Resource allocation games: a compromise stable extension of bankruptcy games

Abstract

This paper presents an extension of the traditional bankruptcy problem. In a resource allocation problem there is a common-pool resource, which needs to be divided among agents. Each agent is characterized by a claim on this pool and an individual linear monetary reward function for assigned resources. Analyzing these problems a new class of transferable utility games is introduced, called resource allocation games. These games are based on the bankruptcy model, as introduced by O’Neill (Math Soc Sci 2:345–371, 1982). It is shown that the properties of totally balancedness and compromise stability can be extended to resource allocation games, although the property of convexity is not maintained in general. Moreover, an explicit expression for the nucleolus of these games is provided.

Appendix

In this Appendix we present the proofs of Lemmas 1, 2, 3, and Theorem 4.

Proof

(Lemma 1) Let $S\subset T\subset N$. We will prove that

$$\begin{aligned} k(S)\le k(T). \end{aligned}$$

First assume that $D(T)=0$. Then $k(T)=1$ and

$$\begin{aligned} E\le \sum _{j\in N\setminus T}d_j. \end{aligned}$$

Hence,

$$\begin{aligned} E\le \sum _{j\in N\setminus T}d_j+\sum _{j\in T\setminus S}d_j=\sum _{j\in N\setminus S}d_j. \end{aligned}$$

This implies $D(S)=0$ and $k(S)=1$.

Next assume that $D(T)>0$. First note that

$$\begin{aligned} D(S)&= \max \left\{ 0,E-\sum _{j\in N\setminus S}d_j\right\} \\&= \max \left\{ 0,E-\sum _{j\in N\setminus T}d_j-\sum _{j\in T\setminus S}d_j\right\} \\&= \max \left\{ 0,D(T)-\sum _{j\in T\setminus S}d_j\right\} . \end{aligned}$$

If $D(S)=0$, then $k(S)=1\le k(T)$ by definition. Therefore, we may assume that $D(S)>0$ and $D(S)=D(T)-\sum _{j\in T\setminus S}d_j$. It follows that

$$\begin{aligned} \sum _{l=1}^{k(S)-1}d(T_l)&= \sum _{l=1}^{k(S)-1}d(S_l)+\sum _{l=1}^{k(S)-1}d(T_l\setminus S_l)\\&\le D(S)+\sum _{l=1}^{k(S)-1}d(T_l\setminus S_l)\\&\le D(S)+\sum _{j\in T\setminus S}d_j\\&= D(T)\\&< \sum _{l=1}^{k(T)}d(T_l) \end{aligned}$$

where the first and last inequality follow from (6). This implies that $k(S)-1< k(T)$ and consequently $k(S)\le k(T)$. $\square$

Proof

(Lemma 2) Let $l\in \{1,\ldots ,m\}$. We will prove that

$$\begin{aligned} \sum _{j\in S_l}x_j^S\le \sum _{j\in S_l}x_j^T. \end{aligned}$$

If $D(S)=0$, then it holds that

$$\begin{aligned} \sum _{j\in S_l}x^S_j=0\le \sum _{j\in S_l}x^T_j. \end{aligned}$$

Assume $D(S)>0$. As before, this implies that $D(S)=D(T)-\sum _{j\in T\setminus S}d_j$. If $l>k(S)$, then $x^S_j=0$ for all $j\in S_l$ and

$$\begin{aligned} \sum _{j\in S_l}x^S_j=0\le \sum _{j\in S_l}x^T_j. \end{aligned}$$

If $l<k(T)$, then it holds that $x^T_j=d_j$ for all $j\in T_l$. Hence, since $S_l\subset T_l$,

$$\begin{aligned} \sum _{j\in S_l}x^S_j\le \sum _{j\in S_l}d_j=\sum _{j\in S_l}x^T_j. \end{aligned}$$

Since $S\subset T$, it follows from Lemma 1 that $k(S)\le k(T)$. Therefore, the only case that remains to be considered is $l=k(S)=k(T)$. In that case

$$\begin{aligned} \sum _{j\in S_l}x^S_j&= D(S)-\sum _{t=1}^{l-1}d(S_t)\\&\le D(S)-\sum _{t=1}^{l-1}d(S_t)+d(T_l\setminus S_l)-\sum _{j\in T_l\setminus S_l}x^T_j+\sum _{t=l+1}^{m}d(T_t\setminus S_t)\\&= D(S)-\sum _{t=1}^{l-1}d(S_t)-\sum _{t=1}^{l-1}d(T_t\setminus S_t)+\sum _{j\in T\setminus S}d_j-\sum _{j\in T_l\setminus S_l}x^T_j\\&= D(S)+\sum _{j\in T\setminus S}d_j -\sum _{t=1}^{l-1}d(T_t)-\sum _{j\in T_l\setminus S_l}x^T_j\\&= D(T)-\sum _{t=1}^{l-1}d(T_t)-\sum _{j\in T_l\setminus S_l}x^T_j\\&= \sum _{j\in T_l}x^T_j-\sum _{j\in T_l\setminus S_l}x^T_j\\&= \sum _{j\in S_l}x^T_j \end{aligned}$$

