On the resolution of cross-liabilities

Abstract

In a variety of systems, in particular in a financial system, entities hold liabilities on each other. The reimbursement abilities are intertwined, thereby potentially generating coordination failures and cascades of defaults calling for orderly resolution. With a single indebted firm, a bankruptcy law organizes such an orderly resolution. With cross-liabilities, a resolution rule should be defined at the system level to account for all those affected, directly or indirectly. This paper investigates such rules assuming their primary goal is to avoid defaults on creditors external to the system, say banks’ defaults on customers’ deposits. I define and characterize the constrained-proportional rule building on two approaches: the minimization of an inequality measure on the reimbursements (made and received) and the axiomatization through desirable properties.

Proofs

Proof of Property 1

Letting $A=\{ i, W_i>0 \}$, the requirements imply: For each i in A, $b_{iN}\ge \ell _{iN}$ and $b_{Ni }\le \ell _{Ni}$. Summing over A implies: $b_{A,N }-b_{N, A } \ge \ell _{A, N} - \ell _{N,A}$. Applying the second identity in (2) to both ${\varvec{b}}$ and ${\varvec{ \ell }}$ yields; $b_{N,N-A }- b_{ N-A, N } \ge \ell _{N,N-A }- \ell _{ N-A, N}$. By contradiction, let $N- A= \{ i, W_i=0 \}$ be nonempty. Hence $W_{N-A} = z_{N-A} + b_{N,N-A }- b_{ N-A, N } \ge z_{N-A} +\ell _{N,N-A }- \ell _{ N-A, N }$. If all firms are strictly solvent, the right hand side is positive, hence $W_{N-A}>0$, a contradiction. This proves $A=N$. Hence each firm has a positive worth so that the total reimbursed is at least equal to the total liabilities $\ell _{N, N}$ and the total received is at most the total claims, $\ell _{N, N}$, which implies that each reimburse exactly its liability: the solution is the exact one and no firm defaults. $\square$

Proof of Proposition 1

We have proved in the text that conditions (3) were necessary for the existence of a solution. Let us prove the converse. consider a directed graph composed of ${{\mathcal {G}}}$ and additional edges from a node, denoted 0, which ’sends’ $z_i$ and ’receives’ i’s worth to each i. The existence of a solution is equivalent to the existence of a circulation in that graph with well defined lower and upper capacities. Formally, the graph has node set $V=\{0\} \cup N=\{0, \cdots , n\}$, the edges are those in ${{\mathcal {G}}}$ plus (0, i), (i, 0) for each i in N with lower and upper capacities, denoted by d(e) and u(e) for edge e, are defined by:

for (0, i) $i\in N$: $d(e)= z_i$, $u(e) =z_{i}$

for (i, 0), $i\in N$: $d(e)= 0$, $u(e) =\infty$

for (i, j) in ${{\mathcal {G}}}$: $d(e)= 0$ and $u(e) =\infty$.

Solutions and circulations satisfying the capacity constraints are in a one-to-one relationship by setting $b_{ij}$ equal to the flow on edge (i, j), $z_i$ to the flow on edge (0, i), the flow on (i, 0) is equal to i’s worth. We can thus use Hoffmann theorem, which provides the following characterization of the graphs for which such a circulation exists. Given a nonempty subset X of the node set V, define the upper capacity of its outgoing edges by $u_{-}(X)= \sum _{e=(i,j), i\in X, j\notin X} u(e)$ and the lower capacity of its ingoing edges by $d_{+}(X)= \sum _{e=(i,j), i\notin X, j \in X} d(e)$. Hoffmann theorem states: either there exists no circulation satisfying the capacity constraints or

$$\begin{aligned} u_{-}(X)\ge d_{+}(X) \hbox { for any subset }X. \end{aligned}$$

(14)

Given X nonempty subset of V, let us compute $u_{-}(X)$ and $d_{+}(X)$. Denote $I= N\cap X$. Consider two cases.

Case 1: $0\notin X$. Surely I is non-empty (since $X\ne \emptyset$). Since the edge (i, 0) for $i\in I$ is outgoing from X and has an infinite upper capacity, $u_{-}(X) =\infty$: (14) is satisfied.

Case 2: $0\in X$. The outgoing edges from X to $V-X$ are either (0, j) for $j\notin I$, each with an upper-capacity equal to $z_j$, or (i, j) in ${{\mathcal {G}}}$ with $i\in I$ and $j \notin I$ and an infinite upper-capacity. The total upper-capacity is thus finite only if no debtor of $j \in N-I$ belongs to I, i.e. if $D({N-I} )\subset {N-I}$, and in that case: $u_{-}(X)=z_{N-I}$. The lower capacity of the ingoing edges are null so that $d_{+}(X)=0$. Hence a circulation exists iff $z_{N-I} \ge 0$ for each subset of N such that $D({N-I} )\subset {N-I}$. This proves (3), denoting $N-I$ by A.

Consider a strictly feasible problem. Choose positive $\epsilon$ small enough so that $z_A \ge n \epsilon$ for each subset A such that $D(A)\subset A$. Modify the lower bounds of the extended graph used in the above proof by setting $d(e)= \epsilon$ for each edge (i, j) in ${{\mathcal {G}}}$. Inequalities (14) are satisfied so that a solution with $b_{ij} \ge \epsilon$ for (i, j) in ${{\mathcal {G}}}$ exists: this proves the existence of positive solutions. Since the aggregate worth at a solution is $z_N$, which is positive, at least one firm has a positive worth.

