Two New Reformulation Convexification Based Hierarchies for 0-1 MIPs

First, we introduce two new reformulation convexification based hierarchies called RTC and RSC for which the rank d continuous relaxations are denoted by ?̂? RTC and ?̂? d RSC, respectively. These two hierarchies are obtained using two different convexification schemes: term convexification in the case of the RTC hierarchy and standard convexification in the case of the RSC hierarchy. Secondly, we compare the strength of these two hierarchies. We will prove that (i) the hierarchy RTC is equivalent to the RLT hierarchy of Sherali-Adams, (ii) the hierarchy RTC dominates the hierarchy RSC, and (iii) the hierarchy RSC is dominated by the Lift-and-Project hierarchy.Thirdly, for every rankd, wewill prove that conv(T∩E t ) ⊆ ?̂? d RTC ⊆ T d and conv(S∩E s ) ⊆ ?̂? d RSC ⊆ S d where the setsT andS are convex, whileE t andE s are two nonconvex sets with empty interior (all these sets depend on the convexification step). The first inclusions allow, in some cases, an explicit characterization (in the space of the original variables) of the RLT relaxations. Finally, we will discuss weak version of both RTC and RSC hierarchies and we will emphasize some connections between them.


Introduction
Let  and  be two integers.Let  and  be the two sets {1, . . ., } and {1, . . .,  + }, respectively.Let  be a subset of R + + representing the set of feasible solutions of a mixed integer linear program.The integer  indicates the number of binary variables.We will assume that the set  is bounded and has the following nonlinear description: −  ≤ 0,  ∈ , ∈ {0, 1} ,  ∈ .
In descriptions (1)-( 4) above the set  contains the indices of the  binary variables describing ; for each index  belonging to  the vectors   and  belong to R  , where  is a positive integer indicating the number of constraints in (1).The th component of the two vectors   and  will be denoted by   and   , respectively.Finally,  defined by the constraints (1), (2), and (3) will denote the continuous (or linear) relaxation of the mixed integer set .
In the sequel, two linear descriptions are said to be equivalent if they define the same polyhedron.A linear description  1 dominates another linear description  2 if the polyhedron defined by  1 is included in the polyhedron defined by  2 .
Optimizing even a linear function over the mixed integer set  is an NP-hard problem in general (see [1][2][3][4]).A way of building strengthened linear relaxations is to use an approach combining reformulation, linearization, and projection such as those proposed in [5][6][7][8][9][10].In such approach, one first reformulates the constraints defining the set of feasible solutions by introducing nonlinearities.Then, the resulting nonlinear system is linearized and projected back onto the original space.
Two important properties characterize the reformulationlinearization approach (also known as Lift-and-Project methods (do not confuse this with the Lift-and-Project hierarchy introduced by Balas et al., see [5])).First, the approach leads to a whole hierarchy (see [10,11]) of relaxations which lie 2 Advances in Operations Research between the continuous relaxation  and the convex hull of the mixed integer set .And for a given hierarchy, a relaxation of higher rank (see [10,11]) is always stronger than a relaxation of lower rank.Secondly, optimizing a linear function over any relaxation of the hierarchy can be done in a polynomial time.
The convexification technique is also widely used to solve nonlinear and nonconvex optimization problems (see [16] and the references therein).Roughly speaking, this technique consists in approximating a nonconvex optimization problem by a convex problem (or a family of convex problems).This can be done by approximating the nonconvex objective function by a convex function and/or by approximating the nonconvex set of feasible solutions by a convex one (see [17][18][19][20][21][22][23][24] and the references therein).In this paper, new hierarchies of continuous relaxations using a reformulation, convexification, and linearization approach will be defined.Theses hierarchies are obtained using two different convexification schemes.
The paper is organized as follows.In the second section, first, we will recall the definition of a reformulationlinearization hierarchies and then give the definition of the Sherali-Adams hierarchy.In the third section, we will define two reformulation-convexification hierarchies: RTC and RSC hierarchies.We will study the main properties of these new hierarchies.In the fourth section, we will study the connections between RTC, RSC, and L&P hierarchies.In the fifth section, we will introduce a weak version of the hierarchies RTC and RSC and emphasize some connections between them.In the last section we make some concluding remarks.

