Centralized Matching. Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group.

Centralized Matching

Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group. Students with schools Men with women, etc. Is there a stable matching – one that cannot be upset by individual negotiations? What procedures or institutions accomplish stable matchings? ►

Preview Marriage/college admissions Problem Agents of both sides are economic agents Deferred Acceptance procedure Dorm House Assignment/School Choice Problem Agents of only one side are economic agents Top-cycle procedure ►

Marriage Problem: Example 1 Women: w1,w2,..,w n Men: m1,m2,…,m n Two-sided problem: a man cannot marry a man First number of each pair gives the ranking of women by the man (1 most preferred) Second number gives the ranking of men by the woman How many matchings are conceivable? n*(n-1)*(n-2)*…*1=n! ► w1w2w3 m11,32,23,1 m23,11,32,2 m32,23,11,3

Stable Two-Sided Matchings A matching is called unstable if there are two men and two women who are matched to each other, respectively, say (m1,w1) and (m2,w2), even though m1 prefers w2 to w1 and w2 prefers m1 to m2. In this case, m1 and w2 would want to divorce and get matched to each other. A matching that is not unstable is called stable. ►

Stable Two-Sided Matchings An example of unstable matching: (m1,w1), (m2,w3), and (m3,w2) M3 and w1 want to divorce their partners and get together with each other ► w1w2w3 M11,32,23,1 M23,11,32,2 m32,23,11,3

Stable Two-Sided Matchings three stable matchings (m1,w1),(m2,w2) and (m3,w3) (men optimal) (m1,w3),(m2,w1) and (m3,w2) (women optimal) (m1,w2), (m2,w3) and (m3,w1) (each gets his/her second choice) ► w1w2w3 M11,32,23,1 M23,11,32,2 m32,23,11,3

Stable Two-Sided Matchings Can we always find stable matchings simply by giving either men or women their first choice? Not always works: two or more members of each group may give the highest rank to the same member of the other group. Do all matching problems have at least one stable matching? ►

Roommate Problem Do all matching problems have at least one stable matching? No. Four students, s1,s2,s3,s4, are divided up into pairs of roommates 1 most preferred, 3 least preferred one-sided matching ► s1s2s3s4 s113 s213 s313 s4

Roommate Problem Nobody wants to share a room with s4 Everybody has somebody who is not s4 who likes her most Three matchings (s1,s2),(s3,s4) (s1,s3),(s2,s4) (s1,s4),(s2,s3) Consider the first one, s2 and s3 want to divorce their roommates and get together ► s1s2s3s4 s113 s213 s313 s4

Roommate Problem Second one: (s1,s3),(s2,s4) S1 and s2 want to divorce their roommates and get together Third one: (s1,s4),(s2,s3) S1 and s3 want to divorce their roommates and get together No stable matching exists. ► s1s2s3s4 s113 s213 s313 s4

Men-Proposed-to-Women Procedure Despite the negative result in one-sided matching, the marriage problem has at least one stable matching. We find one such by using the following men- propose-to-women procedure. ►

Men-Propose-to-Women Procedure Step 1(a). Each man proposes to his most preferred woman Step 1(b). Each woman rejects all except the one whom she most prefers and keeps the most preferred as her suitor … Step k(a). Each man who was rejected in the previous step proposes to the most preferred of those women to whom he has not yet proposed. Step k(b). Each woman keeps as her suitor the man she most prefers among those who have proposed (and not yet rejected) and rejects the rest. … The procedure terminates when all women have received at least one proposal, at which point each woman has one suitor. ►

Men-Propose-to-Women Procedure Use example 1 for illustration. In step 1a, m1 proposes to w1 m2 proposes to w2 m3 proposes to w3 In step 1b, Each woman keeps her only proposer as suitor. The procedure ends at once, with each woman marrying her suitor. ► w1w2w3 m11,32,23,1 m23,11,32,2 m32,23,11,3

Women-Propose-to-Men Procedure: Example 2 w1w2w3 M12,11,23,1 M21,32,33,2 m31,22,13,3 stepw1w2w3 1m2,m3m1

Women-Propose-to-Men Procedure: Example 2 w1w2w3 M12,11,23,1 M21,32,33,2 m31,22,13,3 stepw1w2w3 1m2,m3m1 2m3m1,m2

