Multi-cover inequalities for totally-ordered multiple knapsack sets: theory and computation

Abstract

We propose a method to generate cutting-planes from multiple covers of knapsack constraints. The covers may come from different knapsack inequalities if the weights in the inequalities form a totally-ordered set. Thus, we introduce and study the structure of a totally-ordered multiple knapsack set. The valid multi-cover inequalities we derive for its convex hull have a number of interesting properties. First, they generalize the well-known (1, k)-configuration inequalities. Second, they are not aggregation cuts. Third, they cannot be generated as rank-1 Chvátal-Gomory cuts from the inequality system consisting of the knapsack constraints and all their minimal cover inequalities. We also provide conditions under which the inequalities are facets for the convex hull of the totally-ordered knapsack set, as well as conditions for those inequalities to fully characterize its convex hull. We give an integer program to solve the separation and provide numerical experiments that showcase the strength of these new inequalities

Appendices

Appendix A: Proof of Theorem 3

First, by Step 4 in Algorithm 1, we have the following easy observation.

Observation 3

Let $\{C_1,\dots ,C_k\}$ be a multi-cover and let $\alpha ^T x \le \beta$ be its corresponding S-MCI. If there exist some $t \in \mathbb {N}, \ell \in [n_t]$ and $h' \in [k]$ such that $i_{t, \ell } \in C_{h'}$, then $\{ i_{t,\ell }, \ldots , i_{t, n_t}\} \subseteq C_{h'}$.

Now we can prove Theorem 3.

Proof of Theorem 3

Consider the set of binary points whose support is in one of the following sets:

$$\begin{aligned}&\mathscr {S}_1 := \{C_{h_t} \cup \{i_{t-1, \ell _t}\} \setminus \{i_{t, \ell }\} \mid t = 2, \ldots , \max _{i=1}^n \alpha _i, \ell = 1, \ldots , n_t \}, \end{aligned}$$

(11)

$$\begin{aligned}&\mathscr {S}_2 := \{C_{h_t} \setminus \{i_{1,n_1}\} \mid t = 2, \ldots , \max _{i=1}^n \alpha _i\}, \end{aligned}$$

(12)

$$\begin{aligned}&\mathscr {S}_3 := \{C_{h_1} \setminus \{i_{1, \ell }\} \mid \ell = 1, \ldots , n_1\}, \end{aligned}$$

(13)

$$\begin{aligned}&\mathscr {S}_4 := \{C_{h_1} \cup \{i'\} \setminus \{i_{1, 1}\} \mid i' \notin C\}. \end{aligned}$$

(14)

First, we want to prove that any set in (11)–(14) is not a cover for K. By condition 3, we know $i_{1,n_1} \in C_{h_t}$ for any $t = 2, \ldots , \max _{i=1}^n \alpha _i$, and by condition 4, we know $i_{1,1} \in C_{h_1}$. From Observation 3, we have $\{i_{1,1}, \ldots , i_{1, n_1}\} \subseteq C_{h_1}$. Hence by condition 2, we know that for any $t = 2, \ldots , \max _{i=1}^n \alpha _i, C_{h_t} \setminus \{i_{1,n_1}\}$ is not a cover for K, and for any $\ell = 1, \ldots , n_1, C_{h_1} \setminus \{i_{1,\ell }\}$ is not a cover for K. So any set in $\mathscr {S}_2 \cup \mathscr {S}_3$ is not a cover. Condition 4 directly states that any set in $\mathscr {S}_4$ is not a cover. Furthermore, condition 3 states that $i_{t,1} \in C_{h_t}$, then by Observation 3, we know that for any $\ell \in [n_t], i_{t, \ell } \in C_{h_t}$. Also, condition 3 states that $C_{h_t} \cup \{i_{t-1, \ell _t}\} \setminus \{i_{t, n_t}\}$ is not a cover, so $C_{h_t} \cup \{i_{t-1, \ell _t}\} \setminus \{i_{t,\ell }\}$ is also not a cover of K. Hence any set in $\mathscr {S}_1$ is not a cover of K.

