let ILP be an integer linear program with constraints-matrix $\boldsymbol{\mathrm{M}}\in\mathbb{Z}^{m\times n}$ and cost vector $\boldsymbol{\mathrm{c}}\in\mathbb{Z}^n$,


${\boldsymbol{\mathrm{x}}^\*}\in\mathbb{Z}^n,\,{\boldsymbol{\mathrm{x}}^\*}^T\boldsymbol{\mathrm{c}}\in\mathbb{Z}\le\boldsymbol{\mathrm{x}}^T\boldsymbol{\mathrm{c}}\quad \forall\boldsymbol{\mathrm{x}}\in\mathbb{Z}^n$ the optimal integral solution of ILP.


${\boldsymbol{\mathrm{y}}^\*}\in\mathbb{R}^n,\,{\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\notin\mathbb{Z}\le\boldsymbol{\mathrm{y}}^T\boldsymbol{\mathrm{c}}\quad \forall\boldsymbol{\mathrm{y}}\in\mathbb{R}^n$ the optimal solution of the relaxed ILP, then we have: ${\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\lt{\boldsymbol{\mathrm{x}}^\*}^T\boldsymbol{\mathrm{c}}$



> 
> **Question:**
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> 
> would adding the constraint $\boldsymbol{\mathrm{x}}^T\boldsymbol{\mathrm{c}}\ge\lceil{\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\rceil$ to the ILP constraints be beneficial for finding the optimal integral solution e.g. via cut and branch?
> 
> 
> 



Benefit of adding a trivial constraint to ILPs
This question is addressed in a few OR StackExchange questions:


* <https://or.stackexchange.com/questions/419/feeding-known-lower-bounds-to-solvers>
* <https://or.stackexchange.com/questions/3777/how-to-exploit-known-solution-in-milp>
* <https://or.stackexchange.com/questions/5331/objective-integrality-cuts>


The summary is that explicitly adding such a cut tends to hurt, both because it can cause numerically difficulties and because it hides the distinction among node bounds, but implicitly using such a cut to terminate is basically free and implemented in most solvers.


