GA in Matlab 2014???

Hi Everyone, I'm researching GA in Matlab for My thesis. I have a Problem with it. I solve Problem with 5 Vars: x1; x2; x3; x4; x5 I want to setup lower bound and upper bound for 5 Vars as follow: lb=[0.3 0.4 0.5 1 2] ub=[0.9 0.9 0.9 5 6] I want the x value: x1= 0.3; 0.35; 0.4;.....0.9 x2= 0.4; 0.45; 0.5;.....0.9 x3= 0.5; 0.55; 0.6;.....0.9 x4=1; 2; 3;...5 x5=2; 3; 4;...6 How do you do? Thank you so much.

[Joep]

Sorry needed to rewrite it to read your question.

Hi Everyone, I'm researching GA in Matlab for My thesis. I have a Problem with it. I solve Problem with 5 Vars: x1; x2; x3; x4; x5 I want to setup lower bound and upper bound for 5 Vars as follow:

lb=[0.3 0.4 0.5 1 2] 
ub=[0.9 0.9 0.9 5 6]

I want the x value:

x1= 0.3; 0.35; 0.4;.....0.9 
x2= 0.4; 0.45; 0.4;.....0.9 
x3= 0.4; 0.45; 0.5;.....0.9 
x4=1; 2; 3;...5 
x5=2;3;4...6

How do you do? Thank you so much.

x2 and x3 looks me wrong in your question.

Answer 1

[Mohammad Abouali]

Since you have discrete values for x1 to x5 you need to define it as an integer problem. (I know x1,x2,x3 are not integer input so read to the end).

IntCon in:

x = ga(fitnessfcn,nvars,A,b,[],[],LB,UB,nonlcon,IntCon,options)

IntCon list all the parameters that need integer value. which in your case would be IntCon=1:5 since all your variables need to be expressed as integer.

x4 and x5 are already integer so no problem there. but you need to change x1, x2,and x3.

define:

xx1=0.3:0.05:0.9
xx2=0.4:0.05:0.9
xx3=0.4:0.05:0.9

Now define the lower bound and upper bound as:

LB=[1 1 1 1 2]
UB=[numel(xx1) numel(xx2) numel(xx3) 5 6]

Your original function is defined as fitnessfcn(x1,x2,x3,x4,x5) but now by defining LB,UB, IntCon as mentioned above x1,x2,x3 are fed integer values which you don't want it. for example GA will varry x1 from 1 to 13. You need to change your function definition. Wherever, it was x1 you need to replace it with xx1(x1). This way, correct value is used. Let's say your original function was

fitnessfcn=@(x1,x2,x3,x4,x5) x1+x2+x3+x4+x5;

You need to change it to

fitnessfcn=@(x1,x2,x3,x4,x5) xx1(x1)+xx2(x2)+xx3(x3)+x4+x5;

this way, x1 varies from 1 to 13 but x1=1 will return xx1(1) which is 0.3. Therefore, practically x1=1 is mapped to 0.3, x1=2 is mapped to 0.35 and so on.

[KHANH NGUYEN]

Thank you so much. I'll consider it again.

[Mohammad Abouali]

You are welcome.