On the m-clique free interval subgraphs polytope: polyhedral analysis and applications

Abstract

In this paper we study the m-clique free interval subgraphs. We investigate the facial structure of the polytope defined as the convex hull of the incidence vectors associated with these subgraphs. We also present some facet-defining inequalities to strengthen the associated linear relaxation. As an application, the generalized open-shop problem with disjunctive constraints (GOSDC) is considered. Indeed, by a projection on a set of variables, the m-clique free interval subgraphs represent the solution of an integer linear program solving the GOSDC presented in this paper. Moreover, we propose exact and heuristic separation algorithms, which are exploited into a Branch-and-cut algorithm for solving the GOSDC. Finally, we present and discuss some computational results.

Appendix

Proof of Proposition 2

Let us denote by $az\le \alpha$ inequality $-z_{e}\le 0$ associated with $e\in E$. Let $bz\le \beta$ be a facet defining inequality of $\mathrm {P} _{{\mathcal {I}}}(G,m)$, such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Clearly $I_{0}=\emptyset$ is feasible. Let $I_{0}=\emptyset$ be a solution, and the associated incidence vector $z^{I_{0}}$ verifies $-z_{e}=0$.

Let $e^{\prime }\in E\setminus \{e\}$. The solution $I_{e^{\prime }}=\{e^{\prime }\}$ is feasible and the incidence vector $z^{I_{e^{\prime }}}$ associated with $I_{e^{\prime }}$ verifies $-z_{e} =0$.

Since, $az^{I_{0}}=az^{I_{e^{\prime }}}$, we deduce that $bz^{I_{0}}=bz^{I_{e^{\prime }}}$. This implies that $b(e^{\prime })=0$ for all $e^{\prime }\in E\setminus \{e\}$. Therefore, we set $b(e)=\rho$, and then $b=\rho a$. $\square$

Proof of Proposition 3

Let us denote by $az\le \alpha$ inequality $z_{e}\le 1$ associated with $e\in E$. Let $bz\le \beta$ be a facet defining inequality of $\mathrm {P} _{{\mathcal {I}}}(G,m)$, such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Consider the feasible solution $I_{0}=\{e\}$. And the associated incidence vector $z^{I_{0}}$ verifies $z_{e}=1$.

Let $e^{\prime }\in E\setminus \{e\}$. The solution $I_{e^{\prime }}=\{e^{\prime },e\}$ is feasible and the incidence vector $z^{I_{e^{\prime }}}$ associated with $I_{e^{\prime }}$ verifies $z_{e} =1$.

Since, $az^{I_{0}}=az^{I_{e^{\prime }}}$, we deduce that $bz^{I_{0}}=bz^{I_{e^{\prime }}}$. This implies that $b(e^{\prime })=0$ for all $e^{\prime }\in E\setminus \{e\}$. Therefore, we set $b(e)=\rho$, and then $b=\rho a$. $\square$

Proof of Proposition 4

Let us denote by $az\le \alpha$ the inequality (1). Let $bz\le \beta$ be an inequality that defines a facet of $P_{I}(G,m)$, such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P} _{{\mathcal {I}}}(G,m):bz=\beta \}$. Since $P_{I}(G,m)$ is full dimensional, we need to prove that there exists $\rho$ such that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e_{1},e_{2}\in BC$ be two edges. Clearly, the solutions $I_{1}=BC\setminus \{e_{1}\}$, and $I_{2}=BC\setminus \{{e_{2}}\}$ are feasible. Their incidence vectors satisfy the inequality (1) with equality. Since, $az^{I_{1}}=az^{I_{2}}$, $bz^{I_{1}}=bz^{I_{2}}$, it follows that $b(e_{1})=b(e_{2})$. We set $b(e_{1})=\rho$. As $e_{1}$ and $e_{2}$ are arbitrary in BC, then $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in BC$.

