IBM ILOG CPLEX中Size Limit Exceeded错误求助

我正在用IBM ILOG CPLEX求解包含资源限制、任务依赖的调度问题，代码无语法错误，但持续出现“Size Limit Exceeded”错误，导致程序无法正常运行。

问题背景：我需要调度一组带多种约束的任务，已按照CPLEX文档规范定义变量、目标函数及约束，且问题规模在我认知里并未超出CPLEX的处理能力。我尝试了自身掌握的CPLEX相关调整方法，都没找到解决办法。

调度模型代码 #

int n=...;
 int m=...;  //resources
 range i=0..n-1;
 range j=0..n-1;
 range t=0..n-1;
 range k=0..m-1;
 range z=0..n-1;
 
 int B[k]= ...; //Capacities
 int p[i]= ...; //activity duratio
 int ES[i]=...;
 int LS[i]=...;
 int b[i][k]=...;
 range A=0..n-1; //activities
 range R=0..m-1; //Resources
 range A_plus1 = 0..n-1; // Include n+1 in the range 
 range H= 0..LS[n-1]; //Time horizon
tuple MyTuple { 
    int i; 
    int j; 
} 
{MyTuple} E = ...; // Use your actual data here 
// int V=asSet(1..n);
 dvar boolean x[i][t];
 
 minimize sum(i in 0..n-1) 
           sum(t in ES[i]..LS[i]) t*x[n-1][t];
 subject to 
{
forall(e in E)  
        sum(t in ES[e.j]..LS[e.j])t*x[e.j][t]>= sum(t in ES[e.i]..LS[e.i])t*x[e.i][t]+p[e.i];
 forall(t in H, k in R) 
        sum(i in 0..n-1) 
            b[i][k] * sum(z in maxl(ES[i],t - p[i] + 1) .. minl(LS[i], t))
            x[i][z] <= B[k];
             
forall(i in A_plus1) 
      sum(t in ES[i]..LS[i])x[i][t]==1;
       
 x[0][0]==1;
 //forall(i in A union {n+1})
      //forall(t in {ES[i],LS[i]})               
forall(i in A_plus1)
      forall(t in H : !(t in {ES[i], LS[i]}))
          x[i][t]==0;
           
forall(i in A_plus1) 
      forall(t in H : (t >= ES[i] && t<= LS[i]))  
          x[i][t] == 1; 
}

数据文件 #

n = 32;

m = 4;
 
Activity durations

p = [0, 6, 1, 4, 5, 7, 3, 5, 6, 9, 9, 5, 7, 4, 10, 10, 5, 10, 6, 10, 7, 9, 10, 1, 6, 6, 9, 4, 4, 7, 1,0 ];
 
Resource availability

B = [10, 10, 10, 10];
 
Resource requirements

b = [

[0, 0, 0, 0],

    [1, 1, 1,0],

    [6,0, 2, 1],

    [0,1, 4,3],

    [2, 6, 2, 0],

    [5, 3, 0, 5],

    [6, 4, 4, 0],

    [0, 6, 5, 2],

    [2, 6, 0, 5],

    [6, 6, 6, 0],

    [4, 0, 6, 5],

    [4, 5, 4, 0],

    [2, 3, 0, 3],

    [0, 3, 2, 3],

    [5, 3, 4, 0],

    [2, 5, 4, 0],

    [5, 4, 3, 0],

    [4, 0, 4, 3],

    [7, 1, 0, 4],

    [0, 3, 4, 7],

    [7, 4, 4, 0],

    [0, 4, 6, 7],

    [2, 7, 4, 0],

    [0, 7, 5, 1],

    [2, 0, 4, 3],

    [8, 0, 4, 5],

    [0, 2, 3, 5],

    [2, 6, 3, 0],

    [3, 0, 4, 7],

    [1, 0, 7, 3],

    [6, 2, 5, 0],

    [0, 0, 0, 0]
];

Include data for all activities and resources here
 
Precedence constraints

E = {

    <0, 2>,<0, 4>, <0, 5>, <0, 6>, <0, 7>,<0,16>, <0, 17>, <0, 18>, <0, 19>, <0, 20>, <0, 21>, <0, 22>, <0, 23>, <0, 24>, <0, 25>, <0, 26>, <0, 27>, <0, 28>, <0, 29>, <0, 30>, <0, 31>,

    <1, 14>, <1, 2>,

    <2, 8>, <2, 7>,

    <3, 13>, <3, 12>,

    <4, 11>, <4, 7>,

    <5, 10>,

    <6, 9>,

    <7, 31>,

    <8, 31>,

    <9, 31>,

    <10, 31>,

    <11, 31>,

    <12, 31>,

    <13, 31>,

    <14, 31>,

    <15, 31>,

    <16, 31>,

    <17, 31>,

    <18, 31>,

    <19, 31>,

    <20, 31>,

    <21, 31>,

    <22, 31>,

    <23, 31>,

    <24, 31>,

    <25, 31>,

    <26, 31>,

    <27, 31>,

    <28, 31>,

    <29, 31>,

    <30, 31>,
    

};

  Earliest start times
ES = [0, 0, 0, 6, 0, 0, 0, 0, 5, 1, 3, 7, 5, 10, 10, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20, 0];
 
Latest start times
LS = [0, 0, 10, 6, 8, 8, 8, 15, 14, 11, 11, 15, 13, 16, 10, 10, 15, 10, 0, 10, 13, 11, 10, 10, 19, 14, 14, 11, 16, 16, 19, 20];

内容的提问来源于stack exchange，提问作者hyyperi