On bounded pitch inequalities for the min-knapsack polytope

In the min-knapsack problem one aims at choosing a set of objects with minimum total cost and total profit above a given threshold. In this paper, we study a class of valid inequalities for min-knapsack known as bounded pitch inequalities, which generalize the well-known unweighted cover inequalities. While separating over pitch-1 inequalities is NP-hard, we show that approximate separation over the set of pitch-1 and pitch-2 inequalities can be done in polynomial time. We also investigate integrality gaps of linear relaxations for min-knapsack when these inequalities are added. Among other results, we show that, for any fixed t, the t-th CG closure of the natural linear relaxation has the unbounded integrality gap.

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1 Introduction

The min-knapsack problem (MinKnap)

 mincTx s.t. pTx≥1,x∈{0,1}n (1)

is the variant of the max knapsack problem (MaxKnap

) where, given a cost vector

and a profit vector , we want to minimize the total cost given a lower bound on the total profit. MinKnap is known to be NP-Complete, even when . Moreover, it is easy to see that the classical FPTAS for MaxKnap [9, 13] can be adapted to work for MinKnap, thus completely settling the complexity of MinKnap.

However, pure knapsack problems rarely appear in applications. Hence, one aims at developing techniques that remain valid when less structured constraints are added on top of the original knapsack one. This can be achieved by providing strong linear relaxations for the problem: then, any additional linear constraint can be added to the formulation, providing a good starting point for any branch-and-bound procedure. The most common way to measure the strength of a linear relaxation is by measuring its integrality gap, i.e. the maximum ratio between the optimal solutions of the linear and the integer programs (or of its inverse if the problem is in minimization form) over all the objective functions.

Surprisingly, if we aim at obtaining linear relaxations with few inequalities and bounded integrality gap, MinKnap and MaxKnap seem to be very different. Indeed, the standard linear relaxation for MaxKnap has integrality gap 2, and this can be boosted to by an extended formulation with many variables and constraints, for [2]. Conversely, the standard linear relaxation for MinKnap  has unbounded integrality gap, and this remains true even after rounds of the Lasserre hierarchy [12]. No linear relaxation for MinKnap  with polynomially many constraints and constant integrality gap can be obtained in the original space [6]. It is an open problem whether an extended relaxation with this properties exists. Recent results showed the existence [1] and gave an explicit construction [7] of a linear relaxation for MinKnap  of quasi-polynomial size with integrality gap . This is obtained by giving an approximate formulation for Knapsack Cover inequalities (KC) (see [4] and the references therein). Adding those (exponentially many) inequalities gives an integrality gap of , and can be approximately separated [4]. The bound on the integrality gap is tight, even in the simpler case when . One can then look for other classes of well-behaved inequalities that can be added to further reduce the integrality gap. A prominent family is given by the so called bounded pitch inequalities [3] defined in Section 2. Here, we remark that the pitch is a parameter measuring the complexity of an inequality, and the associated separation problem is NP-Hard already for pitch-. The pitch- inequalities are often known in the literature as unweighted cover inequalities (see e.g. [1]).

In this paper, we study structural properties and separability of bounded pitch inequalities for MinKnap, and the strength of linear relaxations for MinKnap when they are added. Let be the set given by pitch-, pitch-, and inequalities from the linear relaxation of (1). We first show that, for any arbitrarily small precision, we can solve in polynomial time the weak separation problem for the set . Even better, our algorithm either certifies that the given point violates an inequality from , or outputs a point that satisfies all inequalities from and whose objective function value is arbitrarily close to that of . We define such an algorithm as a -oracle in Section 2; see Section 3 for the construction. A major step of our procedure is showing that non-redundant pitch-2 inequalities have a simple structure.

It is then a natural question whether bounded pitch inequalities can help to reduce the integrality gap below . We show that, when , if we add to the linear relaxation of (1) pitch- and pitch- inequalities, the integrality gap is bounded by ; see Section 4.1. However, this is false in general. Indeed, we also prove that KC plus bounded pitch inequalities do not improve upon the integrality gap of ; see Section 4.4. Moreover, bounded pitch alone can be much weaker than KC: we show that, for each fixed , the integrality gap may be unbounded even if all pitch- inequalities are added. Using the relation between bounded pitch and Chvátal-Gomory (CG) closures established in [3], this implies that, for each fixed , the integrality gap of the -th CG closure can be unbounded; see Section 4.2. For an alternative proof that having all KC inequalities bounds the integrality gap to see Section 4.3.

