Pairwise stability and strategy-proofness for college admissions with budget constraints

Transcription

1 Pairwise stability and strategy-proofness for college admissions with budget constraints Azar Abizada University of Rochester First draft: October 22, 2009 This version: January 5, 2012 Abstract We study assignment problem with budget constraints. Examples include assigning students to graduate schools, assigning graduate students as research assistants to faculty members each of whom has fixed research fund, a manager assigning workers to different projects with fixed total benefit, assigning post-doctoral candidates to universities etc. In graduate college admissions, each college has a fixed amount of money to distribute as stipends. Each college also has strict preferences overs groups of students. On the other hand, each student is matched with at most one college and receives a stipend from it. Each student has quasi-linear preferences over collegestipend bundles. Differently from earlier literature, in this paper, we specify a fixed budget for each college. One other different feature of our model is that colleges value money only to the extend that it allows them to enroll better students. We show that introducing budget constraint results in losing some of the previous results in the literature. We define pairwise stability and show that a pairwise stable allocation always exists. We construct an algorithm. The rule defined through this algorithm always selects a pairwise stable allocation. This rule is also strategy-proof for students: no student can ever benefit from misrepresenting his preferences. Finally, we show that starting from an arbitrary allocation, there exist a sequence of allocations, each allocation being obtained from the previous one by satisfying a blocking pair, such that the final allocation is pairwise stable. Department of Economics, University of Rochester. azar.abizada@rochester.edu I would like to thank William Thomson for his guidance, invaluable comments, and unending patience. I am grateful to Paulo Barelli, John Duggan, Srihari Govindan, Bettina Klaus, Asen Kochov, and Gabor Virag for helpful discussions and their useful comments. Finally, I would like to thank Atila Abdulkadiroǧlu, Siwei Chen, Lars Ehlers, Nigar Hashimzade, Çağatay Kayı, Flip Klijn, Fuhito Kojima, Vikram Manjunath, Frank Page, Jay Sethuraman, Utku Ünver, participants of Matching Conference 2010 in Duke University, Midwest Theory and Trade Meeting 2011 in University of Notre Dame, PET 2011 in Indiana University in Bloomington, SED 2011 in University Montreal, 22nd International Game Theory Conference 2011 in Stony Brook University for their useful comments. This research was advanced through the PhD Development and Support Program fellowship granted by the Azerbaijan Diplomatic Academy (ADA). 1

2 JEL Classification: C78, D44 Keywords: pairwise stability, budget constraint, strategy-proofness, responsive preferences. 1 Introduction We consider assignment problems with fixed budget constraints. Natural examples are assigning students to graduate schools when colleges fix their budgets for admissions for the year, assigning graduate students as research assistants to faculty members each of whom has a fixed research fund, a manager assigning workers to different projects where each project has a fixed total benefit, assigning post-doctoral candidates to universities with fixed budgets etc. We discuss all these applications in more detail at the end. For the time being, in order to fix the ideas, we concentrate on one example, graduate college admissions. Every year, students intending to pursue graduate studies apply to graduate programs. Each department with a graduate program can admit up to a certain number of students. In the beginning of the year, each department determines a budget for graduate admissions. The admission committee uses this budget for stipends. In general, a department also specifies a maximal stipend that the committee can offer. Each program has minimal requirements for admission (e.g. a minimum score of 1200 from GRE, or a minimum score of 80 from TOEFL IBT, or a minimum graduation GPA of 3.00 etc.). If a student does not fulfill these requirements, his application is not considered. Each college has strict preferences over all groups of students. Because money left after admissions returns to the department for other uses, the admission committee has no incentive to save money. On the other hand, each student has outside options: staying at home or working. For simplicity, for the rest of the paper, we refer to this option as staying at home, and we normalize the monetary opportunities of a student from staying at home to zero. Each student decides to enroll depending on which colleges offer him admission and with what stipend. For each college, each student has a lowest stipend that he finds acceptable. 1 This amount depends on the student s preferences and it may differ from college to college. An allocation for a problem specifies which student is assigned to which college and how much stipend he receives from it. We look for the rules of matching students to colleges and allocating the budgets of the colleges as stipends, such that the total stipend each college offers does not exceed its budget. We are interested in a property of a rule called pairwise stability. It says the following: 1 If a college offers a stipend less than this lowest amount to that student, then the student would prefer to stay at home rather than going to this college with the offered stipend. 2

3 Let an allocation be selected for some problem. First, no student should prefer staying at home to his assignment. Similarly, no college should be assigned a student who does not fulfill its minimal requirements. Next, suppose that student A is not matched to college K. Also suppose that college K is assigned a student B such that releasing B and admitting A results in an assignment that K prefers to its initial assignment. Suppose that college K can offer a stipend to student A, using the stipend it was offering to student B, in such a way that student A prefers college K with this stipend to his initial assignment. Then college K will be better off admitting student A with that stipend even if it has to release student B. 2 For an allocation to be pairwise stable, there should be no such deviations by a pair of a college and a student. Pairwise stability is a weaker requirement than coalitional stability, which is immunity to deviations by a group of students and colleges. We only consider pairwise deviations because they are the ones likely to occur in real life: a student may contact a college he is not matched to, and if this college could offer him admission with an appropriate stipend, he would attend it. The college would consider whether it benefits from this proposal, and make a decision accordingly. Our requirement is that no such deviations by a college-student pair should be beneficial for both. We focus on deviations by a college-student pair. It is very unlikely that two or more students who are matched to different colleges would be able to identify each other, communicate and agree on stipends with some third college so that all students and the college in the coalition would benefit. Coalitions that include two or more colleges are even less realistic. The college admissions problem is first studied by Gale and Shapley (1962) in a paper in which they propose the now well-known deferred-acceptance algorithm. When the preferences of colleges satisfy certain conditions 3, the deferred-acceptance algorithm selects a coalitional stable 4 allocation. This shows the non-emptiness of the coalitional stable set. The core coincides with the coalitional stable set for this model (Roth (1984, 1985), Roth and Sotomayor (1990) etc.). When money is introduced, under certain restrictions on preferences, competitive allocations exist. In fact, the set of competitive allocations coincide with the stable set and the core (Shapley and Shubik (1971), Crawford and Knoer (1981), Crawford and Kelso (1982), Sotomayor (2002), Sun and Yang (2006)). The authors have 2 Note that, college K can release more than one student, in order to be able to afford the stipend required to admit student A, it prefers its assignment when it admits student A and releases this group of students to its initial assignment 3 This condition is referred to as responsiveness in the literature. 4 Two notions of stability, coalitional and pairwise, coincide under this restriction. 3

