centralized matching. preview many economic problems concern the need to match members of one 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,..,wn

Men: m1,m2,…,mn 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! ►

w1 w2 w3

m1 1,3 2,2 3,1

m2 3,1 1,3 2,2

m3 2,2 3,1 1,3

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. ►


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 ►

w1 w2 w3

M1 1,3 2,2 3,1

M2 3,1 1,3 2,2

m3 2,2 3,1 1,3


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) ►

w1 w2 w3

M1 1,3 2,2 3,1

M2 3,1 1,3 2,2

m3 2,2 3,1 1,3


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 ►

s1 s2 s3 s4

s1 1 3

s2 1 3

s3 1 3

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 ►

s1 s2 s3 s4

s1 1 3

s2 1 3

s3 1 3

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. ►

s1 s2 s3 s4

s1 1 3

s2 1 3

s3 1 3

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. ►

w1 w2 w3

m1 1,3 2,2 3,1

m2 3,1 1,3 2,2

m3 2,2 3,1 1,3

Women-Propose-to-Men Procedure: Example 2

w1 w2 w3

M1 2,1 1,2 3,1

M2 1,3 2,3 3,2

m3 1,2 2,1 3,3

step w1 w2 w3

1 m2,m3 m1


w1 w2 w3

M1 2,1 1,2 3,1

M2 1,3 2,3 3,2

m3 1,2 2,1 3,3

step w1 w2 w3

1 m2,m3 m1

2 m3 m1,m2


w1 w2 w3

M1 2,1 1,2 3,1

M2 1,3 2,3 3,2

m3 1,2 2,1 3,3

step w1 w2 w3

1 m2,m3 m1

2 m3 m1,m2

3 m3 m1 m2

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 ►


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. ►


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. ►


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: h₁ and h₂ How many different ways to assign houses?

Assignment x: assign h₁ to A, and h₂ to B Assignment y: assign h₁ to B, and h₂ 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 h₁ to A and h₂ 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 h₁ to h₂ to h₃. Then he thinks "as long as I get h₁, I don't care whether B gets h₂ and h₃."

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: i₁,i₂,i₃ occupying houses h₁,h₂,h₃, respectively. There is one vacant house: h₄. The seniority order is i₁,i₂,i₃. Their true preferences over the 4 houses are:

i1: h3>h2>h1>h4 i2: h1>h4>h2>h3 i3: h2>h1>h3>h4 ►


Go down the seniority list and ask the students one by one: "Do you want to take one of these free houses?“

i₁ replies: "No, the only free house, h₄, is even worse than my current house, h₁.“

i₂ replies: "Yes, h₄ is better than h₂.“

h₄ is taken. h₂ becomes the only free house. ►

student preferences

i1 h3>h2>h1>h4

i2 h1>h4>h2>h3

i3 h2>h1>h3>h4


Go back to the top of the seniority list and start again.

i₁ replies: "Yes, h₂ is better than h₁.“

h₂ is taken. h₁ becomes the only free house.

Go back to the top of the seniority list (which now has only one student left, namely i₃) and start again.

i₃ replies: "Yes, h₁ is better than h₃.“►

student preferences

i1 h3>h2>h1>h4

i2 h1>h4>h2>h3

i3 h2>h1>h3>h4

Protective, but not Pareto Efficient

Suppose there are 2 students, both are existing tenants: i₁ and i₂ occupy houses h₁ and h₂ respectively. There is one vacant house: h₃.

The students' true preferences over the 3 houses are: i₁:h₂ h₃ h₁ i₂:h₃ h₂ h₁ ≻ ≻ ≻ ≻

The Rochester Mechanism assigns h₃ to i₁ and h₂ to i₂.

But this is not Pareto efficient, as everyone would be happier if i₁ gets h₂ and i₂ gets h₃! ►

Protective, but not Strategy-Proof

i₁:h₂ h₃ h₁ and i₂:h₃ h₂ h₁ (same ≻ ≻ ≻ ≻example)

Suppose i₁ lies and answers, "No, my preferences ranking is actually h₂ h₁ h₃, and hence I don't ≻ ≻want to take h₃."

i₂ takes h₃, frees up h₂. Go back to the top of the seniority list and start

again. Now i₁ has a chance to take h₂. It is not a dominant strategy for i₁ 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, i₅, and one vacant house, h₅.

