Upload
wauna
View
47
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Mechanism Design without Money. Lecture 8. The Match – a success story. Stability – 10 years before Gale Shapley. - PowerPoint PPT Presentation
Citation preview
1
Mechanism Design without Money
Lecture 8
The Match – a success storyStability – 10 years before Gale Shapley
Truthfulness “THE MOST IMPORTANT THING FOR APPLICANTS TO
REMEMBER: Simply rank internship programs based on your TRUE preferences, without consideration for where you believe you might be ranked by these programs. List the program that you want most as rank #1, followed by
your next most-preferred program as rank #2, and so on.” – APPIC
“For over 50 years, most United States medical school seniors have chosen to use a matching program …
“Before such matching programs … medical students often felt pressure, at an unreasonably early stage…
accept offers from residency programs. … This situation was inefficient, chaotic, and unfair…” - Congress
3
Rural hospitals
• The problem with allocation of doctors to rural hospitals.
Theorem: When preferences are strict, the set of doctors employed and positions filled is the same at every stable matching.
Theorem: When preferences are strict, any hospital that does not fill its quota at some stable matching is assigned precisely the same set of students at every stable matching.
4
Couples
• With time and progress, couples became a problem for the NRMP.
• The “leading member” adjustment didn’t work very well…
• This was one of the main reasons the original NRMP algorithm had to be replaced.
5
stable matching with couples
• It turns out that when couples are present, the set of stable matchings may be empty.
• Example:, ,
• Even when stable matchings do exist, there does not have to be a side-optimal stable matching.
6
Detour – complements and substitutes
• Complements: want both A and B (cucumbers and tomatoes to make salad)
• Substitutes: Given A, I want B less (apples and bananas when I need a fruit)
• Substitutes: greedy allocation can’t be bad• Couples create complementarities
7
Where does DA break?
• Suppose we run a doctor’s proposing DA• If a one of the members of the couple gets a
negative answer, both leave
• Problem – if a couple is temporarily assigned and then gets kicked out, a vacancy is created
8
Roth and Peranson (1999)• The new algorithm works as follows:
– First do doctor-proposing deferred-acceptance, with only single doctors involved.
– Then add the couples one by one (in random order) and using similar “proposing” mechanism.
– If any cycles are detected, start over.• If the algorithm terminates, the resulting matching is stable.
• And while supposedly there is no good reason for that, this algorithm always terminates (when using data from previous years or when using random preferences).
• Furthermore, it is generally considered a success story, and it was adopted for other programs as well…
9
Roth and Peranson (1999)• Clearinghouses currently using the algorithm (Taken from Roth’s slides):
10
Some questions
• When is the set of stable matchings (with couples) non-empty? – NPC
• Why does the Roth-Peranson algorithm works?
• What about truthfulness?
11
Large markets again…
• Kojima, Pathak and Roth (working paper) – Using a model similar to Immorlica and Mahdian (2003) and Kojima and Pathak (2009), if the number of couples is then a stable matching exists (and can be reached by the Roth and Peranson algorithm).
• Intuition: Instead of bounding using the probability that there is no rejection cycle (and all rejection chains end with a hospital that had a vacant position). For couples bound using also that there is no rejection path from one member of the couple to the other.
12
Extending Roth’s result
• We will present better bounds and slightly more sophisticated algorithm insure existence of stable matching even if the number of couples is for any .
• No hope for finding a matching with probability 1 – o(1) in case the number of couples is .
Preferences n “strong” doctors ; many “weak” doctors
1.5n residency programs k=n1-e couples, all strong doctors
Each hospital ranks all the strong doctors uniformly at random, and then the weak doctors
Each single ranks all hospitals uniformly at random
Each couple ranks pairs of hospitals uniformly at random, but only considers pairs which are in the same city
Algorithm1. Assign the “strong” singles, according to GS
(doctors propose)
2. Insert couples one by one:a) If inserting couple ci didn’t reject any couple - continueb) If assigning ci caused couple cp to get rejected:
1. Backtrack to the time where we assigned cp. 2. Assign ci before cp and continue
3. Assign all weak doctors (doctors proposing)
Example
Alice
Bob
Charlie
C2
C1
1. Assign Alice Bob and Charlie
2. Assign C1. Charlie gets unassigned
3. Assign Charlie
4. Assign C2. Bob get unassigned
5. Assign Bob. C1 gets unassigned
6. Need to backtrack Redo, with C2 and then C1
H1
H2
H3
H4
H5
H6
H7
C1
Example
Alice
Bob
Charlie
C2
1. Assign Alice Bob and Charlie
2. Assign C2. Bob gets unassigned
3. Assign Bob
4. Try to Assign C1. They don’t get their first choice
5. Assign C1. They get their second choice
H1
H2
H3
H4
H5
H6
H7
Termination success