The Reformulation-Linearization Hierarchies
First, we will introduce the general concept of reformulationlinearization hierarchies.Then, we recall the definition of the well-known Sherali-Adams hierarchy (for more details see [9,10,25]).
Let  be a positive integer.For a finite nonempty set  let  [] be the set of all subsets of  with cardinality , whereas  () is the set of all subset of  with cardinality at most .Sometimes we will need to indicate the cardinality of the sets under consideration, so we will use the notation   to indicate (do not confuse this with the Cartesian product of sets) that the set  has cardinality .
Let   be a set of  elements belonging to the set of binary indices  and let  be a subset from   .We call -factor associated with the sets  and   , denoted by   (,   \ ), the degree  polynomial defined as follows: with the convention that ∏ ∈0   = 1.
Example 1.In the case where  = {1, 2, 3} and  = 2 we have the following 12 nontrivial 2-factors: A rank  reformulation-linearization relaxation (of the mixed integer set  described by ( 1)-( 4)) is defined in three steps.First, the problem is reformulated as a 0-1 polynomial (semialgebraic (a -dimensional semialgebraic set is a solution set of a finite system of polynomial equalities and inequalities; for more details see [26,27])) mixed integer system by multiplying constraints (1)-( 3) with all possible -factors (that is multiplying by   (,   \ ) for all subsets   of  and all  ⊆   ).Then, the nonlinear terms are linearized by replacing them with new variables giving rise to a higher dimensional linear system.The third step consists in projecting back the resulting polyhedron onto the original -space.As observed in [8] the linearization step can be performed in various ways, leading to various hierarchies of relaxations.
The solution set in R + associated with the nonlinear (semialgebraic) description resulting from the reformulation step will be denoted by   and it is defined as follows: where, for each subset   of ,   (  ) is the solution set defined by the following nonlinear system: Starting from this semialgebraic reformulation, various linear relaxations can be constructed depending on the type of linearization considered (for more details see [8,13]).

The Sherali-Adams Hierarchy.
The description of the rank  Sherali-Adams relaxation for the mixed integer set  defined by ( 1)-( 4), denoted by P RLT , is a reformulationlinearization relaxation of rank  where the nonlinear terms appearing in (7) where it is assumed that  0 = 1 and   0 =   for every index  (belonging to  \ ) of a continuous variable.
Example 2. In the case where  = {1, 2, 3} and  =  ∪ {4} we have the following linearization: The resulting higher dimensional linear description will be denoted by   RLT and it is defined as follows: where, for each cardinality  subset   of , the linear description of the polyhedron   RLT (  ) is and where, for every index  belonging to  \ , The above relations (12) are easily obtained by expanding the products involved in the definition of the -factors.
After linearizing the nonlinear terms in (7) using the  variables defined in (8) above, the   RLT description turns out to involve a number of variables and constraints exponential in .The number of variables needed to linearize the nonlinear system (7) is ∑ min{+1,} =1 (   ) +  ∑  =0 (   ) (notice that the variable  0 is not counted here since  0 = 1).Also, it is seen that the number of constraints is ((   ) 2  ( + )).The rank  Sherali-Adams relaxation P RLT is obtained by projecting the polyhedron   RLT onto the subspace R + of the  variables.

Two New Reformulation-Convexification Hierarchies
We will consider two new reformulation-convexification hierarchies.The first one is called reformulation-termconvexification (RTC) hierarchy.It is obtained by convexifying the monomials (also called terms) resulting from the reformulation step.The second hierarchy is called reformulationstandard-convexification (RSC).It is obtained by convexifying the nonlinear factors (a linear combination of monomial products) resulting from the reformulation step.

