CPLEX上不可行的解决方案错误（2D箱包装问题）

Question

[可行解决方案的表示] [1] [1]：https：//i.sstatic.net/83esm.png

大家好，

我目前正在研究CPLEX上3级二维箱包装问题的改编。不幸的是，当我尝试解决问题时，出现一些错误。 该模型的目的是将高度和宽度的项目分配给最小数量的垃圾箱。 这些项目将首先分配给堆栈，然后将堆栈分配给条纹，最后条纹将被分配给垃圾箱（图像附加的示例）。 物品是根据其高度订购的（最高项目将首先使用最小的索引（例如1））。堆栈/条纹具有最高项目的索引（最小的索引）。 原始模型的救援表明，堆栈的物品必须具有相同的宽度。 我正在尝试更改此公式，以便堆栈中的项目可以具有不同的宽度。 我的方法是定义一个新的决策变量（float），该变量将按每个堆栈上每个项目的最大宽度（第二次最后一个约束）的最大宽度，然后限制条纹的宽度，而条纹的宽度不会大于宽度。垃圾箱（最后一个约束）。

在我的描述中，“ bin =手推车”，

我非常感谢您的评论，以了解我在忽略的内容。

引擎日志上的错误 传统回调pi 警告：整数变量的非整体界限。 不可行的行'c160'：0＆lt; = -5。

第160行提及第二个添加的约束。 C160： - _e（1）（2）＃26 + 5 a（2）（2）（2）＃52＆lt; = 0

当我增加宽度容量时，我得到了此错误： 传统回调pi 警告：整数变量的非整体界限。 行'c255'不可行，所有条目在隐含的边界处。

c255: _e(1)(1)#25 + _e(1)(2)#26 + _e(1)(3)#27 + _e(1)(4)#28 + _e(1)(5)#29
   - 20 b(1)(1)#65 <= 0

我尝试给出一个可行的灵魂，但我仍然在新约束上遇到错误。
/*a [1] [1] == 1; a [1] [3] == 1;

a [2] [2] == 1; a [2] [4] == 1;

a [5] [5] == 1;

B [1] [1] == 1; B [1] [2] == 1;

C [1] [1] == 1;

B [5] [5] == 1; C [5] [5] == 1;*/

模型摘要：

int n=...; //number of items
range iitem=1..n;

float heighttrolley=...;    //length of bin j
float widthtrolley=...; //width of bin j

float heightitem[iitem]=...;    //length of item i
float widthitem[iitem]=...;     //width of item i

dvar boolean a[1..n][1..n]; //if item i is assigned to stack j
dvar boolean b[1..n][1..n]; //if stack j is contained in stripe k
dvar boolean c[1..n][1..n]; //if stripe k is contained in bin l
dvar boolean d[1..n-1][2..n][1..n-1]; //if item i contributes to the total height of all stripes in bin l and is contained in stack j
dvar float e[1..n][1..n];   //max width of an item in stack j contained in stripe k


minimize sum (l in 1..n) c[l][l];

subject to{
  
forall(j in 1..n, i in 1..n: i<j)
a[j][i]==0;   
forall(l in 1..n, k in 1..n: k<l)
c[l][k]==0;
forall(l in 1..n-1, i in l+1..n, j in l..n-1: i<l+1 && j<l && j>i-1)  
d[l][i][j]==0;

forall(j, k in iitem)
e[j][k]<=widthtrolley;
        
forall(i in iitem)
sum (j in 1..i) a[j][i]==1; //each item has to be packed once

forall(j in 1..n-1)
sum (i in j+1..n)a[j][i]<=(n-j)*a[j][j];    //items can be assigned to unused stacks

forall(j in 1..n-1, i in iitem: i>j && heightitem[i]+heightitem[j]>heighttrolley)// && widthitem[i]<=widthitem[j])  //total height of any pair of stacked items must not exceed height of bin
a[j][i]==0;

forall(j in 1..n)
sum (k in 1..n) b[k][j]==a[j][j];  // every item j is packed exactly once into a stripe k

forall(k in 2..n, j in 1..k-1)
sum(i in j..n) heightitem[i]*a[j][i]<= sum(i in k..n)heightitem[i]*a[k][i]+(heighttrolley+1)*(1-b[k][j]); //ensure that the height of each stack j never exceeds the height of the stripe k it is contained in (1)

forall(k in 1..n-1, j in k+1..n)
sum(i in j..n) heightitem[i]*a[j][i]<= sum(i in k..n)heightitem[i]*a[k][i]+heighttrolley*(1-b[k][j]);    //ensure that the height of each stack j never exceeds the height of the stripe k it is contained in (2)

forall(k in 1..n)
sum(l in 1..k)c[l][k]==b[k][k]; //force each used stripe k to be packed into exactly one bin

forall(l in 1..n-1)
sum(i in l..n)heightitem[i]*c[l][i]+sum(i in l+1..n)heightitem[i]*sum(j in l..i-1)d[l][i][j]<=heighttrolley*c[l][l];    //bins used height has to be smaller than the height of bin 

forall(l in 1..n-1, i in l+1..n, j in l..i-1){  //force variable d to be set to 1, when item i contributes to the total height of all stripes in bin l and is contained in stack j
a[j][i]+c[l][j]-1<=d[l][i][j];
d[l][i][j]<=(a[j][i]+c[l][j])/2;}

forall(l in 1..n-1)
sum(k in l+1..n)c[l][k]<=(n-l)*c[l][l]; //ensure that no stripes are packed into an unused bin

forall(i,j,k in iitem)              //if item i is assigned to stack j, then e[k][j] has to be at least bigger than the width of item i
widthitem[i]*a[j][i]<=e[k][j];          

forall(k in iitem)
sum(j in iitem)e[k][j]<=widthtrolley*b[k][k];   //for every stripe, the sum of the max width of every stack j on the sripe k must be smaller than the width of the bin 

链接到数据：

Answer 1

您的模型不可行。

例如，如果您将其命名为这样的约束：

forall(j in 1..n, i in 1..n: i<j)
    ct1:a[j][i]==0;   
  forall(l in 1..n, k in 1..n: k<l)
    ct2:c[l][k]==0;
  forall(l in 1..n-1, i in l+1..n, j in l..n-1: i<l+1 && j<l && j>i-1)  
    ct3:d[l][i][j]==0;
  
  forall(j, k in iitem)
    ct4:e[j][k]<=widthtrolley;
            
  forall(i in iitem)
    ct5:sum (j in 1..i) a[j][i]==1; //each item has to be packed once

那么在IDE中，您将获得一系列放松和冲突，可以帮助您调试模型。