OPL_ 对元素求和直到某个元素答案

【问题标题】：OPL_ Summing over elements up to certain elementOPL_ 对元素求和直到某个元素
【发布时间】：2021-11-15 11:50:40
【问题描述】：

假设我有一组星期

{string} weeks ={"Week 1","Week 2","Week 3","Week 4"}

然后我为每个星期定义一个变量

dvar boolean x[weeks];

如果我想，对于所有w in Weeks，将所有周的总和直到第 w 周，你怎么能这样做？我想要做的是以下，它不起作用，因为我无法比较这样的字符串

forall ( w in weeks )
{
  sum ( ww in weeks : ww<=w ) x[ww] >= rhs[w];
}

rhs[w] 就在右手边。

【问题讨论】：

标签： cplex opl

【解决方案1】：

巧合的是，我偶然发现了OPL manual 中的一个页面，描述了函数ord()，这似乎是一个前进的方向。 ord(Set,element) 返回集合中元素的位置。因此问题中的代码应该是

forall ( w in weeks )
{
  sum ( ww in weeks : ord(weeks,ww)<=ord(weeks,w) ) x[ww] >= rhs[w];
}

【讨论】：

【解决方案2】：

其实你可以在 OPL 中比较字符串所以

{string} weeks ={"Week 1","Week 2","Week 3","Week 4"};

float rhs[weeks]=[1,2,3,4];

dvar boolean x[weeks];

subject to
{
  forall ( w in weeks )
   {
     sum ( ww in weeks : ww<=w ) x[ww] >= rhs[w];
   }
}

execute
{
  for(var w1 in weeks) for(var w2 in weeks) if (w1<=w2) writeln(w1," <= ",w2);
}

工作正常并提供

Week 1 <= Week 1
Week 1 <= Week 2
Week 1 <= Week 3
Week 1 <= Week 4
Week 2 <= Week 2
Week 2 <= Week 3
Week 2 <= Week 4
Week 3 <= Week 3
Week 3 <= Week 4
Week 4 <= Week 4

【讨论】：

我想这需要数周。如果我例如有一组天{星期一，星期四，星期三，星期四，星期五}，那么我不能只比较字符串。
正确。那你最好用 ord