运筹优化学习03：Lingo非唯一最优解问题--油气开采构造处最优选择问题

目录

1 基本问题求解

1.1 示例数据：

1.2 模型构建：

1.3 代码实现

1.3.1 Latex代码：

1.3.2 Lingo代码：

1.3.4 解决方案

2 拓展问题求解--柴油发电或电网供电

2.1 约束表达：

2.2 Latex代码：

3 拓展问题--增加新约束

3.1 约束表达：

3.2 Latex代码：

Lingo源码

​3.3 结果展示

4 总结梳理

---

某油田在10个有油气构造处要选择若干个钻探采油,设第j 个构造开采时需花费aj元，投产后预计年收益为cj元，若尤其投资总额为b元，问题应选择哪几个构造开采最为有利。

1 基本问题求解

1.1 示例数据：

构造处	1	2	3	4	5	6	7	8	9	10
	1000	1300	1260	1500	1420	1670	1340	1190	1720	1640
	800	940	920	880	960	1000	870	840	1020	980

B = 7200；

1.2 模型构建：

1.3 代码实现

1.3.1 Latex代码：

max = \sum_{i=1}^{10}x_i * c_i
\sum_{i=1}^{10}x_i * a_i \leq b
x_i \in \{0,1\}
a_i , c_i > 0 

1.3.2 Lingo代码：

model:
sets:
CONSTRUCT/1..10/: C, A, X;
endsets

data:
a = 1000	1300	1260	1500	1420	1670	1340	1190	1720	1640;
c = 800	940	920	880	960	1000	870	840	1020	980;
b = 7200;
enddata

max = @sum(CONSTRUCT : x * c);

@sum(CONSTRUCT : x * a) <= b;
@for(CONSTRUCT : @bin(X));

end

1.3.4 解决方案

Global optimal solution found.   Objective value:                              4720.000   Objective bound:                              4720.000   Infeasibilities:                              0.000000   Extended solver steps:                               0   Total solver iterations:                             0                            Variable           Value        Reduced Cost                               B        7200.000            0.000000                           C( 1)        800.0000            0.000000                           C( 2)        940.0000            0.000000                           C( 3)        920.0000            0.000000                           C( 4)        880.0000            0.000000                           C( 5)        960.0000            0.000000                           C( 6)        1000.000            0.000000                           C( 7)        870.0000            0.000000                           C( 8)        840.0000            0.000000                           C( 9)        1020.000            0.000000                          C( 10)        980.0000            0.000000                           A( 1)        1000.000            0.000000                           A( 2)        1300.000            0.000000                           A( 3)        1260.000            0.000000                           A( 4)        1500.000            0.000000                           A( 5)        1420.000            0.000000                           A( 6)        1670.000            0.000000                           A( 7)        1340.000            0.000000                           A( 8)        1190.000            0.000000                           A( 9)        1720.000            0.000000                          A( 10)        1640.000            0.000000                           X( 1)        0.000000           -800.0000                           X( 2)        1.000000           -940.0000                           X( 3)        1.000000           -920.0000                           X( 4)        1.000000           -880.0000                           X( 5)        1.000000           -960.0000                           X( 6)        0.000000           -1000.000                           X( 7)        0.000000           -870.0000                           X( 8)        0.000000           -840.0000                           X( 9)        1.000000           -1020.000                          X( 10)        0.000000           -980.0000                               Row    Slack or Surplus      Dual Price                               1        4720.000            1.000000                               2        0.000000            0.000000

2 拓展问题求解--柴油发电或电网供电

2.1 约束表达：

定义变量yj标识开采电力来源，若为柴油机发电取值为1；若为电网供电，取值为0

2.2 Latex代码：

\sum_{i=1}^{10} x_i * y_i * d_i \leq f
\sum_{i=1}^{10} 0.3 * x_i * (1 - y_i) * d_i \leq p

3 拓展问题--增加新约束

3.1 约束表达：

约束a     

约束b     

约束c     

约束d    

约束e    

3.2 Latex代码：

0 \leqslant  x_6 - x_8 \leq 1
0 \leqslant  x_5 + x_8 \leq 1
1 \leq x_2 + x_4 \leq 2
x_7 = x_8
x_1 + x_4 + x_6 + x_9 \leq 2

Lingo源码

model:
sets:
CONSTRUCT/1..10/: C, A, X;
endsets

data:
a = 1000	1300	1260	1500	1420	1670	1340	1190	1720	1640;
c = 800	940	920	880	960	1000	870	840	1020	980;
b = 7200;
enddata

max = @sum(CONSTRUCT : x * c);

@sum(CONSTRUCT : x * a) <= b;
@for(CONSTRUCT : @bin(X));

x6 - x8 <= 1;
x6 - x8 >= 0;

x5 + x3 <= 1;
(x4 + x2) >= 1;

x8 = x7;
x1 + x4 + x6 + x9 <= 2; 
end

3.3 结果展示

Local optimal solution found.   Objective value:                              4720.000   Objective bound:                              4720.000   Infeasibilities:                              0.000000   Extended solver steps:                               3   Total solver iterations:                           375                            Variable           Value        Reduced Cost                               B        7200.000            0.000000                              X6        0.000000            0.000000                              X8        0.000000            0.000000                              X5        0.000000            0.000000                              X3        2.000000            0.000000                              X4        0.000000            0.000000                              X2       0.5000000E-03        0.000000                              X7        0.000000            0.000000                              X1        0.000000            0.000000                              X9        0.000000            0.000000                           C( 1)        800.0000            0.000000                           C( 2)        940.0000            0.000000                           C( 3)        920.0000            0.000000                           C( 4)        880.0000            0.000000                           C( 5)        960.0000            0.000000                           C( 6)        1000.000            0.000000                           C( 7)        870.0000            0.000000                           C( 8)        840.0000            0.000000                           C( 9)        1020.000            0.000000                          C( 10)        980.0000            0.000000                           A( 1)        1000.000            0.000000                           A( 2)        1300.000            0.000000                           A( 3)        1260.000            0.000000                           A( 4)        1500.000            0.000000                           A( 5)        1420.000            0.000000                           A( 6)        1670.000            0.000000                           A( 7)        1340.000            0.000000                           A( 8)        1190.000            0.000000                           A( 9)        1720.000            0.000000                          A( 10)        1640.000            0.000000                           X( 1)        1.000000           -800.0000                           X( 2)        1.000000           -940.0000                           X( 3)        0.000000           -920.0000                           X( 4)        0.000000           -880.0000                           X( 5)        1.000000           -960.0000                           X( 6)        1.000000           -1000.000                           X( 7)        0.000000           -870.0000                           X( 8)        0.000000           -840.0000                           X( 9)        1.000000           -1020.000                          X( 10)        0.000000           -980.0000                               Row    Slack or Surplus      Dual Price                               1        4720.000            1.000000                               2        90.00000            0.000000                               3        1.000000            0.000000                               4        0.000000            0.000000                               5        0.000000            0.000000                               6        0.000000            0.000000                               7        0.000000            0.000000                               8        2.000000            0.000000

4 总结梳理

（1）当两个二进制变量存在依赖关系，先决变量取值应不大于后继变量

（2）第一部分和第三部分得到两个相同最优解值，但解的表达形式不一样。应进一步探索，Lingo内部的实现机制，以获取更多形式的最优解。

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