Now we want to show that $\alpha ^T x = \beta$ is the only hyperplane that contains all the incidence vectors of the sets in $\mathscr {S}_1 \cup \mathscr {S}_2 \cup \mathscr {S}_3 \cup \mathscr {S}_4$. Let $u^T x = v$ be a hyperplane that contains all those binary points. Then, from the sets in $\mathscr {S}_3$, we know that $u_{i_{1,1}} = \ldots = u_{i_{1,n_1}}$, and we denote it to be $\kappa$. Since for any $t = 2, \ldots , \max _{i=1}^n \alpha _i$ and $\ell = 1, \ldots , n_t, u(C_{h_t} \cup \{i_{t-1, \ell _t}\} \setminus \{i_{t, \ell }\}) = v$, we know that for any $t = 2, \ldots , \max _{i=1}^n \alpha _i, u_{i_{t,1}} = \ldots = u_{i_{t, n_t}}$. Furthermore, since $u(C_{h_t} \cup \{i_{t-1, \ell _t}\} \setminus \{i_{t, \ell }\}) = u(C_{h_t} \setminus \{i_{1,n_1}\}) = v$, we obtain that for any $t = 2, \ldots , \max _{i=1}^n \alpha _i$ and $\ell = 1, \ldots , n_t, u_{i_{t,\ell }} - u_{i_{t-1, \ell _t}} = u_{i_{1,n_1}} = \kappa$. Lastly, from the points in $\mathscr {S}_3$ and in $\mathscr {S}_4$, we know that $u_{i'} = 0$ for any $i' \notin C$. Hence we obtain that, for any $t = 1, \ldots , \max _{i=1}^n \alpha _i, \ell = 1, \ldots , n_t$, we have $u_{i_{t,\ell }} = \kappa \cdot t$, and for any $i' \notin C, u_{i'} = 0$. Since $\alpha _{i_{t,\ell }} = t$ and $\alpha _{i'} = 0$ for any $t = 1, \ldots , \max _{i=1}^n \alpha _i, \ell = 1, \ldots , n_t, i' \notin C$, and $\{i_{t, 1}, \ldots , i_{t, n_t}\} = \{i \in C \setminus C_0 \mid \alpha _i = t\}$, we know that $u_i = \kappa \cdot \alpha _i$ for any $i \notin C_0$. By condition 1, we have $u = \kappa \cdot \alpha$. Using condition 5, it is simple to check that $v = u(C_{h_t}) - \kappa = \kappa \cdot \alpha (C_{h_t}) - \kappa = \kappa \cdot \beta$ for any $t = 1, \ldots , \max _{i=1}^n \alpha _i$. Thus, we obtain that $(u,v) = \kappa \cdot (\alpha , \beta )$, and this concludes the proof that $\alpha ^T x = \beta$ is the only hyperplane that contains all the incidence vectors of sets in $\mathscr {S}_1 \cup \mathscr {S}_2 \cup \mathscr {S}_3 \cup \mathscr {S}_4$, which are all in the TOMKS K. Since the S-MCI $\alpha ^T x \le \beta$ is a valid inequality for ${\text {conv}}(K)$, we obtain that $\alpha ^T x \le \beta$ is a facet-defining inequality for ${\text {conv}}(K)$. $\square$

Appendix B: Proof of Theorem 4

First, we restate Theorem 1.1 in [15].

Theorem 7

(Theorem 1.1 [15]) If a clutter $\mathscr {L}$ has no $P_4{:}{=} \{\{1,2\}, \{2,3\}, \{3,4\}\}$ minor, then

$$\begin{aligned} {\text {conv}}(\{x \in \mathbb {Z}^n_+ \mid x(L) \ge 1, \forall L \in \mathscr {L}\}) = \{x \in \mathbb {R}^n_+ \mid x(L) \ge 1, \forall L \in \mathscr {L}\} \end{aligned}$$

if and only if $\mathscr {L}$ has no minor isomorphic to one of the following clutters:

• 1. $Q_6 = \{\{1,3,5\}, \{1,4,6\}, \{2,3,6\}, \{2,4,5\}\}$;

• 2. $J_q = \{\{2, \ldots , q\}, \{1, i\} \text { for } i = 2, \ldots , q\}$, $q \ge 3$.

