Supplied objective function must return a scalar value

I have to minimize a function with FMINCON:

f = @(x,y) k*x.*(teta_in-y)

i create this script to apply fmincon:

x0 = [2000,10];
A = [1 0;0 -1];
b = [25;2273];
f = cell(2,1)
f{1} = @(x) k*x.*(teta_in-y)
f{2} = @(x) k*x.*(teta_in-y)
[xmin,fval] =fmincon(f,x0,A,b)

But the programme give me the error: Supplied objective function must return a scalar value

If you have a different way to minimize that function with fmincon i appreciate that

[Dyuman Joshi]

What are the values of k and teta_in?

Why do you use the same function twice?

Also, share the mathematical definition of the objective function you have to minimize.

[EDOARDO GELMI]

K and teta_in are two constants. Previously i had an error which said that i had to use a double in order to solve the problem with fmincon, so i created a cell (i searched on the internet but i wasn't sure about it).

What do you mean by "share the mathematical definition..."? I don't understand your request.

Thank you!

Answer 1

[Star Strider]

The function needs to return one parameter vector. One way to do that is to create a second function to map the inpouts to to such a vector:

ffcn = @(b)f(b(1),b(2));

Then, providing random values for the missing constants ‘k’ and ‘teta_in’ and running it produces —

k = rand

k = 0.0666

teta_in = rand

teta_in = 0.3083

f = @(x,y) k*x.*(teta_in-y);
ffcn  = @(b)f(b(1),b(2));

x0 = [2000,10];
A = [1 0;0 -1];
b = [25;2273];

[xmin,fval] =fmincon(ffcn,x0,A,b)

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.

xmin = 1×2

1.0e+07 * 0.0000 7.7169

fval = -1.2840e+08

.

[EDOARDO GELMI]

I tried to use a solution which was similiar to the one you are offering but maybe it wasn't correct at all, what @(b) stands for? Why i cannot use @(x) f(x(1),x(2))?

I really appreciate your answer, now i try to use it and i'll let you know.

Thanks!

[EDOARDO GELMI]

It works so i really want to thank you. Nevertheless i cannot understand why it didn't work when i used a code like this @(x) f(x(1),x(2)), isn't it the same?

[Dyuman Joshi]

"what @(b) stands for?"

It's a placeholder. You can use any valid variable instead.

"Why i cannot use @(x) f(x(1),x(2))?"

You can. But since you have used x already in defining f before, to avoid confusion a different placeholder is used.

k = rand

k = 0.0208

teta_in = rand

teta_in = 0.9567

f = @(x,y) k*x.*(teta_in-y);
ffcn  = @(x)f(x(1),x(2));

x0 = [2000,10];
A = [1 0;0 -1];
b = [25;2273];

[xmin,fval] =fmincon(ffcn,x0,A,b)

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.

xmin = 1×2

1.0e+15 * 0.0000 1.2777

fval = -6.6576e+14

[EDOARDO GELMI]

Understood, thank you very much for the explanation!

[Torsten]

@Dyuman Joshi

You can. But since you have used x already in defining f before, to avoid confusion a different placeholder is used.

b is also in use :

b = [25;2273];

:-)

[Dyuman Joshi]

@Torsten, I meant x as a placeholder, whereas b is a variable, but your point is valid nonetheless.

Best to use different names (as much as possible) to avoid confusion.

[Matt J]

Understood, thank you very much for the explanation!

@EDOARDO GELMI If so, and everything is clear now, you should Accept-click the answer

[EDOARDO GELMI]

I already did! I cannot find another button

[Star Strider]

Thank you!

In:

ffcn = @(b)f(b(1),b(2));

the ‘b’ argument is local to the function, and does not refer to any other value outside it. (I generally use ‘b’ for parameters because standard optimisantion convention is to define parameters as β, however any variable that makes sense in the context would work.)

.

[EDOARDO GELMI]

It makes sense, thank you very much!

[Star Strider]

As always, my pleasure!