Originally posted by: user1234567

Hello, i'm a beginner in cplex. Can you please give me an example of a jobshop using cplex optimizer, because i just found examples using constraint programming.

I have a mathematical model (milp model) to applicate on cplex. so, i need examples of a jobshop using cplex optmizer. Unfortunately, you just put the examples of jobshop using CP optimizer. There is no example of a jobshop using CPLEX Optimizer ! 

 

Thank you.

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Originally posted by: user1234567

please i need that urgently. 

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Hi,

as can be read at https://fr.slideshare.net/PhilippeLaborie/vug-scheduling28-may2014

CPO can really be good at scheduling.

The jobshop example at CPLEX_Studio1271\opl\examples\opl\sched_jobshop is a very good example

.mod

// --------------------------------------------------------------------------
// Licensed Materials - Property of IBM
//
// 5725-A06 5725-A29 5724-Y48 5724-Y49 5724-Y54 5724-Y55
// Copyright IBM Corporation 1998, 2013. All Rights Reserved.
//
// Note to U.S. Government Users Restricted Rights:
// Use, duplication or disclosure restricted by GSA ADP Schedule
// Contract with IBM Corp.
// --------------------------------------------------------------------------

using CP;

int nbJobs = ...;
int nbMchs = ...;

range Jobs = 0..nbJobs-1;
range Mchs = 0..nbMchs-1;
// Mchs is used both to index machines and operation position in job

tuple Operation {
  int mch; // Machine
  int pt;  // Processing time
};

Operation Ops[j in Jobs][m in Mchs] = ...;

dvar interval itvs[j in Jobs][o in Mchs] size Ops[j][o].pt;
dvar sequence mchs[m in Mchs] in all(j in Jobs, o in Mchs : Ops[j][o].mch == m) itvs[j][o];

execute {
          cp.param.FailLimit = 10000;
}

 

minimize max(j in Jobs) endOf(itvs[j][nbMchs-1]);

subject to {

  forall (m in Mchs)
    noOverlap(mchs[m]);
  forall (j in Jobs, o in 0..nbMchs-2)
    endBeforeStart(itvs[j][o], itvs[j][o+1]);
}

execute {
  for (var j = 0; j <= nbJobs-1; j++) {
    for (var o = 0; o <= nbMchs-1; o++) {
      write(itvs[j][o].start + " ");
    }
    writeln("");
  }
}

.dat

nbJobs = 6;
nbMchs = 6;

Ops = [
 [ <5,4>, <1,3>, <4,3>, <3,2>, <0,1>, <2,2> ],
 [ <1,3>, <0,8>, <5,7>, <2,2>, <4,9>, <3,3> ],
 [ <3,1>, <4,9>, <1,9>, <0,7>, <5,5>, <2,5> ],
 [ <3,8>, <4,2>, <1,1>, <5,7>, <2,8>, <0,9> ],
 [ <1,6>, <3,2>, <4,5>, <5,5>, <0,3>, <2,1> ],
 [ <4,10>, <2,4>, <0,4>, <3,3>, <1,2>, <5,3> ]
];

but you could easily turn this into a MIP:

.mod

int nbJobs = ...;
int nbMchs = ...;

range Jobs = 0..nbJobs-1;
range Mchs = 0..nbMchs-1;
// Mchs is used both to index machines and operation position in job

tuple Operation {
  int mch; // Machine
  int pt;  // Processing time
};

Operation Ops[j in Jobs][m in Mchs] = ...;

 

dvar int+ s[j in Jobs][o in Mchs];
dvar int+ e[j in Jobs][o in Mchs];

 

minimize max(j in Jobs) e[j][nbMchs-1];
subject to {

forall (j in Jobs, o in 0..nbMchs-1) e[j][o]-s[j][o]==Ops[j][o].pt;
 
   
   forall (o1,o2 in Mchs) forall(i,j in Jobs:(Ops[i][o1].mch==Ops[j][o2].mch) && ((i!=j) || (o1!=o2)))
   (s[i][o1]>=e[j][o2]) || (s[j][o2]>=e[i][o1]);
  forall (j in Jobs, o in 0..nbMchs-2)
    (e[j][o]<=s[j][o+1]);
}

execute {
  for (var j = 0; j <= nbJobs-1; j++) {
    for (var o = 0; o <= nbMchs-1; o++) {
      write(s[j][o] + " ");
    }
    writeln("");
  }
}

 

regards

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#DecisionOptimization

Originally posted by: user1234567

Hello Alex and thank you for your reply.

Now i want to add to this code some information related to my problem: 1. I want that cplex determines which operation is assigned to which machine (in a way that it minimizes Cmax). So, I want to add the machine as a decision variable.

