MATLAB: Non Linear optimization problem in MATLAB

MATLABnonlinearoptimization

I have following problem which I am using MATLAB to solve it seems to me pertty Please check Is this correct or NOT?

Question

objective function (objfun2.m)

function f = objfun2(x)
    f = 7.9 * 0.000000001 * 300 * pi * (x(1)^2 - x(2)^2);
end

Non linear constraint (confun2.m)

function [c, ceq] = confun2(x)
%Nonlinear inequality constraints
 sqrt1=sqrt((1.84*100000* (pi/4)*(x(1)^4-x(2)^4))/(7.9*0.000000001*pi*(x(1)^2-x(2)^2)));
 sqrt2= sqrt(1+(x(3)*300^3)/(3 * 1.84 * 100000 *(pi/4)*(x(1)^4-x(2)^4)));
 c = -(((1.875^2)/(300^2 * 2*pi))*sqrt1*sqrt2)+230 ; % since it should have x<m form
% Nonlinear equality constraints
ceq = [];

Main Code

LB=[9.2,6,0.1];
UB=[15,9.8,18];
A=[];
b=[];
Aeq=[];
beq=[];
x0=[10,9,0.15];
% Make a starting guess at the solution
%options = optimoptions(@fmincon,'Algorithm','sqp');
[x,fval] = fmincon(@objfun2,x0,A,b,Aeq,beq,LB,UB,@confun2);

My Solution (Not sure it is correct or NOT)

 x= 9.7603    9.7603   16.8680

How would I check this?

Best Answer

As Matt points out, the solution you found may have issues.

You can see the source of the problem by looking at your nonlinear constraint. There we see you have the difference (x1^2 - x2^2) in the denominator. So when x1 is approximately equal to x2, things go to hell.

You can fix part of that, by factoring the terms in the first radical.

There we see a fraction with

(x1^4 - x2^4)/(x1^2 - x2^2)

this reduces to (x1^2 + x2^2), so no 0/0 there.

The second radical does have a divide by zero, when x1==x2. So again, this will be a problem. You will be able to get almost any value you want for that term, by TINY changes to x1 and x2.

Did the optimizer converge? Hard to tell. I'start the optimization at a different location though, since you still have a singular constraint.

If your goal is to minimize (x1^2-x2^2) you can do so by maximizing x2, as well as minimize x1. (Drop the constants out front! A minimization does not care about them, and at best, they may just introduce numerical problems.)

So your start point should be roughly:

[9.2 9.8 , ?]

I put a question mark in there for x3, because x3 does not appear in the objective. So we can choose a start point for x3 by trying to satisfy the nonlinear constraint.

So now lets look more closely at the constraint. We know that x3 lies in [0.1, 18].

First, lets simplify the constraint.

K1 = 1.875^2/(300^2*2*pi)*sqrt(1.84e5/4/7.9e-9)
K1 =
        15.0018747945489
K2 = 300^3/(3*1.84e5*pi/4)
K2 =
        62.2780212098721

So we can write the entire nonlinear constraint in the form of a function handle:

CON = @(x123) K1*sqrt(x123(1)^2 + x123(2)^2)*sqrt(1 + K2*x123(3)/(x123(1)^4 - x123(2)^4)) - 230;

Will this function ever be greater than zero, when x1 and x2 are at their best locations?

CON([9.2 9.8 .1])
ans =
        -28.653992101973
CON([9.2 9.8 18])
ans =
       -93.8657086630407
ezplot(@(x3) CON([9.2 9.8 x3]),[.1 18])

So at what I chose as the best point for x1 and x2, no value of x3 will allow this constraint to be positive. But if I choose other values for x1 and x2, as long as x1>x2, there are solutions that allow con to be positive.

ezplot(@(x3) CON([9.8 9.7 x3]),[.1 18])
grid on

Things look very different when x1 is slightly less than x2 however.

ezplot(@(x3) CON([9.6 9.7 x3]),[.1 18])
grid on

In fact, it appears there may be no solution at all when x1 is less than x2. I'd need to look more assiduously at the problem to know if that is true.

But it is clearly true that [9.8 9.7 2] is a feasible point. So what does fmincon find?

