Why my work in ILOG CPLEX have no solution?

Question

i'm working on a production mix problem in ILOG CPLEX. But there are some problems that occur. My work is about sawmill production mix (production planning for lumber products that transform from log). I tried to solve my problem using 2 file .dat, one using dummy data and one using real data.

Data processing with dummy data show the solution, but when i'm using real data, result show no solution. So i've tried to change each of my dummy data with my real data (one by one to see what data that caused the problem). And it turned out the data are "LumberRecFac" and "DemandMaks". Then, i tried to change that two data with a several random numbers. And it shows that the maximum data for "LumberRecFac" is 0.5 and the maximum data for "DemandMaks" is 10.9.

Is there something that i missed/wrong in my .mod and .dat? Because i have to run my ILOG CPLEX .mod using my real data.

Here my .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];

}

Here my dummy data

    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];

Here my real data

    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];

Answer 1

your model is not feasible. If you label your constraints then CPLEX will give you some relaxation and conflicts that will help you understand why.

For instance:

    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];

will display a conflict with ct2 and if you remove that then you get a feasible solution