连续变量的GLPK优化问题总是返回零

Question

在连续变量的优化问题中，我使用了GLPK for python（pymprog模块）。结果不是我所期望的。

begin('optimzer')

X = var('X', 5)
k = par('k', [1, 1, 1, 1, 1])

0 <= X[0] <= 100
100 <= X[1] <= 200
400 <= X[2] <= 500
400 <= X[3] <= 500
400 <= X[4] <= 500

sum(k[i]*X[i] for i in range(len(k))) >= 150
minimize(sum(k[i]*X[i] for i in range(len((k)))), 'Profit')

solve()
save(mip='_save.mip')

结果就是这样

Problem:    optimzer
Rows:       1
Columns:    5 (0 integer, 0 binary)
Non-zeros:  5
Status:     INTEGER UNDEFINED
Objective:  Profit = 0 (MINimum)

   No.   Row name        Activity     Lower bound   Upper bound
------ ------------    ------------- ------------- -------------
     1 R1                          0           150               

   No. Column name       Activity     Lower bound   Upper bound
------ ------------    ------------- ------------- -------------
     1 X[0]                        0             0           100 
     2 X[1]                        0           100           200 
     3 X[2]                        0           400           500 
     4 X[3]                        0           400           500 
     5 X[4]                        0           400           500 

Integer feasibility conditions:

KKT.PE: max.abs.err = 0.00e+00 on row 0
        max.rel.err = 0.00e+00 on row 0
        High quality

KKT.PB: max.abs.err = 4.00e+02 on column 3
        max.rel.err = 9.98e-01 on column 3
        SOLUTION IS INFEASIBLE

End of output

如果我使用连续变量，为什么状态为INTEGER UNDEFINED？

如果约束应该> = 150，利润如何才能为0？

Answer 1

这不是真正的MIP模型，而是纯LP模型（glpk似乎有所区别）。您可以通过以下方式保存LP解决方案

save(sol='_save.sol')

这将显示：

Problem:    optimzer
Rows:       1
Columns:    5
Non-zeros:  5
Status:     OPTIMAL
Objective:  Profit = 1300 (MINimum)

   No.   Row name   St   Activity     Lower bound   Upper bound    Marginal
------ ------------ -- ------------- ------------- ------------- -------------
     1 R1           B           1300           150

   No. Column name  St   Activity     Lower bound   Upper bound    Marginal
------ ------------ -- ------------- ------------- ------------- -------------
     1 X[0]         NL             0             0           100             1
     2 X[1]         NL           100           100           200             1
     3 X[2]         NL           400           400           500             1
     4 X[3]         NL           400           400           500             1
     5 X[4]         NL           400           400           500             1

Karush-Kuhn-Tucker optimality conditions:

KKT.PE: max.abs.err = 0.00e+00 on row 0
        max.rel.err = 0.00e+00 on row 0
        High quality

KKT.PB: max.abs.err = 0.00e+00 on row 0
        max.rel.err = 0.00e+00 on row 0
        High quality

KKT.DE: max.abs.err = 0.00e+00 on column 0
        max.rel.err = 0.00e+00 on column 0
        High quality

KKT.DB: max.abs.err = 0.00e+00 on row 0
        max.rel.err = 0.00e+00 on row 0
        High quality

End of output

似乎save(mip='_save.mip')只有在模型中至少有一个离散变量时才有效。