Overlapping multiple object assignments

Abstract

This paper studies an allocation problem with multiple object assignments, indivisible objects, no endowments and no monetary transfers. Agents have complete, transitive and strict preferences over bundles of objects. A rule assigns objects to agents. A single object may be assigned to several agents as long as the agents satisfy a compatibility constraint. If no restrictions are imposed on the compatibility structure, there exists no rule that satisfies Pareto efficiency and compatibility-monotonicity. Imposing two restrictions on the compatibility structure, the class of rules called compatibility-sorting sequential dictatorships can be fully characterized by four different combinations of group-strategyproofness, strategyproofness, Pareto efficiency, non-bossiness, compatibility-monotonicity and compatibility-invariance. It is demonstrated that the characterization in Pápai (J Public Econ Theory 3:258–271, 2001) of sequential dictatorships for the case where assignments are not allowed to overlap is contained as a special case of the main result. Finally, some additional properties are considered and an extension of the model introducing capacity constraints is presented.

Appendices

Appendix 1: Proof of Lemma 4

Lemma 4 will be proven in this appendix. For convenience, the lemma is restated below.

Lemma 4

Suppose $\varphi$ is a strategyproof, Pareto efficient, non-bossy and compatibility-monotonic rule. Let $L\subseteq N$ either be the empty set or the union of one or more maximal compatible sets such that

• $C_{ij}=0$ for all $i\in L$ and all $j\in N{\setminus } L$, and

• there exists some $\check{R}\in \mathcal {R}^n$ such that $\varphi _i(R_{-L}, \check{R}_L)=\varphi _i(R^{\prime }_{-L}, \check{R}_L)$ for all $i\in L$ and all $R,R^{\prime }\in \mathcal {R}^n$.

Let $B\equiv A {\setminus } \bigcup _{i\in L} \varphi _i(R_{-L}, \check{R}_L)$. Then there exists some $K\in \mathcal {K}$ such that $K\subseteq N{\setminus } L$ and $\varphi _i(R_K, R^{\prime }_{-K\cup L}, \check{R}_L)=c(B,R_i)$ for all $i\in K$ and all $R, R^{\prime }\in \mathcal {R}^n$.

Proof

Let $\varphi$ be a strategyproof, Pareto efficient, non-bossy and compatibility-monotonic rule. Let all preference relations in $\bar{R}$ rank B first and $\emptyset$ second. It will first be demonstrated that whenever all agents not in L report $\bar{R}$, there exists some maximal compatible set K such that all agents in K are assigned B. It will then be shown that each agent not contained in K is assigned the empty set and that no such agent can be assigned a non-empty set by unilaterally changing his reported preferences. After this, an inductive argument will be constructed to show that for any set Z containing no agents in K or L, it is impossible for the agents in Z to change their preferences such that some agent in Z will be receive a non-empty assignment. Finally, it will be demonstrated that no matter which preferences the agents in K report, they will be assigned their most preferred subsets of B.

By Pareto efficiency, the set $K\subseteq N {\setminus } L$ containing each agent $i\in N {\setminus } L$ for whom $\varphi _i(\bar{R}_{-L}, \check{R}_L)=B$ is a non-empty maximal compatible set. If $K=N{\setminus } L$, then all agents in $N{\setminus } L$ are compatible and $\varphi _i(R_{-L}, \check{R}_L)=c(B,R_i)$ for all $i\in N {\setminus } L$ and all $R\in \mathcal {R}^n$. Assume that $K\subset N{\setminus } L$. Note that $\varphi _i(\bar{R}_{-L}, \check{R}_L)=\emptyset$ for all $i\in N{\setminus } (K\cup L)$ by feasibility. Consider an agent $j\in N{\setminus } (K\cup L)$, some arbitrary $R_j\in \mathcal {R}$ and recall that $R_j^B$ is defined by letting B be the highest ranked bundle, while letting the relative order of all others bundles be the same as at $R_j$. By strategyproofness, there exists no $R_j\in \mathcal {R}$ such that $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=B$, since $\varphi _j(\bar{R}_{-L}, \check{R}_L)=\emptyset$ and $B\bar{R}_j\emptyset$. Again, strategyproofness implies that $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\varphi _j(R_j^B,\bar{R}_{-L\cup \{j\}}, \check{R}_L)$. Non-bossiness implies that $\varphi (R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\varphi (R_j^B,\bar{R}_{-L\cup \{j\}}, \check{R}_L)$.

It can be shown by contradiction that $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\emptyset$. Assume that $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)\ne \emptyset$. Pareto efficiency implies that any agent $i\in N{\setminus } L\cup \{j\}$ is assigned either B or $\emptyset$. If $M^{\prime }$ is the (possibly empty) set containing each agent in $N{\setminus } L \cup \{j\}$ assigned B at $\varphi (R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)$, then $C_{ij}=1$ for all $i\in M^{\prime }$ by feasilibity. Since it has already been established that $\varphi (R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\varphi (R_j^B,\bar{R}_{-L\cup \{j\}}, \check{R}_L)$, $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\varphi _j(R_j^B,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=B$ by Pareto efficiency, as j could feasibly be assigned B at $\varphi (R_j^B,\bar{R}_{-L\cup \{j\}}, \check{R}_L)$ without affecting the assignment of any other agent. However, this violates the observation above that there exists no $R_j\in \mathcal {R}$ such that $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=B$. Hence, $\varphi _j(R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L)=\emptyset$. Non-bossiness implies that $\varphi (R_j,\bar{R}_{-L\cup \{j\}}, \check{R}_L) = \varphi (\bar{R}_{-L}, \check{R}_L)$. It has then been shown that no agent in $N{\setminus }(K\cup L)$ can affect the allocation by unilaterally changing his reported preferences.

