Optimisation of function containing matrix

I am having trouble optimising this system. I have been trying to use fmincon without success.

C is a 16x16 matrix, mo and v and 1 are 16x1 vectors. I want to choose mo to maximise x = c*mo+v

The constraints are: the components of mo must add to one (I think I've done this one)

and each component of x should be equal, that is x1=x2=...=x16 where x=c*mo+v (I have no idea how to represent this)

fun = @(mo) -C*mo-v;

A=[];
b=[];
Aeq= ones(1,16);
beq= 1; 
lb = zeros(16,1);
ub = ones(16,1);
m0 = 1/16*ones(16,1);
mo = fmincon(fun,m0,A,b,Aeq,beq,lb,ub)

Please let me know where I'm going wrong...

[Fabio Freschi]

Can you share C and v?

[Isabelle Steinmann]

C =

  Columns 1 through 10

    1.0003    0.0245    0.0052    0.0172    0.0273    0.0220    0.0171    0.0208    0.0073    0.0218
    0.0054    0.9994    0.0100    0.0326    0.0329    0.0369    0.0173    0.0155    0.0224    0.0269
    0.0152    0.0097    1.0001    0.0173    0.0374    0.0220    0.0222    0.0208    0.0174    0.0219
    0.0001   -0.0000    0.0000    1.0005    0.0255    0.0054    0.0103    0.0101    0.0007    0.0057
    0.0001   -0.0000    0.0000    0.0056    1.0006    0.0103    0.0052    0.0050    0.0055    0.0204
    0.0001   -0.0003   -0.0000    0.0059    0.0108    1.0007    0.0009   -0.0097    0.0156    0.0054
    0.0001   -0.0001   -0.0000    0.0112    0.0110    0.0255    1.0005    0.0001    0.0359    0.0105
    0.0001   -0.0001   -0.0000    0.0112    0.0110   -0.0041    0.0257    1.0004    0.0262    0.0156
    0.0000   -0.0003   -0.0000    0.0205    0.0008    0.0005    0.0005    0.0002    1.0005    0.0003
    0.0001   -0.0001   -0.0000    0.0157    0.0011    0.0105    0.0056    0.0053    0.0108    1.0004
    0.0001   -0.0002   -0.0000    0.0111    0.0210    0.0157    0.0162    0.0152    0.0213    0.0009
    0.0000   -0.0003   -0.0000    0.0107    0.0154    0.0007    0.0150   -0.0000    0.0155    0.0054
    0.0050   -0.0104   -0.0001    0.0213    0.0212    0.0164    0.0456    0.0151    0.0169    0.0159
   -0.0001   -0.0200   -0.0001    0.0095   -0.0004    0.0092   -0.0102   -0.0052   -0.0006    0.0047
    0.0000   -0.0101   -0.0001    0.0102    0.0054    0.0151    0.0006   -0.0099    0.0002    0.0001
    0.0100   -0.0000    0.0000    0.0158    0.0309    0.0058    0.0007    0.0005    0.0057    0.0158

  Columns 11 through 16

    0.0213    0.0179    0.0317    0.0165    0.0165    0.0064
    0.0068    0.0479    0.0367    0.0270    0.0316    0.0213
    0.0216    0.0228    0.0316    0.0167    0.0164    0.0163
    0.0005    0.0055    0.0002    0.0003    0.0002    0.0155
    0.0007    0.0007    0.0003    0.0003    0.0002    0.0203
    0.0152    0.0307    0.0204    0.0109    0.0107    0.0002
    0.0158    0.0264    0.0008    0.0110    0.0007    0.0104
    0.0158    0.0212    0.0054    0.0009    0.0055    0.0106
    0.0149    0.0205    0.0053    0.0203    0.0053    0.0002
    0.0205    0.0060    0.0056    0.0005    0.0054    0.0152
    1.0009    0.0214    0.0204    0.0010    0.0058    0.0006
    0.0002    1.0008    0.0004    0.0153    0.0200    0.0005
    0.0159    0.0169    1.0008    0.0257    0.0202    0.0009
   -0.0150    0.0041    0.0043    0.9998   -0.0103   -0.0002
    0.0002    0.0151    0.0199    0.0004    1.0003   -0.0000
    0.0154    0.0108    0.0057    0.0104    0.0054    1.0008
    
    v is a zero vector

[Fabio Freschi]

It is better to attach data as mat file

Answer 1

[Fabio Freschi]

While waiting for the data, I try to solve one of the questions. you can set the missing constraint as

x1-x2 = 0

x2-x3 = 0

...

so, let's call M the matrix with +1 -1 along the diagonal

M = diag(ones(16,1),0)+diag(-ones(15,1),1);

The last row has to be removed

% remove last row
M = M(1:end-1,:);

now, x = C*mo+v and we want M*x = 0, so Mx = M*C*mo+M*v = 0 and the new equality constraint is

% new constraints
Aeq2 = M*C;
beq2 = -M*v;

% append
Aeq = [Aeq; Aeq2];
beq = [beq; beq2];

However, the main problem is that your objective function returns a vector and not a scalar quantity. I don't know your problem but is the sum of all x a feasible objective function to maximize?

fun = @(mo) -sum(C*mo+v);

In this case the solution is

>> mo

mo =

    0.0549
    0.0481
    0.0524
    0.0668
    0.0672
    0.0644
    0.0615
    0.0627
    0.0661
    0.0655
    0.0623
    0.0653
    0.0570
    0.0737
    0.0689
    0.0633

>> sum(mo)

ans =

     1

>> C*mo

ans =

    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720
    0.0720

[Isabelle Steinmann]

Will this work though? The variable I am optimising is mo and x is a function of mo.

So the optimisation constraint is Aeq.mo=beq

So that would give mo(1)=...=mo(16) but i was x(1)=...=x(16) where x=C*mo+v

Sorry, I know my choice of variables makes this confusing

[Fabio Freschi]

You are right, sorry for the misunderstanding. I edited the answer

[Fabio Freschi]

Isabelle, is it what you want?

[Isabelle Steinmann]

Yes, that's perfect. Thank you