纸浆：最小化决策变量的标准偏差
在纸浆I中出现的优化问题中，使用以下目标函数：
objective = p.lpSum(vec[r] for r in range(0,len(vec)))
所有变量均为非负整数，因此矢量上的总和给出了我问题的总单位数。 现在，我为这一事实而苦苦挣扎，纸浆只提供了许多解决方案之一，我想缩小解决方案空间的范围，从而有利于决策变量的最小标准偏差的溶液设置。 例如，Say VEC是具有元素6和12的向量。然后7/11、8/10、9/9是同样可行的解决方案，我希望纸浆到达9/9。 然后，该目标
objective = p.lpSum(vec[r]*vec[r] for r in range(0,len(vec)))
显然会创建一个成本函数，这将有助于情况，但是可惜，这是非线性和纸浆会引发错误。 谁能将我指向潜在的解决方案？
In an optimization problem developed in PuLP i use the following objective function:
objective = p.lpSum(vec[r] for r in range(0,len(vec)))
All variables are non-negative integers, hence the sum over the vector gives the total number of units for my problem.
Now i am struggling with the fact, that PuLP only gives one of many solutions and i would like to narrow down the solution space to results that favors the solution set with the smallest standard deviation of the decision variables.
E.g. say vec is a vector with elements 6 and 12. Then 7/11, 8/10, 9/9 are equally feasible solutions and i would like PuLP to arrive at 9/9.
Then the objective
objective = p.lpSum(vec[r]*vec[r] for r in range(0,len(vec)))
would obviously create a cost function, that would help the case, but alas, it is non-linear and PuLP throws an error.
Anyone who can point me to a potential solution?
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您可以最大程度地减少标准偏差（本质上是非线性的），而可以最大程度地减少范围或带宽。沿着：
Instead of minimizing the standard deviation (which is inherently non-linear), you could minimize the range or bandwidth. Along the lines of: