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我对 CPLEX 很陌生,我正在编写一个非常简单的模型,CPLEX 不想满足它。我知道我的模型是“冗长的”,因为我有简单地等于其他变量的变量,但这是我迈向更复杂模型的第一步,所以我想要这种方式。我不明白为什么这会让 CPLEX 感到不安。

我的模型是:

Minimize
obj: v1
Subject To

l1: i_AB1 + i_AC1 - i_AB2 - i_AC3 = 1
l2: - i_AB1 + I_BB = 0
l3: I_CA + I_CC = 0


e5:   i_AB2 + i_BD2 - I_BB = 0

e8:   - i_AC1 - I_CA = 0
e9:   i_AC3 + i_CD3 - I_CC = 0


\Indicator constraints

\For each connection XY

c1:  bAB = 1-> i_AB1 - 1 v1 = 0
c2:  bAB = 1-> i_AB2 - 1 v2 = 0

c5:  bAC = 1-> i_AC1 - 1 v1 = 0
c6:  bAC = 1-> i_AC3 - 1 v3 = 0

c9:  bBD = 1-> i_BD2 - 1 v2 = 0

c13: bCD = 1-> i_CD3 - 1 v3 = 0

c15: bAB = 1
c16: bAC = 1
c17: bBD = 1
c18: bCD = 1


Bounds
    0 <= v1 <= 1000
-1000 <= v2 <= 1000
-1000 <= v3 <= 1000


General

Binaries
bAB bAC bBD bCD
End

这显然没有解决方案(它有,或者至少这是我的意图,但 CPLEX 说不!)。

但是然后我将方程代e8l3并得到我想要的解决方案!这是代码:

Minimize
obj: v1
Subject To


\budget: 

l1: i_AB1 + i_AC1 - i_AB2 - i_AC3 = 1
l2: - i_AB1 + I_BB = 0
l3: - i_AC1  + I_CC = 0

e5:   i_AB2 + i_BD2 - I_BB = 0

\Row C
\e8:   - i_AC1 - I_CA = 0
e9:   i_AC3 + i_CD3 - I_CC = 0


\Indicator constraints

\For each connection XY

c1:  bAB = 1-> i_AB1 - 1 v1 = 0
c2:  bAB = 1-> i_AB2 - 1 v2 = 0

c5:  bAC = 1-> i_AC1 - 1 v1 = 0
c6:  bAC = 1-> i_AC3 - 1 v3 = 0

c9:  bBD = 1-> i_BD2 - 1 v2 = 0

c13: bCD = 1-> i_CD3 - 1 v3 = 0

c15: bAB = 1
c16: bAC = 1
c17: bBD = 1
c18: bCD = 1


Bounds
    0 <= v1 <= 1000
-1000 <= v2 <= 1000
-1000 <= v3 <= 1000


General

Binaries
bAB bAC bBD bCD
End

在我看来,两者都是完全相同的模型。我做错了什么,以至于第一个模型没有解决方案,即使它看起来等同于第二个有解决方案的模型?

顺便说一句,解决方案是:

Populate: phase I 
Tried aggregator 2 times.
MIP Presolve eliminated 4 rows and 4 columns.
Aggregator did 11 substitutions.
All rows and columns eliminated.
Presolve time =    0.00 sec.

Populate: phase II 
Solution status: 129.
Objective value of the incumbent: 1
Incumbent: Column v1:  Value =                 1
Incumbent: Column i_AB1:  Value =                 1
Incumbent: Column i_AC1:  Value =                 1
Incumbent: Column i_AB2:  Value =               0.5
Incumbent: Column i_AC3:  Value =               0.5
Incumbent: Column I_BB:  Value =                 1
Incumbent: Column I_CC:  Value =                 1
Incumbent: Column i_BD2:  Value =               0.5
Incumbent: Column i_CD3:  Value =               0.5
Incumbent: Column bAB:  Value =                 1
Incumbent: Column v2:  Value =               0.5
Incumbent: Column bAC:  Value =                 1
Incumbent: Column v3:  Value =               0.5
Incumbent: Column bBD:  Value =                 1
Incumbent: Column bCD:  Value =                 1

The solution pool contains 1 solutions.
0 solutions were removed due to the solution pool relative gap parameter.
In total, 1 solutions were generated.
The average objective value of the solutions is 1.

Solution        Objective   Number of variables
                value       that differ compared to
                            the incumbent
p1              1         0 / 15

问题本身甚至不是 MIP(因为我在这个初始版本中修复了我的布尔值,但它将是一个正确的 MIP)。这会改变什么吗?我真的不明白问题是什么。

谢谢

4

1 回答 1

1

我认为问题是由于默认情况I_CA下的下限为 0。在LP 格式文档的边界部分中,它说

上限和下限也可以单独输入……默认下限 0(零)和默认上限 +∞ 保持有效,直到明确更改边界。

如果您添加I_CA free到您的第一个 LP 文件的边界部分,那么它变得可行。

于 2016-07-10T18:10:25.867 回答