We are now ready to present our proof of Theorem 4.

Proof (of Theorem 4)

Let ${\mathcal {C}}$ be a minimal cover set of a TOMKS K. We first show that ${\mathcal {C}}$ has no minor isomorphic to $P_4$ or $Q_6$. To see this, let $M = \{\{i_1, i_2\}, \{i_2, i_3\}, \{i_3, i_4\}\}$ be a minor of ${\mathcal {C}}$ isomorphic to $P_4$ and $i_1< \ldots < i_4$. Since K is a TOMKS and $\{i_3, i_4\} \in M$, we know that M should also contain $\{\{i_1, i_3\}, \{i_1, i_4\}, \{i_2, i_4\}\}$. This is a contradiction. Now assume that $M' = \{\{i_1, i_3, i_5\}, \{i_1, i_4, i_6\}, \{i_2, i_3, i_6\}, \{i_2, i_4, i_5\}\}$ is a minor of ${\mathcal {C}}$ isomorphic to $Q_6$ and $i_1< \ldots < i_6$. Then $M'$ must also contain $\{i_2, i_3, i_5\}$, and this also gives us a contradiction. For any other possible ordering of the indices, we can obtain a symmetric argument. This concludes our proof that ${\mathcal {C}}$ has no minor isomorphic to $P_4$ or $Q_6$.

From Theorem 7, we obtain the following statement: If ${\mathcal {C}}$ is the minimal cover set of a TOMKS K, then ${\text {conv}}(\{y \in \mathbb {Z}^n_+ \mid y(C) \ge 1, \forall C \in {\mathcal {C}}\}) = \{y \in \mathbb {R}^n_+ \mid y(C) \ge 1, \forall C \in {\mathcal {C}}\}$ if and only if ${\mathcal {C}}$ has no minor isomorphic to a clutter $J_q$, with $q \ge 3$. Therefore, all extreme points of the polytope $\{y \in [0,1]^n \mid y(C) \ge 1, \forall C \in {\mathcal {C}}\}$ are integral if and only if ${\mathcal {C}}$ has no minor isomorphic to $J_q$. By substituting the variable y with $1-x$, we know that ${\text {conv}}(\{x \in \{0,1\}^n \mid x(C) \le |C| - 1, \forall C \in {\mathcal {C}}\}) = \{x \in [0,1]^n \mid x(C) \le |C| - 1, \forall C \in {\mathcal {C}}\}$ if and only if ${\mathcal {C}}$ has no minor isomorphic to $J_q$. Moreover, it was shown in [2] that ${\text {conv}}(K) = {\text {conv}}(\{x \in \{0,1\}^n \mid x(C) \le |C| - 1, \forall C \in {\mathcal {C}}\})$, and this completes the proof. $\square$

Appendix C: Proof of Theorem 5

Proof

Let $\pi ^T x \le \pi _0$ be a non-trivial facet-defining inequality for ${\text {conv}}(K)$. Then clearly we have that $\pi \in \mathbb {R}^n_+$, $\pi _0 > 0$ (see also [12]). W.l.o.g. we can assume $\pi _0 = 1$. Denote $X{:}{=} \{x \in K \mid \pi ^T x = 1\}$, which has dimension $n-1$ since $\pi ^T x \le 1$ is facet-defining. First, we obtain the following claims. $\square$

Claim 1

$\pi _i = 0$ for all $i = q+1, \ldots , n$.

proof of claim For any $i \in [n] \setminus [q]$, there exists some $x' \in X$ with $x'_i = 0$, since otherwise $X \subseteq \{x \mid x_i = 1, \pi ^T x = 1\},$ which has dimension $n-2$. If ${\text {supp}}(x') \cup \{i\}$ is a cover for K, then it must contain some minimal cover in $J_q$. However, ${\text {supp}}(x')$ does not contain any minimal cover (since $x'$ is feasible), and $i \in [n] \setminus [q]$ is not contained in any minimal cover. Hence we know that $x' + e^i \in K$, which has $\pi ^T (x' + e^i) \le 1$. Because $\pi ^T x' = 1$, we obtain that $\pi _i = 0$.