Let us prove that, conversely, a non-decomposable problem ${\varvec{ \pi }}$ admits a positive solution where at least one firm has a positive worth only if it is strictly feasible. At a solution, the aggregate worth, which is equal to $z_N$, is positive: (3) holds strictly for $A=N$. Let A be a strict subset for which $D(A)\subset A$. From (2), $W_A= z_A-b_{A,N-A}$: Since all debtors of A are in A, the non-decomposability of ${\varvec{ \pi }}$ implies that some firms in A are indebted to firms in $N-A$. Hence, at a positive solution, $b_{A,N-A}>0$ and $W_A< z_A$: $z_A > 0$ must hold at a solution. This proves that ${\varvec{ \pi }}$ is strictly feasible. $\square$

Proof of Proposition 2

Let us first prove the existence and uniqueness of a solution satisfying (4). Consider function $\phi$ defined by $\phi (\delta )=\sum _{j \in T}\max (\delta \ell _{j}, -z_j)$. A cp-solution is associated to $\delta$ in [0, 1] that satisfies: either (a) $\delta <1$ and $\phi (\delta )=z_1$ or (b) $\delta =1$ and $\phi (1)\le z_1$. In case (a), 1’s worth is null and in case (b) each firm receives the maximum compatible with creditor’s priority and minimal rescue: $b_j= -z_j$ if j is insolvent and $b_j=\ell _j$ if j is solvent.

Existence. One computes $\phi (0)= - \sum _{\{j, z_j\ \le 0 \} } z_j$, $\phi (1)= \sum _{\{ j, z_j + \ell _{j} > 0 \} } \ell _{j} - \sum _{\{j, z_j + \ell _{j} \le 0\} } z_j$. By the strict feasibility of problem ${\varvec{ \pi }}$, $\phi (0) <z_1$. Consider two cases. Let $\phi (1) >z_1$. By continuity of $\phi$, there is $\delta <1$ such that $\phi (\delta )=z_1$: case (a) is satisfied. Let $\phi (1) \le z_1$. Case (b) is satisfied for $\delta =1$. This proves the existence of a cp-solution.

The uniqueness of the solution follows from the monotony of function $\phi$. Write $\phi$ as:

$$\begin{aligned} \phi (\delta )= \delta \sum _{\{ j, z_j +\delta \ell _{j} > 0 \} } \ell _{j} - \sum _{\{j, z_j + \delta \ell _{j} \le 0\} } z_j. \end{aligned}$$

$\phi$ is non-decreasing, which implies that case (a) and case (b) cannot hold simultaneously. If $\phi (1)< z_1$, the solution is surely obtained for $\delta =1$ hence is unique. Assume $\phi (1)\ge z_1$. $\phi$ is strictly increasing over values of $\delta$ such that $z_j +\delta \ell _{j} > 0$ for at least one j. This is the case for any positive $\delta$ if one firm has a non-negative external value (since $\ell _{j} > 0$): a unique $\delta$ solves $\phi (\delta )=z_1$. If all external values are negative, define $\underline{\delta }$ the maximum value such that $z_j +\delta \ell _{j} \le 0$ for each j in T. $\phi$ is constant over $[0, \underline{\delta }]$ assuming the value $\phi (0)$, which is strictly less than $z_1$. Therefore, a solution to $\phi (\delta )=z_1$ can only belong to $[ \underline{\delta }, 1 ]$ over which $\phi$ is strictly increasing: this proves its uniqueness.

Let F be invariant to the scale of the liabilities of a constrained debtor and assign solutions satisfying creditors’ priority and minimal rescue. Let us prove that ${\varvec{b}}=F({\varvec{ \pi }})$ is the cp-solution to ${\varvec{ \pi }}$. Let K be the set of solvent firms. We know from the text that minimal rescue determines the payments to the insolvent firms: $b_j=-z_j$ for j not in K and bounds those to the solvent ones: $b_k \le \ell _k$ for k in K. An upper-bound on 1’s total reimbursements is thus $-z_{T-K} +\ell _K$, which leads to consider the following two cases.

Case 1. $z_1 + z_{T-K} -\ell _K>0$. Then 1’s worth is surely positive, hence creditors’ priority implies $b_k \ge \ell _k$. Solvent firms thus receive exactly their claim: ${\varvec{b}}=F({\varvec{ \pi }})$ coincides with the cp-solution supported by $\delta =1$ and $W_1= z_1 + z_{T-K} -\ell _K$.