2 Basics

A MinKnap instance is a binary optimization problem of the form (1), where and we assume , . We will often deal with its natural linear relaxation

 mincTx s.t. pTx≥1,x∈[0,1]n. (2)

The NP-Hardness of MinKnap immediately follows from the fact that MaxKnap is NP-Hard [10], and that a MaxKnap instance

 maxvTx s.t. wTx≤1,x∈{0,1}n. (3)

can be reduced into a MinKnap instance (1) as follows: each is mapped via with ; and for . Note that the reduction is not approximation-preserving.

We say that an inequality with is dominated by a set of inequalities if can be written as a conic combination of inequalities in for some and . is undominated if any set of valid inequalities dominating contains a positive multiple of it.

Consider a family of inequalities valid for (1). We refer to [8] for the definition of weak separation oracle, which is not used in this paper. We say that admits a -oracle if, for each fixed , there exists an algorithm that takes as input a point and, in time polynomial in , either outputs an inequality from that is violated by , or outputs a point , that satisfies all inequalities in . In particular, if contains the linear relaxation of (1), .

Let be a valid inequality for (1), with for all . Its pitch is the minimum such that, for each with , we have . Undominated pitch-1 inequalities are of the form . Note that the map from MaxKnap to MinKnap instances defined above gives a bijection between minimal cover inequalities

 ∑i∈Ixi≤|I|−1

for MaxKnap and undominated pitch-1 inequalities for the corresponding MinKnap instance. Since, given a MaxKnap instance, it is NP-Hard to separate minimal cover inequalities [11], we conclude the following.

Theorem 1.

It is NP-Hard to decide whether a given point satisfies all valid pitch-1 inequalities for a given MinKnap instance.

Given a set , such that , the Knapsack cover inequality associated to is given by

 ∑i∈[n]∖Smin{pi,β}xi≥β (4)

and it is valid for (1).

For a set , we denote by its characteristic vector. An -approximate solution for a minimization integer programming problem is a solution that is feasible, and whose value is at most times the value of the optimal solution. An algorithm is called a polynomial time approximation scheme (PTAS) if for each and any instance of the given problem it returns an -approximate solution in time polynomial in the size of the input. If in addition the running time is polynomial in , then the algorithm is a fully polynomial time approximation scheme (FPTAS).

Given a rational polyhedron with and , the first Chvátal-Gomory (CG) closure [5] of is defined as follows:

 P(1)={x∈Rn: ⌈λ⊤A⌉x≥⌈λ⊤b⌉, ∀λ∈Rm}.

Equivalently, one can consider all such that . For , the -th CG closure of is recursively defined as . The CG closure is an important tool for solving integer programs, see again [5].

3 A (1+ϵ)-oracle for pitch-1 and pitch-2 inequalities

In this section, we show the following:

Theorem 2.

Given a MinKnap instance (1), there exists a -oracle for the set containing: all pitch-1 inequalities, all pitch-2 inequalities and all inequalities from the natural linear relaxation of (1).

We start with a characterization of inequalities of interest for Theorem 2.

Lemma 3.

Let be the set of feasible solutions of a MinKnap instance (1). All pitch- inequalities valid for are implied by the set composed of:

1. Non-negativity constraints for ;

2. All valid pitch- inequalities;

3. All inequalities of the form

 ∑i∈I1xi+2∑i∈I2xi≥2 (5)

where , , , and .

The inequalities in iii) are pitch- and valid.

Proofs of Lemma 5 and Theorem 2 are given in Section 3.1 and Section 3.2, respectively.

3.1 Restricting the set of valid pitch-2 inequalities

We will build on two auxiliary statements in order to prove Lemma 3.

Claim 1.

If and are distinct inequalities valid for and , then the latter inequality is dominated by the former.

Proof.

can be obtained summing nonnegative multiples of and for , which are all valid inequalities.

Claim 2.

Let

 ∑i∈T1xi+2∑i∈T2xi≥2 (6)

be a valid inequality for MinKnap, with and . Then, (6) is dominated by the inequality in iii) with .

Proof.

One readily verifies that Inequality (5) with as above is valid. Suppose now that . Then the integer solution that takes all elements in is feasible for MinKnap, but it does not satisfy (6), a contradiction. Hence . Since , (5) dominates (6) componentwise, and the thesis follows by Claim 1.

