Difficulty in defining Aeq in linprog

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bus14
bus14 on 19 Apr 2019
Commented: bus14 on 19 Apr 2019
Hi community,
As I am now scaling up the problem with having two values for Z and 4 values of Y the dimensional problem becomes more difficult once again. As eye(2),eye(3)... does not work any more for adding values of Y. new Matrix A1[1 1 0 0;0 0 1 1] makes determining Aeq for me rather difficult as the transpose of A1 makes it difficult to calculate.
I tried writing the whole equation out in terms of V as you did above. But, where I am stuck is that Z(1) --> V(5) needs to be multiplied with 2 variables of A1 instead of only one variable as in the previous problem.
l =[0.25; 0.3];
q =[6;5];
s = [1;1.2;1.1;1];
A1 = [1 1 0 0; 0 0 1 1];
x =[20; 25; 28; 26];
d =[120;110];
f =[-s.',(l-q).'];
Aeq =[1 1 0 0;0 0 1 1, A1.'];
beq =x;
lb =[0, 0, 0, 0, 0, 0];
ub =[inf, inf, inf, inf, d(1), d(2)];
sol = linprog(f,[],[],Aeq,beq,lb,ub);
y = [sol(1);sol(2);sol(3);sol(4)]
z= [sol(5); sol(6)]
Hope you can help me out!
  1 Comment
bus14
bus14 on 19 Apr 2019
Objective function of this question is: Min (l-q).'*z-s.'*y where .' indicates the transpose.
the constraints are Y=x-A1.'*z 0<z<d , y.0 in whihc Z=[Z1;Z2] Y=[Y1;Y2;Y3;4]

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