Originally posted by: TRAN Van Ut

Dear my friend, 

I jobs 1, 2,… with processing time p and due (please see the file .dat and in file Z_H.ZIP)Z_H.zip|View Details

Such that all jobs respect of (i.e Cj-dj] <=0 for all j from 1 to n). My code is below:

=

Process = 1.;

p     [Process] = ...

d     [Process] = ...

float+ CProcess];

boolean zProcess[Process];

int+ h;

h;

to

{

      in Processj in Processk in Process)

            [k]<=z[j+zj[k];    

      in 1.nbJobs-1j in i+1. . )

            C[j]>= C[i]+ p[i]-1000*(1-z[i][j]);

      in 1.nbJobs-1j in i+1. . )

            C[i]>= C[j]+ p[j]-1000*z[i][j];               

      j in Process)

            Cj-dj] <=0;                          

      == sum in 1.nbJobs-1, j in i+1.) z[j];

}

In fact, the right result the position of is:  10  9  1  8  7  2  6  5  3  4 and Minh=12

With this model, I got only   10  9  1  8  7  2  6  5  4  3 and Minh=11. But when I checked by hand that  C3-d3=1, it cannot satisfy this constraint: Cj-dj] <=0;

Please help me what was wrong in my code

Thanks in advance for your help

Best regards,

Hi,

I do not have access to your .dat but I wonder whether you had a look at CPO

See example at CPLEX_Studio1271\opl\examples\opl\sched_jobshop

regards

Originally posted by: TRAN Van Ut

Thank you for your answer.

This is the content of .dat file. 

= 10;
= [1, 1, 1, 44, 57, 26, 94, 52, 35, 65];
= [125, 250, 375, 376, 376, 376, 376, 376, 376, 376];

Please help me to check my code!

Thank you for your help!

Regards,

forall(j in Process)

            C[j]-d[j] <=-1;   

instead of

forall(j in Process)

            C[j]-d[j] <=-0;

will give 12 for objective

regards

Originally posted by: TRAN Van Ut

Thank you very much, AlexFleischer!

Have a good day!