4 considered the problem of assigning workers to firms, where total profit (budget) is determined by the productivity of the workers and is shared by two sides. Recently, matching with general contracts has been studied. Sufficient conditions 5 on preferences have been provided guaranteeing the existence of stable allocations (Hatfield and Milgrom (2005), Hatfield and Kojima (2008, 2010)). Although these papers generalize most of the papers on college admission with money, they do not consider the possibility that colleges face fixed budget constraints. We address this issue in this paper. Generally, in graduate college admissions, the budget of a college is fixed and does not depend on how many students the college admits. Nor does it depend on these students are. We establish that in the presence of fixed budget (feasibility) constraints, the preferences of the colleges may fail to satisfy the conditions shown in these papers (the ones on matching with contracts) to be sufficient for the existence of stable allocations. Thus, the results in this literature is not applicable to our problem. We also show that, in the presence of fixed budget constraints, there may be no allocation that is immune to deviations by a coalition of students and a college 6. We say that the preference of a college over groups of students is responsive to its preferences over individual students if (i) adding a student who fulfills its minimal requirements to a group of students makes the college at least as well off, and (ii) adding one student to a group of students that does not include that student, is at least as desirable for the college as adding another student to the same group of students that does not include this student either, if and only if the college finds the first student at least as desirable as the second one. Also, note that assuming that there are maximal stipends that colleges may offer is not a real restriction: the college can still set the maximal stipend equal to its budget. We add maximal stipends to generalize our model and make it more realistic. Our main result is that for each problem, a pairwise stable allocation exits. We show this by means of an algorithm. The rule defined through this algorithm always selects a pairwise stable allocation. We also show that this rule is strategy-proof for the students: no student can ever benefit from misrepresenting his preferences. Our final result is that starting from an arbitrary allocation, there exists a sequence of allocations, each allocation being obtained from the previous one by satisfying a blocking pair, such that the final allocation is pairwise stable. The existence of paths to stable allocations has been studied for matching problems without money (Roth and Vande Vate, 1990; Diamantoudi, Miyagawa, and Xue, 2004; Klaus and Klijn, 2007). We show that starting from an arbitrary allocation, the path algorithm 5 This condition is referred to as (bilateral) substitutability in the literature. 6 To show this result, we modify the example in Mongell and Roth (1986). 4

5 we construct results in a pairwise stable allocation. The rest of the paper is organized as follows. In Section 2 we define the model, our notion of stability together with efficiency and incentive compatibility requirements. In Section 3 we introduce our rule. In Section 4 we state our main results and provide the proofs. In Section 5 we discuss the existence of paths to pairwise stable allocation. In Section 6, we discuss further applications of our model. In Section 7, we discuss generalizations of our model. 2 Model There are a finite set of colleges C = {c 1, c 2,..., c n } and a finite set of students S = {s 1, s 2,..., s m }. Each student may attend college or stay at home. We denote home as c. 7 Each college c has a capacity q c Z +. There are no restrictions on how many students stay home, i.e. q c =. Let q (q c ) c C be the capacity profile. Each college has the option of leaving seats empty. We denote an empty seat as 8 s. Each college c has strict preferences R c over all subsets of students, 2 S. The preferences of a college are responsive to its preferences over individual students if the following two conditions are satisfied (Roth 1985): (i) For each s S, each S S \ {s}, we have S {s} R c S if and only if s Rc s, (ii) For each pair s, s S, each S S \ {s, s }, we have S {s} R c S {s } if and only if s R c s. For simplicity, from now on, we refer to this condition as responsiveness. Let R C (R c ) c C be the preference profile of the colleges. Let R C be the set of all such profiles. Each college c has a fixed budget B c R + that it can distribute as stipend to the students it admits. Let B (B c ) c C be the budget profile. For each college c S there is a maximal stipend, m c, that it can award to any student. Let m = (m c ) c C be the maximal stipend profile. Each student may enroll in at most one college and his assignment is a college-stipend bundle. Then his consumption space is {C { c }} R. Staying at home is the bundle 7 We use subscript c in c for home since we can treat it as a null college to which a student can be assigned. 8 We use subscript s in s for empty seat since we can treat it as a null student who can be assigned to a college. 5

6 ( c, 0). Each student has quasi-linear preferences R s over college-stipend bundles, namely {C { c }} R. Let R be the set of all such preference relations. By (c, x) P s (c, x ) we mean that student s prefers (c, x) to (c, x ). Also, by (c, x) I s (c, x ) we mean that he is indifferent between these two bundles. Let R S (R s ) s S be the preference profile of the students. Let R S be the set of all such profiles. A typical preference relation of a student is shown in Figure 1 c x c c x c x R s c 0 (a) c 0 x (b) Figure 1: (a) The consumption space of a student is shown. We label each axis by a college and staying home. On the horizontal axis, we measure the stipend a student receives from that college. For example, the bullet on the axis labeled with college c represents the bundle (c, x ) which is admission by college c with stipend x. Similarly, the bullet on the axis labeled with college c represents the bundle (c, x) which is admission by college c with stipend x. (b) Preferences of student s are shown. He is indifferent between attending college c with stipend x and staying at home, (c, x) I s ( c, 0). Also student s is indifferent between attending college c with no stipend and staying at home and consuming amount x. For each college c, each student s has a smallest stipend l s c(r s ) R +, which we call his lower bound, that he needs to receive in order to attend college c. This lower bound is derived from his preferences in the following way: l s c(r s ) { 0 if (c, 0) P s ( c, 0) x if (c, x) I s ( c, 0). Let l s (R s ) (l s c(r s )) c C R m + be the lower bound list of student s. A problem is a list π (R C, q, B, m, R S ). Let Π be the set of all problems. 6