Case 2. $z_1 + z_{T-K} -\ell _K <0$. 1 cannot reimburse all the claims of the solvent firms; hence creditors’ priority implies $W_1=0$. Given $\lambda$, $\lambda \le 1$, a scale for 1’s liabilities, let ${\varvec{b}}(\lambda )$ the solution assigned by F in the problem ${\varvec{ \pi }}(\lambda )=({{\varvec{z}}}, (\lambda \ell _j)_{j\in T} )$ and denote ${{\varvec{W}}}(\lambda )$ the worth levels. Let $\Lambda =\{ \lambda , \,W_1(\lambda )=0 \}$ be the set of scales for which 1’s worth is null. $\Lambda$ is non-empty as it contains 1. Invariance to the scale of the liabilities of a constrained debtor implies ${\varvec{b}}(\lambda )={\varvec{b}}$ for $\lambda$ in $\Lambda$. Define $K(\lambda )= \{k, z_k +\lambda \ell _{k} \ge 0\}$ the set of solvent firms in problem ${\varvec{ \pi }}(\lambda )$

Strict feasibility, $z_1 > -z_{T-K}$, implies $z_1 > -z_{T-K}+ \sum _{k\in K}\max ( -z_k, \lambda \ell _{k})$ for $\lambda$ small enough. For such $\lambda$, the problem $({{\varvec{z}}}, (\lambda \ell _j)_{j\in T} )$ satisfies the condition of Case 1: F assigns $b_j(\lambda )= \max (\lambda \ell _{j}, -z_j)$, which is the cp-solution to ${\varvec{ \pi }}$ supported by $\delta =\lambda$ and 1’s worth is positive so that $\lambda$ does not belong to $\Lambda$. Since 1 belongs to $\Lambda$, there is $\lambda ^*$ adherent to both $\Lambda$ and its complement in [0, 1]. The continuity of F at $\lambda ^*$ implies ${\varvec{b}}(\lambda ^*)= {\varvec{b}}$ by right-continuity and that ${\varvec{b}}(\lambda ^*)$ is the cp-solution associated to $\delta =\lambda ^*$ by left-continuity. This proves that ${\varvec{b}}$ is the cp-solution supported by $\delta =\lambda ^*$. $\square$

Proof of Proposition 3

Consider a problem ${\varvec{ \pi }}$ in ${{\mathcal {F}}}$. The objective function f of ${{\mathcal {P}}}$ is separable and strictly convex. Furthermore, the set of positive solutions of ${\varvec{ \pi }}$ has a non-empty interior. It follows that the solution to ${{\mathcal {P}}}$ is unique and characterized by the first-order conditions on the Lagrangian. Denoting by $\alpha _i$ the Kuhn–Tucker multiplier to i’s worth constraint (1), the Lagrangian writes:

$$\begin{aligned} {{\mathcal {L}}}({\varvec{b}}_{{{\mathcal {G}}}})=f({\varvec{b}}_{ {{\mathcal {G}}}}) + \sum _{i \in N } \alpha _i ( b_{iN} -b_{Ni} - z_{i} ). \end{aligned}$$

The first-order conditions with respect to $b_{ij}$ for $(i,j ) \in {{\mathcal {G}}}$ and the complementarity conditions are

$$\begin{aligned} \hbox { for each } (i,j ) \in {{\mathcal {G}}}&:&\frac{ \partial {{{\mathcal {L}}}} }{\partial {b_{ij} }}= \hbox {log}\frac{ b_{ij}}{\ell _{ij}} +\alpha _i - \alpha _j = 0 \end{aligned}$$

(15)

$$\begin{aligned} \hbox { for each } i&:&\alpha _i \ge 0 \hbox { and } \alpha _i ( b_{iN} -b_{Ni} - z_{i} )= 0. \end{aligned}$$

(16)

Taking exponential, (15) is equivalent to

$$\begin{aligned} \hbox {for each } (i,j ) \in {{\mathcal {G}}}:&\, b_{ij}= \delta _i \mu _j \ell _{ij} \end{aligned}$$

(17)

$$\begin{aligned} \hbox {where for each } i:&\delta _i= \exp (-\alpha _i) \hbox { and } \mu _i= \exp \alpha _i. \end{aligned}$$

(18)

Conditions (17) imply bi-proportionality (6) and (18) imply (7), (8) and (9). Conversely, a solution ${\varvec{b}}$ satisfying (6) to (9) solves ${{\mathcal {P}}}$: for such ${\varvec{b}}$, defining $\alpha _i=log (\mu _i)$ one checks that all the sufficient conditions for optimality are satisfied. Since ${{\mathcal {P}}}$ has a unique solution, this proves that a solution satisfying (6) to (9) is unique.

Let us prove the uniqueness of the indices supporting the cp-solution in three steps. In what follows, we say that supporting indices are unique on a subset A of N if the $\mu _j$s are unique on $A\cap C(N)$ and the $\delta _j$s are unique on $A\cap D(N)$.

Step 1: Indices are unique on the set $\{ j \in N,W_j>0\}$, which is non-empty.

This is obvious since, by strict feasibility, there are firms with positive worth (Proposition 1) and their indices are equal to 1 by definition.

Step 2: If indices are unique on A, then they are unique on $A \cup C(A) \cup D(A)$.

This follows from the following points:

(a) for each j in C(A), $\mu _j$ is unique: j is creditor to some i in A so that $\ell _{ij}>0$ where i belongs to $A\cap D(N)$ (since i belongs to D(N)). Hence $\delta _i$ is unique and the equation (6) $b_{ij}= \delta _i \ell _{ij} \mu _j$ determines $\mu _j$.

(b) for each k in D(A), $\delta _k$ is unique: k is debtor to some j in A: $\ell _{kj}>0$ where j belongs to $A\cap C(N)$. Hence $\mu _j$ is unique and the equation $b_{kj}= \delta _k \ell _{kj} \mu _j$ determines $\delta _k$.

