On the exact separation of mixed integer knapsack cuts - Marcos ...

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28 R. Fukasawa, M. Goycooleathe best known feasible solution. In our implementation, we additionally prune nodeswhen (c) it can be shown that every optimal solution in those nodes is dominated.Using dominance to improve the branch and bound search can have an importantimpact on the effectiveness of the search [37]. In fact, lexicographic and cost dominanceallow us to disregard feasible solutions that are not the unique lexicographicallysmallest optimum solution, hence significantly reducing the search space.In general, the problem of detecting if a solution is dominated can be extremelydifficult. Fischetti et al. [27,28] detect domination by means of solving an MIP. Inwhat follows we describe a simple methodology for identifying specific cases of dominationarising in instances of the mixed integer knapsack problem. Note however, thatthe notion of dominance as defined above is much more general, and could be used forgeneral mixed integer programming problems as well [27,28,32]. As a final comment,note that Andonov et al [2] have also used domination in the context of dynamic programmingalgorithms, having had great success in tackling the unbounded knapsackproblem.Throughout this section we will adopt the convention that ∞+r =∞, ∀r ∈ R.3.4.1 Domination between pairs of integer variablesConsider indices i, j ∈ I , and non-zero integers k i , k j .Ifa i k i + a j k j ≥ 0 and c i k i +c j k j < 0 we say that (i, j, k i , k j ) defines an integer cost-domination tuple. If i < j,k i > 0, a i k i + a j k j ≥ 0 and c i k i + c j k j = 0 we say that (i, j, k i , k j ) defines an integerlexicographic-domination tuple. The propositions below show how dominationtuples allow for the easy identification of dominated solutions.Proposition 4 Consider an integer cost/lexicographic-domination tuple (i, j, k i , k j )and let x be a feasible MIKP solution. Define δ ∈ Z n such that δ i = k i , δ j = k j andδ q = 0 for all q ∈{1,...,n}\{i, j}.Ifl i + k i ≤ x i ≤ u i + k i and l j + k j ≤ x j ≤ u j + k j (2)Then x is cost/lexicographically-dominated by x − δ.To see that this proposition is true, it is simply a matter of observing that condition2 implies that (x − δ) is a feasible solution in terms of the bounds, and thata i k i + a j k j ≥ 0 and c i k i + c j k j ≤ 0 imply the domination.The following Theorem follows directly from Proposition 4, and illustrates howdomination tuples can be used to strengthen a branch and bound algorithm.Theorem 1 Consider two integer type variables x i and x j and a domination tuple(i, j, k i , k j ). If in some node of the branch and bound tree we have that l i + k i ≤ x i ≤u i + k i is satisfied by all solutions in that node, then:• If k j > 0 we can impose the constraint x j ≤ l j + k j − 1 in that node,• If k j < 0 we can impose the constraint x j ≥ u j + k j + 1 in that node,and in either case we will not cut off the unique non-dominated optimal solution tothe original MIP.123
On the exact separation of MIK cuts 29Observe that whenever (c i , c j ) and (a i , a j ) are linearly independent there exist aninfinite number of cost-domination tuples. Likewise, there exist an infinite number oflexicographic-domination tuples in the linear dependent case.The two following propositions state that, in each case, there always exists a minimaldomination tuple. That is, a domination tuple (i, j, k i , k j ) such that all otherdomination tuples (i, j, k ′ i , k′ j ) defined for the same variables, satisfy |k i|≤|k ′ i | and|k j |≤|k ′ j |. The proof follows from the observation that since (c i, c j ) and (a i , a j )are in the same orthant (they have the same sign in each coordinate), the intersectionof the half spaces c i x i + c j x j < 0 and a i x i + a j x j ≥ 0 must also be contained inan orthant. Thus, it easily follows that there must be a minimal domination tuple. Wenow prove this more formally.Proposition 5 If (a i , a j ) and (c i , c j ) are linearly independent, let,D = { (k i , k j ) ∈ Z × Z : k i , k j ̸= 0, a i k i + a j k j ≥ 0 and c i k i + c j k j < 0 }The following two properties hold:(i) D is contained in the interior of ( an orthant ) in N × N. More precisely, forevery (k i , k j ) ∈ D we have that: c i − c ja ja i < 0 if and only if k i > 0 and( )c j − c ia ia j < 0 if and only if k j > 0.(ii) There exists (ki o, ko j ) in D such that (k i, k j ) in D implies |ki o|≤|k i| and |k o j |≤|k j |.Proof Consider (k i , k j ) ∈ D. Observe that since sign(a i ) = sign(c i ) we can multiplyconstraint a i k i + a j k j ≥ 0by− c ja jand add it to c i k i + c j k j < 0 and we get that( ) ( )c i − c ja ja i k i < 0. Likewise c j − c ia ia j k j < 0. Thus, (i) follows.Consider (x i , x j ) and (y i , y j ) in D.Letz i = x i if |x i |≤|y i |. Otherwise, let z i = y i .Let z j = x j if |x j |≤|y j |. Otherwise, let z j = y j . Since D is contained in a pointedcone in one of the four orthants, it is easy to see that (z i , z j ) ∈ D. Thus there mustexist (ki o, ko j) in D satisfying the required condition.⊓⊔Proposition 6 If (a i , a j ) and (c i , c j ) are linearly dependent, let,D ={(k i , k j ) ∈ N × Z : k i , k j ̸= 0, a i k i + a j k j = 0 and c i k i + c j k j = 0}There exists (k o i , ko j ) in D such that (k i, k j ) in D implies |k o i |≤|k i| and |k o j |≤|k j|.Proof Given the linear dependence of (a i , a j ) and (c i , c j ) we have that,D ={(k i , k j ) ∈ N × Z : k i , k j ̸= 0, k ik j=− a ja i}That this set is non-empty follows from the fact that a i and a j are∣rational∣numbers. ∣Let(ki o ∣∣ , ko j ) be the point in D with the smallest ko i . Clearly, |ko j |= a i ∣∣a j |ko i|≤∣ a i ∣∣a |ki j|=|k j | for all (k i , k j ) ∈ D. ⊓⊔123
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