Efficient, fair, and strategy-proof (re)allocation under network constraints

Abstract

A resource available in different types is to be distributed or redistributed among agents differing in their ability to consume these types. Preferences over consumption levels are not observable but known to be single-peaked. We identify the only strategy-proof and efficient allocation rule satisfying a formulation of “fair net trades” (Schmeidler and Vind in Econometrica 40(4):637–642, 1972). This rule coincides with the “generalized uniform rule” (Thomson in Axiomatic analysis of generalized economies with single-peaked preferences, 1995a) when the resource is available in a single type and with the “egalitarian rule” (Bochet et al. in J Econ Theory 148:535–562, 2013) when there are no individual endowments. Moreover, it coincides with the “uniform rule” (Sprumont in Econometrica 59(2):509–519, 1991) when the resource is available in a single type and there are no individual endowments. This rule belongs to a new class of group strategy-proof rules. Our analysis provides a unified treatment of the applications in Bochet et al. (Theor Econ 7:395–423, 2012, J Econ Theory 148:535–562, 2013).

Appendix

Notation: For each $i\in N$, let $\mathbf {e}_i\in \mathbb {R}^N$ denote the vector with a one on the ith coordinate and zeros elsewhere. For each index set I and each pair $z,z'\in \mathbb {R}^I$, let $z\ge z'$ if and only if, for each $i\in I$, $z_i\ge z_i'$. Refer to Sect. 4.1 for notation not introduced in Sect. 2.

1.1 Proof of Proposition 1

(i) The egalitarian trades rule is envy-free.

Let $R\in \mathcal {R}^N$ and $x\equiv \tilde{E}(R)$. By way of contradiction, suppose that there is $\{i,j\}\subseteq N$ such that $(\omega _i+ {x_j-\omega _j}) \mathrel {P_i} x_i$ and a $y\in Z$ such that, for each $k\in N\setminus \{i,j\}$, $y_k=x_k$, and $y_i\mathrel {P_i} x_i$. Thus, $x_i\ne p(R_i)$. If $x_i <p(R_i)$ then, by single-peakedness, $x_j-\omega _j>x_i-\omega _i$. By Remark 2, since Z is convex, it contains the convex hull of $\{x,y\}$. Thus, for each $\alpha \in [0,1]$, $z^\alpha \equiv \alpha y+ (1-\alpha )x$ is in Z. Note that for all sufficiently small values of $\alpha$, $z^\alpha \in P(R)$.¹² Note that, for sufficiently small values of $\alpha$, $x_j-\omega _j>z_j^\alpha -\omega _j\ge z_i^\alpha -\omega _i>x_i-\omega _i$ and, for each $k\in N\setminus \{i,j\}$, $z_k^\alpha -\omega _k=x_k-\omega _k$. Thus, $\alpha$ can be chosen so that $\sum _{k\in N} -(z_k^\alpha -\omega _k)^2>\sum _{k\in N} -(x_k -\omega _k)^2$ and $z^\alpha \in P(R)$. This contradicts $x=\tilde{E}(R)$. An analogous argument reaches the same contradiction when $x_i>p(R_i)$.

(ii) For individual ownership economies, the egalitarian trades rule satisfies the endowment lower bound. Suppose that, for each $t\in T$, $\Omega ^t=0$. Let $R\in \mathcal {R}^N$ and $x\equiv \tilde{E}(R)$. By way of contradiction, suppose there is $i\in N$ such that $\omega _i\mathrel {P_i}x_i$. Thus, by single-peakedness, $x_i\ne p(R_i)$. If $x_i<p(R_i)$, by single-peakedness, $x_i<\omega _i$. We will prove that there is $j\in N$ such that $x_j-\omega _j>x_i-\omega _i$ and a scalar $\varepsilon >0$ such that $x+\varepsilon (\mathbf {e}_i -\mathbf {e}_j)\in Z$. Let K denote the intersection of all sets I such that $i\in I$ and $\sum _I x_k=\sum _{t\in \Gamma (I)} m_t$.¹³ That is,

$$\begin{aligned} K\equiv \bigcap \left\{ I\subseteq N: i\in I, \sum _I x_k=\sum _{t\in \Gamma (K)} m_t\right\} . \end{aligned}$$