Reformulation-Term-Convexification Hierarchy.
A rank  relaxation of the RTC hierarchy is obtained by applying local convexification to each constraint of the nonlinear system defining   as follows.For every subset  from  and every subset  from  \  with at most one element, let  be the following operator: with the convention that ∏ ∈0   = 1 and, for any real , () + is equal to max{0, }.
The convexification scheme (13) assumes that the constraint to which it is applied is of the form ≤.
Let T  be the following convex set: where, for each cardinality  subset   from , the convex set T  (  ) corresponds to the solution set   (  ) defined by the nonlinear system deduced from ( 7) by convexification Advances in Operations Research using scheme (13).The convex nonlinear description of the set T  (  ), for a given , reads where, for every index  belonging to  and any scalar ,  ,   [] and  ,  0 [] denote the convexified forms of the polynomials     (,   \ ) and   (,   \ ), respectively; these are defined using the  operator as follows: Since the set T  is the intersection of convex sets (by construction), then we have the following result.Theorem 3.For every integer  belonging to {1, . . ., }, the set T  as defined by ( 15) is a nonlinear convex relaxation of the mixed integer set .
For every integer , let E   be the set (the lower-script  is used to recall that our set is related to the term convexification scheme): where, for subsets  ∈  () and  ∈ ( \ ) (1) , Since every binary vector from {0, 1} + belongs to E   then we deduce that conv(E   ) coincides with the hypercube [0, 1] + .
Let   RTC be the extended linear description obtained from the set T  ∩ E   using the following steps.Let  and V be two sets of additional variables such that for every -element set   , for every subset  from   and for every subset  from  \   with at most one element.First, in (15), the variable    will replace the nonlinear term: and the variable V   will replace Then, we impose the equality constraint: We will use the notation   or    instead of    when  is empty or when  coincides with the singleton {}, respectively.
Thus, we have where, for each subset   of , the polyhedron   RTC (  ) reads min min and where, for every index  belonging to , The linear description ( 23)-( 31) and (32) are stated using only the variables    .This is possible according to (21).As discussed in Section 5, discarding constraints (21) in the definition of the RTC hierarchy will lead to a weaker hierarchy.
For every integer  belonging to {1, . . ., }, let P RTC be the projection onto the -space of the extended linear description   RTC .The polyhedron P RTC will be called rank- reformulation-term-convexification relaxation of .
In the next theorem we will prove that the hierarchy RTC is equivalent to the hierarchy RLT.Theorem 4. For every integer  belonging to {1, . . ., }, the two linear relaxations P RLT and P RTC are equivalent.
Proof.We will proceed by showing that the two extended linear descriptions   RLT and   RTC are the same up to variable renaming.Let  be an integer belonging to {1, . . ., }.As shown in [8], for every set   belonging to  [] , the constraints min {  :  ∈ } ≥   ,  ⊆   , (33) are implicit in the linear description of   RLT .We claim that the constraints are also implicit in the linear description of   RLT .The argument is obvious for constraints (35).For constraints (36), let  be an index belonging to  \  and let  be a subset from   , and the following constraint is valid for P RLT .Using RLT linearization, we get the constraint Combining this last constraint with constraint (34) we deduce that Now consider the following identifications: for every  ∈  \ ,  ⊆ , || ≤ min (, ) . (40) These identifications imply identifications between the  and  variables through (12) and (32).Thus, the two extended linear descriptions   RLT and   RTC are equivalent.This completes the proof.
Consequently, the rank- RTC relaxation coincides with the convex hull of the mixed integer set .The RTC hierarchy is motivated by the next theorem where it is shown that the projection onto the -space of any RTC relaxation can be sandwiched between two convex sets.The following proposition will be useful.Proposition 5. Let  and  be two disjoint subsets from , such that   belongs to [0, 1] for all  belonging to  ∪ ; then (41) Theorem 6.For every integer  belonging to {1, . . ., }, one has Proof.First, to prove the left inclusion in (42) it is sufficient to prove that the set (T  ∩ E   ) is included in P RTC .Let x be a point belonging to the set T  ∩ E   .Let ( x, t, V) be a vector such that Thus, by definition constraints (31) are fulfilled by the vector ( x, t, V).Since x belongs to T  , then ( x, t, V) satisfies constraints ( 23)- (26).By Proposition 5 the point ( x, t, V) satisfies constraints ( 27)-(30).Since the vector x also belongs to the set E   , then the vector ( x, t, V) also satisfies constraints (21).Consequently, the vector ( x, t, V) belongs to   RTC .Thus, x belongs to P RTC .Now, let us prove the right inclusion in (42).Let x be a point belonging to P RTC .There is a vector t such that ( x, t) belongs to   RTC .Without loss of generality, any constraint defining   RTC can be written as follows: where  + 1 ,  + 2 ,  − 1 , and  − 2 are subsets of the power set of  (the superscript of a coefficient indicates its sign).Since the point ( x, t) satisfies the following inequalities: then we deduce Consequently, the point x satisfies all constraints defining T  : that is, P RTC ⊆ T  .This completes the proof.Characterization (42) will allow us, in some cases (see Corollary 8), to give an explicit characterization of any RLT relaxation (characterization in the -space).Before answering this question we will recall, in the next proposition, the simple result stating that any fractional point belonging to the set E  will belong to either a facet or an edge of the hypercube.This implies that the set E  has empty interior.Proof.Let  be an integer belonging to {1, . . ., }.On the one hand, the sets T  and E   are both subsets from the hypercube.On the other hand, if the set T  has integer vertices then both sets T  ∩E   and T  have the same vertices (vertices of T  ).Thus, the convex envelope of the set T  ∩E   coincides with the set T  , because T  is convex.Consequently, using Theorems 4 and 6 we conclude that This completes the proof.
The following example shows that restricting the set T  to have integer vertices in Corollary 8 is not a sufficient condition to characterize P RLT .Let us consider the following set: Its continuous relaxation  is the shaded region drawn in Figure 1.The set T 1 associated with the set  is the shaded region drawn in Figure 2. A careful analysis of the set T 1 shows that it features the following linear description: The rank-1 RLT relaxation has the same linear description as T 1 .But, as shown in Figure 2, the set T 1 has a fractional vertex.Notice that set (49) coincides with the the rank-1 L&P relaxation.As discussed in Section 5, this equality is not true in general.Finally, the set T 2 (the shaded region in Figure 3) has integer vertices and it coincides with the rank-2 RLT relaxation.