(c) It remains to show that $\mu _j$ is unique on $D(A)\cap C(N)$ and $\delta _j$ is unique on $C(A)\cap D(N)$. Let j belong to $D(A)\cap C(N)$. Since j belongs to D(A), $\delta _j$ is unique (by (b)); as j is also a creditor, the balancedness relation (9), $\delta _j\mu _j=1$, determines $\mu _j$. Similarly, for j in $C(A) \cap D(N)$, $\mu _j$ is unique (by (a)); as j is also a debtor, the balancedness relation determines $\delta _j$.

Step 3: Supporting indices are unique. Since N is finite, steps 1 and 2 imply that there is A such that $\{ j \in N,W_j>0\} \subset A$ and $A \cup C(A) \cup D(A)=A$. A contains all its creditors and debtors. If A was not the whole set N, the network would be decomposable. Thus $A =N$, which proves that the indices supporting ${\varvec{b}}_{{\mathcal {G}}}$ are unique. $\square$

Proof of Proposition 4

Let F be invariant to the scale of the claims of a constrained creditor and assign 1-consistent solutions on ${{\mathcal {B}}}$. Given ${\varvec{ \pi }}=({{\varvec{z}}}, {\varvec{ \ell }}_ {S\times T})$ a problem in ${{\mathcal {B}}}$, let ${\varvec{b}}=F({\varvec{ \pi }})$ and ${{\varvec{W}}}=(W_i)$ the achieved worth levels. To show that ${\varvec{b}}$ is the cp-solution to ${\varvec{ \pi }}$, let us first consider problems where matrix ${\varvec{ \ell }}_ {S\times T}$ has all its elements positive.

1- ${\varvec{ \ell }}_ {S\times T}$ has all its elements positive.

Given i in S, let us apply 1-consistency applied to i. i’s reimbursements, $( b_{ij})_{j \in T}$ is the cp-solution in the reduced problem and the achieved worth of each creditor is that of ${\varvec{b}}$, $W_j$. Denote $J=\{ j \in T, W_j=0\}$. Firms in $T-J$ have a positive worth.

Let $J= \emptyset$. Since all worth levels of firms in T are positive, for each i in S, 1-consistency applied to i implies that there is $\delta _i \le 1$ such that

$$\begin{aligned} \hbox { for each } j \in T: b_{ij}= \delta _i \ell _{i j } \hbox { with } \delta _i= 1 \hbox { if } W_i>0. \end{aligned}$$

Thus ${\varvec{b}}$ is the cp-solution supported by the values ${\varvec{\delta }}_S$ and the $\mu _j$ for j in T all equal to 1.

Let $J\ne \emptyset$. Let us consider scales that keep unchanged the claims of firms in $T-J$ and increase the claims of firms in J but keep null their worth levels, i.e. the scales belong to $\Lambda$ defined by

$$\begin{aligned} \Lambda= & {} \{ {\varvec{\lambda }}=(\lambda _j) _{j\in T}, \lambda _j \ge 1, s.t. \hbox { for each } j \in J: \,W_j({\varvec{\lambda }})\\= & {} 0 \hbox { and for each } j \in T-J: \lambda _j= 1 \} \end{aligned}$$

$\Lambda$ is non-empty as it contains the vector of ones. We show that $\Lambda$ admits a maximal element in the following sense: if ${\varvec{\lambda }}$ belongs to $\Lambda$ and $\lambda _j \ge \lambda ^*_j$ for each j in J, then ${\varvec{\lambda }}={\varvec{\lambda }}^*$. ${\varvec{\lambda }}^*$ constitutes rescue indices for ${\varvec{b}}$.

Given ${\varvec{\lambda }}$, denote by ${\varvec{b}}({\varvec{\lambda }})$ the associated solution and by ${{\varvec{W}}}({\varvec{\lambda }})$ the achieved worth levels.

Step 1: $\Lambda$ is bounded and has a maximal element ${\varvec{\lambda }}^*$. $\square$

Proof

Given ${\varvec{\lambda }}\in \Lambda$, consider the problem scaled by ${\varvec{\lambda }}$. Invariance to the scale of the claims of a constrained creditor implies ${\varvec{b}}({\varvec{\lambda }})={\varvec{b}}$. Consider two cases.

(a) $J \ne T$. Both J and $T-J$ are non-empty: there are k with $W_k >0$ and j with $W_j=0$. 1-consistency applied to i in S implies that i reimburses j at least as much as k per unit: $b_{ij}({\varvec{\lambda }})/ (\lambda _j \ell _{ij} )\ge b_{ik} ({\varvec{\lambda }})/ \ell _{ik}$. Since ${\varvec{b}}({\varvec{\lambda }})={\varvec{b}}$ and ${\varvec{b}}$ is positive, it follows that $\lambda _j$ has an upper-bound for each j in J, hence $\Lambda$ is a bounded set.

(b) $J=T$. All firms in T have a null worth, so there is surely i in S with $W_i>0$. Applying 1-consistency to i, $\delta _i=1$ and i’s reimbursements are never smaller than the liabilities: $b_{ij}({\varvec{\lambda }}) /( \lambda _j \ell _{ij} ) \ge 1$ for each j in T. Since ${\varvec{b}}({\varvec{\lambda }})={\varvec{b}}$, it follows that $\lambda _j$ has an upper-bound for each j in T, hence $\Lambda$ is a bounded set. $\square$

$\Lambda$ is closed since F is continuous. As $\Lambda$ is bounded, there is ${\varvec{\lambda }}^*$ in $\Lambda$ that is maximal.

