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

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kintali narendra
kintali narendra on 18 May 2012
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

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Answers (2)

Matt Kindig
Matt Kindig on 18 May 2012
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)
  2 Comments
kintali narendra
kintali narendra on 18 May 2012
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

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Elizabeth
Elizabeth on 30 Jul 2012
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)

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