where the first and fifth equality hold by Theorem 1. $\square$

Proof

(Lemma 3) 1: Let $d_i\ge E$. Then $D(N_{-i})=0$ and $v^R(N_{-i})=0$. Therefore,

$$\begin{aligned} M_i(v^R) = v^R(N)-v^R(N_{-i}) = v^R(N). \end{aligned}$$

2: Let $d_i<E$. Set $k=k(N)$,

2a: Let $i\in P_1$. We start by showing that $k(N_{-i})=k$. By definition of $k$ we have

$$\begin{aligned} \sum _{l=1}^{k-1}d(N_l)-d_i \le E-d_i < \sum _{l=1}^kd(N_l)-d_i. \end{aligned}$$

Since $i\in N_1 \cup \dots \cup N_{k-1}$, this implies

$$\begin{aligned} \sum _{l=1}^{k-1}d\bigl ((N_{-i})_l\bigr ) \le D(N_{-i})< \sum _{l=1}^kd\bigl ((N_{-i})_l\bigr ). \end{aligned}$$

Consequently, $k(N_{-i})=k$. Hence,

$$\begin{aligned} v^R(N_{-i})&= \sum _{l=1}^{k-1}\beta _ld\bigl ((N_{-i})_l\bigr )+\beta _k \biggl (D(N_{-i})-\sum _{l=1}^{k-1}d\bigl ((N_{-i})_l\bigr )\biggr )\\&= \sum _{l=1}^{k-1}\beta _ld(N_l)-\alpha _id_i+\beta _k\biggl (E-d_i-\sum _{l=1}^{k-1}d(N_l)+d_i\biggr )\\&= v^R(N)-\alpha _id_i. \end{aligned}$$

and

$$\begin{aligned} M_i(v^R)=v^R(N)-v^R(N_{-i})=\alpha _id_i. \end{aligned}$$

2b/2c: Let $i\in P_2\cup P_3$. By Theorem 2 it holds that

$$\begin{aligned} v^R(N_{-i})&= \sum _{l=1}^{k(N_{-i})-1}\beta _ld\bigl ((N_{-i})_l\bigr )+\beta _{k(N_{-i})} \biggl (D(N_{-i})-\sum _{l=1}^{k(N_{-i})-1}d\bigl ((N_{-i})_l\bigr )\biggr )\\&= \sum _{l=1}^{k(N_{-i})-1}\beta _ld(N_l)+\beta _{k(N_{-i})} \biggl (E-d_i-\sum _{l=1}^{k(N_{-i})-1}d(N_l)\biggr ) \end{aligned}$$

while

$$\begin{aligned} v^R(N)=\sum _{l=1}^{k-1}\beta _ld(N_l)+\beta _k \biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr ). \end{aligned}$$

Then

$$\begin{aligned} M_i(v^R)&= v^R(N)-v^R(N_{-i})\\&= \sum _{l=k(N_{-i})}^{k-1}\beta _ld(N_l)-\beta _{k(N_{-i})} \biggl (E-d_i-\sum _{l=1}^{k(N_{-i})-1}d(N_l)\biggr )\\&+\beta _k \biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr ). \end{aligned}$$

If $k(N_{-i})=k$, then this simplifies into

$$\begin{aligned} M_i(v^R)= \beta _kd_i. \end{aligned}$$

If $k(N_{-i})<k$, then

$$\begin{aligned} M_i(v^R)&= \beta _{k(N_{-i})} \biggl (\sum _{l=1}^{k(N_{-i})}d(N_l)-E+d_i\biggr ) +\sum _{l=k(N_{-i})+1}^{k-1}\beta _ld(N_l)\\&+ \beta _k \biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr ). \end{aligned}$$

This finishes 2b/2c. $\square$

Proof

(Theorem 4) Clearly, $m(v^R)\ge 0$ for an RA-game $v^R$. Let $S\in 2^N\setminus \{\emptyset \}$. According to Quant et al. (2005), it suffices to prove that

$$\begin{aligned} v^R(S)\le \max \left\{ 0,v^R(N)-\sum _{j\in N\setminus S}M_j(v^R)\right\} . \end{aligned}$$

(8)