Next, it will be shown by induction on |Z| that $\varphi (R_Z,\bar{R}_{-Z\cup L}, \check{R}_L)= \varphi (\bar{R}_{-L}, \check{R}_L)$ for all $Z\subseteq N{\setminus } (K\cup L)$ and all $R_Z\in \mathcal {R}^{|Z|}$. The following induction basis has already been demonstrated.

Induction basis $\varphi (R_Z,\bar{R}_{-Z\cup L}, \check{R}_L)= \varphi (\bar{R}_{-L}, \check{R}_L)$ for all $Z\subseteq N{\setminus } (K\cup L)$ such that $|Z|=1$ and all $R_Z \in \mathcal {R}^{|Z|}$.

If $|N{\setminus } (K\cup L)|=1$, then the argument is complete. Suppose that $|N{\setminus } (K\cup L)|\ge 2$ and consider the following induction hypothesis.

Induction hypothesis Assume that $\varphi (R_Z,\bar{R}_{-Z\cup L}, \check{R}_L)= \varphi (\bar{R}_{-L}, \check{R}_L)$ for all $Z\subset N{\setminus } (K\cup L)$ such that $|Z|\le l < n - |K\cup L|$ and all $R_Z \in \mathcal {R}^{|Z|}$.

The induction hypothesis holds for $l=1$ by the induction basis.

Induction step It can be proven by contradiction that the induction hypothesis holds for $l+1$ as well. Assume there exists some $H\subseteq N{\setminus } (K\cup L)$ such that $|H|=l+1$ and $\varphi (R_H,\bar{R}_{-H\cup L}, \check{R}_L)=\mu \ne \varphi (\bar{R}_{-L},\check{R}_L)$. By strategyproofness and the induction hypothesis, $\mu (i)\ne B$ for all $i\in H$. To see this, note that $B\bar{R}_i B^{\prime }$ for all $B^{\prime }\subseteq A$ and all $i\in H$. If $\mu (i)=B$ for some $i\in H$, then $\mu (i)\bar{P}_i\varphi _i(R_{H{\setminus }\{i\}}, \bar{R}_{-(H{\setminus }\{i\})\cup L}, \check{R}_L)$. This can be seen by noting that $\varphi _i(R_{H{\setminus }\{i\}}, \bar{R}_{-(H{\setminus }\{i\})\cup L}, \check{R}_L)=\emptyset$ by the induction hypothesis, as $|H{\setminus }\{i\}|=l$. Since this violates strategyproofness, $\mu (i)\ne B$ for all $i\in H$. If $\mu (i)=\emptyset$ for some $i\in H$, then $\varphi (R_H,\bar{R}_{-H\cup L}, \check{R}_L)=\varphi (R_{H{\setminus }\{i\}}, \bar{R}_{-(H{\setminus }\{i\})\cup L}, \check{R}_L) = \mu$ by non-bossiness. Since $|H{\setminus }\{i\}|=l$, the induction hypothesis then implies that $\varphi (R_{H{\setminus }\{i\}}, \bar{R}_{-(H{\setminus }\{i\})\cup L}, \check{R}_L)=\varphi (\bar{R}_{-L}, \check{R}_L)$, which violates the assumption that $\varphi (\bar{R}_{-L}, \check{R}_L)\ne \mu$. Hence, $\mu (i)\ne \emptyset$ and $\mu (i)\ne B$ for all $i\in H$.

Strategyproofness and non-bossiness then imply that $\varphi (R_H,\bar{R}_{-H\cup L}, \check{R}_L)=\varphi (R^B_i,R_{H{\setminus }\{i\}},\bar{R}_{-H\cup L}, \check{R}_L)$ for any $i\in H$. Note that $C_{ii^{\prime }}=0$ for all $i\in K$ and all $i^{\prime }\in H$. Hence, $\mu (i)=\emptyset$ for all $i\in K$ by Pareto efficiency. Suppose there exists some $i\in N{\setminus } (H\cup K \cup L)$ such that $C_{ii^{\prime }}=1$ for all $i^{\prime }\in H$. Note that $B \bar{R}_{i} B^{\prime }$ for all $B^{\prime }\subseteq A$. Pareto efficiency then implies that $\mu (i)=B$. However, note that for any $i^{\prime }\in H$, $\varphi _i(R_{H{\setminus }\{i^{\prime }\}}\bar{R}_{-(H{\setminus } \{i^{\prime }\})\cup L}, \check{R}_L)=\varphi _{i^{\prime }}(R_{H{\setminus }\{i^{\prime }\}}\bar{R}_{-(H{\setminus } \{i^{\prime }\})\cup L}, \check{R}_L)=\emptyset$ by the induction hypothesis. Furthermore, since $B \bar{P}_i \emptyset$, $C_{ii^{\prime }}=1$ and $\emptyset \subset \mu (i^{\prime })$, this violates compatibility-monotonicity. Hence, $\mu (i)=\emptyset$ for all $i\in N{\setminus } (H\cup L)$.