2. The Jobshop is Flexible so that each operation of a job can work on more than one machine.

3. I have two other decision variable Xijk = {1 if machine k is selected for operation Oij, 0 otherwise}, 

and Yij i'j' k = {1 if operation Oij precedes operation Oi'j' on machine k, 0 otherwise}.

 

How can I do this ? Can you help me please ? Thank you very much.

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Then you could Have a look at

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Originally posted by: user1234567

i already had a look at the Flexible Jobshop Scheduling Problem. But, in this example, the user enters the assignment of the different operations on machines.(manually)

I want that it will be automatically (the assignment of operations to machines) as i explained above.

So, please help me to solve point 1: Now i want to add to this code some information related to my problem: 1. I want that cplex determines which operation is assigned to which machine (in a way that it minimizes Cmax). So, I want to add the machine as a decision variable.

 

//.MOD

int nbJobs = ...;
int nbMchs = ...;
int nbOps = ...;

range Jobs = 0..nbJobs-1;
range Mchs = 0..nbMchs-1;
range ops =  0..nbOps;

// Mchs is used both to index machines and operation position in job
tuple Operation {
  int id;
  int jobId;
  int pt;  // Processing time
};
Operation Ops[j in Jobs][k in Mchs] = ...;

 

 
dvar int+ s[j in Jobs][o in Mchs];
dvar int+ c[j in Jobs][o in Mchs];
dvar int+ k;

dvar boolean X[i in Jobs][j in ops][k in Mchs];
dvar boolean Y[i in Jobs][j in ops][i1 in Jobs][j1 in ops][k in Mchs];

minimize max(j in Jobs) c[j][nbMchs-1];
subject to {
forall (j in Jobs, o in 0..nbMchs-1) c[j][o]-s[j][o]==Ops[j][o].pt;

 forall (o1,o2 in Mchs) forall(i,j in Jobs:(Ops[i][o1]==Ops[j][o2]) && ((i!=j) || (o1!=o2)))
   (s[i][o1]>=c[j][o2]) || (s[j][o2]>=c[i][o1]);
  forall (j in Jobs, o in 0..nbMchs-2)
    (c[j][o]<=s[j][o+1]);

 

//C00
forall (i in Jobs, j in ops, k in Mchs){
  {
 
sum (k in Mchs) X[i][j][k]==1;

}

}
   
}

 tuple result {
 int Jobs;
 int Mchs;
 int ops;
 }
 {result} results = {<i,j,k>|i in Jobs, j in ops, k in Mchs: X[i][j][k]==1};
 
 
execute {
writeln (results);

  for (var j = 0; j <= nbJobs-1; j++) {
    for (var o = 0; o <= nbMchs-1; o++) {
      write(s[j][o] + " ");
    }
    writeln("");
  }
} 

 

//.DAT

 

nbJobs = 6;
nbMchs = 6;
nbOps=6;

Ops = [
 [ <1,1,5>, <2,1,3>, <3,1,3>, <4,1,2>, <5,1,1>, <6,1,2> ],
 [ <1,2,3>, <2,2,8>, <3,2,7>, <4,2,2>, <5,2,9>, <6,2,3> ],
 [ <1,3,1>, <2,3,9>, <3,3,9>, <4,3,7>, <5,3,5>, <6,3,5> ],
 [ <1,4,8>, <2,4,2>, <3,4,1>, <4,4,7>, <5,4,8>, <6,4,9> ],
 [ <1,5,6>, <2,5,2>, <3,5,5>, <4,5,5>, <5,5,3>, <6,5,1> ],
 [ <1,6,10>, <2,6,4>, <3,6,4>, <4,6,3>, <5,6,2>, <6,6,3> ]
];

 

 

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I would really use CPO.

In the example I mentioned The flexible job-shop scheduling :

forall (o in Ops)
    alternative(ops[o], all(md in Modes: md.opId==o.id) modes[md]);

could help

regards

#CPLEXOptimizers
#DecisionOptimization

Originally posted by: user1234567

Hello, in the jobshopflex.mod:cp, does the program determine the start and the completion time of each operation of a job on machines.

And where can i see exactly the results which indicates the start and the completion time of each operation of a job on machines in the space which displays the solution? 

and also can u explain to me the results (after compilation) related to: 

dvar interval ops  [Ops]; 
dvar interval modes[md in Modes] optional size md.pt;
dvar sequence mchs[m in Mchs] in all(md in Modes: md.mch == m) modes[md];

(I didn't understand well what is their interest and the result of each decision variable).

