Fair student placement

Abstract

We revisit the concept of fairness in the Student Placement framework. We declare an allocation as $\upalpha$-equitable if no agent can propose an alternative allocation that nobody else might argue to be inequitable. It turns out that $\upalpha$-equity is compatible with efficiency. Our analysis fills a gap in the literature by giving normative support to the allocations improving, in terms of efficiency, the Student Optimal Stable allocation.

Appendix

1.1 Appendix A: Existence of ${\upalpha }$-fair allocations

As we anticipated in Sect. 3 we proceed to prove Theorem 1 in a constructive way. The process can be described as follows: consider a given Student Placement problem $\mathbb {P}$, and a matching $\mu$. By interpreting $\mu$ as an initial endowment for each student, we can understand that $(\mathbb {P}; \mu )$ constitutes a ‘seating market’ where students are allowed to exchange the seats they have been allocated. This market exhibits some similarities with the ‘housing market’ introduced by Shapley and Scarf (1974). Therefore, the tools usually employed to solve housing problems might be useful to explore how exchanges are conducted in the seating markets.

Consider a fixed problem $\mathbb {P}$ and compute its Student Optimal Stable matching, $\mu ^\mathrm{{SO}}$. It can be obtained by applying the Deferred Acceptance algorithm (Gale and Shapley 1962). Associated with the pair $(\mathbb {P}, \mu ^\mathrm{{SO}})$ we describe, for each student, her preferences for exchanging, denoted $E_{i}$ as the linear ordering on I defined as follows:

• (a) For each two students h and k such that $\mu ^\mathrm{{SO}}(h) \ne \mu ^\mathrm{{SO}}(k)$, $h \ E_{i} \ k$ if, and only if, $\mu ^\mathrm{{SO}}(h) \succ _{i} \mu ^\mathrm{{SO}}(k)$; and

• (b) For each two distinct students h and k such that $\mu ^\mathrm{{SO}}(h) = \mu ^\mathrm{{SO}}(k) = s_{j} \in S \cup \{s_{o}\}$, $h \ E_{i} \ k$ if, and only if, $h \ P_{j} \ k$.⁸

Given the preferences for exchanging associated with each student, $E = (E_{i})_{i \in I}$ describes a profile of preferences for exchanging.

Note that the pair (I; E) accommodates the structure of a housing market (Shapley and Scarf 1974). We now describe how to Gale’s algorithm is applied to this problem so to calculate the unique core allocation for this market:

• Step 1. Build a directed graph whose nodes are the agents in I. This graph has n arcs connecting each student with her preferred ‘mate for exchanging,’ i.e., for each $i \in I$, there is an arc from i, pointing the maximal on I according $E_{i}$. Since there is exactly one arc starting at each of the n nodes, this graph must have at least one cycle.⁹ Moreover, no student is involved in two different cycles. Then, each student belonging to a cycle is definitively assigned the seat she is pointing in the cycle,¹⁰ and leaves the market. Let $\mu ^{\upvarepsilon }(i)$ denote the school assigned to i, when she is in a cycle.

  Let $I^{1}$ be the set of students not belonging to a cycle. Then, if $I^{1}$ is empty, the algorithm terminates producing matching $\mu ^{\upvarepsilon }$, previously described. Otherwise, go to step 2.

  $$\begin{aligned} \dots \end{aligned}$$

• Step t. A graph involving the students in $I^{t - 1}$ is generated. The nodes coincide with these students. There is an arc connecting each student in $I^{t - 1}$ to her preferred mate for exchanging, among the ones that are still in the market, according her preferences for exchanging $E_{i}$. As in the previous step, this graph has at least one cycle. Each student involved in a cycle is assigned the seat she is pointing to and leaves the market. Let $I^{t}$ be the set of students in $I^{t - 1}$ being not in a cycle in the graph built in this step. If $I^{t}$ is empty, the algorithm terminates producing matching $\mu ^{\upvarepsilon }$, described throughout steps 1 to t. Otherwise, go to step $t + 1$.

Note that, since for each t such that $I^{t} \ne \emptyset$, $I^{t} \subsetneq I^{t - 1}$, the algorithm ends in finite steps.

Previous to demonstrate that matching $\mu ^{\upvarepsilon }$ is $\upalpha$-fair, we will illustrate how to compute it through an example:

Example 3

Consider problem $\mathbb {P}$ involving 8 students, $I = \{1,\, \ldots ,\, i,\, \ldots ,\, 8\}$ and 4 schools, $S = \{a, b, c, d\}$, having 2 vacant seats each. The students’ preferences are

The priorities of the schools are

We first compute the Student Optimal Stable matching, $\mu ^\mathrm{{SO}}$. It is calculated by applying the Deferred Acceptance algorithm. At each step, each student applies for her preferred school—among the ones not having rejected her previously—and each school rejects the less prioritized students among those sending it an application, so to keep all its seats filled. The next table describes, for each step, the applications that each school receives, and which are accepted by the school.

