Defaulting firms and systemic risks in financial networks: a normative approach

Abstract

We study systemic risk in an interbank market, employing an explicit axiomatization inspired by Eisenberg and Noe (Manag Sci 47(2):236–249, 2001) and Rogers and Veraart (Manag Sci 59(4):882–898, 2013). Instead of focusing on a clearing payment scheme, we characterize the smallest (in the sense of inclusion) set of ex post defaulting firms. This novel approach allows us to analyze the normative implications of the Eisenberg–Noe axioms. We first show that the Absolute Priority axiom, which states that defaulting firms must end up with zero net worth, has no impact on minimal default sets. Second, relaxing the Limited Payments axiom, which can be interpreted as allowing a central planner to transfer resources from rich firms to poor, does not further reduce the minimal default sets, although other default sets are possible. Our normative analysis sheds new light on the possible impacts of clearing mechanisms on default outcomes.

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Notes

  1. 1.

    Recent surveys of contagion in financial markets include Allen and Babus (2009), Summer (2013), Cabrales et al. (2016), Hüser (2015), and Glasserman and Young (2016), among others.

  2. 2.

    Our setup notably differs from general-equilibrium frameworks, whether default occurs at equilibrium (Zame 1993; Dubey et al. 2005; Araujo et al. 1996, 1998; Araujo and Páscoa 2002; Modica et al. 1998; Eichberger et al. 2014, for instance) or not (Kehoe and Levine 1993; Alvarez and Jermann 2000; Bidian and Bejan 2015; Martins da Rocha and Vailakis 2017 among others).

  3. 3.

    Note that since the total number of banks is exogenous, this number is equivalent to the minimal share of defaulting banks.

  4. 4.

    This type of default criterion is common in the bankruptcy literature, see Araujo and Páscoa (2002), Modica et al. (1998), and Eichberger et al. (2014), for instance.

  5. 5.

    All formal proofs are presented in “Appendix”.

  6. 6.

    The notation \(A\subsetneq B\) means that A is a proper subset of B.

  7. 7.

    An alternative solution to construct the minimal set would be to use the fictitious default algorithm of Rogers and Veraart (2013). For the paper to remain self-contained, we provide a stand-alone proof.

  8. 8.

    Since \(0\le \alpha ,\beta \le 1\), if \(L_{i,{\mathcal {N}}}-\beta \sum \nolimits _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}{\overline{X}}_{j,{\mathcal {N}}}\le \alpha e_{i}+\beta \sum \nolimits _{j\in \overline{{\mathcal {D}}}}L_{ji}\) then, \(L_{i,{\mathcal {N}}}-\sum \nolimits _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}{\overline{X}}_{j,{\mathcal {N}}}\le e_{i}+\sum \nolimits _{j\in \overline{{\mathcal {D}}}}L_{ji}\).

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Appendix

Appendix

Appendix A: Proof of Proposition 1

Let \(0\le \alpha ,\beta \le 1\), \(e=(e_{j})_{j\in {\mathcal {N}}}\in {\mathcal {E}}\), and \(L=(L_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\). The set of {LL, AP, P, LP}-solutions to (eL) that are non-empty is a result of Rogers and Veraart (2013).

A.1: Proof of Proposition 1-1

If: Let us assume \({\mathcal {D}}_{0}(e,L)=\emptyset \). The proof is straightforward by showing that L satisfies LL, AP, P, and LP.

Only if: Let us assume that the empty set is an {LL, AP, P, LP}-solution to (eL). There is therefore a CPM \(X\in {\mathcal {L}}\) such that \({\mathcal {D}}(L,X)=\emptyset \). By definition, \(X=L\) and \({\mathcal {D}}_{0}(e,L)=\emptyset \) therefore follow from LL.

A.2: Proof of Proposition 1-2

Let \(X=(X_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\) be a CPM satisfying LL, AP, P, and LP such that \({\mathcal {D}}(L,X)\) is a minimal {LL, AP, P, LP}-solution to (eL). Let us assume that we can consider \(k\in {\mathcal {N}}\) such that \(k\in {\mathcal {D}}_{0}(e,L)\) and \(k\notin {\mathcal {D}}(L,X)\).

Because X satisfies LL, we have \(e_{k}+X_{{\mathcal {N}},k}-X_{k,{\mathcal {N}}}\ge 0\). Because X satisfies P, we have \(\forall i\), such that \(L_{i,{\mathcal {N}}}>0\), \(X_{i,k}=L_{i,k}\frac{X_{i,{\mathcal {N}}}}{L_{i,{\mathcal {N}}}}\le L_{i,k}\), where the inequality holds since X also verifies LP. The P axiom also implies that \(\forall i\), such that \(L_{i,{\mathcal {N}}}=0\), \(X_{i,k}=0\). We deduce by summing the inequalities over i that: \(X_{{\mathcal {N}},k}\le L_{{\mathcal {N}},k}\). By definition of \({\mathcal {D}}(L,X)\), \(X_{k,{\mathcal {N}}}=L_{k,{\mathcal {N}}}\). Combining the latter with the previous inequality gives:

$$\begin{aligned} e_{k}+L_{{\mathcal {N}},k}-L_{k,{\mathcal {N}}}\ge 0, \end{aligned}$$

which contradicts the assumption that \(k\in {\mathcal {D}}_{0}(e,L)\).

Appendix B: Proof of Proposition 2

B.1: A preliminary lemma

Lemma 1

Let \(0\le \alpha ,\beta \le 1\). Let \(e=(e_{j})_{j\in {\mathcal {N}}}\in {\mathcal {E}}\) and \(L=(L_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\). Let the CPM \(X\in {\mathcal {L}}\) satisfying LP, LL, P be such that \({\mathcal {D}}(L,X)\) is a minimal {LP, LL, P}-solution to (eL). There is therefore a CPM \(X'\in {\mathcal {L}}\) satisfying LP, LL, P, AP such that \({\mathcal {D}}(L,X')={\mathcal {D}}(L,X)\).