By the sub-modularity of mapping $I\subseteq N\mapsto \sum _{t\in \Gamma (I)} m_t$, the set K is itself such that $\sum _K x_k=\sum _{t\in \Gamma (K)} m_t$. Moreover, since all collective endowments are zero, $\sum _{t\in \Gamma (K)} m_t=\sum _{k\in K} \omega _k$. Thus, $x_i<\omega _i$ implies there is $j\in K$ such that $x_j >\omega _j$. Since K is minimal with respect to set inclusion among all I such that $\sum _I x_k=\sum _{t\in \Gamma (K)} m_t$, $\sum _{K\setminus \{j\}} x_k< \sum _{t\in \Gamma (K\setminus \{j\})} m_t$. Thus, the feasibility constraint is not tight and i’s assignment can be increased: there is a scalar $\varepsilon >0$ such that $y\equiv x+\varepsilon (\mathbf {e}_i -\mathbf {e}_j)\in Z$. By Remark 2, since Z is convex, it contains the convex hull of $\{x,y\}$. From here on, we construct a $z^\alpha$ in the convex hull of $\{x,y\}$ contradicting $x=\tilde{E}(R)$, exactly as we did in part (i) of the proof of Proposition 1. An analogous argument reaches the same contradiction when $x_i>p(R_i)$. $\square$

1.2 Proof of Theorem 1

For each $i\in N$, let $v_i: \mathbb {R}_+\rightarrow \mathbb {R}$ denote a strictly concave and continuous function. For each $R\in \mathcal {R}^N$, let $\varphi (R)\equiv \arg \max \left\{ \sum _N v_i(x_i): x\in P(R)\right\}$. Thus, $\varphi$ is a separably concave rule.

By way of contradiction, suppose that $\varphi$ is not group strategy-proof. Then, there is a group of agents $M\subseteq N$ and a preference profile $R\in \mathcal {R}^N$ at which the agents in M can gain by jointly misreporting their preferences. Let $R'\in \mathcal {R}^N$ denote the preference profile corresponding to this misreport. Thus, for each $i\in N \setminus M$, $R_i'=R_i$ and, for each $i\in M$, $\varphi _i(R')\mathrel {R_i} \varphi _i(R)$ and there is $i\in M$ such that $\varphi _i(R')\mathrel {P_i} \varphi _i(R)$. Let $p\equiv p(R)$, $p'\equiv p(R')$, $x\equiv \varphi (R)$, and $x'\equiv \varphi (R')$. Note that the following is necessary condition for M to manipulate profitably:

$$\begin{aligned} \text {for each }i\in M, \quad x_i'\mathrel {R_i}x_i. \end{aligned}$$

(6)

Claim 1

Without loss of generality, if $i\in M$, $p_i\ne p_i'$.

Proof

Suppose there is $i\in M$ such that $p_i=p_i'$. Then, by Lemma 3,¹⁴ $P(R_i, R_{-i}')=P(R')$. Thus, from the definition of the separably concave rule, $\varphi (R_i, R_{-i}')=\varphi (R')$. Thus, it would be without loss of generality to assume that $M\setminus \{i\}$ is the group misreporting, not M.

Let C and $C'$ denote the inclusion-minimal min-cuts in the networks (G, p) and $(G,p')$, respectively (see Sect. 4.1.2 for the relevant definitions). Let $T_-\equiv T\cap C$, $T_+\equiv T\setminus T_-$, $N_+\equiv \Gamma (T_-)$, and $N_-\equiv N\setminus N_+$. Similarly, define $T_-', T_+', N_-'$, and $N_+'$ with respect to $C'$. Recall, from Lemma 2 and the relevant definitions, that

$$\begin{aligned} \Gamma (N_-)=T_+, \quad \Gamma (T_-)=N_+, \quad \Gamma (N_-')=T_+', \quad \Gamma (T_-')=N_+'. \end{aligned}$$