Claim 2

$\pi _1 = \ldots = \pi _{p-1} = \pi ([n]) - 1> 0$.

proof of claim Suppose $\pi _i = 0$ for some $i \in [p-1]$. From the minimal cover structure of K, we know that $[n] \setminus \{i\}$ is not a cover for K. Hence $\pi ([n]) = \pi ([n] \setminus \{i\}) \le 1 = \pi _0$, which means that $\pi ^T x \le 1$ is dominated by the bound constraints $x_j \le 1, \forall j \in [n]$, a contradiction. Furthermore, for any $i \in [p-1]$, there exists some point $x' \in X$ with $x'_i = 0$, since otherwise $X \subseteq \{x \mid x_i = 1, \pi ^T x = \pi _0\}$ which has dimension $n-2$. Hence $1 = \pi ^T x' \le \pi ([n] \setminus \{i\}) \le 1$, which gives $\pi _i = \pi ([n]) - 1$, for any $i \in [p-1]$.

Claim 3

If $\pi _p > 0$, then $\pi ([p]) = 1$.

proof of claim Since $\sum _{i=1}^p x_i = p-1$ is not valid for K, and X is $(n-1)$-dimensional, we know that there exists $x' \in X$ with $x'([p]) = p$ or $x'([p]) \le p-2$. If $x'([p]) \le p-2$, then there must exist $i,j \in [p]$ such that $x'_i = x'_j = 0$. From Claim 2 and the assumption of $\pi _p > 0$, we know it is impossible. Hence there exists $x' \in X$ with $x'_i = 1$ for any $i \in [p]$. From the minimal cover set K, we know $x'_j = 0$ for any $j \in [q] \setminus [p]$. Therefore, $1 = \pi ^T x' = \pi ([p]) = 1$.

In the remainder of the proof we consider separately three different cases.

Case 1: $\pi _p = 0$. In this case, we want to show that $\pi ^T x \le 1$ is the same as the CI $\sum _{i=1}^{p-1} x_i + \sum _{i=p+1}^q x_i \le q-2$. First, we prove that $x([q]) - x_p = q-2$ for any $x \in X$. If not, since $[q] \setminus \{p\}$ is a cover, then there exists some $x' \in X$ and $i,j \in [q] \setminus \{p\}$, such that $x'_i = x'_j = 0$. We construct a new point $x''$ from $x'$ by switching the j-th component from 0 to 1, and setting $x''_p = 0$. Since $x''_p = x''_i = 0$, we know that $x'' \in K$. Note that $1 \ge \pi ^T x'' = \pi ^T x' + \pi _j = 1+\pi _j$, thus we have $\pi _j = 0$. Hence $\pi ([n] \setminus \{p, j\}) = \pi ([n])$. However, $[n] \setminus \{p, j\}$ is not a cover, and we obtain $\pi ([n]) \le 1$, which means that the inequality $\pi ^T x \le 1$ is dominated by the bound constraints, a contradiction. Therefore $x([q]) - x_p = q-2$ for any $x \in X$. Since $X = \{x \in K \mid \pi ^T x = 1\}$ is assumed to have dimension $(n-1)$, we know that $\pi ^T x \le 1$ must be the same inequality as the CI: $\sum _{i=1}^{p-1} x_i + \sum _{i=p+1}^q x_i \le q-2$. This concludes Case 1.

