[SCIP] / / LPRoot and Original Problem /

[Gregor Hendel]

Hi Samuel and list,

Am 18.02.2016 um 14:02 schrieb Deleplanque Samuel:
> Dear SCIP team,
>
> I'm working on a typical pricing problem. I've two formulations: one 
> is solved with a Branch & Price and the other one with a typical 
> Branch & Bound, both using SCIP.
> I have a problem with collecting the LP Root for the second one. 
> Indeed, I would like the LP-Relaxation objective value of my problem 
> and not the transformed one, or any other solution obtained in the 
> root node.
You should use SCIPretransformObjval(scip, transformedobjval) to 
transform objective values from the transformed space back to the 
original problem space. By convention, we call values in the transformed 
space the lower and upper bound and their original counterparts dual and 
primal bound, respectively.

In order to obtain the first LP value, you can use the special function 
SCIPgetFirstLPDualboundRoot(scip).

>
> For example, in the trace bellow, the LP-Relaxation objective value 
> for an instance of my formulation is 1.041834e+04 (the MIP optimal is 
> found before any branching).
>
>  time | node  | left  |LP iter|LP it/n| mem |mdpt |frac |vars |cons 
> |cols |rows |cuts |confs|strbr|  dualbound   | primalbound  |  gap
> T 0.0s|     1 |     0 |     0 |     - | 248k|   0 |   - |  10 |  20 | 
>  10 |  20 |   0 |   0 |   0 |      --      | 0.000000e+00 |    Inf
>   0.0s|     1 |     0 |    13 |     - | 246k|   0 |   2 |  10 |  20 | 
>  10 |  20 |   0 |   0 |   0 | 1.041834e+04 | 0.000000e+00 |    Inf
> s 0.0s|     1 |     0 |    13 |     - | 247k|   0 |   2 |  10 |  20 | 
>  10 |  20 |   0 |   0 |   0 | 1.041834e+04 | 3.029395e+03 | 243.91%
>   0.0s|     1 |     0 |    16 |     - | 248k|   0 |   0 |  10 |  20 | 
>  10 |  21 |   1 |   0 |   0 | 9.957000e+03 | 3.029395e+03 | 228.68%
> * 0.0s|     1 |     0 |    16 |     - | 248k|   0 |   - |  10 |  20 | 
>  10 |  21 |   1 |   0 |   0 | 9.957000e+03 | 9.957000e+03 |   0.00%
>
> If I use any function to obtain the LPRoot value 
> (/SCIPgetLPRootObjval/, /SCIPgetLPRootObjval/ or 
> /SCIPgetLowerboundRoot/) I logically get 9.957000e+03. If I use 
> /SCIPgetSolOrigObj/ I obtain the same and not the 1.041834e+04 of the 
> LP relaxation of my original problem.
>
> Also, because I want to compare the two formulations, and then to 
> compare what is comparable, I would like to know what are the 
> different algorithms executed in the root. If I try to understand: the 
> T and s mean two heuristics has been used and * means normally an 
> integral LP-relaxation (but not for my original problem).
Exactly. Use 'display statistics' or SCIPprintStatistics() to print an 
exhaustive plugin summary to see which of the plugins ran, which 
succeeded, and which wasted your precious user time. All these 
statistics are also available through the API, but I would suggest to 
have a look at the tabular output first to get an impression of what is 
important.

A list of the primal heuristics of SCIP and their display characters is 
available with the command 'display heuristics'.

Kind regards,
Gregor

>
> Thanks :)
>
> Samuel Deleplanque
>
> -- 
> /***************************************
> Samuel Deleplanque, PHD*
> *
> *PostDoc at /Université Libre de Bruxelles/*
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