fmincon stopped, Converged to an infeasible point.

Converged to an infeasible point.

fmincon stopped because the size of the current step is less than the default value of the step size tolerance but constraints are not satisfied to within the default value of the constraint tolerance.

if someone knows that this error is all about.. please guide me. Thank you.

Answer 1

[Walter Roberson]

Your constraints cannot be met.

Your cost function does not even use x(527) or x(528). The lower bounds and upper bounds for those are both 0, which forces the values to be 0. But maybe they are used in the constraints? Yes, they are used in row 43 of the constraints, as are a number of other variables. beq(43) = 0, so the sum of a number of values must be 0. Can any of the values be negative?

lb(find(Aeq(43,:)))

shows all zeros, and the only way for a sum of non-negative values to be zero is if they are all zero, so all of those variables must be zero.

But let's check... are there other variables whose lower and upper bounds are the same? Yes, it turns out there are. So we proceed to

Nx = length(x0);
X = sym('x', [1, Nx], 'real');
ident = lb == ub;
X(ident) = lb(ident);
eqcon = Aeq*X.' - beq;

Each entry in eqcon must be 0 for the constraints to be met, by definition of equality constraints. So examine it. It starts out plausible,

x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 + x22 + x23 + x24 - 33/25

x42 + x43 + x44 + x45 + x46 + x47 + x48 - 13/10

but then the third entry has

149/50

which cannot possibly be equal to zero. There are other rows that cannot be satisfied, such as 13/5 or 36/25 -- twelve in total.

What does the third constraint say?

mask = find(Aeq(3,:))
Aeq(3,mask)
beq(3)
lb(mask)
ub(mask)

mask =
    60    61    62
ans =
     1     1     1
ans =
                    1.49
ans =
                      1.49                      1.49                      1.49
ans =
                      1.49                      1.49                      1.49

This tells us that 1*x(60) + 1*x(61) + 1*x(62) == 1.49 and that x(60), x(61) and x(62) all have lower bounds of 1.49 and upperbounds of 1.49 and so all three values must be exactly 1.49 . But row 3 of Aeq with entry 3 of beq tells that the sum of the three must be 1.49. With the values having to be exactly 1.49 each, their sum must be 3*1.49 and that can never equal merely 1.49.

If you have R2017b or later, I recommend that you rewrite everything in terms of the new Problem Based optimization system, see https://www.mathworks.com/help/optim/ug/optim.problemdef.optimizationproblem.optimproblem.html and https://www.mathworks.com/help/optim/ug/optimization-expressions.html

[NIKET shah]

I tried writing it in problem based optimization system. But it uses only linprog and intlingprog as solver.

My objective have a quadratic part. That is:

"an*(x(191)^2)-an*(x(192)^2)-an*(x(309)^2)-an*(x(310)^2)-an*(x(311)^2)-an*(x(312)^2)-an*(x(405)^2)-an*(x(406)^2)-an*(x(407)^2)-an*(x(408)^2);"

how can i include it in "problem based optimization system."

can you guide me through this.

thank you for your time.

[Walter Roberson]

Ah, sorry about that, I thought that kind of approach was supported for fmincon as well.

[NIKET shah]

okay. Thank You can you even help me form this part. Like say i have x(60)+x(61)+x(62)=1.49; I only want one of the following variable to be 1.49 remaining 0. how can i form this part.

any help would be appreciated. Thank You.

[Walter Roberson]

I cannot think of any way to implement that without using nonlinear constraints.

[NIKET shah]

thanks

[NIKET shah]

Hey, I did break the code and trying to solve it part by part. Can you please let me know what is still going wrong.

Thank You.

[Walter Roberson]

Your revised code has

ub(92:94)=;

which is not valid syntax.

I compared the two versions, but I was figure out what value you intended there.

[Walter Roberson]

Using the same strategy as I showed above, we can show that Aeq(2,:) cannot be satisfied with your revised code.

Aeq(2,:) is 1 for the 7 locations 42, 43, 44, 45, 46, 47, 48 . lb() for those variables is the same as ub() for those variables, so those variables must each be exactly the bound, which is 0.2167. With there being seven of those 0.2167 being added, the sum is going to be 1.5169 . The sum of those values is required to be beq(2) = 1.3, which is just a small bit less than 6 * 0.2167 .

I guess that you are trying to express that 6 of the values must be 0.2167 and that the 7th must be 0, but as we discussed above you cannot do that with linear constraints.

[NIKET shah]

thank you..