Reformulation-Standard-Convexification Hierarchy.
Contrary to the RTC hierarchy, in the reformulationstandard-convexification hierarchy (RSC) we convexify each factor obtained after reformulation and not the monomials appearing in each such factors.The term standard extension was introduced by Crama (see [24]) in studying concave envelopes of pseudo-boolean functions.
Let S  be the nonlinear convex set: where, for each cardinality  subset   of , the convex set S  (  ) corresponds to the solution set of   (  ) defined by the nonlinear system deduced from (7) using the following convexification scheme: where  and  are two subsets such that  ⊂ ,  ⊂ , and | ∩ ( \ )| ≤ 1.
As for the  operator, depending on the type of the constraint (≥ or ≤), we use the min expression or the max expression in such a way that the resulting solution set will be convex.
The nonlinear description of the set S  (  ), for a given  and a set   , is thus defined as follows: where, for every index  belonging to  and any scalar , L ,   [] and L ,  0 [] denote the convexified forms of the polynomials     (,   \ ) and   (,   \ ) using the operator L, respectively.
Since the set S  as defined in (51) is the intersection of convex sets then we have the following results.Theorem 9.For every integer  belonging to {1, . . ., }, the set S  as defined by ( 51) is a nonlinear convex relaxation of the mixed integer set .
For every integer , let E   be the set (the lower-script  is used to recall that our set is related to the standard convexification scheme) where for subsets   ∈  [] and  ∈ ( \   ) (1) the set E  (  , ) is the subset from [0, 1] + defined by the following constraints: Since every binary vector from {0, 1} + belongs to E   then the set conv(E   ) coincides with the hypercube [0, 1] + .Notice that the set E   is a subset from E   .As shown below, it is possible to represent S  ∩ E   as a polyhedron in some appropriate extended space.
Let   RSC denote the extended linear description of the set S  ∩E   obtained by using the following steps.First, let  and  be two sets of variables such that, for every -element set   , for every subset  from   , and for every subset  from  \   with at most one element,  ,   replaces in (53) the nonlinear term: and the variable  ,   replaces Then we impose the following equality: In the sequel, instead of  ,   or  ,   we will use  ,   or  ,   when the set  coincides with the singleton {} and  ,  0 or  ,  0 otherwise (recall that the set  has at most one element).
Thus, we obtain the extended linear description: where, for each subset   of , the polyhedron   RSC (  ) reads Notice that linear description (60)-(69) are stated using only the variables .This is possible according to (58).As discussed in the last section, discarding constraints (58) in the definition of the RSC hierarchy will lead to a weaker hierarchy.
For every integer  belonging to {1, . . ., }, let P RSC be the projection onto the -space of the extended linear description   RSC .The continuous relaxation Pd RSC will be called rank- reformulation-standard-convexification relaxation of the mixed integer set .
As for the RTC hierarchy, in the following theorem we will prove that any relaxation of the RSC hierarchy can also be sandwiched between two convex sets.
Theorem 10.For every integer  belonging to {1, . . ., }, we have Proof.Let  be an integer belonging to {1, . . ., }.First, to prove the left inclusion in (70) it is sufficient to prove that the set S  ∩ E   is a subset of P RSC .Let x be a point belonging to the set S  ∩ E   .Let ( x, Ĝ, D) be a vector where the two vectors Ĝ and D are defined as follows: for every -element set   , for every subset  from   , and for every subset  from  \   with at most one element:

