Relaxations and duality for multiobjective integer programming

Abstract

Multiobjective integer programs (MOIPs) simultaneously optimize multiple objective functions over a set of linear constraints and integer variables. In this paper, we present continuous, convex hull and Lagrangian relaxations for MOIPs and examine the relationship among them. The convex hull relaxation is tight at supported solutions, i.e., those that can be derived via a weighted-sum scalarization of the MOIP. At unsupported solutions, the convex hull relaxation is not tight and a Lagrangian relaxation may provide a tighter bound. Using the Lagrangian relaxation, we define a Lagrangian dual of an MOIP that satisfies weak duality and is strong at supported solutions under certain conditions on the primal feasible region. We include a numerical experiment to illustrate that bound sets obtained via Lagrangian duality may yield tighter bounds than those from a convex hull relaxation. Subsequently, we generalize the integer programming value function to MOIPs and use its properties to motivate a set-valued superadditive dual that is strong at supported solutions. We also define a simpler vector-valued superadditive dual that exhibits weak duality but is strongly dual if and only if the primal has a unique nondominated point.

Appendices

Appendix

Appendix A: Proofs of results in Sect. 2.1

The set-ordering in Definition 4 is well-defined for subsets of ${\mathbb {R}}^k$. We extend it to $\pm M_{\infty }$ by defining $-M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}M_{\infty }$ for all nonempty sets $S \subseteq {\mathbb {R}}^k$, $-M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}-M_{\infty }$, and $M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}M_{\infty }$. We further assume that $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\not \preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}-M_{\infty }$, and $M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\not \preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S$ for any $S \subseteq {\mathbb {R}}^k$. The relation “${{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}$” is thus defined on the extended power set of ${\mathbb {R}}^k$ (excluding the empty set) and is transitive thereon, but neither reflexive nor antisymmetric. However, the relation defines a partial order on the family of sets ${\mathcal {E}}$ defined in (1).

Proposition 1

Let $S \subseteq {\mathbb {R}}^k$ be nonempty. If S has points that are nondominated from above, then ${{\,\mathrm{\textrm{Max}}\,}}(S) \in {\mathcal {E}}$. If S has points that are nondominated from below, then ${{\,\mathrm{\textrm{Min}}\,}}(S) \in {\mathcal {E}}$.

Proof

We prove the result for ${{\,\mathrm{\textrm{Max}}\,}}(S)$; the proof for ${{\,\mathrm{\textrm{Min}}\,}}(S)$ is similar and therefore omitted. Suppose S is nonempty and has points that are nondominated from above. Let $s,t \in {{\,\mathrm{\textrm{Max}}\,}}(S)$ with $s \ne t$. By definition of nondominance, $s \not \le t$ and $t \not \le s$. Thus, ${{\,\mathrm{\textrm{Max}}\,}}(A) \in {\mathcal {E}}$. $\square$

Proposition 2

The relation ${{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}$ defines a partial order on ${\mathcal {E}}$.

Proof

We first show that the relation is reflexive. Consider a set $S \in {\mathcal {E}}$. If $S = \pm M_{\infty }$, then $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S$ by definition of $\pm M_{\infty }$. Otherwise, if $s \in S$, then $s \leqq s$ by the reflexivity of $\leqq$. Moreover, if $t \in S$ with $s \not = t$, then $t \not \le s$ because distinct elements of S are incomparable as $S \in {\mathcal {E}}$. Therefore $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S$.

Next, to see that ${{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}$ is antisymmetric, consider $S,T \in {\mathcal {E}}$ with $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}T$ and $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S$. If one of $S,T = \pm M_{\infty }$, then $S = T$. Otherwise, let $S,T \subseteq {\mathbb {R}}^k$ and $s \in S$. Because $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}T$, there exists $t \in T$ such that $s \leqq t$. On the other hand, because $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}S$, there is $s' \in S$ such that $t \leqq s'$. Therefore, $s \leqq t \leqq s'$. But distinct elements of S are incomparable, which implies that $s = s'$, so that $s = s' = t$. Therefore $S \subseteq T$. Similarly, $T \subseteq S$ as well, so that $S = T$.