It can be proven by contradiction that all agents in H are incompatible. Suppose there exist two agents $i,i^{\prime }\in H$ such that $C_{ii^{\prime }}=1$. Recall that $\mu (i)\ne \emptyset$, $\mu (i^{\prime })\ne \emptyset$. Pareto efficiency implies that $\mu (i)P_i\emptyset$ and $\mu (i^{\prime })P_{i^{\prime }}\emptyset$. Furthermore, $\varphi _i(R_{H{\setminus }\{i^{\prime }\}}\bar{R}_{-(H{\setminus } \{i^{\prime }\})\cup L}, \check{R}_L)=\varphi _{i^{\prime }}(R_{H{\setminus }\{i^{\prime }\}}\bar{R}_{-(H{\setminus } \{i^{\prime }\})\cup L}, \check{R}_L)=\emptyset$. Since $\mu (i^{\prime })P_{i^{\prime }}\emptyset$ and $\emptyset \subset \mu (i)$, this violates compatibility-monotonicity. Hence, $C_{ii^{\prime }}=0$ for all $i,i^{\prime }\in H$.

Consider some $j,k \in H$. First note that no agent in $N{\setminus } (H\cup L)$ prefers any $B^{\prime }\subset A$ to $\emptyset$. Since $\mu (i)\ne B$ for all $i\in N$, strategyproofness and non-bossiness imply that $\varphi (R^B_H,\bar{R}_{-H\cup L},\check{R}_L)=\mu$. Let $R^{\prime }_i$ rank B first, $\mu (i)$ second and $\emptyset$ third for all $i\in N$. By strategyproofness and non-bossiness, $\varphi (R^{\prime }_H,\bar{R}_{-H\cup L},\check{R}_L)=\mu$. Let $R^{\prime \prime }$ rank B first, $\mu (j)$ second, $\mu (k)$ third and $\emptyset$ fourth. By strategyproofness, Pareto efficiency and non-bossiness, $\varphi (R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L)=\mu$. To see this, note that $\varphi (R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L) \in \{\mu (j),\mu (k), B\}$ by strategyproofness. If $\varphi (R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L) \in \{\mu (j),B\}$, then $\varphi _j(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L) =\emptyset$ by Pareto efficiency. This violates non-bossiness, since $\varphi _j(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L) = \varphi _j(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k,j\}},\bar{R}_{-H\cup L \cup \{j\}},\check{R}_L) = \emptyset$ by the induction hypothesis, while $\varphi _k(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k,j\}},\bar{R}_{-H\cup L \cup \{j\}},\check{R}_L)=\emptyset \ne \varphi _k(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L)$. Hence, $\varphi _k(R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L)=\mu (k)$ and $\varphi (R^{\prime \prime }_k, R^{\prime }_{H{\setminus } \{k\}},\bar{R}_{-H\cup L}, \check{R}_L)=\mu$ by non-bossiness. In general, non-bossiness is violated whenever there exist some $i,i^{\prime }\in H$ such that i is assigned a non-empty bundle and $i^{\prime }$ is assigned the empty set. Strategyproofness implies that

$$\begin{aligned} \varphi \left( R^{\prime \prime }_{\{j,k\}}, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L\right) =\mu . \end{aligned}$$

(1)

Let $\bar{R}^{\prime }$ rank B first, $\mu (k)$ second, $\mu (j)$ third and $\emptyset$ fourth. Let

$$\begin{aligned} \varphi \left( R^{\prime \prime }_k, \bar{R}^{\prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L\right) =\nu . \end{aligned}$$

By Pareto efficiency, $\nu (j)=\mu (k)$, $\nu (k)=\mu (j)$. Furthermore, $\nu (i)=\mu (i)$ for all $i\in N{\setminus }\{j,k\}$ by Pareto efficiency. Let $\bar{R}^{\prime \prime }$ rank B first, $\mu (j)$ second and $\emptyset$ third. By strategyproofness and non-bossiness, $\varphi (\bar{R}^{\prime \prime }_k, \bar{R}^{\prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\nu$. Recall that $R^{\prime \prime }$ ranks B first, $\mu (j)$ second, $\mu (k)$ third and $\emptyset$ fourth and note that strategyproofness and non-bossiness imply that $\varphi (\bar{R}^{\prime \prime }_k, R^{\prime \prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\nu$. To see this, note that if $\varphi _j(\bar{R}^{\prime \prime }_k, R^{\prime \prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\mu (j)$, then $\varphi _k(\bar{R}^{\prime \prime }_k, R^{\prime \prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\emptyset$ by Pareto efficiency. By the same argument as above, this violates non-bossiness. Hence, $\varphi _k(\bar{R}^{\prime \prime }_k, R^{\prime \prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\nu (k)=\mu (j)$. Recall Eq. (1) and note that $\varphi _k(R^{\prime \prime }_{\{j,k\}}, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)=\mu (k)$, while $\mu (j)P^{\prime \prime }_k\mu (k)$. This implies that $\varphi _k(\bar{R}^{\prime \prime }_k, R^{\prime \prime }_j, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L) P^{\prime \prime }_k \varphi _k(R^{\prime \prime }_{\{j,k\}}, R^{\prime }_{H{\setminus } \{j,k\}},\bar{R}_{-H\cup L},\check{R}_L)$, which violates strategyproofness. Hence, the induction hypothesis holds for $l+1$ as well. This concludes the induction step.

It has thereby been shown by the inductive argument above that $\varphi (R_Z,\bar{R}_{-Z\cup L}, \check{R}_L)= \varphi (\bar{R}_{-L}, \check{R}_L)$ for all $Z\subseteq N{\setminus } (K\cup L)$ and all $R\in \mathcal {R}^n$. Hence, $\varphi _i(\bar{R}_K, R_{-K\cup L},\check{R}_L)=B$ for all $i\in K$ and all $R\in \mathcal {R}^n$.

