How can I illustrate the objective function and their constraints in function file?

Constraint:

How can I write the objective function and their constraints in the "function file" in MATLAB?

In this case: C=5, S=3 and N=6

and Xcsn, Pcsn, Rcsn are the binary decision variables while initialYardcs is the parameters that have value in their matrix for example InitialYardcs=[1,0,0;0,0,0 ... ]

Thanks you for your help and your attention!!!

[Torsten]

Don't you have constraints on P and R ?

Are P and R free variables to be determined by the optimizer ?

[Binh]

Yes, I have constraint for P and R. But in my exercise, we ignore the P and R, and only focus on 3 constraints to analyze the results.

[Torsten]

Choose P_csn = 0 for all c,s and n and R_csn = X_csn. Then you get the minimum value for the objective function, namely 0.

[Binh]

Thank you for your answer. But I don't know how to write it in function file

[John D'Errico]

It sounds like you are looking for a complete tutorial on how to use MATLAB, how to build matrices and work with them, and how to write functions. For us to write that would be a waste of time, since there are tutorials existing on how to use MATLAB, and we would want to spend a lot of time teaching you things, some of which you may already know. Essentially, you need to learn MATLAB.

Start with the MATLAB Onramp tutorials.

[Binh]

Thank you for your advice, I'm still newbie in this.

Answer 1

[Torsten]

I'm sure you can use a loop to define the constraints, and I'm sure you will find out how.

C = 5;
S = 3;
N = 6;
prob = optimproblem;
P = optimvar('P',[C,S,N],'Type','integer','LowerBound',0,'UpperBound',1);
R = optimvar('R',[C,S,N],'Type','integer','LowerBound',0,'UpperBound',1);
X = optimvar('X',[C,S,N],'Type','integer','LowerBound',0,'UpperBound',1);
prob.Objective = sum(P,'All');
Xstart = randi(2,C,S,1)-1

Xstart = 5×3

1 1 1 0 0 0 0 0 1 0 0 0 0 1 0

prob.Constraints.c1 = X(:,:,1)-Xstart-P(:,:,1)+R(:,:,1)==0 ;
prob.Constraints.c2 = X(:,:,2)-X(:,:,1)-P(:,:,2)+R(:,:,2)==0 ;
prob.Constraints.c3 = X(:,:,3)-X(:,:,2)-P(:,:,3)+R(:,:,3)==0 ;
prob.Constraints.c4 = X(:,:,4)-X(:,:,3)-P(:,:,4)+R(:,:,4)==0 ;
prob.Constraints.c5 = X(:,:,5)-X(:,:,4)-P(:,:,5)+R(:,:,5)==0 ;
prob.Constraints.c6 = X(:,:,6)-X(:,:,5)-P(:,:,6)+R(:,:,6)==0 ;
sol = solve(prob)

Solving problem using intlinprog.
LP:                Optimal objective value is 0.000000.                                             


Optimal solution found.

Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0. The intcon variables are integer within tolerance,
options.IntegerTolerance = 1e-05.

sol = struct with fields:
    P: [5×3×6 double]
    R: [5×3×6 double]
    X: [5×3×6 double]

sol.P

ans = 
ans(:,:,1) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,2) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,3) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,4) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,5) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,6) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0

sol.R

ans = 
ans(:,:,1) =

     1     1     1
     0     0     0
     0     0     1
     0     0     0
     0     1     0


ans(:,:,2) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,3) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,4) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,5) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,6) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0

sol.X

ans = 
ans(:,:,1) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,2) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,3) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,4) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,5) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0


ans(:,:,6) =

     0     0     0
     0     0     0
     0     0     0
     0     0     0
     0     0     0