Thread Subject: GA optimization

Hi all,

I have solved the given problem described below by GA but
my results are real numbers. How can I get results in
discrete values? They should be whitin the range:

Ri=(60:1:90);
Ro=(80:1:110);
A=(1:0.5:3);
F=(600:10:1000);
Z=(2:1:10);

Is it possible to get results in discrete values when using
gamultiobjective in GA toolbox at all? If yes could you
please explain it to me how to do it and what should I do.
I'm just matlab user and I'm not good in programming at all
some detailed description as for dumbbell :-) would be
great.

Thank You

Problem:
five design variables: x=[Ri;Ro;A;F;Z]

objective functions:

f(1)= pi*(Ro^2-Ri^2)*A*(Z+1)*Rho; % min weight
f(2)=Jz*omega/(Mh+Mf)*1000; % min time

subject to:

linear constraints:
  Aineq=[-1 0 0 0 0;0 1 0 0 0;1 -1 0 0 0;0 0 -1 0 0;...
0 0 1 0 0;0 0 0 0 1;0 0 0 0 -1;0 0 0 -1 0;0 0 0 1 0];
bineq=[-55;110;-20;-1.5;3;9;-1;0;1000];

nonlinear constraints:
where c=<0
c=[-Lmax+(Z+1)*(A+delta);prz-pmax;s*Ms-Mh;...
-th;prz*Vsr-pmax*Vsrmax;Vsr-Vsrmax];
                                                 
ceq=[];

where:

Rimin=55;
Romax=110;
deltaR=20;
Amin=1.5;
Amax=3;
delta=0.5;
Lmax=30;
Zmax=10;
Vsrmax=10000;
mi=0.5;
Rho=0.000007850;
s=1.5;
Ms=40000;
Mf=3000;
n=250;
pmax=1;
Jz=55;
Fmax=1000;

and

S=pi*(Ro^2-Ri^2);
prz=F/S ;
Rsr=(2/3)*((Ro^3-Ri^3)/(Ro^2-Ri^2));
Vsr=(pi*Rsr*n)/30;
omega=(pi*n)/30;
Mh=2/3*mi*F*Z*((Ro^3-Ri^3)/(Ro^2-Ri^2));

I could be wrong, but if you just rewrote your objective
function to round each variable, wouldn't it find the
optimum integer solution? (It would still return non integer
results but rounding them would give the same solution).

> I have solved the given problem described below by GA but
> my results are real numbers. How can I get results in
> discrete values?

This is called "Integer Programming". Check out the
following link:

http://www.mathworks.com/support/solutions/data/1-
10PDHC.html?solution=1-10PDHC

Hello - you can solve this problem with glcDirect, minlpBB and more in
TOMLAB.

The demo is available from here: http://tomopt.com/tomlab/

Best wishes, Marcus

Here is some quick TOMLAB code. It doesn't look like your formulation
is complete since th is missing:

Anyway, it might be a starting point for you. In general I would
recommend that you code the gradient and use MAD for the 2nd order
derivatives:

function Result = newsok

% Ri=(60:1:90);
% Ro=(80:1:110);
% A=(1:0.5:3);
% F=(600:10:1000);
% Z=(2:1:10);
% x=[Ri;Ro;A;F;Z]
% Need 2 here and divide by 2 in funcs
x_L = [60;80;2;600;2];
x_U = [90;110;6;1000;10];

x_0 = round((x_L+x_U)/2);

c_U = zeros(5,1);

Aineq=[-1 0 0 0 0;0 1 0 0 0;1 -1 0 0 0;0 0 -1 0 0;...
    0 0 1 0 0;0 0 0 0 1;0 0 0 0 -1;0 0 0 -1 0;0 0 0 1 0];
bineq=[-55;110;-20;-1.5;3;9;-1;0;1000];
% x(3) is not twice as high, need to divide column by 2
Aineq(:,3) = Aineq(:,3)/2;
Prob = minlpAssign(@objfun, [], [], [], x_L, x_U, 'NewsOK', x_0, ...
    1:5, [], [], [], Aineq, -inf*ones(length(bineq),1), bineq, ...
    @consfun, [], [], [], [], c_U);
Prob.user.Rho=0.000007850;
Prob.PriLevOpt = 4;
Prob.optParam.IterPrint = 1;
Result = tomRun('minlpBB', Prob, 1);
Result.x_k(3) = Result.x_k(3)/2;

function f = objfun(x,Prob)
x(3) = x(3)/2;
Ri = x(1);
Ro = x(2);
A = x(3);
Z = x(5);
Rho = Prob.user.Rho;
f(1)= pi*(Ro^2-Ri^2)*A*(Z+1)*Rho; % min weight

%f(2)=Jz*omega/(Mh+Mf)*1000; % min time

function c = consfun(x,Prob)
x(3) = x(3)/2;
Ri = x(1);
Ro = x(2);
A = x(3);
F = x(4);
Z = x(5);

% Rimin=55;
% Romax=110;
% deltaR=20;
% Amin=1.5;
% Amax=3;
delta=0.5;
Lmax=30;
% Zmax=10;
Vsrmax=10000;
mi=0.5;
s=1.5;
Ms=40000;
% Mf=3000;
n=250;
pmax=1;
% Jz=55;
% Fmax=1000;
S=pi*(Ro^2-Ri^2);
prz=F/S ;
Rsr=(2/3)*((Ro^3-Ri^3)/(Ro^2-Ri^2));
Vsr=(pi*Rsr*n)/30;
% omega=(pi*n)/30;
Mh=2/3*mi*F*Z*((Ro^3-Ri^3)/(Ro^2-Ri^2));

c=[-Lmax+(Z+1)*(A+delta);prz-pmax;s*Ms-Mh;...
prz*Vsr-pmax*Vsrmax;Vsr-Vsrmax];
% -th; removed since not defined

Best wishes, Marcus
Tomlab Optimization Inc.
http://tomopt.com/tomlab/