Finally, to establish transitivity, suppose $S,T,U \in {\mathcal {E}}$ with $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}T$ and $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. If $S = -M_{\infty }$ or $U = M_{\infty }$, the results holds trivially. If S or $T = M_{\infty }$, then $U= M_{\infty }$ and the result holds. Similarly, if T or U equals $-M_{\infty }$, then so does S and the result follows. Assume, therefore, that $S,T,U \ne \pm M_{\infty }$.

Let $s \in S$. Then, there exists $t \in T$ such that $s \leqq t$ and $u \in U$ such that $t \leqq u$. Therefore $s \leqq u$. On the other hand, suppose there exists $u \in U$ and $s \in S$ such that $u \le s$. Then, there is $t \in T$ such that $s \leqq t$, which implies that $u \le s \leqq t$. This contradicts the hypothesis that $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. So, s and u must be incomparable. Therefore, $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$.

Thus, the relation ${{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}$ is reflexive, antisymmetric and transitive on ${\mathcal {E}}$ and therefore defines a partial order thereon. $\square$

Lemma 2

Let $T \subseteq S \subseteq {\mathbb {R}}^k$ be nonempty sets, and let $U \in {\mathcal {E}}$ such that $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. Then, $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$.

Proof

If $U = M_{\infty }$, the result holds by definition. Therefore, suppose $U \ne M_{\infty }$, and let $t \in T$. Because $t \in S$, there is an element $u \in U$ such that $t \leqq u$. Moreover, given $u \in U$, there is no $t \in T$ such that $u \le t$ because that would contradict $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. Thus, $T {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. $\square$

Lemma 3

Let $S \subseteq {\mathbb {R}}^k$ be a closed set and let $U \in {\mathcal {E}}$ such that $S {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. Then, ${{\,\mathrm{\textrm{Max}}\,}}(S) {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$.

Proof

If $S = \emptyset$, then ${{\,\mathrm{\textrm{Max}}\,}}(S) = -M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. If S is unbounded above, then we must have $U = M_{\infty }$ and ${{\,\mathrm{\textrm{Max}}\,}}(S) = M_{\infty } {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}U$. Suppose, therefore, that S is nonempty and bounded above. The result then follows from Lemma 2 by choosing $T = {{\,\mathrm{\textrm{Max}}\,}}(S)$. $\square$

Appendix B: Omitted proofs from Sect. 3

Proposition 5

If ${\mathcal {Y}}_{\textrm{MOLP}}$ is the set of feasible objective values of (MOLP), then ${{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOIP}}) {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}{{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOLP}})$.

Proof

If (MOIP) is infeasible, then ${{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOIP}}) = -M_{\infty }$ and the result is trivially true. If (MOIP) is unbounded then so is (MOLP) because ${\mathcal {Y}}_{\textrm{MOIP}}\subseteq {\mathcal {Y}}_{\textrm{MOLP}}$. Finally, if $Cx^* \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOIP}})$, then $Cx^*$ is a feasible objective to (MOLP). Therefore either ${{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOLP}}) = M_{\infty }$ or there exists $y \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOLP}})$ such that $Cx^* \leqq y$.

Now suppose $C{\tilde{x}} \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOLP}})$ such that $Cx^* \not = C{\tilde{x}}$. Suppose if possible that $C{\tilde{x}} \le Cx^*$. Then, because $x^* \in {\mathcal {X}}$, it is also feasible to (MOLP), which contradicts the nondominance of ${\tilde{x}}$. Thus, $C{\tilde{x}} \not \le Cx^*$ so that $Cx^*$ and $C{\tilde{x}}$ are incomparable. $\square$

Proposition 6

If ${\mathcal {Y}}_{\textrm{CH}}$ is the set of feasible objective values of (CH), then ${{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{MOIP}}) {{\,\mathrm{\hspace{1.111pt}\underline{\hspace{-1.111pt}\preceq \hspace{-0.83328pt}}\hspace{1.111pt}}\,}}{{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{CH}})$.