Step 2: Given j in J, let ${\varvec{\lambda }}$ coincide with ${\varvec{\lambda }}^*$ except for j’s component, with $\lambda _j$ strictly larger than $\lambda ^*_j$. Then $W_j({\varvec{\lambda }}) >0$.

Proof

By definition of ${\varvec{\lambda }}^*$, ${\varvec{\lambda }}$ does not belong to $\Lambda$ so that surely ${\varvec{b}}({\varvec{\lambda }}) \ne {\varvec{b}}$. ${\varvec{b}}({\varvec{\lambda }})$ is the solution to the same problem as ${\varvec{b}}({\varvec{\lambda }}^*)$ except that j’s claims are scaled by the factor $\lambda _j /\lambda ^*_j$. We have $W_j({\varvec{\lambda }}^*) =0$. Apply tnvariance to the scale of the claims of a constrained creditor. If $W_j({\varvec{\lambda }})$ was null, then the solution ${\varvec{b}}({\varvec{\lambda }})$ would be equal to ${\varvec{b}}({\varvec{\lambda }}^*)$, hence equal to ${\varvec{b}}$. This proves $W_j({\varvec{\lambda }}) >0$. $\square$

Step 3: Rescue indices equal to ${\varvec{\lambda }}^*$ support the cp-solution.

Proof

Consider j in J and ${\varvec{\lambda }}$ as in step 2 and apply 1-consistency to i in S. For $\lambda _j$ close enough to $\lambda ^*_j$, $W_k ({\varvec{\lambda }}) >0$ for each k in $T-J$ by continuity of F. Furthermore, j’s worth level is positive¹⁵ (by step 2). So there is a unique $\delta _i\ge 1$ such that

$$\begin{aligned} \delta _i=\frac{b_{ij} ({\varvec{\lambda }})}{ \lambda _j \ell _{ij} } \le \frac{b_{ik} ({\varvec{\lambda }})}{ \lambda ^*_k \ell _{ik}} \hbox { for each } k \ne j, k \in J \hbox { and } \delta _i=\frac{b_{ik} ({\varvec{\lambda }})}{\ell _{ik}} \hbox { for each } k \in T-J \end{aligned}$$

where $\delta _i=1$ if $W_i({\varvec{\lambda }}) >0$. When $\lambda _j$ tends to $\lambda ^*_j$, ${\varvec{b}}({\varvec{\lambda }})$ tends to ${\varvec{b}}({\varvec{\lambda }}^*)={\varvec{b}}$, hence the $\delta _i$s converge to some $\delta _i^*\le 1$ such that

$$\begin{aligned} \delta _i^*=\frac{b_{ij} ({\varvec{\lambda }}^*)}{ \lambda _j ^*\ell _{ij} } \le \frac{b_{ik} ({\varvec{\lambda }}^*)}{ \lambda ^*_k \ell _{ik}} \hbox { for each } k \in J \hbox { and } \delta _i^*=\frac{b_{ik} ({\varvec{\lambda }}^*)}{\ell _{ik}} \hbox { for each } k \in T-J \end{aligned}$$

where $\delta _i^* = 1$ with an equality if $W_i=W_i({\varvec{\lambda }}^*) >0$. Exchanging the role of j and k in J, the inequality $\frac{b_{ij} ({\varvec{\lambda }}^*)}{ \lambda _j ^*\ell _{ij} } \le \frac{b_{ik} ({\varvec{\lambda }}^*)}{ \lambda ^*_k \ell _{ik}}$ is reversed hence must hold as an equality, hence

$$\begin{aligned} \hbox { for each } i \in S: \, \delta _i^*= \frac{b_{ij}}{ \lambda ^*_j \ell _{ij} } \le 1 \hbox { for each } j \in T \hbox { with } \delta _i^*= 1 \hbox { if } W_i >0. \end{aligned}$$

Since $\lambda ^*_k=1$ $\hbox { for each } k \in T-J$, this proves that ${\varvec{b}}$ is the cp-solution supported by indices ${\varvec{\delta }}^*_{ S}$ and ${\varvec{\lambda }}^*_T$. $\square$

2- General case: ${\varvec{ \ell }}_ {S\times T}$ has null elements. Let us perturb the liabilities matrix ${\varvec{ \ell }}_ {S\times T}$ by setting each null liability equal to $\epsilon$ where $\epsilon$ is positive and small enough so that the problem remains strictly feasible. From 1, F assigns to the perturbed problem its cp-solution. Denote it by ${\varvec{b}}(\epsilon )$, ${{\varvec{W}}}(\epsilon )= (W_k(\epsilon ))_{k \in S\cup T}$ the worth levels and ${\varvec{\delta }}_S(\epsilon )$ and ${\varvec{\mu }}_T(\epsilon )$ the indices. All but the rescue indices are in a bounded set: ${\varvec{b}}(\epsilon )$ are non-negative and satisfy for each i in S: $W_i (\epsilon )= z_i -\sum _{j \in T}b_{ij} (\epsilon )\ge 0$, ${{\varvec{W}}}(\epsilon )$ are non-negative and satisfy $\sum _k W_k(\epsilon )=\sum _k z_k$, and ${\varvec{\delta }}_S(\epsilon )$ belongs to $[0,1]^S$. We prove that the rescue indices are also bounded. Let J denote the subset of j in T for which $\mu _j(\epsilon )$ is bounded. There is a sequence $(\epsilon _p)$ converging to 0 when p tends to $\infty$ such that the sequences ${\varvec{b}}(\epsilon _p)$, ${{\varvec{W}}}(\epsilon _p)$, ${\varvec{\delta }}_S(\epsilon _p)$ and ${\varvec{\mu }}_{J}(\epsilon _p)$ converge. Denote the limits respectively by ${\varvec{b}}^*$, ${{\varvec{W}}}^*$, ${\varvec{\delta }}^*_ S$ and ${\varvec{\mu }}^*_{J}$. Denote by I the subset of S for which $\delta ^*_i$ is null. The following properties hold:

