^2+(s(1,2)-r(2)).^2+(s(1,3)-r(3)).^2-vv.*(s(1,4)-r(4)).^2==0;
ceq(2) = (s(2,1)-r(1)).^2+(s(2,2)-r(2)).^2+(s(2,3)-r(3)).^2-vv.*(s(2,4)-r(4)).^2==0;
ceq(3) = (s(3,1)-r(1)).^2+(s(3,2)-r(2)).^2+(s(3,3)-r(3)).^2-vv.*(s(3,4)-r(4)).^2==0;
ceq(4) = (s(4,1)-r(1)).^2+(s(4,2)-r(2)).^2+(s(4,3)-r(3)).^2-vv.*(s(4,4)-r(4)).^2==0;
c = [] ;
end

r0 = [2.5	5  0	0];
ob =@(r)(r(1)-3).^2+(r(2)-4).^2+(r(3)-0.18).^2;
fun=@(r)@nonlcon;
[x,fval,exitflag,output] = fmincon(ob,r0,[],[],[],[],[],[],fun)

Why i am getting this error? Hope someone can enlighten me.

Error jisuan>@(r)@nonlcon (Line 10)
fun=@(r)@nonlcon;

Error fmincon (Line 654)
        [ctmp,ceqtmp] = feval(confcn{3},X,varargin{:});

Error jisuan (Line 11)
[x,fval,exitflag,output] = fmincon(ob,r0,[],[],[],[],[],[],fun)

Answer 1

[Walter Roberson]

fun=@nonlcon;

[艳晖 刘]

then

Error fmincon (Line 726)
FMINCON requires all values returned by functions to be of data type double.

could you enlighten me what is the reason for that behaviour? ;-(

[Walter Roberson]

format long g
ob = @(r)(r(1)-3).^2+(r(2)-4).^2+(r(3)-0.18).^2;
fun = @nonlcon;
syms R [1 4]
OB = ob(R)

OB = 

[~, CON] = fun(R);
CON.'

ans = 

rsol = solve(CON, R)

rsol = struct with fields:
    R1: [2×1 sym]
    R2: [2×1 sym]
    R3: [2×1 sym]
    R4: [2×1 sym]

OBsym = subs(OB, rsol)

OBsym = 

vpa(OBsym)

ans = 

Rvals = double([rsol.R1, rsol.R2, rsol.R3, rsol.R4])

Rvals = 2×4

20415.3549275606 -20273.4450023606 201.836889 11746.5901166667 -20274.4450023606 20416.3549275606 201.836889 11746.5901166667

r0 = [10000 -10000 100 100];

[x,fval,exitflag,output] = fmincon(ob,r0,[],[],[],[],[],[],fun)

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.

x = 1×4

12949.427373901 -12856.0045345482 118.146542083117 7449.97622233855

fval = 
          333003614.481346

exitflag = 
     1

output = struct with fields:
         iterations: 12
          funcCount: 77
    constrviolation: 7.0093612074852
           stepsize: 584.703209541039
          algorithm: 'interior-point'
      firstorderopt: 0.00869249988681986
       cgiterations: 0
            message: '↵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.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 3.357104e-07,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 3.505205e-08, is less than options.ConstraintTolerance = 1.000000e-06.↵↵'
       bestfeasible: []

ob(x)

ans = 
          333003614.481346

[~, x_eq] = fun(x)

x_eq = 1×4

6.9841560125351 5.45746749639511 7.0093612074852 6.02073603868484

Although fmincon considered the constraints to be satisfied, I don't think I can agree.

ob(Rvals(1,:))

ans = 
           827879675.01335

[~, Rval_eq] = fun(Rvals(1,:))

Rval_eq = 1×4

2.38418579101562e-07 1.19209289550781e-07 -1.19209289550781e-07 2.38418579101562e-07

Whereas the symbolic solution gives much much closer to satisfying.

function [c,ceq] = nonlcon(r)
s = [1 2   0.18     0.0997;
    2	3   0.3     0.102;
    3	4   0.5     0.1045;
    4	5   0.5     0.1064];
vv = 6;
ceq(1) = (s(1,1)-r(1)).^2+(s(1,2)-r(2)).^2+(s(1,3)-r(3)).^2-vv.*(s(1,4)-r(4)).^2;
ceq(2) = (s(2,1)-r(1)).^2+(s(2,2)-r(2)).^2+(s(2,3)-r(3)).^2-vv.*(s(2,4)-r(4)).^2;
ceq(3) = (s(3,1)-r(1)).^2+(s(3,2)-r(2)).^2+(s(3,3)-r(3)).^2-vv.*(s(3,4)-r(4)).^2;
ceq(4) = (s(4,1)-r(1)).^2+(s(4,2)-r(2)).^2+(s(4,3)-r(3)).^2-vv.*(s(4,4)-r(4)).^2;
c = [] ;
end

[艳晖 刘]

！！！

I get it!

thank you!!!

[Walter Roberson]

Note that your constraints have four equations in four variables, so they resolve down to two particular solutions. The only question then is whether one of the sets of solutions is better than the other, and the answer to that is No, they are both the same.

Answer 2

[Fangjun Jiang]

nonlcon() is a defined function so fun=@nonlcon should be enough. It might be easier if the function has only one return variable. The returned "c" is empty anyway so you might just delete "c" in nonlcon()

[Steven Lord]

The returned "c" is empty anyway so you might just delete "c" in nonlcon()

No, that would not work. fmincon imposes some requirements on the nonlinear constraint function you pass into it. The nonlinear constraint function must return two outputs, c and ceq, containing the nonlinear inequality and nonlinear equality constraints on the problem respectively. Returning an empty c and a nonempty ceq indicates you have nonlinear equality constraints but no nonlinear inequality constraints.

If you omitted the c output fmincon would treat the nonlinear equality constraints as nonlinear inequality constraints. As one example solving a problem where the solution vector is required to have a norm of exactly 1 (on a unit circle) can have a different solution than one where the solution vector is required to have a norm no greater than 1 (inside a unit circle.)

[Fangjun Jiang]

@Steven Lord, thank you for the note! I didn't know it and learned this today!