Therefore, $\mu ^\mathrm{{SO}}$ is such that $\mu ^\mathrm{{SO}}(a) = \{1, 2\}$, $\mu ^\mathrm{{SO}}(b) = \{3, 7\}$, $\mu ^\mathrm{{SO}}(c) = \{5, 6\}$ and $\mu ^\mathrm{{SO}}(d) = \{4, 8\}$. Now, we describe the students’ preferences for exchanging. Recall that each student orders I, according the seat each student is allowed under $\mu ^\mathrm{{SO}}$—for instance, since $a \succ _{5} b$, $\mu ^\mathrm{{SO}}(1) = a$ and $\mu ^\mathrm{{SO}}(3) = b$, we have that $1 \ E_{5} \ 3$—and ties are broken in accordance with the school priorities, for instance, since $\mu ^\mathrm{{SO}}(1) = \mu ^\mathrm{{SO}}(2) = a$, and $1 \ P_{a} \ 2$, each student i should prefer to exchange her seat to 1 rather than to 2. Summarizing, the preferences for exchange are gathered in the following table:

Now, we can run Gale’s algorithm. At the first step, each student points her preferred student for exchanging. As Fig. 1 shows, this graph has one cycle involving students 1, 3 and 5. Therefore, each of these students obtains a seat at the school that the student she pointed got at $\mu ^\mathrm{{SO}}$. In other words, $\mu ^{\upvarepsilon }(1) = \mu ^\mathrm{{SO}}(3) = b$; $\mu ^{\upvarepsilon }(3) = \mu ^\mathrm{{SO}}(5) = c$; and $\mu ^{\upvarepsilon }(5) = \mu ^\mathrm{{SO}}(1) = a$.

Fig. 1

Gale’s algorithm, first step

Once the first step is concluded, and some students leave the market, the second step is similar to the previous one, taking into account that no (remaining) student can select any student that left the allocative process. Now, as showed in Fig. 2, students 2 and 6 are involved in a cycle. This implies that $\mu ^{\upvarepsilon }(2) = \mu ^\mathrm{{SO}}(6) = c$; and $\mu ^{\upvarepsilon }(6) = \mu ^\mathrm{{SO}}(2) = a$.

Fig. 2

Gale’s algorithm, second step

Figure 3 captures the graph constructed at the third step. Now, students 7 and 8 exchange the seats they have been allocated at $\mu ^\mathrm{{SO}}$.

Fig. 3

Gale’s algorithm, third step

Once step 3 is concluded, the unique remaining student is 4. The application of Gale’s algorithm indicates that this student must keep the seat that $\mu ^\mathrm{{SO}}$ assigned to her.

Note that, as illustrated in Example 3, when running Gale’s algorithm, all the students involved in a cycle at the first step obtain a seat at their preferred school. Similarly, all the students involved in a cycle at the second step are allocated a seat at their preferred school, provided that some seats are still unavailable because they have been already allocated at the previous step, and so on. This implies that $\mu ^{\upvarepsilon }$ is an efficient matching.

We can now formally prove Theorem 1.

Proof

Let $\mathbb {P}$ be a problem and $\mu ^\mathrm{{SO}}$ its Student Optimal Stable matching. Note that, when describing each student’s preferences for exchanging we have that for any two students i and h, $h \ E_{i} \ i$ whenever

• 1. $\mu ^\mathrm{{SO}}(h) \succ _{i} \mu ^\mathrm{{SO}}(i)$; or

• 2. $\mu ^\mathrm{{SO}}(h) = \mu ^\mathrm{{SO}}(i) = s_{j} \in S \cup \{s_{o}\}$ and $h \ P_{j} \ i$.

In particular, this implies that if $\mu ^\mathrm{{SO}}(h) = \mu ^\mathrm{{SO}}(i) = s_{j} \in S \cup \{s_{o}\}$ and $h \ P_{j} \ i$, for each student, say k, $h \ E_{k} \ i$. Therefore, two agents having a seat at the same school under $\mu ^\mathrm{{SO}}$ will not obtain a definitive seat, when running Gale’s algorithm, at the same step.