The seniority order is i₁,i₂,i₃,i₄ and then i₅.

Their true preferences over the 5 houses are: ►

student preferences

i1 h₃ h₄ h₅ h₁≻ ≻ ≻ ≻h₂

i2 h₄ h₅ h₂ h₃≻ ≻ ≻ ≻h₁

i3 h₅ h₃ h₄ h₂≻ ≻ ≻ ≻h₁

i4 h₃ h₅ h₄ h₂≻ ≻ ≻ ≻h₁

i5 h₄ h₅ h₃ h₁≻ ≻ ≻ ≻h₂


i₁ is tentatively matched to h₃. i₂ is tentatively matched to h₄. i₃ is tentatively matched to h₅.

i₄, "The only two remaining houses are h₁ and h₂, and both are worse than my current house, h₄. So I shall call for protection.“

i₄ seizes back h₄, and leaves the process. ►

student preferences

i1 h₃ h₄ h₅ h₁≻ ≻ ≻ ≻h₂

i2 h₄ h₅ h₂ h₃≻ ≻ ≻ ≻h₁

i3 h₅ h₃ h₄ h₂≻ ≻ ≻ ≻h₁

i4 h₃ h₅ h₄ h₂≻ ≻ ≻ ≻h₁

i5 h₄ h₅ h₃ h₁≻ ≻ ≻ ≻h₂


i₂ was the conflicting student, so we re-do the process, starting from i₂.

i₂ is tentatively matched to h₅.

i₃,"The only remaining houses are h₁ and h₂, and both are worse than my current house, h₃. So I shall call for protection.“

i₃ seizes back h₃, and leaves the process. ►

student preferences

I1 h₃ h₄ h₅ h₁≻ ≻ ≻ ≻h₂

I2 h₄ h₅ h₂ h₃≻ ≻ ≻ ≻h₁

I3 h₅ h₃ h₄ h₂≻ ≻ ≻ ≻h₁

i4 h₃ h₅ h₄ h₂≻ ≻ ≻ ≻h₁

i5 h₄ h₅ h₃ h₁≻ ≻ ≻ ≻h₂


i₁ was the conflicting student, so we re-do the process, starting from i₁.

i₁ is tentatively matched to h₅. i₂ is tentatively matched to h₂. i₅ is tentatively matched to h₁.

The process ends, all tentative matches are finalized. ►

student preferences

i1 h₃ h₄ h₅ h₁≻ ≻ ≻ ≻h₂

i2 h₄ h₅ h₂ h₃≻ ≻ ≻ ≻h₁

i3 h₅ h₃ h₄ h₂≻ ≻ ≻ ≻h₁

i4 h₃ h₅ h₄ h₂≻ ≻ ≻ ≻h₁

i5 h₄ h₅ h₃ h₁≻ ≻ ≻ ≻h₂

MIT Mechanism: Example Let us keep to the same

example. The MIT mechanism will

match (i₁,i₂,i₃,i₄,i₅) to (h₅,h₂,h₃,h₄,h₁).

Compare this assignment to the following one: (h₃,h₂,h₅,h₄,h₁); i.e., i₁ and i₃ swap their houses.

i₂,i₄, and i₅ like the second assignment at least as good as the first--they are actually indifferent.

i₁ and i₃ like the second assignment strictly better than the first. ►

student preferences

i1 h₃ h₄ h₅ h₁≻ ≻ ≻ ≻h₂

i2 h₄ h₅ h₂ h₃≻ ≻ ≻ ≻h₁

i3 h₅ h₃ h₄ h₂≻ ≻ ≻ ≻h₁

i4 h₃ h₅ h₄ h₂≻ ≻ ≻ ≻h₁

i5 h₄ h₅ h₃ h₁≻ ≻ ≻ ≻h₂

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. i₁,i₂,i₃ and i₄ occupy houses h₁,h₂,h₃ and h₄, respectively.

Suppose there are one new applicant i₅, and 3 vacant houses, h₅,h₆ and h₇.