Case 2: $\pi _p > 0$, and there exists $x' \in X$ with $x'_p = 0$, such that $x'_i = 1$ for any $i \in {\text {supp}}(\pi ) \cap \{p+1, \ldots , q\}$. In this case consider such point $x'$. From Claim 2, we have $\pi ([n] \setminus \{i\}) = 1$ for any $i \in [p-1]$, thus we know that $x'_i = 1$ for any $i \in [p-1]$. Hence $\pi ^T x' = \pi ([n]) - \pi _p = 1$, from Claim 2 we have $\pi _1 = \ldots = \pi _{p-1} = \pi _p$. Also by Claim 3, we have $\pi _1 = \ldots = \pi _p = \frac{1}{p}$. So the original inequality $\pi ^T x \le 1$ is just $(\frac{1}{p}, \ldots , \frac{1}{p}, \pi _{p+1}, \ldots , \pi _q, 0, \ldots , 0)^T x \le 1$, and from Claim 2, we obtain that $\sum _{i=p+1}^q \pi _i = 1-\frac{1}{p} \cdot (p-1) = \frac{1}{p}$. Multiplying each CI $\sum _{i=1}^p x_i + x_j \le p$ by the non-negative number $\pi _j$ for each $j = p+1, \ldots , q$, and summing them up, we obtain that our facet-defining inequality $\pi ^T x \le 1$ is dominated by the CIs $\sum _{i=1}^p x_i + x_j \le p \text { for } j=p+1, \ldots , q$. This means that $\pi ^T x \le 1$ coincides with one of these CIs. This concludes Case 2.

Case 3: $\pi _p > 0$, and for any $x' \in X$ with $x'_p = 0$, $x'_j = 0$ for some index $j \in {\text {supp}}(\pi ) \cap \{p+1, \ldots , q\}$. In this case, we have the following claim.

Claim 4

$\pi _{p+1} = \ldots = \pi _q = \pi _1 - \pi _p >0$.

proof of claim First, we show that $\pi _i > 0$ for any $i \in [q] \setminus [p]$. If not, we have $i \in [q] \setminus [p]$ such that $\pi _i = 0$. Arbitrarily pick a point $x' \in X$ with $x'_p = 0$. By the assumption of this case, then there must exist $j \in [q] \setminus [p]$ such that $x'_j = 0$ and $\pi _j > 0$. Then we construct one point $x''$ from $x'$ by setting the i-th component to 0. Since $\pi _i = 0$, we have $\pi ^T x'' = \pi ^T x' = 1$, where $x''_p = x'_p = 0$, $x''_j = x'_j = 0$, $x''_i = 0$. Note that $x'' + e^j$ is also a feasible point in K. However, $1 =\pi ^T x'' < \pi ^T (x'' + e^j)$, a contradiction.

Next, we show that $\pi _i = \pi _1 - \pi _p$ for any $i \in [q] \setminus [p]$. Note that since $[q] \setminus \{p\}$ is a minimal cover, we know that $[q] \setminus \{p, i\}$ is not a cover for any $i \in [q] \setminus [p]$. Hence $\pi ([q] \setminus \{p, i\}) \le 1$. Then from Claim 1 and 2, we obtain $\pi _p + \pi _i \ge \pi _1$, for any $i \in [q] \setminus [p]$. If for some $i' \in [q] \setminus [p]$, $\pi _p + \pi _{i'} > \pi _1$, then $\pi ([q] \setminus \{p, i'\}) < 1$. Now we argue that, in this case, $\sum _{i=1}^p x_i + x_{i'} = p$ for any $x \in X$. Assuming ${\bar{x}}_{i_1} = {\bar{x}}_{i_2} = 0$ for some ${\bar{x}} \in X$ and $i_1, i_2 \in [p] \cup \{i'\}$. By Claim 2 and the fact that $\pi ^T {\bar{x}} = 1$ and $\pi _i > 0$ for any $i \in [p] \cup \{i'\}$, we know $\{i_1, i_2\} = \{p, i'\}$. So $1 = \pi ^T {\bar{x}} \le \pi ([q] \setminus \{p, i'\}) < 1$, which gives the contradiction. Thus $\pi ^T x \le 1$ coincides with the CI $\sum _{i=1}^p x_i + x_{i'} \le p$. However, we have shown that $\pi _i > 0$ for any $i \in [q] \setminus [p]$, and this gives a contradiction because $p \le q-2$. Therefore, for any $i \in [q] \setminus [p]$, we have $\pi _p + \pi _i = \pi _1$.

