Multi-commodity rationing problems with maxmin payoffs

Abstract

We address a multi-dimensional extension of standard rationing problems in which several commodities have to be shared among a set of agents who exhibit maxmin preferences on the results they obtain. In this context we investigate efficiency and introduce a property of stability which is supported on a transferable utility game. We also propose a procedure to construct rules for obtaining stable allocations for the special case where all commodities have the same weight.

Appendix (Proofs)

Proof of Theorem 3.2

First, if $y \in T_{\alpha }(c,E)$ then, $\sum _{i\in N} p^i_{\alpha }(y)= v_{\alpha }(N)$, and therefore, $y$ is an optimal solution to $P_{\alpha }(N)$.

$$\begin{aligned} \begin{array}{llll} &{} \displaystyle \max _x \sum _{i\in N} p^i_{\alpha }(x) \\ &{} s.t.\, \sum _{i\in N} x_j^i\le E_j,\quad j\in M&{} \quad \quad ( P_{\alpha }(N)) \\ &{} \quad \quad 0\le x_j^i\le c_j^i,\quad i\in N, j\in M\\ \end{array} \end{aligned}$$

Conversely, let $y$ be an optimal solution to problem $P_{\alpha }(N)$, and consider $x$, such that $x_j^i= \frac{\alpha _k^iy_k^i}{\alpha _j^i}$, where $\alpha _k^iy_k^i= min_j\{\alpha _j^iy_j^i\}$. It is easy to see that $x_j^i\le y_j^i$ for all $i\in N$ and $j\in M$. Therefore, the global allocation $x$ is feasible for $P_{\alpha }(N)$. Moreover, since $p^i_{\alpha }(x)= p^i_{\alpha }(y)$ for all $i\in N,\; x$ is also optimal. Note that for each $i\in N,\; \alpha _j^ix_j^i= \alpha _k^i x_k^i$ for all $j,k\in M$.

We will prove that $x \in T_{\alpha }(c,E)$. Suppose on the contrary that there exists $T\subset N$, such that $\sum _{i\in T} p^i_{\alpha }(x)< v_{\alpha }(T)$.³ It follows that there exists $y_T\ne x_T$ that solves problem $P_{\alpha }(T)$. Now, replace $x_T$ with $y_T$ in $x$. We first prove that the new allocation, $\bar{x}=(x_{N\setminus T}, y_T)$, is feasible in $P_{\alpha }(N)$. For $i\not \in T,\; x_j^i\le min\{c_j^i, E_j\}= d_j^i$. Let $k$ be such that $\alpha _k^i d_k^i= min_j\{\alpha _j^i d_j^i\}$. Hence, $\alpha _j^i x_j^i = \alpha _k^i x_k^i\le \alpha _k^i d_k^i$, and it follows that $x_j^i\le r_j^i$ for all $i\in N, j\in M$.

Now, for each $j$,

• (a) if $w^{\alpha }_j(T)=0$, then $y^i_j=0$ for all $i\in T$. Therefore, $\sum _{i\in T}y_j^i+\sum _{i\not \in T} x_j^i\le E_j$.

• (b) if $w^{\alpha }_j(T)=E_j-\sum _{i\not \in T} r_j^i$, then $\sum _{i\in T}y_j^i+\sum _{i\not \in T} x_j^i\le E_j-\sum _{i\not \in T} r_j^i+ \sum _{i\not \in T} r_j^i= E_j$.

However, $\sum _{i\in N} p_{\alpha }^i(x)=\sum _{i\in T}p_{\alpha }^i(y)+\sum _{i\not \in T} p_{\alpha }^i(x)> v_{\alpha }(N)$ and this contradicts, $x$ being an optimal solution to $v_{\alpha }(N)$. Therefore, $x \in T_{\alpha }(c,E)$, and since $x_j^i\le y_j^i$ for all $i\in N$ and $j\in M$, we also have $y \in T_{\alpha }(c,E)$. $\square$

Proof of Corollary 3.3

Consider an $\alpha$-maxmin stable global allocation $x$ an suppose to the contrary that it is not $\alpha$-maxmin efficient. It follows that there exists another global allocation, $y$, such that $p_{\alpha }^i(y)\ge p_{\alpha }^i(x)$ for all $i\in N$, with $p_{\alpha }^r(y)> p_{\alpha }^r(x)$, for at least one $r\in N$. Therefore $\sum _{i\in N}p_{\alpha }^i(y)> \sum _{i\in N}p_{\alpha }^i(x)$, and this contradicts the fact that $x$ solves $P_{\alpha }(N)$. $\square$

