Problem in solving fmincon (2024)

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Matt J

2012-05-31 23:03:12 UTC

Post by Fevzi
Hello,
I want to solve a nonlinear constrained optimization problem. I prepared 3 m-files which are for constraints, objective function and calling optimization toolbox.
lb = [12.5;0.000460;0.005];
ub = [75;0.000786;0.160];
A =[];
Aeq= [];
b = [];
beq = [];
x0= [32;0.00058;0.1];
[c,ceq] = microcons(x)
function [c,ceq] = microcons(x)
b1 = -152.103;
b2 = 2.16514;
b3 = 1113.81;
b4 = 204245;
b5 = -0.0179843;
b6 = -3391.78;
b7 = -248075399;
b8 = -7.60179;
b9 = 1242.21;
b10 = -45237.6;
w = 0.500;
n = w/x(3);
D = 0.4;
N = (x(1)*1000)/(pi*D);
vL = 12.5;
vU = 75;
fzL = 0.000460;
fzU = 0.000786;
DocL = 0.005;
DocU = 0.160;
TlL = 0;
TlU = 20.5;
nL = 10000;
nU = 60000;
Tl = b1+x(1)*b2+x(3)*b3+x(2)*b4+(x(1)^2)*b5+(x(3)^2)*b6+(x(2)^2)*b7+x(1)*x(3)*b8+x(1)*x(2)*b9+x(2)*x(3)*b10;
c = [vL-x(1);x(1)-vU;fzL-x(2);x(2)-fzU;DocL-x(3);x(3)-DocU;TlL-Tl;Tl-TlU;nL-N;N-nU];
ceq = [(n*x(3))-w];
function f = microobj(x)
b1 = -152.103;
b2 = 2.16514;
b3 = 1113.81;
b4 = 204245;
b5 = -0.0179843;
b6 = -3391.78;
b7 = -248075399;
b8 = -7.60179;
b9 = 1242.21;
b10 = -45237.6;
ae = 0.7;
D = 0.4;
A = 3.2;
k = (A-D)/(2*ae*D);
L = ((4*k*(k+1)*ae*D) + (2*ae*D));
dL = (k-1)*sqrt(2)*ae*D;
Lc = L + dL;
Lr = (sqrt(2)*(A-D))/2;
w = 0.500;
n = w/x(3);
N = (x(1)*1000)/(pi*D);
z = 2;
frCut = x(2)*z*N;
Tl = b1+x(1)*b2+x(3)*b3+x(2)*b4+(x(1)^2)*b5+(x(3)^2)*b6+(x(2)^2)*b7+x(1)*x(3)*b8+x(1)*x(2)*b9+x(2)*x(3)*b10;
Tm = (Lc/frCut)*n;
c0 = 1;
CM = c0* Tm;
Ts = 3;
CS = c0*Ts;
Tc = 3;
CR = c0*Tc*(ceil(Tm/Tl));
tC = 50;
CT = tC*(ceil(Tm/Tl));
f = CM + CS + CR + CT;
However, the tool box ignores "TlL = 0; TlU = 20.5;" limits and finds a negative Tl value. (When the solution is found, the value of 7th row in c vector is positive instead of negative. When I calculated Tl value using x1, x2, x3 and b1 to b10 values, Tl value is sufficiently large negative number). Then I wrote "TlL = 0.0000000001;" (1*10^-10) instead of "TlL = 0;", toolbox finds a reasonable result. Finally I increased the decimals to test whether toolbox always consider the TlL limit or not. When I tested the program for "TlL = 0.0000000000000001;" (1*10^-16), the program again gives a problematic result. I couldn't understand the reason of this problem. If it was a problem with # of decimal digits, it should not give a reasonable results for 1*(10^(-15)) or 1*(10^(-13))

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You haven't shown us the magnitude of the negative number, so there's no reason yet to think the negatives are significantly large. Was the constraint violated by an amount greater than the TolCon tolerance parameter?

Also, you also haven't shown us the exitflag output of FMINCON. It might tell you that convergence failed for some reason.

Finally, if it's important to you to satisfy the bounds all the time then try using the sqp algorithm, or the interior-point algorithm with the AlwaysHonorConstraints option active.

Sargondjani

2012-06-01 05:18:10 UTC

just small advice on programming: i think it is more convenient to pass all your parameters into the two m-files (as inputs) instead of defining them inside your m-file. you can do it like this:

m-file:
function y=fun_obj(x,par1,par2...)

and then you define the values of par1, etc. before calling mfincon. you can call fmincon like this:

fmincon(@(x)fun_obj(x,par1,par,..),A,b,.....)

since you have quite a few parameters you can also store them in a structure:
par.b1=...
par.b2=...

and then your m-file can just just par as an input:
function y=fun_ob(x,par)