$$\begin{aligned} \hbox { for } i\in I&(a): W^*_i=0&(b): b_{ij}^* =0\hbox { for each } j \in J \end{aligned}$$

(19)

$$\begin{aligned} \hbox { for } i \in S-I&(a):\ell _{ij} =0 \hbox { for each } j \in T-J&(b): b^*_{ij}= \delta _i^* \ell _{ij} \mu _j^* \hbox { for each } j \in J \nonumber \\ \end{aligned}$$

(20)

To prove (19)-(a), note that for $i\in I$, $\delta ^*_i=0$ implies $\delta _i(\epsilon _p) <1$ for p large enough hence $W_i(\epsilon _p) =0$. Other properties follow from the fact that $b_{ij} (\epsilon _p)= \delta _i(\epsilon _p) \ell _{ij} (\epsilon _p) \mu _j(\epsilon _p)$ and that $b_{ij} (\epsilon _p)$ converges to $b^*_{ij}$, $\delta _i(\epsilon _p)$ to $\delta ^*_i$ and $\ell _{ij} (\epsilon _p)$ to $\ell _{ij}$. This proves (19)-(b) and (20)-(b) since for $j \in J$, $\mu _j(\epsilon _p)$ converges so $b_{ij}^* = \delta _i^* \ell _{ij} \mu _j^*$ and $\delta _i^* =0$ for i in I. Finally, to prove (20)-(a), use that for $j \in T-J$ and $i\in I$: $\mu _j(\epsilon _p)$ is unbounded and $\delta ^*_i$ is positive so $b_{ij} (\epsilon _p)$ can converge to the finite value $b^*_{ij}$ only if $\ell _{ij} =0$.

Assume J is not the whole set T. The rescue indices of j in $T-J$ are unbounded hence j’s worth is null. (20)-(a) implies that the debtors of $T-J$ are in I: $D(T-J)\subset I$. Thus the strict feasibility of the problem implies $z_I+ z_{T- J} >0$ and $T-J$ receives transfers from I only. Since furthermore, from (19)-(b), the transfers from I to J are null, we obtain $W^*_I + W^*_{T- J}= z_I+ z_{T- J}$, which contradicts the fact that the worths of firms in I and in $T-J$ are null. This proves that J is the whole set T.

This implies that I is empty: for i in I all its reimbursements are null by (19)-(b) hence its worth is equal to $z_i$, which is positive, contradicting (19)-(a). This proves that (20)-(b) holds for each i in S and j in T: ${\varvec{b}}^*$ is associated with finite indices for which all the conditions required at a cp-solution are satisfied. This ends the proof. $\square$

Proof of Proposition 5

Let F be splitting-invariant and assign bipartite-consistent solutions. Let ${\varvec{ \pi }}=({{\varvec{z}}}, {\varvec{ \ell }})$ be a problem in ${{\mathcal {F}}}$, ${\varvec{b}}= F({\varvec{ \pi }})$ and ${{\varvec{W}}}$ the worths achieved at ${\varvec{b}}$. The proof first assumes the network to be complete.

1-Complete liabilities network: $\ell _{ij} >0$ for each $i\ne j$.

Step 1: Let S be a strict subset of N. The reduced bipartite problem $({{\varvec{z}}}, {\varvec{ \ell }})_{S\times N-S,{\varvec{b}}}$ belongs to ${{\mathcal {F}}}$.

Proof

The result follows from Proposition 1: The reduced problem is non-decomposable since liabilities are strictly positive and $b_ {S\times N-S}$ is a positive solution for which one firm at least has a positive worth (as worth levels are identical to those reached at ${\varvec{b}}$).  $\square$

Step 2: ${\varvec{b}}$ is bi-proportional to ${\varvec{ \ell }}$ supported by indices satisfying (7) and (8).