Next, it will be shown that no matter which preferences the agents in K report, they will be assigned their most preferred subsets of B. Fix a set of objects $B_i \subseteq B$ for each $i\in K$. Let all preference relations in $R^{\prime }_i$ rank $B_i$ first, B second if $B_i\ne B$ and $\emptyset$ after B. Suppose there exist some $i\in K$ and $R\in \mathcal {R}^n$ such that $\varphi _i(R^{\prime }_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)\ne B_i$. Then strategyproofness implies that $\varphi _i(R^{\prime }_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)=B$, which violates Pareto efficiency. It violates Pareto efficiency because the allocation at which all agents in $K{\setminus }\{i\}$ are assigned B and i is assigned $B_i$ is feasible and Pareto dominates the allocation at which every agent in K is assigned B. Since $B_i$ is an arbitrary set of objects, there exists, for each $B^{\prime }\subseteq B$ and each $i\in K$, some $\hat{R}_i$ such that $\varphi _i(\hat{R}_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)=B^{\prime }$. Strategyproofness then implies that $\varphi _i(R_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)=c(B,R_i)$ for all $i\in K$ and all $R\in \mathcal {R}^n$. Furthermore, $\varphi _j(R_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)=B$ for each $j\in K {\setminus } \{i\}$ by compatibility-monotonicity, since $\varphi _i(R_i, \bar{R}_{K{\setminus }\{i\}}, R_{-K\cup L},\check{R}_L)\subseteq \varphi _i(\bar{R}_{K}, R_{-K\cup L},\check{R}_L)$ and $C_{ij}=1$. Consecutively replace $\bar{R}_j$ with $R_j$ for each $j\in K{\setminus }\{i\}$. By the same argument as above, strategyproofness and compatibility-monotonicity imply that $\varphi _i(R_{-L},\check{R}_L)=c(B,R_i)$ for all $i\in K$ and all $R\in \mathcal {R}^n$. Equivalently, $\varphi _i(R_K, R^{\prime }_{-K\cup L}, \check{R}_L)=c(B,R_i)$ for all $i\in K$ and all $R, R^{\prime }\in \mathcal {R}^n$. $\square$

Appendix 2: Capacity constraints and additional properties

This appendix contains all technical details, formal definitions and proofs relating to Sect. 6. All propositions will be restated in the appendix for convenience.

Up until this point, N and A have been fixed. To study the properties introduced in Sect. 6, the model must be amended to let both $N\subseteq \mathcal {N}$ and $A\subseteq \mathcal {A}$ be variable, where $\mathcal {N}$ is the agent space and $\mathcal {A}$ is the object space. That is, $\mathcal {N}$ is the set of all potential agents and $\mathcal {A}$ is the set of all potential objects. In this section, a rule is a collection of functions $\varphi =\{\varphi ^{N,A} : \mathcal {R}^n \rightarrow \mathcal {M} \mid N\subseteq \mathcal {N}, A \subseteq \mathcal {A}\}$ rather than a single function. In other words, the rule $\varphi$ selects an allocation $\varphi ^{N,A}(R)$ for each pair $\{N,A\}\in \mathcal {P(N)}\times \mathcal {P(A)}$ and each $R\in \mathcal {R}^n$. For convenience, $\varphi ^{N,A}(R)$ is denoted by $\varphi (R,N,A)$. Consistency and population-monotonicity are formally defined as follows.

Definition 10

A rule $\varphi$ is consistent if for all $N\subseteq \mathcal {N}$, all $A\subseteq \mathcal {A}$, all $R\in \mathcal {R}$, all non-empty $S\subseteq N$ and all $i\in S$, $\varphi _i(R,N,A)=\varphi _i(R_S,S,\bigcup _{j\in S}\varphi _j(R,N,K))$.

Definition 11

A rule is population-monotonic if for all $N\subseteq \mathcal {N}$, all $A\subseteq \mathcal {A}$, all non-empty $S\subseteq N$ and all $R\in \mathcal {R}^n$, either

• $\varphi _i(R_S,S,A)R_i\varphi _i(R,N,A)$ for all $i\in S$, or

• $\varphi _i(R,N,A) R_i \varphi _i(R_S,S,A)$ for all $i\in S$.

Next, let $\mathcal {F}^N$ be the set of all permutations F of N, where F is the defined as the bijective function $F:N \rightarrow \{1,2, \ldots , n\}$. Rather than a single function, a priority structure is redefined as a collection of functions $f=\{f^N : \mathcal {R}^n\rightarrow \mathcal {F}^N \mid N\in \mathcal {N}\}$ in this section. This means that a priority structure now selects an ordering of all agents $f^N(R)$ for each $N\in \mathcal {N}$ and each $R\in \mathcal {R}^n$. Denote the priority of agent $i\in N$ at $f^N(R)$ by $f_{R,N}(i)$.

Furthermore, the definitions of sequential rules, sequential dictatorships and compatibility-sorting sequential dictatorships are amended such that the requirements in Definitions 7, 8 and 9 must be satisfied for each pair $\{N,A\}\in \mathcal {P(N)}\times \mathcal {P(A)}$. For example, if f is an s-hierarchy tree, then it is necessary that $f_{R,N}^{-1}(1)=f_{R^{\prime },N}^{-1}(1)$ for all $N\subseteq \mathcal {N}$ and all $R,R^{\prime }\in \mathcal {R}^n$. However, it is not necessarily the case that $f_{R,N}^{-1}(1)=f_{R_S,S}^{-1}(1)$ when $S\ne N$. While serial dictatorships have been discussed in the paper, they have not yet been formally defined.