Proof of Lemma 1

Let \(X=(X_{ij})_{i,j\in {\mathcal {N}}}\) satisfy LP, LL, P such that \({\mathcal {D}}(L,X)\) is a minimal {LP, LL, P}-solution to (eL). If \({\mathcal {D}}(L,X)=\emptyset \), AP is trivially satisfied and the lemma is proved with \(X'=X\). Then, assume \({\mathcal {D}}(L,X)\ne \emptyset \). In the remainder of the proof, we use the following notation: \({\mathcal {D}}={\mathcal {D}}(L,X)\) and \(\overline{{\mathcal {D}}}=\overline{{\mathcal {D}}}(L,X)\). Since X satisfies LP,

$$\begin{aligned} \forall i\in {\mathcal {N}},X_{i,{\mathcal {N}}}\le L_{i,{\mathcal {N}}}. \end{aligned}$$
(2)

Moreover, because X satisfies LL and P, we can use \({\mathcal {N}}={\mathcal {D}}\cup \overline{{\mathcal {D}}}\) to obtain:

$$\begin{aligned}&\forall i\in {\mathcal {D}},X_{i,{\mathcal {N}}}-\beta .\sum _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}X_{j,{\mathcal {N}}}\le \alpha .e_{i}+\beta .\sum _{j\in \overline{{\mathcal {D}}}}L_{ji},\nonumber \\&\forall i\in \overline{{\mathcal {D}}},L_{i,{\mathcal {N}}}-\sum _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}X_{j,{\mathcal {N}}}\le e_{i}+\sum _{j\in \overline{{\mathcal {D}}}}L_{ji}. \end{aligned}$$
(3)

To prove the lemma and find the \(X^{\prime }\) we are looking for, we start by characterizing this CPM on the set of defaulting firms. We can easily deduce the CPM for non-defaulting firms, as these firms pay their liabilities in full. We consider the following linear program:

$$\begin{aligned}&T=\arg \max _{(X_{i,{\mathcal {N}}}^{\prime })_{i\in {\mathcal {D}}}} \sum _{i\in {\mathcal {D}}}X_{i,{\mathcal {N}}}^{\prime } \end{aligned}$$
(4)
$$\begin{aligned}&\text {subject to:}\nonumber \\&\quad \left\{ \begin{array}{l} \forall i\in {\mathcal {D}},\ 0\le X_{i,{\mathcal {N}}}^{\prime }\le L_{i,{\mathcal {N}}},\\ \forall i\in {\mathcal {D}},\ X_{i,{\mathcal {N}}}^{\prime }-\beta .\sum \limits _{j\in {\mathcal {D}}} \frac{L_{ji}}{L_{j,{\mathcal {N}}}}X_{j,{\mathcal {N}}}^{\prime } \le \alpha .e_{i}+\beta .\sum \limits _{j\in \overline{{\mathcal {D}}}}L_{ji},\\ \forall i\in \overline{{\mathcal {D}}},{\hat{A}}~\ L_{i,{\mathcal {N}}}-\sum \limits _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j, {\mathcal {N}}}}X_{j,{\mathcal {N}}}^{\prime }\le e_{i}+\sum \limits _{j\in \overline{{\mathcal {D}}}}L_{ji}. \end{array}\right. \end{aligned}$$
(5)

The constraints in Eq. (5) of the program (4)–(5) are linear and define a compact set. Hence, the program (4)–(5) has a non-empty set of solutions. Let \({\overline{X}}=({\overline{X}}_{i,{\mathcal {N}}})_{i\in {\mathcal {D}}}\) be such a solution.

We now extend this definition from \({\mathcal {D}}\) to \({\mathcal {N}}\). Let \(\overline{X^{N}}=(\overline{X_{ij}^{N}})_{i,j\in {\mathcal {N}}}\) be defined by \(\forall i\in {\mathcal {D}}\), \(\overline{X_{ij}^{N}}={\overline{X}}_{i,{\mathcal {N}}}\frac{L_{ij}}{L_{i,{\mathcal {N}}}}\), and \(\forall i\in \overline{{\mathcal {D}}}\), \(\overline{X_{ij}^{N}}=L_{ij}\).

We can check that \(\forall i\in {\mathcal {D}}\), \({\overline{X}}_{i,{\mathcal {N}}}=L_{i,{\mathcal {N}}}\) or \({\overline{X}}_{i,{\mathcal {N}}}-\beta \sum \nolimits _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}{\overline{X}}_{j,{\mathcal {N}}}=\alpha e_{i}+\beta \sum \nolimits _{j\in \overline{{\mathcal {D}}}}L_{ji}\) [otherwise, increasing \({\overline{X}}_{i,{\mathcal {N}}}\) by a quantity small enough to ensure that the constraints of Eq. (5) are still satisfied is possible, contradicting the assumption that \({\overline{X}}\) is a solution to program (4)–(5)]. If \(\exists i\in {\mathcal {D}},{\overline{X}}_{i,{\mathcal {N}}}=L_{i,{\mathcal {N}}}\), then \(\overline{X^{N}}\) is such that \({\mathcal {D}}(L,\overline{X^{N}})\subsetneq {\mathcal {D}}\) and the constraints of Eq. (5) imply that \(\overline{X^{N}}\) satisfies LP, P, and LL.Footnote 8 This contradicts the fact that \({\mathcal {D}}\) is a minimal {LP, LL, P}-solution to (eL). We must therefore have \(\forall i\in {\mathcal {D}}\), \({\overline{X}}_{i,{\mathcal {N}}}-\beta \sum \nolimits _{j\in {\mathcal {D}}}\frac{L_{ji}}{L_{j,{\mathcal {N}}}}{\overline{X}}_{j,{\mathcal {N}}}=\alpha e_{i}+\beta \sum \nolimits _{j\in \overline{{\mathcal {D}}}}L_{ji}\). This implies that \(\overline{X^{N}}\) satisfies AP and the lemma is proved with \(X'=\overline{X^{N}}\). \(\square \)

B.2: Proof of Proposition 2

Let \(0\le \alpha ,\beta \le 1\). Let \(e=(e_{j})_{j\in {\mathcal {N}}}\in {\mathcal {E}}\) and \(L=(L_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\).

