具有连续变量的 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 GLPK for python（pymprog 模块）。结果出乎我的意料。

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')

结果是这样的

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？

在应该 >= 150 的给定约束下，利润如何为 0？

Answer 1

这并不是真正的 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') 仅在模型中至少存在一个离散变量时才有效。