(7)

Let $\hat{N}\equiv N_+\cap N_-'$, $\hat{T}\equiv \Gamma (\hat{N})\cap T_-$, $\tilde{N}\equiv N_+'\cap N_-$, and $\tilde{T}\equiv \Gamma (\tilde{N})\cap T_-'$. Note the following:

• (a) By Lemma 3, $\hat{N}\subseteq N_+$ implies $x|_{\hat{N}}\ge p|_{\hat{N}}$ and $\hat{N}\subseteq N_-'$ implies $x'|_{\hat{N}}\le p'|_{\hat{N}}$.

• (b) By (a) and $p|_{\hat{N}\setminus M}=p'|_{\hat{N}\setminus M }$, $x|_{\hat{N}\setminus M}\ge x'|_{\hat{N}\setminus M }$.

• (c) By (6) and single-peakedness, $x|_{\hat{N}\cap M}\ge x'|_{\hat{N}\cap M}$.

• (d) By (b) and (c), $\sum _{\hat{N}} x_k\ge \sum _{\hat{N}} x_k'$. By Lemma 3, at x, agents in $\hat{N}\subseteq N_+$ only receive the resource from $T_-$. Thus, they can jointly receive at most $\sum _{\hat{T}} m_j$ by the definition of $\hat{T}$. Thus, $\sum _{\hat{N}} x_k\le \sum _{\hat{T}} m_j$. Thus,

  $$\begin{aligned} \begin{array}{c} \sum _{\hat{N}} x_k' \le \sum _{\hat{N}} x_k\le \sum _{\hat{T}} m_j. \end{array} \end{aligned}$$

• (e) By Lemma 3, at $x'$, all of the resource in $\hat{T}\subseteq \Gamma (\hat{N})\subseteq \Gamma (N_-')= T_+'$ are consumed by agents in $N_-'$. By (7), since $N_-'\setminus \hat{N}\subseteq N_-$, $\Gamma (N_-'\setminus \hat{N})\subseteq \Gamma (N_-)=T_+$. Thus, no agent in $N_-'\setminus \hat{N}$ can consume the resource in types in $\hat{T}\subseteq T_-$ (because $T_-\cap T_+=\varnothing$). Thus, at $x'$, all stocks of the resource of types in $\hat{T}$ have to be consumed by agents in $\hat{N}$. Thus,

  $$\begin{aligned} \begin{array}{c} \sum _{\hat{T}} m_j\le \sum _{\hat{N}}x_k'. \end{array} \end{aligned}$$

(a$'$) By Lemma 3, $\tilde{N}\subseteq N_+'$ implies $x'|_{\tilde{N}}\ge p'|_{\tilde{N}}$ and $\tilde{N}\subseteq N_-$ implies $x|_{\tilde{N}}\le p|_{\tilde{N}}$.

(b$'$) By (a$'$) and $p'|_{\tilde{N}\setminus M}=p|_{\tilde{N}\setminus M }$, $x'|_{\tilde{N}\setminus M}\ge x|_{\tilde{N}\setminus M }$.

(c$'$) By (6) and single-peakedness, $x'|_{\tilde{N}\cap M}\ge x|_{\tilde{N}\cap M}$.

(d$'$) By (b$'$) and (c$'$), $\sum _{\tilde{N}} x_k\le \sum _{\tilde{N}} x_k'$. By Lemma 3, at $x'$, agents in $\tilde{N}\subseteq N_+'$ only receive the resource from $T_-'$. Thus, they can jointly receive at most $\sum _{\tilde{T}} m_j$ by the definition of $\tilde{T}$. Thus, $\sum _{\tilde{N}} x_k'\le \sum _{\tilde{T}} m_j$. Thus,

$$\begin{aligned} \begin{array}{c} \sum _{\tilde{N}} x_k \le \sum _{\tilde{N}} x_k'\le \sum _{\tilde{T}} m_j. \end{array} \end{aligned}$$