Define CONFUN as:

function [c, ceq]  = confun(x123)
K1 = 1.875^2/(300^2*2*pi)*sqrt(1.84e5/4/7.9e-9);
K2 = 300^3/(3*1.84e5*pi/4);
c = K1*sqrt(x123(1)^2+x123(2)^2)*sqrt(1+K2*x123(3)/(x123(1)^4-x123(2)^4))-230;
c = -c;
ceq = [];

I negated CON in there.

LB=[9.2,6,0.1];
UB=[15,9.8,18];
A=[];
b=[];
Aeq=[];
beq=[];
x0=[9.8,9.7,2];
[x,fval,exitflag] = fmincon(@(x123) x123(1)^2 - x123(2)^2,x0,A,b,Aeq,beq,LB,UB,@CONFUN)
Local minimum possible. Constraints satisfied.
fmincon stopped because the size of the current step is less than
the default value of the step size tolerance and constraints are 
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
x =
          9.53501396574518          9.53501396574518          2.80228092041589
fval =
       2.8421709430404e-14
exitflag =
     2

Fmincon wants to push to the boundary where x1 == x2. The constraint is satisfied at x.

CON(x)
ans =
        1144008644.36073

But, clearly, the solution is a bit of a singular one, since it lies right along the edge where x1==x2, so we have an effective divide by zero.

Related Solutions

MATLAB: How to minimize a nonlinear function

There are issues with the definition of 'myfun'. Compare it with the following to see the errors.

function f = myfun(x)
x1 = x(1);
x2 = x(2);
x3 = x(3);
f = (1./(1+(x1-x2).^2)) + sin(0.5*pi.*x2.*x3) + exp(((-x1+x)./x2)-2).^2;
end

MATLAB: Constraint optimization problem, fmincon, converged to an infeasible point.-

Your last 4 constraints, [0.2-x(1); x(1)-6*(1+0.2*S); 0.2*(1-0.1*S)-x(2); x(2)-6]

are all linear constraints of one variable each. You should convert them to upperbound and lowerbound constraints instead, which will make the search much more efficient.

A = []; b = []; 
S=0.0092;
lb = [0.2, 0.2*(1-0.1*S)];
ub = [6*(1+0.2*S), 6];
[x, fval] = fmincon(@objfun, x0, A, b, Aeq, beq, lb, ub, @confun, options);

Now look at your second constraint, (P/(x(1)+x(2)*sqrt(2)))-17.5*(1+0.1*S)) . With a brief examination you can see that since x(1) and x(2) are both positive, x(1)+x(2)*sqrt(2) will never be 0, so you can rewrite this as a linear constraint

(-(0.1) * (17.5) * S - 17.5) * x1 + (-sqrt(2) * 0.1 * 17.5 * S - sqrt(2) * 17.5) * x2 + P <= 0

which can be coded in the linear constraints as

A = [-(0.1) * (17.5) * S - 17.5), (-sqrt(2) * 0.1 * 17.5 * S - sqrt(2) * 17.5)];
b = [-P];

and you are now down to two non-linear constraints.

Putting these together,

E=2;
h=2;
S = 0.0092;
P = 25*(1+0.1*S);
A = [-(0.1) * (17.5) * S - 17.5), (-sqrt(2) * 0.1 * 17.5 * S - sqrt(2) * 17.5)];
b = [-P];
Aeq = []; beq = [];
lb = [0.2, 0.2*(1-0.1*S)];
ub = [6*(1+0.2*S), 6];
x0 = [3 3]; %Starting guess
options = optimset('Algorithm','active-set');
objfun = @(x) ((P*h)/E)*(1/(x(1)+sqrt(2)*x(2)));
[x, fval] = fmincon(objfun, x0, A, b, Aeq, beq, lb, ub, @(x) confun(x, S, P), options);
%Solution x and function value fval
yopt=x(1); zopt=x(2);

with no separate file for objfun, and with

function [c, ceq] = confun(x, S, P)
%Non-linear constraints
%all the other constraints moved to bounds constraints or linear constraints
    c = [(P * (x(2)+x(1)*sqrt(2)) / (x(1)^2*sqrt(2) + 2*x(1)*x(2))) - 17.5;
         (P/x(1)^2*sqrt(2) + 2*x(1)*x(2)) - 1.2];
    ceq=[];
end

I make no promises that this will allow the equation to be solved. I think I have found there is no solution with those constraints, but the proof of that will wait for another posting.

Best Answer

Related Solutions

MATLAB: How to minimize a nonlinear function

MATLAB: Constraint optimization problem, fmincon, converged to an infeasible point.-

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