7 A matching is a correspondence µ : S C S C c, such that s S, µ(s) 1, c C, µ(c) q c, s S and c C, if µ(s) = c, then s µ(c), Let x c s R + be the stipend offered by college c to student s. Let x = (x µ(s 1) s 1, x µ(s 2) s 2,..., x µ(s m) ) R S + be the stipend list. An allocation for π is a matching µ together with a stipend list x. We can also represent an allocation (µ, x) as a list of triples {(s, µ(s), x µ(s) s ) s S }. An allocation (µ, x) is feasible if it satisfies the following requirements: c C, s S, s µ(c) x µ(s) s x c s B c, m µ(s) s m Let A(π) denote the set of all feasible allocations for π. A rule φ : Π π Π A(π) associates with each problem an allocation for it. Next, we introduce some properties for rules. Let φ be a rule. Let π Π be a problem. Student s is admissible for college c if s P c s. For each c C, let S c be the set of admissible students for c. Bundle (c, x) is acceptable for student s if (c, x) R s ( c, 0). Our first requirement has two parts: first, no college should enroll a student who does not fulfill its minimal requirements. Second, no student s assignment should be worse for him than staying at home. An allocation (µ, x) A(π) is individually rational for π if (i) for each c C, each s µ(c) is admissible for c, and (ii) for each s S, the bundle (µ(s), x µ(s) s ) is acceptable for student s. Let IR(π) be the set of all individually rational allocations for π. Individual rationality: For each π Π, φ(π) IR(π). 7

8 Let an allocation be selected for a problem. Suppose there are a college and a student that are not matched to each other. Also suppose that there is a stipend that the college can afford by releasing some of the students initially matched to it and using the money saved in this way as stipend for student such that the student prefers this college with this stipend to his initial assignment, and the college prefers the new updated class 9 Formally, A college-student pair (c, s) blocks allocation (µ, x) A(π), if µ(s) c, and there are S µ(c) and x R + such that (1) (µ(c) \ S) {s} P c µ(c), (2) 1 + µ(c)\{ S} q c, (3) x min{m c, s S (4) (c, x ) P s (µ(s), x µ(s) s ). x c s + Bc s µ(c) x c s }, An allocation is pairwise stable for π if it is individually rational and there is no college-student pair that blocks it. Let PS(π) be the set of all pairwise stable allocations for π. Pairwise stability: For each π Π, φ(π) P S(π). Next we define two efficiency requirements. First, an allocation is chosen only if there is no other allocation making at least one college or one student better off without making anyone else worse off. Formally, An allocation (µ, x) is Pareto-efficient if there is no ( µ, x) A(π) such that for each s S, we have ( µ(s), x µ(s) s ) R s (µ(s), x µ(s) s ), for each c C, we have µ(c) R c µ(c), 9 The one in which this student is admitted to and some of the students are released from initial class. 8

9 or and either - there is s S such that ( µ(s ), x µ(s ) s ) P s (µ(s ), x µ(s ) s ), - there is c C such that µ(c ) P c µ(c ). Let PE(π) be the set of all Pareto-efficient allocations for π. Pareto-efficiency: For each π Π, φ(π) P E(π). Our second and weaker efficiency requirement says that an allocation is chosen only if there is no other allocation making every student and every college better off. Formally, An allocation (µ, x) is weakly Pareto-efficient for π if there is no ( µ, x) A(π) such that and - for each s S, we have ( µ(s), x µ(s) s ) P s (µ(s), x µ(s) s ), - for each c C, we have µ(c) P c µ(c). Let WPE(π) be the set of all weak Pareto-efficient allocations for π. Weak Pareto-efficiency: For each π Π, φ(π) W P E(π). Our final requirement is a strong incentive compatibility property. No student should ever benefit from misrepresenting his preferences. Formally, Strategy-proofness: For each π Π, each s S, and each R s R we have φ s (R C, q, B, m, R) R s φ(r C, q, B, m, (R s, R s )). 3 Rule For each student s S, let s be an order on C c, such that for each s S, each c C, we have c s c. For each s S, let Γ s be the set of all possible orders on C c. Let ( ) s S be a profile of the orders. Also, let σ : S N be an order on S. For each s S, 9

10 let σ(s) denote the order of s. Let Σ be the set of all possible orders on S. Let π be a problem. We define a rule by means of the following algorithm: 10 Imagine an empty room. Initially, all colleges are in the room and all students are outside of the room. At each step, the student with the lowest order (student s with smallest σ(s)) among the ones who are outside of the room enters the room. Each time a student enters the room, each college that finds this student admissible and has an empty seat or an initially admitted student/s it may release, offers him admission with the highest stipend it can possibly offer to that student at that step 11. Each student compares the offers he receives, if any, together with the option of staying at home, and tentatively accepts the one he prefers. If a student is indifferent between offers, we use a predetermined order of that student (which is s for student s) to break ties [Note that the offers made at each step are tentative and can be rejected or withdrawn at any later step.]. If a college releases some of the initially admitted students in order to offer to new entering student highest stipend college possible can at that step (a college will do that only if it prefers the new entering student to them.), those students leave the room and will re-enter the room at some later step. The algorithm proceeds in this way until no student is outside of the room. The algorithm is defined formally below. Best-Admitted-First Algorithm, BAF σ. We refer to a member of S C as an agent. Let there be a room, I. For each c C and each s S, let M c µ(s) be the set of subsets of students assigned to c at (µ, x) such that c prefers s to each of those subsets. Formally, M µ s (c) { S µ(c) s.t. s P c S}. The maximum stipend that college c can offer to student s at (µ, x) is x c s(µ, x) max {min{m c, x c S M s µ s + (c) Bc x s }}. s S s µ(c) Step 0: Let I 0 C be the set of agents who are in the room at step 0. Let O 0 (S C) \ I 0 10 The algorithm has a flavor of both Gale and Shapley s Student Proposing Deferred Acceptance algorithm, and the algorithms designed in Roth and Vande Vate, 1990; Klaus and Klijn, Note that, this stipend may be less than the maximal stipend college can offer. 10