Proof

The proof is similar to that of Proposition 5 and is omitted. $\square$

Proposition 7

Let $x^*$ be an efficient solution of (CH). Then, $x^*$ is an efficient solution of (MOIP) if and only if $x^*$ is integral.

Proof

If $x^*$ is efficient for (MOIP), then it must be feasible to (MOIP) and therefore integral. To see the opposite containment, suppose $x^*$ is efficient for (CH) and integral. Then, $x^*$ is an integral point in the feasible region of (CH) and therefore $x^{*} \in {\mathcal {X}}_{\textrm{MOIP}}$. Suppose if possible that $x^*$ is not efficient for (MOIP). Proposition 6 implies that for every feasible x to (MOIP), there is a y feasible to (CH) such that $Cx \leqq Cy$. In particular, if $x^*$ is not efficient for (MOIP), then there is a y feasible to (CH) such that $Cx^*\le Cy$. However, this contradicts the hypothesis that $x^{*}$ was efficient to (CH). Thus, $x^*$ must be efficient for (MOIP). $\square$

Proposition 8

If (CH) has efficient solutions, then it must have an integral efficient solution.

Proof

Suppose that $\text {conv}({\mathcal {X}}_{\textrm{MOIP}})$ has a vertex. If an MOLP has efficient solutions, then at least one must occur at a vertex of the feasible region [34, Theorem 4.3.8(ii)]. In particular, (CH) has at least one efficient vertex. Because the vertices of the feasible region of (CH) are all integral, this implies that there is at least one integral efficient solution to (CH).

Now suppose that $\text {conv}({\mathcal {X}}_{\textrm{MOIP}})$ does not have vertices. Then, by [34, Theorem 4.3.8(iii)], it has a nonempty face F such that every $x \in F$ is an efficient solution. Because F is a face of $\text {conv}({\mathcal {X}}_{\textrm{MOIP}})$, there is a matrix B and a vector d such that $F = \{x \in \text {conv}({\mathcal {X}}_{\textrm{MOIP}})\ \vert \ Bx = d\}$ and such that $Bx \leqq d$ for all $x \in {\mathcal {X}}_{\textrm{MOIP}}$. If $x^* \in F$, then $x^* = \sum _{i = 1}^{p} t^ix^i$ for some positive integer p, $t^1, t^2, \ldots ,t^p \in [0,1]$ and $x^1,x^2, \ldots , x^p\in {\mathcal {X}}_{\textrm{MOIP}}$ with $\sum _{i = 1}^{p}t^i = 1$ because $F \subseteq \text {conv}({\mathcal {X}}_{\textrm{MOIP}})$. Then,

$$\begin{aligned} Bx^* = B\Bigg (\sum _{i = 1}^{p}t^ix^i\Bigg )=\sum _{i = 1}^{p}t^iBx^i = d. \end{aligned}$$

Because $Bx^i\leqq d$ for each i, this implies that $Bx^i = d$ for each i. So, $x^i\in F$ and therefore (CH) has an integral efficient solution. $\square$

Proposition 9

([8, 42]) Let $x^*$ be an efficient solution of (MOIP). Then, $x^*$ is a supported solution if and only if it is an efficient solution of (CH).

Proof

Let $x^*$ be an efficient solution of (CH). By Lemma 1, there is a scalarizing vector $\mu \in {\mathbb {R}}_{>}$ such that $x^*$ is optimal to

On the other hand, because $x^*$ is feasible to (MOIP), it is feasible to the IP

Note that (CH$_\mu$) is the convex-hull relaxation of the single-objective IP (MOIP$_\mu$). Thus, $x^*$ must be optimal to (MOIP$_\mu$). Therefore, $x^*$ is a supported efficient solution to (MOIP).

