C++ 为什么我在ILOG CPLEX的工作没有解决方案？

我正在处理ILOG CPLEX中的生产组合问题。但也出现了一些问题。我的工作是关于锯木厂生产组合（从原木转变为木材产品的生产计划）。我尝试使用2个file.dat来解决我的问题，一个使用虚拟数据，另一个使用真实数据

使用虚拟数据进行数据处理会显示解决方案，但当我使用真实数据时，结果不会显示解决方案。因此，我尝试用真实数据更改每个虚拟数据（逐个查看导致问题的数据）。结果证明数据是“LumberRecFac”和“DemandMaks”。然后，我尝试用几个随机数来改变这两个数据。它表明，“LumberRecFac”的最大数据为0.5，“DemandMaks”的最大数据为10.9

我的.mod和.dat中是否有遗漏/错误的内容？因为我必须使用真实数据运行我的ILOG CPLEX.mod

这里是我的.mod

 //product mix
//i= (RST)
//d,l (Log)
//p (Cutting_Pattern)

int i=...;
int d=...;
int l=...;
int p=...;

range RST= 1..i; //Lumber Product(Raw Sawn Timber)
range Diameter_Log=1..d; //Diameter Log
range Panjang_Log=1..l; //Length log
range Cutting_Pattern=1..p; //Cutting Pattern

//set tuple
tuple DMdl{ //diameter and length
int d;
int l;
}

tuple DMdlp{ //diameter, length, and cutting pattern
int d;
int l;
int p;
}

tuple Cdpi{ //conversion factor (from log to lumber products)
int d;
int p;
int i;
}

//set
setof (DMdl) Log={<d,l> | d in Diameter_Log, l in Panjang_Log};
setof (DMdlp) Log_LCR={<d,l,p> | d in Diameter_Log, l in Panjang_Log, p in Cutting_Pattern};
setof (Cdpi) KonversiLog_LCR={<d,p,i> | d in Diameter_Log, p in Cutting_Pattern, i in RST};

//import data
int HargaJual[RST]=...; //Selling price of RST
int BiayaLogBaru[Log]=...; //Cost of new log
int BiayaLogLama[Log]=...; //Cost of old log
int BiayaProduksiLog=...; //Production cost
int BiayaSetup=...; //Setup cost
float LumberRecFac[KonversiLog_LCR]=...; //Conversion Factor (Lumber Recovery Ratio)
float DemandMaks[RST]=...; //Demand
int BM=...; //Big Constant (Big M)
int Tmax=...; //Max Production Time
int ProcessTime[Cutting_Pattern]=...; //Process time for one log
float SetupTime[Cutting_Pattern]=...; //Setup time
float TotalLogLamaSimpan[Log]=...; //Quantity of old log

//decision variable
dvar float+ Pi[RST]; //Quantity of lumber products
dvar float+ VF[RST]; //Quantity of selling lumber products
dvar boolean CPp[Cutting_Pattern]; 
dvar float TotalLogBaruBeli[Log]; //Quantity of new lod that have to buy
dvar float+ LogProduksi[Log_LCR]; //Total quantity of log to produce
dvar float+ LogBaruDipakai[Log]; //Total quantity of new log that processed
dvar float+ LogLamaDipakai[Log]; //Total quantity of old log that processed
dvar float+ TotalLogSimpan[Log]; //Total inventor of new log 

//objective function
dexpr float sales=sum(i in RST) VF[i]*HargaJual[i];
dexpr float purchased=sum(d in Diameter_Log, l in Panjang_Log) (TotalLogBaruBeli[<d,l>]*BiayaLogBaru[<d,l>])+(LogLamaDipakai[<d,l>]*BiayaLogLama[<d,l>]);
dexpr float production=sum(d in Diameter_Log, l in Panjang_Log) LogProduksi[<d,l,p>]*BiayaProduksiLog;
dexpr float setup=sum(p in Cutting_Pattern) BiayaSetup*CPp[p];

maximize sales-(purchased+production+setup);