(e$'$) By Lemma 3, at x, all of the resource in $\tilde{T}\subseteq \Gamma (\tilde{N})\subseteq \Gamma (N_-)= T_+$ is consumed by agents in $N_-$. By (7), since $N_-'\setminus \tilde{N}\subseteq N_-'$, $\Gamma (N_-'\setminus \tilde{N})\subseteq \Gamma (N_-')=T_+'$. Thus, no agent in $N_-'\setminus \tilde{N}$ can consume the types of the resource in $\tilde{T}\subseteq T_-'$ (because $T_-'\cap T_+'=\varnothing$). Thus, at x, all of the stocks of the resource of types in $\tilde{T}$ have to be consumed by agents in $\tilde{N}$. Thus,

$$\begin{aligned} \begin{array}{c} \sum _{\tilde{T}} m_j\le \sum _{\tilde{N}}x_k. \end{array} \end{aligned}$$

Combining the final inequalities in (d) and (e) with (b), and (c),

$$\begin{aligned}&x|_{\hat{N}}=x'|_{\hat{N}} \text { and, at both } x \text { and } x', \nonumber \\&\quad \text { agents in } \hat{N}\text { consume all of the resource from } \hat{T}\text {and only that.} \nonumber \\&\quad \text { Thus, } x|_{\hat{N}},x'|_{\hat{N}}\in Z(\hat{N},\hat{T}) \text {by Lemma 1}. \end{aligned}$$

(8)

Combining the final inequalities in (d$'$) and (e$'$) with (b$'$), and (c$'$),

$$\begin{aligned}&x|_{\tilde{N}}=x'|_{\tilde{N}} \text { and, at both } x \text { and } x', \nonumber \\&\quad \text {agents in } \tilde{N}\text { consume all of the resource from } \tilde{T}\text { and only that.} \nonumber \\&\quad \text {Thus, } x|_{\tilde{N}},x'|_{\tilde{N}}\in Z(\tilde{N},\tilde{T}) \text { by Lemma 1.} \end{aligned}$$

(9)

Claim 2

• (i) $z=x|_{N_-\cap N_-'}$ maximizes $\sum _{N_-\cap N_-'} v_k$ over all $z\in Z(N_-\cap N_-', T_+\cap T_+')$ such that $z\le p|_{N_-\cap N_-'}$.

• (ii) $z=x'|_{N_-\cap N_-'}$ maximizes $\sum _{N_-\cap N_-'} v_k$ over all $z\in Z(N_-\cap N_-', T_+\cap T_+')$ such that $z\le p'|_{N_-\cap N_-'}$.

• (iii) $z=x|_{N_+\cap N_+'}$ maximizes $\sum _{N_+\cap N_+'} v_k$ over all $z\in Z(N_+\cap N_+', T_-\cap T_-')$ such that $z\ge p|_{N_+\cap N_+'}$.

• (iv) $z=x'|_{N_+\cap N_+'}$ maximizes $\sum _{N_+\cap N_+'} v_k$ over all $z\in Z(N_+\cap N_+', T_-\cap T_-')$ such that $z\ge p'|_{N_+\cap N_+'}$.

Proof

We start by proving (i). By Lemma 3, $P(R)=P(R)|_{N_-}\times P(R)|_{N_+}$. Thus, $x|_{N_-}$ maximizes $\sum _{N_-} v_k$ over $P(R)|_{N_-}$. Equivalently, by Lemma 3, $x|_{N_-}$ maximizes $\sum _{N_-} v_k$ over all $z\in Z(N_-, T_+)$ such that $z\le p_{N_-}$. Note that $N_-$ is partitioned into $N_-\cap N_-'$ and $\tilde{N}$. Since $Z(N_-, T_+)\supseteq Z(N_- \setminus \tilde{N}, T_+\setminus \tilde{T}) \times Z(\tilde{N}, \tilde{T})$, $x|_{\tilde{N}}\in Z(\tilde{N}, \tilde{T})$ by (9), and $x|_{N_- \setminus \tilde{N}}\in Z(N_- \setminus \tilde{N}, T_+\setminus \tilde{T})$, $x|_{N_- \setminus \tilde{N}}$ maximizes $\sum _{N_-\setminus \tilde{N}} v_k$ over all $z\in Z(N_- \setminus \tilde{N}, T_+\setminus \tilde{T})$ such that $z\le p|_{N_-\setminus \tilde{N}}$. However, $N_- \setminus \tilde{N}=N_-\cap N_-'$ and $T_+\setminus \tilde{T}=T_+\cap T_+'$, thus establishing statement (i). The proof of statement (iii) is similar. The proofs of statements (ii) and (iv) are fully analogous, using (8) instead of (9).