Let $\pi _1 =: \lambda$, then Claim 1 gives $\pi _{q+1} = \ldots = \pi _n = 0$, Claim 2 gives $\pi _1 = \ldots = \pi _{p-1} = \lambda$, Claim 3 gives $\pi _p = 1-(p-1)\lambda$, and Claim 4 gives $\pi _{p+1} = \ldots = \pi _q = p \lambda - 1$. Also from Claim 2, we have $\pi ([n]) - \lambda = 1$, hence: $\lambda = (q-p)(p \lambda - 1)$, which gives $\lambda = \frac{q-p}{p(q-p)-1}$. Therefore, the original facet-defining inequality $\pi ^T x \le 1$ coincides with $(q-p) \sum _{i=1}^{p-1} x_i + (q-p-1)x_p + \sum _{i=p+1}^q x_i \le p(q-p)-1$. Note that for the multi-cover $\{\{1, \ldots , p-1, p+1, \ldots , q\}, \{1,\ldots , p,q\}\}$, the inequality has the same structure as the one given in Example 3, and the corresponding MCI is $(q-p) \sum _{i=1}^{p-1} x_i + (q-p-1)x_p + \sum _{i=p+1}^q x_i \le p(q-p)-1$. This concludes Case 3 and finishes the proof. $\square$

Separation formulation in Sect. 5

Let $\mathscr {C}$ be a cover-family. From the definition of multi-cover, $\mathscr {C}$ is a multi-cover if and only its skeleton $\mathscr {S}$ satisfies the following property: For any $T \subseteq \cup _{S \in {\mathscr {S}}} S$, there exists some $S' \in {\mathscr {S}}$ such that T is comparable with $S'$. Henceforth all skeletons are assumed to have such property. Now for a given skeleton $\mathscr {S} {:}{=} \{S_1, \ldots , S_k\}$ and a fractional solution ${\tilde{x}}$, we consider the following MIP. We let $S {:}{=} \cup _{i=1}^k S_i$ and ${\bar{S}}_h {:}{=} S \setminus S_h$ for any $h \in [k]$. Here $\Pi (s){:}{=} \{s' \in S \mid \exists h \in [k], \text {s.t.}\ s \in S_h, s' \notin S_h, s' > s\}$, and M is the only pre-fixed parameter for this formulation, which represents the selected upper bound for all the coefficients of the MCI. Then, the separation problem of MCIs with skeleton $\mathscr {S}$ can be solved exactly using the MIP ($\text {Sep}- {\mathscr {S}}$ ).

Theorem 8

Given a TOMKS K, a given skeleton $\mathscr {S}$, and a point ${\tilde{x}}$, there exists a multi-cover $\mathscr {C}$ whose skeleton is $\mathscr {S}$ and an associated MCI that separates ${\tilde{x}}$ from K, if and only if ($\text {Sep}- {\mathscr {S}}$ ) has optimal value less than 1.

Proof

In ($\text {Sep}- {\mathscr {S}}$ ), the binary $u^s_i$ denotes whether or not the index i of variables corresponds to the index s in the skeleton; $\alpha ^s_i$ denotes the MCI coefficient of variable $x_i$, when $u^s_i$ = 1; The binary $w_i$ denotes whether or not variable index i appears in the intersection of the multi-cover; $\gamma _i$ denotes the MCI coefficient of variable $x_i$, when $w_i$ = 1; For any $h \in [k]$ and $j \in [m]$, the binary $\lambda ^h_j$ denotes whether or not the cover $C_h$ corresponding to the skeleton element $S_h$ violates the j-th knapsack constraint of the problem; $t + \sum _{i \in [n]} \gamma _i$ represents the maximum value of $\alpha (C_h)$, $h \in [k]$. Therefore, our separating MCI is represented by the inequality:

$$\begin{aligned} \sum _{i \in [n]} (\gamma _i + \sum _{s \in S} \alpha ^s_i) x_i \le t + \sum _{i \in [n]} \gamma _i - 1. \end{aligned}$$

