Econ 145: Practice Questions I
Suggested Answer
February 8, 2010
1. Consider a room assignment problem with 4 students and 4 rooms
{
h
1
, . . . , h
4
}
.
Room
h
i
is initially
assigned to student
i.
Suppose that the preferences of the students are as follows
h
2
Â
1
h
4
Â
1
h
1
Â
1
h
3
h
2
Â
2
h
1
Â
2
h
3
Â
2
h
4
h
1
Â
3
h
3
Â
3
h
4
Â
3
h
2
h
3
Â
4
h
4
Â
4
h
1
Â
4
h
2
where
Â
i
is student
i
’s (strict) preference over the rooms. Answer following questions.
(a) Is the initial allocation e
ﬃ
cient? Explain why.
Answer:
We know that the TTC algorithm with an e
ﬃ
cient initial allocation leads the same allocation (See
Problem Set I Problem 1 (a)).
So we here simply apply the TTC algorithm, then compare the
resulting allocation with the initial allocation (
h
1
, h
2
, h
3
, h
4
).
Step 1:
We have initial assignment (
h
1
, h
2
, h
3
, h
4
)/ Each student wants to obtain
h
1
→
by 1
h
2
h
2
→
by 2
h
2
h
3
→
by 3
h
1
h
4
→
by 4
h
3
Then, we can
fi
nd a trading cycle
h
2
→
by 2
h
2
(self trading cycle)
So the TTC algorithm assigns
³
h
1
, h
assigned
2
, h
3
, h
4
´
.
Step 2:
We have unassigned rooms
{
h
1
, h
3
, h
4
}
.
Each remaining student (student 1, 3, and 4) wants to obtain
h
1
→
by 1
h
4
h
3
→
by 3
h
1
h
4
→
by 4
h
3
1
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Then, we can
fi
nd a trading cycle
h
1
→
by 1
h
4
→
by 4
h
3
→
by 3
h
1
·
So, the TTC algorithm assigns
³
h
assigned
4
, h
assigned
2
, h
assigned
1
, h
assigned
3
´
.
Clearly, the above allocation is di
ff
erent from initial allocation (
h
1
, h
2
, h
3
, h
3
). Thus, initial allocation
is not e
ﬃ
cient.
(b) Find all e
ﬃ
cient allocations. Explain how you found all the e
ﬃ
cient allocations.
Answer:
As we discussed Problem Set I Problem 2 (a), there are two methods to
fi
nd all e
ﬃ
cient allocations
Method I:
TTC Algorithm with All Possible Initial Assignments
Method II:
Serial Dictatorship with All Possible Orders
We here apply Method II.
Rewriting preferences
h
2
Â
1
h
4
Â
1
h
1
Â
1
h
3
h
2
Â
2
h
1
Â
2
h
3
Â
2
h
4
h
1
Â
3
h
3
Â
3
h
4
Â
3
h
2
h
3
Â
4
h
4
Â
4
h
1
Â
4
h
2
We generate all possible order of serial dictatorships. There are 4 students and 4
×
3
×
2
×
1 = 24
ways of orders.
Order of Dictatorships
Allocation
Type
Order of Dictatorships
Allocation
Type
1
→
2
→
3
→
4
(
h
2
, h
1
, h
3
, h
4
)
I
2
→
1
→
3
→
4
(
h
4
, h
2
, h
1
, h
3
)
V
1
→
2
→
4
→
3
(
h
2
, h
1
, h
4
, h
3
)
II
2
→
1
→
4
→
3
(
h
4
, h
2
, h
1
, h
3
)
V
1
→
3
→
2
→
4
(
h
2
, h
3
, h
1
, h
4
)
III
2
→
3
→
1
→
4
(
h
4
, h
2
, h
1
, h
3
)
V
1
→
3
→
4
→
2
(
h
2
, h
4
, h
1
, h
3
)
IV
2
→
3
→
4
→
1
(
h
4
, h
2
, h
1
, h
3
)
V
1
→
4
→
2
→
3
(
h
2
, h
1
, h
4
, h
3
)
II
2
→
4
→
1
→
3
(
h
4
, h
2
, h
1
, h
3
)
V
1
→
4
→
3
→
2
(
h
2
, h
4
, h
1
, h
3
)
IV
2
→
4
→
3
→
1
(
h
4
, h
2
, h
1
, h
3
)
V
Order of Dictatorships
Allocation
Type
Order of Dictatorships
Allocation
Type
3
→
1
→
2
→
4
(
h
2
, h
3
, h
1
, h
4
)
III
4
→
1
→
2
→
3
(
h
2
, h
1
, h
4
, h
3
)
II
3
→
1
→
4
→
2
(
h
2
, h
4
, h
1
, h
3
)
IV
4
→
1
→
3
→
2
(
h
2
, h
4
, h
1
, h
3
)
IV
3
→
2
→
1
→
4
(
h
4
, h
2
, h
1
, h
3
)
V
4
→
2
→
1
→
3
(
h
4
, h
2
, h
1
, h
3
)
V
3
→
2
→
4
→
1
(
h
4
, h
2
, h
1
, h
3
)
V
4
→
2
→
3
→
1
(
h
4
, h
2
, h
1
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