By Claim 2 (i),

$$\begin{aligned} x|_{N_-\cap N_-'}&= \begin{array}{c} \arg \max \sum _{N_-\cap N_-'} v_k(z_k)\quad \text {such that} \end{array} \\ (\text {C1})&\quad x|_{N_-\cap N_-'}\in Z(N_-\cap N_-', T_+\cap T_+'),\\ (\text {C2})&\quad \text {for each }i\in N_-\cap N_-', \quad z_i\le p_i. \end{aligned}$$

By Lemma 3, if $i\in N_-\cap N_-'$, $x_i\le p_i$ and $x_i'\le p_i'$. Thus, if $i\in N_-\cap N_-'\cap M$ and $p_i'<x_i$, $x_i'\le p_i'<x_i\le p_i$. Thus, by single-peakedness, $x_i \mathrel {P_i}x_i'$, contradicting (6). Thus, in fact,

$$\begin{aligned} \text {for each } i\in N_-\cap N_-'\cap M, \quad x_i\le p_i'. \end{aligned}$$

(10)

By Claim 1, for each $i\in N_-\cap N_-'\cap M$, $p_i\ne p_i'$. By (6) and single-peakedness, for each $i\in N_-\cap N_-'\cap M$, $x_i\le x_i'$ and, by Lemma 3, $x_i' \le p_i'$. Thus, by (10), the following sets partition $N_-\cap N_-'\cap M$:

$$\begin{aligned} I_1&\equiv \{i\in N_-\cap N_-'\cap M: x_i'=x_i=p_i<p_i'\},\\ I_2&\equiv \{i\in N_-\cap N_-'\cap M:x_i<p_i<x_i'\le p_i'\},\\ I_3&\equiv \{i\in N_-\cap N_-'\cap M:x_i\le x_i' \le p_i< p_i', x_i<p_i\},\\ I_4&\equiv \{i\in N_-\cap N_-'\cap M:x_i\le x_i' \le p_i'< p_i\}. \end{aligned}$$

Consider the following alternative constraint to (C2):

$$\begin{aligned} (\text {C2}')\quad&\text {for each }i\in [N_-\cap N_-']\setminus [I_2\cup I_3], \quad z_i\le p_i'=p_i \text { and }\\&\text {for each }i\in [N_-\cap N_-']\cap [I_2\cup I_3], \quad z_i \le p_i'. \end{aligned}$$

Since for each $i\in I_2\cup I_3$, $p_i<p_i'$, $x|_{N_-\cap N_-'}$ satisfies (C1) and (C2$'$). However, all of the constraints in (C2) involving $i\in I_2\cup I_3$ are not tight at $x|_{N_-\cap N_-'}$ to begin with. That is, $x_i<p_i<p_i'$. Thus, relaxing them further does not change the maximizer. Thus,

$$\begin{aligned} x|_{N_-\cap N_-'} \hbox { maximizes } \sum _{N_-\cap N_-'} v_k \hbox { under constraints (C1) and (C2') }. \end{aligned}$$

