我想在我的VRP中创建非线性违规成本。我已经创建了整个VRP的时间窗口,其中有这些决策变量。
dvar float+ w[N][D]; // violation time for late arrivals for every node and every day
这些决策变量都是有效的,但现在我想与违规成本决策变量建立联系,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
我想强制执行
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;
但这些约束条件还是有问题:
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
它对所有约束条件都设置了a5==1。
此外,w在模型中的用法为:。
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。
如果你想让w为负值,有时你应该有以下内容
dvar int w[N][D];
而不是
dvar int+ w[N][D];
另外,你可以用逻辑约束来代替硬编码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);
}