Alternative formulations for the Set Packing Problem and their application to the Winner Determination Problem

Abstract

An alternative formulation for the set packing problem in a higher dimension is presented. The addition of a new family of binary variables allows us to find new valid inequalities, some of which are shown to be facets of the polytope in the higher dimension. We also consider the Winner Determination Problem, which is equivalent to the set packing problem and whose special structure allows us to easily implement these valid inequalities in a very easy way. The computational experiments illustrate the performance of the valid inequalities and obtain good results.

Appendix: Regarding the procedure in Sherali et al. for the Winner Determination Problem

The general procedure in Sherali et al. (1998) generates tighter relaxations for 0–1 mixed integer problems in a higher dimensional space. This procedure consists of an iterative algorithm which leads up to the convex hull representations in the higher space after a finite number of steps.

Basically, the procedure developed in Sherali et al. (1998) can be summarized as follows. Consider

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and T={y∈ℝⁿ:g _i y−g _0i ≥0, for i=1,…,p}. Consider P={1,…,p} and $\overline{P}=\{n\ \mbox{copies of}\ P\}$. At any level of relaxation d∈{1,…,n}, we construct a higher dimensional relaxation $\overline{X}_{d}$ by considering T-factors of order d defined as follows:

$$g(J)=\prod_{i \in J} (g_i y-g_{0i}) \quad \mbox{for each distinct}\ J\subseteq \overline{P}, \ |J|=d.$$

The Reformulation-Linearization Technique then generates the relaxation $\overline{X}_{d}$ at level d as follows:

• (a) Reformulation. Multiply each inequality defining the feasible region (17) by each g(J) of order d, and apply the identity $y_{j}^{2}=y_{j}$ for all j∈{1,…,n}.

• (b) Linearization. Linearize the resulting polynomial program by substituting

  $$x_J=\prod_{j \in J} y_j \quad \forall J \subseteq N, \qquad v_{Jk}= u_k \prod_{j \in J} y_j \quad \forall J \subseteq N.$$

The sequence that defines the projections $\overline{X}_{P_{d}}=\{(y,u) : (y,u,x,v) \in \overline{X}_{d}\}$ for all d=1,…,n ($\overline{X}_{P_{0}}=\overline{X}_{0}$ is the ordinary linear programming relaxation), satisfies the following relationship:

$$\overline{X}_0 \equiv \overline{X}_{P_0} \supseteq \overline{X}_{P_0} \supseteq \cdots \supseteq \overline{X}_{P_n}=\operatorname{conv}(X),$$

where conv (X) denotes the convex hull of X.

In general, the lifting procedure consists of n consecutive liftings. However, during the progress of the procedure, it is possible that the addition of the k-th new family of variables does not lead to new valid inequalities and that this equivalence relation continues through all levels of relaxations. Hence, the k-th level relaxation itself produces the convex hull representation in this case.

The following result illustrates the family of constraints generated by this procedure for the Winner Determination Problem.

Proposition 5.1

The method of Sherali et al. for the feasible region (12)–(13) at the k-th level provides the following set of constraints:

In addition,

$$ x_{s_1s_2}=0, \quad \mathit{for\ any\ } s_1,s_2 \in S\ \mathit{such\ that\ } s_1 \cap s_2 \ne \emptyset.$$

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Remark 5.1

In the above constraints, we can assume without loss of generality that: (i) p _i ≠p _j , and (ii) ${ s_{p_{i}} \cap s_{p_{j}} = \emptyset}$, ∀i,j∈{1,…,k}, because by using (18), the inequalities that consider $p_{i_{0}}$ and $p_{j_{0}}$, such that, $p_{i_{0}}=p_{j_{0}}$ or $s_{p_{i_{0}}} \cap s_{p_{j_{0}}} \ne \emptyset$ for some i ₀, j ₀∈{1,…,k} can be obtained as constraints generated by this procedure in the (k−1)-level.

Proof

Let T be defined by:

$$T= \biggl\{y_s \mbox{ with $s \in S$ and}\ 1- \sum_{s \ni p} y_s \ge 0, \ \forall p \in P, y_s \ge 0\biggr\}.$$

Therefore, we obtain the following constrains for the different levels.

Level 1.

By applying the Sherali et al. method, the constraints for the second level are obtained from the product of the constraints for the first level with $y_{s_{p_{2}}}$ and with $1- \sum_{s_{p_{2}} \ni p_{2}} y_{s_{p_{2}}}$. Therefore, we distinguish the following cases:

Level 2.

