如何解决CPLEX/Docplex 中的单源和多目标问题?
我们能否将 CPLEX/Docplex 用于网络中的单源和多目标问题?例如,在单次行程中将车辆从一个来源路由到多个目的地,从而最大限度地缩短总行程时间。
解决方法
嗨,
您可以查看
中的 tsp OPL 示例examples/opl/models/TravelingSalesmanProblem
如果你想做单源和多目标,你可以改变目标
minimize sum (<i,j> in Edges) dist[<i,j>]*x[<i,j>];
到
minimize sum (<i,j>]-max(j in Cities:<1,j> in Edges) dist[<1,j>]*x[<1,j>];
您在 https://raw.githubusercontent.com/IBMDecisionOptimization/cplex_code_examples/master/docplex/tsp.py 中的 python docplex 中也有相同的示例
对于您的特定实例:
using CP;
execute
{
cp.param.timelimit=10;
}
{string} nodes={"O","A","B","C","D","E","T"};
tuple edge
{
key string o;
key string d;
int time;
}
{edge} edges with o,d in nodes=
{
<"O",40>,<"O",60>,50>,<"A",10>,<"C",20>,70>,<"B",55>,<"D","T",<"E",80>
};
string origin="O";
{string} targets={"A","C"};
int bigvalue=1000;
int repeat=3;
range ranks=1..repeat;
{edge} edgeswithsym=edges union {<d,o,t> | <o,d,t> in edges};
{edge} transitions=edgeswithsym union {<o,bigvalue> | o,d in nodes: <o,d> not in edgeswithsym};
tuple triplet {int id1; int id2; int value;};
{triplet} M = {<ord(nodes,tr.o)+1,ord(nodes,tr.d)+1,tr.time> | tr in transitions};
assert card(transitions)==card(nodes)*(card(nodes));
// Interval for visiting a node
dvar interval itvs[nodes][ranks] optional;
// Sequence means visits
dvar sequence seq in all(n in nodes,r in ranks)itvs[n][r] types
all(n in nodes,r in ranks)(ord(nodes,n)+1);
// First we want to minimize the time for latest visit
// Second we want to minimize present intervals
minimize
staticLex(max(t in targets) min(r in ranks) startOf(itvs[t,r],bigvalue),sum(n in nodes,r in ranks) presenceOf(itvs[n][r]));
subject to
{
// We visit origin at time 0
startOf(itvs[origin,1])==0;
// origin and destinations should be present
presenceOf(itvs[origin,1])==1;
forall(t in targets) presenceOf(itvs[t,1])==1;
// noOverlap constraint will enforce time matrix
noOverlap(seq,M,true);
// break sym
forall(n in nodes) forall(r in ranks:r!=1)
presenceOf(itvs[n,r-1])>=presenceOf(itvs[n,r]);
}
int nbActiveIntervals=sum(n in nodes,r in ranks) presenceOf(itvs[n][r]);
range R=1..nbActiveIntervals;
// Display solution
execute {
writeln(origin);
var s=seq.first();
for(var i in R)
{
if (i!=nbActiveIntervals)
writeln(Opl.item(nodes,-1+Opl.typeOfNext(seq,s,bigvalue,0)));
s=seq.next(s);
}
}
/*
gives
O
A
B
C
*/
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。