3
$\begingroup$

I am using OR-Tools for an employees scheduling problem. I have a dictionary that is defined as an IntVar, where such dictionary contains the number of hours that will be assigned to each employee, ranging from zero to availability_for_day:

intvars[ID_VAR] = model.NewIntVar(0, availability_for_day, ID_VAR)
model.Add(sum[intvars[intv] for intv in intvars.keys()]<= availability_for_day

The ID_VAR keys for intvars are an f-string combination between dates (yyyy-mm-dd) and employee_id. I want to add a constraint that allows me to set a minimum number of employees that are assigned for a task for each day. Therefore, my question is: how do I count how many intvars[intv] values are greater than zero? Then, how do I set the number of intvars[intv] that are greater than zero to be equal or greater than a minimum number of employees that I want to allocate?

So far, I have tried using another IntVar as a counter, but I could not make it work.

$\endgroup$

2 Answers 2

4
$\begingroup$

You can a look at this doc.

You will create one Boolean variable per integer variable, and sum them.

$\endgroup$
2
  • $\begingroup$ I followed the documentation you linked, and created a Boolean variable for each employee, and such variable is "linked" to the Int variable value (it is true only if the int var is greater than zero). However, when I try doing this: model.Add(sum( [ boolvars[boolv] for boolv in boolvars.keys() ]) )<= num_min_resources I get this error: "NotImplementedError evaluating a LinearExpr instance as a Boolean is not implemented" Any idea why is that happening? $\endgroup$
    – luca_dix
    Commented Jun 5, 2023 at 14:39
  • 1
    $\begingroup$ your code is wrong. One ')' is misplaced. $\endgroup$ Commented Jun 5, 2023 at 15:31
0
$\begingroup$

It's like bin packing problem. Just define set of binary variables $x_{d,s}$ where $d,s$ is same index as your intvar keys -(date d, staff s). Then add constraints
$ x_{d,s} \le H_{d,s} \le A_{d,s}x_{d,s} \\ \forall d,s $
where $H_{d,s} $ is your intvar and $A_{d,s}$ is available hours per $s$

$ L_d\le \sum_s x_{d,s} \ \forall d$ where $L$ is min staff required for date $d$

$\endgroup$
3
  • $\begingroup$ The API is different with CP-SAT. It uses indicator constraints instead of big-M. $\endgroup$ Commented Jun 1, 2023 at 20:18
  • $\begingroup$ @LaurentPerron I don't use CP-SAT, hoping the OP will get it. Most solvers will wave indicator constraints $\endgroup$ Commented Jun 1, 2023 at 21:31
  • $\begingroup$ yes, actually CP-SAT being SAT based use indicator constraints natively. bigM are recreated on the fly for the underlying (and optional) simplex. $\endgroup$ Commented Jun 1, 2023 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.