How to select the best result and the best variables from them

[RafaHPSUnicamp]

Hi, I just wanted to know how to select the best results from the Playtpus code that I adapted from LINGO software. Because the Playtpus gives me a lot of good results. For example, I wanted ONLY ONE result, and the best result. How I can do that from this code?

#variaveis

S_preco = 1069.23
F_preco = 1071.09
OB_preco = 2006.66
OR_preco = 2669.21
B_preco = 2540.47
S_custo = 533.17
F_custo = 569.89
OB_custo = 1384.39
OR_custo = 1466.34
B_custo = 2389.89

S_total = 2329278

S_percentoleo = 0.2057
C_percentoleo = 0.0064

OBF_percentoleo = 0.22

OR_massamolar = 873*0.000001
M_massamolar = 32*0.000001
B_massamolar = 292*0.000001
G_massamolar = 92*0.000001

S_capacidade = 3600000
OR_capacidade = 367200
B_capacidade = 887760*(880/1000)

S_demanda = 80638
F_demanda = 398984
OB_demanda = 164700
OR_demanda = 164700
B_demanda = 77634

from platypus import NSGAII, Problem, Real

def belegundu(vars):
    S_comercio = vars[0]
    F_comercio = vars[1]
    OB_comercio = vars[2]
    OR_comercio = vars[3]
    B_total = vars[4]
    S_insumo = vars[5]
    C_insumo = vars[6]
    OB_total = vars[7]
    OR_total = vars[8]
    OR_biodiesel = vars[9]
    MOL = vars[10]
    M_insumo = vars[11]
    G_comercio = vars[12]
    objs = [1*(S_comercio*S_preco - S_comercio*S_custo + F_comercio*F_preco - F_comercio*F_custo + OB_comercio*OB_preco - OB_comercio*OB_custo + OR_comercio*OR_preco - OR_comercio*OR_custo + B_total*B_preco - B_total*B_custo)]
    constrs = [
        S_total - S_comercio - S_insumo,
        S_insumo - C_insumo - F_comercio - OB_total,
        C_insumo - 0.04*S_insumo,
        OB_total - (F_comercio*OBF_percentoleo)/(1 - OBF_percentoleo),
        OB_total - OB_comercio - OR_total,
        OR_total - OR_comercio - OR_biodiesel,
        OR_biodiesel - MOL*OR_massamolar,
        M_insumo - 3*MOL*M_massamolar,
        B_total - 3*MOL*B_massamolar,
        G_comercio - MOL*G_massamolar,
        S_insumo - S_capacidade,
        OR_total - OR_capacidade,
        B_total - B_capacidade,
        S_comercio - S_demanda,
        F_comercio** - F_demanda,
        OB_comercio - OB_demanda,
        OR_comercio - OR_demanda,
        B_total - B_demanda
    ]
    return objs, constrs

problem = Problem(13, 1, 18)
problem.types[:] = [Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278), Real(0, 2329278)]
problem.constraints[0] = "==0"
problem.constraints[1] = "==0"
problem.constraints[2] = "==0"
problem.constraints[3] = "==0"
problem.constraints[4] = "==0"
problem.constraints[5] = "==0"
problem.constraints[6] = "==0"
problem.constraints[7] = "==0"
problem.constraints[8] = "==0"
problem.constraints[9] = "==0"
problem.constraints[10] = "<=0"
problem.constraints[11] = "<=0"
problem.constraints[12] = "<=0"
problem.constraints[13] = ">=0"
problem.constraints[14] = ">=0"
problem.constraints[15] = ">=0"
problem.constraints[16] = ">=0"
problem.constraints[17] = ">=0"
problem.function = belegundu
problem.directions[:] = Problem.MAXIMIZE

algorithm = NSGAII(problem)
algorithm.run(1000)

feasible_solutions = [s for s in algorithm.result if s.feasible]

for solution in algorithm.result:
    print(solution.objectives)

for solution in algorithm.result:
    print(solution.variables)

Remember that I adapted from this LINGO code:

MAX = L_soja + L_farelo + L_oleobruto + L_oleorefinado + L_biodiesel;

L_soja = S_comercio*S_preco - S_comercio*S_custo;
L_farelo = F_comercio*F_preco - F_comercio*F_custo;
L_oleobruto = OB_comercio*OB_preco - OB_comercio*OB_custo;
L_oleorefinado = OR_comercio*OR_preco - OR_comercio*OR_custo;
L_biodiesel = B_total*B_preco - B_total*B_custo;

!variaveis de preço e de custo (em R$/t);

S_preco = 1069.23;
F_preco = 1071.09;
OB_preco = 2006.66;
OR_preco = 2669.21;
B_preco = 2540.47;
S_custo = 533.17;
F_custo = 569.89;
OB_custo = 1384.39;
OR_custo = 1466.34;
B_custo = 2389.89;

!quantidade inicial de soja;

S_total = 2329278; !t;

!produção dos subprodutos;

S_total = S_comercio + S_insumo;
S_insumo = C_insumo + F_comercio + OB_total; !t;
C_insumo = 0.04*S_insumo;

S_percentoleo = 0.2057;
C_percentoleo = 0.0064;

OBF_percentoleo = 0.22;
OB_total = (F_comercio*OBF_percentoleo)/(1 - OBF_percentoleo);

!balanceamento das massas do óleo;

OB_total = OB_comercio + OR_total;

!balanceamento das massas do óleo refinado;

OR_total = OR_comercio + OR_biodiesel;

!estequiometria da reação do biodiesel (massas molares em kg/mol);

OR_biodiesel = MOL*OR_massamolar;
M_insumo = 3*MOL*M_massamolar;
B_total = 3*MOL*B_massamolar;
G_comercio = MOL*G_massamolar;

SUM_REAGENTES = OR_biodiesel + M_insumo;
SUM_PRODUTOS = B_total + G_comercio;

!constantes da reação estequiometrica (convertido de g/mol para t/mol) ;

OR_massamolar = 873*0.000001;
M_massamolar = 32*0.000001;
B_massamolar = 292*0.000001;
G_massamolar = 92*0.000001;

!restrições (capacidade maxima);

S_insumo <= S_capacidade;
OR_total <= OR_capacidade;
B_total <= B_capacidade;

!restrições (demanda mínima);

S_comercio >= S_demanda;
F_comercio >= F_demanda;
OB_comercio >= OB_demanda;
OR_comercio >= OR_demanda;
B_total >= B_demanda;

!constantes das restrições (em t);

S_capacidade = 3600000;
OR_capacidade = 367200;
B_capacidade = 887760*(880/1000); !Convertido de metros cubicos (m³) para kg, e consequentemente em toneladas;

S_demanda = 80638;
F_demanda = 398984;
OB_demanda = 164700;
OR_demanda = 164700;
B_demanda = 77634;

[EveCharbie]

@RafaHPSUnicamp did someone answer you?

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This issue is stale and will be closed soon. If you feel this issue is still relevant, please comment to keep it active. Please also consider working on a fix and submitting a PR.

[EveCharbie]

Just commenting so that the issue is not closed since we did not get an answer.

[github-actions]

This issue is stale and will be closed soon. If you feel this issue is still relevant, please comment to keep it active. Please also consider working on a fix and submitting a PR.