11 be the set of agents who are outside of the room at the end of step 0 (Obviously, O 0 S.) Let (µ 0, x 0 ) be the allocation at step 0 at which no student is assigned to any college. Since no student is assigned during this step, for each c C, we have µ 0 (c) =, and for each s S, we have µ 0 (s) = s. Step 1: Pick a student with the lowest order in O 0 according to σ. Let s 1 min s O 0 σ(s) and let him enter the room, i.e. let I 1 I 0 s 1. College c can offer admission to student s 1 at step 1 if either µ 0 (c) < q c or M µ 0 s 1 (c). Let C 1 (s 1 ) be the set of colleges that can offer admission to s 1 at step 1. Let c C 1 (s 1 ) be such that for each c {c C 1 (s 1 ) such that for each c C 1 (s 1 )\c, (c, x c s 1 (µ 0, x 0 )) R s1 ( c, x c s 1 (µ 0, x 0 ))}, we have c s1 c. Student s 1 compares the offer (c, x c s t (µ 0, x 0 )) with ( c, 0) and tentatively accepts the one he prefers. If he is indifferent between them, he chooses to attend the college 12. If s 1 tentatively accepts the offer of college c, and if college c needs to release some students from µ 0 (c) in order to be able to offer x c s 1 (µ 0, x 0 ), then among the possible sets of students it might release, it releases the set it prefers the least 13. If college c releases any students from µ 0 (c) in order to admit student s 1, those students leave the room. Let RL 1 be the set of students who are released by the college when s 1 is admitted. Let (µ 1, x 1 ) be the allocation obtained after s 1 makes his decision. Let Ī1 I 1 \ RL 1 be the set of agents in the room at the end of step 1. Let O 1 (S C) \ Ī1. If O 1 =, then the algorithm stops, and (µ 1, x 1 ) is the final allocation. If O 1, algorithm proceeds to the next step. Step t = 2, 3,...: Pick a student with the lowest order in O t 1 according to σ. Let s t min σ(s) 14 and let him enter the room, i.e. let I t I t 1 s t. s O 0 College c can offer admission to student s t at step t if either µ t 1 (c) < q c or M µ t 1 s t (c). Let C t (s t ) be the set of colleges that can offer admission to s t at step t. Let c C 1 (s 1 ) be such that for each c {c C t (s t ) such that for each c C t (s t )\c, (c, x c s t (µ t 1, x t 1 )) R st ( c, x c s t (µ t 1, x t 1 ))}, we have c st c. Student s t compares the offer (c, x c s t (µ t 1, x t 1 )) with ( c, 0) and tentatively accepts the one he prefers. If he is indifferent between them, he chooses to attend the college. If s 1 tentatively accepts the offer of college c, and if college c needs to release some students from µ t 1 (c) in order to be able to offer x c s t (µ t 1, x t 1 ), then among the possible sets of students it might release, it 12 This is due to our assumption on tie-breakers: for each s S, each c C, we have c s c 13 Later in Lemma 4 we will show that college releases at most one student. 14 Note that, s t can be same student as s k for some k < t. 11

12 releases the set it prefers the least. If college c releases any students from µ t 1 (c) in order to admit student s t, those students leave the room. Let RL t be the set of students who are released by the college when s t is admitted. Let (µ t, x t ) be the allocation obtained after s t makes his decision. Let Īt I t \ RL t be the set of agents in the room at the end of step t. Let O t (S C) \ Īt. If O t =, then the algorithm stops, and (µ t, x t ) is the final allocation. If O t, algorithm proceeds to the next step. The algorithm continues in this way until Step l at which there is no agent in O l, i.e. O l (C S) =. Then, BAF σ (π) (µ l, x l ) is the final allocation. The Best-Admitted-First Algorithm terminates in a finite number of steps: At each step a student who is outside of the room, enters the room. If the entering student is admitted by some college c C, then the welfare of c increases and the welfare of the other colleges in C \ {c} remains the same. If a student prefers to stay home, then the welfare of all the colleges remains the same. Therefore, at each step, the welfare of each college is weakly increasing. Note that even if a college releases some students at some step in order to admit the entering student, it does so only if it prefers this new student to that group of students, which implies that its welfare is not decreasing. Since the total welfare that a college can get is bounded, and there are finitely many students, the algorithm terminates in finitely many steps. Remark 1: For each σ, σ Σ, each Γ, each π Π, BAF σ (π) = BAF σ. Example 1 (BAF rule): Let C = {c 1, c 2 } and S = {s 1, s 2, s 3, s 4 }. Let σ(s 2 ) < σ(s 3 ) < σ(s 4 ) < σ(s 1 ). Let c 1 s1 c 2, c 2 s2 c 1, c 1 s3 c 2, and c 1 s4 c 2. For each s S, each c C, c s. Let B = (7, 7), m = (5, 6) and q = (2, 2). Preferences of students are as follows c 1 5 c 1 c 2 c 6 R s1 c 2 c 4 R s2 12

13 c 1 c 2 2 R s3 c 1 2 c 2 1 R s4 c 1 c Therefore, l s 1 (R s1 ) = (0, 0), l s 2 (R s2 ) = (1, 0), l s 3 (R s3 ) = (0, 1) and l s 4 (R s4 ) = (2, 1). Preferences of colleges are as follows R c1 R c2 s 1 s 3 s 2 s 3 s 1 s 2 s 2 s 4 s 1 s 2 s 1 s 2 s 3 s 3 s 4 s 3 s 1 s 3 s 2 s 2 s 3 s 4 s 1 s 4 s 4 s 1 Step 0: Let I 0 {c 1, c 2 } and O 0 {s 1, s 2, s 3, s 4 }. Let (µ 2, x 2 ) {c 1, c 2, s 1, s 2, s 3, s 4 } Step 1: Let the student with the lowest order in O 0 enter the room. Since s 2 min s O 0 σ(s), he enters the room, i.e. I 1 {c 1, c 2, s 2 }. College c 1 offers admission to student s 2 with x c 1 s 2 (µ 0, x 0 ) = 5, and college c 2 offers admission to student s 2 with x c 2 s 2 (µ 0, x 0 ) = 6. Student s 1 compares the offers he receives together with ( c, 0). Since (c 1, 5) P s2 (c 2, 6) P s2 ( c, 0), he tentatively accepts the offer of c 1. The allocation at the end of Step 1 is (µ 1, x 1 ) {c 2, (s 2, c 1, 5), s 1, s 3, s 4 }. Let Ī1 {c 1, c 2, s 2 }, and therefore O 1 = (S C) \ Ī1 = {s 1, s 3, s 4 }. Step 2: Let the student with the lowest order in O 1 enter the room. Since s 3 min s O 1 σ(s), he enters the room, i.e. I 2 {c 1, c 2, s 2, s 3 }. Since s 3 P c1 s 2, college c 1 offers admission to student s 3 with x c 1 s 3 (µ 1, x 1 ) = 5 (i.e. since c 1 prefers s 3 to s 2, and after admitting student s 2 it only has amount of 2 left, then in order to offer student s 3 the maximum it can, college c 1 should release student s 2 ), and college c 2 offers admission to student s 3 with x c 2 s 3 (µ 1, x 1 ) = 6. Student s 3 compares the offers he receives together with ( c, 0). Since (c 1, 5) P s3 (c 2, 6) P s3 ( c, 0), he tentatively accepts the offer of college c 1 and student s 2 is released by college c 1, i.e. RL 2 = {s 2 }. The allocation at the end of Step 2 13