Proof

There is a firm, say 1, with $W_1>0$. Define $\delta _1=1$ and ${\varvec{\mu }}$ by $\mu _1=1$ and $\mu _k=\frac{ b_{1k}}{\ell _{1k}}$ for $k>1$. By 1-consistency $\mu _k\ge 1$. Applying bipartite-consistency to $S=\{1,j\}$ implies that there is $\delta _j$ such that

$$\begin{aligned} \hbox { for each } k, l \notin \{1,j\}: \frac{ b_{jk}}{\ell _{jk}}/ \frac{ b_{1k}}{\ell _{1k}}=\delta _j \hbox { with }\delta _j \le 1 \hbox { with } =1 \hbox { if }W_j>0. \end{aligned}$$

(21)

Since $\frac{ b_{1k}}{\ell _{1k}}/ \mu _k =1$ by definition of $\mu _k$, we obtain

$$\begin{aligned} \hbox { for each } j \hbox { and for each } k \ne 1 : b_{jk}= \delta _j \mu _k {\ell _{jk}}\end{aligned}$$

(22)

where the indices satisfy the inequality requirements. It remains to show that the identities in (22) also hold for $k=1$. Applying bipartite consistency to $S=\{j,k\}$ $1 \ne j, 1\ne k$ yields

$$\begin{aligned} \hbox { for each } l \notin \{1,j,k\}: \frac{ b_{j 1}}{\ell _{j 1}}/ \frac{ b_{k1}}{\ell _{k1}} = \frac{ b_{jl}}{\ell _{jl}}/ \frac{ b_{kl}}{\ell _{kl}}. \end{aligned}$$

(23)

From (22), the right hand side of (23) is equal to $\delta _j/\delta _k$. Hence $\frac{ b_{j 1}}{\ell _{j 1}}/ \delta _j$ is independent of j. Denote this common value by c. The matrix ${\varvec{b}}$ is of the form

$$\begin{aligned} {\varvec{b}}= \left( \begin{array}{ccccc}0 &{}\quad \ell _{12} \mu _2 &{}\quad \cdots &{}\quad \ell _{1k} \mu _k &{}\quad \cdots . \\ c\delta _2 \ell _{21}&{}\quad 0 &{}\quad \cdots &{}\quad \delta _2 \ell _{2k} \mu _k &{}\quad \cdots \\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ c \delta _i \ell _{i1} &{}\quad \cdots &{}\quad \cdots &{}\quad \delta _i \ell _{ik} \mu _k &{}\quad \cdots \\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \end{array}\right) \end{aligned}$$

To prove that ${\varvec{b}}$ is bi-proportional to ${\varvec{ \ell }}$, it remains to show $c=1$. Consider two cases.

Case 1: $W_i>0$ for some $i>1$ for example for 2. By 1-consistency applied to 2, 2 reimburses fully 1 so that $c\delta _2 =1$. By bi-consistency applied to ${\{1,2\}}$, the reimbursement ratios of 1 and 2 to $\{3,\cdots , n\}$ are equal, which implies $\delta _2=1$, hence $c=1$.

Case 2: $W_i=0$ for each $i>1$. Let us split 1 and denote by a the value of the liabilities between $1 ^{\prime }$ and $1 ^{\prime \prime }$. Splitting-invariance implies that the worth levels of $1^{\prime }$ and $1^{\prime \prime }$ are half that of 1, hence positive at the solution ${\varvec{b}}'$ in the split model. Displaying only the reimbursements of $1^{\prime }$ and $1^{\prime \prime }$ and 2, ${\varvec{b}}'$ is of the form

$$\begin{aligned} {\varvec{b}}'_{\{ 1 ^{\prime }, 1 ^{\prime \prime }, 2 \} \times N }= \left( \begin{array}{cccccc} 0 &{} a &{} \frac{1}{2}\ell _{12} \mu _2 &{} \cdots &{} \frac{1}{2} \ell _{1k} \mu _k &{} \cdots \\ a &{} 0 &{} \frac{1}{2}\ell _{12} \mu _2 &{} \cdots &{} \frac{1}{2} \ell _{1k} \mu _k &{} \cdots \\ \frac{1}{2} c\delta _2 \ell _{21}&{} \frac{1}{2} c\delta _2 \ell _{21} &{} 0 &{} \cdots &{} \delta _2 \ell _{2k} \mu _k &{} \cdots \\ \end{array}\right) \end{aligned}$$

Bipartite-consistency applied to $S={\{1^{\prime },2\}}$ implies that the reimbursement ratios of $1^{\prime }$ and 2 to ${\{1^{\prime \prime },3,...\}}$ are proportional between each other. The reimbursement ratios of $1 '$ are respectively 1 and $\mu _k$ for $k>2$ and those of 2 are $c\delta _2$ and $\delta _2 \mu _k$ for $k>2$. Hence their proportionality requires $c=1$.

Step 3: The balancedness condition is met: $\delta _i\mu _i=1$ for each i.

Proof

Let us split i and denote with a prime subscript $^{\prime }$ the solution and the supporting indices in the split model (which exist by step 2).

We first show

$$\begin{aligned} \begin{aligned} \hbox { for each } j \in N-\{i\}: \delta _j=\delta '_j \hbox { and } \mu _j=\mu '_j\\ \delta _i=\frac{ (\delta '_{i'} + \delta '_{i''} ) }{2} \hbox { and } \mu _{i}= \frac{ (\mu '_{i'}+ \mu '_{i''}) }{2} \end{aligned}\end{aligned}$$

(24)

Splitting-invariance implies that adding the reimbursements in ${\varvec{b}}'$ made by $i'$ and $i^{\prime \prime }$ and the payments to them yields solution ${\varvec{b}}$. Hence

$$\begin{aligned} \hbox { for each } j \hbox { and } k \in N-\{i\}:&b_{jk}= \delta '_j\ell _{jk} \mu '_k \\ \hbox { for each } k \in N-\{i\}:&b_{ik}= \frac{ ( \delta '_{i'} + \delta '_{i''} ) }{2} \ell _{ik} \mu '_k \hbox { and } b_{ki}= \delta '_k \frac{ (\mu '_{i'}+ \mu '_{i''}) }{2} \ell _{ki} \end{aligned}$$

The worth levels of $k\ne i$ are identical at ${\varvec{b}}'$ and ${\varvec{b}}$ and those of $i'$ and $i''$ are half those of i. Thus their nullity or positivity are identical in the two allocations. This implies that the indices $(\delta '_j)_{j\ne i}, \frac{ (\delta '_{i'} + \delta '_{i''} ) }{2}$ and $(\mu '_j )_{j\ne i}, \frac{ (\mu '_{i'}+ \mu '_{i''}) }{2}$ support the allocation ${\varvec{b}}$. The uniqueness of indices implies (24).