 

Thank you in advance.

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#DecisionOptimization

Originally posted by: user1234567

Hello, can you please help me to correct the error in my code because the result of the decicion variables:  s[o in Oper][k in Mchs]  (the start time of operation j on machine k) and

c[o in Oper][k in Mchs] (the end time of the operation j on machine k) are not correct !!!

 

Here is the .MOD and the .DAT files:

//.MOD

using CP;

tuple paramsT{
    int nbJobs;
    int nbMchs;
    int nbOper;
   
};

int M=10000;

paramsT Params = ...;    
int nbJobs = Params.nbJobs;
int nbOper = Params.nbOper;
int nbMchs = Params.nbMchs;

range Jobs = 1..nbJobs;
range Oper=1..nbOper;
range Mchs = 1..nbMchs; 

tuple Operation {
  int id;    // Operation id
  int jobId; // Job id
  int pos;   // Position in job
};

tuple Mode {
  int opId; // Operation id
  int mch;  // Machine
  int pt;   // Processing time
};

{Operation} Ops   = ...;
{Mode}      Modes = ...;

// Position of last operation of job j
int jlast[j in Jobs] = max(o in Ops: o.jobId==j) o.pos;
dvar interval ops  [Ops]; 
dvar interval modes[md in Modes] optional size md.pt;

dvar sequence mchs[m in Mchs] in all(md in Modes: md.mch == m) modes[md];

dvar int+ s[o in Oper][k in Mchs];
dvar int+ c[o in Oper][k in Mchs];

dexpr int Cmax  = max(j in Jobs, o in Ops: o.pos==jlast[j]) endOf(ops[o]);

execute {
          cp.param.FailLimit = 10000;
}

minimize Cmax;
subject to {

forall ( k in Mchs, o in Oper,  md in Modes: md.mch==k && md.opId==o, op in Ops: op.id==md.opId) 
c[o][k]-s[o][k]==md.pt;

   
   forall (k1 in Mchs, k2 in Mchs, i in Oper, j in Oper, md1 in Modes, md2 in Modes: md1.mch != md2.mch && md1.mch==k1 && md2.mch==k2, op1 in Ops, op2 in Ops: op1.jobId != op2.jobId && op1.jobId==i && op2.jobId==j)
   
   (s[i][k1]>=c[j][k2]) || (s[j][k2]>=c[i][k1]);

 

  

 
    // Precedence constraints between consecutive operations o,z of a job j
  forall (j in Jobs, m in Mchs, o in Ops, z in Ops: o.jobId==j && z.jobId==j && z.pos==1+o.pos){
    endBeforeStart(ops[o],ops[z]); 
}

    //Only operations on the same machine can be consecutive operations.
    //******************************************************************

    // Alternative machines for a given operation o
  forall (o in Ops) 
    alternative(ops[o], all(md in Modes: md.opId==o.id) modes[md]);
       
    // Operations on a given machine m cannot overlap
  forall (m in Mchs)
    noOverlap(mchs[m]);
}

 
   
execute {
  for (var m in Modes) {
    if (modes[m].present)
      writeln("Operation " + m.opId + " on machine " + m.mch + " starting at " + modes[m].start+ "ends at"+ modes[m].end); 
                 
  }

}

tuple solutionT{
    int operation;
    int machine;
    int start;
    int end;
};
{solutionT} solution = {<m.opId, m.mch, startOf(modes[m]), endOf(modes[m])> | m in Modes : startOf(modes[m]) != 0 };

 

//.DAT

Params = <3, 4, 9>;

Ops = {
  <1,1,0>, 
  <2,1,1>,
  <3,1,2>,
  <4,1,3>,
  
  <5,2,0>, 
  <6,2,1>,
  <7,2,2>
   
  <8,3,0>, 
  <9,3,1>,
};

Modes = {
  <1,1,2>,
  <1,2,3>,
 
  
  <2,1,4>,  

  <3,2,3>,
  <3,3,9>,
 
  
  <4,1,5>,
  
  
  <5,2,4>,
  <5,4,7>, 
    
  <6,1,2>,
  <6,2,4>,
  
  <7,1,4>,
  <7,2,12>,
  
  
  <8,1,4>, 
  <8,3,6>, 
   
  <9,4,15>, 
   
};

 

please i need the solution of the start times and end times soon. I have a deadline to give back this work. Thank you.

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Originally posted by: kcapt

Hi,

I know it was been some years but I have the exact same problem. Did you find the answer? Thanks!

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It is better to start a new thread and describe exactly what your problem is. If possible also attach your model and data so that others have a chance to figure out what may be wrong.

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