Definition 12

An s-hierarchy network f associated with some rule $\varphi$ is an exogenous s-hierarchy tree associated with $\varphi$ if for all $N,S\subseteq \mathcal {N}$,

• $f^N(R)=f^N(R^{\prime })$ for all $R,R^{\prime }\in \mathcal {R}^n$, and

• for all $i,j\in N \cap S$, $f_{R_S,S}(i)<f_{R_S,S}(j)$ if and only if $f_{R,N}(i)<f_{R,N}(j)$.

A rule $\varphi$ is a serial dictatorship if there exists an exogenous s-hierarchy tree associated with $\varphi$. Under a serial dictatorship, it is impossible for any agent to affect the priority structure. For a given $N\subseteq \mathcal {N}$, the agent with priority k at some preference profile is the agent with priority k at all preference profiles. If, for some $N,S\subseteq \mathcal {N}$, there are two agents $i,j\in N\cap S$ such that i has higher priority than j when the set of all agents is given by N, then i has higher priority than j when the set of all agents is given by S as well. Note that a serial dictatorship is a sequential dictatorship. A serial dictatorship can be a compatibility-sorting sequential dictatorship, but not all serial dictatorships are compatibility-sorting sequential dictatorships. Similarly, not all compatibility-sorting sequential dictatorships are serial dictatorships. The first result in Sect. 6 is that neither serial dictatorships nor compatibility-sorting sequential dictatorships are consistent or population-monotonic.

Proposition 2

Serial dictatorships and compatibility-sorting sequential dictatorships are not consistent or population-monotonic.

Proof

Suppose the set of all agents is given by $N=\{1,2,3\}$ and that the set of all objects is given by A. Let f be an exogenous s-hierarchy tree associated with some rule $\varphi$. Since $f^S(R)=f^S(R^{\prime })$ for all $S\subseteq \mathcal {N}$ and all $R,R^{\prime }\in \mathcal {R}^{|S|}$, $f^S(R)$ can be denoted by $f^S$ for all $S\subseteq \mathcal {N}$ and all $R\in \mathcal {R}^{|S|}$. Suppose $\mathcal {K}=\{K_1,K_2\}$, where $K_1=\{1,3\}$ and $K_2=\{2\}$. Let $S=\{2,3\}$, $f^N=(1,2,3)$, and let all preference relations in R rank A first. By the definition of an exogenous s-hierarchy tree, $f^S=(2,3)$. Since $\varphi$ is a serial dictatorship, $\varphi _2(R,N,A)=\emptyset$, $\varphi _2(R_S,S,A)=A$, $\varphi _3(R,N,A)=A$ and $\varphi _3(R_S,S,A)=\emptyset$. Note that $\varphi _2(R_S,S,A)P_2\varphi _2(R,N,A)$, but $\varphi _3(R,N,A)P_3 \varphi _3(R_S,S,A)$. This violates both consistency and population-monotonicity. Hence, serial dictatorships are not consistent or population-monotonic.

Next, suppose the set of all agents is given by $N=\{1,2,3\}$ and that the set of all objects is given by A. Furthermore, suppose that $\mathcal {K}=\{K_1,K_3\}$, where $K_1=\{1,2\}$ and $K_3=\{3\}$. Let f be defined such that $f^N(R)=(1,2,3)$ for all $R\in \mathcal {R}^3$ and $f^S(R_S)=(i,j)$ for all $R_S\in \mathcal {R}^2$ whenever $S\subset N$, $|S|=2$ and $i>j$. Then f is a compatibility-sorted s-hierarchy tree associated with some rule $\varphi$, since every priority structure is necessarily compatibility-sorted when there are only two agents. Consider such a compatibility-sorting sequential dictatorship $\varphi$. Let $S=\{2,3\}$ and let R rank A first. Then $\varphi _2(R,N,A)=A$, $\varphi _2(R,S,A)= \emptyset$, $\varphi _3(R,N,A)=\emptyset$ and $\varphi _3(R,S,A)=A$. This implies that $\varphi _2(R,N,A)P_2\varphi _2(R,S,A)$, while $\varphi _3(R,S,A) P_3 \varphi _2(R,N,A)$, which violates both consistency and population-monotonicity. Hence compatibility-sorting sequential dictatorships are not consistent or population-monotonic.$\square$

Next, resource-monotonicity is defined formally.

Definition 13

A rule $\varphi$ is resource-monotonic if for all $N\subseteq \mathcal {N}$, all $i\in N$, all $A,A^{\prime }\subseteq \mathcal {A}$ and all $R\in \mathcal {R}^{n}$, $\varphi _i(R,N,A)R_i\varphi _i(R,N,A^{\prime })$ whenever $A^{\prime }\subseteq A$.

The second result in Sect. 6 is that there exists no Pareto efficient and resource-monotonic rule, regardless of whether assignments are allowed to overlap.

Proposition 3

No Pareto efficient rule is resource-monotonic, even under the restriction that $C_{ij}=0$ for all $i,j\in N, i\ne j$.