Case 1: $(1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}}) y_{s_{p_{2}}}$ for any p ₁∈P, $s_{p_{2}} \in S$. For the case $p_{1} \notin s_{p_{2}}$, we obtain

$$\biggl(1-\sum_{s_{p_1} \ni p_1} y_{s_{p_1}}\biggr) y_{s_{p_2}}=y_{s_{p_2}}- \sum_{s_{p_1} \ni p_1} x_{s_{p_1}s_{p_2}} \ge 0.$$

Observe that for the case $p_{1} \in s_{p_{1}}\cap s_{p_{2}}$, we have that

$$\biggl(1-\sum_{s_{p_1} \ni p_1} y_{s_{p_1}}\biggr) y_{s_{p_2}}=y_{s_{p_2}}-y_{s_{p_2}}^2-\sum_{s_{p_1} \ni p_1}x_{s_{p_1}s_{p_2}}= -\sum_{s_{p_1} \ni p_1,s_{p_1}\ne s_{p_2}}x_{s_{p_1}s_{p_2}}\ge 0.$$

Hence, $\sum_{s_{p_{1}} \ni p_{1}} x_{s_{p_{1}}s_{p_{2}}} \le 0$, and it implies that $x_{s_{p_{1}}s_{p_{2}}}=0$, $\forall s_{p_{1}},s_{p_{2}}\in S$ with $p_{1} \in s_{p_{1}} \cap s_{p_{2}}$, or equivalently,

$$x_{s_{p_1}s_{p_2}}=0, \quad \forall s_{p_1},s_{p_2}\in S \mbox{ with }s_{p_1} \cap s_{p_2} \neq \emptyset.$$

Case 2: $(1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}}) (1-\sum_{s_{p_{2}} \ni p_{2}} y_{s_{p_{2}}})$ for any p ₁,p ₂∈P.

$$\biggl(1-\sum_{s_{p_1} \ni p_1} y_{s_{p_1}}\biggr)\biggl(1-\sum_{s_{p_2} \ni p_2} y_{s_{p_2}}\biggr)=1-\sum_{s_{p_1} \ni p_1} y_{s_{p_1}}-\sum_{s_{p_2} \ni p_2}y_{s_{p_2}}+ \sum_{s_{p_1} \ni p_1, s_{p_2} \ni p_2}x_{s_{p_1}s_{p_2}}\ge 0.$$

Analogously to the second level, by Sherali et al. method, the constraints for the third level are obtained from the product of the constraints for the second level with $y_{s_{p_{3}}}$ and with $1-\sum_{s_{p_{3}} \ni p_{3}} y_{s_{p_{3}}}$, for convenience in the notation, we will multiply by $1- \sum_{s_{p_{1}} \ni p_{1}}y_{s_{p_{1}}}$. Therefore, we highlight the following cases:

Level 3.

Case 1: $(y_{s_{p_{2}}}-\sum_{s_{p_{1}} \ni p_{1}} x_{s_{p_{1}}s_{p_{2}}})y_{s_{p_{3}}}$ for any p ₁∈P and $s_{p_{2}},s_{p_{3}} \in S$.

$$\biggl(y_{s_{p_2}}- \sum_{s_{p_1} \ni p_1}x_{s_{p_1}s_{p_2}}\biggr)y_{s_{p_3}}= x_{s_{p_2}s_{p_3}}-\sum_{s_{p_1}\ni p_1} y_{s_{p_1}s_{p_2}s_{p_3}} \ge 0.$$

Case 2: $(1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}}-\sum_{s_{p_{2}} \ni p_{2}} y_{s_{p_{2}}}+\sum_{s_{p_{1}} \ni p_{1}, \atop s_{p_{2}} \ni p_{2}}x_{s_{p_{1}}s_{p_{2}}})y_{s_{p_{3}}}$ for any p ₁,p ₂∈P, $s_{p_{3}}\in P$.

Case 3: $(y_{s_{p_{3}}}-\sum_{s_{p_{2}} \ni p_{2}} x_{s_{p_{2}}s_{p_{3}}}) (1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}})$ for any p ₁,p ₂∈P, $s_{p_{3}} \in P$.

This constraint has already obtained in the Case 3.2.

Case 4: $(1-\sum_{s_{p_{2}} \ni p_{2}} y_{s_{p_{2}}}-\sum_{s_{p_{3}} \ni p_{3}} y_{s_{p_{3}}}+\sum_{s_{p_{2}} \ni p_{2}, \atop s_{p_{3}} \ni p_{3}}x_{s_{p_{2}}s_{p_{3}}}) (1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}})$ for any p ₁,p ₂,p ₃∈P.

We can see that the expressions for the first, second and third level fit into the general expression given in the theorem for k=1,2,3, respectively. Now, we assume that the general expressions given in the theorem are valid for the (k−1)-th level and we will prove that it is also valid for k-th level. To do so, we first prove the following equality, for any j=2,…,k, $s_{p_{j}}, \ldots, s_{p_{k}} \in S$ and p ₁,…,p _k ∈P:

Indeed,

and

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Hence, the constraints for the k-th level can be obtained from the products of each one of the constraints for the (k−1)-th level with $y_{s_{p_{k}}}$ and the product of the constraint

$$1+ \sum_{t=2}^k (-1)^{t-1}\sum\limits_{\{q_2,\ldots,q_t\} \subseteq \{p_2,\ldots,p_k\},\atop s_{q_2} \ni q_2,\ldots,s_{q_t} \ni q_t} x_{s_{q_2}\cdots s_{q_t}} \ge 0,$$

with $1-\sum_{s_{p_{1}} \ni p_{1}} y_{s_{p_{1}}}$. Therefore, the first type of products gives us the constraints (19) for j=1,…,k and the second provides the following constraint:

and the result follows. □