MaplePrimes - answers and comments on Question, cardinal constraint in a {-1,0,1} Linear Program

Not linear

Robert Israel — Mon, 10 Jan 2011 00:11:24 Z

Maple is exactly right: the constraint is not linear. Well, you could this: for each i, have a new variable v[i] with constraints v[i] >= w[i] and v[i] >= -w[i], and then you can have a constraint add(v[i], i = 1 .. NC) <= 4.

Note that I said <= 4, not = 4, because v[i] >= w[i] and v[i] >= -w[i] just forces v[i] >= abs(w[i]), not v[i] = abs(w[i]). In fact, the feasible region ceases to be convex if you require the number of 1's and -1's to be equal to 4: in a convex problem, if w[i] = -1 and w[i] = 1 are allowed then w[i] = 0.

Second thought

Robert Israel — Mon, 10 Jan 2011 21:24:15 Z

On second thought, it is possible to make the number of 1's and -1's equal to 4 with integer linear programming. For each i, you have two new binary variables a[i] and b[i] with the constraints w[i] = a[i] - b[i], a[i] +b[i] <= 1 (so the possibilities are w[i] = -1 with a[i]=0, b[i]=1, w[i]=0 with a[i]=b[i]=0, w[i]=1 with a[i]=1, b[i]=0), and then
add(a[i] + b[i],i=1 .. NC) = 4.

Thanx for that Robert! In the end I did like

alex_01 — Mon, 10 Jan 2011 22:35:26 Z

Thanx for that Robert! In the end I did like this:

PS:=4:
seq(v[i] >= w[i], i = 1 .. NC), seq(v[i] >= -w[i], i = 1 .. NC),
add(v[i], i = 1 .. NC) <= 2*PS, add(w[i], i = 1 .. NC) = 0;

Then I get 4 long (1) and 4 short (-1) positions and the rest 0.

Thanx Robert :-) Works great! You are

alex_01 — Mon, 10 Jan 2011 01:59:49 Z

Thanx Robert :-) Works great!
You are stable as usual!