//constraint
subject to{
//constraint 1 : raw material inventory
    forall (d in Diameter_Log, l in Panjang_Log)
        TotalLogBaruBeli[<d,l>]==LogBaruDipakai[<d,l>]+TotalLogSimpan[<d,l>];

    forall (d in Diameter_Log, l in Panjang_Log)
       TotalLogLamaSimpan[<d,l>]>=LogLamaDipakai[<d,l>];

//constraint 2 : log processing
    forall (l in Panjang_Log, i in RST) 
        sum(d in Diameter_Log, p in Cutting_Pattern) LogProduksi[<d,l,p>]*LumberRecFac[<d,p,i>]==Pi[i];

    forall (d in Diameter_Log, l in Panjang_Log)
        sum(p in Cutting_Pattern )LogProduksi[<d,l,p>]==LogLamaDipakai[<d,l>]+LogBaruDipakai[<d,l>];

    forall (p in Cutting_Pattern)
        sum(d in Diameter_Log, l in Panjang_Log) LogProduksi[<d,l,p>]<=BM*CPp[p];

    forall (p in Cutting_Pattern)
        sum(d in Diameter_Log, l in Panjang_Log) (LogProduksi[<d,l,p>]*ProcessTime[p])+(SetupTime[p]*CPp[p])<=Tmax;

//constraint 3 : production management and demand satisfaction
    forall (i in RST)
        Pi[i]>=VF[i];

    forall (i in RST)
        VF[i]==DemandMaks[i];

}

这里是我的真实数据

    i = 3;
d = 2;
l = 2;
p = 1;

HargaJual= [1500 1400 1600];

BiayaLogLama=[100 250 
200 400];

BiayaLogBaru=[150 250 
300 450];

BiayaProduksiLog= 400;

BiayaSetup= 5;

LumberRecFac= [0.5  0.5 0.5
0.5  0.5    0.5];

DemandMaks= [9.8 10.9 10.8];

BM= 10000;

Tmax= 48;

ProcessTime= [1];

SetupTime= [0.1];

TotalLogLamaSimpan=[2   1
1   2];

    i = 10;
d = 2;
l = 5;
p = 1;

HargaJual= [3114984 
43347890 
22956482 
7775850 
15380010 
16984110
8703344 
3500008 
3288741 
2525224 
];

BiayaLogLama=[2328042 2834346 3035619 3044953 3199186
4446927 4872151 4924974 6006590 6637329];

BiayaLogBaru=[2360389   3291645 3212665 3231400 3453636
4456572 4884165 5057298 6261820 6733965];

BiayaProduksiLog=  17435453;

BiayaSetup= 5;

LumberRecFac= [0.0127 0.0145 0.0982 0.1353 0.4127 0.0001 0.2800 0.0107 0.1348 0.0166 
0.6842 0.2105 0.0643 0.2359 0.4730 0.0497 0.3728 0.0187 0.2393 0.0434];

DemandMaks= [188.110
75.259
121.711
253.759
1206.444
5.248
180.735
0
164.510
0];

BM= 1000000;

Tmax= 480;

ProcessTime= [1];

SetupTime= [0.1];

TotalLogLamaSimpan=[0.030   2.770
3.776   5.530
2.993   3.880
1.678   2.970
1.588   0.140];

---

你的模式不可行。如果你给你的约束贴上标签，那么CPLEX将给你一些放松和冲突，这将帮助你理解为什么

例如：

    forall (d in Diameter_Log, l in Panjang_Log)
       ct1:TotalLogLamaSimpan[<d,l>]>=LogLamaDipakai[<d,l>];

//constraint 2 : log processing
    forall (l in Panjang_Log, i in RST) 
        ct2:sum(d in Diameter_Log, p in Cutting_Pattern) LogProduksi[<d,l,p>]*LumberRecFac[<d,p,i>]==Pi[i];

forall（直径为d的盘江原木，直径为l的盘江原木）
ct1:Totaloglamasimpan[]>=LogLamaDipakai[]；
//约束2：日志处理
forall（盘江日志中的l，RST中的i）
ct2：总和（直径为d，切割模式为p）LogProduksi[]*LumberRecFac[]==Pi[i]；

---

将显示与ct2的冲突，如果删除该冲突，则可以得到可行的解决方案

您的示例代码很长。尝试简化优化问题（去掉变量、约束），以确定问题的核心。