Consider the following alternative to constraint (C2$'$):

$$\begin{aligned} (\text {C2}'')\quad&\text {for each }i\in [N_-\cap N_-']\setminus [I_2\cup I_3 \cup I_4], \quad z_i\le p_i'=p_i \text { and }\\&\text {for each }i\in [N_-\cap N_-']\cap [I_2\cup I_3\cup I_4], \quad z_i \le p_i'. \end{aligned}$$

Since, for each $i\in I_4$, $x_i\le p_i'<p_i$, $x_{N_-\cap N_-'}$ satisfies (C1) and (C2$''$). Thus, in going from (C2$'$) to (C2$''$), tightening the constraints, we do not exclude the optimum for the relaxed constraints in (C2$'$). Thus,

$$\begin{aligned} x_{N_-\cap N_-'} \hbox { maximizes } \sum _{N_-\cap N_-'} v_k \hbox { under constraints (C1) and (C2'')}. \end{aligned}$$

Consider the following alternative to constraint $(\hbox {C2}''$):

$$\begin{aligned} (\text {C2}''')\quad&\text {for each }i\in [N_-\cap N_-']\setminus [I_1\cup I_2\cup I_3 \cup I_4], \quad z_i\le p_i'=p_i \text { and }\\&\text {for each }i\in [N_-\cap N_-']\cap [I_1\cup I_2\cup I_3\cup I_4], \quad z_i \le p_i'. \end{aligned}$$

Since, for each $i\in I_1$, $x_i\le p_i<p_i'$, $x|_{N_-\cap N_-'}$ satisfies (C1) and(C2$'''$). However, by assumption for each such i, $x_i'=x_i$. Thus,

$$\begin{aligned} x|_{N_-\cap N_-'} \hbox { maximizes } \sum _{N_-\cap N_-'} v_k \hbox { under constraints (C1) and (C2''')}. \end{aligned}$$

However, since $x'|_{N_-\cap N_-'} \in Z(N_-\cap N_-', T_+\cap T_+')$ and $M\cap [N_-\cap N_-']=I_1\cup I_2\cup I_3 \cup I_4$, by Claim 2 (ii),

$$\begin{aligned} x'|_{N_-\cap N_-'} \hbox { maximizes } \sum _{N_-\cap N_-'} v_k \hbox { under constraints (C1) and (C2''')}. \end{aligned}$$

Thus, $x|_{N_-}$ and $x|_{N_-}'$ maximize $\sum _{N_-\cap N_-'} v_k$ over $\mathcal {P}\equiv \{ z\in \mathbb {R}_+^{N_-\cap N_-'}: z \text { satisfies (C1) and (C2''')}\}$. Since $\sum _{N_-\cap N_-'} v_k$ is strictly concave and $\mathcal {P}$ is compact and convex, maximizing $\sum _{N_-\cap N_-'} v_k$ over $\mathcal {P}$ yields a unique maximizer. Thus,

$$\begin{aligned} x'|_{N_-\cap N_-'}=x|_{N_-\cap N_-'}. \end{aligned}$$

(11)

An analogous argument establishes that

$$\begin{aligned} x'|_{N_+\cap N_+'}=x|_{N_+\cap N_+'}. \end{aligned}$$

(12)

Conclusion. By (8), (9), (11), and (12), $x=x'$. Thus (6), a necessary condition for group M to manipulate, yields $x=x'$. Thus, at each preference profile, each group of agents can do no better than by reporting its preferences truthfully. Thus, $\varphi$ is group strategy-proof. $\square$

1.3 Proof of Lemma 4

Let all of the notation be as in the statement of the Lemma 4 and $p\equiv p(R)$.