Ĝ𝐽,𝐽
By definition, x belongs to S  ; then ( x, Ĝ, D) satisfies constraints (60)-(63).By Proposition 5 the point ( x, Ĝ, D) also satisfies constraints (64)-(69).Since the vector x also belongs to the set E   , then the vector ( x, Ĝ, D) also satisfies constraints (58).Thus, the vector ( x, Ĝ, D) belongs to   RSC .Consequently, x belongs to P RSC .This completes the first part of the proof.Now, let us show the right inclusion in (70).Let x be a point belonging to P RSC .There is a vector Ĝ such that ( x, Ĝ) belongs to   RSC .Without loss of generality, each constraint in (60) and (61) can be rewritten as follows: Without loss of generality, we can assume that   is nonnegative (the argument we will use holds also in the case where   is nonpositive).The point ( x, Ĝ) satisfies also the constraints: Thus, we have This means that the point x satisfies constraints (53).Thus, x belongs to S  .This completes the proof.

Links between RTC, RSC, and L&P Extended Linear Descriptions
We will focus, in this section, on the connections between the extended linear descriptions of the RTC, RSC, and L&P relaxations.First, we will compare the strength of the RTC and RSC hierarchies.We will prove that for every rank  the relaxation P RTC dominates the relaxation P RSC .Let  = { 1 , . . .,   } be a set of indices and let  be an integer less than or equal to ||.Let   be the subset from  defined as follows: The following lemmas will be useful to prove the next theorem.
Lemma 11.Let  be { 1 , . . .,   } a subset of indices from .Let  be the set { 0 }∪, where  0 does not belong to .Let us consider the variables  satisfying Then, we have the following equality:

Advances in Operations Research
Proof.Let  be the set { 1 , . . .,   } and let  be the set { 0 } ∪ , where  0 does not belong to .Let us consider the variables  satisfying (76)-(77).First, notice that the set is a partition of the power set of .Thus, Using (77) we obtain Proof.Let  be an integer belonging to {1, . . ., }.Let  be a finite set and let   be a -element subset from .It is a wellknown fact that the linear transformation (82) is a bijection (see [10]) and its inverse is given by First, the variables  ,  satisfy inequalities (85) because of (88) and ( 83) and the fact that the variables  ,  are all nonnegative.Then, using Lemma 11, which is legitimate because the variables  ,  are assumed nonnegative, we deduce the inequalities (86).Finally, notice that for every subset  from   we have Using inequalities (84) we deduce that which is equivalent to Thus, for sets  and   the variable  ,  satisfies inequalities (87) and this completes the proof.Theorem 13.For every integer  belonging to {1, . . ., } we have Proof.Let  be an integer belonging to {1, . . ., }.We will prove that the extended linear description  with  ,  0 we conclude that the point ( x, Ĝ) satisfies all constraints (60)-(63).This completes the proof.Now, we will compare the strength of the two hierarchies RSC and L&P.As shown in [8], the L&P hierarchy can be obtained from the semialgebraic set (6) using a suitable linearization (for more details, see [8]).Precisely, any rank- L&P relaxation is a rank- reformulation-linearization relaxation where the linearization is performed using the following substitutions: for every subset   from , let (96) The linearized system we obtain reads Let   L&P be the extended linear description (97).As before, let P L&P be its projection onto the -space.
In the next theorem we will prove that the hierarchy L&P dominates the hierarchy RSC.RLT is included in P L&P (see [8]).Thus, it follows using Theorem 14 that P RTC is included in P RSC .

