Econ 4415: Problem Set #4 Qingmin Liu Columbia University April 8, 2021 Abstract Due on 10PM at April 14, 2015. See the syllabus for problem set policies. 1 Deferred Acceptance Algorithm (DA) Consider...

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Answered Same DayApr 12, 2021

Answer To: Econ 4415: Problem Set #4 Qingmin Liu Columbia University April 8, 2021 Abstract Due on 10PM at...

Rajeswari answered on Apr 13 2021
163 Votes
80352 assignment
Given that
m1 : w1; w2;
m2 : w1;
m3 : w2; w1;
w1 : m3; m2; m1;
w2 : m1;
m3:
1) Hence individually rational matchings would be
I. (m1,w2) (m2,w1), (m3, Or
II. (m1,w1) (m2, or
III. (m1,w2), (m2,,(m3, w1)
There cannot be any other individual rational matchings.
2) All stable matchings would be the matching where each individual does not want to switch on to other.
Of the rational matchings given above, (m1,w1) (m2, phi) (m3,w2) is stable as long as m2 is content with no wife, from men point of view. But women w1 prefers m3 to m1 and w2 prefers m1 to m3. So they want to switch over
(m3,w1) (m2, (m1, w2) would be ideal for women’s side. Hence as such no stable matchings in this case
3) Here man proposes to woman. M1 proposes to w1, but w1 may reject...
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