Hence the optimal value of ($\text {Sep}- {\mathscr {S}}$ ) is strictly less than 1 if and only if the MCI is a separating inequality. Now we verify that the constraints in ($\text {Sep}- {\mathscr {S}}$ ) are correct. For each index $s \in S$, since it only corresponds to one variable index $i \in [n]$, we have constraint $\sum _{i \in [n]} u^s_i = 1, \forall s \in S$. Similarly, for each index $i \in [n]$, since it either appears in the intersection of the covers, or it corresponds to a single index $s \in S$, or it is not contained in any cover, we have constraint $\sum _{s \in S} u^s_i + w_i \le 1$, $\forall i \in [n]$. Constraint $\alpha ^s_i \ge \alpha ^{s'}_j + (1+M) u^s_i -M$, $\forall j > i$, $s \in S$, $s' \in \Pi (s)$ formulates the Step 4 of Algorithm 1, and constraints $\gamma _i \ge \sum _{j>i} \sum _{s \in {\bar{S}}_h} \alpha ^s_j + (1+n |{\bar{S}}_h| M)w^h_i - n|{\bar{S}}_h| M$, $\forall i \in [n]$, $h \in [k]$, $\gamma _i \ge \sum _{s \in {\bar{S}}_h} \alpha ^s_j + |{\bar{S}}_h| M w^h_i - |{\bar{S}}_h| M$, $\forall j < i, h \in [k]$ as well as $w_i = \sum _{h \in [k]} w^h_i$ formulate the Step 6 of Algorithm 1. Constraints $u^s_i + \sum _{j<i} u^{s+1}_j \le 1$, $\forall s = 1, \ldots , |S| - 1$ formulate the bijective relation between skeleton and the discrepancy family of multi-cover: if index $i \in [n]$ corresponds to the skeleton index $s \in S$, then any skeleton index $s' > s$ only corresponds to index $j > i$. The remaining constraints are easy to interpret. $\square$

Numerical results report

The following two tables present the detailed results for the optimality gap obtained from solving different linear relaxation problems.