We now prove $\delta '_{i'} = \delta '_{i''} = \delta _{i}$ and $\mu '_{i'} = \mu '_{i''} =\mu _{i}$. By splitting-invariance, $b'_{{i'} k }= \frac{b_{{i} k }}{2}$, which writes $\delta '_{i'} \frac{\ell _{ik} }{2} \mu _k = \delta _{i} \frac{\ell _{ik} }{2} \mu _k$ hence $\delta '_{i'} = \delta _{i}$. Similarly the identity $b'_{ k {i'} }= \frac{b_{ k i}}{2}$ writes $\delta _{k} \frac{\ell _{k i} }{2} \mu '_{i'} = \delta _{k} \frac{\ell _{k {i} } }{2} \mu _k$ hence $\mu '_{i'} = \mu _{i}$. Since the same argument applies to $i''$, we finally obtain $\delta '_{i'} = \delta '_{i''} = \delta _{i}$ and $\mu '_{i'} = \mu '_{i''} =\mu _{i}$, as desired.

We end the proof: as the reimbursements between $i'$ and $i^{\prime \prime }$ are equal to their common liability, $b'_{{i'} i^{\prime \prime } }=\ell _{i'i^{\prime \prime }}$, we must have $\delta '_{i'} \mu '_{i''} =1$ hence $\delta _{i} \mu _{i} =1$. $\square$

s2- General liabilities network If the network is not complete, change each null $\ell _{ij}$, $i\ne j$ into $\epsilon$ positive. The perturbed problem is still strictly feasible. Consider the associated worth levels ${{\varvec{W}}}(\epsilon )$ and indices ${\varvec{\delta }}(\epsilon ), {\varvec{\mu }}(\epsilon )$. Arguing as in the proof of Proposition 4, the worth levels and reimbursement indices, ${{\varvec{W}}}(\epsilon )$ and ${\varvec{\delta }}(\epsilon )$, are in a bounded set, so they are converging for a subsequence $(\epsilon _p)$ converging to 0. Denote their limits respectively by ${{\varvec{W}}}^*$ and ${\varvec{\delta }}^*$. We cannot exclude that ${\varvec{b}}(\epsilon )$ and ${\varvec{\mu }}(\epsilon )$ are unbounded. Let I be the set of i for which $\delta ^*_i$ is null. We can assume $\delta _i(\epsilon _p)<1$ for each i in I, possibly by deleting the first values in the sequence. The identity $\delta _i(\epsilon )\mu _i(\epsilon )=1$ implies that $\mu ^*_i(\epsilon _p)$ tends to $\infty$ for i in I and tends to $1/ \delta ^*_i$ for i not in I. The following equations hold at the limit:

$$\begin{aligned} i\in I:&\, (a) \, W_i(\epsilon _p)=0&(b) \lim \limits _{p \rightarrow \infty } b_{ik} (\epsilon _p)=0 \hbox { for } k \in N-I \end{aligned}$$

(25)

$$\begin{aligned} j\in N- I:&\, (a) \lim \limits _{p \rightarrow \infty } b_{jk} (\epsilon _p) =\infty , k \in I \hbox { with } \ell _{jk} >0&(b) \lim \limits _{p \rightarrow \infty } b_{jk} (\epsilon _p) =\frac{ \delta _i ^* }{ \delta _k^* } \ell _{jk}, k \in N- I \nonumber \\ \end{aligned}$$

(26)

If I is empty, then (26)-(b) applies to each pair, so the sequence $( {\varvec{b}}(\epsilon _p))$ converges and the limit satisfies all the conditions required on a cp-solution. It thus suffices to show that I is empty to conclude the proof.

Assume by contradiction $I\ne \emptyset$. Let us prove $D(I) \subset I$. Let j not in I. From (25)-(b) and (26)-(b), j’s received payments are bounded; hence $W_j \ge 0$ implies that

j’s reimbursements are bounded as well. From equation (26)-(a), $b_{jk}(\epsilon _p)$ is bounded for $k \in I$ only if $\ell _{jk} =0$, i.e. j is not indebted to any k in I. This proves $D(I) \subset I$. It thus suffices to show $z_I\le 0$ to obtain a contradiction with the strict feasibility of the problem. From (25)-(a), $W_I(\epsilon _p)=0$, which writes $z_I+ b_{N-I, I}(\epsilon _p) -b_{I, N-I}(\epsilon _p)=0$. Since $b_{I,N-I} (\epsilon _p)$ tends to 0 from (25)-(b), $z_I+ b_{N, I}(\epsilon _p)$ tends to zero, surely $z_I\le 0$, the desired contradiction. $\square$