For $D(S)=0$ it holds that $v^R(S)=0$ and inequality (8) is easily verified. Let $D(S)>0$. This implies that $v^R(S)>0$ and $d_i<E$ for all $i\in N\setminus S$. It remains to prove that

$$\begin{aligned} v^R(N)-\sum _{j\in N\setminus S}M_j(v^R) \ge v^R(S) \end{aligned}$$

(9)

Set $k=k(N)$. First let $\{N_l\setminus S_l|l\in \{k,\ldots ,m\}\}=\emptyset$. This tells us that

$$\begin{aligned} D(S)=E-\sum _{j\in N\setminus S}d_j=E-\sum _{l=1}^{k-1}d(N_l\setminus S_l), \end{aligned}$$

which implies that $k(S)=k$. Hence,

$$\begin{aligned} v^R(N)-\sum _{j\in N\setminus S}M_j(v^R)&= v^R(N)-\sum _{j\in N_l\setminus S_l:l<k}M_j(v^R)\\&= \sum _{l=1}^{k-1}\beta _ld(N_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&- \sum _{l=1}^{k-1}\beta _ld(N_l\setminus S_l)\\&= \sum _{l=1}^{k-1}\beta _ld(S_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&= \sum _{l=1}^{k-1}\beta _ld(S_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l\setminus S_l)-\sum _{l=1}^{k-1}d(S_l)\biggr )\\&= \sum _{l=1}^{k-1}\beta _ld(S_l)+\beta _k\biggl (D(S)-\sum _{l=1}^{k-1}d(S_l)\biggr )\\&= v^R(S). \end{aligned}$$

Here, the second and last equality follow from Theorem 2 and Lemma 3(2a).

Secondly, let $\{N_l\setminus S_l|l\in \{k,\ldots ,m\}\}\not =\emptyset$ and assume that $k(N_{-i})=k$ for all $i\in \bigcup \limits _{l=k}^{m}(N_l\setminus S_l)$. We will prove inequality (9) by showing

$$\begin{aligned} v^R(N)-\sum _{j\in N_l\setminus S_l:l<k}M_j(v^R)-v^R(S) \ge \sum _{j\in N_l\setminus S_l:l\ge k}M_j(v^R). \end{aligned}$$

For this,

$$\begin{aligned}&v^R(N)-\sum _{j\in N_l\setminus S_l:l<k}M_j(v^R)-v^R(S)\\&= \sum _{l=1}^{k-1}\beta _ld(N_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )- \sum _{l=1}^{k-1}\beta _ld(N_l\setminus S_l) -\sum _{l=1}^{k(S)-1}\beta _ld(S_l)\\&-\beta _{k(S)}\biggl (E-\sum _{j\in N\setminus S}d_j-\sum _{l=1}^{k(S)-1}d(S_l)\biggr )\\&= \beta _{k(S)}\biggl (\sum _{l=1}^{k(S)}d(S_l)-E+\sum _{j\in N\setminus S}d_j\biggr ) +\sum _{l=k(S)+1}^{k-1}\beta _ld(S_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&\ge \beta _k\biggl (\sum _{l=1}^{k(S)}d(S_l)-E+\sum _{j\in N\setminus S}d_j+ \sum _{l=k(S)+1}^{k-1}d(S_l)+E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&= \beta _k\biggl (\sum _{l=k}^md(N_l\setminus S_l)\biggr )\\&= \beta _k\biggl (\sum _{j\in N_l\setminus S_l:l\ge k}d_j\biggr )\\&= \sum _{j\in N_l\setminus S_l:l\ge k}M_j(v^R). \end{aligned}$$

Here, the first equality follows from Theorem 2 and Lemma 3(2a). The inequality holds because $\beta _k<\beta _{l}$ for all $l\in \{k(S),\ldots ,k-1\}$. The last equality follows from the fact that $M_i(v^R)=\beta _kd_i$ for all $i\in \bigcup \limits _{l=k}^{m}(N_l\setminus S_l)$.

Thirdly, let $\{N_l\setminus S_l|l\in \{k,\ldots ,m\}\}\not =\emptyset$ and let agent $i^*\in \bigcup \limits _{l=k}^{m}(N_l\setminus S_l)$ be such that $k(N_{-i^*})<k$. Without loss of generality, assume that $d_{i^*}\ge d_j$ for all $j\in \bigcup \limits _{l=k}^m(N_l\setminus S_l)$. We will prove (9) by showing

$$\begin{aligned} v^R(N)-\sum _{j\in N_l\setminus S_l:l<k}M_j(v^R)-v^R(S)-M_{i^*}(v^R)\ge \sum \limits _{j\in N_l\setminus S_l:l\ge k,j\not =i^*}M_j(v^R). \end{aligned}$$