  1. (1)

    Let \({\mathcal {D}}\) be a minimal {LP, LL, P, AP}-solution to (eL). Let us show that \({\mathcal {D}}\) is a minimal {LP, LL, P}-solution to (eL).

    1. (a)

      Let us show that \({\mathcal {D}}\) is an {LP, LL, P}-solution to (eL). By definition, a CPM \(X\in {\mathcal {L}}\) exists such that:

      • X satisfies LP, LL, P, AP,

      • \({\mathcal {D}}={\mathcal {D}}(L,X)\).

      Then, by definition, \({\mathcal {D}}\) is an {LP, LL, P}-solution to (eL).

    2. (b)

      Let us show that \({\mathcal {D}}\) is a minimal {LP, LL, P}-solution to (eL). In order to obtain a contradiction, let \({\mathcal {D}}'\subsetneq {\mathcal {D}}\) be an {LP, LL, P}-solution to (eL). The CPM \(X\in {\mathcal {L}}\) then exists such that:

      • X satisfies LP, LL, P,

      • \({\mathcal {D}}'={\mathcal {D}}(L,X)\).

      Then, by Lemma 1, there is a CPM \(X'\in {\mathcal {L}}\) such that:

      • \(X'\) satisfies LP, LL, P, AP,

      • \({\mathcal {D}}'={\mathcal {D}}(L,X')\).

      Then, \({\mathcal {D}}'\) is an {LP, LL, P, AP}-solution to (eL), contradicting the assumption that \({\mathcal {D}}\) is a minimal {LP, LL, P, AP}-solution to (eL).

  2. (2)

    Let \({\mathcal {D}}\) be a minimal {LP, LL, P}-solution to (eL). Let us show that \({\mathcal {D}}\) is a minimal {LP, LL, P, AP}-solution to (eL).

    1. (a)

      Let us show that \({\mathcal {D}}\) is an {LP, LL, P, AP}-solution to (eL). As \({\mathcal {D}}\) is a minimal {LP, LL, P}-solution to (eL), this implies, by definition, that a CPM \(X\in {\mathcal {L}}\) exists such that:

      • X satisfies LP, LL, P,

      • \({\mathcal {D}}={\mathcal {D}}(L,X)\).

      Then, from Lemma 1, there is a CPM \(X'\in {\mathcal {L}}\) satisfying LP, LL, P, AP such that \({\mathcal {D}}(L,X')={\mathcal {D}}(L,X)\). Hence, by definition, \({\mathcal {D}}\) is an {LP, LL, P, AP}-solution to (eL).

    2. (b)

      Let us show that \({\mathcal {D}}\) is a minimal {LP, LL, P, AP}-solution to (eL). Conversely, let \({\mathcal {D}}'\subsetneq {\mathcal {D}}\) be an {LP, LL, P, AP}-solution to (eL). By definition, there is a CPM \(X\in {\mathcal {L}}\) such that:

      • X satisfies LP, LL, P, AP,

      • \({\mathcal {D}}'={\mathcal {D}}(L,X)\).

      Then, by definition, \({\mathcal {D}}'\) is an {LP, LL, P}-solution to (eL), contradicting the fact that \({\mathcal {D}}\) is a minimal {LP, LL, P}-solution to (eL).

Appendix C: Proof of Proposition 3

Let \(0\le \alpha ,\beta \le 1\). Let \(e=(e_{j})_{j\in {\mathcal {N}}}\in {\mathcal {E}}\) and \(L=(L_{i,j})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\).

The proposition that the set of minimal {LL, P, LP}-solutions to (eL) is non-empty is a straightforward corollary of Rogers and Veraart (2013), Proposition 2, and the definition of minimality. Let us prove the uniqueness of the minimal {LL, P, LP}-solution to (eL).

Let us assume, on the contrary, that there are two CPMs, \(X^{1}=(X_{ij}^{1})_{i,j\in {\mathcal {N}}}\) and \(X^{2}=(X_{ij}^{2})_{i,j\in {\mathcal {N}}}\), satisfying LL, P, and LP such that \({{\mathcal {D}}}(L,X^{1})\) and \({{\mathcal {D}}}(L,X^{2})\) are minimal {LL, P, LP}-solutions to (eL) and \({{\mathcal {D}}}(L,X^{1})\ne {{\mathcal {D}}}(L,X^{2})\). Let us define \(D^{0}={{\mathcal {D}}}(L,X^{1})\cap {{\mathcal {D}}}(L,X^{2})\), \(D^{1}={{\mathcal {D}}}(L,X^{1}){\setminus }{{\mathcal {D}}}(L,X^{2})\), \(D^{2}={{\mathcal {D}}}(L,X^{2}){\setminus }{{\mathcal {D}}}(L,X^{1})\), and \({\overline{N}}=N{\setminus }({{\mathcal {D}}}(L,X^{1})\cup {{\mathcal {D}}}(L,X^{2}))\). From the definition of minimality, we must have \(D^{1}\ne \emptyset \) and \(D^{2}\ne \emptyset \).