We first prove that $f_-$ is sub-modular. Let $I_1,I_2 \subseteq N_-$. Let $J_1\subseteq I_1$ and $J_2\subseteq I_2$ attain the minimum in the definition of $f_-$, for $I_1$ and $I_2$, respectively. Then, by the definition of $f_-$, we obtain the first equality and the last inequality below and, by the sub-modularity of $I\mapsto \sum _{\Gamma (I)\cap T_+}m_j$, we obtain the first inequality below:

$$\begin{aligned} f_-(I_1)+f_-(I_2)&= \sum _{\Gamma (J_1)\cap T_+}m_j +\sum _{I_1\setminus J_1} p_i+\sum _{\Gamma (J_2)\cap T_+}m_j +\sum _{I_2\setminus J_2} p_i\\&\ge \sum _{\Gamma (J_1\cup J_2)\cap T_+}m_j +\sum _{I_1\setminus J_1} p_i+\sum _{\Gamma (J_1\cap J_2) \cap T_+}m_j +\sum _{I_2\setminus J_2} p_i\\&= \sum _{\Gamma (J_1\cup J_2)\cap T_+}m_j+ \sum _{[I_1\cup I_2]\setminus [J_1\cup J_2]}p_i\\&\quad +\sum _{\Gamma (J_1\cap J_2)\cap T_+}m_j + \sum _{[I_1\cap I_2]\setminus [J_1\cap J_2]}p_i\\&\ge f_-(I_1\cup I_2)+f_-(I_1\cap I_2). \end{aligned}$$

Thus, $f_-$ is sub-modular.

It remains to show that $f_-$ is non-decreasing with respect to set inclusion. Suppose that $I_1, I_2 \subseteq N_-$ are such that $I_1\subseteq I_2$. Let $J_1\subseteq I_1$ and $J_2\subseteq I_2$ attain the minimum in the definition of $f_-$, for $I_1$ and $I_2$, respectively. Let $J\equiv I_2\setminus I_1$ and $K\equiv J_2\cap J_1$. Then,

$$\begin{aligned} f_-(I_2)= & {} \sum _{\Gamma (J_2)} m_k+\sum _{I_2\setminus J_2} p_k=\sum _{\Gamma (J_2)} m_k\\&+\sum _{[I_1\cup J]\setminus J_2} p_k\ge \sum _{\Gamma (K)} m_k+\sum _{I_1\setminus K} p_k+\sum _{J\setminus J_2} p_k\ge f_-(I_1). \end{aligned}$$

Thus, $f_-$ is non-decreasing with respect to set inclusion. The fact $f_-(\varnothing )=0$ follows immediately from its definition. The proofs for $f_+$ are analogous.

Next, we establish that $P(R)|_{N_-}=\mathbf {B}(f_-)$. Let $y\in \mathbf {B}(f_-)$. Then, for each $I\subseteq N_-$, $\sum _I y_i \le f_-(I)\le \sum _{\Gamma (I)\cap T_+} m_j$ and $\sum _{N_-}y_i =f_-(N_-)=\sum _{\Gamma (N_-) \cap T_+}m_j=\sum _{T_+}m_j$. For each $i\in N_-$, $y_i\le f_-(\{i\})\le p_i$. Thus, by Lemma 1, $y\in Z(N_-,T_+)$ and $y\le p_{N_-}$. Thus, by Lemma 3, $y\in P(R)|_{N_-}$. Conversely, let $y\in P(R)|_{N_-}$. Then, for each $I\subseteq N_-$,

$$\begin{aligned} \hbox {for each } J\subseteq I, \quad \sum _I y_i =\sum _J y_i+\sum _{I\setminus J} y_i\le \sum _{\Gamma (J)\cap T_+}m_j +\sum _{I\setminus J}p_i. \end{aligned}$$

Thus, minimizing with respect to $J\subseteq I$, $\sum _I y_i \le f_-(I)$. Additionally, $\sum _{N_-}y_i= \sum _{T_+} m_j$. Thus, $y \in \mathbf {B}(f_-)$.