Women-Propose-to-Men Procedure: Example 2 w1w2w3 M12,11,23,1 M21,32,33,2 m31,22,13,3 stepw1w2w3 1m2,m3m1 2m3m1,m2 3m3m1m2 Done! ►

Women-Propose-to-Men Procedure The procedure ends at once, with each woman marrying her suitor. The outcome is stable, since any woman preferred by a man to his current wife must have already rejected him in favor of her current partner. So we cannot find a pair of unmatched man and woman who want to divorce their partners and get marry to each other ►

Deferred Acceptance Procedure Among all stable matchings, one is weakly preferred by every man, and one by every woman. There is a “men-optimal” and a “women-optimal” stable matching. The “men-optimal” stable matching is the one obtained under the men-propose-to-women procedure The “women-optimal” stable matching is the one obtained under the women-propose-to-men procedure Additional merit: It is dominant strategy for each proposer to truthfully report his/her preferences ►

Deferred Acceptance Procedure The receivers may benefit from manipulation through misreporting their preferences Underreporting their capacities Pre-arrangement Preferential treatment: “I want those students that want us most!” However, the benefits of such manipulation are getting smaller as the market becomes larger ►

Assignment of Dorm House A school choice problem Like college admissions problem, it is a two sided matching However, school places are services to the public. Schools are not economic agents who have preferences. Schools are like policemen, cannot choose who to provide service But schools do have priority—like serving those in the neighborhood over outside, those with siblings attending the same school over others, etc. We illustrate this new problem with the dorm house assignment problem ►

Assignment of Dorm House Imagine yourself as a university administrator, and you need to design a mechanism to assign students to dorm houses. The mechanism, if possible, should satisfy the following three conditions. ►

Assignment of Dorm House Strategy-proofness--A mechanism is strategy proof if each student revealing his preferences truthfully constitutes an equilibrium. Clearly, the second price sealed bid auction is strategy proof. Protection of existing tenants--Any new assignment is not allowed to make any existing tenants worse off. ►

Assignment of Dorm House Pareto efficiency--An assignment y is a Pareto improvement of assignment x if: (1) Every student likes y at least as good as x. (2) At least one student likes y strictly better than x. An assignment x is Pareto efficient if it cannot be improved upon. That is, it is Pareto efficient if an assignment y that satisfies the above two conditions cannot be found. ►

Serial Dictatorship: Example 1 2 students: A and B 2 houses: h ₁ and h ₂ How many different ways to assign houses? Assignment x: assign h ₁ to A, and h ₂ to B Assignment y: assign h ₁ to B, and h ₂ to A There need not be conflict. Suppose A prefers x to y, and B prefers x to y too. So the administrator can just choose x. You cannot make anybody even happier. ►

Serial Dictatorship: Example 2 Still the same two students but now A prefers x to y and B prefers y to x. Thus there is a conflict. One solution frequently used is to rank students by "seniority" (or other criteria). Suppose A is more "senior" than B. Then do what A wants, i.e., adopt assignment x and assign h ₁ to A and h ₂ to B. In general, for any number of students and any number of houses, we can use this "serial dictatorship" to assign houses. ►

Serial Dictatorship Suppose each student cares what house is assigned to him, but does not care what house is assigned to each other student. For instance, Suppose A prefers h ₁ to h ₂ to h ₃. Then he thinks "as long as I get h ₁, I don't care whether B gets h ₂ and h ₃." If that is the case, then serial dictatorship satisfies the Pareto efficiency and strategy proofness conditions. Consider the more complicated case where there are existing tenants, and their interest should be respected. ►

Harvard Mechanism Here, we first review a few commonly used mechanism, and show that each has its own problems. Harvard mechanism, also used a Carnegie-Mellon, Duke, Pennsylvania,..., etc. Harvard mechanism is as follows: Serial dictatorship, except that every existing tenant has the right not to participate. If he elects not to participate, he keeps his house. If he elects to participate, he gives up his house first, and then participate as any other non-existing tenant student. ►

Harvard Mechanism There is no guarantee that, once you give up your current house and participate, you'll be able to get a house better than your current one. Although they may not like their current houses, many existing tenants do not participate; they fear that they may get a house that is even worse. Existing tenants are unhappy; they don't feel well protected. New applicants are unhappy: too few houses are freed up and become available for re-assignment. ►

