Originally posted by: AndyHam

I am seeking advice here. Any help will be very appreciated^^

This is a daily production planning problem. Due to sequence dependent setup and time-based pm, I am exploring CP approach.
Suppose there is a set of jobs with different quantities and processing times. The goal is to achieve a daily output target.
The following is a result. The time unit is hour so I converted the "End" hour into "Day".

Then, we can calculate the daily output.

Now, I am trying to make the factory produce 16 units every day. Any shortage amount needs to be summed up and penalized accordingly. Here are my questions:
(1) How to map interval variables value into integer variables in order to calculate the daily output.
(2) When an interval of job spans two days, I need to credit the proportional amount of output to each day. For instance, job k starts at 17 and ends at 40 with 6 units. Then, Day1 should get 2 and Day2 get 4 units.

using CP;
tuple t_Plan {
    key int id;
    int group;
    int qty;
    int pt;    
};
{t_Plan} Plan={
<1,    1,  5,  2>,
<2, 2,  10, 1>,
<3, 3,  7,  2>,
<4,    1,  7,  2>,
<5,    2,  8,  1>,
<6,    3,  8,  2>,
<7,    1,  9,  2>,
<8,    2,  1,  1>,
<9,    3,  3,  2>,
<10,1,  6,  2>,
<11,2,  9,  1>,
<12,3,  5,  2>,
<13,1,  6,  2>,
<14,2,  8,  1>,
<15,3,  4,  2>
};

dvar interval itvPlan[p in Plan] size p.qty*p.pt;
dvar sequence seqMch in all(p in Plan) itvPlan[p];
minimize max(p in Plan) endOf(itvPlan[p]);
subject to {
    noOverlap (seqMch);
}

execute{
for (var p in Plan)      
  writeln( p.id + "\t" + p.qty + "\t" + itvPlan[p].start + "\t" + itvPlan[p].end );
}
 

Thanks,
Andy

Originally posted by: PhilippeLaborie

If the number of days is not too large compared to the time granularity, I would use 'overlapLength' expressions, like in this model:

int outmin = 16;
int nDays = 8;
range Days = 1..nDays;

int start[d in Days] = (d-1)*24;
int end  [d in Days] = d*24;

using CP;

dvar interval itvPlan[j in Plan]  in 0..end[nDays] size j.qty*j.pt;
dvar sequence seqMch in all(p in Plan) itvPlan[p];

dexpr int shortage[d in Days] = maxl(0, outmin - sum(j in Plan) overlapLength(itvPlan[j],start[d],end[d]) div j.pt);

minimize sum(d in Days) shortage[d];
subject to {
  noOverlap(seqMch); // Assuming single machine
}

Originally posted by: AndyHam

This is great! I tried 90 days planning with this suggestion, it worked well. 
I few times used the overlapLength to calculate the overlap length of two interval variables, but I did not know the function can mix with integer variables.
Another learning!