In order to prove Proposition 4.1, we first prove the following lemma

Lemma 6.1

For all $S\subseteq N$, the values $v_{\alpha }(S)$ can be obtained by solving the following linear problem:

$$\begin{aligned} \begin{array}{llll} v_{\alpha }(S) =&{} \displaystyle \max _z \sum _{i\in S} z^i \\ &{} s.t. \sum _{i\in S}\frac{1}{\alpha _i^j} z^i\le w^{\alpha }_j(S),\quad \forall j\in M&{} \quad (R_{\alpha }(S))\\ &{} \quad \quad 0\le z^i\le k^i,\quad i\in S\\ \end{array} \end{aligned}$$

where $k^i=min _{j\in M}\{\alpha _j^ic_j^i\}$.

Proof of Lemma 6.1

For $S\subseteq N$, consider problem $P_{\alpha }(S)$

$$\begin{aligned} \begin{array}{llll} v_{\alpha }(S) =&{} \displaystyle \max _x \sum _{i\in S} p_{\alpha }^i(x) \\ s.t. &{}\sum _{i\in S} x_j^i\le w^{\alpha }_j(S),\quad j\in M&{} \quad \quad (P_{\alpha }(S)) \\ &{} \quad \quad 0\le x_j^i\le c_j^i,\quad i\in S, j\in M\\ \end{array} \end{aligned}$$

By first doing $y_j^i=\alpha _j^ix_j^i$ for $i\in N, j\in M$, we obtain

$$\begin{aligned} \begin{array}{llll} v_{\alpha }(S) =&{} \displaystyle \max _y \sum _{i\in S} min_j\{y_j^i\} \\ &{} s.t. \sum _{i\in S}\frac{1}{\alpha _i^j} y_j^i\le w^{\alpha }_j(S),\quad \forall j\in M\\ &{} \quad \quad 0\le y_j^i\le \alpha _j^ic_j^i,\quad i\in S, j\in M\\ \end{array} \end{aligned}$$

Now, by considering $z^i= min_j\{y_j^i\}$, and as a consequence of the comprehensiveness of the sets $A_j=\{y_j\in \mathbb {R}^n|\, \sum _{i\in S} \frac{1}{\alpha _i^j}y_j^i\le w_j^{\alpha }(S), j\in M,0\le y_j^i\le \alpha _j^ic_j^i, i\in S, j\in M\}$, the optimal value can be found in the intersection, $\bigcap _{j\in M}A_j=\{z\in \mathbb {R}^n|\, \sum _{i\in S} \frac{1}{\alpha _i^j}z^i\le w^{\alpha }_j(S), j\in M,\, 0\le z^i\le \alpha _j^ic_j^i, i\in S, j\in M\}$, and the result follows. $\square$

Proof of Proposition 4.1

Let $x^*$ be global allocation for $(c,E)$, and $z_i^*=p^i(x^*)$. It follows from Lemma 6.1 that $x^*$ is a maxmin stable global allocation if and only if the corresponding $z^*$ is an optimal solution to problem $R_{\alpha }(N)$ for $\alpha =1,\; R(N)$:

$$\begin{aligned} \begin{array}{llll} &{} \displaystyle \max _z \sum _{i\in N} z^i \\ &{} s.t. \,\,\sum _{i\in N} z^i\le E_1\\ &{} \quad \quad 0\le z^i\le k^i,\quad i\in N,\\ \end{array} \end{aligned}$$

where $k^i=min _{j\in M}\{c_j^i\}$.

• (a) If $\sum _{i\in N} k^i\le E_1$, then the optimal value of $R(N)$ is $\sum k_i$ and is attained at $z=k$. Hence $x^*$ is maxmin stable if and only if $z^*=k$. Let $x^*$ be maxmin efficient and suppose to the contrary that $x^*$ is not maxmin stable. In this case $\sum z_i^*< \sum k_i\le E_1$. Since $z_i^*\le k_i$ for all $i\in N$, then for some $r\in N,\; z_r^*<k_r$. Take $\bar{z}_r= k_r\;\bar{z}_i=z_i$ for $i\ne r,\; \bar{z}$ is feasible for $R(N)$. Consider $\bar{x}$ such that $\bar{x}_r^j= k_r$ for all $j\in M$, $\bar{x}_i= x_i$ for $i\ne r$. It follows that $p^i(\bar{x})\ge p^i(x^*)$ with an strict inequality. This contradicts $x^*$ being maxmin efficient.