Weak RTC and Weak RSC Hierarchies
In this section, we will introduce a weak version of the RTC and RSC hierarchies.For both weak hierarchies the rank  extended linear description is obtained by reformulation, convexification using min and max and then linearizing using two distinct sets of variables.More precisely, the rank  extended linear description of the weak-RTC hierarchy, denoted by   WRTC , is defined as   RTC except that we discard equality constraints (21) (see Section 3.1).Thus, both sets of variables    and V   will appear in the description of   WRTC .Similarly, the rank  extended linear description of the weak-RSC hierarchy, denoted by   WRSC , is defined as   RSC except that we discard equality constraints (58) between the two sets of variables  ,   and  ,   (see Section 3.2).
To emphasize some connections (small instances are sufficient to reveal these connections.We wish to emphasize that this is not a computational investigation) between the weak hierarchies we will use the computational results shown in Table 1.The values computed are the minimum value of rank 1 RTC, L&P, RSC, WRTC, and WRSC relaxations for five instances of the multiple constraints knapsack problem.Each instance has 5 constraints and 3 variables.The instances have been generated using the Chu and Beasley procedure given in [28].The constraint matrix coefficients are integers and randomly chosen in {0, . . ., 1000}.The right-hand-side coefficient of the th constraint is set to 0.95 ∑ where   is a real number randomly chosen from the interval [0, 1].
Although RTC is stronger than RSC the WRTC and WRSC hierarchies are not comparable in strength as shown by the instances inst-1 and inst-2.The WRTC hierarchy may be stronger than L&P hierarchy as shown by the instances inst-2 and inst-4.Also, the computational results shown in the Table 1 are coherent with the theoretical results proved before: (i) the hierarchy RTC is stronger than both L&P and RSC hierarchies; (ii) the hierarchy L&P is stronger than hierarchy RSC; (iii) the hierarchies RTC and RSC are stronger than WRTC and WRSC, respectively.

Conclusion
In this paper, we introduced two new hierarchies called RTC and RSC for which the rank  continuous relaxations were denoted by P RTC and P RSC , respectively.These two hierarchies are obtained using a reformulation-convexificationlinearization procedure.The hierarchy RTC is obtained using a term convexification scheme and the RSC hierarchy is obtained using a standard convexification scheme.Then we compared the strength of these two hierarchies.We proved that (i) the hierarchy RTC is equivalent to the RLT hierarchy of Sherali-Adams, (ii) the hierarchy RTC dominates the hierarchy RSC, and (iii) the hierarchy RSC is dominated by the Lift-and-Project hierarchy.Next, for every rank , we proved that conv(T  ∩ E   ) ⊆ P RTC ⊆ T  and conv(S  ∩ E   ) ⊆ P RSC ⊆ S  , where the sets T  and S  are convex, while E   and E   are two nonconvex sets with empty interior.The first inclusions allow, in some cases, an explicit characterization of RLT relaxations.That is a convex nonlinear description of any RLT relaxation in the -space.Finally, we discussed weak version of both RTC and RSC hierarchies and emphasized some connections between them using small numerical examples.
We conclude with some open questions.First, one may ask whether the hierarchy RSC is equivalent to the hierarchy WRSC or not.Also, does the rank  RSC relaxation coincide with the convex envelope of the set ? Finally, is it possible to obtain stronger hierarchies using the exposed reformulationconvexification-linearization approach?Extending this work to more general nonlinear optimization problems will be the subject of a future work.
are linearized by introducing a new set of variables   and    defined by |\|  ,   ,  ⊆   .
(,   \) and   (,   \), respectively; these are related to the   and    variables as follows: ,   and  ,  0 denote the linearized forms of the convexified form of the polynomials
in , every subset  from   , and for every index  from  we have belongs to   RSC .Consequently, the point x belongs to P RSC .This completes the proof.As a byproduct of Theorem 14 we obtain an indirect proof of Theorem 13.Indeed, for any rank , we know from Theorem 4 that P Proof.Let  be an integer belonging to the set {1, . . ., }.Let x be a point belonging to P L&P .There exists a variable Ẑ such that ( x, Ẑ) belongs to   L&P .Notice that, for every -element subset  + ≤  ,   ≤ min { min ℎ∈∪{} { ℎ } , min RTC is equivalent to P RLT .It is a well-known fact that P