33. A country with a parliamentary government has two parties that capture 100% of the vote between them. Each party is awarded seats in proportion to the number of votes received.
33.
(a) One quota will be rounded up and the other down to obtain the Webster apportionment. The quota that is rounded up will have a fractional part greater than 0.5 and will be greater than the fractional part of the quota that is rounded down. The Hamilton method will give the party whose quota has the larger fractional part an additional seat. Thus, the apportionments will be identical.
(b) These paradoxes never occur with the Webster method, which gives the same apportionment in this case.
(c) The Hamilton method, which always satisfies the quota condition, gives the same apportionment.
A-34
Seats | Whigs | Tories | Liberals | Centrists | ||||
Priority | Seat # | Priority | Seat # | Priority | Seat # | Priority | Seat # | |
1 | 5,525,381 | 1 | 3,470,152 | 3 | 3,864,226 | 2 | 201,203 | |
2 | 1,841,794 | 4 | 1,145,717 | 6 | 1,288,075 | 5 | 67,068 | |
3 | 1,105,076 | 7 | 694,030 | 10 | 772,845 | 9 | 40,241 | |
4 | 789,340 | 8 | 495,736 | 14 | 552,032 | 12 | 28,743 | |
5 | 613,931 | 11 | 385,572 | 17 | 429,358 | 15 | 22,356 | |
6 | 502,307 | 13 | 315,468 | 21 | 351,293 | 19 | 18,291 | |
7 | 425,029 | 16 | 266,935 | 24 | 297,248 | 22 | 15,477 | |
8 | 368,359 | 18 | 231,343 | 28 | 257,615 | 26 | 13,414 | |
9 | 325,022 | 20 | 204,127 | 227,307 | 29 | 11,835 | ||
10 | 290,810 | 23 | 182,640 | 203,380 | 10,590 | |||
11 | 263,113 | 25 | 165,245 | 184,011 | 9,581 | |||
12 | 240,234 | 27 | 150,876 | 168,010 | 8,748 | |||
13 | 221,015 | 30 | 138,806 | 154,569 | 8,048 | |||
14 | 204,644 | 128,524 | 143,119 | 7,452 | ||||
15 | 190,530 | 119,660 | 133,249 | 6,938 |
The Whigs get 13 of the first 30 seats, the Liberals get 9, the Tories get 8, and the Centrists get none of the first 30 seats.
(d) No, each of these methods is capable of producing a different apportionment. For example, suppose that one party receives 99.9% of the vote and the other receives 0.1%. If the house size is 100, the Jefferson method would apportion all 100 seats to the dominant party. (So would the Hamilton and Webster methods.) The Hill-Huntington method would apportion the dominant party 99 seats, and 1 seat to the other party. If the dominant party received 99.4% and the other party received 0.6%, the Jefferson method would still apportion all of the seats to the dominant party, but the Hamilton and Webster methods would give 1 seat to the other party.