Conversely, suppose $x^*$ is a supported efficient solution of (MOIP). Then, there exists a scalarizing vector $\mu \in {\mathbb {R}}^k_>$ such that $x^*$ is an optimal solution to (MOIP$_\mu$). Once again, because (CH$_\mu$) is the convex hull relaxation of (MOIP$_\mu$), $x^*$ is optimal to (CH$_\mu$) as well. It follows from Lemma 1 that $x^*$ is efficient for (CH). $\square$

Appendix C: An example to illustrate Theorem 7

This example illustrates Theorem 7. We consider an MOIP whose feasible region does not satisfy condition (15) and show that the Lagrangian dual for this problem is not strong at a supported solution.

Example 6

Consider the problem

$$\begin{aligned} \begin{aligned} \begin{aligned} \max&\begin{bmatrix} 1 &{} \quad 0 \\ 0 &{}\quad 1\end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix} \quad \text {s.t. } 2x_1 + 4x_2 \leqq 5,\ 4x_1 + 2x_2 \leqq 5,\ x_1, x_2 \in \{0,1\}. \end{aligned} \end{aligned} \end{aligned}$$

For this problem, ${\mathcal {X}} = {\mathcal {Y}} = \{(0,0)^\top ,(1,0)^\top ,(0,1)^\top \}$ and the supported nondominated points are $(1,0)^\top$ and $(0,1)^\top$. Set $Q = \{0,1\}^2$, $A^1 = \begin{bmatrix} 2 &{}\quad 4\\ 4 &{}\quad 2 \end{bmatrix}$, and $b^1 = (5, 5)^\top$. Then,

$$\begin{aligned} \begin{aligned} \text {conv}(Q \cap \{x \in {\mathbb {R}}^n\ \vert \ A^1x \leqq b^1\}) \subset \text {conv}(Q) \cap \{x \in {\mathbb {R}}^n\ \vert \ A^1x \leqq b^1\}, \end{aligned} \end{aligned}$$

where the containment is strict. Enumerating $x \in Q$, we have the Lagrangian dual

$$\begin{aligned} \begin{aligned} {{\,\mathrm{\textrm{Min}}\,}}({\mathcal {Y}}_{\textrm{LD}}) = {{\,\mathrm{\textrm{Min}}\,}}\Bigg ( \bigcup \limits _{\varLambda \geqq 0} {{\,\mathrm{\textrm{Max}}\,}}\Bigg \{&\begin{pmatrix} 5\lambda _{11} + 5\lambda _{12}\\ 5\lambda _{21} + 5\lambda _{22} \end{pmatrix}, \begin{pmatrix} 3\lambda _{11} + \lambda _{12}\\ 1 + \lambda _{21} + 3\lambda _{22} \end{pmatrix}, \\&\begin{pmatrix} 1 +\lambda _{11} + 3\lambda _{12}\\ 3\lambda _{21} + \lambda _{22} \end{pmatrix}, \begin{pmatrix} 1-\lambda _{11} -\lambda _{12}\\ 1-\lambda _{21} -\lambda _{22} \end{pmatrix} \Bigg \} \Bigg ). \end{aligned} \end{aligned}$$

