Help with LP

[zhli12]

Hi,

I'm having some trouble with using ProblemLP for linear programming. I tried to implement an example found in https://github.com/KardinalAI/coin_cbc/blob/master/examples/knapsack.rs, which should be a valid linear programming problem. However, I got some errors from cbc_cmd solver.

Attached below are the codes I use. The example I wanted to solve is a maximization problem, but it seems cbc_cmb supports only minimization problems, so I reverted the coefficients.

`// A = [2 8 4 2 5]
// x = [x x x x x]^T
// c^T = [5 3 2 7 4]

fn main() {
let mut A: CooMat = CooMat::::new(
(1 , 5),
vec![0, 0, 0, 0, 0],
vec![0, 1, 2, 3, 4],
vec![-2.0,-8.0,-4.0,-2.0,-5.0]
);

let mut x: Option<Vec<f64>> = Some(vec![1.0,0.0,0.0,1.0,1.0]);
let mut c = vec![-5.0, -3.0, -2.0, -7.0, -4.0];
let mut b = vec![-9.0];
let mut u = vec![0.0, 0.0, 0.0, 0.0, 0.0];
let mut l = vec![-2.0,-2.0,-2.0,-2.0,-2.0];
let mut p = vec![false, false, false, false, false];
let mut problem: numopt::problem::milp::ProblemMilp =  numopt::problem::milp::ProblemMilp::new(
    c,
    A,
    b,
    l,
    u,
    p,
    x
); 
let sol = numopt::solver::cbc_cmd::SolverCbcCmd::read_sol_file(
    "file", 
    &problem, 
    true);
for item in sol.iter(){
    println!("{:?}", item);
}`

The solution I got is
(Error, ProblemSol { x: [0.0, 0.0, 0.0, 0.0, 0.0], lam: [0.0], nu: [], mu: [0.0, 0.0, 0.0, 0.0, 0.0], pi: [0.0, 0.0, 0.0, 0.0, 0.0] })

However, this should have a non-zero solution. I'm not sure where the problem occurs and I'd really appreciate any help.

[ttinoco]

Hi, sorry for the late reply. I just saw this.
The problem you mention is not of the same form as ProblemMilp.
I also don't understand the need to use "read_sol_file". This should happen automatically "under the hood".
The problem you mention is of the form:

maximize c.T*x
subject to A*x <= b
and all x binary.

In contrast, ProblemMilp has the form

minimize c.T*x
subject to A*x = b
           l <= x <= y
and all or some x integer.

Hence, to represent the Knapsack problem you mention on the form of ProblemMilp, a transformation is required.
In particular, a new variable, say "y", needs to be introduced so we can can represent Ax <= b as [A -1](x,y) = 0 and then y <= b.
Alternative, you an use numopt's modeling layer, and define variables and construct expressions, instead of the lower level ProblemMilp data structure. See the example I just introduced here, which shows both approaches.
Let me know if you have further questions.