**Practice: (Apportionment)**

Prepared by:

Joseph Malkevitch

Mathematics Department

York College (CUNY)

Jamaica, NY 11451

email:

__malkevitch@york.cuny.edu__

web page:

__http://york.cuny.edu/~malk/__

1. a. Compare the apportionments obtained for Hamilton, Jefferson, Adams, and Webster for a house size of 250 seats where the populations of 6 states are:

A = 6,936,000

B = 2,091,000

C = 1,646,000

D = 988,000

E = 685,000

F = 154,000

b. Redo this problem with house size = h =12. Use the "table method" to find the Jefferson, Adams, and Webster apportionments. You might enjoy checking that you get the same answers when you use the "rounding rule" approach.

2. Determine how many seats each party gets for a house size of 36 with the following votes for the parties:

A = 27,774

B = 25, 178

C = 19,951

D = 14,610

E = 9,225

F = 3,292

using:

i. Adams

ii. Dean

iii. Huntington-Hill

iv. Webster

v. Jefferson (D'Hondt)

vi. Hamilton

3. Suppose 101 seats are to be distributed to 3 parties using Hamilton's Method:

The three parties involved get the following numbers of votes:

A = 50,600

B = 40,650

C = 9,750

Do you notice anything "unintuitive" about the result?

What is the apportionment for this example using Webster's method?

4. Suppose there is an election where 5 parties are involved.

The exact quota of the parties is:

A = 65.91

B = .53

C = .521

D = .52

E = .519

(the house size 68)

How many seats does each of the parties get using Webster?

Now, imagine the h = the house size stays at 68 but the exact quotas for the 5 states changes to:

A = 66.075

B = .485

C = .481

D = .48

E = .479

What is the Webster apportionment?

Does a comparison with the previous case seem "intuitive?"

Also, for both examples determine the Hamilton apportionment.

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