(n, m, seed)	LP	MCI	E-MCI	L-MCI	CI	ECI	LCI
(20, 1, 1)	1.05	0.81	0.58	0.48	0.87	0.59	0.59
(20, 1, 2)	3.7	1.92	0.17	0	2.55	0.82	0.82
(20, 1, 3)	0.87	0.24	0.24	0.22	0.68	0.63	0.63
(20, 1, 4)	1.78	0	0	0	1.00	0.16	0
(20, 1, 5)	6.15	1.06	1.06	0.14	2.79	2.07	2.07
(20, 1, 6)	9.90	5.69	0.22	0	6.47	1.01	1.01
(20, 1, 7)	7.29	0	0	0	0	0	0
(20, 1, 8)	3.06	0.41	0.40	0.29	1.99	0.54	0.54
(20, 1, 9)	1.48	0.65	0	0	0.87	0.49	0.49
(20, 1, 10)	4.03	0	0	0	2.81	0.15	0.15
Average	3.93	1.08	0.27	0.11	2.00	0.65	0.63
(20, 2, 1)	3.37	1.79	0.89	0.60	1.83	1.50	0.83
(20, 2, 2)	12.50	7.18	3.16	0	8.68	4.98	0
(20, 2, 3)	1.45	0.27	0.05	0	0.81	0.10	0.08
(20, 2, 4)	3.96	1.66	0.65	0.10	2.79	0.83	0.83
(20, 2, 5)	3.15	1.65	1.39	1.03	2.04	1.65	1.65
(20, 2, 6)	5.43	1.73	0	0	4.34	0	0
(20, 2, 7)	9.63	3.69	0.23	0	6.36	0.67	0.67
(20, 2, 8)	2.49	0.50	0.46	0	1.52	1.06	1.06
(20, 2, 9)	14.64	8.82	2.66	0	10.67	2.83	2.83
(20, 2, 10)	3.45	1.67	0.73	0.41	3.24	3.24	3.24
Average	6.01	2.90	1.02	0.21	4.23	1.69	1.12
(20, 3, 1)	5.53	3.27	0	0	3.27	0	0
(20, 3, 2)	4.46	2.88	1.64	0	3.88	3.08	3.08
(20, 3, 3)	25.31	19.53	0	0	19.80	0	0
(20, 3, 4)	3.77	3.03	1.74	1.64	3.60	2.00	2.00
(20, 3, 5)	4.97	2.32	1.46	0	4.12	1.67	1.67
(20, 3, 6)	1.87	0	0	0	0.28	0.12	0.12
(20, 3, 7)	3.72	0.88	0	0	3.15	1.46	1.46
(20, 3, 8)	3.31	1.66	0.41	0	1.96	0.83	0.83
(20, 3, 9)	1.29	0.87	0.44	0	1.08	0.62	0.16
(20, 3, 10)	6.11	4.42	1.58	1.09	5.53	1.90	1.90
Average	6.03	3.89	0.73	0.27	4.67	1.17	1.12
(30, 1, 1)	7.95	5.29	0.64	0	6.05	0.76	0.76
(30, 1, 2)	2.48	0.91	0.20	0	1.38	0.25	0.25
(30, 1, 3)	24.75	23.35	0.70	0	24.24	0.70	0.70
(30, 1, 4)	2.84	1.37	1.27	0.81	1.48	1.48	1.48
(30, 1, 5)	1.66	0	0	0	0	0	0
(30, 1, 6)	9.45	6.74	1.53	0.77	7.60	1.84	1.84
(30, 1, 7)	1.43	1.27	1.17	0.83	1.31	1.25	1.25
(30, 1, 8)	3.81	2.84	0.42	0	3.17	0.47	0.22
(30, 1, 9)	6.20	4.66	0	0	5.23	0	0
(30, 1, 10)	2.83	1.34	0	0	1.91	0	0
Average	6.34	4.78	0.59	0.24	5.24	0.68	0.65

(n, m, seed)	LP	MCI	E-MCI	L-MCI	CI	ECI	LCI
(30, 2, 1)	5.73	4.42	0.67	0	4.78	0.69	0.69
(30, 2, 2)	5.14	3.25	3.25	0.30	4.05	3.78	3.78
(30, 2, 3)	9.94	7.20	0	0	8.11	0	0
(30, 2, 4)	3.02	0.63	0	0	1.38	0	0
(30, 2, 5)	2.98	2.46	0	0	2.46	0	0
(30, 2, 6)	4.93	2.78	0	0	3.42	0	0
(30, 2, 7)	3.40	2.10	1.99	0	2.80	2.80	0.74
(30, 2, 8)	5.86	5.06	1.35	0.20	5.45	1.35	1.35
(30, 2, 9)	3.04	0	0	0	0.86	0.58	0
(30, 2, 10)	3.59	1.61	1.61	1.37	1.77	1.73	1.73
Average	4.76	2.95	0.89	0.19	3.51	1.09	0.83
(30, 3, 1)	3.76	2.94	1.19	0	3.14	1.32	0.71
(30, 3, 2)	4.13	3.26	0.37	0	3.68	0.38	0.19
(30, 3, 3)	14.64	13.62	3.92	1.67	13.98	4.04	4.04
(30, 3, 4)	1.57	1.49	1.37	0.89	1.55	1.41	1.41
(30, 3, 5)	1.09	1.00	0.60	0.37	1.05	0.61	0.42
(30, 3, 6)	1.46	0	0	0	0	0	0
(30, 3, 7)	0.99	0.88	0.11	0	0.92	0.11	0.11
(30, 3, 8)	1.23	0.55	0	0	0.80	0	0
(30, 3, 9)	0.99	0.77	0.41	0.19	0.83	0.51	0.51
(30, 3, 10)	1.46	0.53	0.37	0	0.70	0.60	0.60
Average	3.13	2.50	0.83	0.31	2.67	0.90	0.80