Proof

Suppose the set of all agents is given by $N=\{1,2,3\}$, that the set of all objects is given by $A=\{a,b,c\}$ and that $C_{12}=0$. Let $\varphi$ be a Pareto efficient rule. Furthermore, let $R_1$ rank A first, $\{a,b\}$ second and $\emptyset$ third, let $R_2$ rank $\{b,c\}$ first and $\emptyset$ second and let $R_3$ rank $\emptyset$ first. There are only two Pareto efficient allocations when R is reported; $\mu$ where $\mu (1)=A$ and $\mu (2)=\mu (3)=\emptyset$ and $\nu$ where $\nu (1)=\nu (3)=\emptyset$ and $\nu (2)=\{b,c\}$. Suppose that $\varphi (R,N,A)=\mu$. Let $A^{\prime }=\{b,c\}$ and note that $\varphi _2(R,N,A^{\prime })=\{b,c\}$ by Pareto efficiency. Since $\{b,c\}P_2 \emptyset$ and $\{b,c\}\subset A$, this violates resource-monotonicity. Suppose that $\varphi (R,N,A)=\nu$. Let $A^{\prime }=\{a,b\}$ and note that $\varphi _1(R,N,A^{\prime })=\{a,b\}$ by Pareto efficiency. Since $\{a,b\}P_1 \emptyset$ and $\{a,b\}\subset A$, this violates resource-monotonicity.$\square$

To allow for capacity constraints, some new notation is necessary. Each object $a\in A$ has a capacity $q_a \in [1,+\infty ) \subseteq \mathbb {N}$. The capacity of an object determines how many agents may be assigned the object. Agents are not necessarily homogeneous in the sense that they expend the same amount of capacity of an object they are assigned. The capacity requirement of an agent i is denoted by $r_i\in [1,+\infty ) \subseteq \mathbb {N}$ and the capacity requirement of a set of agents S is denoted by $r(S)\equiv \sum _{i\in S}r_i$. A feasible allocation is defined as in Sect. 2, with the additional requirement that the capacity requirement of the set of agents assigned an object always be weakly smaller than the capacity of the object. Formally, an allocation $\mu$ is feasible if for all $a\in A$, $r(\mu ^{-1}(a)) \le q_a$ and $C_{ij}=1$ for all $i,j \in \mu ^{-1}(a)$. The following set is introduced for ease of notation. For a given priority structure f and a given rule $\varphi$, denote

$$\begin{aligned} Y_R(a,i) \equiv \varphi _a^{-1}(R) \cap \{i^{\prime } \in N \mid f_R(i^{\prime }) < f_R(i) \}. \end{aligned}$$

$Y_R(a,i)$ is the set containing each agent $i^{\prime }\in N$ satisfying the following conditions.

• (1) Agent $i^{\prime }$ is assigned a at $\varphi (R)$.

• (2) Agent $i^{\prime }$ has strictly higher priority than i at f(R).

The definition of $S_R(i)$ must be amended to take the capacity constraint into account. It is now defined as follows.

$$\begin{aligned} S_R(i) \equiv \left\{ a \in A \mid \left( r(Y_R(a,i)\cup \{i\}) \le q_a \right) \wedge \left( \exists K\in \mathcal {K} Y_R(a,i) \cup \{i\} \subseteq K\right) \right\} \end{aligned}$$

As before, $S_R(i)$ is the set of objects that could feasibly be assigned to agent i without changing the assignments of any agents with higher priority than i, given some rule $\varphi$, some priority structure f and some reported preference profile R. Taking only agents with higher priority than i at f(R) into account, i could be assigned any subset of $S_R(i)$ without violating the compatibility constraint or the capacity constraint. It is now the set containing each object $a\in A$ satisfying the following conditions.

• (1) All agents that both are assigned a and have strictly higher priority than i are compatible with i.

• (2) The capacity requirement of the set containing i and all agents that both are assigned a and have strictly higher priority than i is weakly lower than the capacity of a.

For example, suppose that $f_R(i)=1$, $f_R(j)=2$, $r_i=1$, $r_j=2$, $\varphi _i(R)=A^{\prime }\subset A$, $q_a=2$ for all $a\in A^{\prime \prime }\subset A^{\prime }$ and $q_a=3$ for all $a\in A {\setminus } A^{\prime \prime }$. Then $S_R(j)=A{\setminus } A^{\prime \prime }$ if $C_{ij}=1$ and $S_R(j)=A{\setminus } A^{\prime }$ if $C_{ij}=0$.

With this amendment to $S_R(i)$, the definitions of s-hierarchy networks and s-hierarchy trees in Sect. 2 can remain unchanged. The interpretation of a sequential rule is different in that it may not assign an agent i an object a if the capacity requirement of the set containing i and all agents with higher priority than i that are assigned a is strictly larger than the capacity of a. The third result in Sect. 6 is that sequential dictatorships satisfy the same properties in this more general model as they do in the model introduced in Sect. 3.

Proposition 4

When capacity constraints are imposed, sequential dictatorships are group-strategyproof, strategyproof, Pareto efficient and non-bossy.

Proof

For a given priority structure f and a given rule $\varphi$, denote

$$\begin{aligned} Q_R(i) \equiv \{a\in A \mid r(Y_R(a,i)\cup \{i\})>q_a\}. \end{aligned}$$

Under a sequential rule, $Q_R(i)$ can be thought of as the set of objects unattainable by i due to the capacity constraint. Consider the union of $\{i\}$ and the set of agents with higher priority than i at f(R) that are assigned a at $\varphi (R)$. $Q_R(i)$ is such that for all $a\in Q_R(i)$, the capacity requirement of this union exceeds the capacity of a. Note that under a sequential rule, objects in $A{\setminus } Q_R(i)$ may still be unattainable by i due to the compatibility constraint.