The solutions $I_{3}=BC\setminus \{(1,4)\}$, and $I_{4}=I_{3}\cup \{(2,4)\}$, $I_{5}=I_{3}\cup \{(5,4)\}$, and $I_{6}=I_{3}\cup \{(5,7)\}$ are feasible. Their incidence vectors satisfy the inequality (1) with equality. Since, $az^{I_{3}}=az^{I_{4}}=az^{I_{4}}=az^{I_{5}}=az^{I_{6}}$, $bz^{I_{3}}=bz^{I_{4}}=bz^{I_{4}}=bz^{I_{5}}=bz^{I_{6}}$, it follows that $b((2,4))=b((5,4))=b((5,7))=0$. By symmetry, $b(e)=0$, for all $e\in {\overline{BC}}_{h}^{4}\cup {\overline{BC}}_{h}^{5}\cup {\overline{BC}} _{\triangle }$.

The solutions $I_{7}=BC\setminus \{(4,7)\}$ and $I_{8}=I_{7}\cup \{(1,7)\}$ are feasible and verify the inequality (1) with equality. Since, $az^{I_{7}} =az^{I_{8}}$, $bz^{I_{7}}=bz^{I_{8}}$, it follows that $b((1,7))=0$. By symmetry, we have $b((1,5))=b((1,6))=0$.

Let $e_{3}\in E\setminus (BC\cup {\overline{BC}})$. The solutions $I_{3}$ and $I_{9}=I_{3}\cup \{e_{3}\}$ are feasible and verify the inequality (1) with equality. Since, $az^{I_{3}}= az^{I_{9}}$, $bz^{I_{3} }=bz^{I_{9}}$, which implies that $b(e_{3})=0$. By symmetry, we have $b(e)=0$ for all $e\in E\setminus (BC\cup {\overline{BC}})$. Thus $b=\rho a$. $\square$

Proof of Proposition 6

Let us denote by $az\le \alpha$ the inequality (3) associated with e. Let $bz\le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e_{1},e_{2}\in E_{u}^{a}\setminus \{(4,7)\}$. The solutions $I_{1} =E_{u}\setminus \{e_{1}\}$ and $I_{2}=E_{u}\setminus \{e_{2}\}$ are feasible and the incidence vectors $z^{I_{1}}$ and $z^{I_{2}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{1}}=az^{I_{2}}$. Hence, $bz^{I_{1}}=bz^{I_{2}}$. This implies that $b(e_{1})=b(e_{2})$. We set $b(e_{1})=\rho$. As $e_{1}$ and $e_{2}$ are arbitrary in $E_{u} ^{a}\setminus \{(4,7)\}$, by symmetry, we have $b(e^{\prime })=b(e)=\rho$ for all $e\in E_{u}^{a}\setminus \{(4,7)\}$.

The solution $I_{3}=E_{u}^{a}\setminus \{(4,7),(1,6)\}$ and $I_{4}=(I_{3} \cup \{(2,5)\})\setminus \{(4,5)\}$ are feasible and the incidence vectors $z^{I_{3}}$ and $z^{I_{4}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{3}}=az^{I_{4}}$. Hence, $bz^{I_{3}}=bz^{I_{4}}$. This implies that $b((4,5))=b((2,5))$. Thus, from the previous results we deduce that $b((2,5))=b(e)$ for all $e\in E_{u}^{a}\setminus \{(4,7)\}$.

The solution $I_{5}=(I_{3}\cup \{(2,6)\})\setminus \{(5,6)\}$ is feasible, such that the incidence vectors $z^{I_{5}}$ and $z^{I_{3}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{5}} =az^{I_{3}}$. Hence, $bz^{I_{5}}=bz^{I_{6}}$. This implies that $b((2,6))=b((5,6))$. Indeed, from the previous results we have $b((2,6))=b(e)$ for all $e\in E_{u}^{a}\setminus \{(4,7)\}$.