• (b) If $\sum _{i\in N} k^i> E_1$, then the optimal value of $R(N)$ is $E_1$. Hence $x^*$ is maxmin stable if and only if $\sum _{i\in N} p^i(x)=E_1$. Let $x^*$ be maxmin efficient and suppose to the contrary that $x^*$ is not maxmin stable. Let $z_i^*=p^i(x^*)$. In this case $\sum z_i^*< E_1< \sum k_i$. Since $z_i^*\le k_i$ for all $i\in N$, then for some $r\in N$, $z_r^*<k_r$. Take $\bar{z}_r= z_r^*+\varepsilon$ such that $\bar{z}_r\le k_r$ and $\sum \bar{z}_i \le E_1$, $\bar{z}_i=z_i$ for $i\ne r,\; \bar{z}$ is feasible for $R(N)$. Consider the global allocation $\bar{x}$ such that $\bar{x}_r^j= \bar{z}_r$ for all $j\in M$, $\bar{x}_i= x_i$ for $i\ne r$. It follows that $p^i(\bar{x})\ge p^i(x^*)$ with an strict inequality. This contradicts $x^*$ being maxmin efficient.

On the other hand, maxmin stable global allocations are maxmin efficient as stated in Corollary 3.3, and the result follows. $\square$

Proof of Proposition 4.3

For the case $\alpha _j^i=1,\; d_j^i= min \{c_j^i, E_j\},\; d_k^i = min_j\{d_j^i\}= min\{E_1, k^i\},\; r^i_j=r^i=min\{E_1, k^i\}$ does not depend on $j$, and therefore, $w_j(S)= max \{0, E_j-\sum _{i\not \in S} r^i\}$

The problem $R(S)$ is then

$$\begin{aligned} \begin{array}{llll} v(S) =&{} \max \sum _{i\in S} z^i \\ &{} s.t. \sum _{i\in S} z^i\le min_j \{ w_j(S)\}\\ &{} \quad \quad 0\le z^i\le k^i,\quad i\in S\\ \end{array} \end{aligned}$$

Now, $min_j \{ w_j(S)\}= Max \{0, E_1- \sum _{i\not \in S} r^i\}$, and therefore, for each $S\subseteq N$,

• (a) If $E_1-\sum _{i\in N\setminus S}r^i\le 0$, then $min_j \{ w_j(S)\}=0$, and $v(S)=0$.

• (b) If $E_1-\sum _{i\in N\setminus S}r^i\ge 0$, then

  $$\begin{aligned} \begin{array}{llll} v(S) =&{} \max \sum _{i\in S} z^i \\ &{} s.t. \sum _{i\in S} z^i\le E_1-\sum _{i\in N\setminus S}r^i\\ &{} \quad \quad 0\le z^i\le k^i,\quad i\in S\\ \end{array} \end{aligned}$$

  and the solutions to this linear problem are

  • (b1) If $\sum _{i\in S}k^i \ge E_1-\sum _{i\in N\setminus S}r^i$, then $v(S)= E_1-\sum _{i\in N\setminus S}r^i$.

  • (b2) If $\sum _{i\in S}k^i \le E_1-\sum _{i\in N\setminus S}r^i$, then $v(S)= \sum _{i\in S}k^i$.

It follows that

$$\begin{aligned} v_{\alpha }(S)= max\left\{ 0, min\left\{ \,E_1-\sum _{i\not \in S} r^i, \, \sum _{i \in S} k^i\right\} \right\} . \end{aligned}$$

From this expression, if $E_1\ge \sum _{i\in N}k^i$, then for any coalition $S\subseteq N,\; E_1\ge \sum _{i\in N\setminus S}k^i+\sum _{i\in S}k^i\ge \sum _{i\in N\setminus S}r^i+ \sum _{i\in S}k^i$, hence $E_1 - \sum _{i\in N\setminus S}r^i\ge \sum _{i\in S}k^i$, and $v(S)= \sum _{i\in S}k^i$.

On the other hand, when $E_1\le \sum _{i\in N}k^i$, we will prove that $E_1 - \sum _{i\in N\setminus S}r^i\le \sum _{i\in S}k^i$, for any $S\subset N$.

• (a) If $\exists i\in S,\; k^i>E_1$, then $k^i>r^i=E_1$ and $\sum _{i\in S}k^i> E_1$.

• (b) If $\exists i\in N\setminus S,\; k^i>E_1$, then $k^i>r^i=E_1$ and $\sum _{i\in N\setminus S}r^i\ge E_1$.

• (c) If $k^i\le E_1$ for all $i\in N$, then $E_1\le \sum _{i\in N}k^i= \sum _{i\in N\setminus S}r^i+ \sum _{i\in S}k^i$.

The result follows. $\square$