Homework 4 (1)
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Jan 9, 2024
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MATH & POLITICS HOMEWORK 4
Question 9.2
Consider the odd method - Find a divisor d and divide it into the population of each state to
obtain a modified quota
pi/d
for state
i
. Then round these numbers to the nearest odd whole
number.
a)
Does the odd method satisfy population monotonicity?
-
Since rounding to the nearest odd whole number (integer) is a rounding function,
this is a divisor method. All divisor methods satisfy population monotonicity, so
the answer is yes.
b)
Does the odd method satisfy the quota rule?
-
The Balinski theorem states that a method cannot satisfy both population
monotony and the quota rule. Therefore, the answer is no.
c)
What is the apportionment according to the odd method for the census data h=4, n=2, p1
= 123,456,789, and p2 = 123,456,788?
-
Apportionment : a
1
= 1, a
2
= 3
d)
What difficulty would you encounter if you attempted the odd method on the data from
the 2000 USD census with n=50 and h=435?
-
The sum of 2 odd numbers must be even (ex: 1 + 1 = 2, 3 + 3 = 6.) Therefore, the
sum of n=50 odd numbers would be even as well. This means it is impossible for
the sum to be 435, an odd number.
Problem 9.6
Consider a hybrid apportionment method that we call the Hamster method - use the method of
Webster unless a quota violation results. In that case, use the method of Hamilton instead.
a)
Does the method satisfy the quota rule?
-
Yes. In the case that a quota violation occurs due to the Webster method, the
Hamilton method corrects it.
b)
Does the method satisfy the order-preserving property?
-
Yes. The order preserving property states that when ai > aj, it follows that pi > pj.
This is satisfied because if the Hamster method assigns state i more seats than
state j, then either the Webster or Hamilton method created this apportionment.
The result is that the population of state i is regarded as greater than the
population of state j. Thus, the method is order-preserving.
c)
Is the method population monotone?
-
No. The population monotone criterion states that whenever a’i < ai, and a’j < aj,
it follows that either p’i < pi or p’j < pj. In this case, the impossibility theorem
states that no method can satisfy both the quota rule and population monotonicity.
Therefore, since (a) is satisfied, ( c) cannot be.
Problem 9.8
Call an apportionment method nonzero if it never assigns zero seats to any state. Assuming that h
≥ n, which divisor methods that we have encountered satisfy the nonzero criterion?
-
Since both Jefferson and Webster methods round small modified quotas down to 0, they
are not nonzero. However, the Adams method, the Hill Method, and the Dean method
round up numbers between 0 ~ 1 to 1, so they do satisfy the nonzero criterion
Problem 11.2
Suppose that State A has a population of 7,000,000 while state B has a population of 4,000,000,
and suppose that they are to share 4 representatives between them..
a)
If State A gets 3 representatives and State B gets 1 representative, which state has a more
favorable average district size?
-
Average district size =
???????𝑖?? ?? ?????
?????? ?? ?????
-
State A Average district size =
= 2,333,333
7,000,000
3
-
State B Average district size =
= 4,000,000
4,000,000
1
-
Since 2333,333 < 4,000,000, State A has a more favorable average district size
b)
If State A gets 2 representatives and States B gets 2 representatives, which state has a
more favorable average district size?
-
Average district size =
???????𝑖?? ?? ?????
?????? ?? ?????
-
State A Average district size =
= 3,500,000
7,000,000
2
-
State B Average district size =
= 2,000,000
2
-
Since 2,000,000 < 3,500,000 State B has a more favorable average district size
c)
Of the two apportionments suggested in parts (a) and (b), which gives average district
sizes that are closest together?
-
In part (a), the sizes have a difference of 4,000,000 - 2,333,333 = 1666667
-
In part (b), the sizes have a difference of 3,500,000 - 2,000,000 = 1500000
-
Part (b) gives average district sizes that are closest together
d)
Of the two apportionments suggested in parts (a) and (b), which gives degrees of
representation that are closest together?
-
Degree of representation =
1
??????? ?𝑖???𝑖?? ?𝑖𝑧?
-
Difference in Part (a) =
-
=
1
2,333,333 1
4,000,000 5
28,000,000 -
Difference in Part (b) =
-
=
1
2,000,000 1
3,500,000 6
28,000,000 -
<
, so part (a) gives degrees of representation that are closest
5
28,000,000 6
28,000,000 together
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