Rochester Mechanism All students have to participate. At the beginning, only vacant houses are free houses. Go down the seniority list, ask the students one by one. "Do you want to take one of these free houses?“ If a new applicant answers "Yes": Let him takes his choice. Delete his name from the seniority list. Eliminate his choice from the pool of free houses. Continue to move down the seniority list. If an existing tenant answers "Yes": Let him takes his choice. Delete his name from the seniority list. Eliminate his choice from the pool of free houses. Add his current house into the pool of free houses. Go back to top of the seniority list. Start from the third bullet again. The process ends when we get to the bottom of the seniority list (i.e., no student is left) or when no houses are left. ►

Rochester Mechanism: An example Suppose there are 3 students, all are existing tenants: i ₁,i ₂,i ₃ occupying houses h ₁,h ₂,h ₃, respectively. There is one vacant house: h ₄. The seniority order is i ₁,i ₂,i ₃. Their true preferences over the 4 houses are: i1: h3>h2>h1>h4 i2: h1>h4>h2>h3 i3: h2>h1>h3>h4 ►

Rochester Mechanism: An example Go down the seniority list and ask the students one by one: "Do you want to take one of these free houses?“ i ₁ replies: "No, the only free house, h ₄, is even worse than my current house, h ₁.“ i ₂ replies: "Yes, h ₄ is better than h ₂.“ h ₄ is taken. h ₂ becomes the only free house. ► studentpreferences i1h3>h2>h1>h4 i2h1>h4>h2>h3 i3h2>h1>h3>h4

Rochester Mechanism: An example Go back to the top of the seniority list and start again. i ₁ replies: "Yes, h ₂ is better than h ₁.“ h ₂ is taken. h ₁ becomes the only free house. Go back to the top of the seniority list (which now has only one student left, namely i ₃ ) and start again. i ₃ replies: "Yes, h ₁ is better than h ₃.“ ► studentpreferences i1h3>h2>h1>h4 i2h1>h4>h2>h3 i3h2>h1>h3>h4

Protective, but not Pareto Efficient Suppose there are 2 students, both are existing tenants: i ₁ and i ₂ occupy houses h ₁ and h ₂ respectively. There is one vacant house: h ₃. The students' true preferences over the 3 houses are: i ₁ :h ₂≻ h ₃≻ h ₁ i ₂ :h ₃≻ h ₂≻ h ₁ The Rochester Mechanism assigns h ₃ to i ₁ and h ₂ to i ₂. But this is not Pareto efficient, as everyone would be happier if i ₁ gets h ₂ and i ₂ gets h ₃ ! ►

Protective, but not Strategy-Proof i ₁ :h ₂≻ h ₃≻ h ₁ and i ₂ :h ₃≻ h ₂≻ h ₁ (same example) Suppose i ₁ lies and answers, "No, my preferences ranking is actually h ₂≻ h ₁≻ h ₃, and hence I don't want to take h ₃." i ₂ takes h ₃, frees up h ₂. Go back to the top of the seniority list and start again. Now i ₁ has a chance to take h ₂. It is not a dominant strategy for i ₁ to tell the truth. The Rochester Mechanism hence is not strategy- proof. ►

MIT Mechanism Serial dictatorship, except that every existing tenant has the right to call for protection. Every student has to participate. The most senior student is tentatively matched to her favorite house. The second most senior student tentatively matched to her favorite house among the remaining houses. Keep going down the seniority list, until a conflict occurs. ►

MIT Mechanism A conflict occurs if: It is the turn of an existing tenant, he claims that all the remaining houses are worse than his current house, and hence calls for protection. When a conflict occurs: someone more "senior" (called the conflicting student) must have been tentatively matched to this existing tenant's current house; the existing tenant seizes back his current house, and leave the process; re-do the process, starting from the conflicting student. ►

MIT Mechanism: Example Suppose there is one new applicant, i ₅, and one vacant house, h ₅. The seniority order is i ₁,i ₂,i ₃,i ₄ and then i ₅. Their true preferences over the 5 houses are: ► student preferences i1 h ₃≻ h ₄≻ h ₅≻ h ₁≻ h ₂ i2 h ₄≻ h ₅≻ h ₂≻ h ₃≻ h ₁ i3 h ₅≻ h ₃≻ h ₄≻ h ₂≻ h ₁ i4 h ₃≻ h ₅≻ h ₄≻ h ₂≻ h ₁ i5 h ₄≻ h ₅≻ h ₃≻ h ₁≻ h ₂

