Originally posted by: user1234567

Hello, i have a problem with the decision variables c and s. It only displays the first 4 values of the operations. The other values are 0 !!!

Please help me.

I want to use these decision variables later in the constraints. So, i should get correct values of these decision variables. Thank you.

Hi,

in order to debug you could label your constraint:

forall (i in Jobs, j in Oper, o in Ops,md in Modes: o.id == md.opId && md.opId ==j && md.jobId==i)

ctc:c[i][j] == endOf (ops[o]);

and then within the IDE if you click on ctc you will see

regards

Originally posted by: user1234567

yes thank you. But why Cplex does not display the other values of c and s. where is the problem exactly ? And how can i correct it ?

In the "Journal de script", I get the expected and correct results of start and completion times. So i added c and s as decision variables; and i want them to contain these correct values. 

I see

Well you get  those values because they are possible.

If you change

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

 

into

minimize staticLex(max(j in Jobs, o in Ops: o.pos==jlast[j]) endOf(ops[o]),-sum(i in Jobs,j in Oper) s[i][j]);

and

dvar int c[i in Jobs][j in Oper] in 0..1000;
dvar int s[i in Jobs][j in Oper] in 0..1000;

then you will get for s

s = [[7 9 11 14]
             [0 1000 0 1000]
             [1000 1000 1000 381]];

regards

Originally posted by: user1234567

Sorry! The problem is not resolved even when i change the code.

In "Journal de Script", i get this result:

// solution with objective 15; -247
Operation 1of job1 on machine 3 starting at 7and ends at9
Operation 2of job1 on machine 1 starting at 9and ends at11
Operation 3of job1 on machine 2 starting at 11and ends at14
Operation 4of job1 on machine 1 starting at 14and ends at15
Operation 5of job2 on machine 3 starting at 1and ends at3
Operation 6of job2 on machine 2 starting at 3and ends at5
Operation 7of job2 on machine 1 starting at 5and ends at6
Operation 8of job2 on machine 1 starting at 6and ends at9
Operation 9of job3 on machine 1 starting at 0and ends at5
Operation 10of job3 on machine 2 starting at 5and ends at10
Operation 11of job3 on machine 2 starting at 10and ends at11
Operation 12of job3 on machine 3 starting at 11and ends at13

In the left part of "Navigateur de problèmes", i get this solution of s and c: 

s = [[7 9 11 14]
             [0 206 0 0]
             [0 0 0 0]];

c = [[9 11 14 15]
             [0 0 0 0]
             [0 0 0 0]];

Where i should get exactly these results (as displayed in journal de script exactly):

s = [[7 9 11 14]
             [1 3 5 6]
             [0 5 10 11]];

c = [[9 11 14 15]
             [3 5 6 9]
             [5 10 11 13]];

This is what i meant exactly. I don't know how to correct this ! Please help me.

Ok you see

but if you click on c you will get

regards