G - a graph on the couples, such that there is an edge from ci to cp if ci causes cp to get unassignedWant: a topological sort on G
Defining such a graph is hard:- Whether ci gets cp to be unassigned depends on the
rest of the system (who else is assigned right now)- Worst case bounds are not enough: Every couple
can potentially unassign many other couples – no topological sort exists
If no couple ever gets unassigned – the algorithm ends successfully
Example:(c2-husband, H2), (c2-wife, H3) Blob(c2,0)
Blobs and the dependency graph
From blobs to edges; if: 1. (H,d’) Blob(cp,0)2. (H,d) Blob(ci,0)3. Hospital H prefers d to d’Then there is an edge ci cp
Alice
Bob
C2
C1
H1
H2
H3
H4
H5
H6
(c1-husband, H5) Blob (c1,0) , and H5 prefers BobThere is an edge from c2 c1
(Bob,H5) Blob(c2,0)
Blob(ci,0) = set of hospitals doctor pairs which ci can effect, if they were inserted as the first couple
Not everyone can be first: controlling dependencies using high order blobs
If a couple is not inserted first, they might see a different picture, e.g. because the hospital they would naturally get into is now taken by a couple who was there before them
Blob(ci,r) = set of hospitals doctor pairs which ci can effect, if they were inserted as the first couple and an adversary would be allowed to close r hospitals
We add an edge ci cp for high order blobs as well:(H,d’)B(ci,r) and (H,d)B(ci,r) with H preferring d to d’
High order blob example
C1
Alice
Bob
Charlie
C2
H1
H2
H3
H4
H5
H6
H7
Example:(c1-husband, H5), (c1-wife, H6), Blob(c1,0)(c1-husband, H1), (c1-wife, H7), Blob(c1,1)
Proof intuitionBuild the dependency graph G based on blobs of order r=3/
Prove that G is a DAG, and can be topologically sorted
Main Lemma: Blobs of size 3/ are conservative enough, such that no couple will kick someone outside of their blob
Show that our algorithm implicitly generates the blob graph and approximates a topological sort of it
Truthfulness in the Match
“Programs should be ranked in sequence, according to the applicant's true preferences... It is highly unlikely that
either applicants or programs will be able to influence the outcome of the Match in their favor by submitting a list
that differs from their true preferences.” NRMP
Problem: S1 can cheat
H2
s1
c2
s2
Hospital Preferences Capacity is 1
s2
H2
H4
c=(c1,c2)
H3,H4
H1,H2
Student Preferences s1
H4
H1
H3
H2
H1
c1
s1
H4
s2
s1
c2
H3
c1
s1
s1
H1
H2
H3
c1 H1 , c2 H2 , s1 H3 , s2 H4
s1 H1 , s2 H2 , c1 H3 , c2 H4The only stable
marriage
A linear fraction of couples
• The number of married people is proportional to the number of people
• Simulations show decent success rates for a constant fraction of couples
• Is there a way to insert the couples, to get a stable matching?
Thm : consider a random market, with n singles, n couples and more than 20n hospitals.With constant probability, there is no stable outcome
Proof idea – isolated markets
1. Find a small structure, which prevents a stable outcome
• A few hospitals and doctors, which (if left alone) can not form a stable outcome
2. Show that this small structure exists with constant probability
3. Show that no one outside the structure ever enters a hospital in the isolated market
Instable structure• For a single s and a
couple c, with probability O(1/n2) we have the structure
• If the structure occurs – no “local” stable outcome
• There are n singles and n couples, so with constant probability this structure will occur
H1
…
s
…
c1
…
H2
…
c2
…
s
…
S
H2
H1
…
c=(c1,c2)
H1,H2
whatever
Doctor Preferences Hospital Preferences
Capacity is 1
Isolated market
• The only solution is to insert someone else to h1,h2 thus avoiding the problem
• There is an excess of positions, so if a doctor goes to h1,h2 there are hospitals which are left free. We need to show that the doctor prefers them
• A quantitative version of the Rural Hospital Theorem– Define a probabilistic process, show it’s a martingale, use
Azuma’s inequality
29
Extensions• Roommate problems, multi-sided matching
• Many-to-one with discrete money and substitutable preferences (Crawford and Knoer, 1981; Kelso and Crawford, 1982)
• Many-to-many with responsive preferences (Roth, 1984)• Matching with contracts (Hatfield and Milgrom, 2005)• Many-to-many matching with contracts (Echenique and Oveido,
2006)• Matching in supply chains (Ostrovsky, 2008)• Matching in networks with bilateral contracts (Hatfield, Kominers,
Nichifor, Ostrovsky and Westkamp, working paper)• Matching with minimum quotas, regional caps, etc. (Biro, Fleiner,
Irving and Manlove, 2010, Kamada and Kojima, 2013)
30
Related topics
• Roth and Vande Vate (1990) – Random paths to stability
• Jackson and Watts (2002)• Ausubel and Milgrom (2000) on package
bidding
31
Questions?
32
Extra Slides
33
Chicken
34
Road example
A B
1 hour
1 hour
N minutes
N minutes
• 50 people want to get from A to B• There are two roads, each one has two segments. One takes
an hour, and the other one takes the number of people on it
35
Nash in road example
• In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes
A B
1 hour
1 hour
N minutes
N minutes
36
Braess’ paradox
• Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes
A B
1 hour
1 hour
N minutes
N minutes
Free