MIT Mechanism: Example i ₁ is tentatively matched to h ₃. i ₂ is tentatively matched to h ₄. i ₃ is tentatively matched to h ₅. i ₄, "The only two remaining houses are h ₁ and h ₂, and both are worse than my current house, h ₄. So I shall call for protection.“ i ₄ seizes back h ₄, and leaves the process. ► student preferences i1 h ₃≻ h ₄≻ h ₅≻ h ₁≻ h ₂ i2 h ₄≻ h ₅≻ h ₂≻ h ₃≻ h ₁ i3 h ₅≻ h ₃≻ h ₄≻ h ₂≻ h ₁ i4 h ₃≻ h ₅≻ h ₄≻ h ₂≻ h ₁ i5 h ₄≻ h ₅≻ h ₃≻ h ₁≻ h ₂

MIT Mechanism: Example i ₂ was the conflicting student, so we re-do the process, starting from i ₂. i ₂ is tentatively matched to h ₅. i ₃,"The only remaining houses are h ₁ and h ₂, and both are worse than my current house, h ₃. So I shall call for protection.“ i ₃ seizes back h ₃, and leaves the process. ► student preferences I1 h ₃≻ h ₄≻ h ₅≻ h ₁≻ h ₂ I2 h ₄≻ h ₅≻ h ₂≻ h ₃≻ h ₁ I3 h ₅≻ h ₃≻ h ₄≻ h ₂≻ h ₁ i4 h ₃≻ h ₅≻ h ₄≻ h ₂≻ h ₁ i5 h ₄≻ h ₅≻ h ₃≻ h ₁≻ h ₂

MIT Mechanism: Example i ₁ was the conflicting student, so we re-do the process, starting from i ₁. i ₁ is tentatively matched to h ₅. i ₂ is tentatively matched to h ₂. i ₅ is tentatively matched to h ₁. The process ends, all tentative matches are finalized. ► student preferences i1 h ₃≻ h ₄≻ h ₅≻ h ₁≻ h ₂ i2 h ₄≻ h ₅≻ h ₂≻ h ₃≻ h ₁ i3 h ₅≻ h ₃≻ h ₄≻ h ₂≻ h ₁ i4 h ₃≻ h ₅≻ h ₄≻ h ₂≻ h ₁ i5 h ₄≻ h ₅≻ h ₃≻ h ₁≻ h ₂

MIT Mechanism: Example Let us keep to the same example. The MIT mechanism will match (i ₁,i ₂,i ₃,i ₄,i ₅ ) to (h ₅,h ₂,h ₃,h ₄,h ₁ ). Compare this assignment to the following one: (h ₃,h ₂,h ₅,h ₄,h ₁ ); i.e., i ₁ and i ₃ swap their houses. i ₂,i ₄, and i ₅ like the second assignment at least as good as the first--they are actually indifferent. i ₁ and i ₃ like the second assignment strictly better than the first. ► student preferences i1 h ₃≻ h ₄≻ h ₅≻ h ₁≻ h ₂ i2 h ₄≻ h ₅≻ h ₂≻ h ₃≻ h ₁ i3 h ₅≻ h ₃≻ h ₄≻ h ₂≻ h ₁ i4 h ₃≻ h ₅≻ h ₄≻ h ₂≻ h ₁ i5 h ₄≻ h ₅≻ h ₃≻ h ₁≻ h ₂

Top Cycle Mechanism Can we design a mechanism that is Pareto efficient, strategy-proof, and protective of existing tenants? The top cycle mechanism. Suppose there are 4 existing tenants. i ₁,i ₂,i ₃ and i ₄ occupy houses h ₁,h ₂,h ₃ and h ₄, respectively. Suppose there are one new applicant i ₅, and 3 vacant houses, h ₅,h ₆ and h ₇. The seniority order is i ₁,i ₂,i ₃,i ₄, and then i ₅. ► preferences i1 h ₂≻ h ₆≻ h ₅≻ h ₁≻ h ₄≻ h ₃≻ h ₇ i2 h ₇≻ h ₁≻ h ₆≻ h ₅≻ h ₄≻ h ₃≻ h ₂ i3 h ₂≻ h ₁≻ h ₄≻ h ₇≻ h ₃≻ h ₆≻ h ₅ i4 h ₂≻ h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₇≻ h ₅ i5 h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆

Top Cycle Mechanism Every student points an arrow to his favorite house. Every currently occupied house points an arrow to its current occupant. Every vacant house points an arrow to the most senior student. Identify cycles. For each cycle, assign the house to the student who has an arrow pointing to it. Repeat the exercise for the remaining houses and students until no houses or no students are left ►

Top Cycle Mechanism: First Cycles preferences i1 h ₂≻ h ₆≻ h ₅≻ h ₁≻ h ₄≻ h ₃≻ h ₇ i2 h ₇≻ h ₁≻ h ₆≻ h ₅≻ h ₄≻ h ₃≻ h ₂ i3 h ₂≻ h ₁≻ h ₄≻ h ₇≻ h ₃≻ h ₆≻ h ₅ i4 h ₂≻ h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₇≻ h ₅ i5 h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆ i4i4 h2h2 i2i2 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h7h7 i1i1 h1h1

Top Cycle Mechanism: First Cycles preferences i1 h ₂≻ h ₆≻ h ₅≻ h ₁≻ h ₄≻ h ₃≻ h ₇ i2 h ₇≻ h ₁≻ h ₆≻ h ₅≻ h ₄≻ h ₃≻ h ₂ i3 h ₂≻ h ₁≻ h ₄≻ h ₇≻ h ₃≻ h ₆≻ h ₅ i4 h ₂≻ h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₇≻ h ₅ i5 h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆ i4i4 h2h2 i2i2 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h7h7 i1i1 h1h1 one cycle i ₁ →h ₂ →i ₂ →h ₇ →i ₁ assign h ₂ to i ₁ and h ₇ to i ₂.

Top Cycle Mechanism: Second Cycles The remaining students have the following preferences: i ₃ :h ₁≻ h ₄≻ h ₃≻ h ₆≻ h ₅ i ₄ :h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₅ i ₅ :h ₄≻ h ₃≻ h ₁≻ h ₅≻ h ₆ Every (remaining) student points an arrow to his favorite (remaining) house. Every currently occupied house points an arrow to its current occupant. Every vacant house points an arrow to the most senior (remaining) student. ►

Top Cycle Mechanism: Second Cycles preferences i1 h ₂≻ h ₆≻ h ₅≻ h ₁≻ h ₄≻ h ₃≻ h ₇ i2 h ₇≻ h ₁≻ h ₆≻ h ₅≻ h ₄≻ h ₃≻ h ₂ i3 h ₂≻ h ₁≻ h ₄≻ h ₇≻ h ₃≻ h ₆≻ h ₅ i4 h ₂≻ h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₇≻ h ₅ i5 h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆ i4i4 h2h2 i2i2 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h7h7 i1i1 h1h1 one cycle i ₁ →h ₂ →i ₂ →h ₇ →i ₁ assign h ₂ to i ₁ and h ₇ to i ₂.

Top Cycle Mechanism: Second Cycles preferences i1 h ₂≻ h ₆≻ h ₅≻ h ₁≻ h ₄≻ h ₃≻ h ₇ i2 h ₇≻ h ₁≻ h ₆≻ h ₅≻ h ₄≻ h ₃≻ h ₂ i3 h ₂≻ h ₁≻ h ₄≻ h ₇≻ h ₃≻ h ₆≻ h ₅ i4 h ₂≻ h ₄≻ h ₃≻ h ₆≻ h ₁≻ h ₇≻ h ₅ i5 h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆ i4i4 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h1h1

Top Cycle Mechanism: Second Cycles i4i4 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h1h1 From figure 2, we can identify two cycles: i ₃ →h ₁ →i ₃ and i ₄ →h ₄ →i ₄. Eliminate the cycles by: assigning h ₁ to i ₃ and h ₄ to i ₄. i ₅ :h ₃≻ h ₅≻ h ₆ ►

