linprog

Dear all, I am dealing with linprog function and have some difficulties with it. The function I need to optimize is: f=1.44-0.05x-0.04y, subject to x>=0, y>=0, y>= -x +10, y<=-x+12 , y<=(1/3)x.

Where do I include the constant value from the objective function 1.44?

Thanks a lot.

Nina

Answer 1

[Andrei Bobrov]

Removed first variant (17:20 MDT[09:20 EDT])

Hi Nina! Adjustment for the right answer (ADD 13.09.2011 14:20 MDT [06:20EDT])

f = [ 0.05; 0.04];
A = [-1 -1; 1 1];
b = [-10; 12;];
Aeq = [-1/3 1];
beq = 0;
lb = [0; 0;];
[x,fval,exitflag,output,lambda] = linprog(f,A,b,Aeq,beq,lb)

ADD2 13.09.2011 (14:25 MDT [06:25EDT])

f = [ 0.05; 0.04;];
A = [-1 -1 ; 1 1 ; -1/3 1];
b = [-10; 12; 0];
lb = [0; 0;];
[x,fval,exitflag,output,lambda] = linprog(f,A,b,[],[],lb)

Answer 2

[Nina]

Thanks Andrei, but it's wrong :(

The right solution should be 7.5 for x and 2.5 for y.

Why do you put the whole equation on the left side (in matrix A) and don't leavi it for b?

Answer 3

[Aurele Turnes]

If you have R2017b, you can use the new problem-base approach. It will take care of the constant value for you: https://www.mathworks.com/help/optim/problem-based-lp-milp.html