14 is (µ 2, x 2 ) {c 2, (s 3, c 1, 5), s 2, s 1, s 4 }. Let Ī2 I 2 \ RL 2 = {c 1, c 2, s 3 }, and therefore O 2 = (S C) \ Ī2 = {s 1, s 2, s 4 }. Step 3: Let the student with the lowest order in O 2 enter the room. Since s 2 min s O 2 σ(s), he enters the room, i.e. I 3 {c 1, c 2, s 2, s 3 }. Since s 3 P c1 s 2, college c 1 offers admission to student s 2 with x c 1 s 2 (µ 2, x 2 ) = 2, and college c 2 offers admission to student s 2 with x c 2 s 2 (µ 2, x 2 ) = 6. Student s 2 compares the offers he receives together with ( c, 0). Since (c 2, 6) I s2 (c 1, 2) P s2 ( c, 0) and c 2 s2 c 1, he tentatively accepts the offer of college c 2. The allocation at the end of Step 3 is (µ 3, x 3 ) {(s 2, c 2, 6), (s 3, c 1, 5), s 1, s 4 }. Let Ī 3 I 3 \ RL 3 = {c 1, c 2, s 2, s 3 }, and therefore O 3 = (S C) \ Ī3 = {s 1, s 4 }. Step 4: Let the student with the lowest order in O 3 enter the room. Since s 4 min σ(s), s O 3 he enters the room, i.e. I 4 {c 1, c 2, s 2, s 3, s 4 }. Since s P c1 s 4, college c 1 does not offer admission to student s 4, and since s 2 P c2 s 4 college c 2 offers admission to student s 4 with x c 2 s 4 (µ 3, x 3 ) = 1. Student s 4 compares the offers he receives together with ( c, 0). Since (c 2, 1) I s4 ( c, 0) and c 2 s4 c, he tentatively accepts the offer of college c 2. The allocation at the end of Step 4 is (µ 4, x 4 ) {(s 3, c 1, 5), (s 2, c 2, 6), (s 4, c 2, 1), s 1 }. Let Ī4 I 4 \ RL 4 = {c 1, c 2, s 2, s 3, s 4 }, and therefore O 4 = (S C) \ Ī4 = {s 1 }. Step 5: Let the student with the lowest order in O 4 enter the room. Since s 1 min s O 4 σ(s), he enters the room, i.e. I 5 {c 1, c 2, s 1, s 2, s 3, s 4 }. Since s 1 P c1 s 3, college c 1 offers admission to student s 1 with x c 1 s 1 (µ 4, x 4 ) = 5 (i.e. since college c 1 prefers student s 1 to student s 3, and after admitting student s 3 it only has amount of 2 left, then in order to offer student s 1 the maximum it can, college c 1 should release student s 3 ), and since both s 2 P c2 s 1 and s 4 P c2 s 1, and µ 4 (c 2 ) = q c2 = 2, college c 2 does not offer any admissions. Student s 1 compares the offers he receives together with ( c, 0). Since (c 1, 5) P s1 ( c, 0), he tentatively accepts the offer of college c 1 and as a result, student s 3 is released by college c 1, i.e. RL 5 = {s 3 }. The allocation at the end of Step 5 is (µ 5, x 5 ) {(s 1, c 1, 5), (s 2, c 2, 6), (s 4, c 2, 1), s 3 }. Let Ī 5 I 5 \ RL 5 = {c 1, c 2, s 1, s 2, s 4 }, and therefore O 5 = (S C) \ Ī5 = {s 3 }. Step 6: Let the student with the lowest order in O 5 enter the room. Since s 3 min s O 5 σ(s), he enters the room, i.e. I 6 {c 1, c 2, s 1, s 2, s 3, s 4 }. Since s 1 P c1 s 3, college c 1 offers admission to student s 3 with x c 1 s 3 (µ 5, x 5 ) = 2, and since s 2 P c2 s 3 but s 3 P c2 s 4, college c 2 offers admission to student s 3 with x c 2 s 3 (µ 5, x 5 ) = 1 (i.e. since college c 2 prefers student s 3 to student s 4 but also 14