Top Cycle Mechanism: Third Cycles h3h3 i5i5 h5h5 h6h6 From the figure, we can identify two cycles: i ₃ →h ₁ →i ₃ and i ₄ →h ₄ →i ₄. Eliminate the cycles by: assigning h ₁ to i ₃ and h ₄ to i ₄. i ₅ : h ₄≻ h ₃≻ h ₇≻ h ₁≻ h ₂≻ h ₅≻ h ₆ But h ₄, h ₇, h ₁, and h ₂ are gone Hence, i ₅ : h ₃≻ h ₅≻ h ₆ ►

Top Cycle Mechanism: Third Cycles h3h3 i5i5 h5h5 h6h6 Hence, i ₅ : h ₃≻ h ₅≻ h ₆ Every (remaining) student points an arrow to his favorite (remaining) house. Every currently occupied house points an arrow to its current occupant. Every vacant house points an arrow to the most senior (remaining) student. ►

Top Cycle Mechanism: Third Cycles h3h3 i5i5 h5h5 h6h6 From this figure, we can identify a cycle: i ₅ →h ₃ →i ₅. Eliminate the cycle by assigning h ₃ to i ₅. Since no students are left, we are done!! ►

Top Cycle Mechanism is Efficient For those first-cycle students: they get their favorite houses, you can't make them strictly happier. For those second-cycle students; they get their favorite (remaining) houses, you can't make them strictly happier without making some first-cycle students strictly less happy. For those third-cycle students: thy get their favorite (remaining) houses, you can't make them strictly happier without making some first- or second-cycle students strictly less happy. … ►

Top Cycle Mechanism is Protective Suppose I am an existing tenant, then my current house always points an arrow to me. If I leave the game earlier than my current house, I must get a new house better than my current one. If my current house is in any cycle and I am around, I must also in the same cycle as well. In that cycle, I am pointing an arrow to a house at least as good as my current house. So I will be assigned a house at least as good as my current house. ►

Top Cycle Mechanism is Strategy- Proof Suppose you are a second- cycle student, say i ₃. Can you benefit from leaving immediately in the first round? In the first round, the only house that points an arrow at you is your current house, h ₃. In the first round, no one points an arrow to h ₃. The only way for you to leave in the first round is to point an arrow to h ₃ instead. But you are guaranteed to get a house as good as your current house, h ₃, anyway. So why hurry? ► i4i4 h2h2 i2i2 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h7h7 i1i1 h1h1

Top Cycle Mechanism is Strategy- Proof Here is another way to see it: If i ₃ is to benefit from leaving immediately in the first round, he must be pointing an arrow to either h ₂ or h ₇. But neither would stop "i ₁ →h ₂ →i ₂ →h ₇ →i ₁ " from forming a cycle. No matter what i ₃ does, h ₂ and h ₇ will be gone after the first round. ► i4i4 h2h2 i2i2 i3i3 h3h3 h4h4 i5i5 h5h5 h6h6 h7h7 i1i1 h1h1

Top Cycle Mechanism is Strategy- Proof Suppose you are a second-cycle student, say i ₃. Can you benefit from leaving later than in the second round? In the second round, your favorite (remaining) house is h ₁. There is no reason to delay to later cycles. ►

The Top Cycle Mechanism: Summary Students often do not care about which houses other students get. Whenever this is the case, the top cycle mechanism will lead to an assignment that is Pareto efficient, protective of existing tenants, and strategy proof. Strategy proofness means that no matter who you are, it is always a dominant strategy fro you to tell your true preferences; no matter who you are, you can never benefit from lying. ►

Summary Two sided matching Many such real world problem—college admissions In many cases, price mechanism may not be used (i.e., college admissions) Two central concepts: Stability Strategy proof (don’t benefit from lying) A stable matching always exists It can be obtained via a centralized procedure: deferred acceptance procedure ►

Summary Deferred acceptance procedure has been used in the US in assigning medical interns: interns and hospitals Market design economists were retained to design centralized mechanisms Redesign of the school choice systems in Boston and New York City a couple of years ago. Design of kidney exchange —a priceless “market” that allows family members of several patients to sway their kidneys with one another “Market design” or “economic engineering” ►

Centralized Matching. Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group.

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Presentation on theme: "Centralized Matching. Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group."— Presentation transcript:

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Centralized Matching. Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group.

Similar presentations

Presentation on theme: "Centralized Matching. Preview Many economic problems concern the need to match members of one group of agents with one or more members of a second group."— Presentation transcript:

Similar presentations

About project

Feedback