Let us define the CPM \(X^{M}=(X_{ij}^{M})_{i,j\in {\mathcal {N}}}\) as \(\forall i,j\in {\mathcal {N}},X_{ij}^{M}=\max \left( X_{ij}^{1},X_{ij}^{2}\right) \).

From the definition of defaulting firms and from LP, we have \({{\mathcal {D}}}(L,X^{M})=D^{0}\).

Because \(X^{1}\) and \(X^{2}\) satisfy P, it is straightforward to check that \(X^{M}\) satisfies P. It is also straightforward to check that when \(X^{1}\) and \(X^{2}\) satisfy LP then \(X^{M}\) also satisfies LP.

Let us show that \(X^{M}\) satisfies LL.

  1. 1.

    Let \(i\in D^{0}\). When \(X^{1}\) and \(X^{2}\) satisfy LL, this implies:

    $$\begin{aligned} \beta \left( X_{D^{0},i}^{1}+X_{D^{1},i}^{1}+X_{D^{2},i}^{1} +X_{{\overline{N}},i}^{1}\right) +\alpha e_{i}\ge X_{i,{\mathcal {N}}}^{1}, \end{aligned}$$

    and

    $$\begin{aligned} \beta \left( X_{D^{0},i}^{2}+X_{D^{1},i}^{2}+X_{D^{2},i}^{2} +X_{{\overline{N}},i}^{2}\right) +\alpha e_{i}\ge X_{i,{\mathcal {N}}}^{2}. \end{aligned}$$

    By definition of \(X^{M}\) and P, we have \(X_{D^{0},i}^{M}\ge X_{D^{0},i}^{1}\) and \(X_{D^{0},i}^{M}\ge X_{D^{0},i}^{2}\), \(X_{D^{1},i}^{M}=X_{D^{1},i}^{2}=L_{D^{1},i}>X_{D^{1},i}^{1}\), \(X_{D^{2},i}^{M}=X_{D^{2},i}^{1}=L_{D^{2},i}>X_{D^{2},i}^{2}\), and \(X_{{\overline{N}},i}^{M}=X_{{\overline{N}},i}^{1}=X_{{\overline{N}},i}^{2}=L_{{\overline{N}},i}\). Also, from P, we have \(X_{i,{\mathcal {N}}}^{M}=X_{i,{\mathcal {N}}}^{1}\) or \(X_{i,{\mathcal {N}}}^{M}=X_{i,{\mathcal {N}}}^{2}\). Hence,

    $$\begin{aligned} \beta \left( X_{D^{0},i}^{M}+X_{D^{1},i}^{M}+X_{D^{2},i}^{M} +X_{{\overline{N}},i}^{M}\right) +\alpha e_{i}\ge X_{i,{\mathcal {N}}}^{M}. \end{aligned}$$
  2. 2.

    Let \(i\in D^{1}\). \(X^{2}\) satisfying LL implies:

    $$\begin{aligned} \beta \left( X_{D^{0},i}^{2}+X_{D^{1},i}^{2}+X_{D^{2},i}^{2} +X_{{\overline{N}},i}^{2}\right) +\alpha e_{i}\ge X_{i,{\mathcal {N}}}^{2}. \end{aligned}$$

    By definition of \(X^{M}\) and P, we have \(X_{D^{0},i}^{M}\ge X_{D^{0},i}^{2}\), \(X_{D^{1},i}^{M}=X_{D^{1},i}^{2}=L_{D^{1},i}\), \(X_{D^{2},i}^{M}>X_{D^{2},i}^{2}\), \(X_{{\overline{N}},i}^{M}=X_{{\overline{N}},i}^{2}=L_{{\overline{N}},i}\), and \(X_{i,{\mathcal {N}}}^{2}=X_{i,{\mathcal {N}}}^{M}=L_{i,{\mathcal {N}}}\). Hence,

    $$\begin{aligned} \beta \left( X_{D^{0},i}^{M}+X_{D^{1},i}^{M}+X_{D^{2},i}^{M} +X_{{\overline{N}},i}^{M}\right) +\alpha e_{i}\ge X_{i,{\mathcal {N}}}^{M}. \end{aligned}$$

The proof for \(i\in D^{2}\) and \(i\in {\overline{N}}\) is identical and is not presented here.

We have shown that \(X^{M}\) also satisfies LL.

\({\mathcal {D}}(L,X^{M})\) is then strictly included in \({\mathcal {D}}(L,X^{1})\) and \({\mathcal {D}}(L,X^{2})\) and \(X^{M}\) satisfies LL, P, and LP. There is hence a contradiction with \({\mathcal {D}}(L,X^{1})\ne {\mathcal {D}}(L,X^{2})\) being minimal {LL, P, LP}-solutions to (eL).

Appendix D: Proof of Proposition 4

Let \(\alpha =\beta =1\). Let \(e=(e_{j})_{j\in {\mathcal {N}}}\in {\mathcal {E}}\) and \(L=(L_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\). Let \({\mathcal {D}}^{1}\subseteq {\mathcal {N}}\) be a minimal {LP, LL, P}-solution to (eL). From Proposition 2, \({\mathcal {D}}^{1}\) is a minimal {LP, LL, P, AP}-solution to (eL). Assume, in contradiction with this proposition, that \({\mathcal {D}}^{2}\subsetneq {\mathcal {D}}^{1}\) is an {LL, P}-solution to (eL).