Sequential dictatorships are group-strategyproof This is a proof by contradiction. Consider a sequential dictatorship $\varphi$ and assume that it is not group-strategyproof. Then there exist some $R,R^{\prime }\in \mathcal {R}^{n}$ and some $M\subseteq N$ such that $\varphi _i(R^{\prime }_M,R_{-M}) R_i \varphi _i(R)$ for all $i\in M$ and $\varphi _j(R^{\prime }_M,R_{-M}) P_j \varphi _j(R)$ for at least one $j\in M$. Consider one such $j\in M$. If $f_R(j)=1$, then $Q_R(j)=Q_{R^{\prime }}(j)$ for all $R,R^{\prime }\in \mathcal {R}^n$. By the definition of a sequential dictatorship, $\varphi _j(R)=c(A{\setminus } Q_R(j),R_j)$ and $\varphi _j(R^{\prime }_M,R_{-M})=c(A{\setminus } Q_R(j),R^{\prime }_j)$. Since $c(A{\setminus } Q_R(j),R^{\prime }_j)P_j c(A{\setminus } Q_R(j),R_j)$ is a contradiction, $f_R(j)\ge 2$. In general, $c(S,R^{\prime }_i)P_i c(S,R_i)$ is a contradiction since both $c(S,R^{\prime }_i)$ and $c(S,R_i)$ are subsets of S, and no subset of S can be strictly preferred to the most preferred subset of S. It can be demonstrated that $\varphi _k(R^{\prime }_M,R_{-M}) = \varphi _k(R)$ for all $k\in N$ such that $f_R(k)<f_R(j)$ by induction on $f_R(k)$.

Induction basis Let $f_R(k)=1$ and note that $f_R^{-1}(1)=f_{R^{\prime }}^{-1}(1)$ and $Q_R(k)=Q_{R^{\prime }}(k)$ for all $R,R^{\prime }\in \mathcal {R}^n$ by the definition of a sequential dictatorship. If $k\notin M$, then $\varphi _k(R)=\varphi _k(R^{\prime }_M,R_{-M})=c(A{\setminus } Q_R(k),R_k)$ by the definition of a sequential rule. If $k\in M$, then $\varphi _k(R^{\prime }_M,R_{-M})=c(A{\setminus } Q_R(k),R^{\prime }_k)$ and $\varphi _k(R)=c(A{\setminus } Q_R(k),R_k)$. By assumption, $c(A{\setminus } Q_R(k),R^{\prime }_k)R_k c(A{\setminus } Q_R(k),R_k)$. Since $c(A{\setminus } Q_R(k),R^{\prime }_k)P_k c(A{\setminus } Q_R(k),R_k)$ is a contradiction and preferences are strict, $\varphi _k(R^{\prime }_M,R_{-M})=\varphi _k(R)$ in both cases.

Suppose that $f_R(j)=2$, then it follows from the induction basis that $S_{(R^{\prime }_M,R_{-M})}(j)=S_R(j)$. To see this, note that if $C_{jk}=1$, then $S_R(j)=A{\setminus } Q_R(j)$ and if $C_{jk}=0$, then $S_R(j)=A{\setminus }(\varphi _k(R)\cup Q_R(j))$. Here, $Q_R(j)$ is the set containing each object $a\in \varphi _k(R)$ for which $q_a<r_k+r_j$ and each object $a\notin \varphi _k(R)$ for which $q_a<r_j$. Since k is assigned the same set of objects at both $\varphi (R^{\prime }_M,R_{-M})$ and $\varphi (R)$, $f_{(R^{\prime }_M,R_{-M})}(j)=f_R(j)$ and $S_{(R^{\prime }_M,R_{-M})}(j)=S_R(j)$ both when $C_{jk}=1$ and $C_{jk}=0$. By the definition of a sequential dictatorship, $\varphi _j(R^{\prime }_M,R_{-M})=c(S_R(j),R^{\prime }_j)$ and $\varphi _j(R)=c(S_R(j),R_j)$. Since $c(S_R(j),R^{\prime }_j)P_j c(S_R(j),R_j)$ is a contradiction, $f_R(j)\ge 3$. Thus, $\varphi _l(R)=\varphi _l(R^{\prime }_M, R_{-M})$ for all l such that $f_R(l)\le 2$.

Induction hypothesis Let $f_R(k)=t$. Assume that $\varphi _l(R)=\varphi _l(R^{\prime }_M, R_{-M})$ for all $l\in N$ such that $f_R(l)\le t$.

Induction step Let $t<f_R(j)-1$. It follows from the induction hypothesis and the definition of a sequential dictatorship that $f_R^{-1}(t+1)=f^{-1}_{(R^{\prime }_M, R_{-M})}(t+1)\equiv i$ and, by extension, that $S_R(i)=S_{(R^{\prime }_M,R_{-M})}(i)$. If $i\notin M$, then $\varphi _i(R)=c(S_R(i),R_i)=c(S_{(R^{\prime }_M,R_{-M})}(i),R_i) =\varphi _i(R^{\prime }_M,R_{-M})$. If $i\in M$, then $\varphi _i(R)=c(S_R(i),R_i)$ and $\varphi _i(R^{\prime }_M,R_{-M})=c(S_R(i),R^{\prime }_i)$. By assumption, $c(S_R(i),R^{\prime }_i)R_i c(S_R(i),R_i)$. Since $c(S_R(i),R^{\prime }_i)P_i c(S_R(i),R_i)$ is a contradiction and preferences are strict, $\varphi _i(R^{\prime }_M,R_{-M})=\varphi _i(R)$.

Hence, the induction hypothesis holds for all $k\in N$ such that $f_R(k)<f_R(j)$. In other words, $\varphi _k(R^{\prime }_M,R_{-M}) = \varphi _k(R)$ for all k such that $f_R(k)<f_R(j)$. By this result and the definition of a sequential dictatorship, $f_{(R^{\prime }_M,R_{-M})}(j)=f_R(j)$. This implies that $S_{(R^{\prime }_M,R_{-M})}(j)=S_R(j)$ and $c(S_{(R^{\prime }_M,R_{-M})}(j),R^{\prime }_j) = c(S_R(j),R^{\prime }_j) = \varphi _j(R^{\prime }_M,R_{-M})$. Since $j\in M$ and $\varphi _j(R^{\prime }_M,R_{-M}) P_j \varphi _j(R)$ by assumption, it follows that $c(S_R(j),R^{\prime }_j) P_j c(S_R(j),R_j)$, which is a contradiction. Sequential dictatorships are thus group-strategyproof.

