! LINGO model for sequencing jobs in a flow shop.
  Assume that same sequence is used on all machines;
SETS:
 JOB;
 JINP( JOB, JOB):Y;
 MACH;
 MXP( MACH, JOB): PT, FT;
ENDSETS
DATA:
  MACH = A..C; ! The set of machines;
  JOB = 1..6;  ! The set of jobs;
  PT = 9 3 4 8 2 1   ! Processing time on machine 1;
       3 7 6 2 4 5   ! Processing time on machine 2;
       6 5 2 4 8 3;  ! on machine 2;
! Keywords: Sequencing, Scheduling, Flow shop scheduling;
ENDDATA
! Variables:
   Y(j,p) = 1 if job j is assigned to position p,
   FT(m,p) = finish time on machine m of job in position p;

 @FOR( JINP(j,p): @BIN(Y(j,p))); ! The Y's = 0 or 1;

 nm = @SIZE(MACH);
 nj = @SIZE(JOB);
 !MIN = @SUM( JOB(p): FT(nm, p));  MIN = FT(nm,nj); ! Minimize makespan;

 ! Each job must be assigned a position;
 @FOR( JOB(j):
   @SUM( JOB(p): Y(j,p)) = 1;
     );
 ! Each position gets a job;
  @FOR( JOB(p):
    @SUM( JOB(j): Y(j,p)) = 1;
      );

 ! Finish time on machine m of job in position p;
 ! Jobs on machine 1 have no predecessors;
  FT(1,1) >= @SUM( JOB(j): PT(1,j)*Y(j,1));
  @FOR( JOB( p) | p #GT# 1:
 ! Job in position p cannot start until job in position p-1 finishes;
    FT(1,p) >= FT(1,p-1) + @SUM( JOB(j): PT(1,j)*Y(j,p));
      );

 @FOR( MACH(m) | m #GT# 1:
 ! Job in position p on machine m cannot start until it finishes on m-1;
  FT(m,1) >= FT(m-1,1) + @SUM( JOB(j): PT(m,j)*Y(j,1));

  @FOR( JOB( p) | p #GT# 1:
 ! Job in position p cannot start on m until previous job
    finishes on m;
    FT(m,p) >= FT(m,p-1) + @SUM( JOB(j): PT(m,j)*Y(j,p));
 ! Job in position p on machine m cannot start until it finishes on m-1;
    FT(m,p) >= FT(m-1,p) + @SUM( JOB(j): PT(m,j)*Y(j,p));
      );
     );