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我想在我的 VRP 中创建非线性违规成本。我已经创建了带有时间窗口的整个 VRP，其中我有这些决策变量：

dvar float+ w[N][D]; // violation time for late arrivals for every node and every day

这些 dvar 正在工作，但现在我想与违规成本决策变量建立链接，这些变量是：

dvar boolean a1[N][D];// no violation 
dvar boolean a2[N][D];// soft violation of 0-5 minutes
dvar boolean a3[N][D];// soft violation of 6-10 minutes
dvar boolean a4[N][D];// soft violation of 11 -15 minutes 
dvar boolean a5[N][D];// soft violation of 15+ minutes

我想强制：

a1[N][D] to be 1, when w[N][D] <=0, 0 otherwise
a2[N][D] to be 1, when w[N][D] >0 & <=5, 0 otherwise
a3[N][D] to be 1, when w[N][D] >5 & <=10, 0 otherwise
a4[N][D] to be 1, when w[N][D] >10 & <=15, 0 otherwise
a5[N][D] to be 1, when w[N][D] >=16, 0 otherwise

然后我有：

  forall(i in N, d in D)
  (a1[i][d] + a2[i][d] + a3[i][d] + a4[i][d] + a5[i][d]) == 1;

但是，这些约束仍然存在问题。

forall(i in N, d in D)
  (a1[i][d] + a2[i][d] + a3[i][d] + a4[i][d] + a5[i][d]) == 1; //sum of all a's = 1

forall(i in N, d in D)
  w[n][d]<= (5*a2[i][d]) + 1000*(1-a2[i][d]); // a2 == 1 when w[n][d]>0 & <=5

forall(i in N, d in D)
  (6*a3[i][d] - 1000*(a3[i][d]-1))<= w[i][d]; // a3

forall(i in N, d in D)
   w[i][d] <= (10*a3[i][d]) + 1000*(1-a3[i][d]); // a3

forall(i in N, d in D)
  (11*a4[i][d] - 1000*(a4[i][d]-1))<= w[i][d]; //a4

forall(i in N, d in D)
   w[i][d] <= (15*a4[i][d]) + 1000*(1-a4[i][d]); // a4

forall(i in N, d in D)
  (16*a5[i][d] - 1000*(a5[i][d]-1))<= w[i][d]; //a5

它为所有约束设置 a5==1 。

此外，w 在模型中用作：

forall (i in N, d in D:q[i][d]>=1)
y[i][d] - w[i][d] <= sl[i][d]; // late arrival time soft

其中 y[i][d] 是到达时间变量

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1 回答 1

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如果你想让 w 有时是负数，你应该有

dvar int w[N][D];

代替

dvar int+ w[N][D];

另外，您可以使用逻辑约束而不是带有硬编码 1000 的大 M

range N=1..2;
range D=1..3;

dvar boolean a1[N][D];
dvar int w[N][D] in -10..10;

subject to
{
// a1[N][D] to be 1, when w[N][D] <=0, 0 otherwise
forall(n in N,d in D) a1[n][d]==(w[n][n] <=0);
}
于 2020-05-09T15:16:27.540 回答