Matlab Optimization

Question

Hello I am working with Matlab Optimization Solver and I am having problems with the program. I am getting this message fmincon stopped because the objective function value is less than the default value of the objective function limit and constraints are satisfied to within the default value of the constraint tolerance. Also I am getting the following message. Warning: Matrix is close to singular or badly scaled. The values the program is send out are not right. I need to min x and z. The function is as follows

  x*sqrt((16+y^2))+z*sqrt((1+y^2)).

Any help would be greatly appreciated. The follow constraints are as follows.

20*sqrt((16+y^2))-100000*y*x
80*sqrt((1+y^2))-100000y*z 
1<=y
y<=3
(x,z)transpose >0

function [c,ceq]=confun(x)
c=[(20*((16+x(2)^2)^.5))-100000*(x(2)*x(3));  (80*1+x(2)^2)^.5)-100000*x(2)*x(3); -x(2)+3; -x(2)+1; x(3)>0; x(1)>0];
ceq=[];

function f=obj(x)
f=x(1)*((16+x(2)^2)^.5)+x(3)*((1+x(2)^2)^.5);

x0=[1;1;1];
[x,fval]=fmincon(@obj,x0,[],[],[],[],[],[],@confun)

The first two lines aren't constraints (no =, <, > etc..)? What are those? (i.e. 20*sqrt((... — Matthew Gunn
This isn't the complete code, can you post a runnable version? — kmac

kmac kmac · Accepted Answer · 2015-12-02T04:49:31

Your main problem is that you put binary tests returning 1 on success in your non-linear constraint function, which expects <= 0 on success.

Your second problem is putting linear constraints in the nonlinear constraint function. This 'works', but fmincon can do some tricks with known-linear constraints, and you're not getting the benefit.

Below is the runnable example showing the results of

The broken nonlinear-only constraints
The fixed nonlinear-only constraints
The combination linear/nonlinear constraints

code:

function example()
  % initial conditions
  x0=[1;1;1];

  % linear constraint system (Ax <= b)
  A = [0 -1 0;
       0 1 0;
       -1 0 0;
       0 0 -1];
  b = [-1 3 0 0]';

  display('Run with broken nonlinear-only constraints');
  [x,fval]=fmincon(@obj,x0,[],[],[],[],[],[],@conBroken)
  display('Run with fixed nonlinear-only constraints');
  [x,fval]=fmincon(@obj,x0,[],[],[],[],[],[],@conFixed)
  display('Run with linear/nonlinear constraints');
  [x,fval]=fmincon(@obj,x0,A,b,[],[],[],[],@conNonlin)

% oops, x(...)>0 is 1 upon success
function [c,ceq]=conBroken(x)
  c=[(20*((16+x(2)^2)^.5))-100000*(x(2)*x(3));  (80*1+x(2)^2)^.5-100000*x(2)*x(3); -x(2)+3; -x(2)+1; x(3)>0; x(1)>0];
  ceq=[];

% good, -x(...) <= 0 when x>0
function [c,ceq]=conFixed(x)
  c=[(20*((16+x(2)^2)^.5))-100000*(x(2)*x(3));  (80*1+x(2)^2)^.5-100000*x(2)*x(3); -x(2)+3; -x(2)+1; -x(3); -x(1)];
  ceq=[];

% moved linear constraints to where they should be
function [c,ceq]=conNonlin(x)
  c=[(20*((16+x(2)^2)^.5))-100000*(x(2)*x(3)); (80*1+x(2)^2)^.5-100000*x(2)*x(3);];
  ceq=[];

% objective function
function f=obj(x)
  f=x(1)*((16+x(2)^2)^.5)+x(3)*((1+x(2)^2)^.5);

Which gives the following results:

Run with broken nonlinear-only constraints -- violates the x>0 constraint

Solver stopped prematurely.

fmincon stopped because it exceeded the function evaluation limit,
options.MaxFunEvals = 3000 (the default value).

x =
  -23.3480
    6.3806
    0.0470

fval =
 -175.5250

Run with fixed nonlinear-only constraints -- moving slowly in the right direction

Solver stopped prematurely.

fmincon stopped because it exceeded the function evaluation limit,
options.MaxFunEvals = 3000 (the default value).

x =
    0.1893
    1.8200
    0.5598

fval =
    1.9947

Run with linear/nonlinear constraints -- works!

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the function tolerance, and
constraints are satisfied to within the default value of the constraint tolerance.

x =
    0.0000
    1.9993
    0.0004

fval =
    0.0010

The lesson is: always provide your linear constraints separately, so the magic can happen.

Matlab Optimization

1 Answers