The solution $I_{6}=(I_{3}\cup \{(3,6)\})\setminus \{(5,6)\}$ is feasible and the incidence vector $z^{I_{3}}$ and $z^{I_{6}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{3}} =az^{I_{6}}$. Hence, $bz^{I_{3}}=bz^{I_{6}}$. This implies that $b((3,6))=b((5,6))$. Indeed, from the previous results we can deduce that $b((3,6))=b(e)$ for all $e\in E_{u}^{a}\setminus \{(4,7)\}$.

The solutions $I_{7}=E_{u}^{a}\setminus \{(4,7)\cup (1,6)\}$ and $I_{8} =(I_{7}\cup \{(4,7)\}\cup E_{u}^{i})$ are feasible and the incidence vectors $z^{I_{7}}$, and $z^{I_{8}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{7}}=az^{I_{8}}$. Hence $bz^{I_{7}}=bz^{I_{8}}$. This implies that $b(e)=0$ for all $e\in E_{u}^{i}$.

Let $e\in \{(2,7),(3,7),(1,7),(5,7),(6,7)\}$. The solutions $I_{1}$ and $I_{9}=I_{1}\cup \{e\}$ are feasible and the incidence vectors $z^{I_{1}}$, $z^{I_{9}}$ verify the inequality (3) with equality. Moreover, we have $az^{I_{1}}=$ $az^{I_{9}}$. Hence, $bz^{I_{1}}=$ $bz^{I_{9}}$. This implies that $b((2,7))=$ $b((3,7))=$ $b((1,7))=$ $b((5,7))=$ $b((6,7))=0$. Thus $b=\rho a$. $\square$

Proof of Proposition 7

Let us denote by $az\le \alpha$ inequality (4) associated with e. Let $bz \le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{ z \in \mathrm {P}_{{\mathcal {I}}} (G,m): az = \alpha \} \subseteq \{ z \in \mathrm {P}_{{\mathcal {I}}} (G,m): bz = \beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e_{1},e_{2}\in E_{u}^{t}\cup E_{u}^{c}$ be two edges, where $e_{1}\ne e_{2}$. We consider the edge sets $I_{1}=E_{u}\cup \{e_{1}\}$ and $I_{2} =E_{u}\cup \{e_{2}\}$ where the incidence vectors $z^{I_{1}}$ and $z^{I_{2}}$ are solutions of $\mathrm {P}_{{\mathcal {I}}}(G,m)$ and satisfy the inequality (4) with equality. Moreover, we have $az^{I_{1}}=az^{I_{2}}$. Thus, $bz^{I_{1}}=bz^{I_{2}}$. This implies that $b(e_{1})=b(e_{2})$. As $e_{1},e_{2}$ are arbitrary, then $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in E_{u}^{t}\cup E_{u}^{c}$.

Let $e_{3}\in E_{u}^{a}$. The solution $I_{3}=E_{u}\setminus \{e_{3}\}$ is feasible and the incidence vector $z^{I_{3}}$ verifies the inequality (4) with equality. Moreover, we have $az^{I_{1}}=az^{I_{3}}$. Hence, $bz^{I_{1}}=bz^{I_{3}}$. This implies that $b(e_{3})=-b(e_{1})$. Thus, by symmetry we have $b(e^{\prime })=-b(e)$ for all $e\in E_{u}^{t}\cup E_{u}^{c},e^{\prime }\in E_{u}^{a}$.

The solutions $I_{4}=E_{u}\setminus \{(1,6)\}$, $I_{5}=(E_{u}\setminus \{(1,6)\})\cup E_{u}^{i}$, $I_{6}=I_{5}\setminus \{(1,5)\}$, $I_{7} =I_{6}\setminus \{(1,4)\}$ and $I_{8}=I_{7}\setminus \{(1,3)\}$ are feasible. The incidence vectors $z^{I_{4}}$, $z^{I_{5}}$, $z^{I_{6}}$, $z^{I_{7}}$ and $z^{I_{8}}$ verify the inequality (3) with equality. Since $az^{I_{4}}=az^{I_{5}}=az^{I_{6}}=az^{I_{7}}=az^{I_{8}}$, $bz^{I_{4}}=bz^{I_{5}}=bz^{I_{6}}= bz^{I_{7}}=bz^{I_{8}}$, it follows that $b(e)=0$ for all $e\in E_{u}^{i}$.

