Fairness Criteria for Allocating Indivisible Chores:
Connections and Efficiencies^††thanks: A preliminary version of this paper appeared in Proceedings of AAMAS 2020 [37].

Fair division is a central matter of concern in economics, multiagent systems, and artificial intelligence [18, 16, 6]. Over the years, there emerges a tremendous demand for fair division when a set of indivisible resources, such as classrooms, tasks, and properties, are divided among a group of agents. This field has attracted the attention of researchers and most results are established when resources are considered as goods that bring positive utility to agents. However, in real-life division problems, the resources to be allocated can also be chores which, instead of positive utility, bring non-positive utility or cost to agents. For example, one might need to assign tasks among workers, teaching load among teachers, sharing noxious facilities among communities, and so on. Compared to goods, fairly dividing chores is relatively under-developed. At first glance, dividing chores is similar to dividing goods. However, in general, chores allocation is not covered by goods allocation and results established on goods do not necessarily hold on chores. Existing works have already pointed out this difference in the context of envy-freeness [14, 15, 19] and equitability [29, 30]. As an example, Freeman et al. [29] indicate that, when allocating goods, a leximin¹¹1A leximin solution selects the allocation that maximizes the utility of the least well-off agent, subject to maximizing the utility of the second least, and so on. allocation is Pareto optimal and equitable up to any item²²2Equitability requires that any pair of agents are equally happy with their bundles. In equitability up to any item allocations, the violation of equitability can be eliminated by removing any single item from the happier (in goods allocation)/ less happy agent (in chores allocation)., however, a leximin solution does not guarantee equitability up to any item in chores allocation.

Among the variety of fairness notions in the literature, envy-freeness (EF) is one of the most compelling, which has drawn research attention over the past few decades [28, 17, 26]. In an envy-free allocation, no agent envies another agent. Unfortunately, the existence of an envy-free allocation cannot be guaranteed in general when the items are indivisible. A canonical example is that one needs to assign one chore to two agents and the chore has a positive cost for either agent. Clearly, the agent who receives the chore will envy the other. In addition, deciding the existence of an EF allocation is computationally intractable, even for two agents with identical preference. Given this predicament, recent studies mainly devote to relaxations of envy-freeness. One direct relaxation is known as envy-free up to one item (EF1) [35, 20]. In an EF1 allocation, one agent may be jealous of another, but by removing one chore from the bundle of the envious agent, envy can be eliminated. A similar but stricter notion is envy-free up to any item (EFX) [22]. In such an allocation, envy can be eliminated by removing any positive-cost chore from the envious agent’s bundle. Another fairness notion, maximin share (MMS) [20, 3], generalizes the idea of “cut-and-choose” protocol in cake cutting. The maximin share is obtained by minimizing the maximum cost of a bundle of an allocation over all allocations. The last fairness notion we consider is called pairwise maximin share (PMMS) [22], which is similar to maximin share but different from MMS in that each agent partitions the combined bundle of himself and any other agent into two bundles and then receives the one with the larger cost.

The existing research on envy-freeness and its relaxations concentrates on algorithmic features of fairness criteria, such as their existence and (approximation) algorithms for finding them. Relatively little research studies the connections between these fairness criteria themselves, or the trade-off between these fairness criteria and the system efficiency, known as the price of fairness.

When allocating goods, Amanatidis et al. [2] compare the above four relaxations of envy-freeness and provide results on the approximation guarantee of one to another. However, these connections are unclear in allocating chores. On the price of fairness, Bei et al. [11] study allocation of indivisible goods and focus on the notions for which the corresponding fair allocations are guaranteed to exist, such as EF1, maximin Nash welfare³³3Nash welfare is the product of agents’ utilities., and leximin. Caragiannis et al. [21] study the price of fairness for both chores and goods, and focus on the classical fairness notions, namely, EF, proportionality⁴⁴4An allocation of goods (resp. chores) is proportional if the value (resp. cost) of every agent’s bundle is at least (resp. at most) one $n$-th fraction of his value (resp. cost) for all items. and equitability. To the best of our knowledge, no existing work covers the price of fairness for the aforementioned four relaxations (which we will call additive relaxations from time to time in this paper) of envy-freeness in chores allocation.

In this paper, we fill these gaps by investigating the four relaxations of envy-freeness on two aspects. On the one hand, we study the connections between these criteria and, in particular, we consider the following questions: Does one fairness criterion implies another? To what extent can one criterion guarantee for another? On the other hand, we study the trade-off between fairness and efficiency (or social cost defined as the sum of costs of the individual agents). Specifically, for each fairness criterion, we investigate its price of fairness, which is defined as the supremum ratio of the minimum social cost of a fair allocation to the minimum social cost of any allocation.

