Defining variable model as x != 0 in JuMP

Specific Domains Optimization (Mathematical)
1

I am trying to run my first example of optimization using JuMP.

I have a problem with the definition of the values of x. I want to do x != 0 but it gives me an error. It only works for x >= 0 or x <= 0.

using JuMP
import Ipopt
model = Model(Ipopt.Optimizer)
@variable(model, y >= 0)
@variable(model, x != 0) # x >= 0 works
@NLobjective(model, Max, (x^2)*y)
@NLconstraint(model, x^2 + y^2 == 1)
print(model)
optimize!(model)
@show value(x) # sqrt(2/3)
@show value(y) # sqrt(1/3)

2

We may be able to model your problem as a mixed-integer problem. Can you be more specific? Why do you need x \neq 0?

3

It’s a weird thing to ask for a continuous variable to be not equal to a value. If x == 0 is optimal and you forbid it with a constraint, then 0.00000000000000001 is probably the next best value.

It is not possible in JuMP just because it’s not a thing one does in continuous optimization.
You’re not getting x==0 as a solution anyway.

4

Well, i’s possible that I am not thinking right. I know that the results are two points P1(sqrt(2/3); sqrt(1/3)) and P2(-sqrt(2/3); sqrt(1/3)). If I make x >= 0 I only have as a result P1 and if I do x <= 0 I have P2. If I don’t give any restriction to x I have a wrong result. So I thought That the solution could be x != 0.

5

You don’t have to add a bound constraint. Just leave it as a free variable.

What do you mean by wrong result?

6

I tried something like @variable(model, x) but the results are wrong: (0;0).

7

You need to check the problem status
Solutions · JuMP?

Your problem is nonconvex, so ipopt struggled to solve it.

Provide a different starting point

https://jump.dev/JuMP.jl/stable/manual/variables/#Start-values

8

Well, I tried several starting points 0.2, 1, 2, 10 but I don’t get the negative value for x. I tried for ex. @variable(model, x, start = 10).

9

You’ll have to set a negative starting point close to the solution. You’ll also have to provide a starting point for y that makes the solution feasible.

10

I tried

@variable(model, y >= 0, start = 1)
@variable(model, x, start = -1)

But I got only one point:

julia> @show value(x) # sqrt(2/3)
value(x) = -0.8164965800409512
-0.8164965800409512

julia> @show value(y) # sqrt(1/3)
value(y) = 0.5773502704448548
0.5773502704448548
11

Ipopt only finds a single solution. It does not find all solutions.

12

Thanks. Can you suggest an alternative solver for this problem?

13

To find all solutions? None of the JuMP solvers support this.

Why do you need it?

14

I am only studying a simple example, which I know the solutions previously. Is there an alternative package in Julia that gives all solutions?

I tried Wolfram Alpha with maximize x^2*y on x^2+y^2=1 and I got immediately the two solution points!

15

Try: https://github.com/JuliaIntervals/IntervalConstraintProgramming.jl

In general, mathematical optimization is the wrong tool to compute all solutions. Almost all solvers focus on providing an optimal solution, rather than all optimal solutions.

16

But Wolfram Alpha is based on Mathematica… Maybe the reason is that the solution is symbolic, not numerical.

17

Gurobi can find all solutions, I believe for quadratic problems as well, so need to split the higher order terms