Proof of Lemma 3. The fact that an inequality of the form (5) is pitch- and valid is immediate. Because of Claim 2, it is enough to show the thesis with (5) replaced by (6). Consider a pitch- inequality valid for :

 ∑i∈Twixi≥1, (7)

where is the support of the inequality, . Without loss of generality one can assume that for some and . Since (7) is pitch- we have that for all . We can also assume , since otherwise is valid and dominates (7) by Claim 1.

Let be the maximum index such that . Note that such exists, since, if , then (7) is a pitch-1 inequality. If , then, by Claim 1, (7) is dominated by the valid pitch- inequality

 ∑i∈[j]xi+2h∑i=j+1xi≥2, (8)

which again is of the type (6). Hence and again via Claim 1, (7) is dominated by

 w1x1+j∑i=2(1−w1)xi+h∑i=j+1xi≥1, (9)

since for all , so one has . Thus, we can assume that (7) has the form (9). Note that inequality

 h∑i=2xi≥1 (10)

is a valid pitch- inequality, since we observed . Therefore, (7) is implied by (8) and (10), taken with the coefficients and respectively. Recalling that (8) is a valid pitch-2 inequality of the form (6) concludes the proof.

3.2 A (1+ϵ)-oracle

We will prove Theorem 2 in a sequence of intermediate steps. Our argument extends the weak separation of KC inequalities in [4].

Let be the point we want to separate. Note that it suffices to show how to separate over inequalities i)-ii)-iii) from Lemma 3. Separating over i) is trivial. We first show how to separate over iii).

Claim 3.

For , let be the optimal solution to the following IP , and its value:

 min∑i∈[n]: pi<α¯xizi+2∑i∈[n]: pi≥α¯xizi s.t. ∑i∈[n]pi(1−zi)≤1−α,z∈{0,1}n. (11)

If , then violates Inequality (5) with , otherwise does not violate any Inequality (5) with .

Proof.

Fix a feasible solution to (11), and let . Then:

 β:=β(I)=1−∑i∈[n]∖Ipi=1−∑i∈[n]pi(1−¯zi)≥α.

Hence:

 ∑i∈I: pi<β¯xi+2∑i∈I: pi≥β¯xi=∑i∈[n]: pi<β¯xi¯zi+2∑i∈[n]: pi≥β¯xi≤∑i∈[n]: pi<α¯xi¯zi+2∑i∈[n]: pi≥α¯xi¯zi=v(¯z),

where the central inequality holds at equality if . Hence, if , the inequality with from (11) is violated by . Else, all inequalities from (11) with are satisfied.

Note that is a MinKnap instance, hence we can use the appropriate FPTAS to find, for each , an -approximate solution for it.

Since all data are rationals, we can assume there exists such that, for each , for some .

Claim 4.

Let and, for , let be the solution output by the FPTAS for problem and its objective function value. If for some , then violates Inequality (5) with . Else, satisfies all inequalities in Lemma 3.iii).

Proof.

Let for some . If then by Claim 3 violates the corresponding inequality (5). Otherwise, , and is feasible for any pitch- Inequality (5) induced by with .

Now let with . There exists such that (with ). Let . The set of feasible solutions of contains that of , and . Hence, and consequently implies . Thus, for separating all inequalities in Lemma 3.iii), it suffices to check (11) for all as in the statement of the claim.

The following claim follows in a similar fashion to the previous one by observing that, for , (11) separates over undominated pitch- inequalities.

Claim 5.

Let , and be the solution output by the FPTAS for problem , and its objective function value. If , then violates the pitch- inequality with support . Else, satisfies all valid pitch- inequalities.

Next claim shows how to round a point in the unit cube that almost satisfies all pitch- and pitch- inequalities, to one that satisfies them and is still contained in the unit cube.

Claim 6.

Let be such that satisfies all inequalities from Lemma 3, and define as follows: for . Then and satisfies all inequalities from Lemma 3.

Proof.

Clearly . Let . If , , hence satisfies all pitch- inequalities. Thus, . Consider a pitch- inequality of the form (5), and note that the left-hand side of the inequality computed in is lower bounded by , where is the coefficient of . First assume there exists . Then . Similarly, if , then . In both cases, satisfies the pitch- inequality. Hence, we can assume . Then:

 ∑i∈Iαi¯xi≥21+ϵ, from which % we deduce
 ∑i∈I∖{j}αi¯xi≥21+ϵ−¯xj≥21+ϵ−1=1−ϵ1+ϵ and
 ∑i∈Iαi¯yi=∑i∈I∖{j}αi¯yi+1=1+ϵ1−ϵ∑i∈I∖{j}αi¯xi+1≥2,

as required. A similar (simpler) argument shows that also satisfies all pitch- inequalities .