Sequential dictatorships are Pareto efficient Let $\varphi$ be a sequential dictatorship and define $f_R^{-1}(j)\equiv i_j$ for all $j \le |N|$. It can be shown by induction on t that there exist no $t\ge 1$ and $\mu \in \mathcal {M}$ such that $\mu (i_s) R_{i_s}\varphi _{i_s}(R)$ for all $s\le t$ and $\mu (i_s) P_{i_s}\varphi _{i_s}(R)$ for some $s\le t$.

Induction basis There exists no $\mu \in \mathcal {M}$ such that $\mu (i_1) P_{i_1} \varphi _{i_1}(R)$, since $\varphi _{i_1}(R) = c(A{\setminus } Q_R(j), R_{i_1})$.

Induction hypothesis Let $t\ge 1$. Assume there exists no $\mu \in \mathcal {M}$ such that $\mu (i_s) R_{i_s}\varphi _{i_s}(R)$ for all $s\le t$ and $\mu (i_s) P_{i_s}\varphi _{i_s}(R)$ for some $s\le t$.

Induction step By the definition of a sequential rule, $\varphi _{i_{t+1}}(R) = c(S_R(i_{t+1}), R_{i_{t+1}})$. To reach a contradiction, assume that there exists some allocation $\mu \in \mathcal {M}$ such that $\mu (i_{t+1}) P_{i_{t+1}} \varphi _{i_{t+1}}(R)$ and $\mu (i_s) R_{i_s} \varphi _{i_s}(R)$ for all $s\le t+1$. Note that $\mu (i_{t+1}) P_{i_{t+1}} c(S_R(i_{t+1}), R_{i_{t+1}})$ implies that $\mu (i_{t+1})\nsubseteq S_R(i_{t+1})$. This implies that there must exist at least one $i_j \in N$ such that $j\le t$ and $\mu (i_j) \ne \varphi _{i_j}(R)$. This is either because there exists some $a\in \mu (i_{t+1})$ such that $a\in \mu (i_j) \cap Q_R(i_{t+1})$ or because $\mu (i_{t+1})\cap \mu (i_j)\ne \emptyset$ and $C_{i_{t+1}i_{j}}=0$. By assumption, $\mu (i_j) R_{i_j} \varphi _{i_j}(R)$. Since preferences are strict, $\mu (i_j) P_{i_j} \varphi _{i_j}(R)$. This, together with the assumption that $\mu (i_s) R_{i_s} \varphi _{i_s}(R)$ for all $s\le t+1$, contradicts the induction hypothesis. Hence, there exists no allocation $\mu \in \mathcal {M}$ such that $\mu (i_{t+1}) P_{i_{t+1}} \varphi _{i_{t+1}}(R)$ and $\mu (i_s) R_{i_s} \varphi _{i_s}(R)$ for all $s\le t+1$.

It has thus been shown that there exist no $t\ge 1$ and $\mu \in \mathcal {M}$ such that $\mu (i_s) R_{i_s}\varphi _{i_s}(R)$ for all $s\le t$ and $\mu (i_s) P_{i_s}\varphi _{i_s}(R)$ for some $s\le t$. This implies that there exists no $\mu \in \mathcal {M}$ such that $\mu (i) R_{i}\varphi _{i}(R)$ for all $i\in N$ and $\mu (i) P_{i}\varphi _{i}(R)$ for some $i\in N$. Hence, $\varphi$ is Pareto efficient.

Sequential dictatorships are strategyproof and non-bossy This follows immediately from group-strategyproofness and Lemma 1.$\square$

The final result in Sect. 6 is that compatibility-sorting sequential dictatorships are no longer compatibility-monotonic when capacity constraints are introduced.

Proposition 5

When capacity constraints are imposed, compatibility-sorting sequential dictatorships are not compatibility-monotonic.

Proof

Suppose that $N=\{1,2,3\}$, $r_1=r_2=r_3=1$, $C_{ij}=1$ for all $i,j \in N$, $A=\{a,b\}$ and $q_a=q_b=1$. Consider some compatibility-sorting sequential dictatorship $\varphi$ and a compatibility-sorted s-hierarchy tree f associated with $\varphi$, where $f_R=(1,2,3)$ for all $R\in \mathcal {R}$. Let $R_1$ rank $\{a\}$ first and $\emptyset$ second, let $R_2$ rank A first and $\emptyset$ second and let $R_3$ rank b first. Then $\varphi _1(R)=\{a\}$, $\varphi _2(R)=\emptyset$ and $\varphi _3(R)=\{b\}$. Next, let $R_1^{\prime }$ rank $\emptyset$ first. Then $\varphi _1(R_1^{\prime },R_{-1})=\emptyset$, $\varphi _2(R_1^{\prime },R_{-1})=A$ and $\varphi _3(R_1^{\prime },R_{-1})=\emptyset$. Note that $C_{13}=1$ and $\varphi _3(R) P_3 \varphi _3(R_1^{\prime },R_{-1})$, while $\varphi _1(R^{\prime }_1,R_{-1})=\emptyset \subset \{a\}=\varphi _1(R)$. Hence, $\varphi$ is not compatibility-monotonic. $\square$