LP:与正变量相对应的正本减少的成本? [英] LP: postive reduced costs corresponding to positive variables?

问题描述

我有下一个LP问题

Maximize

1000 x1 + 500 x2 - 500 x5 - 250 x6 

Subject To

 c1: x1 + x2 - x3 - x4  = 0

 c2: - x3 + x5  = 0

 c3: - x4 + x6  = 0

With these Bounds

 0 <= x1 <= 10

 0 <= x2 <= 15

 0 <= x5 <= 15

 0 <= x6 <= 5

通过使用Cplex对偶算法解决此问题，我得到了 6250的最佳解决方案.但检查变量的成本降低后，我得到了下一个结果

By solving this problem with Cplex dual algorithm I get an optimal solution of 6250. But checking the reduced costs of the variables I get the next results

Variable   value    reduced cost
1          10.0          500.0 
1           0.0         -0.0 
2          5.0          -0.0 
3          5.0          -0.0
4          5.0          -0.0 
5          5.0         250.0 

是否可以使正值变量的成本降低正数?因为降低的成本值表示在最佳解决方案中变量的值变为正值之前必须将相应变量的目标函数系数提高多少，所以正的降低的成本对正值变量意味着什么?

Is it possible to have a positive reduced cost on a positive valued variable? Because the reduced cost value indicates how much the objective function coefficient on the corresponding variable must be improved before the value of the variable will be positive in the optimal solution, what does a positive reduced cost means on a postive valued variable?

推荐答案

变量1在解决方案中被列出两次?

Variable 1 is listed twice in the solution?

请注意，您需要区分下限的非基本和上限的非基本.降低的成本表示当相应的界线改变一个单位时，物镜可以改变多少.

Note that you need to distinguish between nonbasic at lower bound and nonbasic at upper bound. The reduced cost indicates how much the objective can change when the corresponding bound changes by one unit.

还请注意，大多数教科书侧重于特殊情况 x> = 0 ，而实际求解器同时支持上下限: L< = x< = U .

Note also that most textbooks focus on the special case x >= 0 while practical solvers support both lower and upper bounds: L <= x <= U.

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