15 prefers student s 2 to student s 3, and after admitting students s 2 and s 4 it only has amount of 0 left and no empty seat, then in order to offer student s 3 the maximum it can and a seat, college c 1 should release student s 4 ). Student s 3 compares the offers he receives together with ( c, 0). Since (c 1, 2) P s3 (c 2, 1) I s3 ( c, 0), student s 3 tentatively accepts the offer of college c 1. The allocation at the end of Step 6 is (µ 6, x 6 ) {(s 1, c 1, 5), (s 2, c 2, 6), (s 3, c 1, 2), (s 4, c 2, 1)}. Let Ī6 I 6 \ RL 6 = {c 1, c 2, s 1, s 2, s 3, s 4 }, and therefore O 6 = (S C) \ Ī6 =. Since O 6 =, algorithm stops. The final allocation BAF σ (π) (µ 7, x 7 ) = {(s 1, c 1, 5), (s 2, c 2, 6), (s 3, c 1, 2), (s 4, c 2, 1)} 4 Main Results Our first main result is that for each problem, the pairwise stable set is non-empty. Before stating it formally, we introduce several lemmas. Lemma 1 Let σ Σ and Γ. Let π Π be a problem. Let (µ, x) BAF σ (π) =. For each c C, there is at most one student in µ(c) who receives stipend x (0, m c ). Proof. Suppose by contradiction that there are a c C and a pair s, s µ(c), such that x c s, x c s (0, mc ). Without loss of generality, suppose that s P c s. Let t s be the step at which s was admitted by c for the last time. Similarly, let t s be the step at which s was admitted by c for the last time. We distinguish several cases. Case 1: t s < t s. We know that x c s,t < s mc. College c doesn t offer stipend m c to student s at step t s because (i) it doesn t have enough money, B c x c s,t s 1 < m c, s µ ts 1 (c) and (ii) there is no ŝ µ ts (c) with x c ŝ > 0, such that s P c ŝ (i.e. there is no student whom c could have released in order to offer a stipend higher than x c s,t to s). But since the welfare of c is weakly increasing at each step, it will not have any additional money at any later step that it could possibly offer to s. 15 But then, when s enters the room at step t s, c can not offer him a stipend higher than 0. If it can, then either c has some positive amount of money that it offers to s or s µ ts 1 (c) with x c s > 0, such that s P c s (i.e. there is a student whom c can release in order to offer a positive stipend s ). But since s P c s, then either 15 Even if c releases a student, it does so to admit a better one, and pays him the maximum it can, which is the stipend of the released student. 15

16 one of those cases will imply that c didn t offer the maximum it could to student s at t s. Therefore, x c s,t s = xc s = 0. But then xc s = 0, contradicts our assumption that xc s (0, mc ). Case 2: t s > t s. We know that x c s < mc. College c doesn t offer stipend m c to student s at step t s because (i) it doesn t have enough money, B c x c < s,t s 1 mc, s µ ts 1 (c) and (ii) there is no ŝ µ ts (c) with x c ŝ > 0, such that s P c ŝ (i.e. there is no student whom c could have released in order to offer a stipend higher than x c s to s ). But since the welfare of c is weakly increasing at each step, it will not have any additional money at any later step that it could possibly offer to s. When s enters the room at step t s, college c offers him x c s,t = s xc s < m c. If c does not release s to admit s, then since s P c s and x c s > 0, college c does not offer maximum it can to s at step t s. Then, c should release s to admit s. But since s µ(c) and t s is the last step at which c admits s, we obtain a contradiction. Therefore, for each c C, there can not be more than one student who receives stipend x (0, m c ). Lemma 2 Let σ Σ and Γ. Let π Π be a problem. Let (µ, x) BAF σ (π). For each c C, each pair s, s µ(c), if s P c s then x c s x c s. Proof. Suppose by contradiction that there are a college c C and a pair of students s, s µ(c) such that s P c s and x c s < x c s. We distinguish several cases. Case 1: x c s (0, mc ). By Lemma 1 there is at most one student with stipend x (0, m c ). Thus, by our assumption, x c s = 0. Let t s be the step at which s was admitted by c for the last time. Similarly, let t s be the step at which s was admitted by c for the last time. We distinguish several subcases. Subcase 1: t s < t s. We know that x c s,ts = 0. College c doesn t offer a positive stipend to student s at step t s because both it doesn t have enough money, B c x c s,t s 1 = 0, s µ ts 1 (c) and there is no ŝ µ ts 1 (c) with x c ŝ > 0, such that s P c ŝ (i.e. there is no students whom c could possibly release in order to offer a stipend higher than x c s,t to s). But since the welfare of c is weakly increasing at each step, it will not have any additional money at any later step that it could possibly offer to s. But then, when s enters the room at step t s, college c can not offer him a stipend higher than 0. If it can, then either c has some positive amount of money that it offers to s or s µ ts 1 (c) with x c s > 0, such that s P c s (i.e. there is a student 16

17 whom c can release in order to offer a positive stipend s ). But since s P c s, then either one of those cases will imply that c didn t offer the maximum it could to student s at t s. Therefore, x c s,t s = xc s = 0. But then xc s = 0, contradicts our assumption that xc s (0, mc ). Subcase 2: t s > t s. Since x c s < mc, then c didn t offer s stipend m c at step t s because both it doesn t have enough money, B c x c < s,t s 1 mc, and there is no s µ ts 1 (c) ŝ µ ts 1 (c) with x c ŝ > xc s,t s, such that s P c ŝ (i.e. there is no students whom c could possibly release in order to offer a stipend higher than x c s to s ). But since the welfare of c is weakly increasing at each step, it will not have any additional money at any later step that it could possibly offer to s. When student s enters the room at step t s, college c admits him with x c s,t s. Since ts is the step at which c admits s for the last time, we have x c s,t = s xc s = 0. Since s P c s and x c s > 0, college c should have released student s to admit s with the maximum stipend c could offer him. Since t s is the step when c admits s for the last time, releasing s will result in contradiction. But then, not releasing s implies that c does not offer the maximum it can to s at step t s which is a contradiction to the way the BAF Algorithm works. Case 2: x c s = mc and x c s [0, m c ). This can be shown in the similar way as the Case 1. Thus, we showed that for each c C, each pair s, s µ(c), if s P c s then x c s x c s. Lemma 3 Let σ Σ and Γ. Let π Π be a problem. Let (µ, x) BAF σ (π). For each c C, each s µ(c), the stipend s receives from c, x c s, is one of the following three possibilities (i) x c s = m c (ii) x c s = 0 (iii) x c s = x c (0, m c ) Proof. We distinguish several cases. Case 1: There is s µ(c) such that x c s = x c (0, m c ). Then by Lemmas 1 and 2, for each s µ(c), such that s P c s, we have x c s s µ(c), such that s P c s, we have x c s = = mc. Also by Lemmas 1 and 2, for each

