Fmincon - Variable dependent constraints

Hello,

I'm sorry if this question has already been answered.

I want to find the minimum of a function of three variables f(x,y,z), where the constraints of the variables are defined as follows (for example) :

• 1 < x < 2
• x/20 < y < x/10
• 2y < z < x/5

I would be greatful if someone could help me about how to define this by using the fmincon function.

Thanks in advance!

Answer 1

[Walter Roberson]

lb = [1, 1/20, 2/20]; ub = [2, 2/10, 2/5];
A = [
      1/20, -1, 0
     -1/10,  1, 0
         0,  2, -1
      -1/5,  0, 1
    ]
b = [
    0
    0
    0
    0
    ]

Aeq = []; beq = [];
xyz = fmincon(@OBJECTIVE, A, b, Aeq, beq, lb, ub)

Unless, that is, you need your endpoints to be strict inequalities. If you do, b would end up containing some +/- eps(realmin)

[BraF]

Hi Walter,

Thanks a lot ! Concerning the inequalities, it's okay if they're not strict so that's all good !

I've got some trouble understanding how the matrix A was constructed, and since the constraints above where just an example, I've got some difficulties re-creating it.

I have (omitting strict inequalities) :

• 0.05 < x < 0.5
• x/30 < y < x/10
• 5y < z < x/4

What does A become ? Thanks again!

P.S. Lb and ub become : lb = [0.05, 0.05/30, 5/30], ub = [0.5, 0.5/10, 0.5/4], right ?

[Walter Roberson]

x/30 < y < x/10

rewrite that into two parts,

x/30 < y
y < x/10

Now move the right sides to the left by subtraction, and reorder into consistent < order

x/30 - y < y - y
y - x/10 < x/10 - x/10

so

  1/30*x - 1*y < 0
 -1/10*x + 1*y < 0 

and read off the x y z coefficients to get the A matrix entries

 1/30, -1, 0
 -1/10, 1, 0

and the b entries are the constants to the right of the < -- so [0; 0]

[BraF]

Everything is clear now, thanks a lot to both of you!

And just to make sure, lb and ub are correctly defined now (with these new constraints) ?

I've got lb = [0.05, 0.05/30, 5/30], ub = [0.5, 0.5/10, 0.5/4].

[Walter Roberson]

You have x/30 < y < x/10 so if the lower bound for x is 0.05 then the lower bound for y is (as you calculated) 0.05/30 . But then you have 5y < z < x/4 so you should take 5 times the y lower bound as the z lower bound, which would get you to 5*0.05/30

Upper bound, x is max 0.5, y is up to x/10 which would be 0.5/10 (as you calculated). z upper bound is up to x/4 and x is up to 0.5, so z upper bound would be 0.5/4 (as you calculated)

So you were mostly right, but your lower bound on z was wrong.

Answer 2

[Matt J]

If you wish, you can download prob2matrices() and use the problem-based framework to help set up the linear constraints.

x=optimvar('x','LowerBound',1, 'UpperBound',2);
y=optimvar('y');
z=optimvar('z');

con.xyleft= x/20<=y;
con.xyright= y<=x/10;
con.yzleft= 2*y <= z;
con.yzright= z<=x/5;

[S,idx]=prob2matrices({x,y,z}, 'Constraints',con);

xyz=fmincon(@objective, [x0,y0,z0],  S.Aineq,S.bineq,S.Aeq,S.beq,S.lb,S.ub)