Moreover, since no student leaves the market at some step if she does not belong to a cycle, and each student points her best ‘student for exchanging’ according her preferences, we have that, for each i,

$$\begin{aligned} \mu ^{\upvarepsilon }(i) \succsim _{i} \mu ^\mathrm{{SO}}(i). \end{aligned}$$

Now, assume that $\mu ^{\upvarepsilon }$ is not $\upalpha$-fair. Since, as previously reported, it is efficient, it must fail to be $\upalpha$-equitable. Then, there should be a student i and a matching $\mu ^{\prime }$ such that $(i; \mu ^{\prime })$ constitutes an admissible $\upvarepsilon$-objection to $\mu ^{\upvarepsilon }$. This implies that there is $s_{j} = \mu ^{\prime }(i) \succ _{i} \mu ^{\upvarepsilon }(i) \succsim _{i} \mu ^\mathrm{{SO}}(i)$, and thus $s_{j} \in S$. Therefore, by the Blocking Lemma (Martínez et al. 2010, Theorem 1), matching $\mu$ fails to be equitable, which contradicts that $\upvarepsilon$-objection $(i; \mu ^{\prime })$ to $\mu ^{\upvarepsilon }$ was admissible. $\square$

1.2 Appendix B: Identifying the set of ${\upalpha }$-fair allocations

We now deal with proving Theorem 2. It establishes that, for a given problem $\mathbb {P}$, matching $\mu$ is $\upalpha$-fair if, and only if, it is efficient and, for each student i,

$$\begin{aligned} \mu (i) \succsim _{i} \mu ^\mathrm{{SO}}(i). \end{aligned}$$

(3)

Note that, since $\upalpha$-equity implies efficiency by definition, we only need to concentrate on the fulfillment of Condition (3) above.

Proof

For a given problem $\mathbb {P}$, let $\mu ^\mathrm{{SO}}$ be its Student Optimal Stable matching. Let $\mu$ be an $\upalpha$-fair matching. Therefore, it is efficient. Assume $\mu$ does not fulfill Condition (3) above. Then, there should be a student i such that

$$\begin{aligned} \mu ^\mathrm{{SO}}(i) \succ _{i} \mu (i). \end{aligned}$$

(4)

Note that this implies that there is some $s_{j} \in S$ such that $s_{j} = \mu ^\mathrm{{SO}}(i)$. Otherwise, $\mu$ is dominated by matching $\mu ^{\prime }$ such that $\mu ^{\prime }(i) = s_{o}$ and, for each student $h \ne i$, $\mu ^{\prime }(h) = \mu (h)$.

Efficiency also implies that $|\mu (s_{j})| = q_{j}$. Otherwise, $\mu$ is dominated by matching $\mu ^{\prime \prime }$ such that $\mu ^{\prime \prime }(i) = s_{j}$ and, for each student $h \ne i$, $\mu ^{\prime \prime }(h) = \mu (h)$. Moreover, $\upalpha$-equity of $\mu$ implies that for each $h \in \mu (s_{j})$, $h \ P_{j} \ i$. Note that, otherwise, $(i; \mu ^\mathrm{{SO}})$ constitutes an admissible $\upvarepsilon$-objection to $\mu$.

Since $i \in \mu ^\mathrm{{SO}}(s_{j}) {\setminus } \mu (s_{j})$ and $|\mu (s_{j})| = q_{j}$, there should be a student $h \in \mu (s_{j}) {\setminus } \mu ^\mathrm{{SO}}(s_{j})$. Since $h \ P_{j} \ i$ and $\mu ^\mathrm{{SO}}$ is equitable, it must be the case that $\mu ^\mathrm{{SO}}(h) \succ _{h} \mu (h)$, i.e. Condition (4) is also fulfilled for agent h.

By applying an iterative reasoning, and taking into account that the number of schools is finite, there is an ordered set of students $\{i_{t}\}_{t = 1}^{T}$ and schools $\{s^{t}\}_{t = 1}^{T}$, with $T \le m$, such that for each t (modulo T),

• (a) $\mu (i_{t}) = s^{t};$

• (b) $\mu ^\mathrm{{SO}}(i_{t}) = s^{t + 1}$; and

• (c) $s^{t + 1} \succ _{i_{t}} s^{t}$.

Note that the above implies that $\mu$ fails to be efficient. In fact, it is dominated by matching $\mu ^{\prime }$ such that for each $i \in \{i_{t}\}_{t = 1}^{T}$, $\mu ^{\prime }(i) = \mu ^\mathrm{{SO}}(i)$ and, for each $h \notin \{i_{t}\}_{t = 1}^{T}$, $\mu ^{\prime }(h) = \mu (h)$. A contradiction.

Now, consider $\mu$, an efficient matching that satisfies Condition (3). Assume that $\mu$ fails to be $\upalpha$-fair. Then, there should be a student i and matching $\mu ^{\prime }$ constituting an admissible $\upvarepsilon$-objection to $\mu$. This implies that

$$\begin{aligned} \mu ^{\prime }(i) \succ _{i} \mu (i) \succsim _{i} \mu ^\mathrm{{SO}}(i). \end{aligned}$$

(5)

Note that, in particular, Condition (5) implies that there is some $s_{j} \in S$ such that $s_{j} = \mu ^{\prime }(i)$. Therefore, by the Blocking Lemma (Martínez et al. 2010), $\mu ^{\prime }$ fails to be equitable. Therefore, $(i; \mu ^{\prime })$ fails to be admissible as an $\upvarepsilon$-objection to $\mu$. A contradiction. $\square$