18 Case 2: There is no s µ(c) such that x c s = x c (0, m c ). If B c = 0, then by feasibility, for each s µ(c), we have x c s = 0. If B c > 0 and µ(c), then there is at least one student s µ(c), such that x c s = m c. If B c µ(c) m c, then for each s µ(c), we have x c s = m c. If B c < µ(c) m c, then by feasibility, there is s µ(c) with x c s < mc. Then, by Lemma 2 and the assumption that there is no s µ(c) with x c s (0, m c ), we have x c s = 0. Theorem 1 The pairwise stable set is non-empty. Best-Admitted-First algorithm always produces a pairwise stable allocation. Proof. Suppose by contradiction that there are a college-student pair (c, s) such that µ(s) c, S 2 µ(c), and x R +, such that (1) (µ(c) \ S) {s} P c µ(c), (2) 1 + µ(c)\{ S} q c, (3) x min{m c, s S (4) (c, x ) P s (µ(s), x µ(s) s ). x c s + Bc s µ(c) x c s }, Let t s be the step at which s enters the room for the last time. cases. We distinguish several Case 1: S =. Then, there is no s µ(c) with x c s > 0 such that s P c s and µ(c) < q c. Then x min{m c, B c x c s }. At step ts, college c offered admission to student s with stipend x B c (µ(s), x µ(s) s s µ(c) s µ(c) x c s. Since s / µ(c), then s received a better offer from µ(s), i.e. ) R s (c, x). But since x x, it is not possible that (c, x ) R s (µ(s), x µ(s) s ). We obtain a contradiction. Case 2: S = 1. Call this student s ( S = s ). By Lemma 3, x c s following three amounts (i) x c s = m c (ii) x c s = 0 (iii) x c s = x c (0, m c ) equals to one of the 18

19 x c s,t s Subcase 2.1: x c s = mc. By Lemma 2, at step t s, college c offers student s stipend = mc. Since s / µ(c), student s rejects the offer. But since x m c, it cannot be that (c, x ) R s (µ(s), x µ(s) s ). We obtain a contradiction. Subcase 2.2: x c s stipend x c s,t s = xc (0, m c ). By Lemma 2, at step t s, college c offers student s xc s. Since s / µ(c), student s rejects the offer. But since x x c, it cannot be that (c, x ) R s (µ(s), x s µ(s) ). We obtain a contradiction. Subcase 2.3: x c s = 0. Then, µ(c) = q c. By Lemma 2, at step t s, college c offers s stipend x c s,t s 0. Since s / µ(c), student s rejected the offer of college c. But since x = 0, it cannot be that (c, x ) R s (µ(s), x s µ(s) ). We obtain a contradiction. Case 3: S > 1. First note that, by responsiveness of P c, for each s S we have (µ(c) \ S) {s} P c µ(c) (µ(c) \ {s }) {s} P c µ(c) Claim: there is s S such that, min{m c, x c s + Bc x c s } x. s µ(c) Proof of Claim: Note that by Lemma 3, each ŝ µ(c) receives a stipend x c ŝ equal to one of the following three amounts (i) x c s = m c (ii) x c s = 0 (iii) x c s = x c (0, m c ) that is Subcase 3.1: There is s S such that x c s = mc. Then releasing any such student, would allow college c to offer x to s. Subcase 3.2: There is no s S such that x c s = mc, and there is s µ(c) such that x c s = x c. Then by Lemma 1, s is the only student in S receiving a positive stipend. Once again, releasing any such student, would allow college c to offer x to s. Subcase 3.3: For each s S, x c s = 0. Then releasing any such student, would allow college c to offer 0 to s and release a seat for him. This ends the proof of the claim. The claim means that if there is S with S > 1 that c can release in order to admit s and benefit from this action, then there is s S µ(c), releasing whom in order to admit s will also benefit c. But then we are back to the Case 2. 19

20 Another important property of a rule is strategy-proofness. We show next that BAF rule enjoys this property. Lemma 4 Let π Π be a problem. At each step of the BAF algorithm, in order to admit the new entering student, a college (needs to) releases at most one student. Proof. Suppose by contradiction that there are a college c C and a step t at which µ t 1 (c) \ µ t (c) > 1. Let s be the student who enters the room and is admitted by college c at step t. Without loss of generality, suppose that µ t 1 (c) \ µ t (c) = 2. Let µ t 1 (c) \ µ t (c) = {s, s }. Since c releases {s, s } to admit s, then (µ t 1 (c) \ {s, s }) {s} P c µ t 1 (c). By responsiveness of P c, we know that both (µ t 1 (c) \ {s }) {s} P c µ t 1 (c) and (µ t 1 (c) \ {s }) {s} P c µ t 1 (c) should also hold. Next, by Lemma 3, we know that both s and s are receiving one of the following stipends (i) x c s = m c (ii) x c s = 0 (iii) x c s = x c (0, m c ) and by Lemma 1, at most one of them receive stipend as in (iii). We distinguish several cases: Case 1: Either s or s (or both) receive stipend m c. Then x c s,t m c. Since c can offer to s at most m c, we have x c s,t = m c. Among the possible ways of offering m c to s, college c selects the one that maximizes its welfare. Therefore, it only releases the student among s and s who receives stipend m c. If both receive m c, then c releases the least preferred student among s and s. Case 2: Neither s and s receive stipend m c, and one of them receives x c (0, m c ). Without loss of generality, suppose that x c s,t 1 = xc (0, m c ). Then, by lemma 1 and our assumption that neither s and s receive m c, we have x c s,t 1 = 0. Therefore, the stipend it offers to s equals to x c s,t 1 (since xc s,t 1 = 0). Among the possible ways of offering xc s,t, to s, college c selects the one that maximizes its welfare, namely it is to release only s. Case 3: Both s and s receive stipend 0. Then, the only benefit of releasing students s and s for c is to empty a seat. But then, in order to admit s, college c only needs to release one seat. Among the possible ways of emptying a seat to offer admission to s, college c selects the one that maximizes its welfare. Therefore, releasing either s or s is enough. 20