If \({\mathcal {D}}^{2}=\emptyset \), then LP is trivially satisfied and \({\mathcal {D}}^{2}\) is an {LP, LL, P}-solution to (eL), contradicting the assumption that \({\mathcal {D}}^{1}\) is a minimal {LP, LL, P}-solution to (eL). Now let us assume that \({\mathcal {D}}^{2}\ne \emptyset \). Let \(X^{1}=(X_{ij}^{1})_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\) be such that:

  • \(X^{1}\) satisfies LP, LL, P, AP,

  • \({\mathcal {D}}^{1}={\mathcal {D}}(L,X^{1})\).

\(X^{1}\) exists since, by assumption, \({\mathcal {D}}^{1}\) is a minimal {LP, LL, P, AP}-solution to (eL).

Let \({\mathcal {X}}\) be the set of clearing payment matrices \(X\in {\mathcal {L}}\) such that:

  • X satisfies LL, P,

  • \({\mathcal {D}}^{2}={\mathcal {D}}(L,X)\).

If we assume that \({\mathcal {D}}^{2}\) is an {LP, LL, P}-solution to (eL), then \({\mathcal {X}}\ne \emptyset \). For any \(X=(X_{ij})_{i,j\in {\mathcal {N}}}\in {\mathcal {X}}\), let us define \({\mathcal {D}}^{2+}(X)=\{i\in {\mathcal {D}}^{2},X_{i,{\mathcal {N}}}>X_{i,{\mathcal {N}}}^{1}\}\), \({\mathcal {D}}^{2-}(X)=\{i\in {\mathcal {D}}^{2},X_{i,{\mathcal {N}}}<X_{i,{\mathcal {N}}}^{1}\}\), \({\mathcal {D}}^{2=}(X)=\{i\in {\mathcal {D}}^{2},X_{i,{\mathcal {N}}}=X_{i,{\mathcal {N}}}^{1}\}\). Let us have \(X^{2}\in {\mathcal {X}}\) such that \(\forall X\in {\mathcal {X}},\lnot ({\mathcal {D}}^{2-}(X)\subsetneq {\mathcal {D}}^{2-}(X^{2}))\). By definition, \({\mathcal {D}}^{2+}(X^{2})\cup {\mathcal {D}}^{2-}(X^{2})\cup {\mathcal {D}}^{2=}(X^{2})={\mathcal {D}}^{2}\). Moreover, if \({\mathcal {D}}^{2+}(X^{2})=\emptyset \), \(X^{2}\) obviously satisfies LP which, together with the assumption that \(X^{2}\) satisfies P and LL, contradicts the assumption that \({\mathcal {D}}^{1}\) is a minimal {LP, LL, P}-solution to (eL). Hence,

$$\begin{aligned} {\mathcal {D}}^{2+}(X^{2})\ne \emptyset . \end{aligned}$$
(6)
  1. (1)

    With a proof similar to that of Lemma 1, we can show, with no loss of generality, that \(X^{2}\) is such that \(\forall i\in {\mathcal {D}}^{2-}(X^{2})\), \(X_{{\mathcal {N}},i}^{2}+e_{i}=X_{i,{\mathcal {N}}}^{2}\).

  2. (2)

    Then, summing over \({\mathcal {D}}^{2-}(X^{2})\), we obtain:

    $$\begin{aligned}&L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2-}(X^{2})} +L_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},{\mathcal {D}}^{2-}(X^{2})}\nonumber \\&\quad +X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{2} +X_{{\mathcal {D}}^{2+}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{2} +e_{{\mathcal {D}}^{2-}(X^{2})}\nonumber \\&\quad -X_{{\mathcal {D}}^{2-}(X^{2}),N {\setminus }{\mathcal {D}}^{2-}(X^{2})}^{2}=0. \end{aligned}$$
    (7)

Moreover, since \(X^{1}\) satisfies AP and \({\mathcal {D}}^{2-}(X^{2})\subseteq {\mathcal {D}}^{2}\subsetneq {\mathcal {D}}^{1}\),

$$\begin{aligned}&L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2-}(X^{2})} +X_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},{\mathcal {D}}^{2-}(X^{2})}^{1}\nonumber \\&\quad +X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{1} +X_{{\mathcal {D}}^{2+}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{1} +e_{{\mathcal {D}}^{2-}(X^{2})}\nonumber \\&\quad -X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{2-}(X^{2})}^{1}=0. \end{aligned}$$
(8)

Since \(X^{1}\) and \(X^{2}\) satisfy P, by definition of \({\mathcal {D}}^{2=}(X^{2})\) and \({\mathcal {D}}^{2+}(X^{2})\), \(X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{2}=X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{1}\) and \(X_{{\mathcal {D}}^{2+}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{2}\ge X_{{\mathcal {D}}^{2+}(X^{2}),{\mathcal {D}}^{2-}(X^{2})}^{1}\). Moreover, since \(X^{1}\) satisfies LP, \(L_{{\mathcal {D}}_{1}{\setminus }{\mathcal {D}}_{2},{\mathcal {D}}^{2-}(X^{2})}\ge X_{{\mathcal {D}}_{1}{\setminus }{\mathcal {D}}_{2},{\mathcal {D}}^{2-}(X^{2})}^{1}\). This then means that \(X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{2-}(X^{2})}^{2}\ge X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{2-}(X^{2})}^{1}\). Together with the definition of \({\mathcal {D}}^{2-}(X^{2})\), we must have \(X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{2-}(X^{2})}^{2}=X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{2-}(X^{2})}^{1}=0\). Hence,

$$\begin{aligned} \forall i\in {\mathcal {D}}^{2-}(X^{2}),\forall j\in {\mathcal {N}}{\setminus }{\mathcal {D}}^{2-}(X^{2}),X_{ij}^{1}=X_{ij}^{2}=0. \end{aligned}$$
(9)