Next we establish that $P(R)|_{N_+}=\mathbf {B}(f_+)$. Let $y\in \mathbf {B}(f_+)$. Then, for each $I\subseteq N_+$,

$$\begin{aligned} \sum _I y_i \le f_+(I)\le \sum _{\Gamma (J)\cap T_-} m_j-\sum _{J\setminus I}p_i, \text { for each } J \hbox { with } N_+ \supseteq J\supseteq I. \end{aligned}$$

For $J=I$, $\sum _I y_i \le g(I)\le \sum _{\Gamma (I)\cap T_-} m_j$. We have just shown that, for each $I\subseteq N_+$, $\sum _I y_i \le f_+(I)$. Additionally, $\sum _{N_+}y_i=f_+(N_+)= \sum _{\Gamma (I)\cap T_-} m_j$. Now, let $i\in N_+$. Thus, if $I\equiv N_+\setminus \{i\}$,

$$\begin{aligned}&y_i=\sum _{N_+} y_k-\sum _{N_+\setminus \{i\}}y_k\\&\quad \ge f_+(N_+)-f_+(N_+\setminus \{i\}) \\&\quad = \sum _{ T_-} m_j-\min \{\sum _{\Gamma (J)\cap T_-}m_j-\sum _{J\setminus [N_+\setminus \{i\}]}p_k : J\supseteq N_+\setminus \{i\}\}\\&\quad \ge \sum _{T_-}m_j-[\sum _{\Gamma (N_+)\cap T_-}m_j-\sum _{ N_+\setminus [N_+\setminus \{i\}] }p_j]\\&\quad =p_i \end{aligned}$$

where the last equality follows because $\Gamma (N_+)\supseteq T_-$. Thus, $y\in Z(N_+,T_-)$ and $y\ge p_{N_+}$. Thus, by Lemma 3, $y\in P(R)|_{N_+}$. Conversely, let $y\in P(R)|_{N_+}$ Then, for each $I\subseteq N_+$,

$$\begin{aligned} \text {for each } \ J \ \text { with } \ N_+\supseteq J\supseteq I, \quad \sum _I y_i =\sum _J y_i -\sum _{J\setminus I} y_i\le \sum _{\Gamma (J)\cap T_-}m_j-\sum _{J\setminus I} p_k. \end{aligned}$$

Minimizing with respect to J ($N_+\supseteq J\supseteq I$), $\sum _I y_i \le f_+(I)$. Thus, $y\in \mathbf {B}(f_+)$. $\square$

1.4 Proof of Lemma 6

Let all the notation be as in the statement of Lemma 6.

Suppose that [$i\in N\setminus C,p_i' \ge p_i$]. Since the capacity of a cut in (G, p) is no greater than its capacity in $(G,p')$, $\text {cap}(C', p)\le \text {cap}(C', p')$; since $C'$ is a min-cut in $(G,p')$, $\text {cap}(C,p')\ge \text {cap}(C',p')$; since C is a min-cut in (G, p), $\text {cap}(C',p)\ge \text {cap}(C,p)$. Thus,

$$\begin{aligned} \text {cap}(C,p')\ge \text {cap}(C',p')\ge \text {cap}(C',p)\ge \text {cap}(C,p). \end{aligned}$$

On the other hand, by the definition in (2), $\text {cap}(C,p)=\text {cap}(C,p')$. Thus all of the inequalities above are equalities. Thus, C is a min-cut $(G,p')$ and $C'$ is a min-cut in (G, p). Since C and $C'$ are inclusion minimal, $C\subseteq C'$ and $C\supseteq C'$. Thus, $C=C'$.

Suppose that [$i\in C\cap N,p_i' \le p_i$]. By the definition in (2), for each cut K containing i, $0\le p_i-p_i'=\text {cap}(K, p)-\text {cap}(K,p').$ Thus, the capacity of each cut in $(G,p')$ containing i is $p_i-p_i'$ less than its capacity in (G, p). By the definition in (2), the capacity of each cut in $(G,p')$ not containing i is the same as in (G, p). Thus, since C is a min-cut in (G, p), $\text {cap}(C,p') \le \text {cap}(C', p')$. That is, C is a min-cut in $(G,p')$ so $\text {cap}(C,p') = \text {cap}(C', p')$. Since $C'$ is the inclusion-minimal min-cut in $(G,p')$, $C' \subseteq C$. By the same arguments, $C'$ is a min-cut in (G, p) and, thus, $C'\supseteq C$. Thus, $C=C'$.