21 This completes the proof of the lemma. Before stating our next lemma, we introduce an additional piece of notation. For each c C, each s S, and each step t of the BAF algorithm, let µ t s(c) {s µ t c such that s P c s} Lemma 5 Let π Π be a problem. For each c C, each s S, each step t of the BAF algorithm applied to π, and each k Z +, µ t s(c) µ t+k s (c). Proof. By Lemma 4, we know that in order to admit a student, college need not release more than one student. Therefore, if c admits student a s, it releases at most one of the students whom it had initially admitted. If at some step t, college c releases some student s µ t s(c), then it does so to admit a better student s. Since s P c s P c s, the number of students admitted by c who are preferred to s by c does not decrease. Lemma 6 Let π Π be a problem. For each c C, each s S, each step t of the BAF algorithm applied to π, and each k Z +, x c s,t+k xc s,t. Proof. Suppose by contradiction that there are c C, s S, t N, and k Z +, such that x c s,t+k > xc s,t. m c By feasibility x c s,t < m c. This, together with Lemma 2, imply that B c µ t 1 s (c) < m c. We distinguish several cases. Case 1: x c s,t > 0: Since m c > x c s,t By Lemma 1, at most one student can receive such a stipend. Therefore, for c to be able to offer this stipend to s at step t, either there is s µ t 1 (c) receiving that stipend and s P c s, or there is no student in µ t 1 (c) \ µ t 1 s (c) who receives a positive stipend, and B c µ t 1 s (c) m c = x c s,t. Since, x c s,t+k > xc s,t, we have B c µ t 1 s (c) m c < B c µ t+k 1 (c) m c. This implies that µ t 1 s (c) > µ t+k 1 (c). But by Lemma 5, µ t 1 (c) µ t+k 1 (c). We obtain a contradiction. s s s s Case 2: x c s,t = 0: For c to offer stipend 0 to s at step t, either µ t 1 (c) = q c and there is s µ t 1 (c) receiving 0 such that s P c s, or µ t 1 (c) < q c and there is no student in µ t 1 (c) \ µ t 1 s (c) who receives a positive stipend, and B c µ t 1 s (c) m c 0. Since x c s,t+k > xc s,t, we have B c µ t 1 s (c) m c < B c µ s t+k 1 (c) m c. This implies that µ t 1 s (c) > µ t+k 1 (c). But once again, by Lemma 5, µ t 1 (c) µ t+k 1 (c). We obtain a contradiction. s s s 21

22 Theorem 2 : The BAF rule is strategy-proof. Proof. At each step of the BAF algorithm, one student who is outside of the room enters the room. Each college offers him the maximum it can at that step. To do so, by Lemma 4, the college releases at most one student. Let t be the first step at which s enters the room. Let C t (s) be the set of colleges that can offer admission to s at step t. Let c C t (s) be such that for each c {c C 1 (s) such that for each c C 1 (s)\c, (c, x c s,t) R s ( c, x c s,t)}, we have c s c (Student s weakly prefers (c, x c s,t) to any other offer he receives.). Also, let (c, x c s,t) R s ( c, 0). Then, according to the algorithm, he should accept the offer of college c. Suppose instead that he rejects (c, x c s,t) and tentatively accepts some offer (c, x c s,t) of c C t (s). By Lemma 6, at any later step t > t, student s does not receive a better offer than the one he receives at step t from any college including c. Thus, student s benefits from this misrepresentation, only if, c releases him at some later step t > t in order to admit some other student s, and when s enters the room again at some later step t > t, the best offer it receives is worse than (c, x c s,t) (this implies that college c does not offer him admission at step t, or the offer it makes is strictly worse than (c, x c s,t), i.e. (c, x c s,t) P s (c, x c s,t ).). This implies that xc s,t < xc s,t which in turn implies that x c s,t < mc, which means B c µ t 1 s (c ) m c < m c. By Lemmas 1, 2 and 3, we have x c s,t < mc. By Lemma 4, the reason for x c s,t < xc s,t is that c admitted students each of whom are individually better than s for c. Formally, µ t 1 s (c) 1. Let St c 1 1 µt s (c). Suppose s tentatively accepts the offer (c, x c s,t) at step t. At some step, all students in St c 1 eventually enter the room. By Lemma 6 and since Bc St c 1 mc < m c, we know that c releases s to admit the students in St c 1. And when he enters the room again, c may not be able to offer him as high stipend as x c s,t 16. Therefore even if s misrepresents his preferences, he does not benefit from this. To get the intuition, note that at each step, when a student enters the room, each college that offers him admission does so with the maximum stipend it can. The welfare of each college is nondecreasing and by Lemma 6, the offer each college makes to a student will not be better at any later step. Although, at each step, colleges make offer to the entering student, this 16 Note that c releases s since it can not afford paying m c to new entering student. By the way algorithm performs and Lemma 4, releasing s is the welfare maximizing choice among the possible students whom c could release in order to admit new entering student. Then, by Lemmas 4, 5 and 6, when s enters the room again, it receives strictly worse offer than before 22

23 offer is the highest the student can ask for from them. But given the highest offer student can ask for from each college, the student chooses which college to ask for to admit him. Thus, this process behaves in a similar way as the well-known Student-Proposing Deferred Acceptance Algorithm, which we know is strategy-proof (Dubins and Freedman, 1981; Roth, 1982). Our next example shows that it may not be possible to find an allocation that is not blocked by a coalition of one college and at most two students. The example is a modification of the one in Mongell and Roth (1986). Example 2 (Coalitional stable allocation may not exist.): Let C = {c 1, c 2 } and S = {s 1, s 2, s 3 }. Let c 1 s1 c 2, c 1 s2 c 2, and c 1 s3 c 2. Let B = (5, 11), m = (5, 11) and q = (2, 2). Preferences of students are as follows c 1 5 c 2 10 R s1 c 1 c R s2 c 1 4 c 2 1 R s3 c c c Therefore, l s 1 (R s1 ) = (5, 10), l s 2 (R s2 ) = (1, 2), and l s 3 (R s3 ) = (4, 1). Preferences of colleges are as follows R c1 R c2 s 3 s 2 s 1 s 3 s 3 s 1 s 2 s 2 s 3 s 2 s s 1. s 3 s 2 s Consider the following three allocations that one can reach by starting from an arbitrary feasible allocation and satisfying the possible blocking coalitions: (µ, x) {(s 1, c 2, 10), (s 2, c 1, 0), (s 3, c 1, 5)} 23.