The solutions $I_{9}=E_{u}\cup \{(2,7),(2,4)\}$, $I_{10}=E_{u}\cup \{(2,5),(2,4)\}$, $I_{11}=E_{u}\cup \{(2,4)\}$, $I_{12}=E_{u}\setminus \{(2,3)\}$, $I_{13}=I_{12}\cup \{(2,6)\}$, $I_{14}=I_{12}\cup \{(2,5)\}$, $I_{15}=E_{u}\setminus \{(5,6)\}$ and $I_{16}=I_{15}\cup \{(3,6)\}$ are feasible and verify the inequality (4) with equality. Since $az^{I_{9}}=az^{I_{10}}= az^{I_{11}}=az^{I_{12}}=az^{I_{13}} =az^{I_{14}}=az^{I_{15}}= az^{I_{16}}$, $bz^{I_{9}} =bz^{I_{10}}= bz^{I_{11}}= bz^{I_{12}}= bz^{I_{13}}=bz^{I_{14}} = bz^{I_{15}} = bz^{I_{16}}$,it follows that $b((2,6))=b((2,5))=b((2,7))=b((6,7))=b((3,6))=0$.

Let $e\in E\setminus (E_{u}\cup \bar{E_{u}})$. The solutions $I_{17} =E_{u}\setminus \{(4,7)\}$ and $I_{18}=I_{17}\cup \{e\}$ are feasible and verify the inequality (4) with equality. Since $az^{I_{17}}= az^{I_{18}}$, $bz^{I_{17}}=bz^{I_{18}}$. By symmetry, we have $b(e)=0$.

We set $b(e)=\rho$ for $e\in E_{u}$ and the proof is ended. $\square$

Proof of Proposition 8

Let us denote by $az\le \alpha$ the inequality (5) associated with e. Let $bz\le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e_{1},e_{2}\in E_{net}\setminus E_{net}^{h}$ be two edges, where $e_{1}\ne e_{2}$. We consider the edge sets $I_{1}=E_{net}\setminus \{e_{1}\}$ and $I_{2}=E^{net}\setminus \{e_{2}\}$. Their incidence vectors $z^{I_{1}}$ and $z^{I_{2}}$ are solutions of $\mathrm {P}_{{\mathcal {I}}}(G,m)$ and satisfy the inequality (5) with equality. Moreover, we have $az^{I_{1}} =az^{I_{2}}$ and then $bz^{I_{1}}=bz^{I_{2}}$. This implies that $b(e_{1})=b(e_{2})$. As $e_{1},e_{2}$ are arbitrary, then $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in E_{net}\setminus E_{net}^{h}$.

We consider the edge sets $I_{3}=E_{net}\setminus \{(b,1)\}$ and $I_{4} =E^{net}\setminus (\{(b,1)\}\cup E_{net}^{h}).$Their incidence vectors $z^{I_{3}}$ and $z^{I_{4}}$ are solutions of $\mathrm {P}_{{\mathcal {I}}}(G,m)$ and satisfy the inequality (5) with equality. Moreover, we have $az^{I_{3}}=az^{I_{4}}$. Hence, we have $bz^{I_{3}}=bz^{I_{4}}$. This implies that $b(e)=0$ for all $e\in E_{net}^{h}$.

Let $e_{3}\in {\overline{E}}_{net} \setminus E_{net}^{\bar{h}}$. The solution $I_{5}=E^{net}\cup \{e\}$ is feasible and satisfies the inequality (5) with equality. Moreover, we have $az^{I_{1}}=az^{I_{5}}$ and then $bz^{I_{1}}=bz^{I_{5}}$. This implies that $b(e)=-b(e^{\prime })$ for all $e\in E_{net}\setminus E_{net}^{h}$ and $e^{\prime }\in {\overline{E}}_{net}\setminus E_{net}^{\bar{h}}$.

