efficient solution/non-dominated point missing with MOA.Lexicographic()

[xgandibleux]

For the following 2IP problem:

max z_1  =  x_1  +  x_2  
min z_2  =  x_1  +  3x_2
s.t. 
      2x_1  +   3x_2  <= 30 
      3x_1  +   2x_2  <= 30 
       x_1  +    x_2  <= 5.5 
       x_1  ,    x_2  \in  \mathbb{N}

The lexicographic algorithm returns only one solution, the x=(0,0) corresponding to z=(0,0) is missing.
The code is the following:

using JuMP
import MultiObjectiveAlgorithms as MOA
using GLPK
using Printf


# Setup the model ------------------------------------------------------------
model = Model( )

# Define the variables 
@variable(model, x1≥0, Int)
@variable(model, x2≥0, Int)

# Define the objectives 
@expression(model, fct1, x1 + x2)      # to maximize
@expression(model, fct2, x1 + 3 * x2)  # to minimize
@objective(model, Max, [fct1, (-1) * fct2])

# Define the constraints 
@constraint(model, 2*x1 + 3*x2 ≤ 30)
@constraint(model, 3*x1 + 2*x2 ≤ 30)
@constraint(model,   x1 -   x2 ≤ 5.5)


# Setup the solver (lexicographic method) ------------------------------
set_optimizer(model,()->MOA.Optimizer(GLPK.Optimizer))
set_attribute(model, MOA.Algorithm(), MOA.Lexicographic())

        
# Solve and display results ---------------------------------------------------
optimize!(model)

for i in 1:result_count(model)
    z1_opt = objective_value(model; result = i)[1]
    z2_opt = -1 * objective_value(model; result = i)[2]
    x1_opt = value(x1; result = i)
    x2_opt = value(x2; result = i)   
    @printf( "%2d  :  z = [%3.0f ,%3.0f]  |  x1 =%2.0f  x2 =%2.0f\n", i, z1_opt, z2_opt, x1_opt, x2_opt )    
end

returns:

 1  :  z = [ 12 , 24]  |  x1 = 6  x2 = 6

The decision space and the objective space for this 2IP are:

[xgandibleux]

In looking carefully the implementation of the lexicographic algorithm, the problem that I reported is fixed using the option all_permutations = true as follow:

set_attribute(model, MOA.Algorithm(), MOA.Lexicographic(all_permutations = true))

I would suggest to setup the option to 'true' by default, since invoking 'a lexicographic resolution' it is usual to treat the $p$ objectives of the problem, and not only one.