From Eqs. (7) and (8), we have:

$$\begin{aligned} \forall i\in {\mathcal {N}}{\setminus }{\mathcal {D}}^{2-}(X^{2}),\forall j\in {\mathcal {D}}^{2-}(X^{2}),X_{ij}^{1}=X_{ij}^{2}=0. \end{aligned}$$
(10)
  1. (3)

    Since \(X^{1}\) satisfies AP,

$$\begin{aligned}&X_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus } {\mathcal {D}}^{1}}^{1}+X_{{\mathcal {D}}^{2-}(X^{2}),N {\setminus }{\mathcal {D}}^{1}}^{1}+X_{{\mathcal {D}}^{2=}(X^{2}), N{\setminus }{\mathcal {D}}^{1}}^{1}\\&\quad +X_{{\mathcal {D}}^{2+}(X^{2}), N{\setminus }{\mathcal {D}}^{1}}^{1}-L_{N{\setminus }{\mathcal {D}}^{1}, {\mathcal {D}}^{1}}=e_{N}. \end{aligned}$$

Since \(X^{2}\) satisfies LL,

$$\begin{aligned}&L_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus } {\mathcal {D}}^{1}}+X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus } {\mathcal {D}}^{1}}^{2}+X_{{\mathcal {D}}^{2=}(X^{2}),N{\setminus } {\mathcal {D}}^{1}}^{2}\\&\quad +X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus } {\mathcal {D}}^{1}}^{2}-L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{1}}\le e_{N}. \end{aligned}$$

By definition and the fact that \(X^{1}\) and \(X^{2}\) satisfy P, \(X_{{\mathcal {D}}^{2=}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{1}=X_{{\mathcal {D}}^{2=}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{2}\), \(X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{2}\ge X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{1}\) and since \(X^{1}\) satisfies LP, \(L_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus }{\mathcal {D}}^{1}}\ge X_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus }{\mathcal {D}}^{1}}^{1}\). Moreover, from Eq. (9), \(X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{2}=X_{{\mathcal {D}}^{2-}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{1}=0\). Then,

$$\begin{aligned}&L_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus } {\mathcal {D}}^{1}}+X_{{\mathcal {D}}^{2=}(X^{2}),N{\setminus } {\mathcal {D}}^{1}}^{2}+X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus } {\mathcal {D}}^{1}}^{2}-L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{1}} =e_{N}, \end{aligned}$$
(11)
$$\begin{aligned}&L_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus }{\mathcal {D}}^{1}} =X_{{\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},N{\setminus } {\mathcal {D}}^{1}}^{1}, \end{aligned}$$
(12)
$$\begin{aligned}&X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{2} =X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus }{\mathcal {D}}^{1}}^{1}. \end{aligned}$$
(13)

Because \(X^{1}\) satisfies LP and P and because \(X^{2}\) satisfies P, by definition of \({\mathcal {D}}^{1}\), Eq. (12) implies:

$$\begin{aligned} \forall i\in {\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}, \forall j\in {\mathcal {N}}{\setminus }{\mathcal {D}}^{1},L_{ij}=X_{ij}^{2} =X_{ij}^{1}=0. \end{aligned}$$
(14)

Also, by definition of \({\mathcal {D}}^{2+}(X^{2})\), Eq. (13) implies:

$$\begin{aligned} \forall i\in {\mathcal {D}}^{2+}(X^{2}),\forall j\in {\mathcal {N}} {\setminus }{\mathcal {D}}^{1},X_{ij}^{2}=X_{ij}^{1}=0. \end{aligned}$$
(15)

Moreover, since \(X^{2}\) satisfies LL, Eq. (11) implies:

$$\begin{aligned}&X_{N{\setminus }{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2=}(X^{2})}^{2} +e_{{\mathcal {D}}^{2=}(X^{2})}-X_{{\mathcal {D}}^{2=}(X^{2}),N {\setminus }{\mathcal {D}}^{2=}(X^{2})}^{2} =0,\\&X_{N{\setminus }{\mathcal {D}}^{2+}(X^{2}),{\mathcal {D}}^{2+}(X^{2})}^{2} +e_{{\mathcal {D}}^{2+}(X^{2})}-X_{{\mathcal {D}}^{2+}(X^{2}),N{\setminus } {\mathcal {D}}^{2+}(X^{2})}^{2} =0,\\&X_{N{\setminus }({\mathcal {D}}_{1}{\setminus }{\mathcal {D}}_{2}),{\mathcal {D}}_{1} {\setminus }{\mathcal {D}}_{2}}^{2}+e_{{\mathcal {D}}_{1}{\setminus }{\mathcal {D}}_{2}} -X_{{\mathcal {D}}_{1}{\setminus }{\mathcal {D}}_{2},N{\setminus }({\mathcal {D}}_{1} {\setminus }{\mathcal {D}}_{2})}^{2} =0. \end{aligned}$$
  1. (4)

    We showed above that:

    $$\begin{aligned} X_{N{\setminus }{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2=}(X^{2})}^{2} +e_{{\mathcal {D}}^{2=}(X^{2})}-X_{{\mathcal {D}}^{2=}(X^{2}),N{\setminus } {\mathcal {D}}^{2=}(X^{2})}^{2}=0, \end{aligned}$$

    and by \(X^{1}\) satisfying AP, we have:

    $$\begin{aligned} X_{N{\setminus }{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2=}(X^{2})}^{1} +e_{{\mathcal {D}}^{2=}(X^{2})}-X_{{\mathcal {D}}^{2=}(X^{2}),N {\setminus }{\mathcal {D}}^{2=}(X^{2})}^{1}=0. \end{aligned}$$