In the problem of a fair division of indivisible chores, we have a set $N=\{1,2,\ldots,n\}$ of agents and a set $E=\{e_{1},\ldots,e_{m}\}$ of indivisible chores. As chores are the items with non-positive values, each agent $i\in N$ is associated with a cost function $c_{i}:2^{E}\rightarrow R_{\geq 0}$, which maps any subset of $E$ into a non-negative real number. Throughout this paper, we assume $c_{i}(\emptyset)=0$ and $c_{i}$ is monotone, that is, $c_{i}(S)\leq c_{i}(T)$ for any $S\subseteq T\subseteq E$. We say a (set) function $c(\cdot)$ is:

• •

  Additive, if $c(S)=\sum_{e\in S}c(e)$ for any $S\subseteq E$.

• •

  Submodular,⁶⁶6An equivalent definition is as follows: $c(\cdot)$ is submodular if for any $S\subseteq T\subseteq E$ and $e\in E\setminus T,c(T\cup\{e\})-c(T)\leq c(S\cup\{e\})-c(S)$. if for any $S,T\subseteq E$, $c(S\cup T)+c(S\cap T)\leq c(S)+c(T)$.

• •

  Subadditive, if for any $S,T\subseteq E,c(S\cup T)\leq c(S)+c(T)$.

Clearly, additivity implies submodularity, which in turn implies subadditivity. For simplicity, instead of $c_{i}(\{e_{j}\})$, we use $c_{i}(e_{j})$ to represent the cost of chore $e_{j}$ for agent $i$.

An allocation $\mathbf{A}:=(A_{1},\ldots,A_{n})$ is an $n$-partition of $E$ among agents in $N$, i.e., $A_{i}\cap A_{j}=\emptyset$ for any $i\neq j$ and $\cup_{i\in N}A_{i}=E$. Each subset $S\subseteq E$ also refers to a bundle of chores. For any bundle $S$ and $k\in\mathbb{N}^{+}$, we denote by $\Pi_{k}(S)$ the set of all $k$-partition of $S$, and $|S|$ the number of chores in $S$.

Let us start with EF. According to the definitions, for any $\alpha\geq 1$, $\alpha$-EF is stronger than $\alpha$-EFX and $\alpha$-EF1. The following propositions present the approximation guarantee of $\alpha$-EF for MMS and PMMS.

Regarding tightness, consider the following instance with $n$ agents and $n^{2}$ chores denoted as $\{e_{1},\ldots,e_{n^{2}}\}$. Agents have an identical cost profile and for every $i\in[n]$, $c_{i}(e_{j})=\alpha$ for $1\leq j\leq n$ and $c_{i}(e_{j})=1$ for $n+1\leq j\leq n^{2}$. Consider allocation $\mathbf{B}=(B_{1},\ldots,B_{n})$ with $B_{i}=\{e_{(i-1)n+1},\ldots,e_{in}\}$ for any $i\in N$. It is not hard to verify that allocation $\mathbf{B}$ is $\alpha$-EF. As for $\textnormal{MMS}_{1}(n,E)$, since in total we have $n$ chores with each cost $\alpha$ and $(n-1)n$ chores with each cost 1, then in the partition defining $\textnormal{MMS}_{1}(n,E)$, each bundle contains exactly one chore with cost $\alpha$ and $n-1$ chores with cost 1. Consequently, we have $\textnormal{MMS}_{1}(n,E)=n-1+\alpha$ and the ratio $\frac{c_{1}(B_{1})}{\textnormal{MMS}_{1}(n,E)}=\frac{n\alpha}{n-1+\alpha}$.  $\Box$

Proposition 3.2 indicates that the approximation guarantee of $\alpha$-EF for PMMS is independent of the number of agents. However, according to Proposition 3.1, its approximation guarantee for MMS is affected by the number of agents. Moreover, this guarantee ratio converges to $\alpha$ as $n$ goes to infinity.

We remark that none of EFX, EF1, PMMS and MMS has a bounded guarantee for EF. We show this by a simple example. Consider an instance of two agents and one chore, and the chore has a positive cost for both agents. Assigning the chore to an arbitrary agent results in an allocation that satisfies EFX, EF1, PMMS and MMS, simultaneously. However, since one agent has a positive cost on his own bundle and zero cost on other agents’ bundle, such an allocation has an infinite approximation guarantee for EF.

Next, we consider approximation of EFX and EF1.

