Question 14.71
41. A city has three districts with populations of 100,000, 600,000, and 700,000, respectively. Its council has 20 members, and seats on the council are apportioned by the Hill-Huntington method according to the district populations. Show that there is a tie. Would a tie occur with any of the other apportionment methods that we have considered?
41.
The quotas for the three districts are , and 10, respectively. The Hamilton method simply rounds the first two districts to 1 and 9, respectively, and the Webster and Jefferson methods give the same apportionment. Therefore, these methods do not lead to ties. The Hill-Huntington rounding points are . (It is important to notice that .) The Hill-Huntington roundings of the quotas are therefore 2, 9, and 10, respectively. The apportionment quotient for the first district will be less than (so that it will receive 1 seat), if the divisor is greater than . Using this divisor, the apportionment quotients would be for the first district and , the rounding point between 8 and 9, for the second district. Any divisor large enough to cause the first district’s quota to be rounded down will also cause the second district’s quota to be rounded down. The result would be 1 seat for the first district, 8 seats for the second district, and 10 seats for the third district—a total of
19 seats. The first and second districts are tied for the last seat.