Technical Note—Maintaining the All-Integer Property of an ILP when Using Cuts in Rational Data

If all data for an ILP is integer and a cut is given in rational data, then there exists a transformation of the cut such that all the coefficients and the right hand side constant are integers, thus requiring the slack variable also to be an integer. By repeating this transformation inductively an all-integer problem remains all-integer. As an application of this result, it is shown how to solve an ILP by using transformed strong MILP cuts. As these cuts are superior to fractional cuts, a more efficient algorithm results.