    The same reasoning as above shows that:

    $$\begin{aligned} \forall i\in {\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2},\forall j \in {\mathcal {D}}^{2=}(X^{2}),L_{ij}=X_{ij}^{1}=X_{ij}^{2}=0, \end{aligned}$$
    (16)

    and

    $$\begin{aligned} \forall i\in {\mathcal {D}}^{2+}(X^{2}),\forall j\in {\mathcal {D}}^{2=} (X^{2}),X_{ij}^{2}=X_{ij}^{1}=0. \end{aligned}$$
    (17)
  2. (5)

    Let us consider \({\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})\). Because \(P^{1}\) satisfies AP,

    $$\begin{aligned}&L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}+X_{{\mathcal {D}}^{2-} (X^{2}),{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2})}^{1}\\&\quad +X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1} {\setminus }{\mathcal {D}}^{2})}^{1}\\&\quad -X_{{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}),N{\setminus }({\mathcal {D}}^{2+} (X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}))}^{1} +e_{{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}=0. \end{aligned}$$

    After simplification using Eqs. (9), (10), (14), (15), (16) and (17):

    $$\begin{aligned} L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}+X_{{\mathcal {D}}^{2=} (X^{2}),{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2})}^{1}+e_{{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}=0. \end{aligned}$$

    This implies:

    $$\begin{aligned} e_{{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})} =0. \end{aligned}$$
    (18)

    and

    $$\begin{aligned} L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}+X_{{\mathcal {D}}^{2=} (X^{2}),{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2})}^{1}=0. \end{aligned}$$
    (19)

The same reasoning with \(X^{2}\) gives:

$$\begin{aligned} L_{N{\setminus }{\mathcal {D}}^{1},{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1} {\setminus }{\mathcal {D}}^{2})}+X_{{\mathcal {D}}^{2=}(X^{2}),{\mathcal {D}}^{2+} (X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}^{2}=0. \end{aligned}$$
(20)

Equations (19) and (20) imply

$$\begin{aligned} \forall i\in {\mathcal {N}}{\setminus }{\mathcal {D}}^{1},\forall j\in {\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2}),L_{ij}=X_{ij}^{1}=X_{ij}^{2}=0, \end{aligned}$$
(21)

an

$$\begin{aligned} \forall i\in {\mathcal {D}}^{2=}(X^{2}),\forall j\in {\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2}),X_{ij}^{1}=X_{ij}^{2}=0. \end{aligned}$$
(22)

Since \(X^{1}\) and \(X^{2}\) satisfy LL,

$$\begin{aligned}&\forall i\in {\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}),\nonumber \\&\quad X_{{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2}),i}^{1}-X_{i,{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}^{1}\nonumber \\&\quad =X_{{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus } {\mathcal {D}}^{2}),i}^{2}-X_{i,{\mathcal {D}}^{2+}(X^{2}) \cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}^{2}=0. \end{aligned}$$
(23)

6) From Eq. (6), \({\mathcal {D}}^{2+}(X^{2})\ne \emptyset \) and by definition of \({\mathcal {D}}^{2+}(X^{2})\), \(\forall k\in {\mathcal {D}}^{2+}(X^{2})\), \(X_{k,{\mathcal {N}}}^{2}>X_{k,{\mathcal {N}}}^{1}\ge 0\). Moreover, since \(X^{2}\) satisfies P, \(\forall k\in {\mathcal {D}}^{2+}(X^{2})\), \(L_{k,{\mathcal {N}}}>0\). We can then consider \(k_\mathrm{max}\in {\mathcal {D}}^{2+}(X^{2})\) such that \(\forall k\in {\mathcal {D}}^{2+}(X^{2}),\frac{X_{k_\mathrm{max},N}^{2}}{L_{k_\mathrm{max},N}}\ge \frac{X_{k,{\mathcal {N}}}^{2}}{L_{k,{\mathcal {N}}}}\).

Assume that \(\frac{X_{k_\mathrm{max},N}^{2}}{L_{k_\mathrm{max},N}}\le 1\). This implies that \(X^{2}\) satisfies LP, contradicting the fact that \({\mathcal {D}}^{1}\) is a minimal {LP, P, LL}-solution. Then, consider \(\frac{X_{k_\mathrm{max},N}^{2}}{L_{k_\mathrm{max},N}}>1\).

Now, let us define \(X'=(X_{ij}^{\prime })_{i,j\in {\mathcal {N}}}\in {\mathcal {L}}\) as:

$$\begin{aligned} \forall i,j\in {\mathcal {N}},X_{ij}^{\prime }=\left\{ \begin{array}{ll} X_{ij}^{2}\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}, &{} \quad \text {if}\; i\in {\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}),\\ X_{ij}^{2}, &{} \quad \text {otherwise.} \end{array}\right. \end{aligned}$$

It is straightforward to check that \(X'\) satisfies P.

Let us show that \(X'\) satisfies LL. a) Let \(i\in {\mathcal {N}}{\setminus }({\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}))\). Let us define \(\pi (i)=X_{{\mathcal {N}},i}^{\prime }+e_{i}-X_{i,{\mathcal {N}}}^{\prime }\). By definition of \(X'\), \(\pi (i)=X_{{\mathcal {N}},i}^{\prime }+e_{i}-X_{i,{\mathcal {N}}}^{2}\). From Eqs. (10), (14), (15), (16), and (17), \(\pi (i)=X_{{\mathcal {N}},i}^{2}+e_{i}-X_{i,{\mathcal {N}}}^{2}\). LL is then satisfied by i for \(X'\) as it is for \(X^{2}\) by assumption. b) Let \(i\in {\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})\). Let us define \(\pi (i)=X_{{\mathcal {N}},i}^{\prime }+e_{i}-X_{i,{\mathcal {N}}}^{\prime }\). After simplification using Eqs. (9), (10), (14), (15), (16), (17), (18), (21), and (22), \(\pi (i)=X_{{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}),i}^{\prime }-X_{i,{\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2})}^{\prime }\). By definition of \(X'\): \(\pi (i)=\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}X_{{\mathcal {N}},i}^{2}-X_{i,{\mathcal {N}}}^{2}\). LL is then satisfied by i for \(X'\) as it is for \(X^{2}\) by assumption.

