In a constrained optimization if "beq" is a matrix with a variable, then how to plot the variable in "beq" and the objective function?

I want to have plot of variable (m) value and objective function values (-fval) in "x","y" axis respectively, I can evaluate the values of "fval" but can't plot the desired results.

Following is my working programme:

    fun = @(x)sum((x.*(log(x))));
    A = [];
    b = [];
    x0 = [1/6 1/6 1/6 1/6 1/6 1/6];
    lb = [0,0,0,0,0,0];
    ub = [1,1,1,1,1,1];
    Aeq = [1,1,1,1,1,1;1,2,3,4,5,6];
    for m=1:0.5:6;
    beq = [1;m];
    [x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub);
    plot(m,-fval);
    end

[KSSV]

What difficulty you facing? What is your problem or error?

[Amritansu Ray]

Dear sir, I am not getting any plot after running the MATLAB code, which I have provided.

[Ameer Hamza]

@Amritansu, have you tried the code I gave in the answer?

[Amritansu Ray]

@Ameer, sorry for replying late. I am outside and don't have computer with me. I will surely run the code as early as possible when I am get back home. Thank you in advance for your help.

[Amritansu Ray]

@Ameer,suppose the variable "m" is a triangular fuzzy number(TFN) then to write the codes what can be done? Can you suggest me anything?

[Ameer Hamza]

It depends on how you want to use triangular fuzzy numbers here. If m is just a numeric vector then there is no difference.

[Amritansu Ray]

@Ameer, suppose m=TFN(1.75/3/4). So we will have sub-problems as below and the programme code will have two equation and there will be two solution.The best solution will be considered. Thank you again for your interest.

[Ameer Hamza]

The problem is still not quite clear. You have three values in m, and those are also shown on the x-axis. What are the values on the y-axis? From these two lines, what is the solution which you want?

[Amritansu Ray]

well a TFN or triangular fuzzy number (a/b/c) denotes three values a=most pessimistic,b=most common and c=most optimistic. A TFN is defined as below

 (a/b/c)=(x-a)/(b-a),  when x belongs to [a.b]  
        =(x-c)/(b-c),  when x belongs to [b.c]
        =0,            otherwise

So (1.75,3,4) will have two branches.And as a result the fval has two values and one is better.One thing to remember that here m is not increasing like m=1:0.5:6 ;so only one functional value will be found for each TFN. I hope I can write the codes, otherwise I will seek help from community. Thanks for your queries.

[Ameer Hamza]

To make the things simple, can you tell what value will m take in first and second subproblem? In first subproblem, the x change from [1.75, 3] and y from [0, 1]. So what does it mean to find the solution of the first subproblem? What are the values of m for which we need to optimize the objective function?

[Amritansu Ray]

@Ameer, I will explain later on.

[Amritansu Ray]

@Ameer,when we use beq = [1;m];where m=a triangular fuzzy number then then the given problem is a fuzzy optimisation problem and we generally set fuzzy goals for objective functions and also for the constraint related to fuzzy value. I know I am not able make you understand this but thank you very much for immense help regarding the code.

Answer 1

[Ameer Hamza]

You are not saving all the values of fval. Try this

fun = @(x)sum((x.*(log(x))));
A = [];
b = [];
x0 = [1/6 1/6 1/6 1/6 1/6 1/6];
lb = [0,0,0,0,0,0];
ub = [1,1,1,1,1,1];
Aeq = [1,1,1,1,1,1;1,2,3,4,5,6];
m=1:0.5:6;
fval = zeros(1, numel(m));
for ii=1:numel(m)
  beq = [1;m(ii)];
  [x,fval(ii)] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub);
end
plot(m,-fval);

[Amritansu Ray]

@Ameer, your code is running successfully,thank you very very much.