For this, observe that

$$\begin{aligned}&v^R(N)-\sum _{j\in N_l\setminus S_l:l<k}M_j(v^R)-v^R(S)-M_{i^*}(v^R)\\&= \sum _{l=1}^{k-1}\beta _ld(N_l)+\beta _k\biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )- \sum _{l=1}^{k-1}\beta _ld(N_l\setminus S_l) -\sum _{l=1}^{k(S)-1}\beta _ld(S_l)\\&-\beta _{k(S)}\biggl (E-\sum _{j\in N\setminus S}d_j-\sum _{l=1}^{k(S)-1}d(S_l)\biggr ) -\beta _{k(N_{-i^*})}\biggl (\sum _{l=1}^{k(N_{-i^*})}d(N_l)-E+d_i^*\biggr )\\&-\sum _{l=k(N_{-i^*})+1}^{k-1}\beta _ld(N_l)- \beta _k \biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&= \beta _{k(S)}\biggl (\sum _{l=1}^{k(S)}d(S_l)-E+\sum _{j\in N\setminus S}d_j\biggr )+ \sum _{l=k(S)+1}^{k-1}\beta _ld(S_l)\\&-\beta _{k(N_{-i^*})}\biggl (\sum _{l=1}^{k(N_{-i^*})}d(N_l)-E+d_{i^*}\biggr ) -\sum _{l=k(N_{-i^*})+1}^{k-1}\beta _ld(N_l)\\&= \beta _{k(S)}\biggl (\sum _{l=1}^{k(S)}d(S_l)-E+\sum _{j\in N\setminus S}d_j\biggr )+ \sum _{l=k(S)+1}^{k(N_{-i^*})-1}\beta _ld(S_l)\\&+\beta _{k(N_{-i^*})}\biggl (d(S_{k(N_{-i^*})})-\sum _{l=1}^{k(N_{-i^*})}d(N_l)+E-d_{i^*}\biggr ) \\&-\sum _{l=k(N_{-i^*})+1}^{k-1}\beta _ld(N_l\setminus S_l)\\&\ge \beta _{k(N_{-i^*})}\biggl (\sum _{l=1}^{k(S)}d(S_l)-E+\sum _{j\in N\setminus S}d_j+\sum _{l=k(S)+1}^{k(N_{-i^*})-1}d(S_l)\\&+d(S_{k(N_{-i^*})}) -\sum _{l=1}^{k(N_{-i^*})}d(N_l)+E-d_{i^*}-\sum _{l=k(N_{-i^*})+1}^{k-1}d(N_l\setminus S_l)\biggr )\\&= \beta _{k(N_{-i^*})}\biggl (\sum _{j\in N_l\setminus S_l:l\ge k}d_j-d_{i^*}\biggr )\\&= \beta _{k(N_{-i^*})}\biggl (\sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})=k}}d_j+\sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})<k;j\not =i^*}}d_j\biggr )\\&\ge \sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})=k}}\beta _kd_j+\sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})<k;j\not =i^*}}\beta _{k(N_{-j})}d_j \\&= \sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})=k}}M_j(v^R)+\sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})<k;j\not =i^*}}\beta _{k(N_{-j})}\biggl (\sum _{l=1}^{k(N_{-j})}d(N_l) -E+d_j\\&+\sum _{l=k(N_{-j})+1}^{k-1}d(N_l)+E-\sum _{l=1}^{k-1}d(N_l)\biggr )\\&\ge \sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})=k}}M_j(v^R)+\sum _{\stackrel{j\in N_l\setminus S_l:l\ge k;}{k(N_{-j})<k;j\not =i^*}}\Biggl (\beta _{k(N_{-j})} \biggl (\sum _{l=1}^{k(N_{-j})}d(N_l)-E+d_j\biggr )\\&+\sum _{l=k(N_{-j})+1}^{k-1}\beta _ld(N_l)+\beta _k \biggl (E-\sum _{l=1}^{k-1}d(N_l)\biggr )\Biggr )\\&= \sum _{j\in N_l\setminus S_l:l\ge k,j\not =i^*}M_j(v^R). \end{aligned}$$

The first equality follow from Theorem 2 and Lemma 3(2a,2b). The third equality holds by using Lemma 1 such that $k(N_{-i^*})\ge k(S)$ for $i^*\in N\setminus S$. The first and second inequality are due to the fact that $\beta _{k(N_{-i^*})}<\beta _l$ for all $l\in \{k(S),\ldots ,k(N_{-i^*})-1\}, \beta _{k(N_{-i^*})}>\beta _l$ for all $l\in \{k(N_{-i^*})+1,\ldots ,k-1\}$, and $\beta _{k(N_{-i^*})}\ge \beta _{k(N_{-j})}$ for all $j\in \bigcup _{l=k}^m(N_l\setminus S_l)$. $\square$