Proof of Theorem 2. We can now sum up our -oracle, see Algorithm 1. Correctness and polynomiality follow from the discussion above.

3.3 Separating inequalities of pitch k≥3 with fixed support

Here, we give an example showing that inequalities of pitch- and higher do not have the nice structure of pitch-2. Let

 P={x∈[0,1]7:5x1+6x2+11x3+16x4+17x5+18x6+21x7≥41}. (12)

Inequality is a facet of the first CG closure (although not of the integer hull of ) and thus a valid pitch-. Observe that the coefficient of is higher than in this pitch-, while it is the opposite in (12). Such situations we call inversions and they do not occur in (relevant) pitch-2 inequalities. Inequality is an inverted facet of both the integer hull and the first CG closure.

For later use (in Section 4.4), we observe here that when is fixed, we can efficiently and exactly solve the separation problem over inequalities with support just by solving an LP. Clearly, we are only interested in valid inequalities with and points . Let . We can assume , otherwise there is no valid inequality as above with support . Call massive if . Consider the following LP:

 min∑i∈Iαix∗is.t.∑i∈Jαi≥1 for all massive J⊆Iα≥0 (13)

Note that, for each feasible solution to the previous LP, we have that is a valid inequality for the original MinKnap instance, and conversely that all inequalities with support can be obtained in this way. Hence, let be the optimal solution to the previous LP. If , we obtain an inequality whose support is contained in , that is violated by . The support of the inequality can be extended to by setting for all with . On the other hand, if , satisfies all inequalities with support .

4 Integrality gap for MinKnap with bounded pitch inequalities

4.1 When p=c

Theorem 4.

Consider an instance of MinKnap (1) with . Denote by the linear relaxation of (1) to which all pitch-1 and pitch-2 inequalities have been added. The integrality gap of is at most .

Proof. Let , and let be the optimal integer solution to (1). We can assume , else we are done.

Claim 7.

The support of has size .

Proof.

Let be the size of the support of . If , then is also the optimal fractional solution. Now assume . Remove from the cheapest item as to obtain . We have

 pTx′≥(1−1k)pT¯x>23⋅1.5=1,

contradicting the fact that is the optimal integral solution.

Hence, we can assume that the support of is given by , with . Since , we deduce . Since , we deduce .

Let and . Then .

Proof.

Recall that for we denote its characteristic vector with . If , then is a feasible integral solution of cost strictly less than . Else if , then is a feasible integral solution of cost strictly less than . In both cases we obtain a contradiction.

Because of the previous claim, we can assume w.l.o.g. .

.

Proof.

Suppose . Since for all , there exists such that . Hence is a feasible integer solution of cost at most , a contradiction.

Because of the previous claim, the pitch- inequality is valid. The fractional solution of minimum cost that satisfies this inequality is the one that sets (since ) and all other variables to . This is exactly .

4.2 CG closures of bounded rank of the natural MinKnap relaxation

For , let be the linear relaxation of the (1) given by: the original knapsack inequality; non-negativity constraints; all pitch- inequalities, for .

Lemma 5.

For , the integrality gap of is at least times the integrality gap of .

Proof. Fix , and let be the cost of the optimal integral solution to (1). Let be the integrality gap of . Since is the optimal value of , by the strong duality theorem (and Caratheodory’s theorem), there exist nonnegative multipliers such that the inequality can be obtained as a conic combination of the original knapsack inequality (with multiplier ), non-negativity constraints (with multipliers ), and at most inequalities of pitch at most (with multipliers ). By scaling, we can assume that the rhs of the latter inequalities is . Hence .

Claim 10.

Let be a valid pitch- inequality for (1), and assume w.l.o.g. that . Then inequality is a valid inequality of pitch at most for (1).

Proof.

The inequality is a valid inequality for (1), and by construction it is of pitch at most . If , we obtain and consequently , from which we deduce

 1−d1>1−1t−1=t−2t−1≥12.

If conversely , by Lemma 3 we can assume w.l.o.g. that , and we can conclude .∎

Now consider the conic combination with multipliers given above, where each inequality of pitch- is replaced with the inequality of pitch at most obtained using Claim 10. We obtain an inequality , where one immediately checks that and

 v′′≥α+∑n+1i=1γimax{12,t−2t−1}≥max{12,t−2t−1}(α+∑n+1i=1γi)=max{12,t−2t−1}v′.