Considering the edge sets $I_{6}=E^{net}\cup \{(c,2),(c,3),\ldots ,(c,n),(c,d)\}$, $I_{7}=E^{net}\cup \{(c,2)\}$, $I_{8}=E^{net}\cup \{(d,1),(d,2),\ldots ,(d,n-1)\}$, $I_{9}=E^{net}\cup \{(d,n-1)\}$,$I_{10}=E^{net}\cup \{(a,1)\}$, $I_{11} =E^{net}\cup \{(a,1),(a,c)\}$,$I_{12}=E^{net}\cup \{(a,n)\}$, $I_{13} =E^{net}\cup \{(a,n),(a,d)\}$. These edge sets are solutions and satisfy the inequality (5) with equality. Moreover, we have $az^{I_{6}} =az^{I_{7}}=az^{I_{8}} = az^{I_{9}}=az^{I_{10}}=az^{I_{11}} = az^{I_{12}}=az^{I_{13}}$. Hence, we have $bz^{I_{6}}=bz^{I_{7}} =bz^{I_{8}} = bz^{I_{9}}=bz^{I_{10}}=bz^{I_{11}} = bz^{I_{12}} =bz^{I_{13}}$. Thus, $b(e)=0$ for all $e\in E_{net}^{\bar{h}}$.

Let $e\in E\setminus (E_{net}\cup {\overline{E}}_{net} )$. By considering the feasible solutions $I_{14}=E_{net}\setminus \{(a,b)\}$ and $I_{15}=E_{net} \setminus \{(a,b)\}\cup \{e\}$, we have $az^{I_{14}}=az^{I_{15}}$. Hence, $bz^{I_{14}}=bz^{I_{15}}$. This implies that $b(e)=0$.

We set $b(e)=\rho$ for $e\in E_{net}\setminus E_{net}^{h}$ and the proof is complete. $\square$

Proof of Proposition 9

Let us denote by $az\le \alpha$ the inequality (6) associated with e. Let $bz\le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e_{1},e_{2}\in E_{tent}\setminus E_{tent}^{h}$ be two edges, where $e_{1}\ne e_{2}$. We consider the edge sets $I_{1}=E_{tent}\setminus \{e_{1}\}$ and $I_{2}=E^{tent}\setminus \{e_{2}\}$. Their incidence vectors $z^{I_{1}}$ and $z^{I_{2}}$ are solutions of $\mathrm {P}_{{\mathcal {I}}}(G,m)$ and satisfy the inequality (6) with equality. Moreover, we have $az^{I_{1}}=az^{I_{2}}$ and then $bz^{I_{1}}=bz^{I_{2}}$. This implies that $b(e_{1})=b(e_{2})$. As $e_{1},e_{2}$ are arbitrary, then $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in E_{tent}\setminus E_{tent}^{h}$.

We consider the edge sets $I_{3}=E_{tent}\setminus \{(a,c)\}$, and $I_{4}=E^{tent}\setminus (\{(a,c)\}\cup \{(b,c)\})$ there incidence vectors $z^{I_{3}}$, and $z^{I_{4}}$ are solutions of $\mathrm {P}_{{\mathcal {I}}}(G,m)$ and satisfy the inequality (6) with equality. Moreover, we have $az^{I_{3}}=az^{I_{4}}$ and then $bz^{I_{3}}=bz^{I_{4}}$. This implies by symmetry that $b((b,c))=b((b,2))=b((c,4))=0$.

Let $e_{3}\in {\overline{E}}_{tent} \setminus E_{tent}^{\bar{h}}$. The solution $I_{5}=E^{tent}\cup \{e\}$ is feasible and satisfies the inequality (6) with equality. Moreover, we have $az^{I_{1}}=az^{I_{5}}$. Hence, $bz^{I_{1}}=bz^{I_{5}}$. This implies that $b(e)=-b(e^{\prime })$ for all $e\in E_{tent}\setminus E_{tent}^{h}$ and $e^{\prime }\in \overline{E_{tent}}\setminus E_{tent}^{\bar{h}}$.