Then, there is no $\varLambda \in \mathbb {R}^{2 \times 2}_{\geqq }$ such that $(1,0)^\top \in {{\,\mathrm{\textrm{Min}}\,}}({\mathcal {Y}}_{\textrm{LD}})$. To see this, first suppose that $\varLambda$ was such that $\begin{pmatrix} 5\lambda _{11} + 5\lambda _{12}\\ 5\lambda _{21} + 5\lambda _{22} \end{pmatrix} = (1,0)^\top$. Then, we must have $\lambda _{11} + \lambda _{12} = \frac{1}{5}$ and $\lambda _{21} = \lambda _{22} = 0$. But then, because one of $\lambda _{11},\lambda _{12}$ must be positive, there are positive parameters $\delta _1 = 3\lambda _{11} + \lambda _{12}$ and $\delta _2 = \lambda _{11} + 3\lambda _{22}$ with $\delta _1,\delta _2 \ge \frac{1}{5}$ such that the relaxation (LR($\varLambda$)) is

$$\begin{aligned} \begin{aligned} {{\,\mathrm{\textrm{Max}}\,}}\left\{ \begin{pmatrix}1\\ 0 \end{pmatrix}, \begin{pmatrix}\delta _1\\ 1 \end{pmatrix}, \begin{pmatrix}1+\delta _2\\ 0 \end{pmatrix}, \begin{pmatrix}\frac{4}{5}\\ 1 \end{pmatrix} \right\} . \end{aligned} \end{aligned}$$

Then, $1+\delta _2 \ge \frac{6}{5} > 1$, which implies that $(1,0)\not \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{LR}(\varLambda )})$.

Because $\lambda _{21},\lambda _{22} \ge 0$, there is no feasible $\varLambda$ such that $1+ \lambda _{21} +3\lambda _{22} = 0$. So, $\begin{pmatrix} 3\lambda _{11} + \lambda _{12}\\ 1 + \lambda _{21} + 3\lambda _{22} \end{pmatrix} \not = \begin{pmatrix}1\\ 0 \end{pmatrix}.$

Next, if $\begin{pmatrix} 1 +\lambda _{11} + 3\lambda _{12}\\ 3\lambda _{21} + \lambda _{22} \end{pmatrix} = \begin{pmatrix}1\\ 0 \end{pmatrix}$, then $\varLambda = 0$. However, $(1,0) \not \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{LR}(0)})$ because

$$\begin{aligned} \begin{aligned}{{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{LR}(0)})={{\,\mathrm{\textrm{Max}}\,}}\{(0,0),(0,1),(1,0),(1,1) \} = \{(1,1)\}. \end{aligned} \end{aligned}$$

Finally, if $\varLambda$ was such that $\begin{pmatrix} 1-\lambda _{11}\quad -\lambda _{12}\\ \ 1-\lambda _{21} \quad -\lambda _{22} \end{pmatrix} = \begin{pmatrix} 1\\ 0\end{pmatrix}$ then $\lambda _{11} = \lambda _{12} = 0$ and $\lambda _{21} + \lambda _{22} = 1$. Then, there are positive parameters $\delta _1 = \lambda _{21} + 3\lambda _{22}$ and $\delta _2 = 3\lambda _{21} + \lambda _{22}$ such that $\delta _1, \delta _2 \ge 1$ and

$$\begin{aligned} \begin{aligned} {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{LR}(\varLambda )}) = {{\,\mathrm{\textrm{Max}}\,}}\left\{ \begin{pmatrix} 0\\ 5 \end{pmatrix}, \begin{pmatrix} 0\\ 1+\delta _1 \end{pmatrix}, \begin{pmatrix}1 \\ \delta _2 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \end{pmatrix} \right\} , \end{aligned} \end{aligned}$$

so that $(1,0) \not \in {{\,\mathrm{\textrm{Max}}\,}}({\mathcal {Y}}_{\textrm{LR}(\varLambda )})$ because $\delta _2 \ge 1$. Therefore, there are no Lagrangian relaxations which are tight at $(1,0)^\top$. Moreover, the relaxations are bounded away from $(1,0)^\top$ so that $(1,0)^\top$ cannot be a limit point of ${\mathcal {Y}}_{\textrm{LD}}$.

Thus, condition (15) is not satisfied, and the Lagrangian dual for this problem is not strong at a supported efficient solution. $\square$