[Scip] multiple pricer

Sebastian Ruther Sebastian.Ruther at uon.edu.au
Wed Feb 16 08:05:36 MET 2011


  Hello Gamrath,


I successfully implemented solving the root node if there is only one 
pricing problem. The algorithm terminates correctly and I verified that 
the solution is correct. Btw, I think the termination criteria is that 
no variable was added during the scip_redcost call (so not necessarily 
that the reduced cost where non-negative, correct?) . Also, no column is 
ever generated twice, so that's all good.

However, when I have multiple pricing problems it doesn't terminate 
anymore. The whole thing is setup as follows:
There is only one pricer that holds a map of pricing problems. Each of 
these pricing problems is a struct that hold all the data relevant for 
the pricing problem as well as a subscip pointer and all it's vars and 
cons.
When scip_redcost is called I run through the map and solve the pricing 
problems (for now) sequentially. Each pricing problem gets solved in 
each iteration (for now). And here is the problem: in the first 7 
iteration columns are correctly generated for each pricing problem. 
However, in iterations 8+ each pricing problems generates the same 
column it generated in iteration 7. So there are no new ones at all. I 
used a diff tool to compare the pricing problems of iterations 7 and 8 
and they are exactly the same (so the dual values didn't change). Why 
doesn't it terminate?

Have you experienced something similar?

Thanks
Sebastian

On 29/01/2011 2:50 AM, Gerald Gamrath wrote:
> Hi Sebastian,
>
>> The other option would be to have only one pricer that holds all the
>> pricing problems and their data. Obviously, controlling the sequence of
>> pricing problems would be easy in this case. Can you think of any
>> reasons why I shouldn't do it this way?
> no, that is exactly the way I have already done it in my projects and it
> worked well. I think that this is also the only way to run the pricing
> in parallel, if you do not want to change the SCIP core.
>
> Best,
> Gerald
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