i want my fun value to be zero and also should satisfy the constraints,i uesd fmincon command

MAIN FUNCTION

function [f] = newfrac(x)

wcg=2.1; %gain cross over frequency

bb=4;

cc=60;

r = x(1) + (x(2)*(wcg^(-x(4)))*cos(90*x(4)*(pi/180)))+ (x(3)*(wcg^(x(5)))*cos(90*x(5)*(pi/180)));

s = - x(2)*wcg^(-x(4))*sin(90*x(4)*pi/180)+ x(3)*wcg^(x(5))*sin(90*x(5)*pi/180);

fun1=bb/sqrt((cc*wcg)^2 +1);

fun2=sqrt((r)^2 + (s)^2);

f = 10*log10(fun1*fun2);

end

CONSTRAINTS PROGRAM-TO BE SATISFIED BY ABOVE FUNCTION

function [c, ceq] = confun(x)

wcg=2.1;

wt=10;

ws=0.01;

aa=2;

bb=4;

cc=60;

dd=80;

ff=0;

fipm=80; %phase margin

r = x(1) + ( x(2)*(wcg^(-x(4)))*cos(90*x(4)*(pi/180)) )+(x(3)*(wcg^(x(5)))*cos(90*x(5)*(pi/180)));

% pi/180 give degree value in radians

s = - x(2)*(wcg^(-x(4)))*sin(90*x(4)*(pi/180))+ x(3)*(wcg^(x(5)))*sin(90*x(5)*(pi/180));

ru = -x(2)*x(4)*(wcg^(-x(4)-1))*cos(90*x(4)*(pi/180))+ x(3)*x(5)*(wcg^(x(5)-1))*cos(90*x(5)*(pi/180));

su = -x(2)*x(4)*(wcg^(-x(4)-1))*sin(90*x(4)*(pi/180))+ x(3)*x(5)*(wcg^(x(5)-1))*sin(90*x(5)*(pi/180));

rt = x(1) + x(2)*(wt^((-x(4))))*cos(90*x(4)*(pi/180))+ x(3)*(wt^x(5))*cos(90*x(5)*(pi/180));

st = -x(2)*(wt^(-x(4)))*sin(90*x(4)*(pi/180))+x(3)*(wt^x(5))*sin(90*x(5)*(pi/180));

rs = x(1) + x(2)*(ws^(-x(4)))*cos(90*x(4)*(pi/180))+ x(3)*(ws^x(5))*cos(90*x(5)*(pi/180));

ss = -x(2)*(ws^(-x(4)))*sin(90*x(4)*(pi/180))+ x(3)*(ws^x(5))*sin(90*x(5)*(pi/180));

%1 equation(41)

c1 = (atan(s/r)*180/pi)-(atan(dd*wcg)*180/pi)-(wcg*ff)+180-fipm;

% 180/pi give radian value in degrees

%2 equation(42)

c21 = 1/(1+(s/r)^2);

c22 = (su*r-s*ru)/(r)^2;

c23 = dd/(1+(dd*wcg)^2);

ceq = [c1;c21*c22-c23-ff];

%3 equation(43)

c31 = aa*sqrt((rt^2) + (st^2));

c32 = sqrt((1+aa*rt)^2 + (dd*wt+aa*st)^2);

c3 = (10*log10(c31/c32))+20;

%4 equation(44)

c41 = sqrt((cc*ws)^2 +1);

c42 = sqrt((1+bb*rs)^2+(cc*ws+bb*ss)^2);

c4 = (10*log10(c41/c42))+20;

c=[c3;c4];

end

SIR THESE TWO ARE PROGRAMS I HAVE WRITTEN IN MATLAB,THE COMMANDS I USED ARE

>> f_obj=@(x)newfrac(x)^2;

>> x0=[10;10;10;0.5;0.5];

>> options=optimset('Algorithm','interior-point');

>> [x,fval]=fmincon(f_obj,x0,[],[],[],[],[0;0;0;0;0],[inf;inf;inf;0.9;0.9],@confun,options)

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 were satisfied to within the default value of the constraint tolerance.

criteria details

x =

    1.2403
    3.0557
    0.4837
    0.4822
    0.9000

fval =

104.6186

>> [x,fval]=fmincon(@newfrac,x0,[],[],[],[],[0;0;0;0;0],[inf;inf;inf;0.9;0.9],@confun,options)

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 were satisfied to within the default value of the constraint tolerance.

criteria details

x =

    0.0000
    0.7587
    0.1891
    0.7928
    0.9000

fval =

-22.0702

[kintali narendra]

sorry sir, i dont know this earlier ,i wont repeat

[Walter Roberson]

Duplicate is at http://www.mathworks.com/matlabcentral/answers/38680-what-changes-i-need-to-make-for-better-answer-fmincon

The Question itself has already been merged there, but the Answer needs to be moved there.

Answer 1

[Matt Kindig]

Perhaps your optimization is getting stuck in a local minimum. Try a different set of initial guesses (x0)-- does this help?

Also, in your second optimization (the one with just @newfrac) there is nothing forcing the optimizer to stop once fval=0. You can try to use the absolute value instead, i.e.

    f _obj = @(x) abs(x)
  [x,fval]=fmincon(f_obj,x0,[],[],[],[],[0;0;0;0;0],[inf;inf;inf;0.9;0.9],@confun,options)

[Sargondjani]

dont use abs because this is not differentiable. better use ^2 as he did in first...

[kintali narendra]

sir,i already used ^2,i have posted results also..............

sir ,in the above question i wrote my whole program and the output i am getting ,please do suggest some changes so that i can get better output

Answer 2

[Elizabeth]

Hello sir, I obtained what I assume to be the correct result for your program:

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.
<stopping criteria details>
x =
    0.3348
    2.5203
    0.5623
    0.5791
    0.8251
fval =
  1.5559e-013

I obtained this result my manipulating the logarithm function in you objective function. You can now write it as follows, if you agree with my adjustments:

function [f] = newfrac(x)
wcg=2.1; %gain cross over frequency
bb=4;
cc=60;
r = x(1) + (x(2)*(wcg^(-x(4)))*cos(90*x(4)*(pi/180)))+ (x(3)*(wcg^(x(5)))*cos(90*x(5)*(pi/180)));
s = - x(2)*wcg^(-x(4))*sin(90*x(4)*pi/180)+ x(3)*wcg^(x(5))*sin(90*x(5)*pi/180);
fun1=bb/sqrt((cc*wcg)^2 +1);
fun2=sqrt((r)^2 + (s)^2);      
fun_x=fun1*fun2;
origf=10*log10(fun_x); 
f= 10^origf;
end

PROOF: ***********************************************************************

===> For purposes of our discussion, Let:

    x = fun_x = fun1*fun2
    b = 10 (our base of the log)
    k = 10 (the constant)
    y= origf (the original expression for the objective function)

===> Therefore, the original function f that you had can be written as

    y=k*logb(x) (EQ1)

===> By the properties of logarithmic functions, then

     y= b^[klogb(x)]=x (EQ2)

===> Meanwhile, the following theorem must hold true

    y=k*logb(x) if and only if b^y=x EQ(3)

===> Using EQs (1-3) and substitution, one can see--

     y=k*logb(x)=b^[klogb(x)]

===> Or, in other words,

    y=origf=10*log10(fun_x) EQ(4)
    iff y=10^[10*log10(fun_x)=f (EQ5)

===> and so, we note EQ(4) is equivalent to EQ(5), but EQ(4) is ill-conditioned in comparison to EQ(5). ****************************************************************************8

• In conclusion, the only issue I think you need to correct is to rewrite your objective function in the form of EQ(5)