cant use linprog for maximize

I have my function fx= -0.818x1 + 0.241x2 + 4.737x3 + 0.2x4 + 0.483x5 - 0.02x6 + 0.06x7. And I have some constrains: 1. fx cant be larger than 100

2. 22=<x1=<40

3. 52=<x2=<80

4. 3.5=<x3=<5.9

5. 12=<x4=<22

6. 105=<x5=<150

7. 300=<x6=<700

8. 800=<x7=<1200

9. x4+x3=21

10. x5-x2=50

Please, show me the code of this optimization.

Answer 1

[John D'Errico]

Isn't this basically the same question you asked before? You have added a couple of equality constraints only.

I won't even bother to copy over the rest of your code from your last question. Just pass in these additional two variables now into linprog (where you previously passed in [] as placeholders when you had no equality constraints.)

Aeq = [0 0 1 1 0 0 0;0 -1 0 0 1 0 0];
beq = [21;50];

[Ivan]

man, I have some doubts of my code.

Basically linprog works for minimization. So for max, I have to invert my main function, which I did. But, do i have to invert this lines:

A = [-0.818 0.241 4.737 0.2 0.483 -0.02 0.06];

b = [100];

The question is:

[x,fval] = linprog(-f,A,b,[],[],lb,ub);

or

[x,fval] = linprog(-f,-A,-b,[],[],lb,ub);

[John D'Errico]

READ THE HELP!!!!!!! I'll paste in the relevant lines...

    min f'*x    subject to:   A*x <= b
     x

You changed the sign of f, ONLY because you wanted to maximize the function, not minimize. That changes nothing else though.

However, if you pass in -A and -b, then do you think that you will actually be trying to solve the problem

-A*x <= -b

which is equivalent to

A*x >= b

It seems to me that would not be solving the problem you have posed.

READ THE HELP! THINK!