The seniority order is i₁,i₂,i₃,i₄, and then i₅. ►

preferencesi1 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻

h₇i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻

h₂i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻

h₅i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻

h₅i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻

h₆

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 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₇

i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₂

i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻h₅

i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻h₅

i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻h₆

i4

h2i2

i3

h3

h4i5h5

h6

h7

i1h1

Top Cycle Mechanism: First Cycles

preferences

i1 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₇

i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₂

i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻h₅

i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻h₅

i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻h₆

i4

h2i2

i3

h3

h4i5h5

h6

h7

i1h1

one cycle i₁→h₂→i₂→h₇→i₁assign h₂ to i₁ and h₇ to i₂.

Top Cycle Mechanism: Second Cycles

The remaining students have the following preferences:

i₃:h₁ h₄ h₃ h₆ h₅ i₄:h₄ h₃ h₆ h₁ h₅ ≻ ≻ ≻ ≻ ≻ ≻ ≻ ≻i₅:h₄ h₃ h₁ h₅ h₆≻ ≻ ≻ ≻

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. ►


preferences

i1 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₇

i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₂

i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻h₅

i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻h₅

i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻h₆

i4

h2i2

i3

h3

h4i5h5

h6

h7

i1h1

one cycle i₁→h₂→i₂→h₇→i₁assign h₂ to i₁ and h₇ to i₂.


preferences

i1 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₇

i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻h₂

i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻h₅

i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻h₅

i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻h₆

i4

i3

h3

h4i5h5

h6

h1


preferencesi1 h₂ h₆ h₅ h₁ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻

h₇i2 h₇ h₁ h₆ h₅ h₄ h₃≻ ≻ ≻ ≻ ≻ ≻

h₂i3 h₂ h₁ h₄ h₇ h₃ h₆≻ ≻ ≻ ≻ ≻ ≻

h₅i4 h₂ h₄ h₃ h₆ h₁ h₇≻ ≻ ≻ ≻ ≻ ≻

h₅i5 h₄ h₃ h₇ h₁ h₂ h₅≻ ≻ ≻ ≻ ≻ ≻

h₆ i4

i3

h3

h4i5h5

h6

h1


i4

i3

h3

h4i5h5

h6

h1

From figure 2, we can identify two cycles: i₃→h₁→i₃ and i₄→h₄→i₄.

Eliminate the cycles by: assigning h₁ to i₃ and h₄ to i₄.

i₅:h₃ h₅ h₆ ≻ ≻ ►

Top Cycle Mechanism: Third Cycles

h3

i5h5

h6

From the figure, we can identify two cycles: i₃→h₁→i₃ and i₄→h₄→i₄.

Eliminate the cycles by: assigning h₁ to i₃ and h₄ to i₄.

i₅:h₄ h₃ h₇ h₁ h₂ h₅ h≻ ≻ ≻ ≻ ≻ ≻₆

But h₄, h₇, h₁, and h₂ are gone

Hence, i₅:h₃ h₅ h₆ ≻ ≻ ►


h3

i5h5

h6

Hence, i₅:h₃ h₅ h₆≻ ≻ 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. ►


h3

i5h5

h6

From this figure, we can identify a cycle: i₅→h₃→i₅.

Eliminate the cycle by assigning h₃ to i₅.

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 i₃. 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, h₃.

In the first round, no one points an arrow to h₃. The only way for you to leave in the first round is to point an arrow to h₃ instead.

But you are guaranteed to get a house as good as your current house, h₃, anyway. So why hurry? ► i4

h2i2

i3

h3

h4i5h5

h6

h7

i1h1

Top Cycle Mechanism is Strategy-Proof Here is another way to see

it: If i₃ is to benefit from leaving immediately in the first round, he must be pointing an arrow to either h₂ or h₇.

But neither would stop "i₁→h₂→i₂→h₇→i₁" from forming a cycle.

No matter what i₃ does, h₂ and h₇ will be gone after the first round. ►

i4

h2i2

i3

h3

h4i5h5

h6

h7

i1h1

Top Cycle Mechanism is Strategy-Proof Suppose you are a second-cycle student, say

i₃. Can you benefit from leaving later than in the

second round? In the second round, your favorite (remaining)

house is h₁. 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...

Documents

stable matchings m1

stable twosided matchings

centralized matching

matchings s1

matching problems

onesided matching

w2 women optimal m1

n twosided problem