[SCIP] Bug in the branching rule of the bin packing example ?
Gerald Gamrath
gamrath at zib.de
Thu Mar 23 23:00:00 CET 2017
Dear Xavier,
but does your problem have the property, that each subset of a valid
packing is again a valid packing? And each item must be covered once in
the end?
I would say that you can then ignore pairs of variables for which the
sum over the LP values is larger than 1, because you could post-process
this in the end and reduce this value by removing both items for some
packings.
More specifically, you should choose two items such that those are
together in some of the packings selected by the LP solution and also
some packings are selected that contain exactly one of the items. This
is not the case for your example. If you do Ryan and Foster branching
based in this criterion, each variable should have a value 0 or 1 in the
end.
I hope this makes sense.
Best,
Gerald
Am 23.03.2017 um 18:50 schrieb Xavier Schepler:
> I meant :
>
> After re-optimization, the sum over the variables which cover items 1
> and 3 is still fractional, and its value is *1.5.* (not 0.5)
>
>
>
> 2017-03-23 18:45 GMT+01:00 Xavier Schepler <xavier.schepler at gmail.com
> <mailto:xavier.schepler at gmail.com>>:
>
> Dear Gerald,
>
> Thank you for your response.
>
> Assume that a branching constraint was added, enforcing that items
> 1 and 3 must be in the same bin. The variables that do not satisfy
> this constraint were deleted. The resulting LP is the one in the
> attached file. After re-optimization, the sum over the variables
> which cover items 1 and 3 is still fractional, and its value is
> 0.5. Assume that no improving variable is priced out, during the
> column generation phase. What does prevent the same branching
> constraint to be generated again ?
> This is because, in the case of a set covering problem, the sum
> over the variables which cover items 1 and 3 can be greater than
> one and fractional, which can not happen with a set partitionning
> problem.
> So, my opinion is that, with the set covering formulation, in the
> case of a fractional sum value for a pair of a items, it must be
> checked whether the branching constraint is satisfied (i.e. that
> it is already exists), and only in the case that is is not, to
> create it.
>
> Does it make sense ?
>
> Kind regards,
>
> Xavier
>
>
> 2017-03-23 16:38 GMT+01:00 Gerald Gamrath <gamrath at zib.de
> <mailto:gamrath at zib.de>>:
>
> Dear Xavier,
>
> I would guess that there is a bug in your pricing problem.
>
> Normally, if you branch on items i and j and go into, say, the
> child node where SAME(i,j) should hold, you fix all packings
> locally to 0, which contain exactly one of the two items. But
> additionally, you need to adjust your pricing problem, so that
> no further packings with that property are priced in in this
> subtree, say by adding a constraint x_i = x_j. Note that
> variables are always valid globally, so if you jump through
> the tree and come back to this sub-tree later on, you need to
> repropagate the node again to make sure that variables
> generated in between also stick to the branching decision.
>
> This works fine for set partitioning problems or set covering
> problems if you can ensure that there is always at least one
> optimal solution which would be feasible for the corresponding
> set partitioning problem.
>
> If in your problem, this is not the case and there is no
> optimal solution which fulfills all constraints with equality,
> I fear you cannot use Ryan and Foster branching.
>
> Your fix might cure the symptoms, but I guess the bug itself
> persists. Because if you did some branching on SAME(i,j)
> before, there should not be any incentive to do the same
> branching again. If there is, then the old branching was
> actually not strictly enforced, and as such, you might now
> choose to not do the same branching again, but you might end
> up with a situation where you performed all possible branching
> but are still not feasible, since as it seems, your branchings
> are not (always) respected.
>
> Best,
> Gerald
>
>
> On 23.03.2017 16:09, Xavier Schepler wrote:
>> Dear SCIP developers,
>>
>> I'm working on a variant of the bin-packing problem. I
>> started from the example of a branch-and-price algorithm for
>> the bin packing problem provided with SCIP 3.2.1. It uses the
>> Ryan and Foster branching rule.
>>
>> However, the master problem is a set cover problem, so,
>> potentially, after branching on a pair of item, it may happen
>> that this same pair of items comes up again with a fractional
>> sum value in a child node.
>> To avoid this problem, we must ensure that the pair selected
>> for branching has not been selected before in an ancestor for
>> branching. If it has, another pair must be selected.
>>
>> Before adding a piece of code that checks if a pair was
>> selected before for branching, I had my program branching
>> again and again on the same pair of variable, for one
>> particular instance.
>>
>> The code which computes the fractional values of the pair is
>> now as follows :
>>
>> /* compute weigthts for each order pair */
>> for (v = 0; v < nlpcands; ++v) {
>> assert(lpcands[v] != NULL);
>> /* get variable data which contains the information to which
>> constraints/items the variable belongs */
>> vardata = SCIPvarGetData(lpcands[v]);
>> consids = SCIPvardataGetConsids(vardata);
>> nconsids = SCIPvardataGetNConsids(vardata);
>> assert(nconsids > 0);
>> /* loop over all constraints/items the variable belongs to */
>> for (i = 0; i < nconsids; ++i) {
>> id1 = consids[i];
>> for (j = i + 1; j < nconsids; ++j) {
>> id2 = consids[j];
>> assert(id1 < id2);
>> if (branchpairs[id2][id1]) {
>> continue;
>> }
>> pairweights[id2][id1] += lpcandsfrac[v];
>> assert(SCIPisFeasLE(scip, pairweights[id2][id1], 1.0));
>> assert(SCIPisFeasGE(scip, pairweights[id2][id1], 0.0));
>> }
>> }
>> }
>> branchpairs[id2][id1] is equal to zero if no branching
>> constraints exists on the pair {id2, id1} for the current
>> node, and it is equal to one otherwise.
>> If a branching was done on the pair id2, id1, then
>> pairweights[id2][id1] will be equal to zero, so this pair
>> will not be selected for branching.
>> Am I right ?
>> Kind regards,
>> Xavier Schepler
>>
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