Hence the integrality gap of is

 Cv′≥Cv′′max{12,t−2t−1}

and the thesis follows since the integrality gap of is at most .

Lemma 6.

For a fixed and square integers , consider the MinKnap  instance defined as follows:

 minϵy+√nz+∑ni=1xist(n−√n)y+n2z+∑ni=1xi≥ny,z,x∈{0,1}.

For every fixed , the integrality gap of is .

Proof. Because of Lemma 5, it is enough to show that the integrality gap of is . Clearly, the value of the integral optimal solution of the instance is . We claim that the fractional solution

 (¯y,¯x,¯z)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1,1n−√n+1,…,1n−√n+1n times,2√n⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

is a feasible point of . Since , the thesis follows.

Observe that , hence satisfies the original knapsack inequality.

Now consider a valid pitch-1 inequality whose support contains . Since , satisfies this inequality. Hence, the only pitch-1 inequalities of interest do not have in the support. Note that such inequalities must have in the support, and some of the . Hence, all those inequalities are dominated by the valid pitch- inequalities

 z+∑i∈Ixi≥1∀I⊆[n],|I|=n−√n+1,

which are clearly satisfied by .

Theorem 7.

For a fixed , let CG be the th CG closure of the MinKnap  instance as defined in Lemma 6. The integrality gap of CG is .

Proof. We will use the following fact, proved (for a generic covering problem) in [3]. Let and suppose . Define point , where each component is the minimum between and times the corresponding component of . Then . Now fix . We have therefore that

and the claim follows in a similar fashion to the proof of Lemma 6.

4.3 When all knapsack cover inequalities are added

In this section we consider the min-knapsack formulation with knapsack cover inequalities (we use KC to denote this LP formulation). In [4] it is shown that the integrality gap of KC is 2. In the following we provide a simpler proof.

Let be a feasible fractional solution for KC of cost . Starting from , we show a simple and fast rounding procedure to obtain a feasible integral solution of cost at most .

The rounding procedure:

Let . Set for any . Consider the residual variables . The problem is to assign integral values to the the residual variables. We call this problem the residual problem (RP). By abusing notation, from now on, let denote the fractional solution for KC restricted to residual variables. Consider the following residual relaxation (RR):

 min ∑i∈¯SCixi (14) s.t. ∑i∈¯Sp′ixi≥b′ (15) 0≤xi≤1/2 (16)

where and . Note that satisfies (RR). So if is the optimal solution of (RR) than it follows that

 ∑i∈¯SCix∗i≤∑i∈¯SCi¯xi

Therefore, if it exists an integral solution to (RP) of cost then and we are done. We can rewrite the residual relaxation (RR) in the following equivalent way:

 min ∑i∈¯SCi2yi (17) s.t. ∑i∈¯Sp′i2yi≥b′ (18) 0≤yi≤1 (19)

Clearly the optimal fractional solution to (RR) can be obtained by ordering the variables according to their densities. W.l.o.g., assume that and let be the smallest integer such that and therefore (recall ):

 t∑i=1p′i≥2b′−p′t+1≥b′ (20)

Note that the optimal fractional solution to (RR) picks the first variables integrally and the last potentially fractional (but it could be integral). It follows that:

 t∑i=1Ci2≤∑i∈¯SCix∗i (21)

It follows that the integral solution obtained by setting for and zero otherwise is feasible by and of cost at most twice the optimal fractional of (RR) by (21).

4.4 When all bounded pitch and knapsack cover inequalities are added

Consider the following MinKnap instance with :

 min∑i∈[n]xi+1√n∑j∈[n]zjs.t.∑i∈[n]xi+1n∑j∈[n]zj≥1+ϵnx,z∈{0,1}n. (22)
Lemma 8.

For any fixed and sufficiently large, point with satisfies the natural linear relaxation, all KC and all inequalities of pitch at most valid for (22). Observing that the optimal integral solution is , this gives an IG of .

Proof. We prove the statement by induction. Fix . Note that dominates componentwise the point generated at step , and the latter by induction hypothesis satisfies all inequalities of pitch at most . Let

 ∑i∈Iwixi+∑j∈Jwjzj≥β (23)

be a valid KC or pitch- inequality with support , which gives that and . Observe the following.

Claim 11.

. In addition, or implies .

Proof.

Since all coefficients in (23) are strictly positive and , gives that the feasible solution for (22) is cut off by (23), a contradiction.

Furthermore, if and for some , then is cut off, again a contradiction. Finally, if , and for some , then