For the impossibility result, consider an instance with $n$ agents and $2n$ chores denoted as $\{e_{1},e_{2},\ldots,$ $e_{2n}\}$. Agents have identical cost profile. The cost function of agent 1 is: $c_{1}(e_{1})=p,c_{1}(e_{j})=1,\forall j\geq 2$ where $p\gg 1$. Now consider an allocation $\mathbf{B}=(B_{1},\ldots,B_{n})$ with $B_{i}=\{e_{2i-1},e_{2i}\},\forall i\in N$. It is not hard to see allocation $\mathbf{B}$ is EF1 and except for agent 1, no one else will envy the bundle of others. Thus, we only concern agent 1 when calculate the approximation guarantee for EFX. By removing chore $e_{2}$ from bundle $B_{1}$, $\frac{c_{1}(B_{1}\setminus\{e_{2}\})}{c_{1}(B_{j})}=\frac{p}{2}$ holds for any $j\in N\setminus\{1\}$, and the ratio $\frac{p}{2}\rightarrow\infty$ as $p\rightarrow\infty$.  $\Box$

Next, we consider the approximation guarantee of EF1 for MMS. In allocating goods, Amanatidis et al. [2] present a tight result that an $\alpha$-EF1 allocation is $O(n)$-MMS. In contrast, in allocating chores, $\alpha$-EF1 can have a much better guarantee for MMS.

By the definition of $\alpha$-EF1, for any $j\in N\setminus\{i\}$, $c_{i}(A_{i}\setminus\{{\bar{e}}\})\leq\alpha\cdot c_{i}(A_{j})$ holds. Then, by summing up $j$ over $N\setminus\{i\}$ and adding a term $\alpha c_{i}(A_{i})$ on both sides, the following holds,

$$\alpha\cdot\sum_{j\in N}c_{i}(A_{j})\geq(n-1+\alpha)c_{i}(A_{i})-(n-1)c_{i}({\bar{e}}).$$

(1)

From Lemma 2.1, we have $\textnormal{MMS}_{i}(n,E)\geq\max\{\frac{1}{n}c_{i}(E),c_{i}({\bar{e}})\}$, and by additivity, it holds that

$$n\alpha\textnormal{MMS}_{i}(n,E)\geq(n-1+\alpha)c_{i}(A_{i})-(n-1)\textnormal{MMS}_{i}(n,E).$$

(2)

Inequality (2) is equivalent to $\frac{c_{i}(A_{i})}{\textnormal{MMS}_{i}(n,M)}\leq\frac{n\alpha+n-1}{n-1+\alpha}$, as required.

As for tightness, consider the following instance with $n$ agents and a set $E=\{e_{1},\ldots,e_{n^{2}-n+1}\}$ of $n^{2}-n+1$ chores. Agents have an identical cost profile and for every $i\in[n]$, $c_{i}(e_{1})=\alpha+n-1$, $c_{i}(e_{j})=\alpha$ for any $2\leq j\leq n$ and $c_{i}(e_{j})=1$ for $j\geq n+1$. Now, consider an allocation $\mathbf{B}=\left\{B_{1},\ldots,B_{n}\right\}$ with $B_{1}=\left\{e_{1},\ldots,e_{n}\right\}$ and $B_{j}=\{e_{n+(n-1)(j-2)+1},\ldots,$ $e_{n+(n-1)(j-1)}\}$ for any $j\geq 2$. Then, we have $c_{i}(B_{j})=n-1$ for any $i\in[n]$ and $j\geq 2$. Accordingly, except for agent 1, no one else will violate the condition of $\alpha$-EF1 and MMS. As for agent 1, since $c_{1}(B_{1}\setminus\{e_{1}\})=(n-1)\alpha=\alpha c_{1}(B_{j}),\forall j\geq 2$, then we can claim that allocation $\mathbf{B}$ is $\alpha$-EF1. To calculate $\textnormal{MMS}_{1}(n,E)$, consider an allocation $\mathbf{T}=(T_{1},\ldots,T_{n})$ with $T_{1}=\{e_{1}\}$ and $T_{j}=\{B_{j}\cup\left\{e_{j}\right\}\}$ for any $2\leq j\leq n$. It is not hard to verify that $c_{1}(T_{j})=\alpha+n-1$ for any $j\in N$. Therefore, we have $\textnormal{MMS}_{1}(n,E)=\alpha+n-1$ implying the ratio $\frac{c_{1}(B_{1})}{\textnormal{MMS}_{1}(n,E)}=\frac{n\alpha+n-1}{n-1+\alpha}$, completing the proof.  $\Box$

We now study $\alpha$-EFX in terms of its approximation guarantee for MMS and provide upper and lower bounds for general $\alpha\geq 1$ or $n\geq 2$.

Case 1: $|A_{i}|=1$. Then $e^{*}$ is the unique element in $A_{i}$, and thus $c_{i}(A_{i})=c_{i}(e^{*})$. By Lemma 2.1, $c_{i}(e^{*})\leq\textnormal{MMS}_{i}(n,E)$ holds, and thus, $c_{i}(A_{i})\leq\textnormal{MMS}_{i}(n,E)$.

