每周的目标功能-帮助编写代码-CPLEX

Question

（OPL模型和Lindo模型在代码框中）我需要有关此问题的帮助。我的目标是组织每周（第1、2、3和4周）哪些卡车从始发地运到目的地。在此示例中，总共有6辆卡车，每周必须至少运送一辆。每次装运都有相关的成本，目标是在第一周，然后在第二，第三和第四周将成本降到最低。这是一个字典目标编程问题。我用其他软件（Lindo）制作了扩展模型，并且可以使用，但是用OPL编写时遇到了困难。有人可以帮我吗？

Xijt，i =接送点（1,2,3,4,5），j =目的地（1,2），t =周（1,2,3,4）

! My code in Lindo:
MIN p4
subject to
p1 = 10
p2 = 10
p3 = 100

! Cost 

0 X111 + 0 X211 + 90 X311+  0 X411 + 10 X511 + 0 X121 + 100 X221 + 0 X321 + 50 X421 + 10 X521 -p1 + n1 = 0 ! Week 1 
0 X112 + 0 X212 + 90 X312+  0 X412 + 10 X512 + 0 X122 + 100 X222 + 0 X322 + 50 X422 + 10 X522 -p2 + n2 = 0 ! Week 2 
0 X113 + 0 X213 + 90 X313+  0 X413 + 10 X513 + 0 X123 + 100 X223 + 0 X323 + 50 X423 + 10 X523 -p3 + n3 = 0 ! Week 3 
0 X114 + 0 X214 + 90 X314+  0 X414 + 10 X514 + 0 X124 + 100 X224 + 0 X324 + 50 X424 + 10 X524 -p4 + n4 = 0 ! Week 4 


! Number of trucks

X311 + X312 + X313 + X314  = 1 !In 4 weeks, there is 1 truck from pickup point 3 to destination 1
X511 + X512 + X513 + X514  = 1
X221 + X222 + X223 + X224  = 2
X421 + X422 + X423 + X424  = 1
X521 + X522 + X523 + X524  = 1 


! Week restriction
!At least one truck must be used per week
X311 + X511 + X221 + X421 + X521 >=1
X312 + X512 + X222 + X422 + X522 >=1
X313 + X513 + X223 + X423 + X523 >=1
X314 + X514 + X224 + X424 + X524 >=1


END

INT X311 
INT X312 
INT X313 
INT X314 

INT X511
INT X512 
INT X513 
INT X514

INT X221 
INT X222
INT X223 
INT X224 

INT X421 
INT X422 
INT X423 
INT X424

INT X521
INT X522
INT X523
INT X524 
---------------------------------------------------
// CPLEX
using CP;
// decision variables

{string} Pickup = {"A","B","C","D","E"};
{string} Destination = {"D1" , "D2"};
{string} Weeks = {"W1", "W2", "W3", "W4"};

int Trucks [Pickup][Destination]= [[0,0],[0,2],[1,0],[0,1],[1,1]]; 
int Cost [Pickup]=[140,100,90,50,10]; 

//Decision Variables
dvar int  Delivered [Forest][Destination][Weeks]; //not sure if it is right

//Expressions
dexpr int Week1 = sum (u in Pickup, c in Destination, W1 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W1];
dexpr int Week2 = sum (u in Pickup, c in Destination, W2 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W2];
dexpr int Week3 = sum (u in Pickup, c in Destination, W3 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W3];
dexpr int Week4 = sum (u in Pickup, c in Destination, W4 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W4];



//Objective Function
minimize staticLex (Week1,Week2,Week3,Week4);

//Constraint
subject to {
forall (u in Pickup)
    forall (c in Destination)
        sum (W1 in Weeks)
            Trucks [u][c] >= 1;

forall (u in Forest)
    forall (c in Destination)
        sum (W2 in Weeks)
            Trucks [u][c] >= 1;

forall (u in Pickup)
    forall (c in Destination)
        sum (W3 in Weeks)
            Trucks [u][c] >= 1;

forall (u in Pickup)
    forall (c in Destination)
        sum (W4 in Weeks)
            Trucks [u][c] >= 1;

}


execute Output {
    writeln ("Delivered Plan")
        for (var t in Trucks)
            for (var u in Pickup)
                for (var c in Destination)
                    for (var w in Weeks)
                        if (Delivered [t][u][c][w]>0) {
                        writeln (Delivered [t][u][c][w] + '' + " number of trucks " + '' + t + '' + "  delivered from pickup poit " + '' + u +" to destination " +''+ c + " in the week "+''+ w);

}}

Answer 1

看起来很腥：

//Expressions
dexpr int Week1 = sum (u in Pickup, c in Destination, W1 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W1];
dexpr int Week2 = sum (u in Pickup, c in Destination, W2 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W2];
dexpr int Week3 = sum (u in Pickup, c in Destination, W3 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W3];
dexpr int Week4 = sum (u in Pickup, c in Destination, W4 in Weeks) Trucks[u][c] * Freshness [u] * Delivered[u][c][W4];

//Objective Function
minimize staticLex (Week1,Week2,Week3,Week4);

在single周的表达式中，您要对all周的变量求和。我想您想要的是：

//Expressions
dexpr int Week1 = sum (u in Pickup, c in Destination) Trucks[u][c] * Freshness [u] * Delivered[u][c]["W1"];
dexpr int Week2 = sum (u in Pickup, c in Destination) Trucks[u][c] * Freshness [u] * Delivered[u][c]["W2"];
dexpr int Week3 = sum (u in Pickup, c in Destination) Trucks[u][c] * Freshness [u] * Delivered[u][c]["W3"];
dexpr int Week4 = sum (u in Pickup, c in Destination) Trucks[u][c] * Freshness [u] * Delivered[u][c]["W4"];

注意，Delivered的最后一个索引现在在每个表达式中都是常量。现在，每个表达式只将一个星期的变量求和。