Let $(i,i+3)\in E_{tent}^{\bar{h}}$. The solutions $I_{6}=E^{tent} \cup \{(i,i+3),(i,i+2)\}$ and $I_{7}=E^{tent}\cup \{(i,i+2)\}$ are feasible and satisfy the inequality (6) with equality. Moreover, we have $az^{I_{6}}=az^{I_{7}}$ and then $bz^{I_{6}}=bz^{I_{7}}$. Thus, $b(e)=0$ for all $e\in E_{tent}^{\bar{h}}$.

Let $e\in E\setminus (E_{tent}\cup {\overline{E}}_{tent} )$. By considering the feasible solutions $I_{8}=E_{tent}\setminus \{(a,c)\}$ and $I_{9}=E_{tent} \setminus \{(a,c)\}\cup \{e\}$, we have $az^{I_{8}}=az^{I_{9}}$ and then $bz^{I_{8}}=bz^{I_{9}}$. This implies that $b(e)=0$.

We set $b(e)=\rho$ for $e\in E_{tent}\setminus E_{tent}^{h}$ and the proof is complete. $\square$

Proof of Proposition 12

Remark that if $m=2$, then a solution can be given only by a forest in the subgraph G(K). We deduce that the maximum number of edges in this subgraph is equal to $|V(K)|-1$.

Let us denote by $az\le \alpha$ the inequality (9) associated with e. Let $bz\le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $(u,v),(u,w)\in E(K)$, be two connected edges. Considering a line $T_{uv}$ beginning by (u, v) and finishing by the vertex w and the line $T_{uw}=T_{uv}\setminus \{(u,v)\}\cup \{(u,w)\}$. These two solutions are feasible and satisfy the inequality (9) with equality. Moreover, we have $az^{T_{uv}}=az^{T_{uw}}$. Hence, $bz^{T_{uv}}=bz^{T_{uw}}$. This implies that $b((u,v))=b((u,w))$. By symmetry, we deduce then that $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in E(K)$.

Let $e\in E\setminus E(K)$. By considering the feasible solution $T_{uv}^{\prime }=T_{uv}\cup \{e\}$, we have $az^{T_{uv}}=az^{T_{uv}^{\prime }}$. Hence, $bz^{T_{uv}}=bz^{T_{uv}^{\prime }}$. This implies that $b(e)=0$. $\square$

Proof of Proposition 13

Let us denote by $az\le \alpha$ the inequality (10) associated with e. Let $bz\le \beta$ be a facet defining inequality of $P_{I}(G,m)$ such that $\{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):az=\alpha \}\subseteq \{z\in \mathrm {P}_{{\mathcal {I}}}(G,m):bz=\beta \}$. We show that $b=\rho a$ for some $\rho \in {\mathbb {R}}$.

Let $e,e^{\prime }$ be two edges in E(K). The solution $I_{1}=E(K)\setminus \{e\}$ and $I_{2}=E(K)\setminus \{e^{\prime }\}$ are feasible and verify the inequality (10) with equality. Hence, $az^{I_{1}}=az^{I_{2}}$. Therefore, $bz^{I_{1}}=bz^{I_{2}}$. This implies that $b(e)=b(e^{\prime })$. We set $b(e)=\rho$. As b(e) and $b(e^{\prime })$ are arbitrary, then $b(e)=b(e^{\prime })$ for all $e,e^{\prime }\in E(K)$.

Let $e_{1}$ be an edge in $E\setminus E(K)$. The solution $I_{3}=I_{1} \cup \{e_{1}\}$ and $I_{1}$ are feasible and verify the inequality (10) with equality. Hence, $az^{I_{1}}=az^{I_{3}}$. Therefore $bz^{I_{1}} =bz^{I_{3}}$. This implies that $b(e_{1})=0$. As $b(e_{1})$ is arbitrary we deduce that $b(e)=0$, for all $e\in E\setminus E(K)$. Thus $b=\rho a$. $\square$