Let us show that \(X'\) satisfies LP. (a) The proof is straightforward for all \(i\in {\mathcal {N}}{\setminus }({\mathcal {D}}^{2+}(X^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}))\). (b) Let \(i\in {\mathcal {D}}^{2+}(X^{2})\). \(X_{i,{\mathcal {N}}}^{\prime }=X_{i,{\mathcal {N}}}^{2}\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}\). Then, by definition of \(k_\mathrm{max}\), \(X_{i,{\mathcal {N}}}^{\prime }\le X_{i,{\mathcal {N}}}^{2}\frac{L_{i,{\mathcal {N}}}}{X_{i,{\mathcal {N}}}^{2}}=L_{i,{\mathcal {N}}}\). LP is then satisfied by i. (c) Let \(i\in {\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}\). \(X_{i,{\mathcal {N}}}^{\prime }=X_{i,{\mathcal {N}}}^{2}\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}=L_{i,{\mathcal {N}}}\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}\). Since, \(\frac{L_{k_\mathrm{max},N}}{X_{k_\mathrm{max},N}^{2}}<1\), \(X_{i,{\mathcal {N}}}^{\prime }<L_{i,{\mathcal {N}}}\). LP is then satisfied by i.

It is straightforward to check that since \(({\mathcal {N}}{\setminus }{\mathcal {D}}^{1})\cap ({\mathcal {D}}^{2+}(P^{2})\cup ({\mathcal {D}}^{1}{\setminus }{\mathcal {D}}^{2}))=\emptyset \), \({\mathcal {D}}(L,X')\subseteq {\mathcal {D}}(L,X^{1})\). It is also straightforward to check that \(k_\mathrm{max}\notin {\mathcal {D}}(L,X')\), whereas by definition, \(k_\mathrm{max}\in {\mathcal {D}}^{2+}(X^{2})\subseteq {\mathcal {D}}(L,X^{1})\). Hence, \({\mathcal {D}}(L,X')\subsetneq {\mathcal {D}}(L,X^{1})\), which contradicts the fact that \({\mathcal {D}}^{1}\) is a minimal {LP, P, LL}-solution. This completes the proof.

Appendix E: Proof of Proposition 6

Let \({\mathcal {N}}=\{1,2,3,4\}\), \(e=(1,2,3,4)\), and \(L=\left( \begin{array}{cccc} 0 &{} 1 &{} 2 &{} 1\\ 1 &{} 0 &{} 1 &{} 2\\ 1 &{} 1 &{} 0 &{} 1\\ 0 &{} 0 &{} 2 &{} 0 \end{array}\right) \).

We have \({\mathcal {D}}_{0}(e,L)=\{1\}\).

Let us define the CPM \(X_{1}=\left( \begin{array}{cccc} 0 &{} 6/5 &{} 12/5 &{} 6/5\\ 1 &{} 0 &{} 1 &{} 2\\ 14/5 &{} 14/5 &{} 0 &{} 14/5\\ 0 &{} 0 &{} 2 &{} 0 \end{array}\right) \). \(X_{1}\) satisfies LL, P, and I and \({\mathcal {D}}(L,X_{1})=\{1,3\}\) is a minimal {LL, P, I}-solution to (eL).

Let us define the CPM \(X_{2}=\left( \begin{array}{cccc} 0 &{} 1 &{} 2 &{} 1\\ 1 &{} 0 &{} 1 &{} 2\\ 8/3 &{} 8/3 &{} 0 &{} 8/3\\ 0 &{} 0 &{} 2 &{} 0 \end{array}\right) \). \(X_{2}\) satisfies LL and P and \({\mathcal {D}}(L,X_{2})=\{3\}\) is a minimal {LL, P}-solution to (eL).

Let us define the CPM \(X_{3}=\left( \begin{array}{cccc} 0 &{} 11/15 &{} 22/15 &{} 11/15\\ 14/15 &{} 0 &{} 14/15 &{} 28/15\\ 1 &{} 1 &{} 0 &{} 1\\ 0 &{} 0 &{} 2 &{} 0 \end{array}\right) \). \(X_{3}\) satisfies LL, P, and LP and \({\mathcal {D}}(L,X_{3})=\{1,2\}\) is a minimal {LL, P, LP}-solution to (eL).

\({\mathcal {D}}(L,X_{1})\) is then a minimal {LL, P, I}-solution to (eL) but is not a minimal {LL, P, LP}-solution to (eL), proving Proposition 6-1a. \({\mathcal {D}}(L,X_{1})\) is a minimal {LL, P, I}-solution to (eL) but is not a minimal {LL, P}-solution to (eL) proving Proposition 6-1b. \({\mathcal {D}}(L,X_{2})\) is a minimal {LL, P}-solution to (eL) but is not a minimal {LL, P, I}-solution to (eL), proving Proposition 6-2.

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Houy, N., Jouneau, F. & Le Grand, F. Defaulting firms and systemic risks in financial networks: a normative approach. Econ Theory 70, 503–526 (2020). https://doi.org/10.1007/s00199-019-01217-4

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Keywords

  • Credit risk
  • Systemic risk
  • Clearing system
  • Financial system

JEL Classification

  • G21
  • G32
  • G33