Case 2: $|A_{i}|\geq 2$. By the definition of $\alpha$-EFX, for any $j\in N\setminus\left\{i\right\}$, $c_{i}(A_{i}\setminus\left\{e^{*}\right\})\leq\alpha\cdot c_{i}(A_{j})$. Since $e^{*}\in\arg\min_{e\in A_{i}}c_{i}(e)$ and $|A_{i}|\geq 2$, we have $c_{i}(e^{*})\leq\frac{1}{2}c_{i}(A_{i})$. Then, the following holds,

$$\alpha\cdot c_{i}(A_{j})\geq c_{i}(A_{i})-c_{i}(e^{*})\geq\frac{1}{2}c_{i}(A_{i}),\qquad\forall j\in N\setminus\left\{i\right\}.$$

(3)

By summing up $j$ over $N\setminus\left\{i\right\}$ and adding a term $\alpha c_{i}(A_{i})$ on both sides of inequality (3), the following holds

$$\alpha\cdot c_{i}(E)=\alpha\cdot\sum_{j\in N\setminus\left\{i\right\}}c_{i}(A_{j})+\alpha\cdot c_{i}(A_{i})\geq\frac{n-1+2\alpha}{2}c_{i}(A_{i}).$$

(4)

On the other hand, from Lemma 2.1, we know $\textnormal{MMS}_{i}(n,E)\geq\frac{1}{n}c_{i}(E)$, which combines inequality (4) yielding the ratio

$$\frac{c_{i}(A_{i})}{\textnormal{MMS}_{i}(n,M)}\leq\frac{2n\alpha}{n-1+2\alpha}.$$

Regarding the lower bound $\frac{2n}{n+1}$, consider an instance with $n$ agents and a set $E=\left\{e_{1},e_{2},...,e_{2n}\right\}$ of $2n$ chores. Agents have identical cost profile and $c_{i}(e_{j})=\lceil\frac{j}{2}\rceil$ for any $i,j$. It is not hard to verify that for any $i\in[n]$, $\textnormal{MMS}_{i}(n,E)=n+1$. Then, consider the allocation $\mathbf{B}=(B_{1},...,B_{n})$ with $B_{1}=\left\{e_{2n-1},e_{2n}\right\}$ and $B_{i}=\{e_{i-1},e_{2n-i}\}$ for any $i\geq 2$. Accordingly, we have $c_{i}(B_{j})=n$ for any $i\in[n]$ and $j\geq 2$. Thus, except for agent 1, no one else will violate the condition of MMS and EFX. As for agent 1, since $c_{1}(B_{1}\setminus\{e_{2n}\})=c_{1}(B_{1}\setminus\{e_{2n-1}\})=n$, envy can be eliminated by removing any single chore . Hence, the allocation $\mathbf{B}$ is EFX and its approximation guarantee for MMS equals to $\frac{c_{1}(B_{1})}{\textnormal{MMS}_{1}(n,E)}=\frac{2n}{n+1}$, as required.

Next, for lower bound $\frac{2n\alpha}{2\alpha+2n-3}$, let us consider an instance with $n$ agents and a set $E=\{e_{1},...,e_{2n^{2}-2n}\}$ of $2n^{2}-2n$ chores. We focus on agent 1 with cost function $c_{1}(e_{j})=2\alpha$ for $1\leq j\leq n$ and $c_{1}(e_{j})=1$ for $j\geq n+1$. Consider the allocation $\mathbf{B}=(B_{1},...,B_{n})$ with $B_{1}=\{e_{1},...,e_{n}\},B_{2}=\{e_{n+1},...,e_{3n-2}\}$ and $B_{j}=\{e_{3n-1+(j-3)(2n-1)},\ldots,e_{3n-2+(j-2)(2n-1)}\}$ for any $j\geq 3$. Accordingly, bundle $B_{2}$ contains $2n-2$ chores and $B_{j}$ contains $2n-1$ chores for any $j\geq 3$. For any agent $i\geq 2$, her cost functions is $c_{i}(e)=0$ for $e\in B_{i}$ and $c_{i}(e)=1$ for $e\in E\setminus B_{i}$. Consequently, except for agent 1, no one else violate the condition of MMS and $\alpha$-EFX. As for agent 1, his cost on $B_{2}$ is the smallest over all bundles and $c_{1}(B_{1}\setminus\{e_{1}\})=2\alpha(n-1)=\alpha c_{1}(B_{2})$, as a result, the allocation $\mathbf{B}$ is $\alpha$-EFX. For $\textnormal{MMS}_{1}(n,E)$, it happens that $E$ can be evenly divided into $n$ bundles of the same cost (for agent 1), so we have $\textnormal{MMS}_{1}(n,E)=2\alpha+2n-3$ implying the ratio $\frac{c_{1}(B_{1})}{\textnormal{MMS}_{1}(n,E)}=\frac{2n\alpha}{2\alpha+2n-3}$, completing the proof.  $\Box$