问:CPLEX:如何将 2 个等式写成 1 个常量
Q: CPLEX: How to write 2 equation as 1 constiants
我正在使用 CPLEX 求解 MILP。现在我想用相同的变量写出这两个方程
view the equations
我试着把它写进
(第一个等式为)
ct20 : forall(r1 in request,r2 in request, m1 in deliveries, m2 in deliveries : m1!=m2, k1 in truck, k2 in truck:k1!=k2, j1 in truck, j2 in truck:j1!=j2)
sum(p in plant, k1 in truck:k1!=k2, j1 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k1][j1]) <= sum(p in plant, k2 in truck:k1!=k2, j2 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k2][j2])
(第二个等式为)
&& sum(p in plant, k1 in truck:k1!=k2, j1 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k1][j1]+load_time[p][k1])
<= sum(p in plant, k2 in truck:k1!=k2, j2 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k2][j2])
如何将它们合并为一个,谢谢。
您知道可以使用 {
和 }
对约束进行分组吗?
range r=1..10;
dvar int x[r] in 0..10;
subject to
{
forall(i in r)
{
4<=x[i];
x[i]<=6;
}
}
我正在使用 CPLEX 求解 MILP。现在我想用相同的变量写出这两个方程
view the equations
我试着把它写进
(第一个等式为)
ct20 : forall(r1 in request,r2 in request, m1 in deliveries, m2 in deliveries : m1!=m2, k1 in truck, k2 in truck:k1!=k2, j1 in truck, j2 in truck:j1!=j2)
sum(p in plant, k1 in truck:k1!=k2, j1 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k1][j1]) <= sum(p in plant, k2 in truck:k1!=k2, j2 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k2][j2])
(第二个等式为)
&& sum(p in plant, k1 in truck:k1!=k2, j1 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k1][j1]+load_time[p][k1])
<= sum(p in plant, k2 in truck:k1!=k2, j2 in jobs:j1!=j2)(x[p][r1][m1][k1][j1]*start_load[k2][j2])
如何将它们合并为一个,谢谢。
您知道可以使用 {
和 }
对约束进行分组吗?
range r=1..10;
dvar int x[r] in 0..10;
subject to
{
forall(i in r)
{
4<=x[i];
x[i]<=6;
}
}